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

Optimal. Leaf size=285 \[ \frac{(3 a-b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \text{EllipticF}\left (\tan ^{-1}(\sinh (e+f x)),1-\frac{b}{a}\right )}{3 a^2 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}+\frac{2 (2 a-b) \tanh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 a^2 f}-\frac{2 (2 a-b) \coth (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 a^2 f}-\frac{2 (2 a-b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{3 a^2 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac{\coth (e+f x) \text{csch}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 a f} \]

[Out]

(-2*(2*a - b)*Coth[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(3*a^2*f) - (Coth[e + f*x]*Csch[e + f*x]^2*Sqrt[a + b
*Sinh[e + f*x]^2])/(3*a*f) - (2*(2*a - b)*EllipticE[ArcTan[Sinh[e + f*x]], 1 - b/a]*Sech[e + f*x]*Sqrt[a + b*S
inh[e + f*x]^2])/(3*a^2*f*Sqrt[(Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a]) + ((3*a - b)*EllipticF[ArcTan[Sin
h[e + f*x]], 1 - b/a]*Sech[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(3*a^2*f*Sqrt[(Sech[e + f*x]^2*(a + b*Sinh[e
+ f*x]^2))/a]) + (2*(2*a - b)*Sqrt[a + b*Sinh[e + f*x]^2]*Tanh[e + f*x])/(3*a^2*f)

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Rubi [A]  time = 0.304744, antiderivative size = 285, 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 = {3196, 474, 583, 531, 418, 492, 411} \[ \frac{2 (2 a-b) \tanh (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 a^2 f}-\frac{2 (2 a-b) \coth (e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 a^2 f}+\frac{(3 a-b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} F\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{3 a^2 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac{2 (2 a-b) \text{sech}(e+f x) \sqrt{a+b \sinh ^2(e+f x)} E\left (\tan ^{-1}(\sinh (e+f x))|1-\frac{b}{a}\right )}{3 a^2 f \sqrt{\frac{\text{sech}^2(e+f x) \left (a+b \sinh ^2(e+f x)\right )}{a}}}-\frac{\coth (e+f x) \text{csch}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{3 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(2*a - b)*Coth[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(3*a^2*f) - (Coth[e + f*x]*Csch[e + f*x]^2*Sqrt[a + b
*Sinh[e + f*x]^2])/(3*a*f) - (2*(2*a - b)*EllipticE[ArcTan[Sinh[e + f*x]], 1 - b/a]*Sech[e + f*x]*Sqrt[a + b*S
inh[e + f*x]^2])/(3*a^2*f*Sqrt[(Sech[e + f*x]^2*(a + b*Sinh[e + f*x]^2))/a]) + ((3*a - b)*EllipticF[ArcTan[Sin
h[e + f*x]], 1 - b/a]*Sech[e + f*x]*Sqrt[a + b*Sinh[e + f*x]^2])/(3*a^2*f*Sqrt[(Sech[e + f*x]^2*(a + b*Sinh[e
+ f*x]^2))/a]) + (2*(2*a - b)*Sqrt[a + b*Sinh[e + f*x]^2]*Tanh[e + f*x])/(3*a^2*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 474

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

Mathematica [C]  time = 3.54866, size = 208, normalized size = 0.73 \[ \frac{2 i a (a-b) \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} \text{EllipticF}\left (i (e+f x),\frac{b}{a}\right )-\frac{\coth (e+f x) \text{csch}^2(e+f x) \left (2 \left (4 a^2-5 a b+2 b^2\right ) \cosh (2 (e+f x))-(2 a-b) (2 a-b \cosh (4 (e+f x))-3 b)\right )}{\sqrt{2}}-4 i a (2 a-b) \sqrt{\frac{2 a+b \cosh (2 (e+f x))-b}{a}} E\left (i (e+f x)\left |\frac{b}{a}\right .\right )}{6 a^2 f \sqrt{2 a+b \cosh (2 (e+f x))-b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.187, size = 522, normalized size = 1.8 \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)^(1/2),x)

[Out]

-1/3*(4*(-1/a*b)^(1/2)*a*b*sinh(f*x+e)^6-2*(-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+5*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)^3-2*((a+b*si
nh(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)^
3-4*((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+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)^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)^4-2*(-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-(-1/a*b)^(1/2)*a*b*sinh(f*x+e)^2+(-1/a*b)^(1/2)*a^2)/(
-1/a*b)^(1/2)/a^2/sinh(f*x+e)^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}}{\sqrt{b \sinh \left (f x + e\right )^{2} + a}}\,{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)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\coth \left (f x + e\right )^{4}}{\sqrt{b \sinh \left (f x + e\right )^{2} + a}}, 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)^(1/2),x, algorithm="fricas")

[Out]

integral(coth(f*x + e)^4/sqrt(b*sinh(f*x + e)^2 + a), 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)**(1/2),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}}{\sqrt{b \sinh \left (f x + e\right )^{2} + a}}\,{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)^(1/2),x, algorithm="giac")

[Out]

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