3.360 \(\int \text{sech}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \, dx\)

Optimal. Leaf size=70 \[ \frac{\text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \]

[Out]

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

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Rubi [A]  time = 0.0844919, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {3192, 411} \[ \frac{\text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \]

Antiderivative was successfully verified.

[In]

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

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 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 \text{sech}^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{a+b x^2}}{\left (1+x^2\right )^{3/2}} \, dx,x,\sinh (e+f x)\right )}{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)}}{f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}\\ \end{align*}

Mathematica [C]  time = 0.484483, size = 148, normalized size = 2.11 \[ \frac{-2 i a \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} \text{EllipticF}\left (i (e+f x),\frac{b}{a}\right )+\sqrt{2} \tanh (e+f x) (2 a+b \cosh (2 (e+f x))-b)+2 i a \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} E\left (i (e+f x)\left |\frac{b}{a}\right .\right )}{2 f \sqrt{2 a+b \cosh (2 (e+f x))-b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.101, size = 177, normalized size = 2.5 \begin{align*}{\frac{1}{f\cosh \left ( fx+e \right ) } \left ( \sqrt{-{\frac{b}{a}}}b \left ( \sinh \left ( fx+e \right ) \right ) ^{3}+b\sqrt{{\frac{a+b \left ( \sinh \left ( fx+e \right ) \right ) ^{2}}{a}}}\sqrt{ \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}{\it EllipticF} \left ( \sinh \left ( fx+e \right ) \sqrt{-{\frac{b}{a}}},\sqrt{{\frac{a}{b}}} \right ) -b\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 ) +\sqrt{-{\frac{b}{a}}}a\sinh \left ( fx+e \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(sech(f*x+e)^2*(a+b*sinh(f*x+e)^2)^(1/2),x)

[Out]

((-1/a*b)^(1/2)*b*sinh(f*x+e)^3+b*((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))-b*((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)*a*sinh(f*x+e))/(-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 \sqrt{b \sinh \left (f x + e\right )^{2} + a} \operatorname{sech}\left (f x + e\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(f*x+e)^2*(a+b*sinh(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*sinh(f*x + e)^2 + a)*sech(f*x + e)^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a + b \sinh ^{2}{\left (e + f x \right )}} \operatorname{sech}^{2}{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(f*x+e)**2*(a+b*sinh(f*x+e)**2)**(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(f*x+e)^2*(a+b*sinh(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

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