3.500 \(\int \frac{\coth ^4(e+f x)}{(a+b \sinh ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=341 \[ \frac{(3 a-4 b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \text{EllipticF}\left (\tan ^{-1}(\sinh (e+f x)),1-\frac{b}{a}\right )}{3 a^3 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{(7 a-8 b) \tanh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 a^3 f}-\frac{(7 a-8 b) \coth (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 a^3 f}+\frac{(3 a-4 b) \coth (e+f x) \text{csch}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 a^2 b f}-\frac{(7 a-8 b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{3 a^3 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac{(a-b) \coth (e+f x) \text{csch}^2(e+f x)}{a b f \sqrt{a+b \sinh ^2(e+f x)}} \]

[Out]

-(((a - b)*Coth[e + f*x]*Csch[e + f*x]^2)/(a*b*f*Sqrt[a + b*Sinh[e + f*x]^2])) - ((7*a - 8*b)*Coth[e + f*x]*Sq
rt[a + b*Sinh[e + f*x]^2])/(3*a^3*f) + ((3*a - 4*b)*Coth[e + f*x]*Csch[e + f*x]^2*Sqrt[a + b*Sinh[e + f*x]^2])
/(3*a^2*b*f) - ((7*a - 8*b)*EllipticE[ArcTan[Sinh[e + f*x]], 1 - b/a]*Sech[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2
])/(3*a^3*f*Sqrt[(Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a]) + ((3*a - 4*b)*EllipticF[ArcTan[Sinh[e + f*x]],
1 - b/a]*Sech[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(3*a^3*f*Sqrt[(Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a
]) + ((7*a - 8*b)*Sqrt[a + b*Sinh[e + f*x]^2]*Tanh[e + f*x])/(3*a^3*f)

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Rubi [A]  time = 0.40835, antiderivative size = 341, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {3196, 468, 583, 531, 418, 492, 411} \[ \frac{(7 a-8 b) \tanh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 a^3 f}-\frac{(7 a-8 b) \coth (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 a^3 f}+\frac{(3 a-4 b) \coth (e+f x) \text{csch}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 a^2 b f}+\frac{(3 a-4 b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{3 a^3 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac{(7 a-8 b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{3 a^3 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac{(a-b) \coth (e+f x) \text{csch}^2(e+f x)}{a b f \sqrt{a+b \sinh ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Coth[e + f*x]^4/(a + b*Sinh[e + f*x]^2)^(3/2),x]

[Out]

-(((a - b)*Coth[e + f*x]*Csch[e + f*x]^2)/(a*b*f*Sqrt[a + b*Sinh[e + f*x]^2])) - ((7*a - 8*b)*Coth[e + f*x]*Sq
rt[a + b*Sinh[e + f*x]^2])/(3*a^3*f) + ((3*a - 4*b)*Coth[e + f*x]*Csch[e + f*x]^2*Sqrt[a + b*Sinh[e + f*x]^2])
/(3*a^2*b*f) - ((7*a - 8*b)*EllipticE[ArcTan[Sinh[e + f*x]], 1 - b/a]*Sech[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2
])/(3*a^3*f*Sqrt[(Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a]) + ((3*a - 4*b)*EllipticF[ArcTan[Sinh[e + f*x]],
1 - b/a]*Sech[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(3*a^3*f*Sqrt[(Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a
]) + ((7*a - 8*b)*Sqrt[a + b*Sinh[e + f*x]^2]*Tanh[e + f*x])/(3*a^3*f)

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 468

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

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -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{\coth ^4(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{\left (1+x^2\right )^{3/2}}{x^4 \left (a+b x^2\right )^{3/2}} \, dx,x,\sinh (e+f x)\right )}{f}\\ &=-\frac{(a-b) \coth (e+f x) \text{csch}^2(e+f x)}{a 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{3 a-4 b+(2 a-3 b) x^2}{x^4 \sqrt{1+x^2} \sqrt{a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{a b f}\\ &=-\frac{(a-b) \coth (e+f x) \text{csch}^2(e+f x)}{a b f \sqrt{a+b \sinh ^2(e+f x)}}+\frac{(3 a-4 b) \coth (e+f x) \text{csch}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 a^2 b f}+\frac{\left (\sqrt{\cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{(7 a-8 b) b+(3 a-4 b) b x^2}{x^2 \sqrt{1+x^2} \sqrt{a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{3 a^2 b f}\\ &=-\frac{(a-b) \coth (e+f x) \text{csch}^2(e+f x)}{a b f \sqrt{a+b \sinh ^2(e+f x)}}-\frac{(7 a-8 b) \coth (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 a^3 f}+\frac{(3 a-4 b) \coth (e+f x) \text{csch}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 a^2 b f}-\frac{\left (\sqrt{\cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{-a (3 a-4 b) b-(7 a-8 b) b^2 x^2}{\sqrt{1+x^2} \sqrt{a+b x^2}} \, dx,x,\sinh (e+f x)\right )}{3 a^3 b f}\\ &=-\frac{(a-b) \coth (e+f x) \text{csch}^2(e+f x)}{a b f \sqrt{a+b \sinh ^2(e+f x)}}-\frac{(7 a-8 b) \coth (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 a^3 f}+\frac{(3 a-4 b) \coth (e+f x) \text{csch}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 a^2 b f}+\frac{\left ((3 a-4 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^2 f}+\frac{\left ((7 a-8 b) b \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^3 f}\\ &=-\frac{(a-b) \coth (e+f x) \text{csch}^2(e+f x)}{a b f \sqrt{a+b \sinh ^2(e+f x)}}-\frac{(7 a-8 b) \coth (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 a^3 f}+\frac{(3 a-4 b) \coth (e+f x) \text{csch}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 a^2 b f}+\frac{(3 a-4 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^3 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{(7 a-8 b) \sqrt{a+b \sinh ^2(e+f x)} \tanh (e+f x)}{3 a^3 f}-\frac{\left ((7 a-8 b) \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^3 f}\\ &=-\frac{(a-b) \coth (e+f x) \text{csch}^2(e+f x)}{a b f \sqrt{a+b \sinh ^2(e+f x)}}-\frac{(7 a-8 b) \coth (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 a^3 f}+\frac{(3 a-4 b) \coth (e+f x) \text{csch}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 a^2 b f}-\frac{(7 a-8 b) 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^3 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{(3 a-4 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^3 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{(7 a-8 b) \sqrt{a+b \sinh ^2(e+f x)} \tanh (e+f x)}{3 a^3 f}\\ \end{align*}

Mathematica [C]  time = 3.21993, size = 214, normalized size = 0.63 \[ \frac{8 i a (a-b) \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} \text{EllipticF}\left (i (e+f x),\frac{b}{a}\right )-\frac{\coth (e+f x) \text{csch}^2(e+f x) \left (4 \left (4 a^2-11 a b+8 b^2\right ) \cosh (2 (e+f x))-8 a^2+b (7 a-8 b) \cosh (4 (e+f x))+37 a b-24 b^2\right )}{2 \sqrt{2}}-2 i a (7 a-8 b) \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} E\left (i (e+f x)\left |\frac{b}{a}\right .\right )}{6 a^3 f \sqrt{2 a+b \cosh (2 (e+f x))-b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[e + f*x]^4/(a + b*Sinh[e + f*x]^2)^(3/2),x]

[Out]

(-((-8*a^2 + 37*a*b - 24*b^2 + 4*(4*a^2 - 11*a*b + 8*b^2)*Cosh[2*(e + f*x)] + (7*a - 8*b)*b*Cosh[4*(e + f*x)])
*Coth[e + f*x]*Csch[e + f*x]^2)/(2*Sqrt[2]) - (2*I)*a*(7*a - 8*b)*Sqrt[(2*a - b + b*Cosh[2*(e + f*x)])/a]*Elli
pticE[I*(e + f*x), b/a] + (8*I)*a*(a - b)*Sqrt[(2*a - b + b*Cosh[2*(e + f*x)])/a]*EllipticF[I*(e + f*x), b/a])
/(6*a^3*f*Sqrt[2*a - b + b*Cosh[2*(e + f*x)]])

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Maple [A]  time = 0.195, size = 522, 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(coth(f*x+e)^4/(a+b*sinh(f*x+e)^2)^(3/2),x)

[Out]

-1/3*(7*(-1/a*b)^(1/2)*a*b*sinh(f*x+e)^6-8*(-1/a*b)^(1/2)*b^2*sinh(f*x+e)^6-3*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))*sinh(f*x+e)^3+11*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))*a*sinh(f*x+e)^3-8*((a+b*s
inh(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)
^3-7*((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)^3+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))*b^2*sinh(f*x+e)^3+4*(-1/a*b)^(1/2)*a^2*sinh(f*x+e)^4+3*(-1/a*b)^(1/2)*a*b*sinh(f*x+e)^4-8*(-1/a*b)^(1
/2)*b^2*sinh(f*x+e)^4+5*(-1/a*b)^(1/2)*a^2*sinh(f*x+e)^2-4*(-1/a*b)^(1/2)*a*b*sinh(f*x+e)^2+(-1/a*b)^(1/2)*a^2
)/(-1/a*b)^(1/2)/sinh(f*x+e)^3/a^3/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 \frac{\coth \left (f x + e\right )^{4}}{{\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(coth(f*x+e)^4/(a+b*sinh(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

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

[Out]

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

[Out]

integrate(coth(f*x + e)^4/(b*sinh(f*x + e)^2 + a)^(3/2), x)