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

Optimal. Leaf size=83 \[ \frac{2 a \left (a^2+2 b^2\right ) \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{b^2 \left (a^2+b^2\right )^{3/2}}-\frac{a^2 \cosh (x)}{b \left (a^2+b^2\right ) (a+b \sinh (x))}+\frac{x}{b^2} \]

[Out]

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

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Rubi [A]  time = 0.133505, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {2790, 2735, 2660, 618, 206} \[ \frac{2 a \left (a^2+2 b^2\right ) \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{b^2 \left (a^2+b^2\right )^{3/2}}-\frac{a^2 \cosh (x)}{b \left (a^2+b^2\right ) (a+b \sinh (x))}+\frac{x}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^2/(a + b*Sinh[x])^2,x]

[Out]

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

Rule 2790

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> -Simp[
((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 1)*(a^2 - b^2)), x] - Di
st[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*(2*b*c*d - a*(c^2 + d^2)) + (a^2
*d^2 - 2*a*b*c*d*(m + 2) + b^2*(d^2*(m + 1) + c^2*(m + 2)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e,
f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && 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 ^2(x)}{(a+b \sinh (x))^2} \, dx &=-\frac{a^2 \cosh (x)}{b \left (a^2+b^2\right ) (a+b \sinh (x))}-\frac{i \int \frac{-i a b+i \left (a^2+b^2\right ) \sinh (x)}{a+b \sinh (x)} \, dx}{b \left (a^2+b^2\right )}\\ &=\frac{x}{b^2}-\frac{a^2 \cosh (x)}{b \left (a^2+b^2\right ) (a+b \sinh (x))}-\frac{\left (a \left (a^2+2 b^2\right )\right ) \int \frac{1}{a+b \sinh (x)} \, dx}{b^2 \left (a^2+b^2\right )}\\ &=\frac{x}{b^2}-\frac{a^2 \cosh (x)}{b \left (a^2+b^2\right ) (a+b \sinh (x))}-\frac{\left (2 a \left (a^2+2 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^2 \left (a^2+b^2\right )}\\ &=\frac{x}{b^2}-\frac{a^2 \cosh (x)}{b \left (a^2+b^2\right ) (a+b \sinh (x))}+\frac{\left (4 a \left (a^2+2 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^2 \left (a^2+b^2\right )}\\ &=\frac{x}{b^2}+\frac{2 a \left (a^2+2 b^2\right ) \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{b^2 \left (a^2+b^2\right )^{3/2}}-\frac{a^2 \cosh (x)}{b \left (a^2+b^2\right ) (a+b \sinh (x))}\\ \end{align*}

Mathematica [A]  time = 0.178002, size = 86, normalized size = 1.04 \[ \frac{\frac{2 a \left (a^2+2 b^2\right ) \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{-a^2-b^2}}\right )}{\left (-a^2-b^2\right )^{3/2}}-\frac{a^2 b \cosh (x)}{\left (a^2+b^2\right ) (a+b \sinh (x))}+x}{b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.033, size = 175, normalized size = 2.1 \begin{align*}{\frac{1}{{b}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{1}{{b}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }+2\,{\frac{a\tanh \left ( x/2 \right ) }{ \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}^{2}}{b \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}^{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 ) }-4\,{\frac{a}{ \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)^2/(a+b*sinh(x))^2,x)

[Out]

1/b^2*ln(tanh(1/2*x)+1)-1/b^2*ln(tanh(1/2*x)-1)+2*a/(a*tanh(1/2*x)^2-2*tanh(1/2*x)*b-a)/(a^2+b^2)*tanh(1/2*x)+
2*a^2/b/(a*tanh(1/2*x)^2-2*tanh(1/2*x)*b-a)/(a^2+b^2)-2/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))-4*a/(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)^2/(a+b*sinh(x))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.24456, size = 1245, normalized size = 15. \begin{align*} \frac{2 \, a^{4} b + 2 \, a^{2} b^{3} -{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} x \cosh \left (x\right )^{2} -{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} x \sinh \left (x\right )^{2} +{\left (a^{3} b + 2 \, a b^{3} -{\left (a^{3} b + 2 \, a b^{3}\right )} \cosh \left (x\right )^{2} -{\left (a^{3} b + 2 \, a b^{3}\right )} \sinh \left (x\right )^{2} - 2 \,{\left (a^{4} + 2 \, a^{2} b^{2}\right )} \cosh \left (x\right ) - 2 \,{\left (a^{4} + 2 \, a^{2} b^{2} +{\left (a^{3} b + 2 \, a b^{3}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )\right )} \sqrt{a^{2} + b^{2}} \log \left (\frac{b^{2} \cosh \left (x\right )^{2} + b^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) + 2 \, a^{2} + b^{2} + 2 \,{\left (b^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right ) + 2 \, \sqrt{a^{2} + b^{2}}{\left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}}{b \cosh \left (x\right )^{2} + b \sinh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + 2 \,{\left (b \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) - b}\right ) +{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} x - 2 \,{\left (a^{5} + a^{3} b^{2} +{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} x\right )} \cosh \left (x\right ) - 2 \,{\left (a^{5} + a^{3} b^{2} +{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} x \cosh \left (x\right ) +{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} x\right )} \sinh \left (x\right )}{a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7} -{\left (a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7}\right )} \cosh \left (x\right )^{2} -{\left (a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7}\right )} \sinh \left (x\right )^{2} - 2 \,{\left (a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6}\right )} \cosh \left (x\right ) - 2 \,{\left (a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6} +{\left (a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a^4*b + 2*a^2*b^3 - (a^4*b + 2*a^2*b^3 + b^5)*x*cosh(x)^2 - (a^4*b + 2*a^2*b^3 + b^5)*x*sinh(x)^2 + (a^3*b
+ 2*a*b^3 - (a^3*b + 2*a*b^3)*cosh(x)^2 - (a^3*b + 2*a*b^3)*sinh(x)^2 - 2*(a^4 + 2*a^2*b^2)*cosh(x) - 2*(a^4 +
2*a^2*b^2 + (a^3*b + 2*a*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*co
sh(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)) + (a^4*b + 2*a^2*b^3 + b^5)*x - 2*(a^5 + a^3
*b^2 + (a^5 + 2*a^3*b^2 + a*b^4)*x)*cosh(x) - 2*(a^5 + a^3*b^2 + (a^4*b + 2*a^2*b^3 + b^5)*x*cosh(x) + (a^5 +
2*a^3*b^2 + a*b^4)*x)*sinh(x))/(a^4*b^3 + 2*a^2*b^5 + b^7 - (a^4*b^3 + 2*a^2*b^5 + b^7)*cosh(x)^2 - (a^4*b^3 +
2*a^2*b^5 + b^7)*sinh(x)^2 - 2*(a^5*b^2 + 2*a^3*b^4 + a*b^6)*cosh(x) - 2*(a^5*b^2 + 2*a^3*b^4 + a*b^6 + (a^4*
b^3 + 2*a^2*b^5 + b^7)*cosh(x))*sinh(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)**2/(a+b*sinh(x))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.45168, size = 177, normalized size = 2.13 \begin{align*} -\frac{{\left (a^{3} + 2 \, a 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^{2} + b^{4}\right )} \sqrt{a^{2} + b^{2}}} + \frac{2 \,{\left (a^{3} e^{x} - a^{2} b\right )}}{{\left (a^{2} b^{2} + b^{4}\right )}{\left (b e^{\left (2 \, x\right )} + 2 \, a e^{x} - b\right )}} + \frac{x}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^3 + 2*a*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^2
+ b^4)*sqrt(a^2 + b^2)) + 2*(a^3*e^x - a^2*b)/((a^2*b^2 + b^4)*(b*e^(2*x) + 2*a*e^x - b)) + x/b^2