3.117 \(\int \frac{\sinh ^3(e+f x)}{(a+b \sinh ^2(e+f x))^{5/2}} \, dx\)
Optimal. Leaf size=87 \[ \frac{\sinh ^2(e+f x) \cosh (e+f x)}{3 f (a-b) \left (a+b \cosh ^2(e+f x)-b\right )^{3/2}}-\frac{2 \cosh (e+f x)}{3 f (a-b)^2 \sqrt{a+b \cosh ^2(e+f x)-b}} \]
[Out]
(-2*Cosh[e + f*x])/(3*(a - b)^2*f*Sqrt[a - b + b*Cosh[e + f*x]^2]) + (Cosh[e + f*x]*Sinh[e + f*x]^2)/(3*(a - b
)*f*(a - b + b*Cosh[e + f*x]^2)^(3/2))
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Rubi [A] time = 0.10722, antiderivative size = 87, normalized size of antiderivative = 1.,
number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used =
{3186, 378, 191} \[ \frac{\sinh ^2(e+f x) \cosh (e+f x)}{3 f (a-b) \left (a+b \cosh ^2(e+f x)-b\right )^{3/2}}-\frac{2 \cosh (e+f x)}{3 f (a-b)^2 \sqrt{a+b \cosh ^2(e+f x)-b}} \]
Antiderivative was successfully verified.
[In]
Int[Sinh[e + f*x]^3/(a + b*Sinh[e + f*x]^2)^(5/2),x]
[Out]
(-2*Cosh[e + f*x])/(3*(a - b)^2*f*Sqrt[a - b + b*Cosh[e + f*x]^2]) + (Cosh[e + f*x]*Sinh[e + f*x]^2)/(3*(a - b
)*f*(a - b + b*Cosh[e + f*x]^2)^(3/2))
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 378
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^q)/(a*n*(p + 1)), x] - Dist[(c*q)/(a*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[n*(p + q + 1) + 1, 0] && GtQ[q, 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{\sinh ^3(e+f x)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{1-x^2}{\left (a-b+b x^2\right )^{5/2}} \, dx,x,\cosh (e+f x)\right )}{f}\\ &=\frac{\cosh (e+f x) \sinh ^2(e+f x)}{3 (a-b) f \left (a-b+b \cosh ^2(e+f x)\right )^{3/2}}-\frac{2 \operatorname{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{2 \cosh (e+f x)}{3 (a-b)^2 f \sqrt{a-b+b \cosh ^2(e+f x)}}+\frac{\cosh (e+f x) \sinh ^2(e+f x)}{3 (a-b) f \left (a-b+b \cosh ^2(e+f x)\right )^{3/2}}\\ \end{align*}
Mathematica [A] time = 0.328655, size = 67, normalized size = 0.77 \[ \frac{\sqrt{2} \cosh (e+f x) ((a-3 b) \cosh (2 (e+f x))-5 a+3 b)}{3 f (a-b)^2 (2 a+b \cosh (2 (e+f x))-b)^{3/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[Sinh[e + f*x]^3/(a + b*Sinh[e + f*x]^2)^(5/2),x]
[Out]
(Sqrt[2]*Cosh[e + f*x]*(-5*a + 3*b + (a - 3*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 = 0.077, size = 58, normalized size = 0.7 \begin{align*}{\frac{ \left ( a \left ( \sinh \left ( fx+e \right ) \right ) ^{2}-3\,b \left ( \sinh \left ( fx+e \right ) \right ) ^{2}-2\,a \right ) \cosh \left ( fx+e \right ) }{3\, \left ( a-b \right ) ^{2}f} \left ( a+b \left ( \sinh \left ( fx+e \right ) \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sinh(f*x+e)^3/(a+b*sinh(f*x+e)^2)^(5/2),x)
[Out]
1/3*(a*sinh(f*x+e)^2-3*b*sinh(f*x+e)^2-2*a)*cosh(f*x+e)/(a-b)^2/(a+b*sinh(f*x+e)^2)^(3/2)/f
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Maxima [B] time = 1.77411, size = 1251, normalized size = 14.38 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(f*x+e)^3/(a+b*sinh(f*x+e)^2)^(5/2),x, algorithm="maxima")
[Out]
-1/12*(b^4*e^(-10*f*x - 10*e) - 4*a^3*b + 6*a^2*b^2 - b^4 - (16*a^4 - 32*a^3*b + 6*a^2*b^2 + 10*a*b^3 - 5*b^4)
*e^(-2*f*x - 2*e) + 10*(2*a^3*b - 3*a^2*b^2 + 3*a*b^3 - b^4)*e^(-4*f*x - 4*e) + 10*(3*a^2*b^2 - 3*a*b^3 + b^4)
*e^(-6*f*x - 6*e) + 5*(2*a*b^3 - b^4)*e^(-8*f*x - 8*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) - 1/4*(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
) - 1/4*(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) - 1/12*(b^4 + 5*(2*a*b^3 - b^4)*e^
(-2*f*x - 2*e) + 10*(3*a^2*b^2 - 3*a*b^3 + b^4)*e^(-4*f*x - 4*e) + 10*(2*a^3*b - 3*a^2*b^2 + 3*a*b^3 - b^4)*e^
(-6*f*x - 6*e) - (16*a^4 - 32*a^3*b + 6*a^2*b^2 + 10*a*b^3 - 5*b^4)*e^(-8*f*x - 8*e) - (4*a^3*b - 6*a^2*b^2 +
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 = 3.41105, size = 2873, normalized size = 33.02 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(f*x+e)^3/(a+b*sinh(f*x+e)^2)^(5/2),x, algorithm="fricas")
[Out]
1/3*sqrt(2)*((a - 3*b)*cosh(f*x + e)^6 + 6*(a - 3*b)*cosh(f*x + e)*sinh(f*x + e)^5 + (a - 3*b)*sinh(f*x + e)^6
- 3*(3*a - b)*cosh(f*x + e)^4 + 3*(5*(a - 3*b)*cosh(f*x + e)^2 - 3*a + b)*sinh(f*x + e)^4 + 4*(5*(a - 3*b)*co
sh(f*x + e)^3 - 3*(3*a - b)*cosh(f*x + e))*sinh(f*x + e)^3 - 3*(3*a - b)*cosh(f*x + e)^2 + 3*(5*(a - 3*b)*cosh
(f*x + e)^4 - 6*(3*a - b)*cosh(f*x + e)^2 - 3*a + b)*sinh(f*x + e)^2 + 6*((a - 3*b)*cosh(f*x + e)^5 - 2*(3*a -
b)*cosh(f*x + e)^3 - (3*a - b)*cosh(f*x + e))*sinh(f*x + e) + a - 3*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*cosh(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)] 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)**3/(a+b*sinh(f*x+e)**2)**(5/2),x)
[Out]
Timed out
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Giac [B] time = 1.33895, size = 699, normalized size = 8.03 \begin{align*} -\frac{a - 3 \, b}{3 \,{\left (a^{2} b^{\frac{3}{2}} f - 2 \, a b^{\frac{5}{2}} f + b^{\frac{7}{2}} f\right )}} + \frac{{\left ({\left (\frac{{\left (a^{7} b^{2} f^{3} - 5 \, a^{6} b^{3} f^{3} + 7 \, a^{5} b^{4} f^{3} - 3 \, a^{4} b^{5} f^{3}\right )} e^{\left (2 \, f x + 2 \, e\right )}}{a^{8} b^{2} f^{4} - 4 \, a^{7} b^{3} f^{4} + 6 \, a^{6} b^{4} f^{4} - 4 \, a^{5} b^{5} f^{4} + a^{4} b^{6} f^{4}} - \frac{3 \,{\left (3 \, a^{7} b^{2} f^{3} - 7 \, a^{6} b^{3} f^{3} + 5 \, a^{5} b^{4} f^{3} - a^{4} b^{5} f^{3}\right )}}{a^{8} b^{2} f^{4} - 4 \, a^{7} b^{3} f^{4} + 6 \, a^{6} b^{4} f^{4} - 4 \, a^{5} b^{5} f^{4} + a^{4} b^{6} f^{4}}\right )} e^{\left (2 \, f x + 2 \, e\right )} - \frac{3 \,{\left (3 \, a^{7} b^{2} f^{3} - 7 \, a^{6} b^{3} f^{3} + 5 \, a^{5} b^{4} f^{3} - a^{4} b^{5} f^{3}\right )}}{a^{8} b^{2} f^{4} - 4 \, a^{7} b^{3} f^{4} + 6 \, a^{6} b^{4} f^{4} - 4 \, a^{5} b^{5} f^{4} + a^{4} b^{6} f^{4}}\right )} e^{\left (2 \, f x + 2 \, e\right )} + \frac{a^{7} b^{2} f^{3} - 5 \, a^{6} b^{3} f^{3} + 7 \, a^{5} b^{4} f^{3} - 3 \, a^{4} b^{5} f^{3}}{a^{8} b^{2} f^{4} - 4 \, a^{7} b^{3} f^{4} + 6 \, a^{6} b^{4} f^{4} - 4 \, a^{5} b^{5} f^{4} + a^{4} b^{6} f^{4}}}{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}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sinh(f*x+e)^3/(a+b*sinh(f*x+e)^2)^(5/2),x, algorithm="giac")
[Out]
-1/3*(a - 3*b)/(a^2*b^(3/2)*f - 2*a*b^(5/2)*f + b^(7/2)*f) + 1/3*((((a^7*b^2*f^3 - 5*a^6*b^3*f^3 + 7*a^5*b^4*f
^3 - 3*a^4*b^5*f^3)*e^(2*f*x + 2*e)/(a^8*b^2*f^4 - 4*a^7*b^3*f^4 + 6*a^6*b^4*f^4 - 4*a^5*b^5*f^4 + a^4*b^6*f^4
) - 3*(3*a^7*b^2*f^3 - 7*a^6*b^3*f^3 + 5*a^5*b^4*f^3 - a^4*b^5*f^3)/(a^8*b^2*f^4 - 4*a^7*b^3*f^4 + 6*a^6*b^4*f
^4 - 4*a^5*b^5*f^4 + a^4*b^6*f^4))*e^(2*f*x + 2*e) - 3*(3*a^7*b^2*f^3 - 7*a^6*b^3*f^3 + 5*a^5*b^4*f^3 - a^4*b^
5*f^3)/(a^8*b^2*f^4 - 4*a^7*b^3*f^4 + 6*a^6*b^4*f^4 - 4*a^5*b^5*f^4 + a^4*b^6*f^4))*e^(2*f*x + 2*e) + (a^7*b^2
*f^3 - 5*a^6*b^3*f^3 + 7*a^5*b^4*f^3 - 3*a^4*b^5*f^3)/(a^8*b^2*f^4 - 4*a^7*b^3*f^4 + 6*a^6*b^4*f^4 - 4*a^5*b^5
*f^4 + a^4*b^6*f^4))/(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)