\(\int \frac {(e+f x)^2 \sec (c+d x)}{a+a \sin (c+d x)} \, dx\) [270]

Optimal result

Integrand size = 26, antiderivative size = 278 \[ \int \frac {(e+f x)^2 \sec (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {i (e+f x)^2 \arctan \left (e^{i (c+d x)}\right )}{a d}+\frac {f^2 \text {arctanh}(\sin (c+d x))}{a d^3}+\frac {f^2 \log (\cos (c+d x))}{a d^3}+\frac {i f (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{a d^2}-\frac {i f (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{a d^2}-\frac {f^2 \operatorname {PolyLog}\left (3,-i e^{i (c+d x)}\right )}{a d^3}+\frac {f^2 \operatorname {PolyLog}\left (3,i e^{i (c+d x)}\right )}{a d^3}-\frac {f (e+f x) \sec (c+d x)}{a d^2}-\frac {(e+f x)^2 \sec ^2(c+d x)}{2 a d}+\frac {f (e+f x) \tan (c+d x)}{a d^2}+\frac {(e+f x)^2 \sec (c+d x) \tan (c+d x)}{2 a d} \]

[Out]

-I*(f*x+e)^2*arctan(exp(I*(d*x+c)))/a/d+f^2*arctanh(sin(d*x+c))/a/d^3+f^2*ln(cos(d*x+c))/a/d^3+I*f*(f*x+e)*pol
ylog(2,-I*exp(I*(d*x+c)))/a/d^2-I*f*(f*x+e)*polylog(2,I*exp(I*(d*x+c)))/a/d^2-f^2*polylog(3,-I*exp(I*(d*x+c)))
/a/d^3+f^2*polylog(3,I*exp(I*(d*x+c)))/a/d^3-f*(f*x+e)*sec(d*x+c)/a/d^2-1/2*(f*x+e)^2*sec(d*x+c)^2/a/d+f*(f*x+
e)*tan(d*x+c)/a/d^2+1/2*(f*x+e)^2*sec(d*x+c)*tan(d*x+c)/a/d

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {4627, 4271, 3855, 4266, 2611, 2320, 6724, 4494, 4269, 3556} \[ \int \frac {(e+f x)^2 \sec (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {i (e+f x)^2 \arctan \left (e^{i (c+d x)}\right )}{a d}+\frac {f^2 \text {arctanh}(\sin (c+d x))}{a d^3}-\frac {f^2 \operatorname {PolyLog}\left (3,-i e^{i (c+d x)}\right )}{a d^3}+\frac {f^2 \operatorname {PolyLog}\left (3,i e^{i (c+d x)}\right )}{a d^3}+\frac {f^2 \log (\cos (c+d x))}{a d^3}+\frac {i f (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{a d^2}-\frac {i f (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{a d^2}+\frac {f (e+f x) \tan (c+d x)}{a d^2}-\frac {f (e+f x) \sec (c+d x)}{a d^2}-\frac {(e+f x)^2 \sec ^2(c+d x)}{2 a d}+\frac {(e+f x)^2 \tan (c+d x) \sec (c+d x)}{2 a d} \]

[In]

Int[((e + f*x)^2*Sec[c + d*x])/(a + a*Sin[c + d*x]),x]

[Out]

((-I)*(e + f*x)^2*ArcTan[E^(I*(c + d*x))])/(a*d) + (f^2*ArcTanh[Sin[c + d*x]])/(a*d^3) + (f^2*Log[Cos[c + d*x]
])/(a*d^3) + (I*f*(e + f*x)*PolyLog[2, (-I)*E^(I*(c + d*x))])/(a*d^2) - (I*f*(e + f*x)*PolyLog[2, I*E^(I*(c +
d*x))])/(a*d^2) - (f^2*PolyLog[3, (-I)*E^(I*(c + d*x))])/(a*d^3) + (f^2*PolyLog[3, I*E^(I*(c + d*x))])/(a*d^3)
- (f*(e + f*x)*Sec[c + d*x])/(a*d^2) - ((e + f*x)^2*Sec[c + d*x]^2)/(2*a*d) + (f*(e + f*x)*Tan[c + d*x])/(a*d
^2) + ((e + f*x)^2*Sec[c + d*x]*Tan[c + d*x])/(2*a*d)

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 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 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\int (e+f x)^2 \sec ^3(c+d x) \, dx}{a}-\frac {\int (e+f x)^2 \sec ^2(c+d x) \tan (c+d x) \, dx}{a} \\ & = -\frac {f (e+f x) \sec (c+d x)}{a d^2}-\frac {(e+f x)^2 \sec ^2(c+d x)}{2 a d}+\frac {(e+f x)^2 \sec (c+d x) \tan (c+d x)}{2 a d}+\frac {\int (e+f x)^2 \sec (c+d x) \, dx}{2 a}+\frac {f \int (e+f x) \sec ^2(c+d x) \, dx}{a d}+\frac {f^2 \int \sec (c+d x) \, dx}{a d^2} \\ & = -\frac {i (e+f x)^2 \arctan \left (e^{i (c+d x)}\right )}{a d}+\frac {f^2 \text {arctanh}(\sin (c+d x))}{a d^3}-\frac {f (e+f x) \sec (c+d x)}{a d^2}-\frac {(e+f x)^2 \sec ^2(c+d x)}{2 a d}+\frac {f (e+f x) \tan (c+d x)}{a d^2}+\frac {(e+f x)^2 \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {f \int (e+f x) \log \left (1-i e^{i (c+d x)}\right ) \, dx}{a d}+\frac {f \int (e+f x) \log \left (1+i e^{i (c+d x)}\right ) \, dx}{a d}-\frac {f^2 \int \tan (c+d x) \, dx}{a d^2} \\ & = -\frac {i (e+f x)^2 \arctan \left (e^{i (c+d x)}\right )}{a d}+\frac {f^2 \text {arctanh}(\sin (c+d x))}{a d^3}+\frac {f^2 \log (\cos (c+d x))}{a d^3}+\frac {i f (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{a d^2}-\frac {i f (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{a d^2}-\frac {f (e+f x) \sec (c+d x)}{a d^2}-\frac {(e+f x)^2 \sec ^2(c+d x)}{2 a d}+\frac {f (e+f x) \tan (c+d x)}{a d^2}+\frac {(e+f x)^2 \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {\left (i f^2\right ) \int \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right ) \, dx}{a d^2}+\frac {\left (i f^2\right ) \int \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right ) \, dx}{a d^2} \\ & = -\frac {i (e+f x)^2 \arctan \left (e^{i (c+d x)}\right )}{a d}+\frac {f^2 \text {arctanh}(\sin (c+d x))}{a d^3}+\frac {f^2 \log (\cos (c+d x))}{a d^3}+\frac {i f (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{a d^2}-\frac {i f (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{a d^2}-\frac {f (e+f x) \sec (c+d x)}{a d^2}-\frac {(e+f x)^2 \sec ^2(c+d x)}{2 a d}+\frac {f (e+f x) \tan (c+d x)}{a d^2}+\frac {(e+f x)^2 \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {f^2 \text {Subst}\left (\int \frac {\operatorname {PolyLog}(2,-i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{a d^3}+\frac {f^2 \text {Subst}\left (\int \frac {\operatorname {PolyLog}(2,i x)}{x} \, dx,x,e^{i (c+d x)}\right )}{a d^3} \\ & = -\frac {i (e+f x)^2 \arctan \left (e^{i (c+d x)}\right )}{a d}+\frac {f^2 \text {arctanh}(\sin (c+d x))}{a d^3}+\frac {f^2 \log (\cos (c+d x))}{a d^3}+\frac {i f (e+f x) \operatorname {PolyLog}\left (2,-i e^{i (c+d x)}\right )}{a d^2}-\frac {i f (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{a d^2}-\frac {f^2 \operatorname {PolyLog}\left (3,-i e^{i (c+d x)}\right )}{a d^3}+\frac {f^2 \operatorname {PolyLog}\left (3,i e^{i (c+d x)}\right )}{a d^3}-\frac {f (e+f x) \sec (c+d x)}{a d^2}-\frac {(e+f x)^2 \sec ^2(c+d x)}{2 a d}+\frac {f (e+f x) \tan (c+d x)}{a d^2}+\frac {(e+f x)^2 \sec (c+d x) \tan (c+d x)}{2 a d} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(725\) vs. \(2(278)=556\).

Time = 7.87 (sec) , antiderivative size = 725, normalized size of antiderivative = 2.61 \[ \int \frac {(e+f x)^2 \sec (c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {(e+f x)^3}{\left (-i+e^{i c}\right ) f}+\frac {3 (e+f x)^2 \log \left (1-i e^{-i (c+d x)}\right )}{d}+\frac {6 f \left (i d (e+f x) \operatorname {PolyLog}\left (2,i e^{-i (c+d x)}\right )+f \operatorname {PolyLog}\left (3,i e^{-i (c+d x)}\right )\right )}{d^3}}{6 a}+\frac {x \left (3 e^2+3 e f x+f^2 x^2\right )}{6 a \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right )}-\frac {(\cos (c)+i \sin (c)) \left (d^2 e^2 x \cos (c)+4 f^2 x \cos (c)+d^2 e f x^2 \cos (c)+\frac {1}{3} d^2 f^2 x^3 (\cos (c)-i \sin (c))-i d^2 e^2 x \sin (c)-4 i f^2 x \sin (c)-i d^2 e f x^2 \sin (c)+2 e f \operatorname {PolyLog}(2,-i \cos (c+d x)-\sin (c+d x)) (\cos (c)-i (1+\sin (c)))+2 f^2 x \operatorname {PolyLog}(2,-i \cos (c+d x)-\sin (c+d x)) (\cos (c)-i (1+\sin (c)))-2 d e f x \log (1+i \cos (c+d x)+\sin (c+d x)) (\cos (c)-i \sin (c)) (\cos (c)+i (1+\sin (c)))-d f^2 x^2 \log (1+i \cos (c+d x)+\sin (c+d x)) (\cos (c)-i \sin (c)) (\cos (c)+i (1+\sin (c)))-\frac {\left (d^2 e^2+4 f^2\right ) \log (\cos (c+d x)+i (1+\sin (c+d x))) (\cos (c)-i \sin (c)) (\cos (c)+i (1+\sin (c)))}{d}-\frac {2 f^2 \operatorname {PolyLog}(3,-i \cos (c+d x)-\sin (c+d x)) (\cos (c)-i \sin (c)) (\cos (c)+i (1+\sin (c)))}{d}+\left (d^2 e^2+4 f^2\right ) x (i \cos (c)+\sin (c)) (\cos (c)+i (1+\sin (c)))\right )}{2 a d^2 (\cos (c)+i (1+\sin (c)))}-\frac {(e+f x)^2}{2 a d \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}+\frac {2 \left (e f \sin \left (\frac {d x}{2}\right )+f^2 x \sin \left (\frac {d x}{2}\right )\right )}{a d^2 \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )} \]

[In]

Integrate[((e + f*x)^2*Sec[c + d*x])/(a + a*Sin[c + d*x]),x]

[Out]

-1/6*((e + f*x)^3/((-I + E^(I*c))*f) + (3*(e + f*x)^2*Log[1 - I/E^(I*(c + d*x))])/d + (6*f*(I*d*(e + f*x)*Poly
Log[2, I/E^(I*(c + d*x))] + f*PolyLog[3, I/E^(I*(c + d*x))]))/d^3)/a + (x*(3*e^2 + 3*e*f*x + f^2*x^2))/(6*a*(C
os[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2])) - ((Cos[c] + I*Sin[c])*(d^2*e^2*x*Cos[c] + 4*f^2*x*Cos[c] + d^2*e*f
*x^2*Cos[c] + (d^2*f^2*x^3*(Cos[c] - I*Sin[c]))/3 - I*d^2*e^2*x*Sin[c] - (4*I)*f^2*x*Sin[c] - I*d^2*e*f*x^2*Si
n[c] + 2*e*f*PolyLog[2, (-I)*Cos[c + d*x] - Sin[c + d*x]]*(Cos[c] - I*(1 + Sin[c])) + 2*f^2*x*PolyLog[2, (-I)*
Cos[c + d*x] - Sin[c + d*x]]*(Cos[c] - I*(1 + Sin[c])) - 2*d*e*f*x*Log[1 + I*Cos[c + d*x] + Sin[c + d*x]]*(Cos
[c] - I*Sin[c])*(Cos[c] + I*(1 + Sin[c])) - d*f^2*x^2*Log[1 + I*Cos[c + d*x] + Sin[c + d*x]]*(Cos[c] - I*Sin[c
])*(Cos[c] + I*(1 + Sin[c])) - ((d^2*e^2 + 4*f^2)*Log[Cos[c + d*x] + I*(1 + Sin[c + d*x])]*(Cos[c] - I*Sin[c])
*(Cos[c] + I*(1 + Sin[c])))/d - (2*f^2*PolyLog[3, (-I)*Cos[c + d*x] - Sin[c + d*x]]*(Cos[c] - I*Sin[c])*(Cos[c
] + I*(1 + Sin[c])))/d + (d^2*e^2 + 4*f^2)*x*(I*Cos[c] + Sin[c])*(Cos[c] + I*(1 + Sin[c]))))/(2*a*d^2*(Cos[c]
+ I*(1 + Sin[c]))) - (e + f*x)^2/(2*a*d*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^2) + (2*(e*f*Sin[(d*x)/2] +
f^2*x*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]))

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 616 vs. \(2 (257 ) = 514\).

Time = 0.41 (sec) , antiderivative size = 617, normalized size of antiderivative = 2.22

method	result	size
risch	\(-\frac {i e^{2} \arctan \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{a d}+\frac {f^{2} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) x^{2}}{2 d a}+\frac {e f \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{d a}-\frac {\ln \left (1+i {\mathrm e}^{i \left (d x +c \right )}\right ) f^{2} x^{2}}{2 a d}+\frac {e f \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) c}{d^{2} a}-\frac {2 i f^{2} \arctan \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{3}}+\frac {2 i e f c \arctan \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{2}}+\frac {i e f \,\operatorname {Li}_{2}\left (-i {\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{2}}-\frac {f^{2} \operatorname {Li}_{3}\left (-i {\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{3}}-\frac {\ln \left (1+i {\mathrm e}^{i \left (d x +c \right )}\right ) e f x}{a d}+\frac {f^{2} \ln \left (1+{\mathrm e}^{2 i \left (d x +c \right )}\right )}{a \,d^{3}}-\frac {c^{2} f^{2} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right )}{2 d^{3} a}-\frac {2 f^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{3}}-\frac {\ln \left (1+i {\mathrm e}^{i \left (d x +c \right )}\right ) c e f}{a \,d^{2}}-\frac {i f^{2} c^{2} \arctan \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{3}}+\frac {f^{2} \ln \left (1+i {\mathrm e}^{i \left (d x +c \right )}\right ) c^{2}}{2 a \,d^{3}}-\frac {i f^{2} \operatorname {Li}_{2}\left (i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{a \,d^{2}}+\frac {f^{2} \operatorname {Li}_{3}\left (i {\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{3}}-\frac {i \left (d \,{\mathrm e}^{i \left (d x +c \right )} f^{2} x^{2}+2 d \,{\mathrm e}^{i \left (d x +c \right )} e f x +d \,{\mathrm e}^{i \left (d x +c \right )} e^{2}+2 f^{2} x -2 i f^{2} x \,{\mathrm e}^{i \left (d x +c \right )}+2 e f -2 i e f \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2} a}+\frac {i f^{2} \operatorname {Li}_{2}\left (-i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{a \,d^{2}}-\frac {i e f \,\operatorname {Li}_{2}\left (i {\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{2}}\)	\(617\)

[In]

int((f*x+e)^2*sec(d*x+c)/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

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. 1064 vs. \(2 (248) = 496\).

Time = 0.33 (sec) , antiderivative size = 1064, normalized size of antiderivative = 3.83 \[ \int \frac {(e+f x)^2 \sec (c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \]

[In]

integrate((f*x+e)^2*sec(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/4*(2*d^2*f^2*x^2 + 4*d^2*e*f*x + 2*d^2*e^2 + 4*(d*f^2*x + d*e*f)*cos(d*x + c) + 2*(I*d*f^2*x + I*d*e*f + (I
*d*f^2*x + I*d*e*f)*sin(d*x + c))*dilog(I*cos(d*x + c) + sin(d*x + c)) + 2*(I*d*f^2*x + I*d*e*f + (I*d*f^2*x +
I*d*e*f)*sin(d*x + c))*dilog(I*cos(d*x + c) - sin(d*x + c)) + 2*(-I*d*f^2*x - I*d*e*f + (-I*d*f^2*x - I*d*e*f
)*sin(d*x + c))*dilog(-I*cos(d*x + c) + sin(d*x + c)) + 2*(-I*d*f^2*x - I*d*e*f + (-I*d*f^2*x - I*d*e*f)*sin(d
*x + c))*dilog(-I*cos(d*x + c) - sin(d*x + c)) - (d^2*e^2 - 2*c*d*e*f + (c^2 + 4)*f^2 + (d^2*e^2 - 2*c*d*e*f +
(c^2 + 4)*f^2)*sin(d*x + c))*log(cos(d*x + c) + I*sin(d*x + c) + I) + (d^2*e^2 - 2*c*d*e*f + c^2*f^2 + (d^2*e
^2 - 2*c*d*e*f + c^2*f^2)*sin(d*x + c))*log(cos(d*x + c) - I*sin(d*x + c) + I) - (d^2*f^2*x^2 + 2*d^2*e*f*x +
2*c*d*e*f - c^2*f^2 + (d^2*f^2*x^2 + 2*d^2*e*f*x + 2*c*d*e*f - c^2*f^2)*sin(d*x + c))*log(I*cos(d*x + c) + sin
(d*x + c) + 1) + (d^2*f^2*x^2 + 2*d^2*e*f*x + 2*c*d*e*f - c^2*f^2 + (d^2*f^2*x^2 + 2*d^2*e*f*x + 2*c*d*e*f - c
^2*f^2)*sin(d*x + c))*log(I*cos(d*x + c) - sin(d*x + c) + 1) - (d^2*f^2*x^2 + 2*d^2*e*f*x + 2*c*d*e*f - c^2*f^
2 + (d^2*f^2*x^2 + 2*d^2*e*f*x + 2*c*d*e*f - c^2*f^2)*sin(d*x + c))*log(-I*cos(d*x + c) + sin(d*x + c) + 1) +
(d^2*f^2*x^2 + 2*d^2*e*f*x + 2*c*d*e*f - c^2*f^2 + (d^2*f^2*x^2 + 2*d^2*e*f*x + 2*c*d*e*f - c^2*f^2)*sin(d*x +
c))*log(-I*cos(d*x + c) - sin(d*x + c) + 1) - (d^2*e^2 - 2*c*d*e*f + (c^2 + 4)*f^2 + (d^2*e^2 - 2*c*d*e*f + (
c^2 + 4)*f^2)*sin(d*x + c))*log(-cos(d*x + c) + I*sin(d*x + c) + I) + (d^2*e^2 - 2*c*d*e*f + c^2*f^2 + (d^2*e^
2 - 2*c*d*e*f + c^2*f^2)*sin(d*x + c))*log(-cos(d*x + c) - I*sin(d*x + c) + I) + 2*(f^2*sin(d*x + c) + f^2)*po
lylog(3, I*cos(d*x + c) + sin(d*x + c)) - 2*(f^2*sin(d*x + c) + f^2)*polylog(3, I*cos(d*x + c) - sin(d*x + c))
+ 2*(f^2*sin(d*x + c) + f^2)*polylog(3, -I*cos(d*x + c) + sin(d*x + c)) - 2*(f^2*sin(d*x + c) + f^2)*polylog(
3, -I*cos(d*x + c) - sin(d*x + c)))/(a*d^3*sin(d*x + c) + a*d^3)

Sympy [F]

\[ \int \frac {(e+f x)^2 \sec (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {e^{2} \sec {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {f^{2} x^{2} \sec {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {2 e f x \sec {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]

[In]

integrate((f*x+e)**2*sec(d*x+c)/(a+a*sin(d*x+c)),x)

[Out]

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

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. 1932 vs. \(2 (248) = 496\).

Time = 0.47 (sec) , antiderivative size = 1932, normalized size of antiderivative = 6.95 \[ \int \frac {(e+f x)^2 \sec (c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \]

[In]

integrate((f*x+e)^2*sec(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/4*(2*c*e*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^2*(l
og(sin(d*x + c) + 1)/a - log(sin(d*x + c) - 1)/a - 2/(a*sin(d*x + c) + a)) - 4*(8*(d*x + c)*f^2*cos(2*d*x + 2*
c) + 8*I*(d*x + c)*f^2*sin(2*d*x + 2*c) + 8*d*e*f - 8*c*f^2 - 2*((c^2 + 4)*f^2*cos(2*d*x + 2*c) - 2*(-I*c^2 -
4*I)*f^2*cos(d*x + c) - (-I*c^2 - 4*I)*f^2*sin(2*d*x + 2*c) - 2*(c^2 + 4)*f^2*sin(d*x + c) - (c^2 + 4)*f^2)*ar
ctan2(sin(d*x + c) + 1, cos(d*x + c)) + 2*(c^2*f^2*cos(2*d*x + 2*c) + 2*I*c^2*f^2*cos(d*x + c) + I*c^2*f^2*sin
(2*d*x + 2*c) - 2*c^2*f^2*sin(d*x + c) - c^2*f^2)*arctan2(sin(d*x + c) - 1, cos(d*x + c)) - 2*((d*x + c)^2*f^2
+ 2*(d*e*f - c*f^2)*(d*x + c) - ((d*x + c)^2*f^2 + 2*(d*e*f - c*f^2)*(d*x + c))*cos(2*d*x + 2*c) - 2*(I*(d*x
+ c)^2*f^2 + 2*(I*d*e*f - I*c*f^2)*(d*x + c))*cos(d*x + c) - (I*(d*x + c)^2*f^2 + 2*(I*d*e*f - I*c*f^2)*(d*x +
c))*sin(2*d*x + 2*c) + 2*((d*x + c)^2*f^2 + 2*(d*e*f - c*f^2)*(d*x + c))*sin(d*x + c))*arctan2(cos(d*x + c),
sin(d*x + c) + 1) - 2*((d*x + c)^2*f^2 + 2*(d*e*f - c*f^2)*(d*x + c) - ((d*x + c)^2*f^2 + 2*(d*e*f - c*f^2)*(d
*x + c))*cos(2*d*x + 2*c) - 2*(I*(d*x + c)^2*f^2 + 2*(I*d*e*f - I*c*f^2)*(d*x + c))*cos(d*x + c) - (I*(d*x + c
)^2*f^2 + 2*(I*d*e*f - I*c*f^2)*(d*x + c))*sin(2*d*x + 2*c) + 2*((d*x + c)^2*f^2 + 2*(d*e*f - c*f^2)*(d*x + c)
)*sin(d*x + c))*arctan2(cos(d*x + c), -sin(d*x + c) + 1) + 4*((d*x + c)^2*f^2 - 2*I*d*e*f + (c^2 + 2*I*c)*f^2
+ 2*(d*e*f - (c - I)*f^2)*(d*x + c))*cos(d*x + c) - 4*(d*e*f + (d*x + c)*f^2 - c*f^2 - (d*e*f + (d*x + c)*f^2
- c*f^2)*cos(2*d*x + 2*c) - 2*(I*d*e*f + I*(d*x + c)*f^2 - I*c*f^2)*cos(d*x + c) - (I*d*e*f + I*(d*x + c)*f^2
- I*c*f^2)*sin(2*d*x + 2*c) + 2*(d*e*f + (d*x + c)*f^2 - c*f^2)*sin(d*x + c))*dilog(I*e^(I*d*x + I*c)) + 4*(d*
e*f + (d*x + c)*f^2 - c*f^2 - (d*e*f + (d*x + c)*f^2 - c*f^2)*cos(2*d*x + 2*c) + 2*(-I*d*e*f - I*(d*x + c)*f^2
+ I*c*f^2)*cos(d*x + c) + (-I*d*e*f - I*(d*x + c)*f^2 + I*c*f^2)*sin(2*d*x + 2*c) + 2*(d*e*f + (d*x + c)*f^2
- c*f^2)*sin(d*x + c))*dilog(-I*e^(I*d*x + I*c)) - (I*(d*x + c)^2*f^2 + (I*c^2 + 4*I)*f^2 - 2*(-I*d*e*f + I*c*
f^2)*(d*x + c) + (-I*(d*x + c)^2*f^2 + (-I*c^2 - 4*I)*f^2 - 2*(I*d*e*f - I*c*f^2)*(d*x + c))*cos(2*d*x + 2*c)
+ 2*((d*x + c)^2*f^2 + (c^2 + 4)*f^2 + 2*(d*e*f - c*f^2)*(d*x + c))*cos(d*x + c) + ((d*x + c)^2*f^2 + (c^2 + 4
)*f^2 + 2*(d*e*f - c*f^2)*(d*x + c))*sin(2*d*x + 2*c) - 2*(-I*(d*x + c)^2*f^2 + (-I*c^2 - 4*I)*f^2 + 2*(-I*d*e
*f + I*c*f^2)*(d*x + c))*sin(d*x + c))*log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) - (-I*(d*x +
c)^2*f^2 - I*c^2*f^2 - 2*(I*d*e*f - I*c*f^2)*(d*x + c) + (I*(d*x + c)^2*f^2 + I*c^2*f^2 - 2*(-I*d*e*f + I*c*f^
2)*(d*x + c))*cos(2*d*x + 2*c) - 2*((d*x + c)^2*f^2 + c^2*f^2 + 2*(d*e*f - c*f^2)*(d*x + c))*cos(d*x + c) - ((
d*x + c)^2*f^2 + c^2*f^2 + 2*(d*e*f - c*f^2)*(d*x + c))*sin(2*d*x + 2*c) - 2*(I*(d*x + c)^2*f^2 + I*c^2*f^2 +
2*(I*d*e*f - I*c*f^2)*(d*x + c))*sin(d*x + c))*log(cos(d*x + c)^2 + sin(d*x + c)^2 - 2*sin(d*x + c) + 1) + 4*(
I*f^2*cos(2*d*x + 2*c) - 2*f^2*cos(d*x + c) - f^2*sin(2*d*x + 2*c) - 2*I*f^2*sin(d*x + c) - I*f^2)*polylog(3,
I*e^(I*d*x + I*c)) + 4*(-I*f^2*cos(2*d*x + 2*c) + 2*f^2*cos(d*x + c) + f^2*sin(2*d*x + 2*c) + 2*I*f^2*sin(d*x
+ c) + I*f^2)*polylog(3, -I*e^(I*d*x + I*c)) + 4*(I*(d*x + c)^2*f^2 + 2*d*e*f + (I*c^2 - 2*c)*f^2 + 2*(I*d*e*f
+ (-I*c - 1)*f^2)*(d*x + c))*sin(d*x + c))/(-4*I*a*d^2*cos(2*d*x + 2*c) + 8*a*d^2*cos(d*x + c) + 4*a*d^2*sin(
2*d*x + 2*c) + 8*I*a*d^2*sin(d*x + c) + 4*I*a*d^2))/d

Giac [F]

\[ \int \frac {(e+f x)^2 \sec (c+d x)}{a+a \sin (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{2} \sec \left (d x + c\right )}{a \sin \left (d x + c\right ) + a} \,d x } \]

[In]

integrate((f*x+e)^2*sec(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

integrate((f*x + e)^2*sec(d*x + c)/(a*sin(d*x + c) + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(e+f x)^2 \sec (c+d x)}{a+a \sin (c+d x)} \, dx=\text {Hanged} \]

[In]

int((e + f*x)^2/(cos(c + d*x)*(a + a*sin(c + d*x))),x)

[Out]

\text{Hanged}