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

Optimal. Leaf size=241 \[ -\frac{i \sqrt{\frac{b \sinh ^2(e+f x)}{a}+1} \text{EllipticF}\left (i e+i f x,\frac{b}{a}\right )}{3 b f (a-b) \sqrt{a+b \sinh ^2(e+f x)}}+\frac{(a+b) \sinh (e+f x) \cosh (e+f x)}{3 a f (a-b)^2 \sqrt{a+b \sinh ^2(e+f x)}}+\frac{\sinh (e+f x) \cosh (e+f x)}{3 f (a-b) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac{i (a+b) \sqrt{a+b \sinh ^2(e+f x)} E\left (i e+i f x\left |\frac{b}{a}\right .\right )}{3 a b f (a-b)^2 \sqrt{\frac{b \sinh ^2(e+f x)}{a}+1}} \]

[Out]

(Cosh[e + f*x]*Sinh[e + f*x])/(3*(a - b)*f*(a + b*Sinh[e + f*x]^2)^(3/2)) + ((a + b)*Cosh[e + f*x]*Sinh[e + f*
x])/(3*a*(a - b)^2*f*Sqrt[a + b*Sinh[e + f*x]^2]) + ((I/3)*(a + b)*EllipticE[I*e + I*f*x, b/a]*Sqrt[a + b*Sinh
[e + f*x]^2])/(a*(a - b)^2*b*f*Sqrt[1 + (b*Sinh[e + f*x]^2)/a]) - ((I/3)*EllipticF[I*e + I*f*x, b/a]*Sqrt[1 +
(b*Sinh[e + f*x]^2)/a])/((a - b)*b*f*Sqrt[a + b*Sinh[e + f*x]^2])

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Rubi [A]  time = 0.325695, antiderivative size = 241, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3173, 3172, 3178, 3177, 3183, 3182} \[ \frac{(a+b) \sinh (e+f x) \cosh (e+f x)}{3 a f (a-b)^2 \sqrt{a+b \sinh ^2(e+f x)}}+\frac{\sinh (e+f x) \cosh (e+f x)}{3 f (a-b) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{i \sqrt{\frac{b \sinh ^2(e+f x)}{a}+1} F\left (i e+i f x\left |\frac{b}{a}\right .\right )}{3 b f (a-b) \sqrt{a+b \sinh ^2(e+f x)}}+\frac{i (a+b) \sqrt{a+b \sinh ^2(e+f x)} E\left (i e+i f x\left |\frac{b}{a}\right .\right )}{3 a b f (a-b)^2 \sqrt{\frac{b \sinh ^2(e+f x)}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cosh[e + f*x]*Sinh[e + f*x])/(3*(a - b)*f*(a + b*Sinh[e + f*x]^2)^(3/2)) + ((a + b)*Cosh[e + f*x]*Sinh[e + f*
x])/(3*a*(a - b)^2*f*Sqrt[a + b*Sinh[e + f*x]^2]) + ((I/3)*(a + b)*EllipticE[I*e + I*f*x, b/a]*Sqrt[a + b*Sinh
[e + f*x]^2])/(a*(a - b)^2*b*f*Sqrt[1 + (b*Sinh[e + f*x]^2)/a]) - ((I/3)*EllipticF[I*e + I*f*x, b/a]*Sqrt[1 +
(b*Sinh[e + f*x]^2)/a])/((a - b)*b*f*Sqrt[a + b*Sinh[e + f*x]^2])

Rule 3173

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Sim
p[((A*b - a*B)*Cos[e + f*x]*Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^(p + 1))/(2*a*f*(a + b)*(p + 1)), x] - Dist[1/
(2*a*(a + b)*(p + 1)), Int[(a + b*Sin[e + f*x]^2)^(p + 1)*Simp[a*B - A*(2*a*(p + 1) + b*(2*p + 3)) + 2*(A*b -
a*B)*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && LtQ[p, -1] && NeQ[a + b, 0]

Rule 3172

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Dist[
B/b, Int[Sqrt[a + b*Sin[e + f*x]^2], x], x] + Dist[(A*b - a*B)/b, Int[1/Sqrt[a + b*Sin[e + f*x]^2], x], x] /;
FreeQ[{a, b, e, f, A, B}, x]

Rule 3178

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Dist[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[1 + (b*Sin
[e + f*x]^2)/a], Int[Sqrt[1 + (b*Sin[e + f*x]^2)/a], x], x] /; FreeQ[{a, b, e, f}, x] &&  !GtQ[a, 0]

Rule 3177

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[e + f*x, -(b/a)])/f, x]
/; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]

Rule 3183

Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Dist[Sqrt[1 + (b*Sin[e + f*x]^2)/a]/Sqrt[a +
b*Sin[e + f*x]^2], Int[1/Sqrt[1 + (b*Sin[e + f*x]^2)/a], x], x] /; FreeQ[{a, b, e, f}, x] &&  !GtQ[a, 0]

Rule 3182

Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(1*EllipticF[e + f*x, -(b/a)])/(Sqrt[a]*
f), x] /; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]

Rubi steps

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

Mathematica [A]  time = 1.40341, size = 187, normalized size = 0.78 \[ \frac{-2 i a^2 (a-b) \left (\frac{2 a+b \cosh (2 (e+f x))-b}{a}\right )^{3/2} \text{EllipticF}\left (i (e+f x),\frac{b}{a}\right )+\sqrt{2} b \sinh (2 (e+f x)) \left (4 a^2+b (a+b) \cosh (2 (e+f x))-a b-b^2\right )+2 i a^2 (a+b) \left (\frac{2 a+b \cosh (2 (e+f x))-b}{a}\right )^{3/2} E\left (i (e+f x)\left |\frac{b}{a}\right .\right )}{6 a b f (a-b)^2 (2 a+b \cosh (2 (e+f x))-b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*I)*a^2*(a + b)*((2*a - b + b*Cosh[2*(e + f*x)])/a)^(3/2)*EllipticE[I*(e + f*x), b/a] - (2*I)*a^2*(a - b)*(
(2*a - b + b*Cosh[2*(e + f*x)])/a)^(3/2)*EllipticF[I*(e + f*x), b/a] + Sqrt[2]*b*(4*a^2 - a*b - b^2 + b*(a + b
)*Cosh[2*(e + f*x)])*Sinh[2*(e + f*x)])/(6*a*(a - b)^2*b*f*(2*a - b + b*Cosh[2*(e + f*x)])^(3/2))

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Maple [B]  time = 0.114, size = 598, normalized size = 2.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*((-(-1/a*b)^(1/2)*a*b-(-1/a*b)^(1/2)*b^2)*sinh(f*x+e)*cosh(f*x+e)^4+(-2*(-1/a*b)^(1/2)*a^2+(-1/a*b)^(1/2)
*a*b+(-1/a*b)^(1/2)*b^2)*cosh(f*x+e)^2*sinh(f*x+e)+(cosh(f*x+e)^2)^(1/2)*(b/a*cosh(f*x+e)^2+(a-b)/a)^(1/2)*b*(
a*EllipticF(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))-b*EllipticF(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))+Ellipt
icE(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))*a+b*EllipticE(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2)))*cosh(f*x+e)
^2+a^2*(b/a*cosh(f*x+e)^2+(a-b)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*EllipticF(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2
))-2*a*(b/a*cosh(f*x+e)^2+(a-b)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*EllipticF(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2
))*b+(b/a*cosh(f*x+e)^2+(a-b)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*EllipticF(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))
*b^2+(b/a*cosh(f*x+e)^2+(a-b)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*EllipticE(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))
*a^2-(b/a*cosh(f*x+e)^2+(a-b)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*EllipticE(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))
*b^2)/(-1/a*b)^(1/2)/(a+b*sinh(f*x+e)^2)^(3/2)/(a-b)^2/a/cosh(f*x+e)/f

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh \left (f x + e\right )^{2}}{{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b \sinh \left (f x + e\right )^{2} + a} \sinh \left (f x + e\right )^{2}}{b^{3} \sinh \left (f x + e\right )^{6} + 3 \, a b^{2} \sinh \left (f x + e\right )^{4} + 3 \, a^{2} b \sinh \left (f x + e\right )^{2} + a^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*sinh(f*x + e)^2 + a)*sinh(f*x + e)^2/(b^3*sinh(f*x + e)^6 + 3*a*b^2*sinh(f*x + e)^4 + 3*a^2*b*
sinh(f*x + e)^2 + a^3), x)

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh \left (f x + e\right )^{2}}{{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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