3.172 \(\int \frac {\sinh ^5(c+d x)}{a+b \sinh ^3(c+d x)} \, dx\)

Optimal. Leaf size=295 \[ \frac {2 a \tanh ^{-1}\left (\frac {\sqrt [3]{b}-\sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}+b^{2/3}}}\right )}{3 b^{5/3} d \sqrt {a^{2/3}+b^{2/3}}}+\frac {2 a \tan ^{-1}\left (\frac {(-1)^{5/6} \left (\sqrt [6]{-1} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {-(-1)^{2/3} a^{2/3}-b^{2/3}}}\right )}{3 b^{5/3} d \sqrt {-(-1)^{2/3} a^{2/3}-b^{2/3}}}+\frac {2 a \tan ^{-1}\left (\frac {\sqrt [6]{-1} \left ((-1)^{5/6} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {\sqrt [3]{-1} a^{2/3}-b^{2/3}}}\right )}{3 b^{5/3} d \sqrt {\sqrt [3]{-1} a^{2/3}-b^{2/3}}}+\frac {\sinh (c+d x) \cosh (c+d x)}{2 b d}-\frac {x}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.56, antiderivative size = 295, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3220, 2635, 8, 2660, 618, 206, 204} \[ \frac {2 a \tanh ^{-1}\left (\frac {\sqrt [3]{b}-\sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^{2/3}+b^{2/3}}}\right )}{3 b^{5/3} d \sqrt {a^{2/3}+b^{2/3}}}+\frac {2 a \tan ^{-1}\left (\frac {(-1)^{5/6} \left (\sqrt [6]{-1} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {-(-1)^{2/3} a^{2/3}-b^{2/3}}}\right )}{3 b^{5/3} d \sqrt {-(-1)^{2/3} a^{2/3}-b^{2/3}}}+\frac {2 a \tan ^{-1}\left (\frac {\sqrt [6]{-1} \left ((-1)^{5/6} \sqrt [3]{b}+i \sqrt [3]{a} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {\sqrt [3]{-1} a^{2/3}-b^{2/3}}}\right )}{3 b^{5/3} d \sqrt {\sqrt [3]{-1} a^{2/3}-b^{2/3}}}+\frac {\sinh (c+d x) \cosh (c+d x)}{2 b d}-\frac {x}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[c + d*x]^5/(a + b*Sinh[c + d*x]^3),x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

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

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Mathematica [C]  time = 0.33, size = 299, normalized size = 1.01 \[ \frac {-2 a \text {RootSum}\left [\text {$\#$1}^6 b-3 \text {$\#$1}^4 b+8 \text {$\#$1}^3 a+3 \text {$\#$1}^2 b-b\& ,\frac {2 \text {$\#$1}^4 \log \left (-\text {$\#$1} \sinh \left (\frac {1}{2} (c+d x)\right )+\text {$\#$1} \cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )-\cosh \left (\frac {1}{2} (c+d x)\right )\right )+\text {$\#$1}^4 c+\text {$\#$1}^4 d x-4 \text {$\#$1}^2 \log \left (-\text {$\#$1} \sinh \left (\frac {1}{2} (c+d x)\right )+\text {$\#$1} \cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )-\cosh \left (\frac {1}{2} (c+d x)\right )\right )-2 \text {$\#$1}^2 c-2 \text {$\#$1}^2 d x+2 \log \left (-\text {$\#$1} \sinh \left (\frac {1}{2} (c+d x)\right )+\text {$\#$1} \cosh \left (\frac {1}{2} (c+d x)\right )-\sinh \left (\frac {1}{2} (c+d x)\right )-\cosh \left (\frac {1}{2} (c+d x)\right )\right )+c+d x}{\text {$\#$1}^5 b-2 \text {$\#$1}^3 b+4 \text {$\#$1}^2 a+\text {$\#$1} b}\& \right ]-6 (c+d x)+3 \sinh (2 (c+d x))}{12 b d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[c + d*x]^5/(a + b*Sinh[c + d*x]^3),x]

[Out]

(-6*(c + d*x) - 2*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] - 2*c*#1^2 - 2*d*x*#1^2 - 4*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 + c*#1^4 + d*x*#1^4 + 2*Log
[-Cosh[(c + d*x)/2] - Sinh[(c + d*x)/2] + Cosh[(c + d*x)/2]*#1 - Sinh[(c + d*x)/2]*#1]*#1^4)/(b*#1 + 4*a*#1^2
- 2*b*#1^3 + b*#1^5) & ] + 3*Sinh[2*(c + d*x)])/(12*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)^5/(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 \[ \int \frac {\sinh \left (d x + c\right )^{5}}{b \sinh \left (d x + c\right )^{3} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^5/(a+b*sinh(d*x+c)^3),x, algorithm="giac")

[Out]

integrate(sinh(d*x + c)^5/(b*sinh(d*x + c)^3 + a), x)

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maple [C]  time = 0.11, size = 207, normalized size = 0.70 \[ \frac {1}{2 d b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {1}{2 d b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d b}-\frac {1}{2 d b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {1}{2 d b \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d b}+\frac {4 a \left (\munderset {\textit {\_R} =\RootOf \left (a \,\textit {\_Z}^{6}-3 a \,\textit {\_Z}^{4}-8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}-a \right )}{\sum }\frac {\textit {\_R}^{2} \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a -2 \textit {\_R}^{3} a -4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 d b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(d*x+c)^5/(a+b*sinh(d*x+c)^3),x)

[Out]

1/2/d/b/(tanh(1/2*d*x+1/2*c)-1)^2+1/2/d/b/(tanh(1/2*d*x+1/2*c)-1)+1/2/d/b*ln(tanh(1/2*d*x+1/2*c)-1)-1/2/d/b/(t
anh(1/2*d*x+1/2*c)+1)^2+1/2/d/b/(tanh(1/2*d*x+1/2*c)+1)-1/2/d/b*ln(tanh(1/2*d*x+1/2*c)+1)+4/3/d*a/b*sum(_R^2/(
_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 \[ -\frac {{\left (4 \, d x e^{\left (2 \, d x + 2 \, c\right )} - e^{\left (4 \, d x + 4 \, c\right )} + 1\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, b d} - \frac {1}{32} \, \int \frac {64 \, {\left (a e^{\left (5 \, d x + 5 \, c\right )} - 2 \, a e^{\left (3 \, d x + 3 \, c\right )} + a e^{\left (d x + c\right )}\right )}}{b^{2} e^{\left (6 \, d x + 6 \, c\right )} - 3 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 8 \, a b e^{\left (3 \, d x + 3 \, c\right )} + 3 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} - b^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^5/(a+b*sinh(d*x+c)^3),x, algorithm="maxima")

[Out]

-1/8*(4*d*x*e^(2*d*x + 2*c) - e^(4*d*x + 4*c) + 1)*e^(-2*d*x - 2*c)/(b*d) - 1/32*integrate(64*(a*e^(5*d*x + 5*
c) - 2*a*e^(3*d*x + 3*c) + a*e^(d*x + 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 = 11.48, size = 1114, normalized size = 3.78 \[ \left (\sum _{k=1}^6\ln \left (-\mathrm {root}\left (729\,a^2\,b^{10}\,d^6\,z^6+729\,b^{12}\,d^6\,z^6-243\,a^2\,b^8\,d^4\,z^4+27\,a^4\,b^4\,d^2\,z^2-a^6,z,k\right )\,\left (\mathrm {root}\left (729\,a^2\,b^{10}\,d^6\,z^6+729\,b^{12}\,d^6\,z^6-243\,a^2\,b^8\,d^4\,z^4+27\,a^4\,b^4\,d^2\,z^2-a^6,z,k\right )\,\left (\mathrm {root}\left (729\,a^2\,b^{10}\,d^6\,z^6+729\,b^{12}\,d^6\,z^6-243\,a^2\,b^8\,d^4\,z^4+27\,a^4\,b^4\,d^2\,z^2-a^6,z,k\right )\,\left (\mathrm {root}\left (729\,a^2\,b^{10}\,d^6\,z^6+729\,b^{12}\,d^6\,z^6-243\,a^2\,b^8\,d^4\,z^4+27\,a^4\,b^4\,d^2\,z^2-a^6,z,k\right )\,\left (\frac {\left (8\,a^6\,d^4+4\,a^4\,b^2\,d^4-a^5\,b\,d^4\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^2\,b^{10}\,d^6\,z^6+729\,b^{12}\,d^6\,z^6-243\,a^2\,b^8\,d^4\,z^4+27\,a^4\,b^4\,d^2\,z^2-a^6,z,k\right )}\,5\right )\,663552}{b^7}+\frac {\mathrm {root}\left (729\,a^2\,b^{10}\,d^6\,z^6+729\,b^{12}\,d^6\,z^6-243\,a^2\,b^8\,d^4\,z^4+27\,a^4\,b^4\,d^2\,z^2-a^6,z,k\right )\,\left (-a^4\,b\,d^5+a^5\,d^5\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^2\,b^{10}\,d^6\,z^6+729\,b^{12}\,d^6\,z^6-243\,a^2\,b^8\,d^4\,z^4+27\,a^4\,b^4\,d^2\,z^2-a^6,z,k\right )}\,4+a^3\,b^2\,d^5\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^2\,b^{10}\,d^6\,z^6+729\,b^{12}\,d^6\,z^6-243\,a^2\,b^8\,d^4\,z^4+27\,a^4\,b^4\,d^2\,z^2-a^6,z,k\right )}\,5\right )\,1990656}{b^5}\right )+\frac {\left (4\,a^6\,b\,d^3+a^7\,d^3\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^2\,b^{10}\,d^6\,z^6+729\,b^{12}\,d^6\,z^6-243\,a^2\,b^8\,d^4\,z^4+27\,a^4\,b^4\,d^2\,z^2-a^6,z,k\right )}\,8-a^5\,b^2\,d^3\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^2\,b^{10}\,d^6\,z^6+729\,b^{12}\,d^6\,z^6-243\,a^2\,b^8\,d^4\,z^4+27\,a^4\,b^4\,d^2\,z^2-a^6,z,k\right )}\,5\right )\,442368}{b^9}\right )-\frac {a^6\,d^2\,\left (2\,b-a\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^2\,b^{10}\,d^6\,z^6+729\,b^{12}\,d^6\,z^6-243\,a^2\,b^8\,d^4\,z^4+27\,a^4\,b^4\,d^2\,z^2-a^6,z,k\right )}\,5\right )\,294912}{b^{10}}\right )-\frac {a^7\,d\,\left (8\,a-b\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^2\,b^{10}\,d^6\,z^6+729\,b^{12}\,d^6\,z^6-243\,a^2\,b^8\,d^4\,z^4+27\,a^4\,b^4\,d^2\,z^2-a^6,z,k\right )}\,5\right )\,24576}{b^{12}}\right )-\frac {a^8\,\left (b-a\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^{\mathrm {root}\left (729\,a^2\,b^{10}\,d^6\,z^6+729\,b^{12}\,d^6\,z^6-243\,a^2\,b^8\,d^4\,z^4+27\,a^4\,b^4\,d^2\,z^2-a^6,z,k\right )}\,4\right )\,32768}{b^{14}}\right )\,\mathrm {root}\left (729\,a^2\,b^{10}\,d^6\,z^6+729\,b^{12}\,d^6\,z^6-243\,a^2\,b^8\,d^4\,z^4+27\,a^4\,b^4\,d^2\,z^2-a^6,z,k\right )\right )-\frac {x}{2\,b}-\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,b\,d}+\frac {{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,b\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(c + d*x)^5/(a + b*sinh(c + d*x)^3),x)

[Out]

symsum(log(- root(729*a^2*b^10*d^6*z^6 + 729*b^12*d^6*z^6 - 243*a^2*b^8*d^4*z^4 + 27*a^4*b^4*d^2*z^2 - a^6, z,
k)*(root(729*a^2*b^10*d^6*z^6 + 729*b^12*d^6*z^6 - 243*a^2*b^8*d^4*z^4 + 27*a^4*b^4*d^2*z^2 - a^6, z, k)*(roo
t(729*a^2*b^10*d^6*z^6 + 729*b^12*d^6*z^6 - 243*a^2*b^8*d^4*z^4 + 27*a^4*b^4*d^2*z^2 - a^6, z, k)*(root(729*a^
2*b^10*d^6*z^6 + 729*b^12*d^6*z^6 - 243*a^2*b^8*d^4*z^4 + 27*a^4*b^4*d^2*z^2 - a^6, z, k)*((663552*(8*a^6*d^4
+ 4*a^4*b^2*d^4 - 5*a^5*b*d^4*exp(d*x)*exp(root(729*a^2*b^10*d^6*z^6 + 729*b^12*d^6*z^6 - 243*a^2*b^8*d^4*z^4
+ 27*a^4*b^4*d^2*z^2 - a^6, z, k))))/b^7 + (1990656*root(729*a^2*b^10*d^6*z^6 + 729*b^12*d^6*z^6 - 243*a^2*b^8
*d^4*z^4 + 27*a^4*b^4*d^2*z^2 - a^6, z, k)*(4*a^5*d^5*exp(d*x)*exp(root(729*a^2*b^10*d^6*z^6 + 729*b^12*d^6*z^
6 - 243*a^2*b^8*d^4*z^4 + 27*a^4*b^4*d^2*z^2 - a^6, z, k)) - a^4*b*d^5 + 5*a^3*b^2*d^5*exp(d*x)*exp(root(729*a
^2*b^10*d^6*z^6 + 729*b^12*d^6*z^6 - 243*a^2*b^8*d^4*z^4 + 27*a^4*b^4*d^2*z^2 - a^6, z, k))))/b^5) + (442368*(
4*a^6*b*d^3 + 8*a^7*d^3*exp(d*x)*exp(root(729*a^2*b^10*d^6*z^6 + 729*b^12*d^6*z^6 - 243*a^2*b^8*d^4*z^4 + 27*a
^4*b^4*d^2*z^2 - a^6, z, k)) - 5*a^5*b^2*d^3*exp(d*x)*exp(root(729*a^2*b^10*d^6*z^6 + 729*b^12*d^6*z^6 - 243*a
^2*b^8*d^4*z^4 + 27*a^4*b^4*d^2*z^2 - a^6, z, k))))/b^9) - (294912*a^6*d^2*(2*b - 5*a*exp(d*x)*exp(root(729*a^
2*b^10*d^6*z^6 + 729*b^12*d^6*z^6 - 243*a^2*b^8*d^4*z^4 + 27*a^4*b^4*d^2*z^2 - a^6, z, k))))/b^10) - (24576*a^
7*d*(8*a - 5*b*exp(d*x)*exp(root(729*a^2*b^10*d^6*z^6 + 729*b^12*d^6*z^6 - 243*a^2*b^8*d^4*z^4 + 27*a^4*b^4*d^
2*z^2 - a^6, z, k))))/b^12) - (32768*a^8*(b - 4*a*exp(d*x)*exp(root(729*a^2*b^10*d^6*z^6 + 729*b^12*d^6*z^6 -
243*a^2*b^8*d^4*z^4 + 27*a^4*b^4*d^2*z^2 - a^6, z, k))))/b^14)*root(729*a^2*b^10*d^6*z^6 + 729*b^12*d^6*z^6 -
243*a^2*b^8*d^4*z^4 + 27*a^4*b^4*d^2*z^2 - a^6, z, k), k, 1, 6) - x/(2*b) - exp(- 2*c - 2*d*x)/(8*b*d) + exp(2
*c + 2*d*x)/(8*b*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh ^{5}{\left (c + d x \right )}}{a + b \sinh ^{3}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)**5/(a+b*sinh(d*x+c)**3),x)

[Out]

Integral(sinh(c + d*x)**5/(a + b*sinh(c + d*x)**3), x)

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