3.23 \(\int (c+d x)^3 \text {csch}(a+b x) \, dx\)

Optimal. Leaf size=149 \[ -\frac {6 d^3 \text {Li}_4\left (-e^{a+b x}\right )}{b^4}+\frac {6 d^3 \text {Li}_4\left (e^{a+b x}\right )}{b^4}+\frac {6 d^2 (c+d x) \text {Li}_3\left (-e^{a+b x}\right )}{b^3}-\frac {6 d^2 (c+d x) \text {Li}_3\left (e^{a+b x}\right )}{b^3}-\frac {3 d (c+d x)^2 \text {Li}_2\left (-e^{a+b x}\right )}{b^2}+\frac {3 d (c+d x)^2 \text {Li}_2\left (e^{a+b x}\right )}{b^2}-\frac {2 (c+d x)^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]

[Out]

-2*(d*x+c)^3*arctanh(exp(b*x+a))/b-3*d*(d*x+c)^2*polylog(2,-exp(b*x+a))/b^2+3*d*(d*x+c)^2*polylog(2,exp(b*x+a)
)/b^2+6*d^2*(d*x+c)*polylog(3,-exp(b*x+a))/b^3-6*d^2*(d*x+c)*polylog(3,exp(b*x+a))/b^3-6*d^3*polylog(4,-exp(b*
x+a))/b^4+6*d^3*polylog(4,exp(b*x+a))/b^4

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Rubi [A]  time = 0.14, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {4182, 2531, 6609, 2282, 6589} \[ \frac {6 d^2 (c+d x) \text {PolyLog}\left (3,-e^{a+b x}\right )}{b^3}-\frac {6 d^2 (c+d x) \text {PolyLog}\left (3,e^{a+b x}\right )}{b^3}-\frac {3 d (c+d x)^2 \text {PolyLog}\left (2,-e^{a+b x}\right )}{b^2}+\frac {3 d (c+d x)^2 \text {PolyLog}\left (2,e^{a+b x}\right )}{b^2}-\frac {6 d^3 \text {PolyLog}\left (4,-e^{a+b x}\right )}{b^4}+\frac {6 d^3 \text {PolyLog}\left (4,e^{a+b x}\right )}{b^4}-\frac {2 (c+d x)^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3*Csch[a + b*x],x]

[Out]

(-2*(c + d*x)^3*ArcTanh[E^(a + b*x)])/b - (3*d*(c + d*x)^2*PolyLog[2, -E^(a + b*x)])/b^2 + (3*d*(c + d*x)^2*Po
lyLog[2, E^(a + b*x)])/b^2 + (6*d^2*(c + d*x)*PolyLog[3, -E^(a + b*x)])/b^3 - (6*d^2*(c + d*x)*PolyLog[3, E^(a
+ b*x)])/b^3 - (6*d^3*PolyLog[4, -E^(a + b*x)])/b^4 + (6*d^3*PolyLog[4, E^(a + b*x)])/b^4

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 4182

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*Ar
cTanh[E^(-(I*e) + f*fz*x)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 - E^(-(I*e) + f*
fz*x)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e) + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, x] && IGtQ[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]

Rule 6609

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[((e + f*x)^m*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]), x] - Dist[(f*m)/(b*c*p*Log[F]), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, x] && GtQ[m,
0]

Rubi steps

\begin {align*} \int (c+d x)^3 \text {csch}(a+b x) \, dx &=-\frac {2 (c+d x)^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {(3 d) \int (c+d x)^2 \log \left (1-e^{a+b x}\right ) \, dx}{b}+\frac {(3 d) \int (c+d x)^2 \log \left (1+e^{a+b x}\right ) \, dx}{b}\\ &=-\frac {2 (c+d x)^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {3 d (c+d x)^2 \text {Li}_2\left (-e^{a+b x}\right )}{b^2}+\frac {3 d (c+d x)^2 \text {Li}_2\left (e^{a+b x}\right )}{b^2}+\frac {\left (6 d^2\right ) \int (c+d x) \text {Li}_2\left (-e^{a+b x}\right ) \, dx}{b^2}-\frac {\left (6 d^2\right ) \int (c+d x) \text {Li}_2\left (e^{a+b x}\right ) \, dx}{b^2}\\ &=-\frac {2 (c+d x)^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {3 d (c+d x)^2 \text {Li}_2\left (-e^{a+b x}\right )}{b^2}+\frac {3 d (c+d x)^2 \text {Li}_2\left (e^{a+b x}\right )}{b^2}+\frac {6 d^2 (c+d x) \text {Li}_3\left (-e^{a+b x}\right )}{b^3}-\frac {6 d^2 (c+d x) \text {Li}_3\left (e^{a+b x}\right )}{b^3}-\frac {\left (6 d^3\right ) \int \text {Li}_3\left (-e^{a+b x}\right ) \, dx}{b^3}+\frac {\left (6 d^3\right ) \int \text {Li}_3\left (e^{a+b x}\right ) \, dx}{b^3}\\ &=-\frac {2 (c+d x)^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {3 d (c+d x)^2 \text {Li}_2\left (-e^{a+b x}\right )}{b^2}+\frac {3 d (c+d x)^2 \text {Li}_2\left (e^{a+b x}\right )}{b^2}+\frac {6 d^2 (c+d x) \text {Li}_3\left (-e^{a+b x}\right )}{b^3}-\frac {6 d^2 (c+d x) \text {Li}_3\left (e^{a+b x}\right )}{b^3}-\frac {\left (6 d^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}+\frac {\left (6 d^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^{a+b x}\right )}{b^4}\\ &=-\frac {2 (c+d x)^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {3 d (c+d x)^2 \text {Li}_2\left (-e^{a+b x}\right )}{b^2}+\frac {3 d (c+d x)^2 \text {Li}_2\left (e^{a+b x}\right )}{b^2}+\frac {6 d^2 (c+d x) \text {Li}_3\left (-e^{a+b x}\right )}{b^3}-\frac {6 d^2 (c+d x) \text {Li}_3\left (e^{a+b x}\right )}{b^3}-\frac {6 d^3 \text {Li}_4\left (-e^{a+b x}\right )}{b^4}+\frac {6 d^3 \text {Li}_4\left (e^{a+b x}\right )}{b^4}\\ \end {align*}

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Mathematica [A]  time = 2.65, size = 191, normalized size = 1.28 \[ \frac {-2 b^3 (c+d x)^3 \tanh ^{-1}(\sinh (a+b x)+\cosh (a+b x))-3 d \left (b^2 (c+d x)^2 \text {Li}_2(-\cosh (a+b x)-\sinh (a+b x))-2 b d (c+d x) \text {Li}_3(-\cosh (a+b x)-\sinh (a+b x))+2 d^2 \text {Li}_4(-\cosh (a+b x)-\sinh (a+b x))\right )+3 d \left (b^2 (c+d x)^2 \text {Li}_2(\cosh (a+b x)+\sinh (a+b x))-2 b d (c+d x) \text {Li}_3(\cosh (a+b x)+\sinh (a+b x))+2 d^2 \text {Li}_4(\cosh (a+b x)+\sinh (a+b x))\right )}{b^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x)^3*Csch[a + b*x],x]

[Out]

(-2*b^3*(c + d*x)^3*ArcTanh[Cosh[a + b*x] + Sinh[a + b*x]] - 3*d*(b^2*(c + d*x)^2*PolyLog[2, -Cosh[a + b*x] -
Sinh[a + b*x]] - 2*b*d*(c + d*x)*PolyLog[3, -Cosh[a + b*x] - Sinh[a + b*x]] + 2*d^2*PolyLog[4, -Cosh[a + b*x]
- Sinh[a + b*x]]) + 3*d*(b^2*(c + d*x)^2*PolyLog[2, Cosh[a + b*x] + Sinh[a + b*x]] - 2*b*d*(c + d*x)*PolyLog[3
, Cosh[a + b*x] + Sinh[a + b*x]] + 2*d^2*PolyLog[4, Cosh[a + b*x] + Sinh[a + b*x]]))/b^4

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fricas [C]  time = 0.47, size = 396, normalized size = 2.66 \[ \frac {6 \, d^{3} {\rm polylog}\left (4, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 6 \, d^{3} {\rm polylog}\left (4, -\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) + 3 \, {\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + b^{2} c^{2} d\right )} {\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 3 \, {\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + b^{2} c^{2} d\right )} {\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) - {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + {\left (b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + 3 \, a b^{2} c^{2} d - 3 \, a^{2} b c d^{2} + a^{3} d^{3}\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right ) - 6 \, {\left (b d^{3} x + b c d^{2}\right )} {\rm polylog}\left (3, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 6 \, {\left (b d^{3} x + b c d^{2}\right )} {\rm polylog}\left (3, -\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right )}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*csch(b*x+a),x, algorithm="fricas")

[Out]

(6*d^3*polylog(4, cosh(b*x + a) + sinh(b*x + a)) - 6*d^3*polylog(4, -cosh(b*x + a) - sinh(b*x + a)) + 3*(b^2*d
^3*x^2 + 2*b^2*c*d^2*x + b^2*c^2*d)*dilog(cosh(b*x + a) + sinh(b*x + a)) - 3*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + b^
2*c^2*d)*dilog(-cosh(b*x + a) - sinh(b*x + a)) - (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3)*log
(cosh(b*x + a) + sinh(b*x + a) + 1) + (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*log(cosh(b*x + a) +
sinh(b*x + a) - 1) + (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + a^3*d^3)
*log(-cosh(b*x + a) - sinh(b*x + a) + 1) - 6*(b*d^3*x + b*c*d^2)*polylog(3, cosh(b*x + a) + sinh(b*x + a)) + 6
*(b*d^3*x + b*c*d^2)*polylog(3, -cosh(b*x + a) - sinh(b*x + a)))/b^4

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{3} \operatorname {csch}\left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*csch(b*x+a),x, algorithm="giac")

[Out]

integrate((d*x + c)^3*csch(b*x + a), x)

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maple [B]  time = 0.11, size = 541, normalized size = 3.63 \[ \frac {3 d^{3} \polylog \left (2, {\mathrm e}^{b x +a}\right ) x^{2}}{b^{2}}-\frac {6 d^{3} \polylog \left (3, {\mathrm e}^{b x +a}\right ) x}{b^{3}}+\frac {2 d^{3} a^{3} \arctanh \left ({\mathrm e}^{b x +a}\right )}{b^{4}}+\frac {6 c \,d^{2} \polylog \left (3, -{\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {6 c \,d^{2} \polylog \left (3, {\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {3 c^{2} d \polylog \left (2, -{\mathrm e}^{b x +a}\right )}{b^{2}}+\frac {3 c^{2} d \polylog \left (2, {\mathrm e}^{b x +a}\right )}{b^{2}}-\frac {3 d^{3} \polylog \left (2, -{\mathrm e}^{b x +a}\right ) x^{2}}{b^{2}}+\frac {6 d^{3} \polylog \left (3, -{\mathrm e}^{b x +a}\right ) x}{b^{3}}-\frac {d^{3} \ln \left (1+{\mathrm e}^{b x +a}\right ) x^{3}}{b}-\frac {d^{3} \ln \left (1+{\mathrm e}^{b x +a}\right ) a^{3}}{b^{4}}+\frac {d^{3} \ln \left (1-{\mathrm e}^{b x +a}\right ) x^{3}}{b}+\frac {d^{3} \ln \left (1-{\mathrm e}^{b x +a}\right ) a^{3}}{b^{4}}-\frac {6 c \,d^{2} a^{2} \arctanh \left ({\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {3 c \,d^{2} a^{2} \ln \left (1+{\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {3 c \,d^{2} a^{2} \ln \left (1-{\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {3 c \,d^{2} \ln \left (1-{\mathrm e}^{b x +a}\right ) x^{2}}{b}-\frac {3 c^{2} d \ln \left (1+{\mathrm e}^{b x +a}\right ) x}{b}-\frac {3 c^{2} d \ln \left (1+{\mathrm e}^{b x +a}\right ) a}{b^{2}}+\frac {3 c^{2} d \ln \left (1-{\mathrm e}^{b x +a}\right ) x}{b}+\frac {3 c^{2} d \ln \left (1-{\mathrm e}^{b x +a}\right ) a}{b^{2}}-\frac {3 c \,d^{2} \ln \left (1+{\mathrm e}^{b x +a}\right ) x^{2}}{b}-\frac {6 c \,d^{2} \polylog \left (2, -{\mathrm e}^{b x +a}\right ) x}{b^{2}}+\frac {6 c \,d^{2} \polylog \left (2, {\mathrm e}^{b x +a}\right ) x}{b^{2}}+\frac {6 c^{2} d a \arctanh \left ({\mathrm e}^{b x +a}\right )}{b^{2}}-\frac {6 d^{3} \polylog \left (4, -{\mathrm e}^{b x +a}\right )}{b^{4}}+\frac {6 d^{3} \polylog \left (4, {\mathrm e}^{b x +a}\right )}{b^{4}}-\frac {2 c^{3} \arctanh \left ({\mathrm e}^{b x +a}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^3*csch(b*x+a),x)

[Out]

-6*d^3*polylog(4,-exp(b*x+a))/b^4+6*d^3*polylog(4,exp(b*x+a))/b^4+3/b^2*d^3*polylog(2,exp(b*x+a))*x^2-6/b^3*d^
3*polylog(3,exp(b*x+a))*x+2/b^4*d^3*a^3*arctanh(exp(b*x+a))+6/b^3*c*d^2*polylog(3,-exp(b*x+a))-6/b^3*c*d^2*pol
ylog(3,exp(b*x+a))-3/b^2*c^2*d*polylog(2,-exp(b*x+a))+3/b^2*c^2*d*polylog(2,exp(b*x+a))-1/b*d^3*ln(1+exp(b*x+a
))*x^3-1/b^4*d^3*ln(1+exp(b*x+a))*a^3-3/b^2*d^3*polylog(2,-exp(b*x+a))*x^2+6/b^3*d^3*polylog(3,-exp(b*x+a))*x+
1/b*d^3*ln(1-exp(b*x+a))*x^3+1/b^4*d^3*ln(1-exp(b*x+a))*a^3-6/b^2*c*d^2*polylog(2,-exp(b*x+a))*x+3/b^3*c*d^2*a
^2*ln(1+exp(b*x+a))-3/b^3*c*d^2*a^2*ln(1-exp(b*x+a))+3/b*c*d^2*ln(1-exp(b*x+a))*x^2+6/b^2*c*d^2*polylog(2,exp(
b*x+a))*x-3/b*c^2*d*ln(1+exp(b*x+a))*x-3/b^2*c^2*d*ln(1+exp(b*x+a))*a+3/b*c^2*d*ln(1-exp(b*x+a))*x+3/b^2*c^2*d
*ln(1-exp(b*x+a))*a-3/b*c*d^2*ln(1+exp(b*x+a))*x^2+6/b^2*c^2*d*a*arctanh(exp(b*x+a))-6/b^3*c*d^2*a^2*arctanh(e
xp(b*x+a))-2/b*c^3*arctanh(exp(b*x+a))

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maxima [B]  time = 0.64, size = 333, normalized size = 2.23 \[ -c^{3} {\left (\frac {\log \left (e^{\left (-b x - a\right )} + 1\right )}{b} - \frac {\log \left (e^{\left (-b x - a\right )} - 1\right )}{b}\right )} - \frac {3 \, {\left (b x \log \left (e^{\left (b x + a\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (b x + a\right )}\right )\right )} c^{2} d}{b^{2}} + \frac {3 \, {\left (b x \log \left (-e^{\left (b x + a\right )} + 1\right ) + {\rm Li}_2\left (e^{\left (b x + a\right )}\right )\right )} c^{2} d}{b^{2}} - \frac {3 \, {\left (b^{2} x^{2} \log \left (e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(-e^{\left (b x + a\right )})\right )} c d^{2}}{b^{3}} + \frac {3 \, {\left (b^{2} x^{2} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(e^{\left (b x + a\right )})\right )} c d^{2}}{b^{3}} - \frac {{\left (b^{3} x^{3} \log \left (e^{\left (b x + a\right )} + 1\right ) + 3 \, b^{2} x^{2} {\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 6 \, b x {\rm Li}_{3}(-e^{\left (b x + a\right )}) + 6 \, {\rm Li}_{4}(-e^{\left (b x + a\right )})\right )} d^{3}}{b^{4}} + \frac {{\left (b^{3} x^{3} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 3 \, b^{2} x^{2} {\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 6 \, b x {\rm Li}_{3}(e^{\left (b x + a\right )}) + 6 \, {\rm Li}_{4}(e^{\left (b x + a\right )})\right )} d^{3}}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*csch(b*x+a),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^3/sinh(a + b*x),x)

[Out]

int((c + d*x)^3/sinh(a + b*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right )^{3} \operatorname {csch}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**3*csch(b*x+a),x)

[Out]

Integral((c + d*x)**3*csch(a + b*x), x)

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