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

Optimal. Leaf size=82 \[ -\frac{a+b \text{csch}^{-1}(c x)}{e \sqrt{d+e x^2}}-\frac{b c x \tan ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d} \sqrt{-c^2 x^2-1}}\right )}{\sqrt{d} e \sqrt{-c^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.11148, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {6300, 446, 93, 204} \[ -\frac{a+b \text{csch}^{-1}(c x)}{e \sqrt{d+e x^2}}-\frac{b c x \tan ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d} \sqrt{-c^2 x^2-1}}\right )}{\sqrt{d} e \sqrt{-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6300

Int[((a_.) + ArcCsch[(c_.)*(x_)]*(b_.))*(x_)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)^(p +
1)*(a + b*ArcCsch[c*x]))/(2*e*(p + 1)), x] - Dist[(b*c*x)/(2*e*(p + 1)*Sqrt[-(c^2*x^2)]), Int[(d + e*x^2)^(p
+ 1)/(x*Sqrt[-1 - c^2*x^2]), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.15874, size = 122, normalized size = 1.49 \[ \frac{b c x \sqrt{\frac{1}{c^2 x^2}+1} \sqrt{-d-e x^2} \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{c^2 x^2+1}}{\sqrt{-d-e x^2}}\right )}{\sqrt{d} e \sqrt{c^2 x^2+1} \sqrt{d+e x^2}}-\frac{a+b \text{csch}^{-1}(c x)}{e \sqrt{d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.463, size = 0, normalized size = 0. \begin{align*} \int{x \left ( a+b{\rm arccsch} \left (cx\right ) \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(x*(a+b*arccsch(c*x))/(e*x^2+d)^(3/2),x)

[Out]

int(x*(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*} -{\left (c^{2} \int \frac{x}{{\left (c^{2} e x^{2} + e\right )} \sqrt{c^{2} x^{2} + 1} \sqrt{e x^{2} + d} +{\left (c^{2} e x^{2} + e\right )} \sqrt{e x^{2} + d}}\,{d x} + \frac{\log \left (\sqrt{c^{2} x^{2} + 1} + 1\right )}{\sqrt{e x^{2} + d} e} + \int \frac{{\left (e \log \left (c\right ) - e\right )} c^{2} x^{3} -{\left (c^{2} d - e \log \left (c\right )\right )} x +{\left (c^{2} e x^{3} + e x\right )} \log \left (x\right )}{{\left (c^{2} e^{2} x^{4} +{\left (c^{2} d e + e^{2}\right )} x^{2} + d e\right )} \sqrt{e x^{2} + d}}\,{d x}\right )} b - \frac{a}{\sqrt{e x^{2} + d} e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c^2*integrate(x/((c^2*e*x^2 + e)*sqrt(c^2*x^2 + 1)*sqrt(e*x^2 + d) + (c^2*e*x^2 + e)*sqrt(e*x^2 + d)), x) +
log(sqrt(c^2*x^2 + 1) + 1)/(sqrt(e*x^2 + d)*e) + integrate(((e*log(c) - e)*c^2*x^3 - (c^2*d - e*log(c))*x + (c
^2*e*x^3 + e*x)*log(x))/((c^2*e^2*x^4 + (c^2*d*e + e^2)*x^2 + d*e)*sqrt(e*x^2 + d)), x))*b - a/(sqrt(e*x^2 + d
)*e)

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Fricas [B]  time = 3.04869, size = 822, normalized size = 10.02 \begin{align*} \left [-\frac{4 \, \sqrt{e x^{2} + d} b d \log \left (\frac{c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right ) + 4 \, \sqrt{e x^{2} + d} a d -{\left (b e x^{2} + b d\right )} \sqrt{d} \log \left (\frac{{\left (c^{4} d^{2} + 6 \, c^{2} d e + e^{2}\right )} x^{4} + 8 \,{\left (c^{2} d^{2} + d e\right )} x^{2} + 4 \,{\left ({\left (c^{3} d + c e\right )} x^{3} + 2 \, c d x\right )} \sqrt{e x^{2} + d} \sqrt{d} \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} + 8 \, d^{2}}{x^{4}}\right )}{4 \,{\left (d e^{2} x^{2} + d^{2} e\right )}}, -\frac{2 \, \sqrt{e x^{2} + d} b d \log \left (\frac{c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right ) + 2 \, \sqrt{e x^{2} + d} a d +{\left (b e x^{2} + b d\right )} \sqrt{-d} \arctan \left (\frac{{\left ({\left (c^{3} d + c e\right )} x^{3} + 2 \, c d x\right )} \sqrt{e x^{2} + d} \sqrt{-d} \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}}}{2 \,{\left (c^{2} d e x^{4} +{\left (c^{2} d^{2} + d e\right )} x^{2} + d^{2}\right )}}\right )}{2 \,{\left (d e^{2} x^{2} + d^{2} e\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(4*sqrt(e*x^2 + d)*b*d*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x)) + 4*sqrt(e*x^2 + d)*a*d - (b*e
*x^2 + b*d)*sqrt(d)*log(((c^4*d^2 + 6*c^2*d*e + e^2)*x^4 + 8*(c^2*d^2 + d*e)*x^2 + 4*((c^3*d + c*e)*x^3 + 2*c*
d*x)*sqrt(e*x^2 + d)*sqrt(d)*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 8*d^2)/x^4))/(d*e^2*x^2 + d^2*e), -1/2*(2*sqrt(e*
x^2 + d)*b*d*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x)) + 2*sqrt(e*x^2 + d)*a*d + (b*e*x^2 + b*d)*sqrt
(-d)*arctan(1/2*((c^3*d + c*e)*x^3 + 2*c*d*x)*sqrt(e*x^2 + d)*sqrt(-d)*sqrt((c^2*x^2 + 1)/(c^2*x^2))/(c^2*d*e*
x^4 + (c^2*d^2 + d*e)*x^2 + d^2)))/(d*e^2*x^2 + d^2*e)]

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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(x*(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{{\left (b \operatorname{arcsch}\left (c x\right ) + a\right )} x}{{\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(x*(a+b*arccsch(c*x))/(e*x^2+d)^(3/2),x, algorithm="giac")

[Out]

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