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

Optimal. Leaf size=256 \[ -\frac {3 d^3 \text {Li}_2\left (-e^{a+b x}\right )}{b^4}+\frac {3 d^3 \text {Li}_2\left (e^{a+b x}\right )}{b^4}+\frac {3 d^3 \text {Li}_4\left (-e^{a+b x}\right )}{b^4}-\frac {3 d^3 \text {Li}_4\left (e^{a+b x}\right )}{b^4}-\frac {3 d^2 (c+d x) \text {Li}_3\left (-e^{a+b x}\right )}{b^3}+\frac {3 d^2 (c+d x) \text {Li}_3\left (e^{a+b x}\right )}{b^3}-\frac {6 d^2 (c+d x) \tanh ^{-1}\left (e^{a+b x}\right )}{b^3}+\frac {3 d (c+d x)^2 \text {Li}_2\left (-e^{a+b x}\right )}{2 b^2}-\frac {3 d (c+d x)^2 \text {Li}_2\left (e^{a+b x}\right )}{2 b^2}-\frac {3 d (c+d x)^2 \text {csch}(a+b x)}{2 b^2}+\frac {(c+d x)^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {(c+d x)^3 \coth (a+b x) \text {csch}(a+b x)}{2 b} \]

[Out]

-6*d^2*(d*x+c)*arctanh(exp(b*x+a))/b^3+(d*x+c)^3*arctanh(exp(b*x+a))/b-3/2*d*(d*x+c)^2*csch(b*x+a)/b^2-1/2*(d*
x+c)^3*coth(b*x+a)*csch(b*x+a)/b-3*d^3*polylog(2,-exp(b*x+a))/b^4+3/2*d*(d*x+c)^2*polylog(2,-exp(b*x+a))/b^2+3
*d^3*polylog(2,exp(b*x+a))/b^4-3/2*d*(d*x+c)^2*polylog(2,exp(b*x+a))/b^2-3*d^2*(d*x+c)*polylog(3,-exp(b*x+a))/
b^3+3*d^2*(d*x+c)*polylog(3,exp(b*x+a))/b^3+3*d^3*polylog(4,-exp(b*x+a))/b^4-3*d^3*polylog(4,exp(b*x+a))/b^4

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Rubi [A]  time = 0.28, antiderivative size = 256, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 8, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4186, 4182, 2279, 2391, 2531, 6609, 2282, 6589} \[ -\frac {3 d^2 (c+d x) \text {PolyLog}\left (3,-e^{a+b x}\right )}{b^3}+\frac {3 d^2 (c+d x) \text {PolyLog}\left (3,e^{a+b x}\right )}{b^3}+\frac {3 d (c+d x)^2 \text {PolyLog}\left (2,-e^{a+b x}\right )}{2 b^2}-\frac {3 d (c+d x)^2 \text {PolyLog}\left (2,e^{a+b x}\right )}{2 b^2}-\frac {3 d^3 \text {PolyLog}\left (2,-e^{a+b x}\right )}{b^4}+\frac {3 d^3 \text {PolyLog}\left (2,e^{a+b x}\right )}{b^4}+\frac {3 d^3 \text {PolyLog}\left (4,-e^{a+b x}\right )}{b^4}-\frac {3 d^3 \text {PolyLog}\left (4,e^{a+b x}\right )}{b^4}-\frac {6 d^2 (c+d x) \tanh ^{-1}\left (e^{a+b x}\right )}{b^3}-\frac {3 d (c+d x)^2 \text {csch}(a+b x)}{2 b^2}+\frac {(c+d x)^3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {(c+d x)^3 \coth (a+b x) \text {csch}(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3*Csch[a + b*x]^3,x]

[Out]

(-6*d^2*(c + d*x)*ArcTanh[E^(a + b*x)])/b^3 + ((c + d*x)^3*ArcTanh[E^(a + b*x)])/b - (3*d*(c + d*x)^2*Csch[a +
b*x])/(2*b^2) - ((c + d*x)^3*Coth[a + b*x]*Csch[a + b*x])/(2*b) - (3*d^3*PolyLog[2, -E^(a + b*x)])/b^4 + (3*d
*(c + d*x)^2*PolyLog[2, -E^(a + b*x)])/(2*b^2) + (3*d^3*PolyLog[2, E^(a + b*x)])/b^4 - (3*d*(c + d*x)^2*PolyLo
g[2, E^(a + b*x)])/(2*b^2) - (3*d^2*(c + d*x)*PolyLog[3, -E^(a + b*x)])/b^3 + (3*d^2*(c + d*x)*PolyLog[3, E^(a
+ b*x)])/b^3 + (3*d^3*PolyLog[4, -E^(a + b*x)])/b^4 - (3*d^3*PolyLog[4, E^(a + b*x)])/b^4

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 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 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]

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 4186

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(b^2*(c + d*x)^m*Cot[e
+ f*x]*(b*Csc[e + f*x])^(n - 2))/(f*(n - 1)), x] + (Dist[(b^2*d^2*m*(m - 1))/(f^2*(n - 1)*(n - 2)), Int[(c + d
*x)^(m - 2)*(b*Csc[e + f*x])^(n - 2), x], x] + Dist[(b^2*(n - 2))/(n - 1), Int[(c + d*x)^m*(b*Csc[e + f*x])^(n
- 2), x], x] - Simp[(b^2*d*m*(c + d*x)^(m - 1)*(b*Csc[e + f*x])^(n - 2))/(f^2*(n - 1)*(n - 2)), x]) /; FreeQ[
{b, c, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2] && GtQ[m, 1]

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

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Mathematica [A]  time = 3.31, size = 440, normalized size = 1.72 \[ -\frac {b^3 c^3 \log \left (1-e^{a+b x}\right )-b^3 c^3 \log \left (e^{a+b x}+1\right )+3 b^3 c^2 d x \log \left (1-e^{a+b x}\right )-3 b^3 c^2 d x \log \left (e^{a+b x}+1\right )+3 b^3 c d^2 x^2 \log \left (1-e^{a+b x}\right )-3 b^3 c d^2 x^2 \log \left (e^{a+b x}+1\right )+b^3 d^3 x^3 \log \left (1-e^{a+b x}\right )-b^3 d^3 x^3 \log \left (e^{a+b x}+1\right )-3 d \text {Li}_2\left (-e^{a+b x}\right ) \left (b^2 (c+d x)^2-2 d^2\right )+3 d \text {Li}_2\left (e^{a+b x}\right ) \left (b^2 (c+d x)^2-2 d^2\right )+b^2 (c+d x)^2 \text {csch}(a+b x) (b (c+d x) \coth (a+b x)+3 d)+6 b c d^2 \text {Li}_3\left (-e^{a+b x}\right )-6 b c d^2 \text {Li}_3\left (e^{a+b x}\right )-6 b c d^2 \log \left (1-e^{a+b x}\right )+6 b c d^2 \log \left (e^{a+b x}+1\right )+6 b d^3 x \text {Li}_3\left (-e^{a+b x}\right )-6 b d^3 x \text {Li}_3\left (e^{a+b x}\right )-6 d^3 \text {Li}_4\left (-e^{a+b x}\right )+6 d^3 \text {Li}_4\left (e^{a+b x}\right )-6 b d^3 x \log \left (1-e^{a+b x}\right )+6 b d^3 x \log \left (e^{a+b x}+1\right )}{2 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(b^2*(c + d*x)^2*(3*d + b*(c + d*x)*Coth[a + b*x])*Csch[a + b*x] + b^3*c^3*Log[1 - E^(a + b*x)] - 6*b*c*d
^2*Log[1 - E^(a + b*x)] + 3*b^3*c^2*d*x*Log[1 - E^(a + b*x)] - 6*b*d^3*x*Log[1 - E^(a + b*x)] + 3*b^3*c*d^2*x^
2*Log[1 - E^(a + b*x)] + b^3*d^3*x^3*Log[1 - E^(a + b*x)] - b^3*c^3*Log[1 + E^(a + b*x)] + 6*b*c*d^2*Log[1 + E
^(a + b*x)] - 3*b^3*c^2*d*x*Log[1 + E^(a + b*x)] + 6*b*d^3*x*Log[1 + E^(a + b*x)] - 3*b^3*c*d^2*x^2*Log[1 + E^
(a + b*x)] - b^3*d^3*x^3*Log[1 + E^(a + b*x)] - 3*d*(-2*d^2 + b^2*(c + d*x)^2)*PolyLog[2, -E^(a + b*x)] + 3*d*
(-2*d^2 + b^2*(c + d*x)^2)*PolyLog[2, E^(a + b*x)] + 6*b*c*d^2*PolyLog[3, -E^(a + b*x)] + 6*b*d^3*x*PolyLog[3,
-E^(a + b*x)] - 6*b*c*d^2*PolyLog[3, E^(a + b*x)] - 6*b*d^3*x*PolyLog[3, E^(a + b*x)] - 6*d^3*PolyLog[4, -E^(
a + b*x)] + 6*d^3*PolyLog[4, E^(a + b*x)])/b^4

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fricas [C]  time = 2.22, size = 4008, normalized size = 15.66 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(b^3*d^3*x^3 + b^3*c^3 + 3*b^2*c^2*d + 3*(b^3*c*d^2 + b^2*d^3)*x^2 + 3*(b^3*c^2*d + 2*b^2*c*d^2)*x)*co
sh(b*x + a)^3 + 6*(b^3*d^3*x^3 + b^3*c^3 + 3*b^2*c^2*d + 3*(b^3*c*d^2 + b^2*d^3)*x^2 + 3*(b^3*c^2*d + 2*b^2*c*
d^2)*x)*cosh(b*x + a)*sinh(b*x + a)^2 + 2*(b^3*d^3*x^3 + b^3*c^3 + 3*b^2*c^2*d + 3*(b^3*c*d^2 + b^2*d^3)*x^2 +
3*(b^3*c^2*d + 2*b^2*c*d^2)*x)*sinh(b*x + a)^3 + 2*(b^3*d^3*x^3 + b^3*c^3 - 3*b^2*c^2*d + 3*(b^3*c*d^2 - b^2*
d^3)*x^2 + 3*(b^3*c^2*d - 2*b^2*c*d^2)*x)*cosh(b*x + a) + 3*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + b^2*c^2*d + (b^2*d^
3*x^2 + 2*b^2*c*d^2*x + b^2*c^2*d - 2*d^3)*cosh(b*x + a)^4 + 4*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + b^2*c^2*d - 2*d^
3)*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 - 2*d^3)*sinh(b*x + a)^4 - 2*d^3 -
2*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + b^2*c^2*d - 2*d^3)*cosh(b*x + a)^2 - 2*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + b^2*c^
2*d - 2*d^3 - 3*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + b^2*c^2*d - 2*d^3)*cosh(b*x + a)^2)*sinh(b*x + a)^2 + 4*((b^2*d
^3*x^2 + 2*b^2*c*d^2*x + b^2*c^2*d - 2*d^3)*cosh(b*x + a)^3 - (b^2*d^3*x^2 + 2*b^2*c*d^2*x + b^2*c^2*d - 2*d^3
)*cosh(b*x + a))*sinh(b*x + a))*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 + (b^2*d^3*x^2 + 2*b^2*c*d^2*x + b^2*c^2*d - 2*d^3)*cosh(b*x + a)^4 + 4*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + b^2
*c^2*d - 2*d^3)*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 - 2*d^3)*sinh(b*x + a
)^4 - 2*d^3 - 2*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + b^2*c^2*d - 2*d^3)*cosh(b*x + a)^2 - 2*(b^2*d^3*x^2 + 2*b^2*c*d
^2*x + b^2*c^2*d - 2*d^3 - 3*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + b^2*c^2*d - 2*d^3)*cosh(b*x + a)^2)*sinh(b*x + a)^
2 + 4*((b^2*d^3*x^2 + 2*b^2*c*d^2*x + b^2*c^2*d - 2*d^3)*cosh(b*x + a)^3 - (b^2*d^3*x^2 + 2*b^2*c*d^2*x + b^2*
c^2*d - 2*d^3)*cosh(b*x + a))*sinh(b*x + a))*dilog(-cosh(b*x + a) - sinh(b*x + a)) - (b^3*d^3*x^3 + 3*b^3*c*d^
2*x^2 + b^3*c^3 + (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + b^3*c^3 - 6*b*c*d^2 + 3*(b^3*c^2*d - 2*b*d^3)*x)*cosh(b*x +
a)^4 + 4*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + b^3*c^3 - 6*b*c*d^2 + 3*(b^3*c^2*d - 2*b*d^3)*x)*cosh(b*x + a)*sinh
(b*x + a)^3 + (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + b^3*c^3 - 6*b*c*d^2 + 3*(b^3*c^2*d - 2*b*d^3)*x)*sinh(b*x + a)^
4 - 6*b*c*d^2 - 2*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + b^3*c^3 - 6*b*c*d^2 + 3*(b^3*c^2*d - 2*b*d^3)*x)*cosh(b*x +
a)^2 - 2*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + b^3*c^3 - 6*b*c*d^2 - 3*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + b^3*c^3 -
6*b*c*d^2 + 3*(b^3*c^2*d - 2*b*d^3)*x)*cosh(b*x + a)^2 + 3*(b^3*c^2*d - 2*b*d^3)*x)*sinh(b*x + a)^2 + 3*(b^3*c
^2*d - 2*b*d^3)*x + 4*((b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + b^3*c^3 - 6*b*c*d^2 + 3*(b^3*c^2*d - 2*b*d^3)*x)*cosh(
b*x + a)^3 - (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + b^3*c^3 - 6*b*c*d^2 + 3*(b^3*c^2*d - 2*b*d^3)*x)*cosh(b*x + a))*
sinh(b*x + a))*log(cosh(b*x + a) + sinh(b*x + a) + 1) + (b^3*c^3 - 3*a*b^2*c^2*d + 3*(a^2 - 2)*b*c*d^2 + (b^3*
c^3 - 3*a*b^2*c^2*d + 3*(a^2 - 2)*b*c*d^2 - (a^3 - 6*a)*d^3)*cosh(b*x + a)^4 + 4*(b^3*c^3 - 3*a*b^2*c^2*d + 3*
(a^2 - 2)*b*c*d^2 - (a^3 - 6*a)*d^3)*cosh(b*x + a)*sinh(b*x + a)^3 + (b^3*c^3 - 3*a*b^2*c^2*d + 3*(a^2 - 2)*b*
c*d^2 - (a^3 - 6*a)*d^3)*sinh(b*x + a)^4 - (a^3 - 6*a)*d^3 - 2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*(a^2 - 2)*b*c*d^2
- (a^3 - 6*a)*d^3)*cosh(b*x + a)^2 - 2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*(a^2 - 2)*b*c*d^2 - (a^3 - 6*a)*d^3 - 3*(b
^3*c^3 - 3*a*b^2*c^2*d + 3*(a^2 - 2)*b*c*d^2 - (a^3 - 6*a)*d^3)*cosh(b*x + a)^2)*sinh(b*x + a)^2 + 4*((b^3*c^3
- 3*a*b^2*c^2*d + 3*(a^2 - 2)*b*c*d^2 - (a^3 - 6*a)*d^3)*cosh(b*x + a)^3 - (b^3*c^3 - 3*a*b^2*c^2*d + 3*(a^2
- 2)*b*c*d^2 - (a^3 - 6*a)*d^3)*cosh(b*x + a))*sinh(b*x + a))*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*a*b^2*c^2*d - 3*a^2*b*c*d^2 + (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*a*b^2*c^2*d - 3*a
^2*b*c*d^2 + (a^3 - 6*a)*d^3 + 3*(b^3*c^2*d - 2*b*d^3)*x)*cosh(b*x + a)^4 + 4*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 +
3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + (a^3 - 6*a)*d^3 + 3*(b^3*c^2*d - 2*b*d^3)*x)*cosh(b*x + a)*sinh(b*x + a)^3 +
(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + (a^3 - 6*a)*d^3 + 3*(b^3*c^2*d - 2*b*d^3)*x)*
sinh(b*x + a)^4 + (a^3 - 6*a)*d^3 - 2*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + (a^3 -
6*a)*d^3 + 3*(b^3*c^2*d - 2*b*d^3)*x)*cosh(b*x + a)^2 - 2*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*a*b^2*c^2*d - 3*a
^2*b*c*d^2 + (a^3 - 6*a)*d^3 - 3*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + (a^3 - 6*a)*
d^3 + 3*(b^3*c^2*d - 2*b*d^3)*x)*cosh(b*x + a)^2 + 3*(b^3*c^2*d - 2*b*d^3)*x)*sinh(b*x + a)^2 + 3*(b^3*c^2*d -
2*b*d^3)*x + 4*((b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + (a^3 - 6*a)*d^3 + 3*(b^3*c^2
*d - 2*b*d^3)*x)*cosh(b*x + a)^3 - (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + (a^3 - 6*a
)*d^3 + 3*(b^3*c^2*d - 2*b*d^3)*x)*cosh(b*x + a))*sinh(b*x + a))*log(-cosh(b*x + a) - sinh(b*x + a) + 1) + 6*(
d^3*cosh(b*x + a)^4 + 4*d^3*cosh(b*x + a)*sinh(b*x + a)^3 + d^3*sinh(b*x + a)^4 - 2*d^3*cosh(b*x + a)^2 + d^3
+ 2*(3*d^3*cosh(b*x + a)^2 - d^3)*sinh(b*x + a)^2 + 4*(d^3*cosh(b*x + a)^3 - d^3*cosh(b*x + a))*sinh(b*x + a))
*polylog(4, cosh(b*x + a) + sinh(b*x + a)) - 6*(d^3*cosh(b*x + a)^4 + 4*d^3*cosh(b*x + a)*sinh(b*x + a)^3 + d^
3*sinh(b*x + a)^4 - 2*d^3*cosh(b*x + a)^2 + d^3 + 2*(3*d^3*cosh(b*x + a)^2 - d^3)*sinh(b*x + a)^2 + 4*(d^3*cos
h(b*x + a)^3 - d^3*cosh(b*x + a))*sinh(b*x + a))*polylog(4, -cosh(b*x + a) - sinh(b*x + a)) - 6*(b*d^3*x + (b*
d^3*x + b*c*d^2)*cosh(b*x + a)^4 + 4*(b*d^3*x + b*c*d^2)*cosh(b*x + a)*sinh(b*x + a)^3 + (b*d^3*x + b*c*d^2)*s
inh(b*x + a)^4 + b*c*d^2 - 2*(b*d^3*x + b*c*d^2)*cosh(b*x + a)^2 - 2*(b*d^3*x + b*c*d^2 - 3*(b*d^3*x + b*c*d^2
)*cosh(b*x + a)^2)*sinh(b*x + a)^2 + 4*((b*d^3*x + b*c*d^2)*cosh(b*x + a)^3 - (b*d^3*x + b*c*d^2)*cosh(b*x + a
))*sinh(b*x + a))*polylog(3, cosh(b*x + a) + sinh(b*x + a)) + 6*(b*d^3*x + (b*d^3*x + b*c*d^2)*cosh(b*x + a)^4
+ 4*(b*d^3*x + b*c*d^2)*cosh(b*x + a)*sinh(b*x + a)^3 + (b*d^3*x + b*c*d^2)*sinh(b*x + a)^4 + b*c*d^2 - 2*(b*
d^3*x + b*c*d^2)*cosh(b*x + a)^2 - 2*(b*d^3*x + b*c*d^2 - 3*(b*d^3*x + b*c*d^2)*cosh(b*x + a)^2)*sinh(b*x + a)
^2 + 4*((b*d^3*x + b*c*d^2)*cosh(b*x + a)^3 - (b*d^3*x + b*c*d^2)*cosh(b*x + a))*sinh(b*x + a))*polylog(3, -co
sh(b*x + a) - sinh(b*x + a)) + 2*(b^3*d^3*x^3 + b^3*c^3 - 3*b^2*c^2*d + 3*(b^3*c*d^2 - b^2*d^3)*x^2 + 3*(b^3*d
^3*x^3 + b^3*c^3 + 3*b^2*c^2*d + 3*(b^3*c*d^2 + b^2*d^3)*x^2 + 3*(b^3*c^2*d + 2*b^2*c*d^2)*x)*cosh(b*x + a)^2
+ 3*(b^3*c^2*d - 2*b^2*c*d^2)*x)*sinh(b*x + a))/(b^4*cosh(b*x + a)^4 + 4*b^4*cosh(b*x + a)*sinh(b*x + a)^3 + b
^4*sinh(b*x + a)^4 - 2*b^4*cosh(b*x + a)^2 + b^4 + 2*(3*b^4*cosh(b*x + a)^2 - b^4)*sinh(b*x + a)^2 + 4*(b^4*co
sh(b*x + a)^3 - b^4*cosh(b*x + a))*sinh(b*x + a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.37, size = 876, normalized size = 3.42 \[ \frac {6 d^{3} a \arctanh \left ({\mathrm e}^{b x +a}\right )}{b^{4}}-\frac {6 c \,d^{2} \arctanh \left ({\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {{\mathrm e}^{b x +a} \left (b \,d^{3} x^{3} {\mathrm e}^{2 b x +2 a}+3 b c \,d^{2} x^{2} {\mathrm e}^{2 b x +2 a}+3 b \,c^{2} d x \,{\mathrm e}^{2 b x +2 a}+b \,d^{3} x^{3}+3 d^{3} x^{2} {\mathrm e}^{2 b x +2 a}+b \,c^{3} {\mathrm e}^{2 b x +2 a}+3 b c \,d^{2} x^{2}+6 c \,d^{2} x \,{\mathrm e}^{2 b x +2 a}+3 b \,c^{2} d x +3 c^{2} d \,{\mathrm e}^{2 b x +2 a}-3 d^{3} x^{2}+b \,c^{3}-6 c \,d^{2} x -3 c^{2} d \right )}{b^{2} \left ({\mathrm e}^{2 b x +2 a}-1\right )^{2}}+\frac {3 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 ) a^{3}}{2 b^{4}}-\frac {3 c \,d^{2} \polylog \left (3, -{\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {3 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 )}{2 b^{2}}+\frac {3 d^{3} \polylog \left (2, -{\mathrm e}^{b x +a}\right ) x^{2}}{2 b^{2}}-\frac {3 d^{3} \polylog \left (2, {\mathrm e}^{b x +a}\right ) x^{2}}{2 b^{2}}-\frac {d^{3} \ln \left (1-{\mathrm e}^{b x +a}\right ) x^{3}}{2 b}-\frac {d^{3} \ln \left (1-{\mathrm e}^{b x +a}\right ) a^{3}}{2 b^{4}}+\frac {d^{3} \ln \left (1+{\mathrm e}^{b x +a}\right ) x^{3}}{2 b}-\frac {3 c \,d^{2} \ln \left (1-{\mathrm e}^{b x +a}\right ) x^{2}}{2 b}+\frac {3 c^{2} d \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 ) a}{2 b^{2}}-\frac {3 c^{2} d \ln \left (1-{\mathrm e}^{b x +a}\right ) x}{2 b}-\frac {3 c \,d^{2} a^{2} \ln \left (1+{\mathrm e}^{b x +a}\right )}{2 b^{3}}+\frac {3 c \,d^{2} a^{2} \ln \left (1-{\mathrm e}^{b x +a}\right )}{2 b^{3}}+\frac {3 c \,d^{2} \ln \left (1+{\mathrm e}^{b x +a}\right ) x^{2}}{2 b}-\frac {3 c^{2} d a \arctanh \left ({\mathrm e}^{b x +a}\right )}{b^{2}}+\frac {3 c \,d^{2} a^{2} \arctanh \left ({\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {3 c \,d^{2} \polylog \left (2, -{\mathrm e}^{b x +a}\right ) x}{b^{2}}-\frac {3 d^{3} \polylog \left (4, {\mathrm e}^{b x +a}\right )}{b^{4}}+\frac {3 d^{3} \polylog \left (4, -{\mathrm e}^{b x +a}\right )}{b^{4}}+\frac {c^{3} \arctanh \left ({\mathrm e}^{b x +a}\right )}{b}-\frac {3 c^{2} d \ln \left (1-{\mathrm e}^{b x +a}\right ) a}{2 b^{2}}+\frac {3 d^{3} \ln \left (1-{\mathrm e}^{b x +a}\right ) x}{b^{3}}-\frac {3 d^{3} \polylog \left (2, -{\mathrm e}^{b x +a}\right )}{b^{4}}+\frac {3 d^{3} \polylog \left (2, {\mathrm e}^{b x +a}\right )}{b^{4}}-\frac {3 d^{3} \ln \left (1+{\mathrm e}^{b x +a}\right ) x}{b^{3}}-\frac {3 d^{3} \ln \left (1+{\mathrm e}^{b x +a}\right ) a}{b^{4}}-\frac {3 d^{3} \polylog \left (3, -{\mathrm e}^{b x +a}\right ) x}{b^{3}}-\frac {3 c \,d^{2} \polylog \left (2, {\mathrm e}^{b x +a}\right ) x}{b^{2}}+\frac {3 d^{3} \ln \left (1-{\mathrm e}^{b x +a}\right ) a}{b^{4}}-\frac {d^{3} a^{3} \arctanh \left ({\mathrm e}^{b x +a}\right )}{b^{4}}-\frac {3 c^{2} d \polylog \left (2, {\mathrm e}^{b x +a}\right )}{2 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-exp(b*x+a)*(b*d^3*x^3*exp(2*b*x+2*a)+3*b*c*d^2*x^2*exp(2*b*x+2*a)+3*b*c^2*d*x*exp(2*b*x+2*a)+b*d^3*x^3+3*d^3*
x^2*exp(2*b*x+2*a)+b*c^3*exp(2*b*x+2*a)+3*b*c*d^2*x^2+6*c*d^2*x*exp(2*b*x+2*a)+3*b*c^2*d*x+3*c^2*d*exp(2*b*x+2
*a)-3*d^3*x^2+b*c^3-6*c*d^2*x-3*c^2*d)/b^2/(exp(2*b*x+2*a)-1)^2+1/2/b^4*d^3*ln(1+exp(b*x+a))*a^3-3/2/b^2*c^2*d
*polylog(2,exp(b*x+a))-1/b^4*d^3*a^3*arctanh(exp(b*x+a))-3/b^3*c*d^2*polylog(3,-exp(b*x+a))+3/b^3*c*d^2*polylo
g(3,exp(b*x+a))+3/2/b^2*c^2*d*polylog(2,-exp(b*x+a))+3/2/b^2*d^3*polylog(2,-exp(b*x+a))*x^2-3/b^3*d^3*polylog(
3,-exp(b*x+a))*x-1/2/b*d^3*ln(1-exp(b*x+a))*x^3-1/2/b^4*d^3*ln(1-exp(b*x+a))*a^3-3/2/b^2*d^3*polylog(2,exp(b*x
+a))*x^2+3/b^3*d^3*polylog(3,exp(b*x+a))*x+1/2/b*d^3*ln(1+exp(b*x+a))*x^3-3*d^3*polylog(4,exp(b*x+a))/b^4-3*d^
3*polylog(2,-exp(b*x+a))/b^4+3*d^3*polylog(2,exp(b*x+a))/b^4+3*d^3*polylog(4,-exp(b*x+a))/b^4-3/2/b*c*d^2*ln(1
-exp(b*x+a))*x^2-3/b^2*c*d^2*polylog(2,exp(b*x+a))*x+3/2/b*c^2*d*ln(1+exp(b*x+a))*x+3/2/b^2*c^2*d*ln(1+exp(b*x
+a))*a-3/2/b*c^2*d*ln(1-exp(b*x+a))*x-3/2/b^3*c*d^2*a^2*ln(1+exp(b*x+a))+3/2/b^3*c*d^2*a^2*ln(1-exp(b*x+a))+3/
2/b*c*d^2*ln(1+exp(b*x+a))*x^2-3/b^2*c^2*d*a*arctanh(exp(b*x+a))+3/b^3*c*d^2*a^2*arctanh(exp(b*x+a))+6/b^4*d^3
*a*arctanh(exp(b*x+a))-6/b^3*c*d^2*arctanh(exp(b*x+a))+1/b*c^3*arctanh(exp(b*x+a))-3/2/b^2*c^2*d*ln(1-exp(b*x+
a))*a+3/b^2*c*d^2*polylog(2,-exp(b*x+a))*x+3/b^3*d^3*ln(1-exp(b*x+a))*x-3/b^3*d^3*ln(1+exp(b*x+a))*x-3/b^4*d^3
*ln(1+exp(b*x+a))*a+3/b^4*d^3*ln(1-exp(b*x+a))*a

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maxima [B]  time = 0.54, size = 605, normalized size = 2.36 \[ \frac {1}{2} \, 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} + \frac {2 \, {\left (e^{\left (-b x - a\right )} + e^{\left (-3 \, b x - 3 \, a\right )}\right )}}{b {\left (2 \, e^{\left (-2 \, b x - 2 \, a\right )} - e^{\left (-4 \, b x - 4 \, a\right )} - 1\right )}}\right )} + \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}}{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}}{2 \, b^{3}} - \frac {3 \, c d^{2} \log \left (e^{\left (b x + a\right )} + 1\right )}{b^{3}} + \frac {3 \, c d^{2} \log \left (e^{\left (b x + a\right )} - 1\right )}{b^{3}} - \frac {{\left (b d^{3} x^{3} e^{\left (3 \, a\right )} + 3 \, c^{2} d e^{\left (3 \, a\right )} + 3 \, {\left (b c d^{2} + d^{3}\right )} x^{2} e^{\left (3 \, a\right )} + 3 \, {\left (b c^{2} d + 2 \, c d^{2}\right )} x e^{\left (3 \, a\right )}\right )} e^{\left (3 \, b x\right )} + {\left (b d^{3} x^{3} e^{a} - 3 \, c^{2} d e^{a} + 3 \, {\left (b c d^{2} - d^{3}\right )} x^{2} e^{a} + 3 \, {\left (b c^{2} d - 2 \, c d^{2}\right )} x e^{a}\right )} e^{\left (b x\right )}}{b^{2} e^{\left (4 \, b x + 4 \, a\right )} - 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}} + \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}}{2 \, 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}}{2 \, b^{4}} + \frac {3 \, {\left (b^{2} c^{2} d - 2 \, d^{3}\right )} {\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 )}}{2 \, b^{4}} - \frac {3 \, {\left (b^{2} c^{2} d - 2 \, d^{3}\right )} {\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 )}}{2 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*c^3*(log(e^(-b*x - a) + 1)/b - log(e^(-b*x - a) - 1)/b + 2*(e^(-b*x - a) + e^(-3*b*x - 3*a))/(b*(2*e^(-2*b
*x - 2*a) - e^(-4*b*x - 4*a) - 1))) + 3/2*(b^2*x^2*log(e^(b*x + a) + 1) + 2*b*x*dilog(-e^(b*x + a)) - 2*polylo
g(3, -e^(b*x + a)))*c*d^2/b^3 - 3/2*(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*c*d^2*log(e^(b*x + a) + 1)/b^3 + 3*c*d^2*log(e^(b*x + a) - 1)/b^3 - ((b*d^3*x^3*e^(
3*a) + 3*c^2*d*e^(3*a) + 3*(b*c*d^2 + d^3)*x^2*e^(3*a) + 3*(b*c^2*d + 2*c*d^2)*x*e^(3*a))*e^(3*b*x) + (b*d^3*x
^3*e^a - 3*c^2*d*e^a + 3*(b*c*d^2 - d^3)*x^2*e^a + 3*(b*c^2*d - 2*c*d^2)*x*e^a)*e^(b*x))/(b^2*e^(4*b*x + 4*a)
- 2*b^2*e^(2*b*x + 2*a) + b^2) + 1/2*(b^3*x^3*log(e^(b*x + a) + 1) + 3*b^2*x^2*dilog(-e^(b*x + a)) - 6*b*x*pol
ylog(3, -e^(b*x + a)) + 6*polylog(4, -e^(b*x + a)))*d^3/b^4 - 1/2*(b^3*x^3*log(-e^(b*x + a) + 1) + 3*b^2*x^2*d
ilog(e^(b*x + a)) - 6*b*x*polylog(3, e^(b*x + a)) + 6*polylog(4, e^(b*x + a)))*d^3/b^4 + 3/2*(b^2*c^2*d - 2*d^
3)*(b*x*log(e^(b*x + a) + 1) + dilog(-e^(b*x + a)))/b^4 - 3/2*(b^2*c^2*d - 2*d^3)*(b*x*log(-e^(b*x + a) + 1) +
dilog(e^(b*x + a)))/b^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^3/sinh(a + b*x)^3,x)

[Out]

int((c + d*x)^3/sinh(a + b*x)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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