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

Optimal. Leaf size=177 \[ \frac{\text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \text{EllipticF}\left (\tan ^{-1}(\sinh (e+f x)),1-\frac{b}{a}\right )}{a f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{\tanh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{b f}-\frac{\text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{b f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \]

[Out]

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

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Rubi [A]  time = 0.14809, antiderivative size = 177, 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 = {3192, 422, 418, 492, 411} \[ \frac{\tanh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{b f}+\frac{\text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{a f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac{\text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{b f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3192

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Sin[e + f*x], x]}, Dist[(ff*Sqrt[Cos[e + f*x]^2])/(f*Cos[e + f*x]), 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/2] &&  !IntegerQ
[p]

Rule 422

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[a, Int[1/(Sqrt[a + b*x^2]*Sqrt[c +
d*x^2]), x], x] + Dist[b, Int[x^2/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d}, x] && PosQ[
d/c] && PosQ[b/a]

Rule 418

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticF[ArcT
an[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(a*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /
; FreeQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]

Rule 492

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(x*Sqrt[a + b*x^2])/(b*Sqr
t[c + d*x^2]), x] - Dist[c/b, Int[Sqrt[a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, d/c]

Rule 411

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticE[ArcTan
[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(c*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /;
FreeQ[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]

Rubi steps

\begin{align*} \int \frac{\cosh ^2(e+f x)}{\sqrt{a+b \sinh ^2(e+f x)}} \, dx &=\frac{\left (\sqrt{\cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1+x^2}}{\sqrt{a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{f}\\ &=\frac{\left (\sqrt{\cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+x^2} \sqrt{a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{f}+\frac{\left (\sqrt{\cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1+x^2} \sqrt{a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{f}\\ &=\frac{F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right ) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{a f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{\sqrt{a+b \sinh ^2(e+f x)} \tanh (e+f x)}{b f}-\frac{\left (\sqrt{\cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x^2}}{\left (1+x^2\right )^{3/2}} \, dx,x,\sinh (e+f x)\right )}{b f}\\ &=-\frac{E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right ) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{b f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right ) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{a f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{\sqrt{a+b \sinh ^2(e+f x)} \tanh (e+f x)}{b f}\\ \end{align*}

Mathematica [C]  time = 0.24074, size = 95, normalized size = 0.54 \[ -\frac{i \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} \left ((b-a) \text{EllipticF}\left (i (e+f x),\frac{b}{a}\right )+a E\left (i (e+f x)\left |\frac{b}{a}\right .\right )\right )}{b f \sqrt{2 a+b \cosh (2 (e+f x))-b}} \]

Antiderivative was successfully verified.

[In]

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

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Maple [A]  time = 0.082, size = 86, normalized size = 0.5 \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}}{\it EllipticE} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \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(cosh(f*x+e)^2/(a+b*sinh(f*x+e)^2)^(1/2),x)

[Out]

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

[Out]

integrate(cosh(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{\cosh \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(cosh(f*x+e)^2/(a+b*sinh(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

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

[Out]

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