3.103 \(\int \frac{\sinh ^2(e+f x)}{\sqrt{a+b \sinh ^2(e+f x)}} \, dx\)

Optimal. Leaf size=128 \[ \frac{i a \sqrt{\frac{b \sinh ^2(e+f x)}{a}+1} \text{EllipticF}\left (i e+i f x,\frac{b}{a}\right )}{b f \sqrt{a+b \sinh ^2(e+f x)}}-\frac{i \sqrt{a+b \sinh ^2(e+f x)} E\left (i e+i f x\left |\frac{b}{a}\right .\right )}{b f \sqrt{\frac{b \sinh ^2(e+f x)}{a}+1}} \]

[Out]

((-I)*EllipticE[I*e + I*f*x, b/a]*Sqrt[a + b*Sinh[e + f*x]^2])/(b*f*Sqrt[1 + (b*Sinh[e + f*x]^2)/a]) + (I*a*El
lipticF[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 = 0.135813, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3172, 3178, 3177, 3183, 3182} \[ \frac{i a \sqrt{\frac{b \sinh ^2(e+f x)}{a}+1} F\left (i e+i f x\left |\frac{b}{a}\right .\right )}{b f \sqrt{a+b \sinh ^2(e+f x)}}-\frac{i \sqrt{a+b \sinh ^2(e+f x)} E\left (i e+i f x\left |\frac{b}{a}\right .\right )}{b f \sqrt{\frac{b \sinh ^2(e+f x)}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[e + f*x]^2/Sqrt[a + b*Sinh[e + f*x]^2],x]

[Out]

((-I)*EllipticE[I*e + I*f*x, b/a]*Sqrt[a + b*Sinh[e + f*x]^2])/(b*f*Sqrt[1 + (b*Sinh[e + f*x]^2)/a]) + (I*a*El
lipticF[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 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)}{\sqrt{a+b \sinh ^2(e+f x)}} \, dx &=\frac{\int \sqrt{a+b \sinh ^2(e+f x)} \, dx}{b}-\frac{a \int \frac{1}{\sqrt{a+b \sinh ^2(e+f x)}} \, dx}{b}\\ &=\frac{\sqrt{a+b \sinh ^2(e+f x)} \int \sqrt{1+\frac{b \sinh ^2(e+f x)}{a}} \, dx}{b \sqrt{1+\frac{b \sinh ^2(e+f x)}{a}}}-\frac{\left (a \sqrt{1+\frac{b \sinh ^2(e+f x)}{a}}\right ) \int \frac{1}{\sqrt{1+\frac{b \sinh ^2(e+f x)}{a}}} \, dx}{b \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)}}{b f \sqrt{1+\frac{b \sinh ^2(e+f x)}{a}}}+\frac{i a 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 = 0.256913, size = 89, normalized size = 0.7 \[ -\frac{i \sqrt{2 a+b \cosh (2 (e+f x))-b} \left (E\left (i (e+f x)\left |\frac{b}{a}\right .\right )-\text{EllipticF}\left (i (e+f x),\frac{b}{a}\right )\right )}{b f \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[e + f*x]^2/Sqrt[a + b*Sinh[e + f*x]^2],x]

[Out]

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

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Maple [A]  time = 0.066, size = 113, normalized size = 0.9 \begin{align*} -{\frac{1}{f\cosh \left ( fx+e \right ) }\sqrt{{\frac{a+b \left ( \sinh \left ( fx+e \right ) \right ) ^{2}}{a}}}\sqrt{ \left ( \cosh \left ( fx+e \right ) \right ) ^{2}} \left ({\it EllipticF} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ) -{\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)^(1/2),x)

[Out]

-1/(-1/a*b)^(1/2)*((a+b*sinh(f*x+e)^2)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*(EllipticF(sinh(f*x+e)*(-1/a*b)^(1/2),(a
/b)^(1/2))-EllipticE(sinh(f*x+e)*(-1/a*b)^(1/2),(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}}{\sqrt{b \sinh \left (f x + e\right )^{2} + a}}\,{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)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sinh \left (f x + e\right )^{2}}{\sqrt{b \sinh \left (f x + e\right )^{2} + a}}, 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)^(1/2),x, algorithm="fricas")

[Out]

integral(sinh(f*x + e)^2/sqrt(b*sinh(f*x + e)^2 + a), 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)**(1/2),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}}{\sqrt{b \sinh \left (f x + e\right )^{2} + a}}\,{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)^(1/2),x, algorithm="giac")

[Out]

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