3.392 \(\int \frac {\cosh ^3(e+f x)}{(a+b \sinh ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=73 \[ \frac {2 \sinh (e+f x)}{3 a^2 f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {\sinh (e+f x) \cosh ^2(e+f x)}{3 a f \left (a+b \sinh ^2(e+f x)\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {3190, 378, 191} \[ \frac {2 \sinh (e+f x)}{3 a^2 f \sqrt {a+b \sinh ^2(e+f x)}}+\frac {\sinh (e+f x) \cosh ^2(e+f x)}{3 a f \left (a+b \sinh ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[e + f*x]^3/(a + b*Sinh[e + f*x]^2)^(5/2),x]

[Out]

(Cosh[e + f*x]^2*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 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]

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 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]

Rubi steps

\begin {align*} \int \frac {\cosh ^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 x^2\right )^{5/2}} \, dx,x,\sinh (e+f x)\right )}{f}\\ &=\frac {\cosh ^2(e+f x) \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 {\cosh ^2(e+f x) \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*}

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Mathematica [A]  time = 0.10, size = 50, normalized size = 0.68 \[ \frac {(a+2 b) \sinh ^3(e+f x)+3 a \sinh (e+f x)}{3 a^2 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[e + f*x]^3/(a + b*Sinh[e + f*x]^2)^(5/2),x]

[Out]

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

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fricas [B]  time = 0.99, size = 945, normalized size = 12.95 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(f*x+e)^3/(a+b*sinh(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

1/3*sqrt(2)*((a + 2*b)*cosh(f*x + e)^6 + 6*(a + 2*b)*cosh(f*x + e)*sinh(f*x + e)^5 + (a + 2*b)*sinh(f*x + e)^6
+ 3*(3*a - 2*b)*cosh(f*x + e)^4 + 3*(5*(a + 2*b)*cosh(f*x + e)^2 + 3*a - 2*b)*sinh(f*x + e)^4 + 4*(5*(a + 2*b
)*cosh(f*x + e)^3 + 3*(3*a - 2*b)*cosh(f*x + e))*sinh(f*x + e)^3 - 3*(3*a - 2*b)*cosh(f*x + e)^2 + 3*(5*(a + 2
*b)*cosh(f*x + e)^4 + 6*(3*a - 2*b)*cosh(f*x + e)^2 - 3*a + 2*b)*sinh(f*x + e)^2 + 6*((a + 2*b)*cosh(f*x + e)^
5 + 2*(3*a - 2*b)*cosh(f*x + e)^3 - (3*a - 2*b)*cosh(f*x + e))*sinh(f*x + e) - a - 2*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*sinh(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*cosh(f*x + e)^2 + 8*(7*a^2*b^2*f*co
sh(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*co
sh(f*x + e))*sinh(f*x + e))

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(f*x+e)^3/(a+b*sinh(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Eval
uation time: 0.43Error: Bad Argument Type

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maple [C]  time = 0.14, size = 65, normalized size = 0.89 \[ \frac {\mathit {`\,int/indef0`\,}\left (\frac {\cosh ^{2}\left (f x +e \right )}{\left (b^{2} \left (\sinh ^{4}\left (f x +e \right )\right )+2 a b \left (\sinh ^{2}\left (f x +e \right )\right )+a^{2}\right ) \sqrt {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}}, \sinh \left (f x +e \right )\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(f*x+e)^3/(a+b*sinh(f*x+e)^2)^(5/2),x)

[Out]

`int/indef0`(cosh(f*x+e)^2/(b^2*sinh(f*x+e)^4+2*a*b*sinh(f*x+e)^2+a^2)/(a+b*sinh(f*x+e)^2)^(1/2),sinh(f*x+e))/
f

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maxima [B]  time = 0.52, size = 927, normalized size = 12.70 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(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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mupad [B]  time = 1.83, size = 144, normalized size = 1.97 \[ \frac {2\,{\mathrm {e}}^{e+f\,x}\,\left ({\mathrm {e}}^{2\,e+2\,f\,x}-1\right )\,\sqrt {a+b\,{\left (\frac {{\mathrm {e}}^{e+f\,x}}{2}-\frac {{\mathrm {e}}^{-e-f\,x}}{2}\right )}^2}\,\left (a+2\,b+10\,a\,{\mathrm {e}}^{2\,e+2\,f\,x}+a\,{\mathrm {e}}^{4\,e+4\,f\,x}-4\,b\,{\mathrm {e}}^{2\,e+2\,f\,x}+2\,b\,{\mathrm {e}}^{4\,e+4\,f\,x}\right )}{3\,a^2\,f\,{\left (b+4\,a\,{\mathrm {e}}^{2\,e+2\,f\,x}-2\,b\,{\mathrm {e}}^{2\,e+2\,f\,x}+b\,{\mathrm {e}}^{4\,e+4\,f\,x}\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(e + f*x)^3/(a + b*sinh(e + f*x)^2)^(5/2),x)

[Out]

(2*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)*(a + 2*b + 10*a*exp
(2*e + 2*f*x) + a*exp(4*e + 4*f*x) - 4*b*exp(2*e + 2*f*x) + 2*b*exp(4*e + 4*f*x)))/(3*a^2*f*(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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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(f*x+e)**3/(a+b*sinh(f*x+e)**2)**(5/2),x)

[Out]

Timed out

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