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

Optimal. Leaf size=80 \[ -\frac{a+b \sec ^{-1}(c x)}{e \sqrt{d+e x^2}}-\frac{b c x \tan ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d} \sqrt{c^2 x^2-1}}\right )}{\sqrt{d} e \sqrt{c^2 x^2}} \]

[Out]

-((a + b*ArcSec[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])])/(Sqr
t[d]*e*Sqrt[c^2*x^2])

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Rubi [A]  time = 0.101998, antiderivative size = 80, 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 = {5236, 446, 93, 204} \[ -\frac{a+b \sec ^{-1}(c x)}{e \sqrt{d+e x^2}}-\frac{b c x \tan ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d} \sqrt{c^2 x^2-1}}\right )}{\sqrt{d} e \sqrt{c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b*ArcSec[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])])/(Sqr
t[d]*e*Sqrt[c^2*x^2])

Rule 5236

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*(x_)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)^(p +
1)*(a + b*ArcSec[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[c^2*x^2 - 1]), 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 \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx &=-\frac{a+b \sec ^{-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 \sec ^{-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 \sec ^{-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 \sec ^{-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 = 0.145414, size = 95, normalized size = 1.19 \[ \frac{b c x \sqrt{1-\frac{1}{c^2 x^2}} \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{c^2 x^2-1}}{\sqrt{d+e x^2}}\right )}{\sqrt{d} e \sqrt{c^2 x^2-1}}-\frac{a+b \sec ^{-1}(c x)}{e \sqrt{d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 1.233, size = 0, normalized size = 0. \begin{align*} \int{x \left ( a+b{\rm arcsec} \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*arcsec(c*x))/(e*x^2+d)^(3/2),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.35462, size = 626, normalized size = 7.82 \begin{align*} \left [-\frac{{\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 \, \sqrt{c^{2} x^{2} - 1}{\left ({\left (c^{2} d - e\right )} x^{2} - 2 \, d\right )} \sqrt{e x^{2} + d} \sqrt{-d} + 8 \, d^{2}}{x^{4}}\right ) + 4 \, \sqrt{e x^{2} + d}{\left (b d \operatorname{arcsec}\left (c x\right ) + a d\right )}}{4 \,{\left (d e^{2} x^{2} + d^{2} e\right )}}, -\frac{{\left (b e x^{2} + b d\right )} \sqrt{d} \arctan \left (-\frac{\sqrt{c^{2} x^{2} - 1}{\left ({\left (c^{2} d - e\right )} x^{2} - 2 \, d\right )} \sqrt{e x^{2} + d} \sqrt{d}}{2 \,{\left (c^{2} d e x^{4} +{\left (c^{2} d^{2} - d e\right )} x^{2} - d^{2}\right )}}\right ) + 2 \, \sqrt{e x^{2} + d}{\left (b d \operatorname{arcsec}\left (c x\right ) + a d\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*arcsec(c*x))/(e*x^2+d)^(3/2),x, algorithm="fricas")

[Out]

[-1/4*((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*sqrt(c^2*x^2
- 1)*((c^2*d - e)*x^2 - 2*d)*sqrt(e*x^2 + d)*sqrt(-d) + 8*d^2)/x^4) + 4*sqrt(e*x^2 + d)*(b*d*arcsec(c*x) + a*d
))/(d*e^2*x^2 + d^2*e), -1/2*((b*e*x^2 + b*d)*sqrt(d)*arctan(-1/2*sqrt(c^2*x^2 - 1)*((c^2*d - e)*x^2 - 2*d)*sq
rt(e*x^2 + d)*sqrt(d)/(c^2*d*e*x^4 + (c^2*d^2 - d*e)*x^2 - d^2)) + 2*sqrt(e*x^2 + d)*(b*d*arcsec(c*x) + a*d))/
(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*asec(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{arcsec}\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*arcsec(c*x))/(e*x^2+d)^(3/2),x, algorithm="giac")

[Out]

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