3.395 \(\int \frac{\cosh ^6(e+f x)}{(a+b \sinh ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=330 \[ \frac{(4 a-b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \text{EllipticF}\left (\tan ^{-1}(\sinh (e+f x)),1-\frac{b}{a}\right )}{3 a^2 b^2 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{\left (8 a^2-3 a b-2 b^2\right ) \tanh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 a^2 b^3 f}-\frac{2 (a-b) (2 a+b) \sinh (e+f x) \cosh (e+f x)}{3 a^2 b^2 f \sqrt{a+b \sinh ^2(e+f x)}}-\frac{\left (8 a^2-3 a b-2 b^2\right ) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{3 a^2 b^3 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac{(a-b) \sinh (e+f x) \cosh ^3(e+f x)}{3 a b f \left (a+b \sinh ^2(e+f x)\right )^{3/2}} \]

[Out]

-((a - b)*Cosh[e + f*x]^3*Sinh[e + f*x])/(3*a*b*f*(a + b*Sinh[e + f*x]^2)^(3/2)) - (2*(a - b)*(2*a + b)*Cosh[e
+ f*x]*Sinh[e + f*x])/(3*a^2*b^2*f*Sqrt[a + b*Sinh[e + f*x]^2]) - ((8*a^2 - 3*a*b - 2*b^2)*EllipticE[ArcTan[S
inh[e + f*x]], 1 - b/a]*Sech[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(3*a^2*b^3*f*Sqrt[(Sech[e + f*x]^2*(a + b*S
inh[e + f*x]^2))/a]) + ((4*a - b)*EllipticF[ArcTan[Sinh[e + f*x]], 1 - b/a]*Sech[e + f*x]*Sqrt[a + b*Sinh[e +
f*x]^2])/(3*a^2*b^2*f*Sqrt[(Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a]) + ((8*a^2 - 3*a*b - 2*b^2)*Sqrt[a + b
*Sinh[e + f*x]^2]*Tanh[e + f*x])/(3*a^2*b^3*f)

________________________________________________________________________________________

Rubi [A]  time = 0.331467, antiderivative size = 330, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {3192, 413, 526, 531, 418, 492, 411} \[ \frac{\left (8 a^2-3 a b-2 b^2\right ) \tanh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 a^2 b^3 f}-\frac{2 (a-b) (2 a+b) \sinh (e+f x) \cosh (e+f x)}{3 a^2 b^2 f \sqrt{a+b \sinh ^2(e+f x)}}+\frac{(4 a-b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{3 a^2 b^2 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac{\left (8 a^2-3 a b-2 b^2\right ) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{3 a^2 b^3 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac{(a-b) \sinh (e+f x) \cosh ^3(e+f x)}{3 a b f \left (a+b \sinh ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[e + f*x]^6/(a + b*Sinh[e + f*x]^2)^(5/2),x]

[Out]

-((a - b)*Cosh[e + f*x]^3*Sinh[e + f*x])/(3*a*b*f*(a + b*Sinh[e + f*x]^2)^(3/2)) - (2*(a - b)*(2*a + b)*Cosh[e
+ f*x]*Sinh[e + f*x])/(3*a^2*b^2*f*Sqrt[a + b*Sinh[e + f*x]^2]) - ((8*a^2 - 3*a*b - 2*b^2)*EllipticE[ArcTan[S
inh[e + f*x]], 1 - b/a]*Sech[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(3*a^2*b^3*f*Sqrt[(Sech[e + f*x]^2*(a + b*S
inh[e + f*x]^2))/a]) + ((4*a - b)*EllipticF[ArcTan[Sinh[e + f*x]], 1 - b/a]*Sech[e + f*x]*Sqrt[a + b*Sinh[e +
f*x]^2])/(3*a^2*b^2*f*Sqrt[(Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a]) + ((8*a^2 - 3*a*b - 2*b^2)*Sqrt[a + b
*Sinh[e + f*x]^2]*Tanh[e + f*x])/(3*a^2*b^3*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 413

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((a*d - c*b)*x*(a + b*x^n)^
(p + 1)*(c + d*x^n)^(q - 1))/(a*b*n*(p + 1)), x] - Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)
^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*(p + q) + 1))*x^n, x], x], x] /;
FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q
, x]

Rule 526

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

Rule 531

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Dist[
e, Int[(a + b*x^n)^p*(c + d*x^n)^q, x], x] + Dist[f, Int[x^n*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, n, p, q}, x]

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 ^6(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{\left (1+x^2\right )^{5/2}}{\left (a+b x^2\right )^{5/2}} \, dx,x,\sinh (e+f x)\right )}{f}\\ &=-\frac{(a-b) \cosh ^3(e+f x) \sinh (e+f x)}{3 a b 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{\sqrt{1+x^2} \left (a+2 b+(4 a-b) x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx,x,\sinh (e+f x)\right )}{3 a b f}\\ &=-\frac{(a-b) \cosh ^3(e+f x) \sinh (e+f x)}{3 a b f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{2 (a-b) (2 a+b) \cosh (e+f x) \sinh (e+f x)}{3 a^2 b^2 f \sqrt{a+b \sinh ^2(e+f x)}}-\frac{\left (\sqrt{\cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{-a (4 a-b)+\left (-8 a^2+3 a b+2 b^2\right ) x^2}{\sqrt{1+x^2} \sqrt{a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{3 a^2 b^2 f}\\ &=-\frac{(a-b) \cosh ^3(e+f x) \sinh (e+f x)}{3 a b f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{2 (a-b) (2 a+b) \cosh (e+f x) \sinh (e+f x)}{3 a^2 b^2 f \sqrt{a+b \sinh ^2(e+f x)}}+\frac{\left ((4 a-b) \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 b^2 f}-\frac{\left (\left (-8 a^2+3 a b+2 b^2\right ) \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 )}{3 a^2 b^2 f}\\ &=-\frac{(a-b) \cosh ^3(e+f x) \sinh (e+f x)}{3 a b f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{2 (a-b) (2 a+b) \cosh (e+f x) \sinh (e+f x)}{3 a^2 b^2 f \sqrt{a+b \sinh ^2(e+f x)}}+\frac{(4 a-b) 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 b^2 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{\left (8 a^2-3 a b-2 b^2\right ) \sqrt{a+b \sinh ^2(e+f x)} \tanh (e+f x)}{3 a^2 b^3 f}+\frac{\left (\left (-8 a^2+3 a b+2 b^2\right ) \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 )}{3 a^2 b^3 f}\\ &=-\frac{(a-b) \cosh ^3(e+f x) \sinh (e+f x)}{3 a b f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{2 (a-b) (2 a+b) \cosh (e+f x) \sinh (e+f x)}{3 a^2 b^2 f \sqrt{a+b \sinh ^2(e+f x)}}-\frac{\left (8 a^2-3 a b-2 b^2\right ) 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)}}{3 a^2 b^3 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{(4 a-b) 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 b^2 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{\left (8 a^2-3 a b-2 b^2\right ) \sqrt{a+b \sinh ^2(e+f x)} \tanh (e+f x)}{3 a^2 b^3 f}\\ \end{align*}

Mathematica [C]  time = 2.13341, size = 206, normalized size = 0.62 \[ \frac{\frac{1}{2} (a-b) \left (-2 \sqrt{2} b \sinh (2 (e+f x)) \left (8 a^2+b (5 a+2 b) \cosh (2 (e+f x))+a b-2 b^2\right )+4 i a^2 (8 a+b) \left (\frac{2 a+b \cosh (2 (e+f x))-b}{a}\right )^{3/2} \text{EllipticF}\left (i (e+f x),\frac{b}{a}\right )\right )-2 i a^2 \left (8 a^2-3 a b-2 b^2\right ) \left (\frac{2 a+b \cosh (2 (e+f x))-b}{a}\right )^{3/2} E\left (i (e+f x)\left |\frac{b}{a}\right .\right )}{6 a^2 b^3 f (2 a+b \cosh (2 (e+f x))-b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[e + f*x]^6/(a + b*Sinh[e + f*x]^2)^(5/2),x]

[Out]

((-2*I)*a^2*(8*a^2 - 3*a*b - 2*b^2)*((2*a - b + b*Cosh[2*(e + f*x)])/a)^(3/2)*EllipticE[I*(e + f*x), b/a] + ((
a - b)*((4*I)*a^2*(8*a + b)*((2*a - b + b*Cosh[2*(e + f*x)])/a)^(3/2)*EllipticF[I*(e + f*x), b/a] - 2*Sqrt[2]*
b*(8*a^2 + a*b - 2*b^2 + b*(5*a + 2*b)*Cosh[2*(e + f*x)])*Sinh[2*(e + f*x)]))/2)/(6*a^2*b^3*f*(2*a - b + b*Cos
h[2*(e + f*x)])^(3/2))

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Maple [B]  time = 0.133, size = 812, normalized size = 2.5 \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)^6/(a+b*sinh(f*x+e)^2)^(5/2),x)

[Out]

-1/3*((5*(-1/a*b)^(1/2)*a^2*b-3*(-1/a*b)^(1/2)*a*b^2-2*(-1/a*b)^(1/2)*b^3)*sinh(f*x+e)*cosh(f*x+e)^4+(4*(-1/a*
b)^(1/2)*a^3-6*(-1/a*b)^(1/2)*a^2*b+2*(-1/a*b)^(1/2)*b^3)*cosh(f*x+e)^2*sinh(f*x+e)+(b/a*cosh(f*x+e)^2+(a-b)/a
)^(1/2)*(cosh(f*x+e)^2)^(1/2)*b*(4*EllipticF(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))*a^2-2*EllipticF(sinh(f*x+
e)*(-1/a*b)^(1/2),(a/b)^(1/2))*a*b-2*EllipticF(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))*b^2-8*EllipticE(sinh(f*
x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))*a^2+3*EllipticE(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))*a*b+2*EllipticE(sinh(
f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))*b^2)*cosh(f*x+e)^2+4*(b/a*cosh(f*x+e)^2+(a-b)/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^3-6*(b/a*cosh(f*x+e)^2+(a-b)/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^2*b+2*(b/a*cosh(f*x+e)^2+(a-b)/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^3-8*(b/a*cosh(f*x+e)^2+(a-b)/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^3+11*(b/a*cosh(f*x+e)^2+(a-b)/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*b-(b/a*cosh(f*x+e)^2+(a-b)/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^2-2*(b/a*cosh(f*x+e)^2+(a-b)/a)^(1/2)*(co
sh(f*x+e)^2)^(1/2)*EllipticE(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))*b^3)/a^2/(a+b*sinh(f*x+e)^2)^(3/2)/(-1/a*
b)^(1/2)/b^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 )^{6}}{{\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)^6/(a+b*sinh(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

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

[Out]

integrate(cosh(f*x + e)^6/(b*sinh(f*x + e)^2 + a)^(5/2), x)