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

Optimal. Leaf size=144 \[ -\frac{2 a^3 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}+\frac{4 a b^2 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac{\left (a^2-b^2\right ) \tanh (x)}{\left (a^2+b^2\right )^2}-\frac{2 a b \text{sech}(x)}{\left (a^2+b^2\right )^2}-\frac{a^2 b \cosh (x)}{\left (a^2+b^2\right )^2 (a+b \sinh (x))} \]

[Out]

(-2*a^3*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2) + (4*a*b^2*ArcTanh[(b - a*Tanh[x/2])/Sqr
t[a^2 + b^2]])/(a^2 + b^2)^(5/2) - (2*a*b*Sech[x])/(a^2 + b^2)^2 - (a^2*b*Cosh[x])/((a^2 + b^2)^2*(a + b*Sinh[
x])) - ((a^2 - b^2)*Tanh[x])/(a^2 + b^2)^2

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Rubi [A]  time = 0.248748, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.692, Rules used = {2731, 2664, 12, 2660, 618, 206, 2669, 3767, 8} \[ -\frac{2 a^3 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}+\frac{4 a b^2 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac{\left (a^2-b^2\right ) \tanh (x)}{\left (a^2+b^2\right )^2}-\frac{2 a b \text{sech}(x)}{\left (a^2+b^2\right )^2}-\frac{a^2 b \cosh (x)}{\left (a^2+b^2\right )^2 (a+b \sinh (x))} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^2/(a + b*Sinh[x])^2,x]

[Out]

(-2*a^3*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2) + (4*a*b^2*ArcTanh[(b - a*Tanh[x/2])/Sqr
t[a^2 + b^2]])/(a^2 + b^2)^(5/2) - (2*a*b*Sech[x])/(a^2 + b^2)^2 - (a^2*b*Cosh[x])/((a^2 + b^2)^2*(a + b*Sinh[
x])) - ((a^2 - b^2)*Tanh[x])/(a^2 + b^2)^2

Rule 2731

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Int[ExpandIntegrand
[(Sin[e + f*x]^p*(a + b*Sin[e + f*x])^m)/(1 - Sin[e + f*x]^2)^(p/2), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a
^2 - b^2, 0] && IntegersQ[m, p/2]

Rule 2664

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n +
1))/(d*(n + 1)*(a^2 - b^2)), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1
) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integer
Q[2*n]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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])

Rule 2669

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b*(g*Cos[
e + f*x])^(p + 1))/(f*g*(p + 1)), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x]
&& (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

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

Rubi steps

\begin{align*} \int \frac{\tanh ^2(x)}{(a+b \sinh (x))^2} \, dx &=-\int \left (-\frac{a^2}{\left (a^2+b^2\right ) (a+b \sinh (x))^2}+\frac{2 a b^2}{\left (a^2+b^2\right )^2 (a+b \sinh (x))}+\frac{\text{sech}^2(x) \left (a^2 \left (1-\frac{b^2}{a^2}\right )-2 a b \sinh (x)\right )}{\left (a^2+b^2\right )^2}\right ) \, dx\\ &=-\frac{\int \text{sech}^2(x) \left (a^2 \left (1-\frac{b^2}{a^2}\right )-2 a b \sinh (x)\right ) \, dx}{\left (a^2+b^2\right )^2}-\frac{\left (2 a b^2\right ) \int \frac{1}{a+b \sinh (x)} \, dx}{\left (a^2+b^2\right )^2}+\frac{a^2 \int \frac{1}{(a+b \sinh (x))^2} \, dx}{a^2+b^2}\\ &=-\frac{2 a b \text{sech}(x)}{\left (a^2+b^2\right )^2}-\frac{a^2 b \cosh (x)}{\left (a^2+b^2\right )^2 (a+b \sinh (x))}+\frac{a^2 \int \frac{a}{a+b \sinh (x)} \, dx}{\left (a^2+b^2\right )^2}-\frac{\left (4 a b^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x-a x^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )}{\left (a^2+b^2\right )^2}-\frac{\left (a^2-b^2\right ) \int \text{sech}^2(x) \, dx}{\left (a^2+b^2\right )^2}\\ &=-\frac{2 a b \text{sech}(x)}{\left (a^2+b^2\right )^2}-\frac{a^2 b \cosh (x)}{\left (a^2+b^2\right )^2 (a+b \sinh (x))}+\frac{a^3 \int \frac{1}{a+b \sinh (x)} \, dx}{\left (a^2+b^2\right )^2}+\frac{\left (8 a b^2\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 )}{\left (a^2+b^2\right )^2}-\frac{\left (i \left (a^2-b^2\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-i \tanh (x))}{\left (a^2+b^2\right )^2}\\ &=\frac{4 a b^2 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac{2 a b \text{sech}(x)}{\left (a^2+b^2\right )^2}-\frac{a^2 b \cosh (x)}{\left (a^2+b^2\right )^2 (a+b \sinh (x))}-\frac{\left (a^2-b^2\right ) \tanh (x)}{\left (a^2+b^2\right )^2}+\frac{\left (2 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x-a x^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )}{\left (a^2+b^2\right )^2}\\ &=\frac{4 a b^2 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac{2 a b \text{sech}(x)}{\left (a^2+b^2\right )^2}-\frac{a^2 b \cosh (x)}{\left (a^2+b^2\right )^2 (a+b \sinh (x))}-\frac{\left (a^2-b^2\right ) \tanh (x)}{\left (a^2+b^2\right )^2}-\frac{\left (4 a^3\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 )}{\left (a^2+b^2\right )^2}\\ &=-\frac{2 a^3 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}+\frac{4 a b^2 \tanh ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac{2 a b \text{sech}(x)}{\left (a^2+b^2\right )^2}-\frac{a^2 b \cosh (x)}{\left (a^2+b^2\right )^2 (a+b \sinh (x))}-\frac{\left (a^2-b^2\right ) \tanh (x)}{\left (a^2+b^2\right )^2}\\ \end{align*}

Mathematica [A]  time = 0.264834, size = 100, normalized size = 0.69 \[ \frac{\left (b^2-a^2\right ) \tanh (x)+\frac{2 a \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac{b-a \tanh \left (\frac{x}{2}\right )}{\sqrt{-a^2-b^2}}\right )}{\sqrt{-a^2-b^2}}-\frac{a^2 b \cosh (x)}{a+b \sinh (x)}-2 a b \text{sech}(x)}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.05, size = 142, normalized size = 1. \begin{align*} 2\,{\frac{ \left ( -{a}^{2}+{b}^{2} \right ) \tanh \left ( x/2 \right ) -2\,ab}{ \left ({a}^{4}+2\,{a}^{2}{b}^{2}+{b}^{4} \right ) \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) }}-2\,{\frac{a}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}} \left ({\frac{-{b}^{2}\tanh \left ( x/2 \right ) -ab}{a \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a}}-{\frac{{a}^{2}-2\,{b}^{2}}{\sqrt{{a}^{2}+{b}^{2}}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( x/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.14812, size = 2144, normalized size = 14.89 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*a^4*b + 2*a^2*b^3 - 2*b^5 - 2*(a^5 - a^3*b^2 - 2*a*b^4)*cosh(x)^3 - 2*(a^5 - a^3*b^2 - 2*a*b^4)*sinh(x)^3 +
2*(4*a^4*b + 5*a^2*b^3 + b^5)*cosh(x)^2 + 2*(4*a^4*b + 5*a^2*b^3 + b^5 - 3*(a^5 - a^3*b^2 - 2*a*b^4)*cosh(x))
*sinh(x)^2 + ((a^3*b - 2*a*b^3)*cosh(x)^4 + (a^3*b - 2*a*b^3)*sinh(x)^4 - a^3*b + 2*a*b^3 + 2*(a^4 - 2*a^2*b^2
)*cosh(x)^3 + 2*(a^4 - 2*a^2*b^2 + 2*(a^3*b - 2*a*b^3)*cosh(x))*sinh(x)^3 + 6*((a^3*b - 2*a*b^3)*cosh(x)^2 + (
a^4 - 2*a^2*b^2)*cosh(x))*sinh(x)^2 + 2*(a^4 - 2*a^2*b^2)*cosh(x) + 2*(a^4 - 2*a^2*b^2 + 2*(a^3*b - 2*a*b^3)*c
osh(x)^3 + 3*(a^4 - 2*a^2*b^2)*cosh(x)^2)*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 + a^3*b^2)*cosh(x) - 2*(3*a^5 + 3
*a^3*b^2 + 3*(a^5 - a^3*b^2 - 2*a*b^4)*cosh(x)^2 - 2*(4*a^4*b + 5*a^2*b^3 + b^5)*cosh(x))*sinh(x))/(a^6*b + 3*
a^4*b^3 + 3*a^2*b^5 + b^7 - (a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*cosh(x)^4 - (a^6*b + 3*a^4*b^3 + 3*a^2*b^5 +
b^7)*sinh(x)^4 - 2*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cosh(x)^3 - 2*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6 +
2*(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*cosh(x))*sinh(x)^3 - 6*((a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*cosh(x)
^2 + (a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cosh(x))*sinh(x)^2 - 2*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cosh(x
) - 2*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6 + 2*(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*cosh(x)^3 + 3*(a^7 + 3*a^
5*b^2 + 3*a^3*b^4 + a*b^6)*cosh(x)^2)*sinh(x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh ^{2}{\left (x \right )}}{\left (a + b \sinh{\left (x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(tanh(x)**2/(a + b*sinh(x))**2, x)

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Giac [A]  time = 1.18394, size = 244, normalized size = 1.69 \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^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{a^{2} + b^{2}}} + \frac{2 \,{\left (a^{3} e^{\left (3 \, x\right )} - 2 \, a b^{2} e^{\left (3 \, x\right )} - 4 \, a^{2} b e^{\left (2 \, x\right )} - b^{3} e^{\left (2 \, x\right )} + 3 \, a^{3} e^{x} - 2 \, a^{2} b + b^{3}\right )}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}{\left (b e^{\left (4 \, x\right )} + 2 \, a e^{\left (3 \, x\right )} + 2 \, a e^{x} - b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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