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

Optimal. Leaf size=132 \[ -\frac {12 d^2 (c+d x) \text {Li}_2\left (-i e^{e+f x}\right )}{a f^3}-\frac {6 d (c+d x)^2 \log \left (1+i e^{e+f x}\right )}{a f^2}+\frac {(c+d x)^3 \tanh \left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )}{a f}+\frac {(c+d x)^3}{a f}+\frac {12 d^3 \text {Li}_3\left (-i e^{e+f x}\right )}{a f^4} \]

[Out]

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

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Rubi [A]  time = 0.30, antiderivative size = 132, normalized size of antiderivative = 1.00, 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 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 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 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 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 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 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 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*}

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Mathematica [A]  time = 2.93, size = 206, normalized size = 1.56 \[ \frac {2 \left (\frac {3 d e^e \left (-\frac {2 i d e^{-e} \left (e^e-i\right ) \left (f (c+d x) \text {Li}_2\left (i e^{-e-f x}\right )+d \text {Li}_3\left (i e^{-e-f x}\right )\right )}{f^3}+\frac {\left (e^{-e}+i\right ) (c+d x)^2 \log \left (1-i e^{-e-f x}\right )}{f}+\frac {e^{-e} (c+d x)^3}{3 d}\right )}{-1-i e^e}+\frac {(c+d x)^3 \sinh \left (\frac {f x}{2}\right )}{\left (\cosh \left (\frac {e}{2}\right )+i \sinh \left (\frac {e}{2}\right )\right ) \left (\cosh \left (\frac {1}{2} (e+f x)\right )+i \sinh \left (\frac {1}{2} (e+f x)\right )\right )}\right )}{a f} \]

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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fricas [C]  time = 0.50, size = 363, normalized size = 2.75 \[ \frac {-2 i \, d^{3} e^{3} + 6 i \, c d^{2} e^{2} f - 6 i \, c^{2} d e f^{2} + 2 i \, c^{3} f^{3} + {\left (12 i \, d^{3} f x + 12 i \, c d^{2} f - 12 \, {\left (d^{3} f x + c d^{2} f\right )} e^{\left (f x + e\right )}\right )} {\rm Li}_2\left (-i \, e^{\left (f x + e\right )}\right ) + 2 \, {\left (d^{3} f^{3} x^{3} + 3 \, c d^{2} f^{3} x^{2} + 3 \, c^{2} d f^{3} x + d^{3} e^{3} - 3 \, c d^{2} e^{2} f + 3 \, c^{2} d e f^{2}\right )} e^{\left (f x + e\right )} + {\left (6 i \, d^{3} e^{2} - 12 i \, c d^{2} e f + 6 i \, c^{2} d f^{2} - 6 \, {\left (d^{3} e^{2} - 2 \, c d^{2} e f + c^{2} d f^{2}\right )} e^{\left (f x + e\right )}\right )} \log \left (e^{\left (f x + e\right )} - i\right ) + {\left (6 i \, d^{3} f^{2} x^{2} + 12 i \, c d^{2} f^{2} x - 6 i \, d^{3} e^{2} + 12 i \, c d^{2} e f - 6 \, {\left (d^{3} f^{2} x^{2} + 2 \, c d^{2} f^{2} x - d^{3} e^{2} + 2 \, c d^{2} e f\right )} e^{\left (f x + e\right )}\right )} \log \left (i \, e^{\left (f x + e\right )} + 1\right ) + {\left (12 \, d^{3} e^{\left (f x + e\right )} - 12 i \, d^{3}\right )} {\rm polylog}\left (3, -i \, e^{\left (f x + e\right )}\right )}{a f^{4} e^{\left (f x + e\right )} - i \, a f^{4}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d x + c\right )}^{3}}{i \, a \sinh \left (f x + e\right ) + a}\,{d x} \]

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)

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maple [B]  time = 0.17, size = 435, normalized size = 3.30 \[ \frac {2 i \left (d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}\right )}{f a \left ({\mathrm e}^{f x +e}-i\right )}-\frac {4 d^{3} e^{3}}{a \,f^{4}}-\frac {6 d^{3} e^{2} \ln \left ({\mathrm e}^{f x +e}-i\right )}{a \,f^{4}}+\frac {2 d^{3} x^{3}}{a f}-\frac {6 d^{3} \ln \left (1+i {\mathrm e}^{f x +e}\right ) x^{2}}{a \,f^{2}}+\frac {6 d^{3} \ln \left (1+i {\mathrm e}^{f x +e}\right ) e^{2}}{a \,f^{4}}-\frac {6 d \ln \left ({\mathrm e}^{f x +e}-i\right ) c^{2}}{a \,f^{2}}+\frac {12 d^{2} c e x}{a \,f^{2}}-\frac {12 d^{2} c \ln \left (1+i {\mathrm e}^{f x +e}\right ) x}{a \,f^{2}}-\frac {12 d^{2} c \ln \left (1+i {\mathrm e}^{f x +e}\right ) e}{a \,f^{3}}-\frac {12 d^{2} c e \ln \left ({\mathrm e}^{f x +e}\right )}{a \,f^{3}}+\frac {6 d \ln \left ({\mathrm e}^{f x +e}\right ) c^{2}}{a \,f^{2}}+\frac {6 d^{3} e^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{a \,f^{4}}-\frac {12 d^{3} \polylog \left (2, -i {\mathrm e}^{f x +e}\right ) x}{a \,f^{3}}+\frac {12 d^{2} c e \ln \left ({\mathrm e}^{f x +e}-i\right )}{a \,f^{3}}+\frac {6 d^{2} c \,e^{2}}{a \,f^{3}}-\frac {12 d^{2} c \polylog \left (2, -i {\mathrm e}^{f x +e}\right )}{a \,f^{3}}+\frac {6 d^{2} c \,x^{2}}{a f}-\frac {6 d^{3} e^{2} x}{a \,f^{3}}+\frac {12 d^{3} \polylog \left (3, -i {\mathrm e}^{f x +e}\right )}{a \,f^{4}} \]

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

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maxima [B]  time = 0.57, size = 237, normalized size = 1.80 \[ 6 \, c^{2} d {\left (\frac {x e^{\left (f x + e\right )}}{a f e^{\left (f x + e\right )} - i \, a f} - \frac {\log \left ({\left (e^{\left (f x + e\right )} - i\right )} e^{\left (-e\right )}\right )}{a f^{2}}\right )} - \frac {2 \, c^{3}}{{\left (i \, a e^{\left (-f x - e\right )} - a\right )} f} + \frac {2 i \, d^{3} x^{3} + 6 i \, c d^{2} x^{2}}{a f e^{\left (f x + e\right )} - i \, a f} - \frac {12 \, {\left (f x \log \left (i \, e^{\left (f x + e\right )} + 1\right ) + {\rm Li}_2\left (-i \, e^{\left (f x + e\right )}\right )\right )} c d^{2}}{a f^{3}} - \frac {6 \, {\left (f^{2} x^{2} \log \left (i \, e^{\left (f x + e\right )} + 1\right ) + 2 \, f x {\rm Li}_2\left (-i \, e^{\left (f x + e\right )}\right ) - 2 \, {\rm Li}_{3}(-i \, e^{\left (f x + e\right )})\right )} d^{3}}{a f^{4}} + \frac {2 \, {\left (d^{3} f^{3} x^{3} + 3 \, c d^{2} f^{3} x^{2}\right )}}{a f^{4}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c+d\,x\right )}^3}{a+a\,\mathrm {sinh}\left (e+f\,x\right )\,1{}\mathrm {i}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^3/(a + a*sinh(e + f*x)*1i),x)

[Out]

int((c + d*x)^3/(a + a*sinh(e + f*x)*1i), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {- 2 i c^{3} e^{e} - 6 i c^{2} d x e^{e} - 6 i c d^{2} x^{2} e^{e} - 2 i d^{3} x^{3} e^{e}}{- i a f e^{e} - a f e^{- f x}} - \frac {6 d \left (\int \frac {c^{2} e^{f x}}{e^{e} e^{f x} - i}\, dx + \int \frac {d^{2} x^{2} e^{f x}}{e^{e} e^{f x} - i}\, dx + \int \frac {2 c d x e^{f x}}{e^{e} e^{f x} - i}\, dx\right ) e^{e}}{a f} \]

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]

(-2*I*c**3*exp(e) - 6*I*c**2*d*x*exp(e) - 6*I*c*d**2*x**2*exp(e) - 2*I*d**3*x**3*exp(e))/(-I*a*f*exp(e) - a*f*
exp(-f*x)) - 6*d*(Integral(c**2*exp(f*x)/(exp(e)*exp(f*x) - I), x) + Integral(d**2*x**2*exp(f*x)/(exp(e)*exp(f
*x) - I), x) + Integral(2*c*d*x*exp(f*x)/(exp(e)*exp(f*x) - I), x))*exp(e)/(a*f)

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