3.108 \(\int \frac{(c+d x)^3}{a+i a \sinh (e+f x)} \, dx\)

Optimal. Leaf size=132 \[ -\frac{12 d^2 (c+d x) \text{PolyLog}\left (2,-i e^{e+f x}\right )}{a f^3}+\frac{12 d^3 \text{PolyLog}\left (3,-i e^{e+f x}\right )}{a f^4}-\frac{6 d (c+d x)^2 \log \left (1+i e^{e+f x}\right )}{a f^2}+\frac{(c+d x)^3 \tanh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right )}{a f}+\frac{(c+d x)^3}{a f} \]

[Out]

(c + d*x)^3/(a*f) - (6*d*(c + d*x)^2*Log[1 + I*E^(e + f*x)])/(a*f^2) - (12*d^2*(c + d*x)*PolyLog[2, (-I)*E^(e
+ f*x)])/(a*f^3) + (12*d^3*PolyLog[3, (-I)*E^(e + f*x)])/(a*f^4) + ((c + d*x)^3*Tanh[e/2 + (I/4)*Pi + (f*x)/2]
)/(a*f)

________________________________________________________________________________________

Rubi [A]  time = 0.302292, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {3318, 4184, 3716, 2190, 2531, 2282, 6589} \[ -\frac{12 d^2 (c+d x) \text{PolyLog}\left (2,-i e^{e+f x}\right )}{a f^3}+\frac{12 d^3 \text{PolyLog}\left (3,-i e^{e+f x}\right )}{a f^4}-\frac{6 d (c+d x)^2 \log \left (1+i e^{e+f x}\right )}{a f^2}+\frac{(c+d x)^3 \tanh \left (\frac{e}{2}+\frac{f x}{2}+\frac{i \pi }{4}\right )}{a f}+\frac{(c+d x)^3}{a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c + d*x)^3/(a*f) - (6*d*(c + d*x)^2*Log[1 + I*E^(e + f*x)])/(a*f^2) - (12*d^2*(c + d*x)*PolyLog[2, (-I)*E^(e
+ f*x)])/(a*f^3) + (12*d^3*PolyLog[3, (-I)*E^(e + f*x)])/(a*f^4) + ((c + d*x)^3*Tanh[e/2 + (I/4)*Pi + (f*x)/2]
)/(a*f)

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 2531

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 2282

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 6589

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*} \int \frac{(c+d x)^3}{a+i a \sinh (e+f x)} \, dx &=\frac{\int (c+d x)^3 \csc ^2\left (\frac{1}{2} \left (i e+\frac{\pi }{2}\right )+\frac{i f x}{2}\right ) \, dx}{2 a}\\ &=\frac{(c+d x)^3 \tanh \left (\frac{e}{2}+\frac{i \pi }{4}+\frac{f x}{2}\right )}{a f}-\frac{(3 d) \int (c+d x)^2 \coth \left (\frac{e}{2}-\frac{i \pi }{4}+\frac{f x}{2}\right ) \, dx}{a f}\\ &=\frac{(c+d x)^3}{a f}+\frac{(c+d x)^3 \tanh \left (\frac{e}{2}+\frac{i \pi }{4}+\frac{f x}{2}\right )}{a f}-\frac{(6 i d) \int \frac{e^{2 \left (\frac{e}{2}+\frac{f x}{2}\right )} (c+d x)^2}{1+i e^{2 \left (\frac{e}{2}+\frac{f x}{2}\right )}} \, dx}{a f}\\ &=\frac{(c+d x)^3}{a f}-\frac{6 d (c+d x)^2 \log \left (1+i e^{e+f x}\right )}{a f^2}+\frac{(c+d x)^3 \tanh \left (\frac{e}{2}+\frac{i \pi }{4}+\frac{f x}{2}\right )}{a f}+\frac{\left (12 d^2\right ) \int (c+d x) \log \left (1+i e^{2 \left (\frac{e}{2}+\frac{f x}{2}\right )}\right ) \, dx}{a f^2}\\ &=\frac{(c+d x)^3}{a f}-\frac{6 d (c+d x)^2 \log \left (1+i e^{e+f x}\right )}{a f^2}-\frac{12 d^2 (c+d x) \text{Li}_2\left (-i e^{e+f x}\right )}{a f^3}+\frac{(c+d x)^3 \tanh \left (\frac{e}{2}+\frac{i \pi }{4}+\frac{f x}{2}\right )}{a f}+\frac{\left (12 d^3\right ) \int \text{Li}_2\left (-i e^{2 \left (\frac{e}{2}+\frac{f x}{2}\right )}\right ) \, dx}{a f^3}\\ &=\frac{(c+d x)^3}{a f}-\frac{6 d (c+d x)^2 \log \left (1+i e^{e+f x}\right )}{a f^2}-\frac{12 d^2 (c+d x) \text{Li}_2\left (-i e^{e+f x}\right )}{a f^3}+\frac{(c+d x)^3 \tanh \left (\frac{e}{2}+\frac{i \pi }{4}+\frac{f x}{2}\right )}{a f}+\frac{\left (12 d^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{2 \left (\frac{e}{2}+\frac{f x}{2}\right )}\right )}{a f^4}\\ &=\frac{(c+d x)^3}{a f}-\frac{6 d (c+d x)^2 \log \left (1+i e^{e+f x}\right )}{a f^2}-\frac{12 d^2 (c+d x) \text{Li}_2\left (-i e^{e+f x}\right )}{a f^3}+\frac{12 d^3 \text{Li}_3\left (-i e^{e+f x}\right )}{a f^4}+\frac{(c+d x)^3 \tanh \left (\frac{e}{2}+\frac{i \pi }{4}+\frac{f x}{2}\right )}{a f}\\ \end{align*}

Mathematica [A]  time = 2.89412, size = 206, normalized size = 1.56 \[ \frac{2 \left (\frac{3 d e^e \left (-\frac{2 i d e^{-e} \left (e^e-i\right ) \left (f (c+d x) \text{PolyLog}\left (2,i e^{-e-f x}\right )+d \text{PolyLog}\left (3,i e^{-e-f x}\right )\right )}{f^3}+\frac{\left (e^{-e}+i\right ) (c+d x)^2 \log \left (1-i e^{-e-f x}\right )}{f}+\frac{e^{-e} (c+d x)^3}{3 d}\right )}{-1-i e^e}+\frac{(c+d x)^3 \sinh \left (\frac{f x}{2}\right )}{\left (\cosh \left (\frac{e}{2}\right )+i \sinh \left (\frac{e}{2}\right )\right ) \left (\cosh \left (\frac{1}{2} (e+f x)\right )+i \sinh \left (\frac{1}{2} (e+f x)\right )\right )}\right )}{a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.112, size = 435, normalized size = 3.3 \begin{align*}{\frac{2\,i \left ({d}^{3}{x}^{3}+3\,c{d}^{2}{x}^{2}+3\,{c}^{2}dx+{c}^{3} \right ) }{fa \left ({{\rm e}^{fx+e}}-i \right ) }}-12\,{\frac{c{d}^{2}{\it polylog} \left ( 2,-i{{\rm e}^{fx+e}} \right ) }{a{f}^{3}}}-4\,{\frac{{d}^{3}{e}^{3}}{{f}^{4}a}}+6\,{\frac{d\ln \left ({{\rm e}^{fx+e}} \right ){c}^{2}}{a{f}^{2}}}+6\,{\frac{{d}^{3}{e}^{2}\ln \left ( 1+i{{\rm e}^{fx+e}} \right ) }{{f}^{4}a}}-12\,{\frac{c{d}^{2}\ln \left ( 1+i{{\rm e}^{fx+e}} \right ) x}{a{f}^{2}}}-12\,{\frac{c{d}^{2}\ln \left ( 1+i{{\rm e}^{fx+e}} \right ) e}{a{f}^{3}}}+12\,{\frac{c{d}^{2}ex}{a{f}^{2}}}+2\,{\frac{{d}^{3}{x}^{3}}{fa}}-6\,{\frac{{d}^{3}{e}^{2}x}{a{f}^{3}}}-12\,{\frac{c{d}^{2}e\ln \left ({{\rm e}^{fx+e}} \right ) }{a{f}^{3}}}+6\,{\frac{c{d}^{2}{x}^{2}}{fa}}+6\,{\frac{c{d}^{2}{e}^{2}}{a{f}^{3}}}-6\,{\frac{{d}^{3}{e}^{2}\ln \left ({{\rm e}^{fx+e}}-i \right ) }{{f}^{4}a}}+12\,{\frac{c{d}^{2}e\ln \left ({{\rm e}^{fx+e}}-i \right ) }{a{f}^{3}}}-12\,{\frac{{d}^{3}{\it polylog} \left ( 2,-i{{\rm e}^{fx+e}} \right ) x}{a{f}^{3}}}+12\,{\frac{{d}^{3}{\it polylog} \left ( 3,-i{{\rm e}^{fx+e}} \right ) }{{f}^{4}a}}-6\,{\frac{{d}^{3}\ln \left ( 1+i{{\rm e}^{fx+e}} \right ){x}^{2}}{a{f}^{2}}}-6\,{\frac{d\ln \left ({{\rm e}^{fx+e}}-i \right ){c}^{2}}{a{f}^{2}}}+6\,{\frac{{d}^{3}{e}^{2}\ln \left ({{\rm e}^{fx+e}} \right ) }{{f}^{4}a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^3/(a+I*a*sinh(f*x+e)),x)

[Out]

2*I*(d^3*x^3+3*c*d^2*x^2+3*c^2*d*x+c^3)/f/a/(exp(f*x+e)-I)-12*d^2/f^3/a*c*polylog(2,-I*exp(f*x+e))-4*d^3/f^4/a
*e^3+6*d/f^2/a*ln(exp(f*x+e))*c^2+6*d^3/f^4/a*e^2*ln(1+I*exp(f*x+e))-12*d^2/f^2/a*c*ln(1+I*exp(f*x+e))*x-12*d^
2/f^3/a*c*ln(1+I*exp(f*x+e))*e+12*d^2/f^2/a*c*e*x+2*d^3/f/a*x^3-6*d^3/f^3/a*e^2*x-12*d^2/f^3/a*c*e*ln(exp(f*x+
e))+6*d^2/f/a*c*x^2+6*d^2/f^3/a*c*e^2-6*d^3/f^4/a*e^2*ln(exp(f*x+e)-I)+12*d^2/f^3/a*c*e*ln(exp(f*x+e)-I)-12*d^
3/f^3/a*polylog(2,-I*exp(f*x+e))*x+12*d^3*polylog(3,-I*exp(f*x+e))/a/f^4-6*d^3/f^2/a*ln(1+I*exp(f*x+e))*x^2-6*
d/f^2/a*ln(exp(f*x+e)-I)*c^2+6*d^3/f^4/a*e^2*ln(exp(f*x+e))

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Maxima [B]  time = 1.68317, size = 320, normalized size = 2.42 \begin{align*} 6 \, c^{2} d{\left (\frac{x e^{\left (f x + e\right )}}{a f e^{\left (f x + e\right )} - i \, a f} - \frac{\log \left ({\left (e^{\left (f x + e\right )} - i\right )} e^{\left (-e\right )}\right )}{a f^{2}}\right )} - \frac{2 \, c^{3}}{{\left (i \, a e^{\left (-f x - e\right )} - a\right )} f} + \frac{2 i \, d^{3} x^{3} + 6 i \, c d^{2} x^{2}}{a f e^{\left (f x + e\right )} - i \, a f} - \frac{12 \,{\left (f x \log \left (i \, e^{\left (f x + e\right )} + 1\right ) +{\rm Li}_2\left (-i \, e^{\left (f x + e\right )}\right )\right )} c d^{2}}{a f^{3}} - \frac{6 \,{\left (f^{2} x^{2} \log \left (i \, e^{\left (f x + e\right )} + 1\right ) + 2 \, f x{\rm Li}_2\left (-i \, e^{\left (f x + e\right )}\right ) - 2 \,{\rm Li}_{3}(-i \, e^{\left (f x + e\right )})\right )} d^{3}}{a f^{4}} + \frac{2 \,{\left (d^{3} f^{3} x^{3} + 3 \, c d^{2} f^{3} x^{2}\right )}}{a f^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6*c^2*d*(x*e^(f*x + e)/(a*f*e^(f*x + e) - I*a*f) - log((e^(f*x + e) - I)*e^(-e))/(a*f^2)) - 2*c^3/((I*a*e^(-f*
x - e) - a)*f) + (2*I*d^3*x^3 + 6*I*c*d^2*x^2)/(a*f*e^(f*x + e) - I*a*f) - 12*(f*x*log(I*e^(f*x + e) + 1) + di
log(-I*e^(f*x + e)))*c*d^2/(a*f^3) - 6*(f^2*x^2*log(I*e^(f*x + e) + 1) + 2*f*x*dilog(-I*e^(f*x + e)) - 2*polyl
og(3, -I*e^(f*x + e)))*d^3/(a*f^4) + 2*(d^3*f^3*x^3 + 3*c*d^2*f^3*x^2)/(a*f^4)

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Fricas [C]  time = 2.72744, size = 863, normalized size = 6.54 \begin{align*} \frac{-2 i \, d^{3} e^{3} + 6 i \, c d^{2} e^{2} f - 6 i \, c^{2} d e f^{2} + 2 i \, c^{3} f^{3} +{\left (12 i \, d^{3} f x + 12 i \, c d^{2} f - 12 \,{\left (d^{3} f x + c d^{2} f\right )} e^{\left (f x + e\right )}\right )}{\rm Li}_2\left (-i \, e^{\left (f x + e\right )}\right ) + 2 \,{\left (d^{3} f^{3} x^{3} + 3 \, c d^{2} f^{3} x^{2} + 3 \, c^{2} d f^{3} x + d^{3} e^{3} - 3 \, c d^{2} e^{2} f + 3 \, c^{2} d e f^{2}\right )} e^{\left (f x + e\right )} +{\left (6 i \, d^{3} e^{2} - 12 i \, c d^{2} e f + 6 i \, c^{2} d f^{2} - 6 \,{\left (d^{3} e^{2} - 2 \, c d^{2} e f + c^{2} d f^{2}\right )} e^{\left (f x + e\right )}\right )} \log \left (e^{\left (f x + e\right )} - i\right ) +{\left (6 i \, d^{3} f^{2} x^{2} + 12 i \, c d^{2} f^{2} x - 6 i \, d^{3} e^{2} + 12 i \, c d^{2} e f - 6 \,{\left (d^{3} f^{2} x^{2} + 2 \, c d^{2} f^{2} x - d^{3} e^{2} + 2 \, c d^{2} e f\right )} e^{\left (f x + e\right )}\right )} \log \left (i \, e^{\left (f x + e\right )} + 1\right ) +{\left (12 \, d^{3} e^{\left (f x + e\right )} - 12 i \, d^{3}\right )}{\rm polylog}\left (3, -i \, e^{\left (f x + e\right )}\right )}{a f^{4} e^{\left (f x + e\right )} - i \, a f^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((d*x+c)**3/(a+I*a*sinh(f*x+e)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}^{3}}{i \, a \sinh \left (f x + e\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x + c)^3/(I*a*sinh(f*x + e) + a), x)