3.397 \(\int \frac{\cosh ^2(e+f x)}{(a+b \sinh ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=228 \[ \frac{\text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \text{EllipticF}\left (\tan ^{-1}(\sinh (e+f x)),1-\frac{b}{a}\right )}{3 a^2 f (a-b) \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{(a-2 b) \cosh (e+f x) E\left (\tan ^{-1}\left (\frac{\sqrt{b} \sinh (e+f x)}{\sqrt{a}}\right )|1-\frac{a}{b}\right )}{3 a^{3/2} \sqrt{b} f (a-b) \sqrt{a+b \sinh ^2(e+f x)} \sqrt{\frac{a \cosh ^2(e+f x)}{a+b \sinh ^2(e+f x)}}}+\frac{\sinh (e+f x) \cosh (e+f x)}{3 a f \left (a+b \sinh ^2(e+f x)\right )^{3/2}} \]

[Out]

(Cosh[e + f*x]*Sinh[e + f*x])/(3*a*f*(a + b*Sinh[e + f*x]^2)^(3/2)) + ((a - 2*b)*Cosh[e + f*x]*EllipticE[ArcTa
n[(Sqrt[b]*Sinh[e + f*x])/Sqrt[a]], 1 - a/b])/(3*a^(3/2)*(a - b)*Sqrt[b]*f*Sqrt[(a*Cosh[e + f*x]^2)/(a + b*Sin
h[e + f*x]^2)]*Sqrt[a + b*Sinh[e + f*x]^2]) + (EllipticF[ArcTan[Sinh[e + f*x]], 1 - b/a]*Sech[e + f*x]*Sqrt[a
+ b*Sinh[e + f*x]^2])/(3*a^2*(a - b)*f*Sqrt[(Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a])

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Rubi [A]  time = 0.20429, antiderivative size = 228, 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, 412, 525, 418, 411} \[ \frac{(a-2 b) \cosh (e+f x) E\left (\tan ^{-1}\left (\frac{\sqrt{b} \sinh (e+f x)}{\sqrt{a}}\right )|1-\frac{a}{b}\right )}{3 a^{3/2} \sqrt{b} f (a-b) \sqrt{a+b \sinh ^2(e+f x)} \sqrt{\frac{a \cosh ^2(e+f x)}{a+b \sinh ^2(e+f x)}}}+\frac{\text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{3 a^2 f (a-b) \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{\sinh (e+f x) \cosh (e+f x)}{3 a f \left (a+b \sinh ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[e + f*x]^2/(a + b*Sinh[e + f*x]^2)^(5/2),x]

[Out]

(Cosh[e + f*x]*Sinh[e + f*x])/(3*a*f*(a + b*Sinh[e + f*x]^2)^(3/2)) + ((a - 2*b)*Cosh[e + f*x]*EllipticE[ArcTa
n[(Sqrt[b]*Sinh[e + f*x])/Sqrt[a]], 1 - a/b])/(3*a^(3/2)*(a - b)*Sqrt[b]*f*Sqrt[(a*Cosh[e + f*x]^2)/(a + b*Sin
h[e + f*x]^2)]*Sqrt[a + b*Sinh[e + f*x]^2]) + (EllipticF[ArcTan[Sinh[e + f*x]], 1 - b/a]*Sech[e + f*x]*Sqrt[a
+ b*Sinh[e + f*x]^2])/(3*a^2*(a - b)*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 412

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^q)/(a*n*(p + 1)), x] + Dist[1/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(n*(p
+ 1) + 1) + d*(n*(p + q + 1) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p,
-1] && LtQ[0, q, 1] && IntBinomialQ[a, b, c, d, n, p, q, x]

Rule 525

Int[((e_) + (f_.)*(x_)^2)/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)^(3/2)), x_Symbol] :> Dist[(b*e - a*
f)/(b*c - a*d), Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[Sqrt[a + b
*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ[b/a] && PosQ[d/c]

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 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)}{\left (a+b \sinh ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\left (\sqrt{\cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1+x^2}}{\left (a+b x^2\right )^{5/2}} \, dx,x,\sinh (e+f x)\right )}{f}\\ &=\frac{\cosh (e+f x) \sinh (e+f x)}{3 a f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{\left (\sqrt{\cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{-2-x^2}{\sqrt{1+x^2} \left (a+b x^2\right )^{3/2}} \, dx,x,\sinh (e+f x)\right )}{3 a f}\\ &=\frac{\cosh (e+f x) \sinh (e+f x)}{3 a f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\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 )}{3 a (a-b) f}+\frac{\left ((a-2 b) \sqrt{\cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1+x^2}}{\left (a+b x^2\right )^{3/2}} \, dx,x,\sinh (e+f x)\right )}{3 a (a-b) f}\\ &=\frac{\cosh (e+f x) \sinh (e+f x)}{3 a f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac{(a-2 b) \cosh (e+f x) E\left (\tan ^{-1}\left (\frac{\sqrt{b} \sinh (e+f x)}{\sqrt{a}}\right )|1-\frac{a}{b}\right )}{3 a^{3/2} (a-b) \sqrt{b} f \sqrt{\frac{a \cosh ^2(e+f x)}{a+b \sinh ^2(e+f x)}} \sqrt{a+b \sinh ^2(e+f x)}}+\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)}}{3 a^2 (a-b) f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}\\ \end{align*}

Mathematica [C]  time = 1.40191, size = 193, normalized size = 0.85 \[ \frac{-2 i a^2 (a-b) \left (\frac{2 a+b \cosh (2 (e+f x))-b}{a}\right )^{3/2} \text{EllipticF}\left (i (e+f x),\frac{b}{a}\right )-\sqrt{2} b \sinh (2 (e+f x)) \left (-4 a^2-b (a-2 b) \cosh (2 (e+f x))+7 a b-2 b^2\right )+2 i a^2 (a-2 b) \left (\frac{2 a+b \cosh (2 (e+f x))-b}{a}\right )^{3/2} E\left (i (e+f x)\left |\frac{b}{a}\right .\right )}{6 a^2 b f (a-b) (2 a+b \cosh (2 (e+f x))-b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[e + f*x]^2/(a + b*Sinh[e + f*x]^2)^(5/2),x]

[Out]

((2*I)*a^2*(a - 2*b)*((2*a - b + b*Cosh[2*(e + f*x)])/a)^(3/2)*EllipticE[I*(e + f*x), b/a] - (2*I)*a^2*(a - b)
*((2*a - b + b*Cosh[2*(e + f*x)])/a)^(3/2)*EllipticF[I*(e + f*x), b/a] - Sqrt[2]*b*(-4*a^2 + 7*a*b - 2*b^2 - (
a - 2*b)*b*Cosh[2*(e + f*x)])*Sinh[2*(e + f*x)])/(6*a^2*(a - b)*b*f*(2*a - b + b*Cosh[2*(e + f*x)])^(3/2))

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Maple [B]  time = 0.151, size = 662, normalized size = 2.9 \begin{align*} \text{result too large to display} \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)^(5/2),x)

[Out]

1/3*((-1/a*b)^(1/2)*a*b*sinh(f*x+e)^5-2*(-1/a*b)^(1/2)*b^2*sinh(f*x+e)^5+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))*a*b*sinh(f*x+e)^2-2*((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))*b^2*sinh(f*x+e)^2-((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))*a*b*sinh(f*x+e)^2+2*(
(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))*b^2*sinh(
f*x+e)^2+2*(-1/a*b)^(1/2)*a^2*sinh(f*x+e)^3-2*(-1/a*b)^(1/2)*a*b*sinh(f*x+e)^3-2*(-1/a*b)^(1/2)*b^2*sinh(f*x+e
)^3+2*a^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)
)-2*a*((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))*a^2+2*((
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))*a*b+2*(-1/
a*b)^(1/2)*a^2*sinh(f*x+e)-3*(-1/a*b)^(1/2)*a*b*sinh(f*x+e))/a^2/(a-b)/(a+b*sinh(f*x+e)^2)^(3/2)/(-1/a*b)^(1/2
)/cosh(f*x+e)/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}}{{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}}}\,{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)^(5/2),x, algorithm="maxima")

[Out]

integrate(cosh(f*x + e)^2/(b*sinh(f*x + e)^2 + a)^(5/2), 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} \cosh \left (f x + e\right )^{2}}{b^{3} \sinh \left (f x + e\right )^{6} + 3 \, a b^{2} \sinh \left (f x + e\right )^{4} + 3 \, a^{2} b \sinh \left (f x + e\right )^{2} + a^{3}}, 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)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sinh(f*x + e)^2 + a)*cosh(f*x + e)^2/(b^3*sinh(f*x + e)^6 + 3*a*b^2*sinh(f*x + e)^4 + 3*a^2*b*
sinh(f*x + e)^2 + a^3), 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)**(5/2),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}}{{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}}}\,{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)^(5/2),x, algorithm="giac")

[Out]

integrate(cosh(f*x + e)^2/(b*sinh(f*x + e)^2 + a)^(5/2), x)