\(\int (c+d x)^3 \sec (a+b x) \tan ^2(a+b x) \, dx\) [298]

Optimal result

Integrand size = 22, antiderivative size = 337 \[ \int (c+d x)^3 \sec (a+b x) \tan ^2(a+b x) \, dx=-\frac {6 i d^2 (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b^3}+\frac {i (c+d x)^3 \arctan \left (e^{i (a+b x)}\right )}{b}+\frac {3 i d^3 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^4}-\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{2 b^2}-\frac {3 i d^3 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^4}+\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{2 b^2}+\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^3}-\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^3}+\frac {3 i d^3 \operatorname {PolyLog}\left (4,-i e^{i (a+b x)}\right )}{b^4}-\frac {3 i d^3 \operatorname {PolyLog}\left (4,i e^{i (a+b x)}\right )}{b^4}-\frac {3 d (c+d x)^2 \sec (a+b x)}{2 b^2}+\frac {(c+d x)^3 \sec (a+b x) \tan (a+b x)}{2 b} \]

[Out]

-6*I*d^2*(d*x+c)*arctan(exp(I*(b*x+a)))/b^3+I*(d*x+c)^3*arctan(exp(I*(b*x+a)))/b+3*I*d^3*polylog(2,-I*exp(I*(b
*x+a)))/b^4-3/2*I*d*(d*x+c)^2*polylog(2,-I*exp(I*(b*x+a)))/b^2-3*I*d^3*polylog(2,I*exp(I*(b*x+a)))/b^4+3/2*I*d
*(d*x+c)^2*polylog(2,I*exp(I*(b*x+a)))/b^2+3*d^2*(d*x+c)*polylog(3,-I*exp(I*(b*x+a)))/b^3-3*d^2*(d*x+c)*polylo
g(3,I*exp(I*(b*x+a)))/b^3+3*I*d^3*polylog(4,-I*exp(I*(b*x+a)))/b^4-3*I*d^3*polylog(4,I*exp(I*(b*x+a)))/b^4-3/2
*d*(d*x+c)^2*sec(b*x+a)/b^2+1/2*(d*x+c)^3*sec(b*x+a)*tan(b*x+a)/b

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.00, number of steps used = 25, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {4498, 4266, 2611, 6744, 2320, 6724, 4271, 2317, 2438} \[ \int (c+d x)^3 \sec (a+b x) \tan ^2(a+b x) \, dx=-\frac {6 i d^2 (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b^3}+\frac {i (c+d x)^3 \arctan \left (e^{i (a+b x)}\right )}{b}+\frac {3 i d^3 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^4}-\frac {3 i d^3 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^4}+\frac {3 i d^3 \operatorname {PolyLog}\left (4,-i e^{i (a+b x)}\right )}{b^4}-\frac {3 i d^3 \operatorname {PolyLog}\left (4,i e^{i (a+b x)}\right )}{b^4}+\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^3}-\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^3}-\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{2 b^2}+\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{2 b^2}-\frac {3 d (c+d x)^2 \sec (a+b x)}{2 b^2}+\frac {(c+d x)^3 \tan (a+b x) \sec (a+b x)}{2 b} \]

[In]

Int[(c + d*x)^3*Sec[a + b*x]*Tan[a + b*x]^2,x]

[Out]

((-6*I)*d^2*(c + d*x)*ArcTan[E^(I*(a + b*x))])/b^3 + (I*(c + d*x)^3*ArcTan[E^(I*(a + b*x))])/b + ((3*I)*d^3*Po
lyLog[2, (-I)*E^(I*(a + b*x))])/b^4 - (((3*I)/2)*d*(c + d*x)^2*PolyLog[2, (-I)*E^(I*(a + b*x))])/b^2 - ((3*I)*
d^3*PolyLog[2, I*E^(I*(a + b*x))])/b^4 + (((3*I)/2)*d*(c + d*x)^2*PolyLog[2, I*E^(I*(a + b*x))])/b^2 + (3*d^2*
(c + d*x)*PolyLog[3, (-I)*E^(I*(a + b*x))])/b^3 - (3*d^2*(c + d*x)*PolyLog[3, I*E^(I*(a + b*x))])/b^3 + ((3*I)
*d^3*PolyLog[4, (-I)*E^(I*(a + b*x))])/b^4 - ((3*I)*d^3*PolyLog[4, I*E^(I*(a + b*x))])/b^4 - (3*d*(c + d*x)^2*
Sec[a + b*x])/(2*b^2) + ((c + d*x)^3*Sec[a + b*x]*Tan[a + b*x])/(2*b)

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 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 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 4498

Int[((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]*Tan[(a_.) + (b_.)*(x_)]^(p_), x_Symbol] :> -Int[(c + d*
x)^m*Sec[a + b*x]*Tan[a + b*x]^(p - 2), x] + Int[(c + d*x)^m*Sec[a + b*x]^3*Tan[a + b*x]^(p - 2), x] /; FreeQ[
{a, b, c, d, m}, x] && IGtQ[p/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*} \text {integral}& = -\int (c+d x)^3 \sec (a+b x) \, dx+\int (c+d x)^3 \sec ^3(a+b x) \, dx \\ & = \frac {2 i (c+d x)^3 \arctan \left (e^{i (a+b x)}\right )}{b}-\frac {3 d (c+d x)^2 \sec (a+b x)}{2 b^2}+\frac {(c+d x)^3 \sec (a+b x) \tan (a+b x)}{2 b}+\frac {1}{2} \int (c+d x)^3 \sec (a+b x) \, dx+\frac {(3 d) \int (c+d x)^2 \log \left (1-i e^{i (a+b x)}\right ) \, dx}{b}-\frac {(3 d) \int (c+d x)^2 \log \left (1+i e^{i (a+b x)}\right ) \, dx}{b}+\frac {\left (3 d^2\right ) \int (c+d x) \sec (a+b x) \, dx}{b^2} \\ & = -\frac {6 i d^2 (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b^3}+\frac {i (c+d x)^3 \arctan \left (e^{i (a+b x)}\right )}{b}-\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^2}+\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^2}-\frac {3 d (c+d x)^2 \sec (a+b x)}{2 b^2}+\frac {(c+d x)^3 \sec (a+b x) \tan (a+b x)}{2 b}-\frac {(3 d) \int (c+d x)^2 \log \left (1-i e^{i (a+b x)}\right ) \, dx}{2 b}+\frac {(3 d) \int (c+d x)^2 \log \left (1+i e^{i (a+b x)}\right ) \, dx}{2 b}+\frac {\left (6 i d^2\right ) \int (c+d x) \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right ) \, dx}{b^2}-\frac {\left (6 i d^2\right ) \int (c+d x) \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right ) \, dx}{b^2}-\frac {\left (3 d^3\right ) \int \log \left (1-i e^{i (a+b x)}\right ) \, dx}{b^3}+\frac {\left (3 d^3\right ) \int \log \left (1+i e^{i (a+b x)}\right ) \, dx}{b^3} \\ & = -\frac {6 i d^2 (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b^3}+\frac {i (c+d x)^3 \arctan \left (e^{i (a+b x)}\right )}{b}-\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{2 b^2}+\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{2 b^2}+\frac {6 d^2 (c+d x) \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^3}-\frac {6 d^2 (c+d x) \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^3}-\frac {3 d (c+d x)^2 \sec (a+b x)}{2 b^2}+\frac {(c+d x)^3 \sec (a+b x) \tan (a+b x)}{2 b}-\frac {\left (3 i d^2\right ) \int (c+d x) \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right ) \, dx}{b^2}+\frac {\left (3 i d^2\right ) \int (c+d x) \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right ) \, dx}{b^2}+\frac {\left (3 i d^3\right ) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^4}-\frac {\left (3 i d^3\right ) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^4}-\frac {\left (6 d^3\right ) \int \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right ) \, dx}{b^3}+\frac {\left (6 d^3\right ) \int \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right ) \, dx}{b^3} \\ & = -\frac {6 i d^2 (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b^3}+\frac {i (c+d x)^3 \arctan \left (e^{i (a+b x)}\right )}{b}+\frac {3 i d^3 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^4}-\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{2 b^2}-\frac {3 i d^3 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^4}+\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{2 b^2}+\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^3}-\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^3}-\frac {3 d (c+d x)^2 \sec (a+b x)}{2 b^2}+\frac {(c+d x)^3 \sec (a+b x) \tan (a+b x)}{2 b}+\frac {\left (6 i d^3\right ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(3,-i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^4}-\frac {\left (6 i d^3\right ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(3,i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^4}+\frac {\left (3 d^3\right ) \int \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right ) \, dx}{b^3}-\frac {\left (3 d^3\right ) \int \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right ) \, dx}{b^3} \\ & = -\frac {6 i d^2 (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b^3}+\frac {i (c+d x)^3 \arctan \left (e^{i (a+b x)}\right )}{b}+\frac {3 i d^3 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^4}-\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{2 b^2}-\frac {3 i d^3 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^4}+\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{2 b^2}+\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^3}-\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^3}+\frac {6 i d^3 \operatorname {PolyLog}\left (4,-i e^{i (a+b x)}\right )}{b^4}-\frac {6 i d^3 \operatorname {PolyLog}\left (4,i e^{i (a+b x)}\right )}{b^4}-\frac {3 d (c+d x)^2 \sec (a+b x)}{2 b^2}+\frac {(c+d x)^3 \sec (a+b x) \tan (a+b x)}{2 b}-\frac {\left (3 i d^3\right ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(3,-i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^4}+\frac {\left (3 i d^3\right ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(3,i x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^4} \\ & = -\frac {6 i d^2 (c+d x) \arctan \left (e^{i (a+b x)}\right )}{b^3}+\frac {i (c+d x)^3 \arctan \left (e^{i (a+b x)}\right )}{b}+\frac {3 i d^3 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^4}-\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{2 b^2}-\frac {3 i d^3 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^4}+\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{2 b^2}+\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^3}-\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^3}+\frac {3 i d^3 \operatorname {PolyLog}\left (4,-i e^{i (a+b x)}\right )}{b^4}-\frac {3 i d^3 \operatorname {PolyLog}\left (4,i e^{i (a+b x)}\right )}{b^4}-\frac {3 d (c+d x)^2 \sec (a+b x)}{2 b^2}+\frac {(c+d x)^3 \sec (a+b x) \tan (a+b x)}{2 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.75 (sec) , antiderivative size = 530, normalized size of antiderivative = 1.57 \[ \int (c+d x)^3 \sec (a+b x) \tan ^2(a+b x) \, dx=\frac {2 i b^3 c^3 \arctan \left (e^{i (a+b x)}\right )-12 i b c d^2 \arctan \left (e^{i (a+b x)}\right )-3 b^3 c^2 d x \log \left (1-i e^{i (a+b x)}\right )+6 b d^3 x \log \left (1-i e^{i (a+b x)}\right )-3 b^3 c d^2 x^2 \log \left (1-i e^{i (a+b x)}\right )-b^3 d^3 x^3 \log \left (1-i e^{i (a+b x)}\right )+3 b^3 c^2 d x \log \left (1+i e^{i (a+b x)}\right )-6 b d^3 x \log \left (1+i e^{i (a+b x)}\right )+3 b^3 c d^2 x^2 \log \left (1+i e^{i (a+b x)}\right )+b^3 d^3 x^3 \log \left (1+i e^{i (a+b x)}\right )-3 i d \left (-2 d^2+b^2 (c+d x)^2\right ) \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )+3 i d \left (-2 d^2+b^2 (c+d x)^2\right ) \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )+6 b c d^2 \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )+6 b d^3 x \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )-6 b c d^2 \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )-6 b d^3 x \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )+6 i d^3 \operatorname {PolyLog}\left (4,-i e^{i (a+b x)}\right )-6 i d^3 \operatorname {PolyLog}\left (4,i e^{i (a+b x)}\right )+b^2 (c+d x)^2 \sec (a+b x) (-3 d+b (c+d x) \tan (a+b x))}{2 b^4} \]

[In]

Integrate[(c + d*x)^3*Sec[a + b*x]*Tan[a + b*x]^2,x]

[Out]

((2*I)*b^3*c^3*ArcTan[E^(I*(a + b*x))] - (12*I)*b*c*d^2*ArcTan[E^(I*(a + b*x))] - 3*b^3*c^2*d*x*Log[1 - I*E^(I
*(a + b*x))] + 6*b*d^3*x*Log[1 - I*E^(I*(a + b*x))] - 3*b^3*c*d^2*x^2*Log[1 - I*E^(I*(a + b*x))] - b^3*d^3*x^3
*Log[1 - I*E^(I*(a + b*x))] + 3*b^3*c^2*d*x*Log[1 + I*E^(I*(a + b*x))] - 6*b*d^3*x*Log[1 + I*E^(I*(a + b*x))]
+ 3*b^3*c*d^2*x^2*Log[1 + I*E^(I*(a + b*x))] + b^3*d^3*x^3*Log[1 + I*E^(I*(a + b*x))] - (3*I)*d*(-2*d^2 + b^2*
(c + d*x)^2)*PolyLog[2, (-I)*E^(I*(a + b*x))] + (3*I)*d*(-2*d^2 + b^2*(c + d*x)^2)*PolyLog[2, I*E^(I*(a + b*x)
)] + 6*b*c*d^2*PolyLog[3, (-I)*E^(I*(a + b*x))] + 6*b*d^3*x*PolyLog[3, (-I)*E^(I*(a + b*x))] - 6*b*c*d^2*PolyL
og[3, I*E^(I*(a + b*x))] - 6*b*d^3*x*PolyLog[3, I*E^(I*(a + b*x))] + (6*I)*d^3*PolyLog[4, (-I)*E^(I*(a + b*x))
] - (6*I)*d^3*PolyLog[4, I*E^(I*(a + b*x))] + b^2*(c + d*x)^2*Sec[a + b*x]*(-3*d + b*(c + d*x)*Tan[a + b*x]))/
(2*b^4)

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1126 vs. \(2 (293 ) = 586\).

Time = 2.01 (sec) , antiderivative size = 1127, normalized size of antiderivative = 3.34

method	result	size
risch	\(\text {Expression too large to display}\)	\(1127\)

[In]

int((d*x+c)^3*sec(b*x+a)*tan(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

-3*I*d^3*polylog(2,I*exp(I*(b*x+a)))/b^4-3*I*d^3*polylog(4,I*exp(I*(b*x+a)))/b^4-3/2/b*c*d^2*ln(1-I*exp(I*(b*x
+a)))*x^2-3/2/b^3*a^2*c*d^2*ln(1+I*exp(I*(b*x+a)))-3/2/b*c^2*d*ln(1-I*exp(I*(b*x+a)))*x-3/2/b^2*c^2*d*ln(1-I*e
xp(I*(b*x+a)))*a+3*I/b^3*c*d^2*a^2*arctan(exp(I*(b*x+a)))+3*I*d^3*polylog(2,-I*exp(I*(b*x+a)))/b^4+3*I*d^3*pol
ylog(4,-I*exp(I*(b*x+a)))/b^4+3*I/b^2*c*d^2*polylog(2,I*exp(I*(b*x+a)))*x-3*I/b^2*c*d^2*polylog(2,-I*exp(I*(b*
x+a)))*x-3*I/b^2*c^2*d*a*arctan(exp(I*(b*x+a)))+3/b^3*d^3*polylog(3,-I*exp(I*(b*x+a)))*x+1/2/b*d^3*ln(1+I*exp(
I*(b*x+a)))*x^3-1/2/b*d^3*ln(1-I*exp(I*(b*x+a)))*x^3-3/b^3*d^3*ln(1+I*exp(I*(b*x+a)))*x+3/b^4*d^3*ln(1-I*exp(I
*(b*x+a)))*a+3/b^3*d^3*ln(1-I*exp(I*(b*x+a)))*x-3/b^4*d^3*ln(1+I*exp(I*(b*x+a)))*a-3/b^3*d^3*polylog(3,I*exp(I
*(b*x+a)))*x+3/b^3*c*d^2*polylog(3,-I*exp(I*(b*x+a)))-1/2/b^4*a^3*d^3*ln(1-I*exp(I*(b*x+a)))-3/b^3*c*d^2*polyl
og(3,I*exp(I*(b*x+a)))+I/b*c^3*arctan(exp(I*(b*x+a)))+3/2/b*c*d^2*ln(1+I*exp(I*(b*x+a)))*x^2+3/2/b^3*a^2*c*d^2
*ln(1-I*exp(I*(b*x+a)))+3/2/b*c^2*d*ln(1+I*exp(I*(b*x+a)))*x+3/2/b^2*c^2*d*ln(1+I*exp(I*(b*x+a)))*a-I/b^4*d^3*
a^3*arctan(exp(I*(b*x+a)))+6*I/b^4*d^3*a*arctan(exp(I*(b*x+a)))+3/2*I/b^2*c^2*d*polylog(2,I*exp(I*(b*x+a)))-6*
I/b^3*c*d^2*arctan(exp(I*(b*x+a)))+3/2*I/b^2*d^3*polylog(2,I*exp(I*(b*x+a)))*x^2-3/2*I/b^2*d^3*polylog(2,-I*ex
p(I*(b*x+a)))*x^2-3/2*I/b^2*c^2*d*polylog(2,-I*exp(I*(b*x+a)))-I/b^2/(exp(2*I*(b*x+a))+1)^2*(d^3*x^3*b*exp(3*I
*(b*x+a))+3*c*d^2*x^2*b*exp(3*I*(b*x+a))+3*c^2*d*x*b*exp(3*I*(b*x+a))-d^3*x^3*b*exp(I*(b*x+a))+c^3*b*exp(3*I*(
b*x+a))-3*c*d^2*x^2*b*exp(I*(b*x+a))-3*I*d^3*x^2*exp(3*I*(b*x+a))-3*c^2*d*x*b*exp(I*(b*x+a))-6*I*c*d^2*x*exp(3
*I*(b*x+a))-c^3*b*exp(I*(b*x+a))-3*I*c^2*d*exp(3*I*(b*x+a))-3*I*d^3*x^2*exp(I*(b*x+a))-6*I*c*d^2*x*exp(I*(b*x+
a))-3*I*c^2*d*exp(I*(b*x+a)))+1/2/b^4*a^3*d^3*ln(1+I*exp(I*(b*x+a)))

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1315 vs. \(2 (273) = 546\).

Time = 0.33 (sec) , antiderivative size = 1315, normalized size of antiderivative = 3.90 \[ \int (c+d x)^3 \sec (a+b x) \tan ^2(a+b x) \, dx=\text {Too large to display} \]

[In]

integrate((d*x+c)^3*sec(b*x+a)*tan(b*x+a)^2,x, algorithm="fricas")

[Out]

1/4*(-6*I*d^3*cos(b*x + a)^2*polylog(4, I*cos(b*x + a) + sin(b*x + a)) - 6*I*d^3*cos(b*x + a)^2*polylog(4, I*c
os(b*x + a) - sin(b*x + a)) + 6*I*d^3*cos(b*x + a)^2*polylog(4, -I*cos(b*x + a) + sin(b*x + a)) + 6*I*d^3*cos(
b*x + a)^2*polylog(4, -I*cos(b*x + a) - sin(b*x + a)) - 3*(-I*b^2*d^3*x^2 - 2*I*b^2*c*d^2*x - I*b^2*c^2*d + 2*
I*d^3)*cos(b*x + a)^2*dilog(I*cos(b*x + a) + sin(b*x + a)) - 3*(-I*b^2*d^3*x^2 - 2*I*b^2*c*d^2*x - I*b^2*c^2*d
+ 2*I*d^3)*cos(b*x + a)^2*dilog(I*cos(b*x + a) - sin(b*x + a)) - 3*(I*b^2*d^3*x^2 + 2*I*b^2*c*d^2*x + I*b^2*c
^2*d - 2*I*d^3)*cos(b*x + a)^2*dilog(-I*cos(b*x + a) + sin(b*x + a)) - 3*(I*b^2*d^3*x^2 + 2*I*b^2*c*d^2*x + I*
b^2*c^2*d - 2*I*d^3)*cos(b*x + a)^2*dilog(-I*cos(b*x + a) - sin(b*x + a)) - (b^3*c^3 - 3*a*b^2*c^2*d + 3*(a^2
- 2)*b*c*d^2 - (a^3 - 6*a)*d^3)*cos(b*x + a)^2*log(cos(b*x + a) + I*sin(b*x + a) + I) + (b^3*c^3 - 3*a*b^2*c^2
*d + 3*(a^2 - 2)*b*c*d^2 - (a^3 - 6*a)*d^3)*cos(b*x + a)^2*log(cos(b*x + a) - I*sin(b*x + a) + I) - (b^3*d^3*x
^3 + 3*b^3*c*d^2*x^2 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + (a^3 - 6*a)*d^3 + 3*(b^3*c^2*d - 2*b*d^3)*x)*cos(b*x +
a)^2*log(I*cos(b*x + a) + sin(b*x + a) + 1) + (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 +
(a^3 - 6*a)*d^3 + 3*(b^3*c^2*d - 2*b*d^3)*x)*cos(b*x + a)^2*log(I*cos(b*x + a) - sin(b*x + a) + 1) - (b^3*d^3
*x^3 + 3*b^3*c*d^2*x^2 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + (a^3 - 6*a)*d^3 + 3*(b^3*c^2*d - 2*b*d^3)*x)*cos(b*x
+ a)^2*log(-I*cos(b*x + a) + sin(b*x + a) + 1) + (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^
2 + (a^3 - 6*a)*d^3 + 3*(b^3*c^2*d - 2*b*d^3)*x)*cos(b*x + a)^2*log(-I*cos(b*x + a) - sin(b*x + a) + 1) - (b^3
*c^3 - 3*a*b^2*c^2*d + 3*(a^2 - 2)*b*c*d^2 - (a^3 - 6*a)*d^3)*cos(b*x + a)^2*log(-cos(b*x + a) + I*sin(b*x + a
) + I) + (b^3*c^3 - 3*a*b^2*c^2*d + 3*(a^2 - 2)*b*c*d^2 - (a^3 - 6*a)*d^3)*cos(b*x + a)^2*log(-cos(b*x + a) -
I*sin(b*x + a) + I) + 6*(b*d^3*x + b*c*d^2)*cos(b*x + a)^2*polylog(3, I*cos(b*x + a) + sin(b*x + a)) - 6*(b*d^
3*x + b*c*d^2)*cos(b*x + a)^2*polylog(3, I*cos(b*x + a) - sin(b*x + a)) + 6*(b*d^3*x + b*c*d^2)*cos(b*x + a)^2
*polylog(3, -I*cos(b*x + a) + sin(b*x + a)) - 6*(b*d^3*x + b*c*d^2)*cos(b*x + a)^2*polylog(3, -I*cos(b*x + a)
- sin(b*x + a)) - 6*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + b^2*c^2*d)*cos(b*x + a) + 2*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2
+ 3*b^3*c^2*d*x + b^3*c^3)*sin(b*x + a))/(b^4*cos(b*x + a)^2)

Sympy [F]

\[ \int (c+d x)^3 \sec (a+b x) \tan ^2(a+b x) \, dx=\int \left (c + d x\right )^{3} \tan ^{2}{\left (a + b x \right )} \sec {\left (a + b x \right )}\, dx \]

[In]

integrate((d*x+c)**3*sec(b*x+a)*tan(b*x+a)**2,x)

[Out]

Integral((c + d*x)**3*tan(a + b*x)**2*sec(a + b*x), x)

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 3831 vs. \(2 (273) = 546\).

Time = 1.37 (sec) , antiderivative size = 3831, normalized size of antiderivative = 11.37 \[ \int (c+d x)^3 \sec (a+b x) \tan ^2(a+b x) \, dx=\text {Too large to display} \]

[In]

integrate((d*x+c)^3*sec(b*x+a)*tan(b*x+a)^2,x, algorithm="maxima")

[Out]

-1/4*(c^3*(2*sin(b*x + a)/(sin(b*x + a)^2 - 1) + log(sin(b*x + a) + 1) - log(sin(b*x + a) - 1)) - 3*a*c^2*d*(2
*sin(b*x + a)/(sin(b*x + a)^2 - 1) + log(sin(b*x + a) + 1) - log(sin(b*x + a) - 1))/b + 3*a^2*c*d^2*(2*sin(b*x
+ a)/(sin(b*x + a)^2 - 1) + log(sin(b*x + a) + 1) - log(sin(b*x + a) - 1))/b^2 - a^3*d^3*(2*sin(b*x + a)/(sin
(b*x + a)^2 - 1) + log(sin(b*x + a) + 1) - log(sin(b*x + a) - 1))/b^3 - 4*(2*((b*x + a)^3*d^3 - 6*b*c*d^2 + 6*
a*d^3 + 3*(b*c*d^2 - a*d^3)*(b*x + a)^2 + 3*(b^2*c^2*d - 2*a*b*c*d^2 + (a^2 - 2)*d^3)*(b*x + a) + ((b*x + a)^3
*d^3 - 6*b*c*d^2 + 6*a*d^3 + 3*(b*c*d^2 - a*d^3)*(b*x + a)^2 + 3*(b^2*c^2*d - 2*a*b*c*d^2 + (a^2 - 2)*d^3)*(b*
x + a))*cos(4*b*x + 4*a) + 2*((b*x + a)^3*d^3 - 6*b*c*d^2 + 6*a*d^3 + 3*(b*c*d^2 - a*d^3)*(b*x + a)^2 + 3*(b^2
*c^2*d - 2*a*b*c*d^2 + (a^2 - 2)*d^3)*(b*x + a))*cos(2*b*x + 2*a) + (I*(b*x + a)^3*d^3 - 6*I*b*c*d^2 + 6*I*a*d
^3 + 3*(I*b*c*d^2 - I*a*d^3)*(b*x + a)^2 + 3*(I*b^2*c^2*d - 2*I*a*b*c*d^2 + (I*a^2 - 2*I)*d^3)*(b*x + a))*sin(
4*b*x + 4*a) + 2*(I*(b*x + a)^3*d^3 - 6*I*b*c*d^2 + 6*I*a*d^3 + 3*(I*b*c*d^2 - I*a*d^3)*(b*x + a)^2 + 3*(I*b^2
*c^2*d - 2*I*a*b*c*d^2 + (I*a^2 - 2*I)*d^3)*(b*x + a))*sin(2*b*x + 2*a))*arctan2(cos(b*x + a), sin(b*x + a) +
1) + 2*((b*x + a)^3*d^3 - 6*b*c*d^2 + 6*a*d^3 + 3*(b*c*d^2 - a*d^3)*(b*x + a)^2 + 3*(b^2*c^2*d - 2*a*b*c*d^2 +
(a^2 - 2)*d^3)*(b*x + a) + ((b*x + a)^3*d^3 - 6*b*c*d^2 + 6*a*d^3 + 3*(b*c*d^2 - a*d^3)*(b*x + a)^2 + 3*(b^2*
c^2*d - 2*a*b*c*d^2 + (a^2 - 2)*d^3)*(b*x + a))*cos(4*b*x + 4*a) + 2*((b*x + a)^3*d^3 - 6*b*c*d^2 + 6*a*d^3 +
3*(b*c*d^2 - a*d^3)*(b*x + a)^2 + 3*(b^2*c^2*d - 2*a*b*c*d^2 + (a^2 - 2)*d^3)*(b*x + a))*cos(2*b*x + 2*a) + (I
*(b*x + a)^3*d^3 - 6*I*b*c*d^2 + 6*I*a*d^3 + 3*(I*b*c*d^2 - I*a*d^3)*(b*x + a)^2 + 3*(I*b^2*c^2*d - 2*I*a*b*c*
d^2 + (I*a^2 - 2*I)*d^3)*(b*x + a))*sin(4*b*x + 4*a) + 2*(I*(b*x + a)^3*d^3 - 6*I*b*c*d^2 + 6*I*a*d^3 + 3*(I*b
*c*d^2 - I*a*d^3)*(b*x + a)^2 + 3*(I*b^2*c^2*d - 2*I*a*b*c*d^2 + (I*a^2 - 2*I)*d^3)*(b*x + a))*sin(2*b*x + 2*a
))*arctan2(cos(b*x + a), -sin(b*x + a) + 1) - 4*((b*x + a)^3*d^3 - 3*I*b^2*c^2*d + 6*I*a*b*c*d^2 - 3*I*a^2*d^3
+ 3*(b*c*d^2 - (a + I)*d^3)*(b*x + a)^2 + 3*(b^2*c^2*d - 2*(a + I)*b*c*d^2 + (a^2 + 2*I*a)*d^3)*(b*x + a))*co
s(3*b*x + 3*a) + 4*((b*x + a)^3*d^3 + 3*I*b^2*c^2*d - 6*I*a*b*c*d^2 + 3*I*a^2*d^3 + 3*(b*c*d^2 - (a - I)*d^3)*
(b*x + a)^2 + 3*(b^2*c^2*d - 2*(a - I)*b*c*d^2 + (a^2 - 2*I*a)*d^3)*(b*x + a))*cos(b*x + a) + 6*(b^2*c^2*d - 2
*a*b*c*d^2 + (b*x + a)^2*d^3 + (a^2 - 2)*d^3 + 2*(b*c*d^2 - a*d^3)*(b*x + a) + (b^2*c^2*d - 2*a*b*c*d^2 + (b*x
+ a)^2*d^3 + (a^2 - 2)*d^3 + 2*(b*c*d^2 - a*d^3)*(b*x + a))*cos(4*b*x + 4*a) + 2*(b^2*c^2*d - 2*a*b*c*d^2 + (
b*x + a)^2*d^3 + (a^2 - 2)*d^3 + 2*(b*c*d^2 - a*d^3)*(b*x + a))*cos(2*b*x + 2*a) + (I*b^2*c^2*d - 2*I*a*b*c*d^
2 + I*(b*x + a)^2*d^3 + (I*a^2 - 2*I)*d^3 + 2*(I*b*c*d^2 - I*a*d^3)*(b*x + a))*sin(4*b*x + 4*a) + 2*(I*b^2*c^2
*d - 2*I*a*b*c*d^2 + I*(b*x + a)^2*d^3 + (I*a^2 - 2*I)*d^3 + 2*(I*b*c*d^2 - I*a*d^3)*(b*x + a))*sin(2*b*x + 2*
a))*dilog(I*e^(I*b*x + I*a)) - 6*(b^2*c^2*d - 2*a*b*c*d^2 + (b*x + a)^2*d^3 + (a^2 - 2)*d^3 + 2*(b*c*d^2 - a*d
^3)*(b*x + a) + (b^2*c^2*d - 2*a*b*c*d^2 + (b*x + a)^2*d^3 + (a^2 - 2)*d^3 + 2*(b*c*d^2 - a*d^3)*(b*x + a))*co
s(4*b*x + 4*a) + 2*(b^2*c^2*d - 2*a*b*c*d^2 + (b*x + a)^2*d^3 + (a^2 - 2)*d^3 + 2*(b*c*d^2 - a*d^3)*(b*x + a))
*cos(2*b*x + 2*a) - (-I*b^2*c^2*d + 2*I*a*b*c*d^2 - I*(b*x + a)^2*d^3 + (-I*a^2 + 2*I)*d^3 + 2*(-I*b*c*d^2 + I
*a*d^3)*(b*x + a))*sin(4*b*x + 4*a) - 2*(-I*b^2*c^2*d + 2*I*a*b*c*d^2 - I*(b*x + a)^2*d^3 + (-I*a^2 + 2*I)*d^3
+ 2*(-I*b*c*d^2 + I*a*d^3)*(b*x + a))*sin(2*b*x + 2*a))*dilog(-I*e^(I*b*x + I*a)) - (-I*(b*x + a)^3*d^3 + 6*I
*b*c*d^2 - 6*I*a*d^3 - 3*(I*b*c*d^2 - I*a*d^3)*(b*x + a)^2 - 3*(I*b^2*c^2*d - 2*I*a*b*c*d^2 + (I*a^2 - 2*I)*d^
3)*(b*x + a) + (-I*(b*x + a)^3*d^3 + 6*I*b*c*d^2 - 6*I*a*d^3 - 3*(I*b*c*d^2 - I*a*d^3)*(b*x + a)^2 - 3*(I*b^2*
c^2*d - 2*I*a*b*c*d^2 + (I*a^2 - 2*I)*d^3)*(b*x + a))*cos(4*b*x + 4*a) - 2*(I*(b*x + a)^3*d^3 - 6*I*b*c*d^2 +
6*I*a*d^3 + 3*(I*b*c*d^2 - I*a*d^3)*(b*x + a)^2 + 3*(I*b^2*c^2*d - 2*I*a*b*c*d^2 + (I*a^2 - 2*I)*d^3)*(b*x + a
))*cos(2*b*x + 2*a) + ((b*x + a)^3*d^3 - 6*b*c*d^2 + 6*a*d^3 + 3*(b*c*d^2 - a*d^3)*(b*x + a)^2 + 3*(b^2*c^2*d
- 2*a*b*c*d^2 + (a^2 - 2)*d^3)*(b*x + a))*sin(4*b*x + 4*a) + 2*((b*x + a)^3*d^3 - 6*b*c*d^2 + 6*a*d^3 + 3*(b*c
*d^2 - a*d^3)*(b*x + a)^2 + 3*(b^2*c^2*d - 2*a*b*c*d^2 + (a^2 - 2)*d^3)*(b*x + a))*sin(2*b*x + 2*a))*log(cos(b
*x + a)^2 + sin(b*x + a)^2 + 2*sin(b*x + a) + 1) - (I*(b*x + a)^3*d^3 - 6*I*b*c*d^2 + 6*I*a*d^3 - 3*(-I*b*c*d^
2 + I*a*d^3)*(b*x + a)^2 - 3*(-I*b^2*c^2*d + 2*I*a*b*c*d^2 + (-I*a^2 + 2*I)*d^3)*(b*x + a) + (I*(b*x + a)^3*d^
3 - 6*I*b*c*d^2 + 6*I*a*d^3 - 3*(-I*b*c*d^2 + I*a*d^3)*(b*x + a)^2 - 3*(-I*b^2*c^2*d + 2*I*a*b*c*d^2 + (-I*a^2
+ 2*I)*d^3)*(b*x + a))*cos(4*b*x + 4*a) - 2*(-I*(b*x + a)^3*d^3 + 6*I*b*c*d^2 - 6*I*a*d^3 + 3*(-I*b*c*d^2 + I
*a*d^3)*(b*x + a)^2 + 3*(-I*b^2*c^2*d + 2*I*a*b*c*d^2 + (-I*a^2 + 2*I)*d^3)*(b*x + a))*cos(2*b*x + 2*a) - ((b*
x + a)^3*d^3 - 6*b*c*d^2 + 6*a*d^3 + 3*(b*c*d^2 - a*d^3)*(b*x + a)^2 + 3*(b^2*c^2*d - 2*a*b*c*d^2 + (a^2 - 2)*
d^3)*(b*x + a))*sin(4*b*x + 4*a) - 2*((b*x + a)^3*d^3 - 6*b*c*d^2 + 6*a*d^3 + 3*(b*c*d^2 - a*d^3)*(b*x + a)^2
+ 3*(b^2*c^2*d - 2*a*b*c*d^2 + (a^2 - 2)*d^3)*(b*x + a))*sin(2*b*x + 2*a))*log(cos(b*x + a)^2 + sin(b*x + a)^2
- 2*sin(b*x + a) + 1) - 12*(d^3*cos(4*b*x + 4*a) + 2*d^3*cos(2*b*x + 2*a) + I*d^3*sin(4*b*x + 4*a) + 2*I*d^3*
sin(2*b*x + 2*a) + d^3)*polylog(4, I*e^(I*b*x + I*a)) + 12*(d^3*cos(4*b*x + 4*a) + 2*d^3*cos(2*b*x + 2*a) + I*
d^3*sin(4*b*x + 4*a) + 2*I*d^3*sin(2*b*x + 2*a) + d^3)*polylog(4, -I*e^(I*b*x + I*a)) + 12*(I*b*c*d^2 + I*(b*x
+ a)*d^3 - I*a*d^3 + (I*b*c*d^2 + I*(b*x + a)*d^3 - I*a*d^3)*cos(4*b*x + 4*a) + 2*(I*b*c*d^2 + I*(b*x + a)*d^
3 - I*a*d^3)*cos(2*b*x + 2*a) - (b*c*d^2 + (b*x + a)*d^3 - a*d^3)*sin(4*b*x + 4*a) - 2*(b*c*d^2 + (b*x + a)*d^
3 - a*d^3)*sin(2*b*x + 2*a))*polylog(3, I*e^(I*b*x + I*a)) + 12*(-I*b*c*d^2 - I*(b*x + a)*d^3 + I*a*d^3 + (-I*
b*c*d^2 - I*(b*x + a)*d^3 + I*a*d^3)*cos(4*b*x + 4*a) + 2*(-I*b*c*d^2 - I*(b*x + a)*d^3 + I*a*d^3)*cos(2*b*x +
2*a) + (b*c*d^2 + (b*x + a)*d^3 - a*d^3)*sin(4*b*x + 4*a) + 2*(b*c*d^2 + (b*x + a)*d^3 - a*d^3)*sin(2*b*x + 2
*a))*polylog(3, -I*e^(I*b*x + I*a)) + 4*(-I*(b*x + a)^3*d^3 - 3*b^2*c^2*d + 6*a*b*c*d^2 - 3*a^2*d^3 + 3*(-I*b*
c*d^2 + (I*a - 1)*d^3)*(b*x + a)^2 + 3*(-I*b^2*c^2*d + 2*(I*a - 1)*b*c*d^2 + (-I*a^2 + 2*a)*d^3)*(b*x + a))*si
n(3*b*x + 3*a) + 4*(I*(b*x + a)^3*d^3 - 3*b^2*c^2*d + 6*a*b*c*d^2 - 3*a^2*d^3 + 3*(I*b*c*d^2 + (-I*a - 1)*d^3)
*(b*x + a)^2 + 3*(I*b^2*c^2*d + 2*(-I*a - 1)*b*c*d^2 + (I*a^2 + 2*a)*d^3)*(b*x + a))*sin(b*x + a))/(-4*I*b^3*c
os(4*b*x + 4*a) - 8*I*b^3*cos(2*b*x + 2*a) + 4*b^3*sin(4*b*x + 4*a) + 8*b^3*sin(2*b*x + 2*a) - 4*I*b^3))/b

Giac [F]

\[ \int (c+d x)^3 \sec (a+b x) \tan ^2(a+b x) \, dx=\int { {\left (d x + c\right )}^{3} \sec \left (b x + a\right ) \tan \left (b x + a\right )^{2} \,d x } \]

[In]

integrate((d*x+c)^3*sec(b*x+a)*tan(b*x+a)^2,x, algorithm="giac")

[Out]

integrate((d*x + c)^3*sec(b*x + a)*tan(b*x + a)^2, x)

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^3 \sec (a+b x) \tan ^2(a+b x) \, dx=\text {Hanged} \]

[In]

int((tan(a + b*x)^2*(c + d*x)^3)/cos(a + b*x),x)

[Out]

\text{Hanged}