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

Optimal. Leaf size=130 \[ -\frac{4 i f^2 \text{PolyLog}\left (2,-i e^{c+d x}\right )}{a d^3}-\frac{4 i f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac{i (e+f x)^2 \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{a d}+\frac{i (e+f x)^2}{a d}-\frac{i (e+f x)^3}{3 a f} \]

[Out]

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

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Rubi [A]  time = 0.272064, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.276, Rules used = {5557, 32, 3318, 4184, 3716, 2190, 2279, 2391} \[ -\frac{4 i f^2 \text{PolyLog}\left (2,-i e^{c+d x}\right )}{a d^3}-\frac{4 i f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac{i (e+f x)^2 \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{a d}+\frac{i (e+f x)^2}{a d}-\frac{i (e+f x)^3}{3 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I*(e + f*x)^2)/(a*d) - ((I/3)*(e + f*x)^3)/(a*f) - ((4*I)*f*(e + f*x)*Log[1 + I*E^(c + d*x)])/(a*d^2) - ((4*I
)*f^2*PolyLog[2, (-I)*E^(c + d*x)])/(a*d^3) + (I*(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 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 (c+d x)}{a+i a \sinh (c+d x)} \, dx &=i \int \frac{(e+f x)^2}{a+i a \sinh (c+d x)} \, dx-\frac{i \int (e+f x)^2 \, dx}{a}\\ &=-\frac{i (e+f x)^3}{3 a f}+\frac{i \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{i (e+f x)^3}{3 a f}+\frac{i (e+f x)^2 \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{a d}-\frac{(2 i f) \int (e+f x) \coth \left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \, dx}{a d}\\ &=\frac{i (e+f x)^2}{a d}-\frac{i (e+f x)^3}{3 a f}+\frac{i (e+f x)^2 \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{a d}+\frac{(4 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{i (e+f x)^2}{a d}-\frac{i (e+f x)^3}{3 a f}-\frac{4 i f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac{i (e+f x)^2 \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{a d}+\frac{\left (4 i 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{i (e+f x)^2}{a d}-\frac{i (e+f x)^3}{3 a f}-\frac{4 i f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}+\frac{i (e+f x)^2 \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{a d}+\frac{\left (4 i 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{i (e+f x)^2}{a d}-\frac{i (e+f x)^3}{3 a f}-\frac{4 i f (e+f x) \log \left (1+i e^{c+d x}\right )}{a d^2}-\frac{4 i f^2 \text{Li}_2\left (-i e^{c+d x}\right )}{a d^3}+\frac{i (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 = 2.45513, size = 188, normalized size = 1.45 \[ \frac{\frac{3 \left (4 \left (1+i e^c\right ) f^2 \text{PolyLog}\left (2,i e^{-c-d x}\right )-2 d (e+f x) \left (d (e+f x)+2 \left (1+i e^c\right ) f \log \left (1-i e^{-c-d x}\right )\right )\right )}{\left (e^c-i\right ) d^3}+\frac{6 i \sinh \left (\frac{d x}{2}\right ) (e+f x)^2}{d \left (\cosh \left (\frac{c}{2}\right )+i \sinh \left (\frac{c}{2}\right )\right ) \left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right )}-i x \left (3 e^2+3 e f x+f^2 x^2\right )}{3 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*x*(3*e^2 + 3*e*f*x + f^2*x^2) + (3*(-2*d*(e + f*x)*(d*(e + f*x) + 2*(1 + I*E^c)*f*Log[1 - I*E^(-c - d*x)
]) + 4*(1 + I*E^c)*f^2*PolyLog[2, I*E^(-c - d*x)]))/(d^3*(-I + E^c)) + ((6*I)*(e + f*x)^2*Sinh[(d*x)/2])/(d*(C
osh[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 = 0.069, size = 269, normalized size = 2.1 \begin{align*}{\frac{-{\frac{i}{3}}{x}^{3}{f}^{2}}{a}}-{\frac{ief{x}^{2}}{a}}-{\frac{i{e}^{2}x}{a}}-2\,{\frac{{x}^{2}{f}^{2}+2\,efx+{e}^{2}}{da \left ({{\rm e}^{dx+c}}-i \right ) }}-{\frac{4\,i\ln \left ({{\rm e}^{dx+c}}-i \right ) ef}{a{d}^{2}}}+{\frac{4\,i\ln \left ({{\rm e}^{dx+c}} \right ) ef}{a{d}^{2}}}+{\frac{2\,i{f}^{2}{x}^{2}}{da}}+{\frac{4\,i{f}^{2}cx}{a{d}^{2}}}+{\frac{2\,i{f}^{2}{c}^{2}}{a{d}^{3}}}-{\frac{4\,i{f}^{2}\ln \left ( 1+i{{\rm e}^{dx+c}} \right ) x}{a{d}^{2}}}-{\frac{4\,i{f}^{2}\ln \left ( 1+i{{\rm e}^{dx+c}} \right ) c}{a{d}^{3}}}-{\frac{4\,i{f}^{2}{\it polylog} \left ( 2,-i{{\rm e}^{dx+c}} \right ) }{a{d}^{3}}}+{\frac{4\,i{f}^{2}c\ln \left ({{\rm e}^{dx+c}}-i \right ) }{a{d}^{3}}}-{\frac{4\,i{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)/(a+I*a*sinh(d*x+c)),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{6} \, f^{2}{\left (\frac{2 i \, d x^{3} e^{\left (d x + c\right )} + 2 \, d x^{3} + 12 \, x^{2}}{a d e^{\left (d x + c\right )} - i \, a d} - 24 \, \int \frac{x}{a d e^{\left (d x + c\right )} - i \, a d}\,{d x}\right )} + e f{\left (\frac{-i \, d x^{2} +{\left (d x^{2} e^{c} - 4 \, x e^{c}\right )} e^{\left (d x\right )}}{i \, a d e^{\left (d x + c\right )} + a d} - \frac{4 i \, \log \left ({\left (e^{\left (d x + c\right )} - i\right )} e^{\left (-c\right )}\right )}{a d^{2}}\right )} - e^{2}{\left (\frac{i \,{\left (d x + c\right )}}{a d} + \frac{2}{{\left (a e^{\left (-d x - c\right )} + i \, a\right )} d}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.46449, size = 649, normalized size = 4.99 \begin{align*} -\frac{d^{3} f^{2} x^{3} + 3 \, d^{3} e f x^{2} + 3 \, d^{3} e^{2} x + 6 \, d^{2} e^{2} - 12 \, c d e f + 6 \, c^{2} f^{2} -{\left (-12 i \, f^{2} e^{\left (d x + c\right )} - 12 \, f^{2}\right )}{\rm Li}_2\left (-i \, e^{\left (d x + c\right )}\right ) -{\left (-i \, d^{3} f^{2} x^{3} + 12 i \, c d e f - 6 i \, c^{2} f^{2} +{\left (-3 i \, d^{3} e f + 6 i \, d^{2} f^{2}\right )} x^{2} +{\left (-3 i \, d^{3} e^{2} + 12 i \, d^{2} e f\right )} x\right )} e^{\left (d x + c\right )} +{\left (12 \, d e f - 12 \, c f^{2} -{\left (-12 i \, d e f + 12 i \, c f^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (e^{\left (d x + c\right )} - i\right ) +{\left (12 \, d f^{2} x + 12 \, c f^{2} -{\left (-12 i \, d f^{2} x - 12 i \, c f^{2}\right )} e^{\left (d x + c\right )}\right )} \log \left (i \, e^{\left (d x + c\right )} + 1\right )}{3 \, a d^{3} e^{\left (d x + c\right )} - 3 i \, a d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)/(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 )}{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)/(a+I*a*sinh(d*x+c)),x, algorithm="giac")

[Out]

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