3.177 \(\int \frac {1}{a+b \sinh ^3(c+d x)} \, dx\)

Optimal. Leaf size=280 \[ -\frac {2 \tanh ^{-1}\left (\frac {\sqrt [3]{b}-\sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}+b^{2/3}}}\right )}{3 a^{2/3} d \sqrt {a^{2/3}+b^{2/3}}}-\frac {2 (-1)^{2/3} \tan ^{-1}\left (\frac {\sqrt [6]{-1} \left (\sqrt [6]{-1} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {\sqrt [3]{-1} a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 a^{2/3} d \sqrt {\sqrt [3]{-1} a^{2/3}-(-1)^{2/3} b^{2/3}}}-\frac {2 (-1)^{2/3} \tan ^{-1}\left (\frac {\sqrt [6]{-1} \left ((-1)^{5/6} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {\sqrt [3]{-1} a^{2/3}-b^{2/3}}}\right )}{3 a^{2/3} d \sqrt {\sqrt [3]{-1} a^{2/3}-b^{2/3}}} \]

[Out]

-2/3*(-1)^(2/3)*arctan((-1)^(1/6)*((-1)^(5/6)*b^(1/3)+I*a^(1/3)*tanh(1/2*d*x+1/2*c))/((-1)^(1/3)*a^(2/3)-b^(2/
3))^(1/2))/a^(2/3)/d/((-1)^(1/3)*a^(2/3)-b^(2/3))^(1/2)-2/3*arctanh((b^(1/3)-a^(1/3)*tanh(1/2*d*x+1/2*c))/(a^(
2/3)+b^(2/3))^(1/2))/a^(2/3)/d/(a^(2/3)+b^(2/3))^(1/2)-2/3*(-1)^(2/3)*arctan((-1)^(1/6)*((-1)^(1/6)*b^(1/3)+I*
a^(1/3)*tanh(1/2*d*x+1/2*c))/((-1)^(1/3)*a^(2/3)-(-1)^(2/3)*b^(2/3))^(1/2))/a^(2/3)/d/((-1)^(1/3)*a^(2/3)-(-1)
^(2/3)*b^(2/3))^(1/2)

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Rubi [A]  time = 0.25, antiderivative size = 280, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3213, 2660, 618, 204} \[ -\frac {2 \tanh ^{-1}\left (\frac {\sqrt [3]{b}-\sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}+b^{2/3}}}\right )}{3 a^{2/3} d \sqrt {a^{2/3}+b^{2/3}}}-\frac {2 (-1)^{2/3} \tan ^{-1}\left (\frac {\sqrt [6]{-1} \left (\sqrt [6]{-1} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {\sqrt [3]{-1} a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 a^{2/3} d \sqrt {\sqrt [3]{-1} a^{2/3}-(-1)^{2/3} b^{2/3}}}-\frac {2 (-1)^{2/3} \tan ^{-1}\left (\frac {\sqrt [6]{-1} \left ((-1)^{5/6} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {\sqrt [3]{-1} a^{2/3}-b^{2/3}}}\right )}{3 a^{2/3} d \sqrt {\sqrt [3]{-1} a^{2/3}-b^{2/3}}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sinh[c + d*x]^3)^(-1),x]

[Out]

(-2*(-1)^(2/3)*ArcTan[((-1)^(1/6)*((-1)^(1/6)*b^(1/3) + I*a^(1/3)*Tanh[(c + d*x)/2]))/Sqrt[(-1)^(1/3)*a^(2/3)
- (-1)^(2/3)*b^(2/3)]])/(3*a^(2/3)*Sqrt[(-1)^(1/3)*a^(2/3) - (-1)^(2/3)*b^(2/3)]*d) - (2*(-1)^(2/3)*ArcTan[((-
1)^(1/6)*((-1)^(5/6)*b^(1/3) + I*a^(1/3)*Tanh[(c + d*x)/2]))/Sqrt[(-1)^(1/3)*a^(2/3) - b^(2/3)]])/(3*a^(2/3)*S
qrt[(-1)^(1/3)*a^(2/3) - b^(2/3)]*d) - (2*ArcTanh[(b^(1/3) - a^(1/3)*Tanh[(c + d*x)/2])/Sqrt[a^(2/3) + b^(2/3)
]])/(3*a^(2/3)*Sqrt[a^(2/3) + b^(2/3)]*d)

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 3213

Int[((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Int[ExpandTrig[(a + b*(c*sin[e + f*
x])^n)^p, x], x] /; FreeQ[{a, b, c, e, f, n}, x] && (IGtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))

Rubi steps

\begin {align*} \int \frac {1}{a+b \sinh ^3(c+d x)} \, dx &=\int \left (\frac {\sqrt [6]{-1}}{3 a^{2/3} \left (\sqrt [6]{-1} \sqrt [3]{a}-i \sqrt [3]{b} \sinh (c+d x)\right )}+\frac {\sqrt [6]{-1}}{3 a^{2/3} \left (\sqrt [6]{-1} \sqrt [3]{a}+\sqrt [6]{-1} \sqrt [3]{b} \sinh (c+d x)\right )}+\frac {\sqrt [6]{-1}}{3 a^{2/3} \left (\sqrt [6]{-1} \sqrt [3]{a}+(-1)^{5/6} \sqrt [3]{b} \sinh (c+d x)\right )}\right ) \, dx\\ &=\frac {\sqrt [6]{-1} \int \frac {1}{\sqrt [6]{-1} \sqrt [3]{a}-i \sqrt [3]{b} \sinh (c+d x)} \, dx}{3 a^{2/3}}+\frac {\sqrt [6]{-1} \int \frac {1}{\sqrt [6]{-1} \sqrt [3]{a}+\sqrt [6]{-1} \sqrt [3]{b} \sinh (c+d x)} \, dx}{3 a^{2/3}}+\frac {\sqrt [6]{-1} \int \frac {1}{\sqrt [6]{-1} \sqrt [3]{a}+(-1)^{5/6} \sqrt [3]{b} \sinh (c+d x)} \, dx}{3 a^{2/3}}\\ &=-\frac {\left (2 (-1)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [6]{-1} \sqrt [3]{a}-2 \sqrt [3]{b} x+\sqrt [6]{-1} \sqrt [3]{a} x^2} \, dx,x,\tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{3 a^{2/3} d}-\frac {\left (2 (-1)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [6]{-1} \sqrt [3]{a}+2 \sqrt [3]{-1} \sqrt [3]{b} x+\sqrt [6]{-1} \sqrt [3]{a} x^2} \, dx,x,\tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{3 a^{2/3} d}-\frac {\left (2 (-1)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [6]{-1} \sqrt [3]{a}-2 (-1)^{2/3} \sqrt [3]{b} x+\sqrt [6]{-1} \sqrt [3]{a} x^2} \, dx,x,\tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{3 a^{2/3} d}\\ &=\frac {\left (4 (-1)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (\sqrt [3]{-1} a^{2/3}-b^{2/3}\right )-x^2} \, dx,x,-2 \sqrt [3]{b}+2 \sqrt [6]{-1} \sqrt [3]{a} \tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{3 a^{2/3} d}+\frac {\left (4 (-1)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \sqrt [3]{-1} \left (a^{2/3}+b^{2/3}\right )-x^2} \, dx,x,-2 (-1)^{2/3} \sqrt [3]{b}+2 \sqrt [6]{-1} \sqrt [3]{a} \tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{3 a^{2/3} d}+\frac {\left (4 (-1)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (\sqrt [3]{-1} a^{2/3}-(-1)^{2/3} b^{2/3}\right )-x^2} \, dx,x,2 \sqrt [3]{-1} \sqrt [3]{b}+2 \sqrt [6]{-1} \sqrt [3]{a} \tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{3 a^{2/3} d}\\ &=\frac {2 (-1)^{2/3} \tan ^{-1}\left (\frac {\sqrt [3]{b}-(-1)^{2/3} \sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\sqrt [3]{-1} a^{2/3}-b^{2/3}}}\right )}{3 a^{2/3} \sqrt {\sqrt [3]{-1} a^{2/3}-b^{2/3}} d}-\frac {2 (-1)^{2/3} \tan ^{-1}\left (\frac {\sqrt [3]{-1} \sqrt [3]{b}+(-1)^{2/3} \sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\sqrt [3]{-1} a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 a^{2/3} \sqrt {\sqrt [3]{-1} a^{2/3}-(-1)^{2/3} b^{2/3}} d}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt [3]{b}-\sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}+b^{2/3}}}\right )}{3 a^{2/3} \sqrt {a^{2/3}+b^{2/3}} d}\\ \end {align*}

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Mathematica [C]  time = 0.17, size = 131, normalized size = 0.47 \[ \frac {2 \text {RootSum}\left [\text {$\#$1}^6 b-3 \text {$\#$1}^4 b+8 \text {$\#$1}^3 a+3 \text {$\#$1}^2 b-b\& ,\frac {2 \text {$\#$1} \log \left (-\text {$\#$1} \sinh \left (\frac {1}{2} (c+d x)\right )+\text {$\#$1} \cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )-\cosh \left (\frac {1}{2} (c+d x)\right )\right )+\text {$\#$1} c+\text {$\#$1} d x}{\text {$\#$1}^4 b-2 \text {$\#$1}^2 b+4 \text {$\#$1} a+b}\& \right ]}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sinh[c + d*x]^3)^(-1),x]

[Out]

(2*RootSum[-b + 3*b*#1^2 + 8*a*#1^3 - 3*b*#1^4 + b*#1^6 & , (c*#1 + d*x*#1 + 2*Log[-Cosh[(c + d*x)/2] - Sinh[(
c + d*x)/2] + Cosh[(c + d*x)/2]*#1 - Sinh[(c + d*x)/2]*#1]*#1)/(b + 4*a*#1 - 2*b*#1^2 + b*#1^4) & ])/(3*d)

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sinh(d*x+c)^3),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{b \sinh \left (d x + c\right )^{3} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sinh(d*x+c)^3),x, algorithm="giac")

[Out]

integrate(1/(b*sinh(d*x + c)^3 + a), x)

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maple [C]  time = 0.09, size = 87, normalized size = 0.31 \[ \frac {\munderset {\textit {\_R} =\RootOf \left (a \,\textit {\_Z}^{6}-3 a \,\textit {\_Z}^{4}-8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}-a \right )}{\sum }\frac {\left (-\textit {\_R}^{4}+2 \textit {\_R}^{2}-1\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a -2 \textit {\_R}^{3} a -4 \textit {\_R}^{2} b +\textit {\_R} a}}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sinh(d*x+c)^3),x)

[Out]

1/3/d*sum((-_R^4+2*_R^2-1)/(_R^5*a-2*_R^3*a-4*_R^2*b+_R*a)*ln(tanh(1/2*d*x+1/2*c)-_R),_R=RootOf(_Z^6*a-3*_Z^4*
a-8*_Z^3*b+3*_Z^2*a-a))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{b \sinh \left (d x + c\right )^{3} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sinh(d*x+c)^3),x, algorithm="maxima")

[Out]

integrate(1/(b*sinh(d*x + c)^3 + a), x)

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mupad [B]  time = 9.49, size = 1261, normalized size = 4.50 \[ \sum _{k=1}^6\ln \left (\frac {\left (-4\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^4\,b^2\,d^6\,z^6+729\,a^6\,d^6\,z^6-243\,a^4\,d^4\,z^4+27\,a^2\,d^2\,z^2-1,z,k\right )+d\,x}+\mathrm {root}\left (729\,a^4\,b^2\,d^6\,z^6+729\,a^6\,d^6\,z^6-243\,a^4\,d^4\,z^4+27\,a^2\,d^2\,z^2-1,z,k\right )\,b\,d+{\mathrm {root}\left (729\,a^4\,b^2\,d^6\,z^6+729\,a^6\,d^6\,z^6-243\,a^4\,d^4\,z^4+27\,a^2\,d^2\,z^2-1,z,k\right )}^2\,a\,b\,d^2\,12-\mathrm {root}\left (729\,a^4\,b^2\,d^6\,z^6+729\,a^6\,d^6\,z^6-243\,a^4\,d^4\,z^4+27\,a^2\,d^2\,z^2-1,z,k\right )\,a\,d\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^4\,b^2\,d^6\,z^6+729\,a^6\,d^6\,z^6-243\,a^4\,d^4\,z^4+27\,a^2\,d^2\,z^2-1,z,k\right )+d\,x}\,20+{\mathrm {root}\left (729\,a^4\,b^2\,d^6\,z^6+729\,a^6\,d^6\,z^6-243\,a^4\,d^4\,z^4+27\,a^2\,d^2\,z^2-1,z,k\right )}^2\,a^2\,d^2\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^4\,b^2\,d^6\,z^6+729\,a^6\,d^6\,z^6-243\,a^4\,d^4\,z^4+27\,a^2\,d^2\,z^2-1,z,k\right )+d\,x}\,24+{\mathrm {root}\left (729\,a^4\,b^2\,d^6\,z^6+729\,a^6\,d^6\,z^6-243\,a^4\,d^4\,z^4+27\,a^2\,d^2\,z^2-1,z,k\right )}^3\,a^3\,d^3\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^4\,b^2\,d^6\,z^6+729\,a^6\,d^6\,z^6-243\,a^4\,d^4\,z^4+27\,a^2\,d^2\,z^2-1,z,k\right )+d\,x}\,216+{\mathrm {root}\left (729\,a^4\,b^2\,d^6\,z^6+729\,a^6\,d^6\,z^6-243\,a^4\,d^4\,z^4+27\,a^2\,d^2\,z^2-1,z,k\right )}^4\,a^4\,d^4\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^4\,b^2\,d^6\,z^6+729\,a^6\,d^6\,z^6-243\,a^4\,d^4\,z^4+27\,a^2\,d^2\,z^2-1,z,k\right )+d\,x}\,108-{\mathrm {root}\left (729\,a^4\,b^2\,d^6\,z^6+729\,a^6\,d^6\,z^6-243\,a^4\,d^4\,z^4+27\,a^2\,d^2\,z^2-1,z,k\right )}^5\,a^5\,d^5\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^4\,b^2\,d^6\,z^6+729\,a^6\,d^6\,z^6-243\,a^4\,d^4\,z^4+27\,a^2\,d^2\,z^2-1,z,k\right )+d\,x}\,324+{\mathrm {root}\left (729\,a^4\,b^2\,d^6\,z^6+729\,a^6\,d^6\,z^6-243\,a^4\,d^4\,z^4+27\,a^2\,d^2\,z^2-1,z,k\right )}^3\,a^2\,b\,d^3\,54+{\mathrm {root}\left (729\,a^4\,b^2\,d^6\,z^6+729\,a^6\,d^6\,z^6-243\,a^4\,d^4\,z^4+27\,a^2\,d^2\,z^2-1,z,k\right )}^4\,a^3\,b\,d^4\,108+{\mathrm {root}\left (729\,a^4\,b^2\,d^6\,z^6+729\,a^6\,d^6\,z^6-243\,a^4\,d^4\,z^4+27\,a^2\,d^2\,z^2-1,z,k\right )}^5\,a^4\,b\,d^5\,81-{\mathrm {root}\left (729\,a^4\,b^2\,d^6\,z^6+729\,a^6\,d^6\,z^6-243\,a^4\,d^4\,z^4+27\,a^2\,d^2\,z^2-1,z,k\right )}^4\,a^2\,b^2\,d^4\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^4\,b^2\,d^6\,z^6+729\,a^6\,d^6\,z^6-243\,a^4\,d^4\,z^4+27\,a^2\,d^2\,z^2-1,z,k\right )+d\,x}\,27-{\mathrm {root}\left (729\,a^4\,b^2\,d^6\,z^6+729\,a^6\,d^6\,z^6-243\,a^4\,d^4\,z^4+27\,a^2\,d^2\,z^2-1,z,k\right )}^5\,a^3\,b^2\,d^5\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^4\,b^2\,d^6\,z^6+729\,a^6\,d^6\,z^6-243\,a^4\,d^4\,z^4+27\,a^2\,d^2\,z^2-1,z,k\right )+d\,x}\,405\right )\,24576}{b^5}\right )\,\mathrm {root}\left (729\,a^4\,b^2\,d^6\,z^6+729\,a^6\,d^6\,z^6-243\,a^4\,d^4\,z^4+27\,a^2\,d^2\,z^2-1,z,k\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*sinh(c + d*x)^3),x)

[Out]

symsum(log((24576*(root(729*a^4*b^2*d^6*z^6 + 729*a^6*d^6*z^6 - 243*a^4*d^4*z^4 + 27*a^2*d^2*z^2 - 1, z, k)*b*
d - 4*exp(root(729*a^4*b^2*d^6*z^6 + 729*a^6*d^6*z^6 - 243*a^4*d^4*z^4 + 27*a^2*d^2*z^2 - 1, z, k) + d*x) + 12
*root(729*a^4*b^2*d^6*z^6 + 729*a^6*d^6*z^6 - 243*a^4*d^4*z^4 + 27*a^2*d^2*z^2 - 1, z, k)^2*a*b*d^2 - 20*root(
729*a^4*b^2*d^6*z^6 + 729*a^6*d^6*z^6 - 243*a^4*d^4*z^4 + 27*a^2*d^2*z^2 - 1, z, k)*a*d*exp(root(729*a^4*b^2*d
^6*z^6 + 729*a^6*d^6*z^6 - 243*a^4*d^4*z^4 + 27*a^2*d^2*z^2 - 1, z, k) + d*x) + 24*root(729*a^4*b^2*d^6*z^6 +
729*a^6*d^6*z^6 - 243*a^4*d^4*z^4 + 27*a^2*d^2*z^2 - 1, z, k)^2*a^2*d^2*exp(root(729*a^4*b^2*d^6*z^6 + 729*a^6
*d^6*z^6 - 243*a^4*d^4*z^4 + 27*a^2*d^2*z^2 - 1, z, k) + d*x) + 216*root(729*a^4*b^2*d^6*z^6 + 729*a^6*d^6*z^6
- 243*a^4*d^4*z^4 + 27*a^2*d^2*z^2 - 1, z, k)^3*a^3*d^3*exp(root(729*a^4*b^2*d^6*z^6 + 729*a^6*d^6*z^6 - 243*
a^4*d^4*z^4 + 27*a^2*d^2*z^2 - 1, z, k) + d*x) + 108*root(729*a^4*b^2*d^6*z^6 + 729*a^6*d^6*z^6 - 243*a^4*d^4*
z^4 + 27*a^2*d^2*z^2 - 1, z, k)^4*a^4*d^4*exp(root(729*a^4*b^2*d^6*z^6 + 729*a^6*d^6*z^6 - 243*a^4*d^4*z^4 + 2
7*a^2*d^2*z^2 - 1, z, k) + d*x) - 324*root(729*a^4*b^2*d^6*z^6 + 729*a^6*d^6*z^6 - 243*a^4*d^4*z^4 + 27*a^2*d^
2*z^2 - 1, z, k)^5*a^5*d^5*exp(root(729*a^4*b^2*d^6*z^6 + 729*a^6*d^6*z^6 - 243*a^4*d^4*z^4 + 27*a^2*d^2*z^2 -
1, z, k) + d*x) + 54*root(729*a^4*b^2*d^6*z^6 + 729*a^6*d^6*z^6 - 243*a^4*d^4*z^4 + 27*a^2*d^2*z^2 - 1, z, k)
^3*a^2*b*d^3 + 108*root(729*a^4*b^2*d^6*z^6 + 729*a^6*d^6*z^6 - 243*a^4*d^4*z^4 + 27*a^2*d^2*z^2 - 1, z, k)^4*
a^3*b*d^4 + 81*root(729*a^4*b^2*d^6*z^6 + 729*a^6*d^6*z^6 - 243*a^4*d^4*z^4 + 27*a^2*d^2*z^2 - 1, z, k)^5*a^4*
b*d^5 - 27*root(729*a^4*b^2*d^6*z^6 + 729*a^6*d^6*z^6 - 243*a^4*d^4*z^4 + 27*a^2*d^2*z^2 - 1, z, k)^4*a^2*b^2*
d^4*exp(root(729*a^4*b^2*d^6*z^6 + 729*a^6*d^6*z^6 - 243*a^4*d^4*z^4 + 27*a^2*d^2*z^2 - 1, z, k) + d*x) - 405*
root(729*a^4*b^2*d^6*z^6 + 729*a^6*d^6*z^6 - 243*a^4*d^4*z^4 + 27*a^2*d^2*z^2 - 1, z, k)^5*a^3*b^2*d^5*exp(roo
t(729*a^4*b^2*d^6*z^6 + 729*a^6*d^6*z^6 - 243*a^4*d^4*z^4 + 27*a^2*d^2*z^2 - 1, z, k) + d*x)))/b^5)*root(729*a
^4*b^2*d^6*z^6 + 729*a^6*d^6*z^6 - 243*a^4*d^4*z^4 + 27*a^2*d^2*z^2 - 1, z, k), k, 1, 6)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{a + b \sinh ^{3}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sinh(d*x+c)**3),x)

[Out]

Integral(1/(a + b*sinh(c + d*x)**3), x)

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