3.3.69 \(\int \frac {(e+f x)^3 \sec (c+d x)}{a+a \sin (c+d x)} \, dx\) [269]

Optimal. Leaf size=502 \[ -\frac {3 i f (e+f x)^2}{2 a d^2}-\frac {6 i f^2 (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d^3}-\frac {i (e+f x)^3 \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac {3 f^2 (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{a d^3}+\frac {3 i f^3 \text {Li}_2\left (-i e^{i (c+d x)}\right )}{a d^4}+\frac {3 i f (e+f x)^2 \text {Li}_2\left (-i e^{i (c+d x)}\right )}{2 a d^2}-\frac {3 i f^3 \text {Li}_2\left (i e^{i (c+d x)}\right )}{a d^4}-\frac {3 i f (e+f x)^2 \text {Li}_2\left (i e^{i (c+d x)}\right )}{2 a d^2}-\frac {3 i f^3 \text {Li}_2\left (-e^{2 i (c+d x)}\right )}{2 a d^4}-\frac {3 f^2 (e+f x) \text {Li}_3\left (-i e^{i (c+d x)}\right )}{a d^3}+\frac {3 f^2 (e+f x) \text {Li}_3\left (i e^{i (c+d x)}\right )}{a d^3}-\frac {3 i f^3 \text {Li}_4\left (-i e^{i (c+d x)}\right )}{a d^4}+\frac {3 i f^3 \text {Li}_4\left (i e^{i (c+d x)}\right )}{a d^4}-\frac {3 f (e+f x)^2 \sec (c+d x)}{2 a d^2}-\frac {(e+f x)^3 \sec ^2(c+d x)}{2 a d}+\frac {3 f (e+f x)^2 \tan (c+d x)}{2 a d^2}+\frac {(e+f x)^3 \sec (c+d x) \tan (c+d x)}{2 a d} \]

[Out]

-3/2*I*f*(f*x+e)^2/a/d^2-I*(f*x+e)^3*arctan(exp(I*(d*x+c)))/a/d+3*I*f^3*polylog(2,-I*exp(I*(d*x+c)))/a/d^4+3*f
^2*(f*x+e)*ln(1+exp(2*I*(d*x+c)))/a/d^3-6*I*f^2*(f*x+e)*arctan(exp(I*(d*x+c)))/a/d^3-3*I*f^3*polylog(2,I*exp(I
*(d*x+c)))/a/d^4-3/2*I*f^3*polylog(2,-exp(2*I*(d*x+c)))/a/d^4-3*I*f^3*polylog(4,-I*exp(I*(d*x+c)))/a/d^4-3/2*I
*f*(f*x+e)^2*polylog(2,I*exp(I*(d*x+c)))/a/d^2-3*f^2*(f*x+e)*polylog(3,-I*exp(I*(d*x+c)))/a/d^3+3*f^2*(f*x+e)*
polylog(3,I*exp(I*(d*x+c)))/a/d^3+3/2*I*f*(f*x+e)^2*polylog(2,-I*exp(I*(d*x+c)))/a/d^2+3*I*f^3*polylog(4,I*exp
(I*(d*x+c)))/a/d^4-3/2*f*(f*x+e)^2*sec(d*x+c)/a/d^2-1/2*(f*x+e)^3*sec(d*x+c)^2/a/d+3/2*f*(f*x+e)^2*tan(d*x+c)/
a/d^2+1/2*(f*x+e)^3*sec(d*x+c)*tan(d*x+c)/a/d

________________________________________________________________________________________

Rubi [A]
time = 0.32, antiderivative size = 502, normalized size of antiderivative = 1.00, number of steps used = 22, number of rules used = 13, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4627, 4271, 4266, 2317, 2438, 2611, 6744, 2320, 6724, 4494, 4269, 3800, 2221} \begin {gather*} \frac {3 i f^3 \text {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{a d^4}-\frac {3 i f^3 \text {PolyLog}\left (2,i e^{i (c+d x)}\right )}{a d^4}-\frac {3 i f^3 \text {PolyLog}\left (2,-e^{2 i (c+d x)}\right )}{2 a d^4}-\frac {3 i f^3 \text {PolyLog}\left (4,-i e^{i (c+d x)}\right )}{a d^4}+\frac {3 i f^3 \text {PolyLog}\left (4,i e^{i (c+d x)}\right )}{a d^4}-\frac {3 f^2 (e+f x) \text {PolyLog}\left (3,-i e^{i (c+d x)}\right )}{a d^3}+\frac {3 f^2 (e+f x) \text {PolyLog}\left (3,i e^{i (c+d x)}\right )}{a d^3}+\frac {3 i f (e+f x)^2 \text {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{2 a d^2}-\frac {3 i f (e+f x)^2 \text {PolyLog}\left (2,i e^{i (c+d x)}\right )}{2 a d^2}-\frac {6 i f^2 (e+f x) \text {ArcTan}\left (e^{i (c+d x)}\right )}{a d^3}-\frac {i (e+f x)^3 \text {ArcTan}\left (e^{i (c+d x)}\right )}{a d}+\frac {3 f^2 (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{a d^3}+\frac {3 f (e+f x)^2 \tan (c+d x)}{2 a d^2}-\frac {3 f (e+f x)^2 \sec (c+d x)}{2 a d^2}-\frac {(e+f x)^3 \sec ^2(c+d x)}{2 a d}+\frac {(e+f x)^3 \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {3 i f (e+f x)^2}{2 a d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((e + f*x)^3*Sec[c + d*x])/(a + a*Sin[c + d*x]),x]

[Out]

(((-3*I)/2)*f*(e + f*x)^2)/(a*d^2) - ((6*I)*f^2*(e + f*x)*ArcTan[E^(I*(c + d*x))])/(a*d^3) - (I*(e + f*x)^3*Ar
cTan[E^(I*(c + d*x))])/(a*d) + (3*f^2*(e + f*x)*Log[1 + E^((2*I)*(c + d*x))])/(a*d^3) + ((3*I)*f^3*PolyLog[2,
(-I)*E^(I*(c + d*x))])/(a*d^4) + (((3*I)/2)*f*(e + f*x)^2*PolyLog[2, (-I)*E^(I*(c + d*x))])/(a*d^2) - ((3*I)*f
^3*PolyLog[2, I*E^(I*(c + d*x))])/(a*d^4) - (((3*I)/2)*f*(e + f*x)^2*PolyLog[2, I*E^(I*(c + d*x))])/(a*d^2) -
(((3*I)/2)*f^3*PolyLog[2, -E^((2*I)*(c + d*x))])/(a*d^4) - (3*f^2*(e + f*x)*PolyLog[3, (-I)*E^(I*(c + d*x))])/
(a*d^3) + (3*f^2*(e + f*x)*PolyLog[3, I*E^(I*(c + d*x))])/(a*d^3) - ((3*I)*f^3*PolyLog[4, (-I)*E^(I*(c + d*x))
])/(a*d^4) + ((3*I)*f^3*PolyLog[4, I*E^(I*(c + d*x))])/(a*d^4) - (3*f*(e + f*x)^2*Sec[c + d*x])/(2*a*d^2) - ((
e + f*x)^3*Sec[c + d*x]^2)/(2*a*d) + (3*f*(e + f*x)^2*Tan[c + d*x])/(2*a*d^2) + ((e + f*x)^3*Sec[c + d*x]*Tan[
c + d*x])/(2*a*d)

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 3800

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I*((c + d*x)^(m + 1)/(d*(m + 1))), x
] - Dist[2*I, Int[(c + d*x)^m*(E^(2*I*(e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 4266

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[-2*(c + d*x)^m*(ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))]/f), x] + (-Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 4269

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(c + d*x)^m)*(Cot[e + f*x]/f), x
] + Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 4271

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-b^2)*(c + d*x)^m*Cot[e
+ f*x]*((b*Csc[e + f*x])^(n - 2)/(f*(n - 1))), x] + (Dist[b^2*d^2*m*((m - 1)/(f^2*(n - 1)*(n - 2))), Int[(c +
d*x)^(m - 2)*(b*Csc[e + f*x])^(n - 2), x], x] + Dist[b^2*((n - 2)/(n - 1)), Int[(c + d*x)^m*(b*Csc[e + f*x])^
(n - 2), x], x] - Simp[b^2*d*m*(c + d*x)^(m - 1)*((b*Csc[e + f*x])^(n - 2)/(f^2*(n - 1)*(n - 2))), x]) /; Free
Q[{b, c, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2] && GtQ[m, 1]

Rule 4494

Int[((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.)*Tan[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Simp[
(c + d*x)^m*(Sec[a + b*x]^n/(b*n)), x] - Dist[d*(m/(b*n)), Int[(c + d*x)^(m - 1)*Sec[a + b*x]^n, x], x] /; Fre
eQ[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]

Rule 4627

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sec[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbo
l] :> Dist[1/a, Int[(e + f*x)^m*Sec[c + d*x]^(n + 2), x], x] - Dist[1/b, Int[(e + f*x)^m*Sec[c + d*x]^(n + 1)*
Tan[c + d*x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && IGtQ[m, 0] && EqQ[a^2 - b^2, 0]

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rule 6744

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a + b*x)))^p]/(b*c*p*Log[F])), x] - Dist[f*(m/(b*c*p*Log[F])), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, x] && GtQ[m,
0]

Rubi steps

\begin {align*} \int \frac {(e+f x)^3 \sec (c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {\int (e+f x)^3 \sec ^3(c+d x) \, dx}{a}-\frac {\int (e+f x)^3 \sec ^2(c+d x) \tan (c+d x) \, dx}{a}\\ &=-\frac {3 f (e+f x)^2 \sec (c+d x)}{2 a d^2}-\frac {(e+f x)^3 \sec ^2(c+d x)}{2 a d}+\frac {(e+f x)^3 \sec (c+d x) \tan (c+d x)}{2 a d}+\frac {\int (e+f x)^3 \sec (c+d x) \, dx}{2 a}+\frac {(3 f) \int (e+f x)^2 \sec ^2(c+d x) \, dx}{2 a d}+\frac {\left (3 f^2\right ) \int (e+f x) \sec (c+d x) \, dx}{a d^2}\\ &=-\frac {6 i f^2 (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d^3}-\frac {i (e+f x)^3 \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac {3 f (e+f x)^2 \sec (c+d x)}{2 a d^2}-\frac {(e+f x)^3 \sec ^2(c+d x)}{2 a d}+\frac {3 f (e+f x)^2 \tan (c+d x)}{2 a d^2}+\frac {(e+f x)^3 \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {(3 f) \int (e+f x)^2 \log \left (1-i e^{i (c+d x)}\right ) \, dx}{2 a d}+\frac {(3 f) \int (e+f x)^2 \log \left (1+i e^{i (c+d x)}\right ) \, dx}{2 a d}-\frac {\left (3 f^2\right ) \int (e+f x) \tan (c+d x) \, dx}{a d^2}-\frac {\left (3 f^3\right ) \int \log \left (1-i e^{i (c+d x)}\right ) \, dx}{a d^3}+\frac {\left (3 f^3\right ) \int \log \left (1+i e^{i (c+d x)}\right ) \, dx}{a d^3}\\ &=-\frac {3 i f (e+f x)^2}{2 a d^2}-\frac {6 i f^2 (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d^3}-\frac {i (e+f x)^3 \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac {3 i f (e+f x)^2 \text {Li}_2\left (-i e^{i (c+d x)}\right )}{2 a d^2}-\frac {3 i f (e+f x)^2 \text {Li}_2\left (i e^{i (c+d x)}\right )}{2 a d^2}-\frac {3 f (e+f x)^2 \sec (c+d x)}{2 a d^2}-\frac {(e+f x)^3 \sec ^2(c+d x)}{2 a d}+\frac {3 f (e+f x)^2 \tan (c+d x)}{2 a d^2}+\frac {(e+f x)^3 \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {\left (3 i f^2\right ) \int (e+f x) \text {Li}_2\left (-i e^{i (c+d x)}\right ) \, dx}{a d^2}+\frac {\left (3 i f^2\right ) \int (e+f x) \text {Li}_2\left (i e^{i (c+d x)}\right ) \, dx}{a d^2}+\frac {\left (6 i f^2\right ) \int \frac {e^{2 i (c+d x)} (e+f x)}{1+e^{2 i (c+d x)}} \, dx}{a d^2}+\frac {\left (3 i f^3\right ) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{a d^4}-\frac {\left (3 i f^3\right ) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{a d^4}\\ &=-\frac {3 i f (e+f x)^2}{2 a d^2}-\frac {6 i f^2 (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d^3}-\frac {i (e+f x)^3 \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac {3 f^2 (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{a d^3}+\frac {3 i f^3 \text {Li}_2\left (-i e^{i (c+d x)}\right )}{a d^4}+\frac {3 i f (e+f x)^2 \text {Li}_2\left (-i e^{i (c+d x)}\right )}{2 a d^2}-\frac {3 i f^3 \text {Li}_2\left (i e^{i (c+d x)}\right )}{a d^4}-\frac {3 i f (e+f x)^2 \text {Li}_2\left (i e^{i (c+d x)}\right )}{2 a d^2}-\frac {3 f^2 (e+f x) \text {Li}_3\left (-i e^{i (c+d x)}\right )}{a d^3}+\frac {3 f^2 (e+f x) \text {Li}_3\left (i e^{i (c+d x)}\right )}{a d^3}-\frac {3 f (e+f x)^2 \sec (c+d x)}{2 a d^2}-\frac {(e+f x)^3 \sec ^2(c+d x)}{2 a d}+\frac {3 f (e+f x)^2 \tan (c+d x)}{2 a d^2}+\frac {(e+f x)^3 \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {\left (3 f^3\right ) \int \log \left (1+e^{2 i (c+d x)}\right ) \, dx}{a d^3}+\frac {\left (3 f^3\right ) \int \text {Li}_3\left (-i e^{i (c+d x)}\right ) \, dx}{a d^3}-\frac {\left (3 f^3\right ) \int \text {Li}_3\left (i e^{i (c+d x)}\right ) \, dx}{a d^3}\\ &=-\frac {3 i f (e+f x)^2}{2 a d^2}-\frac {6 i f^2 (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d^3}-\frac {i (e+f x)^3 \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac {3 f^2 (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{a d^3}+\frac {3 i f^3 \text {Li}_2\left (-i e^{i (c+d x)}\right )}{a d^4}+\frac {3 i f (e+f x)^2 \text {Li}_2\left (-i e^{i (c+d x)}\right )}{2 a d^2}-\frac {3 i f^3 \text {Li}_2\left (i e^{i (c+d x)}\right )}{a d^4}-\frac {3 i f (e+f x)^2 \text {Li}_2\left (i e^{i (c+d x)}\right )}{2 a d^2}-\frac {3 f^2 (e+f x) \text {Li}_3\left (-i e^{i (c+d x)}\right )}{a d^3}+\frac {3 f^2 (e+f x) \text {Li}_3\left (i e^{i (c+d x)}\right )}{a d^3}-\frac {3 f (e+f x)^2 \sec (c+d x)}{2 a d^2}-\frac {(e+f x)^3 \sec ^2(c+d x)}{2 a d}+\frac {3 f (e+f x)^2 \tan (c+d x)}{2 a d^2}+\frac {(e+f x)^3 \sec (c+d x) \tan (c+d x)}{2 a d}+\frac {\left (3 i f^3\right ) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i (c+d x)}\right )}{2 a d^4}-\frac {\left (3 i f^3\right ) \text {Subst}\left (\int \frac {\text {Li}_3(-i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{a d^4}+\frac {\left (3 i f^3\right ) \text {Subst}\left (\int \frac {\text {Li}_3(i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{a d^4}\\ &=-\frac {3 i f (e+f x)^2}{2 a d^2}-\frac {6 i f^2 (e+f x) \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d^3}-\frac {i (e+f x)^3 \tan ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac {3 f^2 (e+f x) \log \left (1+e^{2 i (c+d x)}\right )}{a d^3}+\frac {3 i f^3 \text {Li}_2\left (-i e^{i (c+d x)}\right )}{a d^4}+\frac {3 i f (e+f x)^2 \text {Li}_2\left (-i e^{i (c+d x)}\right )}{2 a d^2}-\frac {3 i f^3 \text {Li}_2\left (i e^{i (c+d x)}\right )}{a d^4}-\frac {3 i f (e+f x)^2 \text {Li}_2\left (i e^{i (c+d x)}\right )}{2 a d^2}-\frac {3 i f^3 \text {Li}_2\left (-e^{2 i (c+d x)}\right )}{2 a d^4}-\frac {3 f^2 (e+f x) \text {Li}_3\left (-i e^{i (c+d x)}\right )}{a d^3}+\frac {3 f^2 (e+f x) \text {Li}_3\left (i e^{i (c+d x)}\right )}{a d^3}-\frac {3 i f^3 \text {Li}_4\left (-i e^{i (c+d x)}\right )}{a d^4}+\frac {3 i f^3 \text {Li}_4\left (i e^{i (c+d x)}\right )}{a d^4}-\frac {3 f (e+f x)^2 \sec (c+d x)}{2 a d^2}-\frac {(e+f x)^3 \sec ^2(c+d x)}{2 a d}+\frac {3 f (e+f x)^2 \tan (c+d x)}{2 a d^2}+\frac {(e+f x)^3 \sec (c+d x) \tan (c+d x)}{2 a d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1438\) vs. \(2(502)=1004\).
time = 8.89, size = 1438, normalized size = 2.86 \begin {gather*} \frac {x \left (4 e^3+6 e^2 f x+4 e f^2 x^2+f^3 x^3\right )}{8 a \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right )}-\frac {(\cos (c)+i \sin (c)) \left (-i e^3 x-\frac {3}{2} i e^2 f x^2-i e f^2 x^3-\frac {1}{4} i f^3 x^4+\frac {e^3 \log (1+i \cos (c+d x)-\sin (c+d x))}{d}+\frac {3 e^2 f x \log (1+i \cos (c+d x)-\sin (c+d x))}{d}+\frac {3 e f^2 x^2 \log (1+i \cos (c+d x)-\sin (c+d x))}{d}+\frac {f^3 x^3 \log (1+i \cos (c+d x)-\sin (c+d x))}{d}+\frac {6 i f^3 \text {Li}_4(-i \cos (c+d x)+\sin (c+d x))}{d^4}-\frac {i e^3 \log (1+i \cos (c+d x)-\sin (c+d x)) (\cos (c)-i \sin (c))}{d}-\frac {3 i e^2 f x \log (1+i \cos (c+d x)-\sin (c+d x)) (\cos (c)-i \sin (c))}{d}-\frac {3 i e f^2 x^2 \log (1+i \cos (c+d x)-\sin (c+d x)) (\cos (c)-i \sin (c))}{d}-\frac {i f^3 x^3 \log (1+i \cos (c+d x)-\sin (c+d x)) (\cos (c)-i \sin (c))}{d}+\frac {6 f^3 \text {Li}_4(-i \cos (c+d x)+\sin (c+d x)) (\cos (c)-i \sin (c))}{d^4}+\frac {6 f^2 (e+f x) \text {Li}_3(-i \cos (c+d x)+\sin (c+d x)) (\cos (c)+i (-1+\sin (c))) (\cos (c)-i \sin (c))}{d^3}+\frac {3 f (e+f x)^2 \text {Li}_2(-i \cos (c+d x)+\sin (c+d x)) (\cos (c)-i \sin (c)) (-1-i \cos (c)+\sin (c))}{d^2}\right )}{2 a (\cos (c)+i (-1+\sin (c)))}-\frac {(\cos (c)+i \sin (c)) \left (i d^2 e^3 x+12 i e f^2 x+\frac {3}{2} i d^2 e^2 f x^2+6 i f^3 x^2+i d^2 e f^2 x^3+\frac {1}{4} i d^2 f^3 x^4-3 d e^2 f x \log (1-i \cos (c+d x)+\sin (c+d x))-\frac {12 f^3 x \log (1-i \cos (c+d x)+\sin (c+d x))}{d}-3 d e f^2 x^2 \log (1-i \cos (c+d x)+\sin (c+d x))-d f^3 x^3 \log (1-i \cos (c+d x)+\sin (c+d x))-d e^3 \log (\cos (c+d x)+i (1+\sin (c+d x)))-\frac {12 e f^2 \log (\cos (c+d x)+i (1+\sin (c+d x)))}{d}-\frac {6 i f^3 \text {Li}_4(i \cos (c+d x)-\sin (c+d x))}{d^2}-3 i d e^2 f x \log (1-i \cos (c+d x)+\sin (c+d x)) (\cos (c)-i \sin (c))-\frac {12 i f^3 x \log (1-i \cos (c+d x)+\sin (c+d x)) (\cos (c)-i \sin (c))}{d}-3 i d e f^2 x^2 \log (1-i \cos (c+d x)+\sin (c+d x)) (\cos (c)-i \sin (c))-i d f^3 x^3 \log (1-i \cos (c+d x)+\sin (c+d x)) (\cos (c)-i \sin (c))-i d e^3 \log (\cos (c+d x)+i (1+\sin (c+d x))) (\cos (c)-i \sin (c))-\frac {12 i e f^2 \log (\cos (c+d x)+i (1+\sin (c+d x))) (\cos (c)-i \sin (c))}{d}+\frac {6 f^3 \text {Li}_4(i \cos (c+d x)-\sin (c+d x)) (\cos (c)-i \sin (c))}{d^2}-\frac {6 f^2 (e+f x) \text {Li}_3(i \cos (c+d x)-\sin (c+d x)) (\cos (c)-i \sin (c)) (\cos (c)+i (1+\sin (c)))}{d}+\frac {3 f \left (4 f^2+d^2 (e+f x)^2\right ) \text {Li}_2(i \cos (c+d x)-\sin (c+d x)) (i \cos (c)+\sin (c)) (\cos (c)+i (1+\sin (c)))}{d^2}\right )}{2 a d^2 (\cos (c)+i (1+\sin (c)))}-\frac {(e+f x)^3}{2 a d \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}+\frac {3 \left (e^2 f \sin \left (\frac {d x}{2}\right )+2 e f^2 x \sin \left (\frac {d x}{2}\right )+f^3 x^2 \sin \left (\frac {d x}{2}\right )\right )}{a d^2 \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[((e + f*x)^3*Sec[c + d*x])/(a + a*Sin[c + d*x]),x]

[Out]

(x*(4*e^3 + 6*e^2*f*x + 4*e*f^2*x^2 + f^3*x^3))/(8*a*(Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2])) - ((Cos[c] +
I*Sin[c])*((-I)*e^3*x - ((3*I)/2)*e^2*f*x^2 - I*e*f^2*x^3 - (I/4)*f^3*x^4 + (e^3*Log[1 + I*Cos[c + d*x] - Sin
[c + d*x]])/d + (3*e^2*f*x*Log[1 + I*Cos[c + d*x] - Sin[c + d*x]])/d + (3*e*f^2*x^2*Log[1 + I*Cos[c + d*x] - S
in[c + d*x]])/d + (f^3*x^3*Log[1 + I*Cos[c + d*x] - Sin[c + d*x]])/d + ((6*I)*f^3*PolyLog[4, (-I)*Cos[c + d*x]
+ Sin[c + d*x]])/d^4 - (I*e^3*Log[1 + I*Cos[c + d*x] - Sin[c + d*x]]*(Cos[c] - I*Sin[c]))/d - ((3*I)*e^2*f*x*
Log[1 + I*Cos[c + d*x] - Sin[c + d*x]]*(Cos[c] - I*Sin[c]))/d - ((3*I)*e*f^2*x^2*Log[1 + I*Cos[c + d*x] - Sin[
c + d*x]]*(Cos[c] - I*Sin[c]))/d - (I*f^3*x^3*Log[1 + I*Cos[c + d*x] - Sin[c + d*x]]*(Cos[c] - I*Sin[c]))/d +
(6*f^3*PolyLog[4, (-I)*Cos[c + d*x] + Sin[c + d*x]]*(Cos[c] - I*Sin[c]))/d^4 + (6*f^2*(e + f*x)*PolyLog[3, (-I
)*Cos[c + d*x] + Sin[c + d*x]]*(Cos[c] + I*(-1 + Sin[c]))*(Cos[c] - I*Sin[c]))/d^3 + (3*f*(e + f*x)^2*PolyLog[
2, (-I)*Cos[c + d*x] + Sin[c + d*x]]*(Cos[c] - I*Sin[c])*(-1 - I*Cos[c] + Sin[c]))/d^2))/(2*a*(Cos[c] + I*(-1
+ Sin[c]))) - ((Cos[c] + I*Sin[c])*(I*d^2*e^3*x + (12*I)*e*f^2*x + ((3*I)/2)*d^2*e^2*f*x^2 + (6*I)*f^3*x^2 + I
*d^2*e*f^2*x^3 + (I/4)*d^2*f^3*x^4 - 3*d*e^2*f*x*Log[1 - I*Cos[c + d*x] + Sin[c + d*x]] - (12*f^3*x*Log[1 - I*
Cos[c + d*x] + Sin[c + d*x]])/d - 3*d*e*f^2*x^2*Log[1 - I*Cos[c + d*x] + Sin[c + d*x]] - d*f^3*x^3*Log[1 - I*C
os[c + d*x] + Sin[c + d*x]] - d*e^3*Log[Cos[c + d*x] + I*(1 + Sin[c + d*x])] - (12*e*f^2*Log[Cos[c + d*x] + I*
(1 + Sin[c + d*x])])/d - ((6*I)*f^3*PolyLog[4, I*Cos[c + d*x] - Sin[c + d*x]])/d^2 - (3*I)*d*e^2*f*x*Log[1 - I
*Cos[c + d*x] + Sin[c + d*x]]*(Cos[c] - I*Sin[c]) - ((12*I)*f^3*x*Log[1 - I*Cos[c + d*x] + Sin[c + d*x]]*(Cos[
c] - I*Sin[c]))/d - (3*I)*d*e*f^2*x^2*Log[1 - I*Cos[c + d*x] + Sin[c + d*x]]*(Cos[c] - I*Sin[c]) - I*d*f^3*x^3
*Log[1 - I*Cos[c + d*x] + Sin[c + d*x]]*(Cos[c] - I*Sin[c]) - I*d*e^3*Log[Cos[c + d*x] + I*(1 + Sin[c + d*x])]
*(Cos[c] - I*Sin[c]) - ((12*I)*e*f^2*Log[Cos[c + d*x] + I*(1 + Sin[c + d*x])]*(Cos[c] - I*Sin[c]))/d + (6*f^3*
PolyLog[4, I*Cos[c + d*x] - Sin[c + d*x]]*(Cos[c] - I*Sin[c]))/d^2 - (6*f^2*(e + f*x)*PolyLog[3, I*Cos[c + d*x
] - Sin[c + d*x]]*(Cos[c] - I*Sin[c])*(Cos[c] + I*(1 + Sin[c])))/d + (3*f*(4*f^2 + d^2*(e + f*x)^2)*PolyLog[2,
I*Cos[c + d*x] - Sin[c + d*x]]*(I*Cos[c] + Sin[c])*(Cos[c] + I*(1 + Sin[c])))/d^2))/(2*a*d^2*(Cos[c] + I*(1 +
Sin[c]))) - (e + f*x)^3/(2*a*d*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^2) + (3*(e^2*f*Sin[(d*x)/2] + 2*e*f^
2*x*Sin[(d*x)/2] + f^3*x^2*Sin[(d*x)/2]))/(a*d^2*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2
]))

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1264 vs. \(2 (444 ) = 888\).
time = 0.30, size = 1265, normalized size = 2.52

method	result	size
risch	\(\text {Expression too large to display}\)	\(1265\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^3*sec(d*x+c)/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

3*I*f^3*polylog(4,I*exp(I*(d*x+c)))/a/d^4+6/a/d^3*f^3*ln(1-I*exp(I*(d*x+c)))*x+6/a/d^4*f^3*ln(1-I*exp(I*(d*x+c
)))*c-1/2/a/d^4*f^3*c^3*ln(1+I*exp(I*(d*x+c)))+6/a/d^4*f^3*c*ln(exp(I*(d*x+c)))-6/a/d^4*f^3*c*ln(exp(I*(d*x+c)
)+I)+1/2/a/d^4*f^3*c^3*ln(exp(I*(d*x+c))-I)-6/a/d^3*e*f^2*ln(exp(I*(d*x+c)))+6/a/d^3*e*f^2*ln(exp(I*(d*x+c))+I
)-3/a/d^3*e*f^2*polylog(3,-I*exp(I*(d*x+c)))-3/a/d^3*f^3*polylog(3,-I*exp(I*(d*x+c)))*x-6*I/a/d^4*f^3*polylog(
2,I*exp(I*(d*x+c)))-3*I/a/d^2*f^3*x^2-3*I/a/d^4*c^2*f^3+3/2/a/d^3*ln(1+I*exp(I*(d*x+c)))*c^2*e*f^2-3/2/a/d*ln(
1+I*exp(I*(d*x+c)))*e*f^2*x^2-1/2/a/d*f^3*ln(1+I*exp(I*(d*x+c)))*x^3-I*(d*f^3*x^3*exp(I*(d*x+c))+3*d*e*f^2*x^2
*exp(I*(d*x+c))+3*d*e^2*f*x*exp(I*(d*x+c))+d*e^3*exp(I*(d*x+c))+3*f^3*x^2-3*I*f^3*x^2*exp(I*(d*x+c))+6*e*f^2*x
-6*I*e*f^2*x*exp(I*(d*x+c))+3*e^2*f-3*I*e^2*f*exp(I*(d*x+c)))/d^2/(exp(I*(d*x+c))+I)^2/a+1/2/d/a*ln(exp(I*(d*x
+c))+I)*e^3-1/2/a/d*e^3*ln(exp(I*(d*x+c))-I)+3*I/a/d^2*polylog(2,-I*exp(I*(d*x+c)))*e*f^2*x-3*I/a/d^2*polylog(
2,I*exp(I*(d*x+c)))*e*f^2*x+3/2/d/a*e*f^2*ln(1-I*exp(I*(d*x+c)))*x^2-3/2/d^3/a*e*f^2*ln(1-I*exp(I*(d*x+c)))*c^
2+1/2/d/a*f^3*ln(1-I*exp(I*(d*x+c)))*x^3+1/2/d^4/a*f^3*ln(1-I*exp(I*(d*x+c)))*c^3+3/2/d/a*e^2*f*ln(1-I*exp(I*(
d*x+c)))*x+3/2/d^2/a*e^2*f*ln(1-I*exp(I*(d*x+c)))*c-3/2/d^2/a*e^2*f*c*ln(exp(I*(d*x+c))+I)+3/2/d^3/a*e*f^2*c^2
*ln(exp(I*(d*x+c))+I)+3/2/a/d^2*e^2*f*c*ln(exp(I*(d*x+c))-I)-3/2/a/d*ln(1+I*exp(I*(d*x+c)))*e^2*f*x-3/2/a/d^2*
ln(1+I*exp(I*(d*x+c)))*c*e^2*f-3/2/a/d^3*e*f^2*c^2*ln(exp(I*(d*x+c))-I)-3/2*I/a/d^2*f^3*polylog(2,I*exp(I*(d*x
+c)))*x^2+3/2*I/a/d^2*e^2*f*polylog(2,-I*exp(I*(d*x+c)))-3/2*I/a/d^2*e^2*f*polylog(2,I*exp(I*(d*x+c)))-6*I/a/d
^3*f^3*c*x+3/2*I/a/d^2*f^3*polylog(2,-I*exp(I*(d*x+c)))*x^2+3/d^3/a*e*f^2*polylog(3,I*exp(I*(d*x+c)))-1/2/d^4/
a*f^3*c^3*ln(exp(I*(d*x+c))+I)+3/d^3/a*f^3*polylog(3,I*exp(I*(d*x+c)))*x-3*I*f^3*polylog(4,-I*exp(I*(d*x+c)))/
a/d^4

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 3854 vs. \(2 (421) = 842\).
time = 1.19, size = 3854, normalized size = 7.68 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*sec(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/4*(3*c*e^2*f*(2/(a*d*sin(d*x + c) + a*d) - log(sin(d*x + c) + 1)/(a*d) + log(sin(d*x + c) - 1)/(a*d)) + e^3*
(log(sin(d*x + c) + 1)/a - log(sin(d*x + c) - 1)/a - 2/(a*sin(d*x + c) + a)) - 4*(12*d^2*e^2*f - 24*c*d*e*f^2
+ 12*c^2*f^3 + 2*(3*(c^2 + 4)*d*e*f^2 - (c^3 + 12*c)*f^3 - (3*(c^2 + 4)*d*e*f^2 - (c^3 + 12*c)*f^3)*cos(2*d*x
+ 2*c) + 2*(3*(-I*c^2 - 4*I)*d*e*f^2 + (I*c^3 + 12*I*c)*f^3)*cos(d*x + c) + (3*(-I*c^2 - 4*I)*d*e*f^2 + (I*c^3
+ 12*I*c)*f^3)*sin(2*d*x + 2*c) + 2*(3*(c^2 + 4)*d*e*f^2 - (c^3 + 12*c)*f^3)*sin(d*x + c))*arctan2(sin(d*x +
c) + 1, cos(d*x + c)) - 2*(3*c^2*d*e*f^2 - c^3*f^3 - (3*c^2*d*e*f^2 - c^3*f^3)*cos(2*d*x + 2*c) - 2*(3*I*c^2*d
*e*f^2 - I*c^3*f^3)*cos(d*x + c) - (3*I*c^2*d*e*f^2 - I*c^3*f^3)*sin(2*d*x + 2*c) + 2*(3*c^2*d*e*f^2 - c^3*f^3
)*sin(d*x + c))*arctan2(sin(d*x + c) - 1, cos(d*x + c)) - 2*((d*x + c)^3*f^3 + 3*(d*e*f^2 - c*f^3)*(d*x + c)^2
+ 3*(d^2*e^2*f - 2*c*d*e*f^2 + (c^2 + 4)*f^3)*(d*x + c) - ((d*x + c)^3*f^3 + 3*(d*e*f^2 - c*f^3)*(d*x + c)^2
+ 3*(d^2*e^2*f - 2*c*d*e*f^2 + (c^2 + 4)*f^3)*(d*x + c))*cos(2*d*x + 2*c) - 2*(I*(d*x + c)^3*f^3 + 3*(I*d*e*f^
2 - I*c*f^3)*(d*x + c)^2 + 3*(I*d^2*e^2*f - 2*I*c*d*e*f^2 + (I*c^2 + 4*I)*f^3)*(d*x + c))*cos(d*x + c) - (I*(d
*x + c)^3*f^3 + 3*(I*d*e*f^2 - I*c*f^3)*(d*x + c)^2 + 3*(I*d^2*e^2*f - 2*I*c*d*e*f^2 + (I*c^2 + 4*I)*f^3)*(d*x
+ c))*sin(2*d*x + 2*c) + 2*((d*x + c)^3*f^3 + 3*(d*e*f^2 - c*f^3)*(d*x + c)^2 + 3*(d^2*e^2*f - 2*c*d*e*f^2 +
(c^2 + 4)*f^3)*(d*x + c))*sin(d*x + c))*arctan2(cos(d*x + c), sin(d*x + c) + 1) - 2*((d*x + c)^3*f^3 + 3*(d*e*
f^2 - c*f^3)*(d*x + c)^2 + 3*(d^2*e^2*f - 2*c*d*e*f^2 + c^2*f^3)*(d*x + c) - ((d*x + c)^3*f^3 + 3*(d*e*f^2 - c
*f^3)*(d*x + c)^2 + 3*(d^2*e^2*f - 2*c*d*e*f^2 + c^2*f^3)*(d*x + c))*cos(2*d*x + 2*c) - 2*(I*(d*x + c)^3*f^3 +
3*(I*d*e*f^2 - I*c*f^3)*(d*x + c)^2 + 3*(I*d^2*e^2*f - 2*I*c*d*e*f^2 + I*c^2*f^3)*(d*x + c))*cos(d*x + c) - (
I*(d*x + c)^3*f^3 + 3*(I*d*e*f^2 - I*c*f^3)*(d*x + c)^2 + 3*(I*d^2*e^2*f - 2*I*c*d*e*f^2 + I*c^2*f^3)*(d*x + c
))*sin(2*d*x + 2*c) + 2*((d*x + c)^3*f^3 + 3*(d*e*f^2 - c*f^3)*(d*x + c)^2 + 3*(d^2*e^2*f - 2*c*d*e*f^2 + c^2*
f^3)*(d*x + c))*sin(d*x + c))*arctan2(cos(d*x + c), -sin(d*x + c) + 1) + 12*((d*x + c)^2*f^3 + 2*(d*e*f^2 - c*
f^3)*(d*x + c))*cos(2*d*x + 2*c) + 4*((d*x + c)^3*f^3 - 3*I*d^2*e^2*f + 3*(c^2 + 2*I*c)*d*e*f^2 - (c^3 + 3*I*c
^2)*f^3 + 3*(d*e*f^2 - (c - I)*f^3)*(d*x + c)^2 + 3*(d^2*e^2*f - 2*(c - I)*d*e*f^2 + (c^2 - 2*I*c)*f^3)*(d*x +
c))*cos(d*x + c) - 6*(d^2*e^2*f - 2*c*d*e*f^2 + (d*x + c)^2*f^3 + (c^2 + 4)*f^3 + 2*(d*e*f^2 - c*f^3)*(d*x +
c) - (d^2*e^2*f - 2*c*d*e*f^2 + (d*x + c)^2*f^3 + (c^2 + 4)*f^3 + 2*(d*e*f^2 - c*f^3)*(d*x + c))*cos(2*d*x + 2
*c) - 2*(I*d^2*e^2*f - 2*I*c*d*e*f^2 + I*(d*x + c)^2*f^3 + (I*c^2 + 4*I)*f^3 + 2*(I*d*e*f^2 - I*c*f^3)*(d*x +
c))*cos(d*x + c) - (I*d^2*e^2*f - 2*I*c*d*e*f^2 + I*(d*x + c)^2*f^3 + (I*c^2 + 4*I)*f^3 + 2*(I*d*e*f^2 - I*c*f
^3)*(d*x + c))*sin(2*d*x + 2*c) + 2*(d^2*e^2*f - 2*c*d*e*f^2 + (d*x + c)^2*f^3 + (c^2 + 4)*f^3 + 2*(d*e*f^2 -
c*f^3)*(d*x + c))*sin(d*x + c))*dilog(I*e^(I*d*x + I*c)) + 6*(d^2*e^2*f - 2*c*d*e*f^2 + (d*x + c)^2*f^3 + c^2*
f^3 + 2*(d*e*f^2 - c*f^3)*(d*x + c) - (d^2*e^2*f - 2*c*d*e*f^2 + (d*x + c)^2*f^3 + c^2*f^3 + 2*(d*e*f^2 - c*f^
3)*(d*x + c))*cos(2*d*x + 2*c) + 2*(-I*d^2*e^2*f + 2*I*c*d*e*f^2 - I*(d*x + c)^2*f^3 - I*c^2*f^3 + 2*(-I*d*e*f
^2 + I*c*f^3)*(d*x + c))*cos(d*x + c) + (-I*d^2*e^2*f + 2*I*c*d*e*f^2 - I*(d*x + c)^2*f^3 - I*c^2*f^3 + 2*(-I*
d*e*f^2 + I*c*f^3)*(d*x + c))*sin(2*d*x + 2*c) + 2*(d^2*e^2*f - 2*c*d*e*f^2 + (d*x + c)^2*f^3 + c^2*f^3 + 2*(d
*e*f^2 - c*f^3)*(d*x + c))*sin(d*x + c))*dilog(-I*e^(I*d*x + I*c)) - (I*(d*x + c)^3*f^3 - 3*(-I*c^2 - 4*I)*d*e
*f^2 + (-I*c^3 - 12*I*c)*f^3 - 3*(-I*d*e*f^2 + I*c*f^3)*(d*x + c)^2 - 3*(-I*d^2*e^2*f + 2*I*c*d*e*f^2 + (-I*c^
2 - 4*I)*f^3)*(d*x + c) + (-I*(d*x + c)^3*f^3 - 3*(I*c^2 + 4*I)*d*e*f^2 + (I*c^3 + 12*I*c)*f^3 - 3*(I*d*e*f^2
- I*c*f^3)*(d*x + c)^2 - 3*(I*d^2*e^2*f - 2*I*c*d*e*f^2 + (I*c^2 + 4*I)*f^3)*(d*x + c))*cos(2*d*x + 2*c) + 2*(
(d*x + c)^3*f^3 + 3*(c^2 + 4)*d*e*f^2 - (c^3 + 12*c)*f^3 + 3*(d*e*f^2 - c*f^3)*(d*x + c)^2 + 3*(d^2*e^2*f - 2*
c*d*e*f^2 + (c^2 + 4)*f^3)*(d*x + c))*cos(d*x + c) + ((d*x + c)^3*f^3 + 3*(c^2 + 4)*d*e*f^2 - (c^3 + 12*c)*f^3
+ 3*(d*e*f^2 - c*f^3)*(d*x + c)^2 + 3*(d^2*e^2*f - 2*c*d*e*f^2 + (c^2 + 4)*f^3)*(d*x + c))*sin(2*d*x + 2*c) -
2*(-I*(d*x + c)^3*f^3 + 3*(-I*c^2 - 4*I)*d*e*f^2 + (I*c^3 + 12*I*c)*f^3 + 3*(-I*d*e*f^2 + I*c*f^3)*(d*x + c)^
2 + 3*(-I*d^2*e^2*f + 2*I*c*d*e*f^2 + (-I*c^2 - 4*I)*f^3)*(d*x + c))*sin(d*x + c))*log(cos(d*x + c)^2 + sin(d*
x + c)^2 + 2*sin(d*x + c) + 1) - (-3*I*c^2*d*e*f^2 - I*(d*x + c)^3*f^3 + I*c^3*f^3 - 3*(I*d*e*f^2 - I*c*f^3)*(
d*x + c)^2 - 3*(I*d^2*e^2*f - 2*I*c*d*e*f^2 + I*c^2*f^3)*(d*x + c) + (3*I*c^2*d*e*f^2 + I*(d*x + c)^3*f^3 - I*
c^3*f^3 - 3*(-I*d*e*f^2 + I*c*f^3)*(d*x + c)^2 - 3*(-I*d^2*e^2*f + 2*I*c*d*e*f^2 - I*c^2*f^3)*(d*x + c))*cos(2
*d*x + 2*c) - 2*(3*c^2*d*e*f^2 + (d*x + c)^3*f^3 - c^3*f^3 + 3*(d*e*f^2 - c*f^3)*(d*x + c)^2 + 3*(d^2*e^2*f -
2*c*d*e*f^2 + c^2*f^3)*(d*x + c))*cos(d*x + c) ...

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1871 vs. \(2 (433) = 866\).
time = 0.48, size = 1871, normalized size = 3.73 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*sec(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/4*(2*d^3*f^3*x^3 + 6*d^3*f^2*x^2*e + 6*d^3*f*x*e^2 + 2*d^3*e^3 + 6*(d^2*f^3*x^2 + 2*d^2*f^2*x*e + d^2*f*e^2
)*cos(d*x + c) + 3*(I*d^2*f^3*x^2 + 2*I*d^2*f^2*x*e + I*d^2*f*e^2 + (I*d^2*f^3*x^2 + 2*I*d^2*f^2*x*e + I*d^2*f
*e^2)*sin(d*x + c))*dilog(I*cos(d*x + c) + sin(d*x + c)) + 3*(I*d^2*f^3*x^2 + 2*I*d^2*f^2*x*e + I*d^2*f*e^2 +
4*I*f^3 + (I*d^2*f^3*x^2 + 2*I*d^2*f^2*x*e + I*d^2*f*e^2 + 4*I*f^3)*sin(d*x + c))*dilog(I*cos(d*x + c) - sin(d
*x + c)) + 3*(-I*d^2*f^3*x^2 - 2*I*d^2*f^2*x*e - I*d^2*f*e^2 + (-I*d^2*f^3*x^2 - 2*I*d^2*f^2*x*e - I*d^2*f*e^2
)*sin(d*x + c))*dilog(-I*cos(d*x + c) + sin(d*x + c)) + 3*(-I*d^2*f^3*x^2 - 2*I*d^2*f^2*x*e - I*d^2*f*e^2 - 4*
I*f^3 + (-I*d^2*f^3*x^2 - 2*I*d^2*f^2*x*e - I*d^2*f*e^2 - 4*I*f^3)*sin(d*x + c))*dilog(-I*cos(d*x + c) - sin(d
*x + c)) + (3*c*d^2*f*e^2 - 3*(c^2 + 4)*d*f^2*e + (c^3 + 12*c)*f^3 - d^3*e^3 + (3*c*d^2*f*e^2 - 3*(c^2 + 4)*d*
f^2*e + (c^3 + 12*c)*f^3 - d^3*e^3)*sin(d*x + c))*log(cos(d*x + c) + I*sin(d*x + c) + I) - (c^3*f^3 - 3*c^2*d*
f^2*e + 3*c*d^2*f*e^2 - d^3*e^3 + (c^3*f^3 - 3*c^2*d*f^2*e + 3*c*d^2*f*e^2 - d^3*e^3)*sin(d*x + c))*log(cos(d*
x + c) - I*sin(d*x + c) + I) - (d^3*f^3*x^3 + 12*d*f^3*x + (c^3 + 12*c)*f^3 + 3*(d^3*f*x + c*d^2*f)*e^2 + 3*(d
^3*f^2*x^2 - c^2*d*f^2)*e + (d^3*f^3*x^3 + 12*d*f^3*x + (c^3 + 12*c)*f^3 + 3*(d^3*f*x + c*d^2*f)*e^2 + 3*(d^3*
f^2*x^2 - c^2*d*f^2)*e)*sin(d*x + c))*log(I*cos(d*x + c) + sin(d*x + c) + 1) + (d^3*f^3*x^3 + c^3*f^3 + 3*(d^3
*f*x + c*d^2*f)*e^2 + 3*(d^3*f^2*x^2 - c^2*d*f^2)*e + (d^3*f^3*x^3 + c^3*f^3 + 3*(d^3*f*x + c*d^2*f)*e^2 + 3*(
d^3*f^2*x^2 - c^2*d*f^2)*e)*sin(d*x + c))*log(I*cos(d*x + c) - sin(d*x + c) + 1) - (d^3*f^3*x^3 + 12*d*f^3*x +
(c^3 + 12*c)*f^3 + 3*(d^3*f*x + c*d^2*f)*e^2 + 3*(d^3*f^2*x^2 - c^2*d*f^2)*e + (d^3*f^3*x^3 + 12*d*f^3*x + (c
^3 + 12*c)*f^3 + 3*(d^3*f*x + c*d^2*f)*e^2 + 3*(d^3*f^2*x^2 - c^2*d*f^2)*e)*sin(d*x + c))*log(-I*cos(d*x + c)
+ sin(d*x + c) + 1) + (d^3*f^3*x^3 + c^3*f^3 + 3*(d^3*f*x + c*d^2*f)*e^2 + 3*(d^3*f^2*x^2 - c^2*d*f^2)*e + (d^
3*f^3*x^3 + c^3*f^3 + 3*(d^3*f*x + c*d^2*f)*e^2 + 3*(d^3*f^2*x^2 - c^2*d*f^2)*e)*sin(d*x + c))*log(-I*cos(d*x
+ c) - sin(d*x + c) + 1) + (3*c*d^2*f*e^2 - 3*(c^2 + 4)*d*f^2*e + (c^3 + 12*c)*f^3 - d^3*e^3 + (3*c*d^2*f*e^2
- 3*(c^2 + 4)*d*f^2*e + (c^3 + 12*c)*f^3 - d^3*e^3)*sin(d*x + c))*log(-cos(d*x + c) + I*sin(d*x + c) + I) - (c
^3*f^3 - 3*c^2*d*f^2*e + 3*c*d^2*f*e^2 - d^3*e^3 + (c^3*f^3 - 3*c^2*d*f^2*e + 3*c*d^2*f*e^2 - d^3*e^3)*sin(d*x
+ c))*log(-cos(d*x + c) - I*sin(d*x + c) + I) + 6*(-I*f^3*sin(d*x + c) - I*f^3)*polylog(4, I*cos(d*x + c) + s
in(d*x + c)) + 6*(-I*f^3*sin(d*x + c) - I*f^3)*polylog(4, I*cos(d*x + c) - sin(d*x + c)) + 6*(I*f^3*sin(d*x +
c) + I*f^3)*polylog(4, -I*cos(d*x + c) + sin(d*x + c)) + 6*(I*f^3*sin(d*x + c) + I*f^3)*polylog(4, -I*cos(d*x
+ c) - sin(d*x + c)) + 6*(d*f^3*x + d*f^2*e + (d*f^3*x + d*f^2*e)*sin(d*x + c))*polylog(3, I*cos(d*x + c) + si
n(d*x + c)) - 6*(d*f^3*x + d*f^2*e + (d*f^3*x + d*f^2*e)*sin(d*x + c))*polylog(3, I*cos(d*x + c) - sin(d*x + c
)) + 6*(d*f^3*x + d*f^2*e + (d*f^3*x + d*f^2*e)*sin(d*x + c))*polylog(3, -I*cos(d*x + c) + sin(d*x + c)) - 6*(
d*f^3*x + d*f^2*e + (d*f^3*x + d*f^2*e)*sin(d*x + c))*polylog(3, -I*cos(d*x + c) - sin(d*x + c)))/(a*d^4*sin(d
*x + c) + a*d^4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {e^{3} \sec {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {f^{3} x^{3} \sec {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {3 e f^{2} x^{2} \sec {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {3 e^{2} f x \sec {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)**3*sec(d*x+c)/(a+a*sin(d*x+c)),x)

[Out]

(Integral(e**3*sec(c + d*x)/(sin(c + d*x) + 1), x) + Integral(f**3*x**3*sec(c + d*x)/(sin(c + d*x) + 1), x) +
Integral(3*e*f**2*x**2*sec(c + d*x)/(sin(c + d*x) + 1), x) + Integral(3*e**2*f*x*sec(c + d*x)/(sin(c + d*x) +
1), x))/a

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*sec(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

integrate((f*x + e)^3*sec(d*x + c)/(a*sin(d*x + c) + a), x)

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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e + f*x)^3/(cos(c + d*x)*(a + a*sin(c + d*x))),x)

[Out]

\text{Hanged}

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