3.2.3 \(\int \frac {\sinh ^2(e+f x)}{\sqrt {a+b \sinh ^2(e+f x)}} \, dx\) [103]

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

[Out]

-I*(cos(I*e+I*f*x)^2)^(1/2)/cos(I*e+I*f*x)*EllipticE(sin(I*e+I*f*x),(b/a)^(1/2))*(a+b*sinh(f*x+e)^2)^(1/2)/b/f
/(1+b*sinh(f*x+e)^2/a)^(1/2)+I*a*(cos(I*e+I*f*x)^2)^(1/2)/cos(I*e+I*f*x)*EllipticF(sin(I*e+I*f*x),(b/a)^(1/2))
*(1+b*sinh(f*x+e)^2/a)^(1/2)/b/f/(a+b*sinh(f*x+e)^2)^(1/2)

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Rubi [A]
time = 0.10, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3251, 3257, 3256, 3262, 3261} \begin {gather*} \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}} \end {gather*}

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 3251

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 3256

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

Rule 3257

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 3261

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

Rule 3262

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]

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*}

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Mathematica [A]
time = 0.19, size = 89, normalized size = 0.70 \begin {gather*} -\frac {i \sqrt {2 a-b+b \cosh (2 (e+f x))} \left (E\left (i (e+f x)\left |\frac {b}{a}\right .\right )-F\left (i (e+f x)\left |\frac {b}{a}\right .\right )\right )}{b f \sqrt {\frac {2 a-b+b \cosh (2 (e+f x))}{a}}} \end {gather*}

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 = 0.90, size = 113, normalized size = 0.88

method	result	size
default	\(-\frac {\sqrt {\frac {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}{a}}\, \sqrt {\frac {\cosh \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \left (\EllipticF \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right )-\EllipticE \left (\sinh \left (f x +e \right ) \sqrt {-\frac {b}{a}}, \sqrt {\frac {a}{b}}\right )\right )}{\sqrt {-\frac {b}{a}}\, \cosh \left (f x +e \right ) \sqrt {a +b \left (\sinh ^{2}\left (f x +e \right )\right )}\, f}\)	\(113\)

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,method=_RETURNVERBOSE)

[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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.09, size = 25, normalized size = 0.20 \begin {gather*} {\rm integral}\left (\frac {\sinh \left (f x + e\right )^{2}}{\sqrt {b \sinh \left (f x + e\right )^{2} + a}}, x\right ) \end {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sinh ^{2}{\left (e + f x \right )}}{\sqrt {a + b \sinh ^{2}{\left (e + f x \right )}}}\, dx \end {gather*}

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]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:Unable to divide,
perhaps due to rounding error%%%{32,[4,2,4]%%%}+%%%{%%%{-64,[1]%%%},[4,2,3]%%%}+%%%{%%%{32,[2]%%%},[4,2,2]%%%
}+%%%{%%%{-64,

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {sinh}\left (e+f\,x\right )}^2}{\sqrt {b\,{\mathrm {sinh}\left (e+f\,x\right )}^2+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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