3.507 \(\int \frac{\tanh ^4(e+f x)}{(a+b \sinh ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=333 \[ \frac{(3 a+b) (a+3 b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \text{EllipticF}\left (\tan ^{-1}(\sinh (e+f x)),1-\frac{b}{a}\right )}{3 a f (a-b)^4 \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac{2 (2 a+b) \tanh (e+f x)}{3 f (a-b)^2 \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{b (5 a+3 b) \sinh (e+f x) \cosh (e+f x)}{3 f (a-b)^3 \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac{\tanh (e+f x) \text{sech}^2(e+f x)}{3 f (a-b) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{8 \sqrt{a} \sqrt{b} (a+b) \cosh (e+f x) E\left (\tan ^{-1}\left (\frac{\sqrt{b} \sinh (e+f x)}{\sqrt{a}}\right )|1-\frac{a}{b}\right )}{3 f (a-b)^4 \sqrt{a+b \sinh ^2(e+f x)} \sqrt{\frac{a \cosh ^2(e+f x)}{a+b \sinh ^2(e+f x)}}} \]

[Out]

-(b*(5*a + 3*b)*Cosh[e + f*x]*Sinh[e + f*x])/(3*(a - b)^3*f*(a + b*Sinh[e + f*x]^2)^(3/2)) - (8*Sqrt[a]*Sqrt[b
]*(a + b)*Cosh[e + f*x]*EllipticE[ArcTan[(Sqrt[b]*Sinh[e + f*x])/Sqrt[a]], 1 - a/b])/(3*(a - b)^4*f*Sqrt[(a*Co
sh[e + f*x]^2)/(a + b*Sinh[e + f*x]^2)]*Sqrt[a + b*Sinh[e + f*x]^2]) + ((3*a + b)*(a + 3*b)*EllipticF[ArcTan[S
inh[e + f*x]], 1 - b/a]*Sech[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(3*a*(a - b)^4*f*Sqrt[(Sech[e + f*x]^2*(a +
b*Sinh[e + f*x]^2))/a]) - (2*(2*a + b)*Tanh[e + f*x])/(3*(a - b)^2*f*(a + b*Sinh[e + f*x]^2)^(3/2)) + (Sech[e
+ f*x]^2*Tanh[e + f*x])/(3*(a - b)*f*(a + b*Sinh[e + f*x]^2)^(3/2))

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Rubi [A]  time = 0.399476, antiderivative size = 333, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3196, 470, 527, 525, 418, 411} \[ -\frac{2 (2 a+b) \tanh (e+f x)}{3 f (a-b)^2 \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{b (5 a+3 b) \sinh (e+f x) \cosh (e+f x)}{3 f (a-b)^3 \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac{\tanh (e+f x) \text{sech}^2(e+f x)}{3 f (a-b) \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{8 \sqrt{a} \sqrt{b} (a+b) \cosh (e+f x) E\left (\tan ^{-1}\left (\frac{\sqrt{b} \sinh (e+f x)}{\sqrt{a}}\right )|1-\frac{a}{b}\right )}{3 f (a-b)^4 \sqrt{a+b \sinh ^2(e+f x)} \sqrt{\frac{a \cosh ^2(e+f x)}{a+b \sinh ^2(e+f x)}}}+\frac{(3 a+b) (a+3 b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{3 a f (a-b)^4 \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[e + f*x]^4/(a + b*Sinh[e + f*x]^2)^(5/2),x]

[Out]

-(b*(5*a + 3*b)*Cosh[e + f*x]*Sinh[e + f*x])/(3*(a - b)^3*f*(a + b*Sinh[e + f*x]^2)^(3/2)) - (8*Sqrt[a]*Sqrt[b
]*(a + b)*Cosh[e + f*x]*EllipticE[ArcTan[(Sqrt[b]*Sinh[e + f*x])/Sqrt[a]], 1 - a/b])/(3*(a - b)^4*f*Sqrt[(a*Co
sh[e + f*x]^2)/(a + b*Sinh[e + f*x]^2)]*Sqrt[a + b*Sinh[e + f*x]^2]) + ((3*a + b)*(a + 3*b)*EllipticF[ArcTan[S
inh[e + f*x]], 1 - b/a]*Sech[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(3*a*(a - b)^4*f*Sqrt[(Sech[e + f*x]^2*(a +
b*Sinh[e + f*x]^2))/a]) - (2*(2*a + b)*Tanh[e + f*x])/(3*(a - b)^2*f*(a + b*Sinh[e + f*x]^2)^(3/2)) + (Sech[e
+ f*x]^2*Tanh[e + f*x])/(3*(a - b)*f*(a + b*Sinh[e + f*x]^2)^(3/2))

Rule 3196

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_), 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)/(1 - ff^2*x^2)^((m + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2]
&&  !IntegerQ[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 527

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 + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

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{\tanh ^4(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{x^4}{\left (1+x^2\right )^{5/2} \left (a+b x^2\right )^{5/2}} \, dx,x,\sinh (e+f x)\right )}{f}\\ &=\frac{\text{sech}^2(e+f x) \tanh (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{a+(-3 a-2 b) x^2}{\left (1+x^2\right )^{3/2} \left (a+b x^2\right )^{5/2}} \, dx,x,\sinh (e+f x)\right )}{3 (a-b) f}\\ &=-\frac{2 (2 a+b) \tanh (e+f x)}{3 (a-b)^2 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac{\text{sech}^2(e+f x) \tanh (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{3 a (a+b)-6 b (2 a+b) x^2}{\sqrt{1+x^2} \left (a+b x^2\right )^{5/2}} \, dx,x,\sinh (e+f x)\right )}{3 (a-b) (-a+b) f}\\ &=-\frac{b (5 a+3 b) \cosh (e+f x) \sinh (e+f x)}{3 (a-b)^3 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{2 (2 a+b) \tanh (e+f x)}{3 (a-b)^2 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac{\text{sech}^2(e+f x) \tanh (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{3 a^2 (3 a+5 b)-3 a b (5 a+3 b) x^2}{\sqrt{1+x^2} \left (a+b x^2\right )^{3/2}} \, dx,x,\sinh (e+f x)\right )}{9 a (a-b)^2 (-a+b) f}\\ &=-\frac{b (5 a+3 b) \cosh (e+f x) \sinh (e+f x)}{3 (a-b)^3 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{2 (2 a+b) \tanh (e+f x)}{3 (a-b)^2 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac{\text{sech}^2(e+f x) \tanh (e+f x)}{3 (a-b) f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac{\left (8 a b (a+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-b)^3 (-a+b) f}-\frac{\left ((3 a+b) (a+3 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)^3 (-a+b) f}\\ &=-\frac{b (5 a+3 b) \cosh (e+f x) \sinh (e+f x)}{3 (a-b)^3 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}-\frac{8 \sqrt{a} \sqrt{b} (a+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-b)^4 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{(3 a+b) (a+3 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 (a-b)^4 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac{2 (2 a+b) \tanh (e+f x)}{3 (a-b)^2 f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}+\frac{\text{sech}^2(e+f x) \tanh (e+f x)}{3 (a-b) f \left (a+b \sinh ^2(e+f x)\right )^{3/2}}\\ \end{align*}

Mathematica [C]  time = 3.36955, size = 252, normalized size = 0.76 \[ -\frac{i \left (2 a b \left (\frac{2 a+b \cosh (2 (e+f x))-b}{a}\right )^{3/2} \left (\left (-5 a^2+2 a b+3 b^2\right ) \text{EllipticF}\left (i (e+f x),\frac{b}{a}\right )+8 a (a+b) E\left (i (e+f x)\left |\frac{b}{a}\right .\right )\right )-i \sqrt{2} b \left (2 a b (a-b) \sinh (2 (e+f x))+4 b (a+b) \sinh (2 (e+f x)) (2 a+b \cosh (2 (e+f x))-b)+4 (a+b) \tanh (e+f x) (2 a+b \cosh (2 (e+f x))-b)^2-(a-b) \tanh (e+f x) \text{sech}^2(e+f x) (2 a+b \cosh (2 (e+f x))-b)^2\right )\right )}{6 b f (a-b)^4 (2 a+b \cosh (2 (e+f x))-b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[e + f*x]^4/(a + b*Sinh[e + f*x]^2)^(5/2),x]

[Out]

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

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Maple [A]  time = 0.217, size = 661, normalized size = 2. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(f*x+e)^4/(a+b*sinh(f*x+e)^2)^(5/2),x)

[Out]

1/3*((-8*(-1/a*b)^(1/2)*a*b^2-8*(-1/a*b)^(1/2)*b^3)*sinh(f*x+e)*cosh(f*x+e)^6+(-13*(-1/a*b)^(1/2)*a^2*b+2*(-1/
a*b)^(1/2)*a*b^2+11*(-1/a*b)^(1/2)*b^3)*cosh(f*x+e)^4*sinh(f*x+e)+(-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)+((-1/a*b)^(1/2)*a^3-3*(-1/a*b)^(1/2)*a^2*b+3*(-1/a*b)^(1/2)*
a*b^2-(-1/a*b)^(1/2)*b^3)*sinh(f*x+e)+(b/a*cosh(f*x+e)^2+(a-b)/a)^(1/2)*(cosh(f*x+e)^2)^(1/2)*b*(3*EllipticF(s
inh(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-5*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*b+8*Ellipti
cE(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))*b^2)*cosh(f*x+e)^4+(b/a*cosh(f*x+e)^2+(a-b)/a)^(1/2)*(cosh(f*x+e)^2
)^(1/2)*(3*EllipticF(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))*a^3-EllipticF(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1
/2))*a^2*b-7*EllipticF(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))*a*b^2+5*EllipticF(sinh(f*x+e)*(-1/a*b)^(1/2),(a
/b)^(1/2))*b^3+8*EllipticE(sinh(f*x+e)*(-1/a*b)^(1/2),(a/b)^(1/2))*a^2*b-8*EllipticE(sinh(f*x+e)*(-1/a*b)^(1/2
),(a/b)^(1/2))*b^3)*cosh(f*x+e)^2)/cosh(f*x+e)^3/(-1/a*b)^(1/2)/(a+b*sinh(f*x+e)^2)^(3/2)/(a-b)^4/f

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh \left (f x + e\right )^{4}}{{\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(tanh(f*x+e)^4/(a+b*sinh(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

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

[Out]

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

[Out]

integrate(tanh(f*x + e)^4/(b*sinh(f*x + e)^2 + a)^(5/2), x)