3.111 \(\int \frac{\sinh ^6(e+f x)}{(a+b \sinh ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=341 \[ -\frac{(4 a-b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \text{EllipticF}\left (\tan ^{-1}(\sinh (e+f x)),1-\frac{b}{a}\right )}{3 b^2 f (a-b) \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac{\left (8 a^2-3 a b-2 b^2\right ) \tanh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 b^3 f (a-b)}+\frac{\left (8 a^2-3 a b-2 b^2\right ) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{3 b^3 f (a-b) \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{(4 a-b) \sinh (e+f x) \cosh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 b^2 f (a-b)}-\frac{a \sinh ^3(e+f x) \cosh (e+f x)}{b f (a-b) \sqrt{a+b \sinh ^2(e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.340662, antiderivative size = 341, 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 = {3188, 470, 582, 531, 418, 492, 411} \[ -\frac{\left (8 a^2-3 a b-2 b^2\right ) \tanh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 b^3 f (a-b)}+\frac{\left (8 a^2-3 a b-2 b^2\right ) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{3 b^3 f (a-b) \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{(4 a-b) \sinh (e+f x) \cosh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 b^2 f (a-b)}-\frac{(4 a-b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{3 b^2 f (a-b) \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac{a \sinh ^3(e+f x) \cosh (e+f x)}{b f (a-b) \sqrt{a+b \sinh ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[e + f*x]^6/(a + b*Sinh[e + f*x]^2)^(3/2),x]

[Out]

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

Rule 3188

Int[sin[(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^(m + 1)*Sqrt[Cos[e + f*x]^2])/(f*Cos[e + f*x]), Subst[Int[(x^m*(a + b*ff^2*
x^2)^p)/Sqrt[1 - ff^2*x^2], x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] &&  !In
tegerQ[p]

Rule 470

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

Rule 582

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

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

Mathematica [C]  time = 1.25201, size = 211, normalized size = 0.62 \[ \frac{-2 i \sqrt{2} a \left (8 a^2-7 a b-b^2\right ) \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} \text{EllipticF}\left (i (e+f x),\frac{b}{a}\right )-b \sinh (2 (e+f x)) \left (-8 a^2+b (b-a) \cosh (2 (e+f x))+3 a b-b^2\right )+2 i \sqrt{2} a \left (8 a^2-3 a b-2 b^2\right ) \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} E\left (i (e+f x)\left |\frac{b}{a}\right .\right )}{6 b^3 f (a-b) \sqrt{4 a+2 b \cosh (2 (e+f x))-2 b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[e + f*x]^6/(a + b*Sinh[e + f*x]^2)^(3/2),x]

[Out]

((2*I)*Sqrt[2]*a*(8*a^2 - 3*a*b - 2*b^2)*Sqrt[(2*a - b + b*Cosh[2*(e + f*x)])/a]*EllipticE[I*(e + f*x), b/a] -
(2*I)*Sqrt[2]*a*(8*a^2 - 7*a*b - b^2)*Sqrt[(2*a - b + b*Cosh[2*(e + f*x)])/a]*EllipticF[I*(e + f*x), b/a] - b
*(-8*a^2 + 3*a*b - b^2 + b*(-a + b)*Cosh[2*(e + f*x)])*Sinh[2*(e + f*x)])/(6*(a - b)*b^3*f*Sqrt[4*a - 2*b + 2*
b*Cosh[2*(e + f*x)]])

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Maple [A]  time = 0.102, size = 500, normalized size = 1.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(f*x+e)^6/(a+b*sinh(f*x+e)^2)^(3/2),x)

[Out]

1/3*((-1/a*b)^(1/2)*a*b*sinh(f*x+e)^5-(-1/a*b)^(1/2)*b^2*sinh(f*x+e)^5+4*(-1/a*b)^(1/2)*a^2*sinh(f*x+e)^3-(-1/
a*b)^(1/2)*b^2*sinh(f*x+e)^3+4*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-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^2-8*((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+3*((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*((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+4*(-1/a*b)^(1/2)*a^2*sinh(f*x+e)-(-1/a*b)^(1/2)*a*b*sinh(f*x+e))/b^2/(a-b)/(-1/a*b)^(1/2)/cos
h(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{\sinh \left (f x + e\right )^{6}}{{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(f*x+e)^6/(a+b*sinh(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(sinh(f*x + e)^6/(b*sinh(f*x + e)^2 + a)^(3/2), 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} \sinh \left (f x + e\right )^{6}}{b^{2} \sinh \left (f x + e\right )^{4} + 2 \, a b \sinh \left (f x + e\right )^{2} + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(f*x+e)^6/(a+b*sinh(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sinh(f*x + e)^2 + a)*sinh(f*x + e)^6/(b^2*sinh(f*x + e)^4 + 2*a*b*sinh(f*x + e)^2 + a^2), 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(sinh(f*x+e)**6/(a+b*sinh(f*x+e)**2)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh \left (f x + e\right )^{6}}{{\left (b \sinh \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(f*x+e)^6/(a+b*sinh(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate(sinh(f*x + e)^6/(b*sinh(f*x + e)^2 + a)^(3/2), x)