3.154 \(\int \frac{a+b \text{csch}^{-1}(c x)}{(d+e x^2)^{3/2}} \, dx\)

Optimal. Leaf size=111 \[ \frac{x \left (a+b \text{csch}^{-1}(c x)\right )}{d \sqrt{d+e x^2}}-\frac{b x \sqrt{d+e x^2} \text{EllipticF}\left (\tan ^{-1}(c x),1-\frac{e}{c^2 d}\right )}{d^2 \sqrt{-c^2 x^2} \sqrt{-c^2 x^2-1} \sqrt{\frac{d+e x^2}{d \left (c^2 x^2+1\right )}}} \]

[Out]

(x*(a + b*ArcCsch[c*x]))/(d*Sqrt[d + e*x^2]) - (b*x*Sqrt[d + e*x^2]*EllipticF[ArcTan[c*x], 1 - e/(c^2*d)])/(d^
2*Sqrt[-(c^2*x^2)]*Sqrt[-1 - c^2*x^2]*Sqrt[(d + e*x^2)/(d*(1 + c^2*x^2))])

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Rubi [A]  time = 0.075112, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {191, 6292, 12, 418} \[ \frac{x \left (a+b \text{csch}^{-1}(c x)\right )}{d \sqrt{d+e x^2}}-\frac{b x \sqrt{d+e x^2} F\left (\tan ^{-1}(c x)|1-\frac{e}{c^2 d}\right )}{d^2 \sqrt{-c^2 x^2} \sqrt{-c^2 x^2-1} \sqrt{\frac{d+e x^2}{d \left (c^2 x^2+1\right )}}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCsch[c*x])/(d + e*x^2)^(3/2),x]

[Out]

(x*(a + b*ArcCsch[c*x]))/(d*Sqrt[d + e*x^2]) - (b*x*Sqrt[d + e*x^2]*EllipticF[ArcTan[c*x], 1 - e/(c^2*d)])/(d^
2*Sqrt[-(c^2*x^2)]*Sqrt[-1 - c^2*x^2]*Sqrt[(d + e*x^2)/(d*(1 + c^2*x^2))])

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 6292

Int[((a_.) + ArcCsch[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(d + e*x^
2)^p, x]}, Dist[a + b*ArcCsch[c*x], u, x] - Dist[(b*c*x)/Sqrt[-(c^2*x^2)], Int[SimplifyIntegrand[u/(x*Sqrt[-1
- c^2*x^2]), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && (IGtQ[p, 0] || ILtQ[p + 1/2, 0])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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]

Rubi steps

\begin{align*} \int \frac{a+b \text{csch}^{-1}(c x)}{\left (d+e x^2\right )^{3/2}} \, dx &=\frac{x \left (a+b \text{csch}^{-1}(c x)\right )}{d \sqrt{d+e x^2}}-\frac{(b c x) \int \frac{1}{d \sqrt{-1-c^2 x^2} \sqrt{d+e x^2}} \, dx}{\sqrt{-c^2 x^2}}\\ &=\frac{x \left (a+b \text{csch}^{-1}(c x)\right )}{d \sqrt{d+e x^2}}-\frac{(b c x) \int \frac{1}{\sqrt{-1-c^2 x^2} \sqrt{d+e x^2}} \, dx}{d \sqrt{-c^2 x^2}}\\ &=\frac{x \left (a+b \text{csch}^{-1}(c x)\right )}{d \sqrt{d+e x^2}}-\frac{b x \sqrt{d+e x^2} F\left (\tan ^{-1}(c x)|1-\frac{e}{c^2 d}\right )}{d^2 \sqrt{-c^2 x^2} \sqrt{-1-c^2 x^2} \sqrt{\frac{d+e x^2}{d \left (1+c^2 x^2\right )}}}\\ \end{align*}

Mathematica [A]  time = 0.192663, size = 113, normalized size = 1.02 \[ \frac{b c x \sqrt{\frac{1}{c^2 x^2}+1} \sqrt{\frac{e x^2}{d}+1} \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{-c^2} x\right ),\frac{e}{c^2 d}\right )}{\sqrt{-c^2} d \sqrt{c^2 x^2+1} \sqrt{d+e x^2}}+\frac{x \left (a+b \text{csch}^{-1}(c x)\right )}{d \sqrt{d+e x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcCsch[c*x])/(d + e*x^2)^(3/2),x]

[Out]

(x*(a + b*ArcCsch[c*x]))/(d*Sqrt[d + e*x^2]) + (b*c*Sqrt[1 + 1/(c^2*x^2)]*x*Sqrt[1 + (e*x^2)/d]*EllipticF[ArcS
in[Sqrt[-c^2]*x], e/(c^2*d)])/(Sqrt[-c^2]*d*Sqrt[1 + c^2*x^2]*Sqrt[d + e*x^2])

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Maple [F]  time = 0.451, size = 0, normalized size = 0. \begin{align*} \int{(a+b{\rm arccsch} \left (cx\right )) \left ( e{x}^{2}+d \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccsch(c*x))/(e*x^2+d)^(3/2),x)

[Out]

int((a+b*arccsch(c*x))/(e*x^2+d)^(3/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} b \int \frac{\log \left (\sqrt{\frac{1}{c^{2} x^{2}} + 1} + \frac{1}{c x}\right )}{{\left (e x^{2} + d\right )}^{\frac{3}{2}}}\,{d x} + \frac{a x}{\sqrt{e x^{2} + d} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsch(c*x))/(e*x^2+d)^(3/2),x, algorithm="maxima")

[Out]

b*integrate(log(sqrt(1/(c^2*x^2) + 1) + 1/(c*x))/(e*x^2 + d)^(3/2), x) + a*x/(sqrt(e*x^2 + d)*d)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{e x^{2} + d}{\left (b \operatorname{arcsch}\left (c x\right ) + a\right )}}{e^{2} x^{4} + 2 \, d e x^{2} + d^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsch(c*x))/(e*x^2+d)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(e*x^2 + d)*(b*arccsch(c*x) + a)/(e^2*x^4 + 2*d*e*x^2 + d^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((a+b*acsch(c*x))/(e*x**2+d)**(3/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsch(c*x))/(e*x^2+d)^(3/2),x, algorithm="giac")

[Out]

integrate((b*arccsch(c*x) + a)/(e*x^2 + d)^(3/2), x)