3.108 \(\int \frac{\sinh (e+f x)}{(a+b \sinh ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=36 \[ \frac{\cosh (e+f x)}{f (a-b) \sqrt{a+b \cosh ^2(e+f x)-b}} \]

[Out]

Cosh[e + f*x]/((a - b)*f*Sqrt[a - b + b*Cosh[e + f*x]^2])

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Rubi [A]  time = 0.0526382, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {3186, 191} \[ \frac{\cosh (e+f x)}{f (a-b) \sqrt{a+b \cosh ^2(e+f x)-b}} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[e + f*x]/(a + b*Sinh[e + f*x]^2)^(3/2),x]

[Out]

Cosh[e + f*x]/((a - b)*f*Sqrt[a - b + b*Cosh[e + f*x]^2])

Rule 3186

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

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

\begin{align*} \int \frac{\sinh (e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (a-b+b x^2\right )^{3/2}} \, dx,x,\cosh (e+f x)\right )}{f}\\ &=\frac{\cosh (e+f x)}{(a-b) f \sqrt{a-b+b \cosh ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.15519, size = 43, normalized size = 1.19 \[ \frac{\sqrt{2} \cosh (e+f x)}{f (a-b) \sqrt{2 a+b \cosh (2 (e+f x))-b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[e + f*x]/(a + b*Sinh[e + f*x]^2)^(3/2),x]

[Out]

(Sqrt[2]*Cosh[e + f*x])/((a - b)*f*Sqrt[2*a - b + b*Cosh[2*(e + f*x)]])

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Maple [A]  time = 0.051, size = 32, normalized size = 0.9 \begin{align*}{\frac{\cosh \left ( fx+e \right ) }{ \left ( a-b \right ) f}{\frac{1}{\sqrt{a+b \left ( \sinh \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(f*x+e)/(a+b*sinh(f*x+e)^2)^(3/2),x)

[Out]

cosh(f*x+e)/(a-b)/(a+b*sinh(f*x+e)^2)^(1/2)/f

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Maxima [B]  time = 1.60599, size = 319, normalized size = 8.86 \begin{align*} \frac{b^{2} e^{\left (-6 \, f x - 6 \, e\right )} + 2 \, a b - b^{2} +{\left (8 \, a^{2} - 8 \, a b + 3 \, b^{2}\right )} e^{\left (-2 \, f x - 2 \, e\right )} + 3 \,{\left (2 \, a b - b^{2}\right )} e^{\left (-4 \, f x - 4 \, e\right )}}{2 \,{\left (a^{2} - a b\right )}{\left (2 \,{\left (2 \, a - b\right )} e^{\left (-2 \, f x - 2 \, e\right )} + b e^{\left (-4 \, f x - 4 \, e\right )} + b\right )}^{\frac{3}{2}} f} + \frac{b^{2} + 3 \,{\left (2 \, a b - b^{2}\right )} e^{\left (-2 \, f x - 2 \, e\right )} +{\left (8 \, a^{2} - 8 \, a b + 3 \, b^{2}\right )} e^{\left (-4 \, f x - 4 \, e\right )} +{\left (2 \, a b - b^{2}\right )} e^{\left (-6 \, f x - 6 \, e\right )}}{2 \,{\left (a^{2} - a b\right )}{\left (2 \,{\left (2 \, a - b\right )} e^{\left (-2 \, f x - 2 \, e\right )} + b e^{\left (-4 \, f x - 4 \, e\right )} + b\right )}^{\frac{3}{2}} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(f*x+e)/(a+b*sinh(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

1/2*(b^2*e^(-6*f*x - 6*e) + 2*a*b - b^2 + (8*a^2 - 8*a*b + 3*b^2)*e^(-2*f*x - 2*e) + 3*(2*a*b - b^2)*e^(-4*f*x
- 4*e))/((a^2 - a*b)*(2*(2*a - b)*e^(-2*f*x - 2*e) + b*e^(-4*f*x - 4*e) + b)^(3/2)*f) + 1/2*(b^2 + 3*(2*a*b -
b^2)*e^(-2*f*x - 2*e) + (8*a^2 - 8*a*b + 3*b^2)*e^(-4*f*x - 4*e) + (2*a*b - b^2)*e^(-6*f*x - 6*e))/((a^2 - a*
b)*(2*(2*a - b)*e^(-2*f*x - 2*e) + b*e^(-4*f*x - 4*e) + b)^(3/2)*f)

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Fricas [B]  time = 2.17215, size = 738, normalized size = 20.5 \begin{align*} \frac{\sqrt{2}{\left (\cosh \left (f x + e\right )^{2} + 2 \, \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + \sinh \left (f x + e\right )^{2} + 1\right )} \sqrt{\frac{b \cosh \left (f x + e\right )^{2} + b \sinh \left (f x + e\right )^{2} + 2 \, a - b}{\cosh \left (f x + e\right )^{2} - 2 \, \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + \sinh \left (f x + e\right )^{2}}}}{{\left (a b - b^{2}\right )} f \cosh \left (f x + e\right )^{4} + 4 \,{\left (a b - b^{2}\right )} f \cosh \left (f x + e\right ) \sinh \left (f x + e\right )^{3} +{\left (a b - b^{2}\right )} f \sinh \left (f x + e\right )^{4} + 2 \,{\left (2 \, a^{2} - 3 \, a b + b^{2}\right )} f \cosh \left (f x + e\right )^{2} + 2 \,{\left (3 \,{\left (a b - b^{2}\right )} f \cosh \left (f x + e\right )^{2} +{\left (2 \, a^{2} - 3 \, a b + b^{2}\right )} f\right )} \sinh \left (f x + e\right )^{2} +{\left (a b - b^{2}\right )} f + 4 \,{\left ({\left (a b - b^{2}\right )} f \cosh \left (f x + e\right )^{3} +{\left (2 \, a^{2} - 3 \, a b + b^{2}\right )} f \cosh \left (f x + e\right )\right )} \sinh \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(f*x+e)/(a+b*sinh(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

sqrt(2)*(cosh(f*x + e)^2 + 2*cosh(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2 + 1)*sqrt((b*cosh(f*x + e)^2 + b*si
nh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*cosh(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2))/((a*b - b^2)*f*c
osh(f*x + e)^4 + 4*(a*b - b^2)*f*cosh(f*x + e)*sinh(f*x + e)^3 + (a*b - b^2)*f*sinh(f*x + e)^4 + 2*(2*a^2 - 3*
a*b + b^2)*f*cosh(f*x + e)^2 + 2*(3*(a*b - b^2)*f*cosh(f*x + e)^2 + (2*a^2 - 3*a*b + b^2)*f)*sinh(f*x + e)^2 +
(a*b - b^2)*f + 4*((a*b - b^2)*f*cosh(f*x + e)^3 + (2*a^2 - 3*a*b + b^2)*f*cosh(f*x + e))*sinh(f*x + e))

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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(f*x+e)/(a+b*sinh(f*x+e)**2)**(3/2),x)

[Out]

Timed out

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Giac [B]  time = 1.22891, size = 207, normalized size = 5.75 \begin{align*} -\frac{\sqrt{b} f}{256 \,{\left (a b^{4} - b^{5}\right )}} + \frac{\frac{{\left (a^{3} f - a^{2} b f\right )} e^{\left (2 \, f x + 2 \, e\right )}}{a^{4} b^{3} - 2 \, a^{3} b^{4} + a^{2} b^{5}} + \frac{a^{3} f - a^{2} b f}{a^{4} b^{3} - 2 \, a^{3} b^{4} + a^{2} b^{5}}}{256 \, \sqrt{b e^{\left (4 \, f x + 4 \, e\right )} + 4 \, a e^{\left (2 \, f x + 2 \, e\right )} - 2 \, b e^{\left (2 \, f x + 2 \, e\right )} + b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(f*x+e)/(a+b*sinh(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

-1/256*sqrt(b)*f/(a*b^4 - b^5) + 1/256*((a^3*f - a^2*b*f)*e^(2*f*x + 2*e)/(a^4*b^3 - 2*a^3*b^4 + a^2*b^5) + (a
^3*f - a^2*b*f)/(a^4*b^3 - 2*a^3*b^4 + a^2*b^5))/sqrt(b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x +
2*e) + b)