3.1.80 \(\int \frac {\sinh ^4(x)}{(a+b \sinh (x))^2} \, dx\) [80]

Optimal. Leaf size=162 \[ \frac {\left (6 a^2-b^2\right ) x}{2 b^4}+\frac {2 a^3 \left (3 a^2+4 b^2\right ) \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{b^4 \left (a^2+b^2\right )^{3/2}}-\frac {a \left (3 a^2+2 b^2\right ) \cosh (x)}{b^3 \left (a^2+b^2\right )}+\frac {\left (3 a^2+b^2\right ) \cosh (x) \sinh (x)}{2 b^2 \left (a^2+b^2\right )}-\frac {a^2 \cosh (x) \sinh ^2(x)}{b \left (a^2+b^2\right ) (a+b \sinh (x))} \]

[Out]

1/2*(6*a^2-b^2)*x/b^4+2*a^3*(3*a^2+4*b^2)*arctanh((b-a*tanh(1/2*x))/(a^2+b^2)^(1/2))/b^4/(a^2+b^2)^(3/2)-a*(3*
a^2+2*b^2)*cosh(x)/b^3/(a^2+b^2)+1/2*(3*a^2+b^2)*cosh(x)*sinh(x)/b^2/(a^2+b^2)-a^2*cosh(x)*sinh(x)^2/b/(a^2+b^
2)/(a+b*sinh(x))

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Rubi [A]
time = 0.30, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {2871, 3128, 3102, 2814, 2739, 632, 212} \begin {gather*} -\frac {a^2 \sinh ^2(x) \cosh (x)}{b \left (a^2+b^2\right ) (a+b \sinh (x))}+\frac {\left (3 a^2+b^2\right ) \sinh (x) \cosh (x)}{2 b^2 \left (a^2+b^2\right )}+\frac {x \left (6 a^2-b^2\right )}{2 b^4}-\frac {a \left (3 a^2+2 b^2\right ) \cosh (x)}{b^3 \left (a^2+b^2\right )}+\frac {2 a^3 \left (3 a^2+4 b^2\right ) \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{b^4 \left (a^2+b^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^4/(a + b*Sinh[x])^2,x]

[Out]

((6*a^2 - b^2)*x)/(2*b^4) + (2*a^3*(3*a^2 + 4*b^2)*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]])/(b^4*(a^2 + b^2
)^(3/2)) - (a*(3*a^2 + 2*b^2)*Cosh[x])/(b^3*(a^2 + b^2)) + ((3*a^2 + b^2)*Cosh[x]*Sinh[x])/(2*b^2*(a^2 + b^2))
- (a^2*Cosh[x]*Sinh[x]^2)/(b*(a^2 + b^2)*(a + b*Sinh[x]))

Rule 212

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

Rule 632

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 2739

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 2814

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*(x/d)
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2871

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/
(d*f*(n + 1)*(c^2 - d^2))), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e
+ f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 +
c*d*(a^2 + b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - d^2) - m*(b*c - a*d)^2 +
d*n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(
b*(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 3128

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e
+ f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*
x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n +
2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d
^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rubi steps

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

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Mathematica [A]
time = 0.32, size = 118, normalized size = 0.73 \begin {gather*} \frac {-2 \left (-6 a^2+b^2\right ) x+\frac {8 a^3 \left (3 a^2+4 b^2\right ) \text {ArcTan}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\left (-a^2-b^2\right )^{3/2}}-8 a b \cosh (x)-\frac {4 a^4 b \cosh (x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}+b^2 \sinh (2 x)}{4 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]^4/(a + b*Sinh[x])^2,x]

[Out]

(-2*(-6*a^2 + b^2)*x + (8*a^3*(3*a^2 + 4*b^2)*ArcTan[(b - a*Tanh[x/2])/Sqrt[-a^2 - b^2]])/(-a^2 - b^2)^(3/2) -
8*a*b*Cosh[x] - (4*a^4*b*Cosh[x])/((a^2 + b^2)*(a + b*Sinh[x])) + b^2*Sinh[2*x])/(4*b^4)

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Maple [A]
time = 0.66, size = 218, normalized size = 1.35

method	result	size
default	\(-\frac {1}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {-b +4 a}{2 b^{3} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {\left (6 a^{2}-b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 b^{4}}+\frac {1}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {-b -4 a}{2 b^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {\left (-6 a^{2}+b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 b^{4}}+\frac {2 a^{3} \left (\frac {\frac {b^{2} \tanh \left (\frac {x}{2}\right )}{a^{2}+b^{2}}+\frac {a b}{a^{2}+b^{2}}}{a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-2 b \tanh \left (\frac {x}{2}\right )-a}-\frac {\left (3 a^{2}+4 b^{2}\right ) \arctanh \left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{b^{4}}\)	\(218\)
risch	\(\frac {3 x \,a^{2}}{b^{4}}-\frac {x}{2 b^{2}}+\frac {{\mathrm e}^{2 x}}{8 b^{2}}-\frac {a \,{\mathrm e}^{x}}{b^{3}}-\frac {a \,{\mathrm e}^{-x}}{b^{3}}-\frac {{\mathrm e}^{-2 x}}{8 b^{2}}+\frac {2 a^{4} \left (a \,{\mathrm e}^{x}-b \right )}{b^{4} \left (a^{2}+b^{2}\right ) \left (b \,{\mathrm e}^{2 x}+2 a \,{\mathrm e}^{x}-b \right )}+\frac {3 a^{5} \ln \left ({\mathrm e}^{x}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}+a^{4}+2 a^{2} b^{2}+b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}} b^{4}}+\frac {4 a^{3} \ln \left ({\mathrm e}^{x}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}+a^{4}+2 a^{2} b^{2}+b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}} b^{2}}-\frac {3 a^{5} \ln \left ({\mathrm e}^{x}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}-a^{4}-2 a^{2} b^{2}-b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}} b^{4}}-\frac {4 a^{3} \ln \left ({\mathrm e}^{x}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}-a^{4}-2 a^{2} b^{2}-b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}} b^{2}}\)	\(343\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^4/(a+b*sinh(x))^2,x,method=_RETURNVERBOSE)

[Out]

-1/2/b^2/(tanh(1/2*x)+1)^2-1/2*(-b+4*a)/b^3/(tanh(1/2*x)+1)+1/2*(6*a^2-b^2)/b^4*ln(tanh(1/2*x)+1)+1/2/b^2/(tan
h(1/2*x)-1)^2-1/2*(-b-4*a)/b^3/(tanh(1/2*x)-1)+1/2/b^4*(-6*a^2+b^2)*ln(tanh(1/2*x)-1)+2/b^4*a^3*((b^2/(a^2+b^2
)*tanh(1/2*x)+a*b/(a^2+b^2))/(a*tanh(1/2*x)^2-2*b*tanh(1/2*x)-a)-(3*a^2+4*b^2)/(a^2+b^2)^(3/2)*arctanh(1/2*(2*
a*tanh(1/2*x)-2*b)/(a^2+b^2)^(1/2)))

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Maxima [A]
time = 0.51, size = 256, normalized size = 1.58 \begin {gather*} -\frac {{\left (3 \, a^{2} + 4 \, b^{2}\right )} a^{3} \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{2} b^{4} + b^{6}\right )} \sqrt {a^{2} + b^{2}}} + \frac {a^{2} b^{3} + b^{5} - 6 \, {\left (a^{3} b^{2} + a b^{4}\right )} e^{\left (-x\right )} - {\left (32 \, a^{4} b + 17 \, a^{2} b^{3} + b^{5}\right )} e^{\left (-2 \, x\right )} - 8 \, {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} e^{\left (-3 \, x\right )}}{8 \, {\left ({\left (a^{2} b^{5} + b^{7}\right )} e^{\left (-2 \, x\right )} + 2 \, {\left (a^{3} b^{4} + a b^{6}\right )} e^{\left (-3 \, x\right )} - {\left (a^{2} b^{5} + b^{7}\right )} e^{\left (-4 \, x\right )}\right )}} - \frac {8 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )}}{8 \, b^{3}} + \frac {{\left (6 \, a^{2} - b^{2}\right )} x}{2 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^4/(a+b*sinh(x))^2,x, algorithm="maxima")

[Out]

-(3*a^2 + 4*b^2)*a^3*log((b*e^(-x) - a - sqrt(a^2 + b^2))/(b*e^(-x) - a + sqrt(a^2 + b^2)))/((a^2*b^4 + b^6)*s
qrt(a^2 + b^2)) + 1/8*(a^2*b^3 + b^5 - 6*(a^3*b^2 + a*b^4)*e^(-x) - (32*a^4*b + 17*a^2*b^3 + b^5)*e^(-2*x) - 8
*(2*a^5 - a^3*b^2 - a*b^4)*e^(-3*x))/((a^2*b^5 + b^7)*e^(-2*x) + 2*(a^3*b^4 + a*b^6)*e^(-3*x) - (a^2*b^5 + b^7
)*e^(-4*x)) - 1/8*(8*a*e^(-x) + b*e^(-2*x))/b^3 + 1/2*(6*a^2 - b^2)*x/b^4

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 1769 vs. \(2 (154) = 308\).
time = 0.56, size = 1769, normalized size = 10.92 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^4/(a+b*sinh(x))^2,x, algorithm="fricas")

[Out]

1/8*(a^4*b^3 + 2*a^2*b^5 + b^7 + (a^4*b^3 + 2*a^2*b^5 + b^7)*cosh(x)^6 + (a^4*b^3 + 2*a^2*b^5 + b^7)*sinh(x)^6
- 6*(a^5*b^2 + 2*a^3*b^4 + a*b^6)*cosh(x)^5 - 6*(a^5*b^2 + 2*a^3*b^4 + a*b^6 - (a^4*b^3 + 2*a^2*b^5 + b^7)*co
sh(x))*sinh(x)^5 - (16*a^6*b + 33*a^4*b^3 + 18*a^2*b^5 + b^7 - 4*(6*a^6*b + 11*a^4*b^3 + 4*a^2*b^5 - b^7)*x)*c
osh(x)^4 - (16*a^6*b + 33*a^4*b^3 + 18*a^2*b^5 + b^7 - 15*(a^4*b^3 + 2*a^2*b^5 + b^7)*cosh(x)^2 - 4*(6*a^6*b +
11*a^4*b^3 + 4*a^2*b^5 - b^7)*x + 30*(a^5*b^2 + 2*a^3*b^4 + a*b^6)*cosh(x))*sinh(x)^4 + 8*(2*a^7 + 2*a^5*b^2
+ (6*a^7 + 11*a^5*b^2 + 4*a^3*b^4 - a*b^6)*x)*cosh(x)^3 + 4*(4*a^7 + 4*a^5*b^2 + 5*(a^4*b^3 + 2*a^2*b^5 + b^7)
*cosh(x)^3 - 15*(a^5*b^2 + 2*a^3*b^4 + a*b^6)*cosh(x)^2 + 2*(6*a^7 + 11*a^5*b^2 + 4*a^3*b^4 - a*b^6)*x - (16*a
^6*b + 33*a^4*b^3 + 18*a^2*b^5 + b^7 - 4*(6*a^6*b + 11*a^4*b^3 + 4*a^2*b^5 - b^7)*x)*cosh(x))*sinh(x)^3 - (32*
a^6*b + 49*a^4*b^3 + 18*a^2*b^5 + b^7 + 4*(6*a^6*b + 11*a^4*b^3 + 4*a^2*b^5 - b^7)*x)*cosh(x)^2 - (32*a^6*b +
49*a^4*b^3 + 18*a^2*b^5 + b^7 - 15*(a^4*b^3 + 2*a^2*b^5 + b^7)*cosh(x)^4 + 60*(a^5*b^2 + 2*a^3*b^4 + a*b^6)*co
sh(x)^3 + 6*(16*a^6*b + 33*a^4*b^3 + 18*a^2*b^5 + b^7 - 4*(6*a^6*b + 11*a^4*b^3 + 4*a^2*b^5 - b^7)*x)*cosh(x)^
2 + 4*(6*a^6*b + 11*a^4*b^3 + 4*a^2*b^5 - b^7)*x - 24*(2*a^7 + 2*a^5*b^2 + (6*a^7 + 11*a^5*b^2 + 4*a^3*b^4 - a
*b^6)*x)*cosh(x))*sinh(x)^2 + 8*((3*a^5*b + 4*a^3*b^3)*cosh(x)^4 + (3*a^5*b + 4*a^3*b^3)*sinh(x)^4 + 2*(3*a^6
+ 4*a^4*b^2)*cosh(x)^3 + 2*(3*a^6 + 4*a^4*b^2 + 2*(3*a^5*b + 4*a^3*b^3)*cosh(x))*sinh(x)^3 - (3*a^5*b + 4*a^3*
b^3)*cosh(x)^2 - (3*a^5*b + 4*a^3*b^3 - 6*(3*a^5*b + 4*a^3*b^3)*cosh(x)^2 - 6*(3*a^6 + 4*a^4*b^2)*cosh(x))*sin
h(x)^2 + 2*(2*(3*a^5*b + 4*a^3*b^3)*cosh(x)^3 + 3*(3*a^6 + 4*a^4*b^2)*cosh(x)^2 - (3*a^5*b + 4*a^3*b^3)*cosh(x
))*sinh(x))*sqrt(a^2 + b^2)*log((b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a*b*cosh(x) + 2*a^2 + b^2 + 2*(b^2*cosh(x)
+ a*b)*sinh(x) + 2*sqrt(a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a))/(b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(
b*cosh(x) + a)*sinh(x) - b)) + 6*(a^5*b^2 + 2*a^3*b^4 + a*b^6)*cosh(x) + 2*(3*a^5*b^2 + 6*a^3*b^4 + 3*a*b^6 +
3*(a^4*b^3 + 2*a^2*b^5 + b^7)*cosh(x)^5 - 15*(a^5*b^2 + 2*a^3*b^4 + a*b^6)*cosh(x)^4 - 2*(16*a^6*b + 33*a^4*b^
3 + 18*a^2*b^5 + b^7 - 4*(6*a^6*b + 11*a^4*b^3 + 4*a^2*b^5 - b^7)*x)*cosh(x)^3 + 12*(2*a^7 + 2*a^5*b^2 + (6*a^
7 + 11*a^5*b^2 + 4*a^3*b^4 - a*b^6)*x)*cosh(x)^2 - (32*a^6*b + 49*a^4*b^3 + 18*a^2*b^5 + b^7 + 4*(6*a^6*b + 11
*a^4*b^3 + 4*a^2*b^5 - b^7)*x)*cosh(x))*sinh(x))/((a^4*b^5 + 2*a^2*b^7 + b^9)*cosh(x)^4 + (a^4*b^5 + 2*a^2*b^7
+ b^9)*sinh(x)^4 + 2*(a^5*b^4 + 2*a^3*b^6 + a*b^8)*cosh(x)^3 + 2*(a^5*b^4 + 2*a^3*b^6 + a*b^8 + 2*(a^4*b^5 +
2*a^2*b^7 + b^9)*cosh(x))*sinh(x)^3 - (a^4*b^5 + 2*a^2*b^7 + b^9)*cosh(x)^2 - (a^4*b^5 + 2*a^2*b^7 + b^9 - 6*(
a^4*b^5 + 2*a^2*b^7 + b^9)*cosh(x)^2 - 6*(a^5*b^4 + 2*a^3*b^6 + a*b^8)*cosh(x))*sinh(x)^2 + 2*(2*(a^4*b^5 + 2*
a^2*b^7 + b^9)*cosh(x)^3 + 3*(a^5*b^4 + 2*a^3*b^6 + a*b^8)*cosh(x)^2 - (a^4*b^5 + 2*a^2*b^7 + b^9)*cosh(x))*si
nh(x))

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)**4/(a+b*sinh(x))**2,x)

[Out]

Timed out

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Giac [A]
time = 0.42, size = 235, normalized size = 1.45 \begin {gather*} -\frac {{\left (3 \, a^{5} + 4 \, a^{3} b^{2}\right )} \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{2} b^{4} + b^{6}\right )} \sqrt {a^{2} + b^{2}}} + \frac {{\left (6 \, a^{2} - b^{2}\right )} x}{2 \, b^{4}} + \frac {b^{2} e^{\left (2 \, x\right )} - 8 \, a b e^{x}}{8 \, b^{4}} + \frac {{\left (a^{2} b^{3} + b^{5} + 8 \, {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} e^{\left (3 \, x\right )} - {\left (32 \, a^{4} b + 17 \, a^{2} b^{3} + b^{5}\right )} e^{\left (2 \, x\right )} + 6 \, {\left (a^{3} b^{2} + a b^{4}\right )} e^{x}\right )} e^{\left (-2 \, x\right )}}{8 \, {\left (a^{2} + b^{2}\right )} {\left (b e^{\left (2 \, x\right )} + 2 \, a e^{x} - b\right )} b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^4/(a+b*sinh(x))^2,x, algorithm="giac")

[Out]

-(3*a^5 + 4*a^3*b^2)*log(abs(2*b*e^x + 2*a - 2*sqrt(a^2 + b^2))/abs(2*b*e^x + 2*a + 2*sqrt(a^2 + b^2)))/((a^2*
b^4 + b^6)*sqrt(a^2 + b^2)) + 1/2*(6*a^2 - b^2)*x/b^4 + 1/8*(b^2*e^(2*x) - 8*a*b*e^x)/b^4 + 1/8*(a^2*b^3 + b^5
+ 8*(2*a^5 - a^3*b^2 - a*b^4)*e^(3*x) - (32*a^4*b + 17*a^2*b^3 + b^5)*e^(2*x) + 6*(a^3*b^2 + a*b^4)*e^x)*e^(-
2*x)/((a^2 + b^2)*(b*e^(2*x) + 2*a*e^x - b)*b^4)

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Mupad [B]
time = 0.86, size = 305, normalized size = 1.88 \begin {gather*} \frac {{\mathrm {e}}^{2\,x}}{8\,b^2}-\frac {{\mathrm {e}}^{-2\,x}}{8\,b^2}-\frac {\frac {2\,a^4}{b^2\,\left (a^2\,b+b^3\right )}-\frac {2\,a^5\,{\mathrm {e}}^x}{b^3\,\left (a^2\,b+b^3\right )}}{2\,a\,{\mathrm {e}}^x-b+b\,{\mathrm {e}}^{2\,x}}+\frac {x\,\left (6\,a^2-b^2\right )}{2\,b^4}-\frac {a\,{\mathrm {e}}^x}{b^3}-\frac {a\,{\mathrm {e}}^{-x}}{b^3}-\frac {a^3\,\ln \left (\frac {2\,{\mathrm {e}}^x\,\left (3\,a^5+4\,a^3\,b^2\right )}{a^2\,b^5+b^7}-\frac {2\,a^3\,\left (b-a\,{\mathrm {e}}^x\right )\,\left (3\,a^2+4\,b^2\right )}{b^5\,{\left (a^2+b^2\right )}^{3/2}}\right )\,\left (3\,a^2+4\,b^2\right )}{b^4\,{\left (a^2+b^2\right )}^{3/2}}+\frac {a^3\,\ln \left (\frac {2\,{\mathrm {e}}^x\,\left (3\,a^5+4\,a^3\,b^2\right )}{a^2\,b^5+b^7}+\frac {2\,a^3\,\left (b-a\,{\mathrm {e}}^x\right )\,\left (3\,a^2+4\,b^2\right )}{b^5\,{\left (a^2+b^2\right )}^{3/2}}\right )\,\left (3\,a^2+4\,b^2\right )}{b^4\,{\left (a^2+b^2\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^4/(a + b*sinh(x))^2,x)

[Out]

exp(2*x)/(8*b^2) - exp(-2*x)/(8*b^2) - ((2*a^4)/(b^2*(a^2*b + b^3)) - (2*a^5*exp(x))/(b^3*(a^2*b + b^3)))/(2*a
*exp(x) - b + b*exp(2*x)) + (x*(6*a^2 - b^2))/(2*b^4) - (a*exp(x))/b^3 - (a*exp(-x))/b^3 - (a^3*log((2*exp(x)*
(3*a^5 + 4*a^3*b^2))/(b^7 + a^2*b^5) - (2*a^3*(b - a*exp(x))*(3*a^2 + 4*b^2))/(b^5*(a^2 + b^2)^(3/2)))*(3*a^2
+ 4*b^2))/(b^4*(a^2 + b^2)^(3/2)) + (a^3*log((2*exp(x)*(3*a^5 + 4*a^3*b^2))/(b^7 + a^2*b^5) + (2*a^3*(b - a*ex
p(x))*(3*a^2 + 4*b^2))/(b^5*(a^2 + b^2)^(3/2)))*(3*a^2 + 4*b^2))/(b^4*(a^2 + b^2)^(3/2))

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