3.194 \(\int \frac{(e+f x)^2 \sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx\)

Optimal. Leaf size=184 \[ \frac{4 f^2 \text{PolyLog}\left (2,-i e^{c+d x}\right )}{a d^3}+\frac{4 f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac{2 i f (e+f x) \sinh (c+d x)}{a d^2}-\frac{2 i f^2 \cosh (c+d x)}{a d^3}-\frac{i (e+f x)^2 \cosh (c+d x)}{a d}-\frac{(e+f x)^2 \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{a d}-\frac{(e+f x)^2}{a d}+\frac{(e+f x)^3}{3 a f} \]

[Out]

-((e + f*x)^2/(a*d)) + (e + f*x)^3/(3*a*f) - ((2*I)*f^2*Cosh[c + d*x])/(a*d^3) - (I*(e + f*x)^2*Cosh[c + d*x])
/(a*d) + (4*f*(e + f*x)*Log[1 + I*E^(c + d*x)])/(a*d^2) + (4*f^2*PolyLog[2, (-I)*E^(c + d*x)])/(a*d^3) + ((2*I
)*f*(e + f*x)*Sinh[c + d*x])/(a*d^2) - ((e + f*x)^2*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/(a*d)

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Rubi [A]  time = 0.390758, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 10, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.323, Rules used = {5557, 3296, 2638, 32, 3318, 4184, 3716, 2190, 2279, 2391} \[ \frac{4 f^2 \text{PolyLog}\left (2,-i e^{c+d x}\right )}{a d^3}+\frac{4 f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac{2 i f (e+f x) \sinh (c+d x)}{a d^2}-\frac{2 i f^2 \cosh (c+d x)}{a d^3}-\frac{i (e+f x)^2 \cosh (c+d x)}{a d}-\frac{(e+f x)^2 \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{a d}-\frac{(e+f x)^2}{a d}+\frac{(e+f x)^3}{3 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((e + f*x)^2/(a*d)) + (e + f*x)^3/(3*a*f) - ((2*I)*f^2*Cosh[c + d*x])/(a*d^3) - (I*(e + f*x)^2*Cosh[c + d*x])
/(a*d) + (4*f*(e + f*x)*Log[1 + I*E^(c + d*x)])/(a*d^2) + (4*f^2*PolyLog[2, (-I)*E^(c + d*x)])/(a*d^3) + ((2*I
)*f*(e + f*x)*Sinh[c + d*x])/(a*d^2) - ((e + f*x)^2*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/(a*d)

Rule 5557

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

Rule 3296

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

Rule 2638

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

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 3318

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

Rule 4184

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 3716

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(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*fz*x)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 2279

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 2391

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]

Rubi steps

\begin{align*} \int \frac{(e+f x)^2 \sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx &=i \int \frac{(e+f x)^2 \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx-\frac{i \int (e+f x)^2 \sinh (c+d x) \, dx}{a}\\ &=-\frac{i (e+f x)^2 \cosh (c+d x)}{a d}+\frac{\int (e+f x)^2 \, dx}{a}+\frac{(2 i f) \int (e+f x) \cosh (c+d x) \, dx}{a d}-\int \frac{(e+f x)^2}{a+i a \sinh (c+d x)} \, dx\\ &=\frac{(e+f x)^3}{3 a f}-\frac{i (e+f x)^2 \cosh (c+d x)}{a d}+\frac{2 i f (e+f x) \sinh (c+d x)}{a d^2}-\frac{\int (e+f x)^2 \csc ^2\left (\frac{1}{2} \left (i c+\frac{\pi }{2}\right )+\frac{i d x}{2}\right ) \, dx}{2 a}-\frac{\left (2 i f^2\right ) \int \sinh (c+d x) \, dx}{a d^2}\\ &=\frac{(e+f x)^3}{3 a f}-\frac{2 i f^2 \cosh (c+d x)}{a d^3}-\frac{i (e+f x)^2 \cosh (c+d x)}{a d}+\frac{2 i f (e+f x) \sinh (c+d x)}{a d^2}-\frac{(e+f x)^2 \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{a d}+\frac{(2 f) \int (e+f x) \coth \left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \, dx}{a d}\\ &=-\frac{(e+f x)^2}{a d}+\frac{(e+f x)^3}{3 a f}-\frac{2 i f^2 \cosh (c+d x)}{a d^3}-\frac{i (e+f x)^2 \cosh (c+d x)}{a d}+\frac{2 i f (e+f x) \sinh (c+d x)}{a d^2}-\frac{(e+f x)^2 \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{a d}+\frac{(4 i f) \int \frac{e^{2 \left (\frac{c}{2}+\frac{d x}{2}\right )} (e+f x)}{1+i e^{2 \left (\frac{c}{2}+\frac{d x}{2}\right )}} \, dx}{a d}\\ &=-\frac{(e+f x)^2}{a d}+\frac{(e+f x)^3}{3 a f}-\frac{2 i f^2 \cosh (c+d x)}{a d^3}-\frac{i (e+f x)^2 \cosh (c+d x)}{a d}+\frac{4 f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac{2 i f (e+f x) \sinh (c+d x)}{a d^2}-\frac{(e+f x)^2 \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{a d}-\frac{\left (4 f^2\right ) \int \log \left (1+i e^{2 \left (\frac{c}{2}+\frac{d x}{2}\right )}\right ) \, dx}{a d^2}\\ &=-\frac{(e+f x)^2}{a d}+\frac{(e+f x)^3}{3 a f}-\frac{2 i f^2 \cosh (c+d x)}{a d^3}-\frac{i (e+f x)^2 \cosh (c+d x)}{a d}+\frac{4 f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac{2 i f (e+f x) \sinh (c+d x)}{a d^2}-\frac{(e+f x)^2 \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{a d}-\frac{\left (4 f^2\right ) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{2 \left (\frac{c}{2}+\frac{d x}{2}\right )}\right )}{a d^3}\\ &=-\frac{(e+f x)^2}{a d}+\frac{(e+f x)^3}{3 a f}-\frac{2 i f^2 \cosh (c+d x)}{a d^3}-\frac{i (e+f x)^2 \cosh (c+d x)}{a d}+\frac{4 f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac{4 f^2 \text{Li}_2\left (-i e^{c+d x}\right )}{a d^3}+\frac{2 i f (e+f x) \sinh (c+d x)}{a d^2}-\frac{(e+f x)^2 \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{a d}\\ \end{align*}

Mathematica [A]  time = 3.4604, size = 260, normalized size = 1.41 \[ \frac{\frac{6 \left (\frac{d (e+f x) \left (2 \left (e^c-i\right ) f \log \left (1-i e^{-c-d x}\right )-i d (e+f x)\right )}{e^c-i}-2 f^2 \text{PolyLog}\left (2,i e^{-c-d x}\right )\right )}{d^3}-\frac{3 i \cosh (d x) \left (\cosh (c) \left (d^2 (e+f x)^2+2 f^2\right )-2 d f \sinh (c) (e+f x)\right )}{d^3}-\frac{3 i \sinh (d x) \left (\sinh (c) \left (d^2 (e+f x)^2+2 f^2\right )-2 d f \cosh (c) (e+f x)\right )}{d^3}-\frac{6 \sinh \left (\frac{d x}{2}\right ) (e+f x)^2}{d \left (\cosh \left (\frac{c}{2}\right )+i \sinh \left (\frac{c}{2}\right )\right ) \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )}+x \left (3 e^2+3 e f x+f^2 x^2\right )}{3 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(3*e^2 + 3*e*f*x + f^2*x^2) + (6*((d*(e + f*x)*((-I)*d*(e + f*x) + 2*(-I + E^c)*f*Log[1 - I*E^(-c - d*x)]))
/(-I + E^c) - 2*f^2*PolyLog[2, I*E^(-c - d*x)]))/d^3 - ((3*I)*Cosh[d*x]*((2*f^2 + d^2*(e + f*x)^2)*Cosh[c] - 2
*d*f*(e + f*x)*Sinh[c]))/d^3 - ((3*I)*(-2*d*f*(e + f*x)*Cosh[c] + (2*f^2 + d^2*(e + f*x)^2)*Sinh[c])*Sinh[d*x]
)/d^3 - (6*(e + f*x)^2*Sinh[(d*x)/2])/(d*(Cosh[c/2] + I*Sinh[c/2])*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])))
/(3*a)

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Maple [B]  time = 0.13, size = 374, normalized size = 2. \begin{align*}{\frac{{x}^{3}{f}^{2}}{3\,a}}+{\frac{ef{x}^{2}}{a}}+{\frac{{e}^{2}x}{a}}-{\frac{2\,i \left ({x}^{2}{f}^{2}+2\,efx+{e}^{2} \right ) }{da \left ({{\rm e}^{dx+c}}-i \right ) }}-{\frac{{\frac{i}{2}} \left ({f}^{2}{x}^{2}{d}^{2}+2\,{d}^{2}efx+{d}^{2}{e}^{2}-2\,d{f}^{2}x-2\,efd+2\,{f}^{2} \right ){{\rm e}^{dx+c}}}{a{d}^{3}}}-{\frac{{\frac{i}{2}} \left ({f}^{2}{x}^{2}{d}^{2}+2\,{d}^{2}efx+{d}^{2}{e}^{2}+2\,d{f}^{2}x+2\,efd+2\,{f}^{2} \right ){{\rm e}^{-dx-c}}}{a{d}^{3}}}+4\,{\frac{f\ln \left ({{\rm e}^{dx+c}}-i \right ) e}{a{d}^{2}}}-4\,{\frac{f\ln \left ({{\rm e}^{dx+c}} \right ) e}{a{d}^{2}}}-2\,{\frac{{x}^{2}{f}^{2}}{da}}-4\,{\frac{{f}^{2}cx}{a{d}^{2}}}-2\,{\frac{{c}^{2}{f}^{2}}{a{d}^{3}}}+4\,{\frac{{f}^{2}\ln \left ( 1+i{{\rm e}^{dx+c}} \right ) x}{a{d}^{2}}}+4\,{\frac{{f}^{2}\ln \left ( 1+i{{\rm e}^{dx+c}} \right ) c}{a{d}^{3}}}+4\,{\frac{{f}^{2}{\it polylog} \left ( 2,-i{{\rm e}^{dx+c}} \right ) }{a{d}^{3}}}-4\,{\frac{{f}^{2}c\ln \left ({{\rm e}^{dx+c}}-i \right ) }{a{d}^{3}}}+4\,{\frac{{f}^{2}c\ln \left ({{\rm e}^{dx+c}} \right ) }{a{d}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^2*sinh(d*x+c)^2/(a+I*a*sinh(d*x+c)),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{2} \, e f{\left (\frac{4 \, x e^{\left (d x + c\right )}}{a d e^{\left (d x + c\right )} - i \, a d} - \frac{-2 i \, d^{2} x^{2} e^{c} - 2 i \, d x e^{c} -{\left (2 i \, d x e^{\left (3 \, c\right )} - 2 i \, e^{\left (3 \, c\right )}\right )} e^{\left (2 \, d x\right )} + 2 \,{\left (d^{2} x^{2} e^{\left (2 \, c\right )} - 3 \, d x e^{\left (2 \, c\right )} + e^{\left (2 \, c\right )}\right )} e^{\left (d x\right )} - 2 \,{\left (d x + 1\right )} e^{\left (-d x\right )} - 2 i \, e^{c}}{a d^{2} e^{\left (d x + 2 \, c\right )} - i \, a d^{2} e^{c}} - \frac{8 \, \log \left ({\left (e^{\left (d x + c\right )} - i\right )} e^{\left (-c\right )}\right )}{a d^{2}}\right )} + \frac{1}{12} \, f^{2}{\left (\frac{-4 i \, d^{3} x^{3} - 30 i \, d^{2} x^{2} - 12 i \, d x -{\left (6 i \, d^{2} x^{2} e^{\left (2 \, c\right )} - 12 i \, d x e^{\left (2 \, c\right )} + 12 i \, e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )} + 2 \,{\left (2 \, d^{3} x^{3} e^{c} - 3 \, d^{2} x^{2} e^{c} + 6 \, d x e^{c} - 6 \, e^{c}\right )} e^{\left (d x\right )} - 12 i}{a d^{3} e^{\left (d x + c\right )} - i \, a d^{3}} + 48 i \, \int \frac{x}{a d e^{\left (d x + c\right )} - i \, a d}\,{d x}\right )} + \frac{1}{4} \, e^{2}{\left (\frac{4 \,{\left (d x + c\right )}}{a d} + \frac{2 \,{\left (-5 i \, e^{\left (-d x - c\right )} + 1\right )}}{{\left (i \, a e^{\left (-d x - c\right )} + a e^{\left (-2 \, d x - 2 \, c\right )}\right )} d} - \frac{2 i \, e^{\left (-d x - c\right )}}{a d}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*e*f*(4*x*e^(d*x + c)/(a*d*e^(d*x + c) - I*a*d) - (-2*I*d^2*x^2*e^c - 2*I*d*x*e^c - (2*I*d*x*e^(3*c) - 2*I
*e^(3*c))*e^(2*d*x) + 2*(d^2*x^2*e^(2*c) - 3*d*x*e^(2*c) + e^(2*c))*e^(d*x) - 2*(d*x + 1)*e^(-d*x) - 2*I*e^c)/
(a*d^2*e^(d*x + 2*c) - I*a*d^2*e^c) - 8*log((e^(d*x + c) - I)*e^(-c))/(a*d^2)) + 1/12*f^2*((-4*I*d^3*x^3 - 30*
I*d^2*x^2 - 12*I*d*x - (6*I*d^2*x^2*e^(2*c) - 12*I*d*x*e^(2*c) + 12*I*e^(2*c))*e^(2*d*x) + 2*(2*d^3*x^3*e^c -
3*d^2*x^2*e^c + 6*d*x*e^c - 6*e^c)*e^(d*x) - 12*I)/(a*d^3*e^(d*x + c) - I*a*d^3) + 48*I*integrate(x/(a*d*e^(d*
x + c) - I*a*d), x)) + 1/4*e^2*(4*(d*x + c)/(a*d) + 2*(-5*I*e^(-d*x - c) + 1)/((I*a*e^(-d*x - c) + a*e^(-2*d*x
- 2*c))*d) - 2*I*e^(-d*x - c)/(a*d))

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Fricas [B]  time = 2.72295, size = 1123, normalized size = 6.1 \begin{align*} -\frac{3 \, d^{2} f^{2} x^{2} + 3 \, d^{2} e^{2} + 6 \, d e f + 6 \, f^{2} + 6 \,{\left (d^{2} e f + d f^{2}\right )} x - 24 \,{\left (f^{2} e^{\left (2 \, d x + 2 \, c\right )} - i \, f^{2} e^{\left (d x + c\right )}\right )}{\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) -{\left (-3 i \, d^{2} f^{2} x^{2} - 3 i \, d^{2} e^{2} + 6 i \, d e f - 6 i \, f^{2} +{\left (-6 i \, d^{2} e f + 6 i \, d f^{2}\right )} x\right )} e^{\left (3 \, d x + 3 \, c\right )} -{\left (2 \, d^{3} f^{2} x^{3} - 3 \, d^{2} e^{2} - 6 \,{\left (4 \, c - 1\right )} d e f + 6 \,{\left (2 \, c^{2} - 1\right )} f^{2} + 3 \,{\left (2 \, d^{3} e f - 5 \, d^{2} f^{2}\right )} x^{2} + 6 \,{\left (d^{3} e^{2} - 5 \, d^{2} e f + d f^{2}\right )} x\right )} e^{\left (2 \, d x + 2 \, c\right )} -{\left (-2 i \, d^{3} f^{2} x^{3} - 15 i \, d^{2} e^{2} +{\left (24 i \, c - 6 i\right )} d e f +{\left (-12 i \, c^{2} - 6 i\right )} f^{2} +{\left (-6 i \, d^{3} e f - 3 i \, d^{2} f^{2}\right )} x^{2} +{\left (-6 i \, d^{3} e^{2} - 6 i \, d^{2} e f - 6 i \, d f^{2}\right )} x\right )} e^{\left (d x + c\right )} -{\left (24 \,{\left (d e f - c f^{2}\right )} e^{\left (2 \, d x + 2 \, c\right )} +{\left (-24 i \, d e f + 24 i \, c f^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} - i\right ) -{\left (24 \,{\left (d f^{2} x + c f^{2}\right )} e^{\left (2 \, d x + 2 \, c\right )} +{\left (-24 i \, d f^{2} x - 24 i \, c f^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (i \, e^{\left (d x + c\right )} + 1\right )}{6 \, a d^{3} e^{\left (2 \, d x + 2 \, c\right )} - 6 i \, a d^{3} e^{\left (d x + c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(3*d^2*f^2*x^2 + 3*d^2*e^2 + 6*d*e*f + 6*f^2 + 6*(d^2*e*f + d*f^2)*x - 24*(f^2*e^(2*d*x + 2*c) - I*f^2*e^(d*x
+ c))*dilog(-I*e^(d*x + c)) - (-3*I*d^2*f^2*x^2 - 3*I*d^2*e^2 + 6*I*d*e*f - 6*I*f^2 + (-6*I*d^2*e*f + 6*I*d*f
^2)*x)*e^(3*d*x + 3*c) - (2*d^3*f^2*x^3 - 3*d^2*e^2 - 6*(4*c - 1)*d*e*f + 6*(2*c^2 - 1)*f^2 + 3*(2*d^3*e*f - 5
*d^2*f^2)*x^2 + 6*(d^3*e^2 - 5*d^2*e*f + d*f^2)*x)*e^(2*d*x + 2*c) - (-2*I*d^3*f^2*x^3 - 15*I*d^2*e^2 + (24*I*
c - 6*I)*d*e*f + (-12*I*c^2 - 6*I)*f^2 + (-6*I*d^3*e*f - 3*I*d^2*f^2)*x^2 + (-6*I*d^3*e^2 - 6*I*d^2*e*f - 6*I*
d*f^2)*x)*e^(d*x + c) - (24*(d*e*f - c*f^2)*e^(2*d*x + 2*c) + (-24*I*d*e*f + 24*I*c*f^2)*e^(d*x + c))*log(e^(d
*x + c) - I) - (24*(d*f^2*x + c*f^2)*e^(2*d*x + 2*c) + (-24*I*d*f^2*x - 24*I*c*f^2)*e^(d*x + c))*log(I*e^(d*x
+ c) + 1))/(6*a*d^3*e^(2*d*x + 2*c) - 6*I*a*d^3*e^(d*x + c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)**2*sinh(d*x+c)**2/(a+I*a*sinh(d*x+c)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )}^{2} \sinh \left (d x + c\right )^{2}}{i \, a \sinh \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((f*x + e)^2*sinh(d*x + c)^2/(I*a*sinh(d*x + c) + a), x)