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

Optimal. Leaf size=50 \[ \frac{x}{b}-\frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh (c+d x)}{\sqrt{a}}\right )}{b d \sqrt{a-b}} \]

[Out]

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

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Rubi [A]  time = 0.0871054, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3171, 3181, 208} \[ \frac{x}{b}-\frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh (c+d x)}{\sqrt{a}}\right )}{b d \sqrt{a-b}} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[c + d*x]^2/(a + b*Sinh[c + d*x]^2),x]

[Out]

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

Rule 3171

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

Rule 3181

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[1/(a + (a + b)*ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x]

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

Mathematica [A]  time = 0.131342, size = 50, normalized size = 1. \[ \frac{-\frac{\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh (c+d x)}{\sqrt{a}}\right )}{\sqrt{a-b}}+c+d x}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.033, size = 312, normalized size = 6.2 \begin{align*}{\frac{1}{bd}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{1}{bd}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{a}{bd}{\it Artanh} \left ({a\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ){\frac{1}{\sqrt{ \left ( 2\,\sqrt{-b \left ( a-b \right ) }+a-2\,b \right ) a}}}} \right ){\frac{1}{\sqrt{ \left ( 2\,\sqrt{-b \left ( a-b \right ) }+a-2\,b \right ) a}}}}+{\frac{a}{d}{\it Artanh} \left ({a\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ){\frac{1}{\sqrt{ \left ( 2\,\sqrt{-b \left ( a-b \right ) }+a-2\,b \right ) a}}}} \right ){\frac{1}{\sqrt{-b \left ( a-b \right ) }}}{\frac{1}{\sqrt{ \left ( 2\,\sqrt{-b \left ( a-b \right ) }+a-2\,b \right ) a}}}}+{\frac{a}{bd}\arctan \left ({a\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ){\frac{1}{\sqrt{ \left ( 2\,\sqrt{-b \left ( a-b \right ) }-a+2\,b \right ) a}}}} \right ){\frac{1}{\sqrt{ \left ( 2\,\sqrt{-b \left ( a-b \right ) }-a+2\,b \right ) a}}}}+{\frac{a}{d}\arctan \left ({a\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ){\frac{1}{\sqrt{ \left ( 2\,\sqrt{-b \left ( a-b \right ) }-a+2\,b \right ) a}}}} \right ){\frac{1}{\sqrt{-b \left ( a-b \right ) }}}{\frac{1}{\sqrt{ \left ( 2\,\sqrt{-b \left ( a-b \right ) }-a+2\,b \right ) a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d/b*ln(tanh(1/2*d*x+1/2*c)+1)-1/d/b*ln(tanh(1/2*d*x+1/2*c)-1)-1/d*a/b/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2)*a
rctanh(a*tanh(1/2*d*x+1/2*c)/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2))+1/d*a/(-b*(a-b))^(1/2)/((2*(-b*(a-b))^(1/2)
+a-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*(-b*(a-b))^(1/2)+a-2*b)*a)^(1/2))+1/d*a/b/((2*(-b*(a-b))^(1
/2)-a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*(-b*(a-b))^(1/2)-a+2*b)*a)^(1/2))+1/d*a/(-b*(a-b))^(1/2)/
((2*(-b*(a-b))^(1/2)-a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*(-b*(a-b))^(1/2)-a+2*b)*a)^(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(d*x+c)^2/(a+b*sinh(d*x+c)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.01965, size = 1122, normalized size = 22.44 \begin{align*} \left [\frac{2 \, d x + \sqrt{\frac{a}{a - b}} \log \left (\frac{b^{2} \cosh \left (d x + c\right )^{4} + 4 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + b^{2} \sinh \left (d x + c\right )^{4} + 2 \,{\left (2 \, a b - b^{2}\right )} \cosh \left (d x + c\right )^{2} + 2 \,{\left (3 \, b^{2} \cosh \left (d x + c\right )^{2} + 2 \, a b - b^{2}\right )} \sinh \left (d x + c\right )^{2} + 8 \, a^{2} - 8 \, a b + b^{2} + 4 \,{\left (b^{2} \cosh \left (d x + c\right )^{3} +{\left (2 \, a b - b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + 4 \,{\left ({\left (a b - b^{2}\right )} \cosh \left (d x + c\right )^{2} + 2 \,{\left (a b - b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) +{\left (a b - b^{2}\right )} \sinh \left (d x + c\right )^{2} + 2 \, a^{2} - 3 \, a b + b^{2}\right )} \sqrt{\frac{a}{a - b}}}{b \cosh \left (d x + c\right )^{4} + 4 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + b \sinh \left (d x + c\right )^{4} + 2 \,{\left (2 \, a - b\right )} \cosh \left (d x + c\right )^{2} + 2 \,{\left (3 \, b \cosh \left (d x + c\right )^{2} + 2 \, a - b\right )} \sinh \left (d x + c\right )^{2} + 4 \,{\left (b \cosh \left (d x + c\right )^{3} +{\left (2 \, a - b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + b}\right )}{2 \, b d}, \frac{d x - \sqrt{-\frac{a}{a - b}} \arctan \left (\frac{{\left (b \cosh \left (d x + c\right )^{2} + 2 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + b \sinh \left (d x + c\right )^{2} + 2 \, a - b\right )} \sqrt{-\frac{a}{a - b}}}{2 \, a}\right )}{b d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(d*x+c)**2/(a+b*sinh(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.26475, size = 88, normalized size = 1.76 \begin{align*} -\frac{a \arctan \left (\frac{b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a - b}{2 \, \sqrt{-a^{2} + a b}}\right )}{\sqrt{-a^{2} + a b} b d} + \frac{d x + c}{b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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