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

Optimal. Leaf size=39 \[ \frac{x}{b}-\frac{\sqrt{a+b} \tanh ^{-1}\left (\frac{\sqrt{a} \tanh (x)}{\sqrt{a+b}}\right )}{\sqrt{a} b} \]

[Out]

x/b - (Sqrt[a + b]*ArcTanh[(Sqrt[a]*Tanh[x])/Sqrt[a + b]])/(Sqrt[a]*b)

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Rubi [A]  time = 0.0667251, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {3191, 391, 206, 208} \[ \frac{x}{b}-\frac{\sqrt{a+b} \tanh ^{-1}\left (\frac{\sqrt{a} \tanh (x)}{\sqrt{a+b}}\right )}{\sqrt{a} b} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]^2/(a + b*Cosh[x]^2),x]

[Out]

x/b - (Sqrt[a + b]*ArcTanh[(Sqrt[a]*Tanh[x])/Sqrt[a + b]])/(Sqrt[a]*b)

Rule 3191

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, T
an[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rule 391

Int[1/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x^n),
x], x] - Dist[d/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 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 208

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

Rubi steps

\begin{align*} \int \frac{\sinh ^2(x)}{a+b \cosh ^2(x)} \, dx &=-\operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right ) \left (a-(a+b) x^2\right )} \, dx,x,\coth (x)\right )\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\coth (x)\right )}{b}-\frac{(a+b) \operatorname{Subst}\left (\int \frac{1}{a+(-a-b) x^2} \, dx,x,\coth (x)\right )}{b}\\ &=\frac{x}{b}-\frac{\sqrt{a+b} \tanh ^{-1}\left (\frac{\sqrt{a} \tanh (x)}{\sqrt{a+b}}\right )}{\sqrt{a} b}\\ \end{align*}

Mathematica [A]  time = 0.0569631, size = 36, normalized size = 0.92 \[ \frac{x-\frac{\sqrt{a+b} \tanh ^{-1}\left (\frac{\sqrt{a} \tanh (x)}{\sqrt{a+b}}\right )}{\sqrt{a}}}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x - (Sqrt[a + b]*ArcTanh[(Sqrt[a]*Tanh[x])/Sqrt[a + b]])/Sqrt[a])/b

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Maple [B]  time = 0.03, size = 183, normalized size = 4.7 \begin{align*}{\frac{1}{b}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{1}{2\,b}\sqrt{a}\ln \left ( \sqrt{a+b} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+2\,\sqrt{a}\tanh \left ( x/2 \right ) +\sqrt{a+b} \right ){\frac{1}{\sqrt{a+b}}}}+{\frac{1}{2\,b}\sqrt{a}\ln \left ( \sqrt{a+b} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}-2\,\sqrt{a}\tanh \left ( x/2 \right ) +\sqrt{a+b} \right ){\frac{1}{\sqrt{a+b}}}}-{\frac{1}{2}\ln \left ( \sqrt{a+b} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+2\,\sqrt{a}\tanh \left ( x/2 \right ) +\sqrt{a+b} \right ){\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{a+b}}}}+{\frac{1}{2}\ln \left ( \sqrt{a+b} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}-2\,\sqrt{a}\tanh \left ( x/2 \right ) +\sqrt{a+b} \right ){\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{a+b}}}}-{\frac{1}{b}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*ln(tanh(1/2*x)+1)-1/2/b*a^(1/2)/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*x)^2+2*a^(1/2)*tanh(1/2*x)+(a+b)^(1/2)
)+1/2/b*a^(1/2)/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*x)^2-2*a^(1/2)*tanh(1/2*x)+(a+b)^(1/2))-1/2/a^(1/2)/(a+b)^
(1/2)*ln((a+b)^(1/2)*tanh(1/2*x)^2+2*a^(1/2)*tanh(1/2*x)+(a+b)^(1/2))+1/2/a^(1/2)/(a+b)^(1/2)*ln((a+b)^(1/2)*t
anh(1/2*x)^2-2*a^(1/2)*tanh(1/2*x)+(a+b)^(1/2))-1/b*ln(tanh(1/2*x)-1)

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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*cosh(x)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.98057, size = 845, normalized size = 21.67 \begin{align*} \left [\frac{\sqrt{\frac{a + b}{a}} \log \left (\frac{b^{2} \cosh \left (x\right )^{4} + 4 \, b^{2} \cosh \left (x\right ) \sinh \left (x\right )^{3} + b^{2} \sinh \left (x\right )^{4} + 2 \,{\left (2 \, a b + b^{2}\right )} \cosh \left (x\right )^{2} + 2 \,{\left (3 \, b^{2} \cosh \left (x\right )^{2} + 2 \, a b + b^{2}\right )} \sinh \left (x\right )^{2} + 8 \, a^{2} + 8 \, a b + b^{2} + 4 \,{\left (b^{2} \cosh \left (x\right )^{3} +{\left (2 \, a b + b^{2}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right ) + 4 \,{\left (a b \cosh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) \sinh \left (x\right ) + a b \sinh \left (x\right )^{2} + 2 \, a^{2} + a b\right )} \sqrt{\frac{a + b}{a}}}{b \cosh \left (x\right )^{4} + 4 \, b \cosh \left (x\right ) \sinh \left (x\right )^{3} + b \sinh \left (x\right )^{4} + 2 \,{\left (2 \, a + b\right )} \cosh \left (x\right )^{2} + 2 \,{\left (3 \, b \cosh \left (x\right )^{2} + 2 \, a + b\right )} \sinh \left (x\right )^{2} + 4 \,{\left (b \cosh \left (x\right )^{3} +{\left (2 \, a + b\right )} \cosh \left (x\right )\right )} \sinh \left (x\right ) + b}\right ) + 2 \, x}{2 \, b}, -\frac{\sqrt{-\frac{a + b}{a}} \arctan \left (\frac{{\left (b \cosh \left (x\right )^{2} + 2 \, b \cosh \left (x\right ) \sinh \left (x\right ) + b \sinh \left (x\right )^{2} + 2 \, a + b\right )} \sqrt{-\frac{a + b}{a}}}{2 \,{\left (a + b\right )}}\right ) - x}{b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt((a + b)/a)*log((b^2*cosh(x)^4 + 4*b^2*cosh(x)*sinh(x)^3 + b^2*sinh(x)^4 + 2*(2*a*b + b^2)*cosh(x)^2
+ 2*(3*b^2*cosh(x)^2 + 2*a*b + b^2)*sinh(x)^2 + 8*a^2 + 8*a*b + b^2 + 4*(b^2*cosh(x)^3 + (2*a*b + b^2)*cosh(x
))*sinh(x) + 4*(a*b*cosh(x)^2 + 2*a*b*cosh(x)*sinh(x) + a*b*sinh(x)^2 + 2*a^2 + a*b)*sqrt((a + b)/a))/(b*cosh(
x)^4 + 4*b*cosh(x)*sinh(x)^3 + b*sinh(x)^4 + 2*(2*a + b)*cosh(x)^2 + 2*(3*b*cosh(x)^2 + 2*a + b)*sinh(x)^2 + 4
*(b*cosh(x)^3 + (2*a + b)*cosh(x))*sinh(x) + b)) + 2*x)/b, -(sqrt(-(a + b)/a)*arctan(1/2*(b*cosh(x)^2 + 2*b*co
sh(x)*sinh(x) + b*sinh(x)^2 + 2*a + b)*sqrt(-(a + b)/a)/(a + b)) - x)/b]

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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*cosh(x)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.2753, size = 70, normalized size = 1.79 \begin{align*} -\frac{{\left (a + b\right )} \arctan \left (\frac{b e^{\left (2 \, x\right )} + 2 \, a + b}{2 \, \sqrt{-a^{2} - a b}}\right )}{\sqrt{-a^{2} - a b} b} + \frac{x}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a + b)*arctan(1/2*(b*e^(2*x) + 2*a + b)/sqrt(-a^2 - a*b))/(sqrt(-a^2 - a*b)*b) + x/b