3.2.18 \(\int \frac {\sinh (e+f x)}{(a+b \sinh ^2(e+f x))^{5/2}} \, dx\) [118]

Optimal. Leaf size=79 \[ \frac {\cosh (e+f x)}{3 (a-b) f \left (a-b+b \cosh ^2(e+f x)\right )^{3/2}}+\frac {2 \cosh (e+f x)}{3 (a-b)^2 f \sqrt {a-b+b \cosh ^2(e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3265, 198, 197} \begin {gather*} \frac {2 \cosh (e+f x)}{3 f (a-b)^2 \sqrt {a+b \cosh ^2(e+f x)-b}}+\frac {\cosh (e+f x)}{3 f (a-b) \left (a+b \cosh ^2(e+f x)-b\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 197

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]

Rule 198

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 3265

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]

Rubi steps

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

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Mathematica [A]
time = 0.12, size = 63, normalized size = 0.80 \begin {gather*} \frac {2 \sqrt {2} \cosh (e+f x) (3 a-2 b+b \cosh (2 (e+f x)))}{3 (a-b)^2 f (2 a-b+b \cosh (2 (e+f x)))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 1.39, size = 57, normalized size = 0.72

method	result	size
default	\(\frac {\left (2 b \left (\sinh ^{2}\left (f x +e \right )\right )+3 a -b \right ) \cosh \left (f x +e \right )}{3 \left (a +b \left (\sinh ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}} \left (a^{2}-2 a b +b^{2}\right ) f}\)	\(57\)
risch	\(\text {Expression too large to display}\)	\(394373\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 499 vs. \(2 (75) = 150\).
time = 0.52, size = 499, normalized size = 6.32 \begin {gather*} \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(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] Leaf count of result is larger than twice the leaf count of optimal. 1186 vs. \(2 (71) = 142\).
time = 0.71, size = 1186, normalized size = 15.01 \begin {gather*} \frac {2 \, \sqrt {2} {\left (b \cosh \left (f x + e\right )^{6} + 6 \, b \cosh \left (f x + e\right ) \sinh \left (f x + e\right )^{5} + b \sinh \left (f x + e\right )^{6} + 3 \, {\left (2 \, a - b\right )} \cosh \left (f x + e\right )^{4} + 3 \, {\left (5 \, b \cosh \left (f x + e\right )^{2} + 2 \, a - b\right )} \sinh \left (f x + e\right )^{4} + 4 \, {\left (5 \, b \cosh \left (f x + e\right )^{3} + 3 \, {\left (2 \, a - b\right )} \cosh \left (f x + e\right )\right )} \sinh \left (f x + e\right )^{3} + 3 \, {\left (2 \, a - b\right )} \cosh \left (f x + e\right )^{2} + 3 \, {\left (5 \, b \cosh \left (f x + e\right )^{4} + 6 \, {\left (2 \, a - b\right )} \cosh \left (f x + e\right )^{2} + 2 \, a - b\right )} \sinh \left (f x + e\right )^{2} + 6 \, {\left (b \cosh \left (f x + e\right )^{5} + 2 \, {\left (2 \, a - b\right )} \cosh \left (f x + e\right )^{3} + {\left (2 \, a - b\right )} \cosh \left (f x + e\right )\right )} \sinh \left (f x + e\right ) + b\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}}}}{3 \, {\left ({\left (a^{2} b^{2} - 2 \, a b^{3} + b^{4}\right )} f \cosh \left (f x + e\right )^{8} + 8 \, {\left (a^{2} b^{2} - 2 \, a b^{3} + b^{4}\right )} f \cosh \left (f x + e\right ) \sinh \left (f x + e\right )^{7} + {\left (a^{2} b^{2} - 2 \, a b^{3} + b^{4}\right )} f \sinh \left (f x + e\right )^{8} + 4 \, {\left (2 \, a^{3} b - 5 \, a^{2} b^{2} + 4 \, a b^{3} - b^{4}\right )} f \cosh \left (f x + e\right )^{6} + 4 \, {\left (7 \, {\left (a^{2} b^{2} - 2 \, a b^{3} + b^{4}\right )} f \cosh \left (f x + e\right )^{2} + {\left (2 \, a^{3} b - 5 \, a^{2} b^{2} + 4 \, a b^{3} - b^{4}\right )} f\right )} \sinh \left (f x + e\right )^{6} + 2 \, {\left (8 \, a^{4} - 24 \, a^{3} b + 27 \, a^{2} b^{2} - 14 \, a b^{3} + 3 \, b^{4}\right )} f \cosh \left (f x + e\right )^{4} + 8 \, {\left (7 \, {\left (a^{2} b^{2} - 2 \, a b^{3} + b^{4}\right )} f \cosh \left (f x + e\right )^{3} + 3 \, {\left (2 \, a^{3} b - 5 \, a^{2} b^{2} + 4 \, a b^{3} - b^{4}\right )} f \cosh \left (f x + e\right )\right )} \sinh \left (f x + e\right )^{5} + 2 \, {\left (35 \, {\left (a^{2} b^{2} - 2 \, a b^{3} + b^{4}\right )} f \cosh \left (f x + e\right )^{4} + 30 \, {\left (2 \, a^{3} b - 5 \, a^{2} b^{2} + 4 \, a b^{3} - b^{4}\right )} f \cosh \left (f x + e\right )^{2} + {\left (8 \, a^{4} - 24 \, a^{3} b + 27 \, a^{2} b^{2} - 14 \, a b^{3} + 3 \, b^{4}\right )} f\right )} \sinh \left (f x + e\right )^{4} + 4 \, {\left (2 \, a^{3} b - 5 \, a^{2} b^{2} + 4 \, a b^{3} - b^{4}\right )} f \cosh \left (f x + e\right )^{2} + 8 \, {\left (7 \, {\left (a^{2} b^{2} - 2 \, a b^{3} + b^{4}\right )} f \cosh \left (f x + e\right )^{5} + 10 \, {\left (2 \, a^{3} b - 5 \, a^{2} b^{2} + 4 \, a b^{3} - b^{4}\right )} f \cosh \left (f x + e\right )^{3} + {\left (8 \, a^{4} - 24 \, a^{3} b + 27 \, a^{2} b^{2} - 14 \, a b^{3} + 3 \, b^{4}\right )} f \cosh \left (f x + e\right )\right )} \sinh \left (f x + e\right )^{3} + 4 \, {\left (7 \, {\left (a^{2} b^{2} - 2 \, a b^{3} + b^{4}\right )} f \cosh \left (f x + e\right )^{6} + 15 \, {\left (2 \, a^{3} b - 5 \, a^{2} b^{2} + 4 \, a b^{3} - b^{4}\right )} f \cosh \left (f x + e\right )^{4} + 3 \, {\left (8 \, a^{4} - 24 \, a^{3} b + 27 \, a^{2} b^{2} - 14 \, a b^{3} + 3 \, b^{4}\right )} f \cosh \left (f x + e\right )^{2} + {\left (2 \, a^{3} b - 5 \, a^{2} b^{2} + 4 \, a b^{3} - b^{4}\right )} f\right )} \sinh \left (f x + e\right )^{2} + {\left (a^{2} b^{2} - 2 \, a b^{3} + b^{4}\right )} f + 8 \, {\left ({\left (a^{2} b^{2} - 2 \, a b^{3} + b^{4}\right )} f \cosh \left (f x + e\right )^{7} + 3 \, {\left (2 \, a^{3} b - 5 \, a^{2} b^{2} + 4 \, a b^{3} - b^{4}\right )} f \cosh \left (f x + e\right )^{5} + {\left (8 \, a^{4} - 24 \, a^{3} b + 27 \, a^{2} b^{2} - 14 \, a b^{3} + 3 \, b^{4}\right )} f \cosh \left (f x + e\right )^{3} + {\left (2 \, a^{3} b - 5 \, a^{2} b^{2} + 4 \, a b^{3} - b^{4}\right )} f \cosh \left (f x + e\right )\right )} \sinh \left (f x + e\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(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 - 2*a*b^3 + b^4)*f*cosh(f*x + e)^8 + 8*(a^2*b^2 - 2*a*b^3 + b^4)*f*cosh(f*x
+ e)*sinh(f*x + e)^7 + (a^2*b^2 - 2*a*b^3 + b^4)*f*sinh(f*x + e)^8 + 4*(2*a^3*b - 5*a^2*b^2 + 4*a*b^3 - b^4)*f
*cosh(f*x + e)^6 + 4*(7*(a^2*b^2 - 2*a*b^3 + b^4)*f*cosh(f*x + e)^2 + (2*a^3*b - 5*a^2*b^2 + 4*a*b^3 - b^4)*f)
*sinh(f*x + e)^6 + 2*(8*a^4 - 24*a^3*b + 27*a^2*b^2 - 14*a*b^3 + 3*b^4)*f*cosh(f*x + e)^4 + 8*(7*(a^2*b^2 - 2*
a*b^3 + b^4)*f*cosh(f*x + e)^3 + 3*(2*a^3*b - 5*a^2*b^2 + 4*a*b^3 - b^4)*f*cosh(f*x + e))*sinh(f*x + e)^5 + 2*
(35*(a^2*b^2 - 2*a*b^3 + b^4)*f*cosh(f*x + e)^4 + 30*(2*a^3*b - 5*a^2*b^2 + 4*a*b^3 - b^4)*f*cosh(f*x + e)^2 +
(8*a^4 - 24*a^3*b + 27*a^2*b^2 - 14*a*b^3 + 3*b^4)*f)*sinh(f*x + e)^4 + 4*(2*a^3*b - 5*a^2*b^2 + 4*a*b^3 - b^
4)*f*cosh(f*x + e)^2 + 8*(7*(a^2*b^2 - 2*a*b^3 + b^4)*f*cosh(f*x + e)^5 + 10*(2*a^3*b - 5*a^2*b^2 + 4*a*b^3 -
b^4)*f*cosh(f*x + e)^3 + (8*a^4 - 24*a^3*b + 27*a^2*b^2 - 14*a*b^3 + 3*b^4)*f*cosh(f*x + e))*sinh(f*x + e)^3 +
4*(7*(a^2*b^2 - 2*a*b^3 + b^4)*f*cosh(f*x + e)^6 + 15*(2*a^3*b - 5*a^2*b^2 + 4*a*b^3 - b^4)*f*cosh(f*x + e)^4
+ 3*(8*a^4 - 24*a^3*b + 27*a^2*b^2 - 14*a*b^3 + 3*b^4)*f*cosh(f*x + e)^2 + (2*a^3*b - 5*a^2*b^2 + 4*a*b^3 - b
^4)*f)*sinh(f*x + e)^2 + (a^2*b^2 - 2*a*b^3 + b^4)*f + 8*((a^2*b^2 - 2*a*b^3 + b^4)*f*cosh(f*x + e)^7 + 3*(2*a
^3*b - 5*a^2*b^2 + 4*a*b^3 - b^4)*f*cosh(f*x + e)^5 + (8*a^4 - 24*a^3*b + 27*a^2*b^2 - 14*a*b^3 + 3*b^4)*f*cos
h(f*x + e)^3 + (2*a^3*b - 5*a^2*b^2 + 4*a*b^3 - b^4)*f*cosh(f*x + e))*sinh(f*x + e))

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(f*x+e)/(a+b*sinh(f*x+e)**2)**(5/2),x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 254 vs. \(2 (71) = 142\).
time = 0.89, size = 254, normalized size = 3.22 \begin {gather*} \frac {2 \, {\left (\frac {a^{2} b e^{\left (6 \, e\right )}}{a^{4} e^{\left (6 \, e\right )} - 2 \, a^{3} b e^{\left (6 \, e\right )} + a^{2} b^{2} e^{\left (6 \, e\right )}} + {\left ({\left (\frac {a^{2} b e^{\left (2 \, f x + 12 \, e\right )}}{a^{4} e^{\left (6 \, e\right )} - 2 \, a^{3} b e^{\left (6 \, e\right )} + a^{2} b^{2} e^{\left (6 \, e\right )}} + \frac {3 \, {\left (2 \, a^{3} e^{\left (10 \, e\right )} - a^{2} b e^{\left (10 \, e\right )}\right )}}{a^{4} e^{\left (6 \, e\right )} - 2 \, a^{3} b e^{\left (6 \, e\right )} + a^{2} b^{2} e^{\left (6 \, e\right )}}\right )} e^{\left (2 \, f x\right )} + \frac {3 \, {\left (2 \, a^{3} e^{\left (8 \, e\right )} - a^{2} b e^{\left (8 \, e\right )}\right )}}{a^{4} e^{\left (6 \, e\right )} - 2 \, a^{3} b e^{\left (6 \, e\right )} + a^{2} b^{2} e^{\left (6 \, e\right )}}\right )} e^{\left (2 \, f x\right )}\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}} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(a^2*b*e^(6*e)/(a^4*e^(6*e) - 2*a^3*b*e^(6*e) + a^2*b^2*e^(6*e)) + ((a^2*b*e^(2*f*x + 12*e)/(a^4*e^(6*e) -
2*a^3*b*e^(6*e) + a^2*b^2*e^(6*e)) + 3*(2*a^3*e^(10*e) - a^2*b*e^(10*e))/(a^4*e^(6*e) - 2*a^3*b*e^(6*e) + a^2
*b^2*e^(6*e)))*e^(2*f*x) + 3*(2*a^3*e^(8*e) - a^2*b*e^(8*e))/(a^4*e^(6*e) - 2*a^3*b*e^(6*e) + a^2*b^2*e^(6*e))
)*e^(2*f*x))/((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)*f)

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Mupad [B]
time = 1.29, size = 133, normalized size = 1.68 \begin {gather*} \frac {4\,{\mathrm {e}}^{e+f\,x}\,\left ({\mathrm {e}}^{2\,e+2\,f\,x}+1\right )\,\sqrt {a+b\,{\left (\frac {{\mathrm {e}}^{e+f\,x}}{2}-\frac {{\mathrm {e}}^{-e-f\,x}}{2}\right )}^2}\,\left (b+6\,a\,{\mathrm {e}}^{2\,e+2\,f\,x}-4\,b\,{\mathrm {e}}^{2\,e+2\,f\,x}+b\,{\mathrm {e}}^{4\,e+4\,f\,x}\right )}{3\,f\,{\left (a-b\right )}^2\,{\left (b+4\,a\,{\mathrm {e}}^{2\,e+2\,f\,x}-2\,b\,{\mathrm {e}}^{2\,e+2\,f\,x}+b\,{\mathrm {e}}^{4\,e+4\,f\,x}\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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