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

Optimal. Leaf size=87 \[ \frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d} \sqrt{1-c^2 x^2}}\right )}{\sqrt{d} e}-\frac{a+b \text{sech}^{-1}(c x)}{e \sqrt{d+e x^2}} \]

[Out]

-((a + b*ArcSech[c*x])/(e*Sqrt[d + e*x^2])) + (b*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcTanh[Sqrt[d + e*x^2]/(S
qrt[d]*Sqrt[1 - c^2*x^2])])/(Sqrt[d]*e)

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Rubi [A]  time = 0.245075, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {6299, 517, 446, 93, 207} \[ \frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d} \sqrt{1-c^2 x^2}}\right )}{\sqrt{d} e}-\frac{a+b \text{sech}^{-1}(c x)}{e \sqrt{d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b*ArcSech[c*x])/(e*Sqrt[d + e*x^2])) + (b*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcTanh[Sqrt[d + e*x^2]/(S
qrt[d]*Sqrt[1 - c^2*x^2])])/(Sqrt[d]*e)

Rule 6299

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

Rule 517

Int[(u_.)*((c_) + (d_.)*(x_)^(n_.))^(q_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b2_.)*(x_)^(non2_.))^
(p_.), x_Symbol] :> Int[u*(a1*a2 + b1*b2*x^n)^p*(c + d*x^n)^q, x] /; FreeQ[{a1, b1, a2, b2, c, d, n, p, q}, x]
&& EqQ[non2, n/2] && EqQ[a2*b1 + a1*b2, 0] && (IntegerQ[p] || (GtQ[a1, 0] && GtQ[a2, 0]))

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 207

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

Rubi steps

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

Mathematica [A]  time = 0.59144, size = 135, normalized size = 1.55 \[ -\frac{a+b \text{sech}^{-1}(c x)}{e \sqrt{d+e x^2}}-\frac{b \sqrt{\frac{1-c x}{c x+1}} \sqrt{1-c^2 x^2} \sqrt{-d-e x^2} \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{1-c^2 x^2}}{\sqrt{-d-e x^2}}\right )}{\sqrt{d} e (c x-1) \sqrt{d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.918, size = 0, normalized size = 0. \begin{align*} \int{x \left ( a+b{\rm arcsech} \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*arcsech(c*x))/(e*x^2+d)^(3/2),x)

[Out]

int(x*(a+b*arcsech(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{x \log \left (\sqrt{\frac{1}{c x} + 1} \sqrt{\frac{1}{c x} - 1} + \frac{1}{c x}\right )}{{\left (e x^{2} + d\right )}^{\frac{3}{2}}}\,{d x} - \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*arcsech(c*x))/(e*x^2+d)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.39633, size = 829, normalized size = 9.53 \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*arcsech(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)*s
qrt(-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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \left (a + b \operatorname{asech}{\left (c x \right )}\right )}{\left (d + e x^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*(a + b*asech(c*x))/(d + e*x**2)**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arsech}\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*arcsech(c*x))/(e*x^2+d)^(3/2),x, algorithm="giac")

[Out]

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