3.173 \(\int \frac {\sinh ^4(c+d x)}{a+b \sinh ^3(c+d x)} \, dx\)

Optimal. Leaf size=303 \[ -\frac {2 a^{2/3} \tanh ^{-1}\left (\frac {\sqrt [3]{b}-\sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}+b^{2/3}}}\right )}{3 b^{4/3} d \sqrt {a^{2/3}+b^{2/3}}}-\frac {2 a^{2/3} \tan ^{-1}\left (\frac {\sqrt [6]{-1} \left (\sqrt [6]{-1} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {\sqrt [3]{-1} a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 b^{4/3} d \sqrt {\sqrt [3]{-1} a^{2/3}-(-1)^{2/3} b^{2/3}}}+\frac {2 \sqrt [3]{-1} a^{2/3} \tan ^{-1}\left (\frac {\sqrt [6]{-1} \left ((-1)^{5/6} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {\sqrt [3]{-1} a^{2/3}-b^{2/3}}}\right )}{3 b^{4/3} d \sqrt {\sqrt [3]{-1} a^{2/3}-b^{2/3}}}+\frac {\cosh (c+d x)}{b d} \]

[Out]

cosh(d*x+c)/b/d+2/3*(-1)^(1/3)*a^(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))/b^(4/3)/d/((-1)^(1/3)*a^(2/3)-b^(2/3))^(1/2)-2/3*a^(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))/b^(4/3)/d/(a^(2/3)+b^(2/3))^(1/2)-2/3*a^(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))/b^(4/3
)/d/((-1)^(1/3)*a^(2/3)-(-1)^(2/3)*b^(2/3))^(1/2)

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Rubi [A]  time = 0.54, antiderivative size = 303, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3220, 2638, 2660, 618, 204} \[ -\frac {2 a^{2/3} \tanh ^{-1}\left (\frac {\sqrt [3]{b}-\sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}+b^{2/3}}}\right )}{3 b^{4/3} d \sqrt {a^{2/3}+b^{2/3}}}-\frac {2 a^{2/3} \tan ^{-1}\left (\frac {\sqrt [6]{-1} \left (\sqrt [6]{-1} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {\sqrt [3]{-1} a^{2/3}-(-1)^{2/3} b^{2/3}}}\right )}{3 b^{4/3} d \sqrt {\sqrt [3]{-1} a^{2/3}-(-1)^{2/3} b^{2/3}}}+\frac {2 \sqrt [3]{-1} a^{2/3} \tan ^{-1}\left (\frac {\sqrt [6]{-1} \left ((-1)^{5/6} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {\sqrt [3]{-1} a^{2/3}-b^{2/3}}}\right )}{3 b^{4/3} d \sqrt {\sqrt [3]{-1} a^{2/3}-b^{2/3}}}+\frac {\cosh (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[c + d*x]^4/(a + b*Sinh[c + d*x]^3),x]

[Out]

(-2*a^(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*Sqrt[(-1)^(1/3)*a^(2/3) - (-1)^(2/3)*b^(2/3)]*b^(4/3)*d) + (2*(-1)^(1/3)*a^(2/3)*ArcTa
n[((-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*Sqrt
[(-1)^(1/3)*a^(2/3) - b^(2/3)]*b^(4/3)*d) - (2*a^(2/3)*ArcTanh[(b^(1/3) - a^(1/3)*Tanh[(c + d*x)/2])/Sqrt[a^(2
/3) + b^(2/3)]])/(3*Sqrt[a^(2/3) + b^(2/3)]*b^(4/3)*d) + Cosh[c + d*x]/(b*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 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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 3220

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> Int[ExpandTr
ig[sin[e + f*x]^m*(a + b*sin[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, e, f}, x] && IntegersQ[m, p] && (EqQ[n, 4]
|| GtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))

Rubi steps

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

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Mathematica [C]  time = 0.36, size = 214, normalized size = 0.71 \[ \frac {3 \cosh (c+d x)-a \text {RootSum}\left [\text {$\#$1}^6 b-3 \text {$\#$1}^4 b+8 \text {$\#$1}^3 a+3 \text {$\#$1}^2 b-b\& ,\frac {2 \text {$\#$1}^2 \log \left (-\text {$\#$1} \sinh \left (\frac {1}{2} (c+d x)\right )+\text {$\#$1} \cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )-\cosh \left (\frac {1}{2} (c+d x)\right )\right )+\text {$\#$1}^2 c+\text {$\#$1}^2 d x-2 \log \left (-\text {$\#$1} \sinh \left (\frac {1}{2} (c+d x)\right )+\text {$\#$1} \cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )-\cosh \left (\frac {1}{2} (c+d x)\right )\right )-c-d x}{\text {$\#$1}^4 b-2 \text {$\#$1}^2 b+4 \text {$\#$1} a+b}\& \right ]}{3 b d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[c + d*x]^4/(a + b*Sinh[c + d*x]^3),x]

[Out]

(3*Cosh[c + d*x] - a*RootSum[-b + 3*b*#1^2 + 8*a*#1^3 - 3*b*#1^4 + b*#1^6 & , (-c - d*x - 2*Log[-Cosh[(c + d*x
)/2] - Sinh[(c + d*x)/2] + Cosh[(c + d*x)/2]*#1 - Sinh[(c + d*x)/2]*#1] + c*#1^2 + d*x*#1^2 + 2*Log[-Cosh[(c +
d*x)/2] - Sinh[(c + d*x)/2] + Cosh[(c + d*x)/2]*#1 - Sinh[(c + d*x)/2]*#1]*#1^2)/(b + 4*a*#1 - 2*b*#1^2 + b*#
1^4) & ])/(3*b*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(sinh(d*x+c)^4/(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 \[ \int \frac {\sinh \left (d x + c\right )^{4}}{b \sinh \left (d x + c\right )^{3} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^4/(a+b*sinh(d*x+c)^3),x, algorithm="giac")

[Out]

integrate(sinh(d*x + c)^4/(b*sinh(d*x + c)^3 + a), x)

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maple [C]  time = 0.10, size = 128, normalized size = 0.42 \[ \frac {1}{d b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 a \left (\munderset {\textit {\_R} =\RootOf \left (a \,\textit {\_Z}^{6}-3 a \,\textit {\_Z}^{4}-8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}-a \right )}{\sum }\frac {\left (\textit {\_R}^{3}-\textit {\_R} \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a -2 \textit {\_R}^{3} a -4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 d b}-\frac {1}{d b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(d*x+c)^4/(a+b*sinh(d*x+c)^3),x)

[Out]

1/d/b/(tanh(1/2*d*x+1/2*c)+1)-2/3/d*a/b*sum((_R^3-_R)/(_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))-1/d/b/(tanh(1/2*d*x+1/2*c)-1)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )} e^{\left (-d x - c\right )}}{2 \, b d} - \frac {1}{16} \, \int \frac {64 \, {\left (a e^{\left (4 \, d x + 4 \, c\right )} - a e^{\left (2 \, d x + 2 \, c\right )}\right )}}{b^{2} e^{\left (6 \, d x + 6 \, c\right )} - 3 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 8 \, a b e^{\left (3 \, d x + 3 \, c\right )} + 3 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} - b^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^4/(a+b*sinh(d*x+c)^3),x, algorithm="maxima")

[Out]

1/2*(e^(2*d*x + 2*c) + 1)*e^(-d*x - c)/(b*d) - 1/16*integrate(64*(a*e^(4*d*x + 4*c) - a*e^(2*d*x + 2*c))/(b^2*
e^(6*d*x + 6*c) - 3*b^2*e^(4*d*x + 4*c) + 8*a*b*e^(3*d*x + 3*c) + 3*b^2*e^(2*d*x + 2*c) - b^2), x)

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mupad [B]  time = 23.50, size = 906, normalized size = 2.99 \[ \left (\sum _{k=1}^6\ln \left (-\mathrm {root}\left (729\,a^2\,b^8\,d^6\,z^6+729\,b^{10}\,d^6\,z^6+243\,a^2\,b^6\,d^4\,z^4-a^4,z,k\right )\,\left (\mathrm {root}\left (729\,a^2\,b^8\,d^6\,z^6+729\,b^{10}\,d^6\,z^6+243\,a^2\,b^6\,d^4\,z^4-a^4,z,k\right )\,\left (\mathrm {root}\left (729\,a^2\,b^8\,d^6\,z^6+729\,b^{10}\,d^6\,z^6+243\,a^2\,b^6\,d^4\,z^4-a^4,z,k\right )\,\left (\mathrm {root}\left (729\,a^2\,b^8\,d^6\,z^6+729\,b^{10}\,d^6\,z^6+243\,a^2\,b^6\,d^4\,z^4-a^4,z,k\right )\,\left (\frac {663552\,\left (4\,a^5\,b\,d^4+16\,a^6\,d^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^2\,b^8\,d^6\,z^6+729\,b^{10}\,d^6\,z^6+243\,a^2\,b^6\,d^4\,z^4-a^4,z,k\right )}+11\,a^4\,b^2\,d^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^2\,b^8\,d^6\,z^6+729\,b^{10}\,d^6\,z^6+243\,a^2\,b^6\,d^4\,z^4-a^4,z,k\right )}\right )}{b^7}+\frac {\mathrm {root}\left (729\,a^2\,b^8\,d^6\,z^6+729\,b^{10}\,d^6\,z^6+243\,a^2\,b^6\,d^4\,z^4-a^4,z,k\right )\,\left (4\,a^5\,d^5\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^2\,b^8\,d^6\,z^6+729\,b^{10}\,d^6\,z^6+243\,a^2\,b^6\,d^4\,z^4-a^4,z,k\right )}-a^4\,b\,d^5+5\,a^3\,b^2\,d^5\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^2\,b^8\,d^6\,z^6+729\,b^{10}\,d^6\,z^6+243\,a^2\,b^6\,d^4\,z^4-a^4,z,k\right )}\right )\,1990656}{b^5}\right )-\frac {221184\,\left (8\,a^6\,d^3+a^4\,b^2\,d^3-25\,a^5\,b\,d^3\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^2\,b^8\,d^6\,z^6+729\,b^{10}\,d^6\,z^6+243\,a^2\,b^6\,d^4\,z^4-a^4,z,k\right )}\right )}{b^8}\right )-\frac {294912\,a^5\,d^2\,\left (b-7\,a\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^2\,b^8\,d^6\,z^6+729\,b^{10}\,d^6\,z^6+243\,a^2\,b^6\,d^4\,z^4-a^4,z,k\right )}\right )}{b^9}\right )-\frac {196608\,a^6\,d\,\left (b-2\,a\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^2\,b^8\,d^6\,z^6+729\,b^{10}\,d^6\,z^6+243\,a^2\,b^6\,d^4\,z^4-a^4,z,k\right )}\right )}{b^{11}}\right )+\frac {8192\,a^6\,\left (8\,a-b\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^2\,b^8\,d^6\,z^6+729\,b^{10}\,d^6\,z^6+243\,a^2\,b^6\,d^4\,z^4-a^4,z,k\right )}\right )}{b^{12}}\right )\,\mathrm {root}\left (729\,a^2\,b^8\,d^6\,z^6+729\,b^{10}\,d^6\,z^6+243\,a^2\,b^6\,d^4\,z^4-a^4,z,k\right )\right )+\frac {{\mathrm {e}}^{c+d\,x}}{2\,b\,d}+\frac {{\mathrm {e}}^{-c-d\,x}}{2\,b\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(c + d*x)^4/(a + b*sinh(c + d*x)^3),x)

[Out]

symsum(log((8192*a^6*(8*a - b*exp(d*x)*exp(root(729*a^2*b^8*d^6*z^6 + 729*b^10*d^6*z^6 + 243*a^2*b^6*d^4*z^4 -
a^4, z, k))))/b^12 - root(729*a^2*b^8*d^6*z^6 + 729*b^10*d^6*z^6 + 243*a^2*b^6*d^4*z^4 - a^4, z, k)*(root(729
*a^2*b^8*d^6*z^6 + 729*b^10*d^6*z^6 + 243*a^2*b^6*d^4*z^4 - a^4, z, k)*(root(729*a^2*b^8*d^6*z^6 + 729*b^10*d^
6*z^6 + 243*a^2*b^6*d^4*z^4 - a^4, z, k)*(root(729*a^2*b^8*d^6*z^6 + 729*b^10*d^6*z^6 + 243*a^2*b^6*d^4*z^4 -
a^4, z, k)*((663552*(4*a^5*b*d^4 + 16*a^6*d^4*exp(d*x)*exp(root(729*a^2*b^8*d^6*z^6 + 729*b^10*d^6*z^6 + 243*a
^2*b^6*d^4*z^4 - a^4, z, k)) + 11*a^4*b^2*d^4*exp(d*x)*exp(root(729*a^2*b^8*d^6*z^6 + 729*b^10*d^6*z^6 + 243*a
^2*b^6*d^4*z^4 - a^4, z, k))))/b^7 + (1990656*root(729*a^2*b^8*d^6*z^6 + 729*b^10*d^6*z^6 + 243*a^2*b^6*d^4*z^
4 - a^4, z, k)*(4*a^5*d^5*exp(d*x)*exp(root(729*a^2*b^8*d^6*z^6 + 729*b^10*d^6*z^6 + 243*a^2*b^6*d^4*z^4 - a^4
, z, k)) - a^4*b*d^5 + 5*a^3*b^2*d^5*exp(d*x)*exp(root(729*a^2*b^8*d^6*z^6 + 729*b^10*d^6*z^6 + 243*a^2*b^6*d^
4*z^4 - a^4, z, k))))/b^5) - (221184*(8*a^6*d^3 + a^4*b^2*d^3 - 25*a^5*b*d^3*exp(d*x)*exp(root(729*a^2*b^8*d^6
*z^6 + 729*b^10*d^6*z^6 + 243*a^2*b^6*d^4*z^4 - a^4, z, k))))/b^8) - (294912*a^5*d^2*(b - 7*a*exp(d*x)*exp(roo
t(729*a^2*b^8*d^6*z^6 + 729*b^10*d^6*z^6 + 243*a^2*b^6*d^4*z^4 - a^4, z, k))))/b^9) - (196608*a^6*d*(b - 2*a*e
xp(d*x)*exp(root(729*a^2*b^8*d^6*z^6 + 729*b^10*d^6*z^6 + 243*a^2*b^6*d^4*z^4 - a^4, z, k))))/b^11))*root(729*
a^2*b^8*d^6*z^6 + 729*b^10*d^6*z^6 + 243*a^2*b^6*d^4*z^4 - a^4, z, k), k, 1, 6) + exp(c + d*x)/(2*b*d) + exp(-
c - d*x)/(2*b*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh ^{4}{\left (c + d x \right )}}{a + b \sinh ^{3}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)**4/(a+b*sinh(d*x+c)**3),x)

[Out]

Integral(sinh(c + d*x)**4/(a + b*sinh(c + d*x)**3), x)

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