3.393 \(\int \frac{\cosh (e+f x)}{(a+b \sinh ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=65 \[ \frac{2 \sinh (e+f x)}{3 a^2 f \sqrt{a+b \sinh ^2(e+f x)}}+\frac{\sinh (e+f x)}{3 a f \left (a+b \sinh ^2(e+f x)\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0579766, antiderivative size = 65, 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 = {3190, 192, 191} \[ \frac{2 \sinh (e+f x)}{3 a^2 f \sqrt{a+b \sinh ^2(e+f x)}}+\frac{\sinh (e+f x)}{3 a f \left (a+b \sinh ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[e + f*x]/(a + b*Sinh[e + f*x]^2)^(5/2),x]

[Out]

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

Rule 3190

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

Rule 192

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

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{\cosh (e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (a+b x^2\right )^{5/2}} \, dx,x,\sinh (e+f x)\right )}{f}\\ &=\frac{\sinh (e+f x)}{3 a f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac{2 \operatorname{Subst}\left (\int \frac{1}{\left (a+b x^2\right )^{3/2}} \, dx,x,\sinh (e+f x)\right )}{3 a f}\\ &=\frac{\sinh (e+f x)}{3 a f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac{2 \sinh (e+f x)}{3 a^2 f \sqrt{a+b \sinh ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.0449284, size = 47, normalized size = 0.72 \[ \frac{\sinh (e+f x) \left (3 a+2 b \sinh ^2(e+f x)\right )}{3 a^2 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[e + f*x]/(a + b*Sinh[e + f*x]^2)^(5/2),x]

[Out]

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

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Maple [A]  time = 0.011, size = 56, normalized size = 0.9 \begin{align*}{\frac{1}{f} \left ({\frac{\sinh \left ( fx+e \right ) }{3\,a} \left ( a+b \left ( \sinh \left ( fx+e \right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}}+{\frac{2\,\sinh \left ( fx+e \right ) }{3\,{a}^{2}}{\frac{1}{\sqrt{a+b \left ( \sinh \left ( fx+e \right ) \right ) ^{2}}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(f*x+e)/(a+b*sinh(f*x+e)^2)^(5/2),x)

[Out]

1/f*(1/3*sinh(f*x+e)/a/(a+b*sinh(f*x+e)^2)^(3/2)+2/3/a^2*sinh(f*x+e)/(a+b*sinh(f*x+e)^2)^(1/2))

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Maxima [B]  time = 1.6551, size = 655, normalized size = 10.08 \begin{align*} \frac{2 \, a^{2} b^{2} - 2 \, a b^{3} + b^{4} + 5 \,{\left (4 \, a^{3} b - 6 \, a^{2} b^{2} + 4 \, a b^{3} - b^{4}\right )} e^{\left (-2 \, f x - 2 \, e\right )} + 2 \,{\left (24 \, a^{4} - 48 \, a^{3} b + 49 \, a^{2} b^{2} - 25 \, a b^{3} + 5 \, b^{4}\right )} e^{\left (-4 \, f x - 4 \, e\right )} + 10 \,{\left (6 \, a^{3} b - 9 \, a^{2} b^{2} + 5 \, a b^{3} - b^{4}\right )} e^{\left (-6 \, f x - 6 \, e\right )} + 5 \,{\left (4 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} e^{\left (-8 \, f x - 8 \, e\right )} +{\left (2 \, a b^{3} - b^{4}\right )} e^{\left (-10 \, f x - 10 \, e\right )}}{3 \,{\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\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{5}{2}} f} - \frac{2 \, a b^{3} - b^{4} + 5 \,{\left (4 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} e^{\left (-2 \, f x - 2 \, e\right )} + 10 \,{\left (6 \, a^{3} b - 9 \, a^{2} b^{2} + 5 \, a b^{3} - b^{4}\right )} e^{\left (-4 \, f x - 4 \, e\right )} + 2 \,{\left (24 \, a^{4} - 48 \, a^{3} b + 49 \, a^{2} b^{2} - 25 \, a b^{3} + 5 \, b^{4}\right )} e^{\left (-6 \, f x - 6 \, e\right )} + 5 \,{\left (4 \, a^{3} b - 6 \, a^{2} b^{2} + 4 \, a b^{3} - b^{4}\right )} e^{\left (-8 \, f x - 8 \, e\right )} +{\left (2 \, a^{2} b^{2} - 2 \, a b^{3} + b^{4}\right )} e^{\left (-10 \, f x - 10 \, e\right )}}{3 \,{\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\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{5}{2}} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(f*x+e)/(a+b*sinh(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

1/3*(2*a^2*b^2 - 2*a*b^3 + b^4 + 5*(4*a^3*b - 6*a^2*b^2 + 4*a*b^3 - b^4)*e^(-2*f*x - 2*e) + 2*(24*a^4 - 48*a^3
*b + 49*a^2*b^2 - 25*a*b^3 + 5*b^4)*e^(-4*f*x - 4*e) + 10*(6*a^3*b - 9*a^2*b^2 + 5*a*b^3 - b^4)*e^(-6*f*x - 6*
e) + 5*(4*a^2*b^2 - 4*a*b^3 + b^4)*e^(-8*f*x - 8*e) + (2*a*b^3 - b^4)*e^(-10*f*x - 10*e))/((a^4 - 2*a^3*b + a^
2*b^2)*(2*(2*a - b)*e^(-2*f*x - 2*e) + b*e^(-4*f*x - 4*e) + b)^(5/2)*f) - 1/3*(2*a*b^3 - b^4 + 5*(4*a^2*b^2 -
4*a*b^3 + b^4)*e^(-2*f*x - 2*e) + 10*(6*a^3*b - 9*a^2*b^2 + 5*a*b^3 - b^4)*e^(-4*f*x - 4*e) + 2*(24*a^4 - 48*a
^3*b + 49*a^2*b^2 - 25*a*b^3 + 5*b^4)*e^(-6*f*x - 6*e) + 5*(4*a^3*b - 6*a^2*b^2 + 4*a*b^3 - b^4)*e^(-8*f*x - 8
*e) + (2*a^2*b^2 - 2*a*b^3 + b^4)*e^(-10*f*x - 10*e))/((a^4 - 2*a^3*b + a^2*b^2)*(2*(2*a - b)*e^(-2*f*x - 2*e)
+ b*e^(-4*f*x - 4*e) + b)^(5/2)*f)

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Fricas [B]  time = 3.09934, size = 2161, normalized size = 33.25 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(f*x+e)/(a+b*sinh(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

2/3*sqrt(2)*(b*cosh(f*x + e)^6 + 6*b*cosh(f*x + e)*sinh(f*x + e)^5 + b*sinh(f*x + e)^6 + 3*(2*a - b)*cosh(f*x
+ e)^4 + 3*(5*b*cosh(f*x + e)^2 + 2*a - b)*sinh(f*x + e)^4 + 4*(5*b*cosh(f*x + e)^3 + 3*(2*a - b)*cosh(f*x + e
))*sinh(f*x + e)^3 - 3*(2*a - b)*cosh(f*x + e)^2 + 3*(5*b*cosh(f*x + e)^4 + 6*(2*a - b)*cosh(f*x + e)^2 - 2*a
+ b)*sinh(f*x + e)^2 + 6*(b*cosh(f*x + e)^5 + 2*(2*a - b)*cosh(f*x + e)^3 - (2*a - b)*cosh(f*x + e))*sinh(f*x
+ e) - b)*sqrt((b*cosh(f*x + e)^2 + b*sinh(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^2*b^2*f*cosh(f*x + e)^8 + 8*a^2*b^2*f*cosh(f*x + e)*sinh(f*x + e)^7 + a^2*b^2*f*sin
h(f*x + e)^8 + 4*(2*a^3*b - a^2*b^2)*f*cosh(f*x + e)^6 + 4*(7*a^2*b^2*f*cosh(f*x + e)^2 + (2*a^3*b - a^2*b^2)*
f)*sinh(f*x + e)^6 + 2*(8*a^4 - 8*a^3*b + 3*a^2*b^2)*f*cosh(f*x + e)^4 + 8*(7*a^2*b^2*f*cosh(f*x + e)^3 + 3*(2
*a^3*b - a^2*b^2)*f*cosh(f*x + e))*sinh(f*x + e)^5 + a^2*b^2*f + 2*(35*a^2*b^2*f*cosh(f*x + e)^4 + 30*(2*a^3*b
- a^2*b^2)*f*cosh(f*x + e)^2 + (8*a^4 - 8*a^3*b + 3*a^2*b^2)*f)*sinh(f*x + e)^4 + 4*(2*a^3*b - a^2*b^2)*f*cos
h(f*x + e)^2 + 8*(7*a^2*b^2*f*cosh(f*x + e)^5 + 10*(2*a^3*b - a^2*b^2)*f*cosh(f*x + e)^3 + (8*a^4 - 8*a^3*b +
3*a^2*b^2)*f*cosh(f*x + e))*sinh(f*x + e)^3 + 4*(7*a^2*b^2*f*cosh(f*x + e)^6 + 15*(2*a^3*b - a^2*b^2)*f*cosh(f
*x + e)^4 + 3*(8*a^4 - 8*a^3*b + 3*a^2*b^2)*f*cosh(f*x + e)^2 + (2*a^3*b - a^2*b^2)*f)*sinh(f*x + e)^2 + 8*(a^
2*b^2*f*cosh(f*x + e)^7 + 3*(2*a^3*b - a^2*b^2)*f*cosh(f*x + e)^5 + (8*a^4 - 8*a^3*b + 3*a^2*b^2)*f*cosh(f*x +
e)^3 + (2*a^3*b - a^2*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(cosh(f*x+e)/(a+b*sinh(f*x+e)**2)**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 1.69479, size = 702, normalized size = 10.8 \begin{align*} \frac{2 \,{\left ({\left ({\left (\frac{{\left (a^{6} b^{3} f - 4 \, a^{5} b^{4} f + 6 \, a^{4} b^{5} f - 4 \, a^{3} b^{6} f + a^{2} b^{7} f\right )} e^{\left (2 \, f x + 2 \, e\right )}}{a^{8} b^{2} f^{2} - 4 \, a^{7} b^{3} f^{2} + 6 \, a^{6} b^{4} f^{2} - 4 \, a^{5} b^{5} f^{2} + a^{4} b^{6} f^{2}} + \frac{3 \,{\left (2 \, a^{7} b^{2} f - 9 \, a^{6} b^{3} f + 16 \, a^{5} b^{4} f - 14 \, a^{4} b^{5} f + 6 \, a^{3} b^{6} f - a^{2} b^{7} f\right )}}{a^{8} b^{2} f^{2} - 4 \, a^{7} b^{3} f^{2} + 6 \, a^{6} b^{4} f^{2} - 4 \, a^{5} b^{5} f^{2} + a^{4} b^{6} f^{2}}\right )} e^{\left (2 \, f x + 2 \, e\right )} - \frac{3 \,{\left (2 \, a^{7} b^{2} f - 9 \, a^{6} b^{3} f + 16 \, a^{5} b^{4} f - 14 \, a^{4} b^{5} f + 6 \, a^{3} b^{6} f - a^{2} b^{7} f\right )}}{a^{8} b^{2} f^{2} - 4 \, a^{7} b^{3} f^{2} + 6 \, a^{6} b^{4} f^{2} - 4 \, a^{5} b^{5} f^{2} + a^{4} b^{6} f^{2}}\right )} e^{\left (2 \, f x + 2 \, e\right )} - \frac{a^{6} b^{3} f - 4 \, a^{5} b^{4} f + 6 \, a^{4} b^{5} f - 4 \, a^{3} b^{6} f + a^{2} b^{7} f}{a^{8} b^{2} f^{2} - 4 \, a^{7} b^{3} f^{2} + 6 \, a^{6} b^{4} f^{2} - 4 \, a^{5} b^{5} f^{2} + a^{4} b^{6} f^{2}}\right )}}{3 \,{\left (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\right )}^{\frac{3}{2}}} + \frac{2}{3 \, a^{2} \sqrt{b} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(f*x+e)/(a+b*sinh(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

2/3*((((a^6*b^3*f - 4*a^5*b^4*f + 6*a^4*b^5*f - 4*a^3*b^6*f + a^2*b^7*f)*e^(2*f*x + 2*e)/(a^8*b^2*f^2 - 4*a^7*
b^3*f^2 + 6*a^6*b^4*f^2 - 4*a^5*b^5*f^2 + a^4*b^6*f^2) + 3*(2*a^7*b^2*f - 9*a^6*b^3*f + 16*a^5*b^4*f - 14*a^4*
b^5*f + 6*a^3*b^6*f - a^2*b^7*f)/(a^8*b^2*f^2 - 4*a^7*b^3*f^2 + 6*a^6*b^4*f^2 - 4*a^5*b^5*f^2 + a^4*b^6*f^2))*
e^(2*f*x + 2*e) - 3*(2*a^7*b^2*f - 9*a^6*b^3*f + 16*a^5*b^4*f - 14*a^4*b^5*f + 6*a^3*b^6*f - a^2*b^7*f)/(a^8*b
^2*f^2 - 4*a^7*b^3*f^2 + 6*a^6*b^4*f^2 - 4*a^5*b^5*f^2 + a^4*b^6*f^2))*e^(2*f*x + 2*e) - (a^6*b^3*f - 4*a^5*b^
4*f + 6*a^4*b^5*f - 4*a^3*b^6*f + a^2*b^7*f)/(a^8*b^2*f^2 - 4*a^7*b^3*f^2 + 6*a^6*b^4*f^2 - 4*a^5*b^5*f^2 + a^
4*b^6*f^2))/(b*e^(4*f*x + 4*e) + 4*a*e^(2*f*x + 2*e) - 2*b*e^(2*f*x + 2*e) + b)^(3/2) + 2/3/(a^2*sqrt(b)*f)