3.80 \(\int \frac{\sinh ^4(x)}{(a+b \sinh (x))^2} \, dx\)

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

[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]))

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Rubi [A]  time = 0.405337, antiderivative size = 162, normalized size of antiderivative = 1., 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 = {2792, 3049, 3023, 2735, 2660, 618, 206} \[ \frac{x \left (6 a^2-b^2\right )}{2 b^4}-\frac{a \left (3 a^2+2 b^2\right ) \cosh (x)}{b^3 \left (a^2+b^2\right )}+\frac{2 a^3 \left (3 a^2+4 b^2\right ) \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{b^4 \left (a^2+b^2\right )^{3/2}}-\frac{a^2 \sinh ^2(x) \cosh (x)}{b \left (a^2+b^2\right ) (a+b \sinh (x))}+\frac{\left (3 a^2+b^2\right ) \sinh (x) \cosh (x)}{2 b^2 \left (a^2+b^2\right )} \]

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 2792

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[((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 3049

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])))

Rule 3023

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 2735

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 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 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 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])

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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [A]  time = 0.407256, size = 118, normalized size = 0.73 \[ \frac{-2 x \left (b^2-6 a^2\right )+\frac{8 a^3 \left (3 a^2+4 b^2\right ) \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{-a^2-b^2}}\right )}{\left (-a^2-b^2\right )^{3/2}}-\frac{4 a^4 b \cosh (x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}-8 a b \cosh (x)+b^2 \sinh (2 x)}{4 b^4} \]

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 = 0.048, size = 296, normalized size = 1.8 \begin{align*} -{\frac{1}{2\,{b}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}}+{\frac{1}{2\,{b}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-2\,{\frac{a}{{b}^{3} \left ( \tanh \left ( x/2 \right ) +1 \right ) }}+3\,{\frac{\ln \left ( \tanh \left ( x/2 \right ) +1 \right ){a}^{2}}{{b}^{4}}}-{\frac{1}{2\,{b}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+{\frac{1}{2\,{b}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-2}}+{\frac{1}{2\,{b}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}+2\,{\frac{a}{{b}^{3} \left ( \tanh \left ( x/2 \right ) -1 \right ) }}-3\,{\frac{\ln \left ( \tanh \left ( x/2 \right ) -1 \right ){a}^{2}}{{b}^{4}}}+{\frac{1}{2\,{b}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }+2\,{\frac{{a}^{3}\tanh \left ( x/2 \right ) }{{b}^{2} \left ( a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) \left ({a}^{2}+{b}^{2} \right ) }}+2\,{\frac{{a}^{4}}{{b}^{3} \left ( a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) \left ({a}^{2}+{b}^{2} \right ) }}-6\,{\frac{{a}^{5}}{{b}^{4} \left ({a}^{2}+{b}^{2} \right ) ^{3/2}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( x/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }-8\,{\frac{{a}^{3}}{{b}^{2} \left ({a}^{2}+{b}^{2} \right ) ^{3/2}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( x/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/b^2/(tanh(1/2*x)+1)^2+1/2/b^2/(tanh(1/2*x)+1)-2/b^3/(tanh(1/2*x)+1)*a+3/b^4*ln(tanh(1/2*x)+1)*a^2-1/2/b^2
*ln(tanh(1/2*x)+1)+1/2/b^2/(tanh(1/2*x)-1)^2+1/2/b^2/(tanh(1/2*x)-1)+2/b^3/(tanh(1/2*x)-1)*a-3/b^4*ln(tanh(1/2
*x)-1)*a^2+1/2/b^2*ln(tanh(1/2*x)-1)+2/b^2*a^3/(a*tanh(1/2*x)^2-2*tanh(1/2*x)*b-a)/(a^2+b^2)*tanh(1/2*x)+2/b^3
*a^4/(a*tanh(1/2*x)^2-2*tanh(1/2*x)*b-a)/(a^2+b^2)-6/b^4*a^5/(a^2+b^2)^(3/2)*arctanh(1/2*(2*a*tanh(1/2*x)-2*b)
/(a^2+b^2)^(1/2))-8/b^2*a^3/(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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.78534, size = 4020, normalized size = 24.81 \begin{align*} \text{result too large to display} \end{align*}

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)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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 = 1.42231, size = 317, normalized size = 1.96 \begin{align*} -\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{align*}

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)