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

Optimal. Leaf size=173 \[ -\frac{i \sqrt{\frac{b \sinh ^2(e+f x)}{a}+1} \text{EllipticF}\left (i e+i f x,\frac{b}{a}\right )}{b f \sqrt{a+b \sinh ^2(e+f x)}}+\frac{\sinh (e+f x) \cosh (e+f x)}{f (a-b) \sqrt{a+b \sinh ^2(e+f x)}}+\frac{i \sqrt{a+b \sinh ^2(e+f x)} E\left (i e+i f x\left |\frac{b}{a}\right .\right )}{b f (a-b) \sqrt{\frac{b \sinh ^2(e+f x)}{a}+1}} \]

[Out]

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

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Rubi [A]  time = 0.212076, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{\sinh (e+f x) \cosh (e+f x)}{f (a-b) \sqrt{a+b \sinh ^2(e+f x)}}-\frac{i \sqrt{\frac{b \sinh ^2(e+f x)}{a}+1} F\left (i e+i f x\left |\frac{b}{a}\right .\right )}{b f \sqrt{a+b \sinh ^2(e+f x)}}+\frac{i \sqrt{a+b \sinh ^2(e+f x)} E\left (i e+i f x\left |\frac{b}{a}\right .\right )}{b f (a-b) \sqrt{\frac{b \sinh ^2(e+f x)}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cosh[e + f*x]*Sinh[e + f*x])/((a - b)*f*Sqrt[a + b*Sinh[e + f*x]^2]) + (I*EllipticE[I*e + I*f*x, b/a]*Sqrt[a
+ b*Sinh[e + f*x]^2])/((a - b)*b*f*Sqrt[1 + (b*Sinh[e + f*x]^2)/a]) - (I*EllipticF[I*e + I*f*x, b/a]*Sqrt[1 +
(b*Sinh[e + f*x]^2)/a])/(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 )^{3/2}} \, dx &=\frac{\cosh (e+f x) \sinh (e+f x)}{(a-b) f \sqrt{a+b \sinh ^2(e+f x)}}-\frac{\int \frac{a+a \sinh ^2(e+f x)}{\sqrt{a+b \sinh ^2(e+f x)}} \, dx}{a (a-b)}\\ &=\frac{\cosh (e+f x) \sinh (e+f x)}{(a-b) f \sqrt{a+b \sinh ^2(e+f x)}}+\frac{\int \frac{1}{\sqrt{a+b \sinh ^2(e+f x)}} \, dx}{b}-\frac{\int \sqrt{a+b \sinh ^2(e+f x)} \, dx}{(a-b) b}\\ &=\frac{\cosh (e+f x) \sinh (e+f x)}{(a-b) f \sqrt{a+b \sinh ^2(e+f x)}}-\frac{\sqrt{a+b \sinh ^2(e+f x)} \int \sqrt{1+\frac{b \sinh ^2(e+f x)}{a}} \, dx}{(a-b) 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}{b \sqrt{a+b \sinh ^2(e+f x)}}\\ &=\frac{\cosh (e+f x) \sinh (e+f x)}{(a-b) f \sqrt{a+b \sinh ^2(e+f x)}}+\frac{i E\left (i e+i f x\left |\frac{b}{a}\right .\right ) \sqrt{a+b \sinh ^2(e+f x)}}{(a-b) 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}}}{b f \sqrt{a+b \sinh ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.481106, size = 151, normalized size = 0.87 \[ \frac{-i \sqrt{2} (a-b) \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} \text{EllipticF}\left (i (e+f x),\frac{b}{a}\right )+i \sqrt{2} a \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} E\left (i (e+f x)\left |\frac{b}{a}\right .\right )+b \sinh (2 (e+f x))}{b f (a-b) \sqrt{4 a+2 b \cosh (2 (e+f x))-2 b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.084, size = 127, normalized size = 0.7 \begin{align*} -{\frac{1}{ \left ( a-b \right ) \cosh \left ( fx+e \right ) f} \left ( -\sqrt{-{\frac{b}{a}}}\sinh \left ( fx+e \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{2}+\sqrt{{\frac{b \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}{a}}+{\frac{a-b}{a}}}\sqrt{ \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}{\it EllipticE} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ) \right ){\frac{1}{\sqrt{-{\frac{b}{a}}}}}{\frac{1}{\sqrt{a+b \left ( \sinh \left ( fx+e \right ) \right ) ^{2}}}}} \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)^(3/2),x)

[Out]

-(-(-1/a*b)^(1/2)*sinh(f*x+e)*cosh(f*x+e)^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-b)/(-1/a*b)^(1/2)/cosh(f*x+e)/(a+b*sinh(f*x+e)^2)^(1/2)/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{3}{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)^(3/2),x, algorithm="maxima")

[Out]

integrate(sinh(f*x + e)^2/(b*sinh(f*x + e)^2 + a)^(3/2), 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^{2} \sinh \left (f x + e\right )^{4} + 2 \, a b \sinh \left (f x + e\right )^{2} + a^{2}}, 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)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sinh(f*x + e)^2 + a)*sinh(f*x + e)^2/(b^2*sinh(f*x + e)^4 + 2*a*b*sinh(f*x + e)^2 + a^2), 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)**(3/2),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{3}{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)^(3/2),x, algorithm="giac")

[Out]

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