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

Optimal. Leaf size=147 \[ -\frac{a+b \text{sech}^{-1}(c x)}{2 e \left (d+e x^2\right )}+\frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )}{2 d e}-\frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{1-c^2 x^2}}{\sqrt{c^2 d+e}}\right )}{2 d \sqrt{e} \sqrt{c^2 d+e}} \]

[Out]

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

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Rubi [A]  time = 0.244675, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {6299, 517, 446, 86, 63, 208} \[ -\frac{a+b \text{sech}^{-1}(c x)}{2 e \left (d+e x^2\right )}+\frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )}{2 d e}-\frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{1-c^2 x^2}}{\sqrt{c^2 d+e}}\right )}{2 d \sqrt{e} \sqrt{c^2 d+e}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a + b*ArcSech[c*x])/(2*e*(d + e*x^2)) + (b*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcTanh[Sqrt[1 - c^2*x^2]])/(2
*d*e) - (b*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcTanh[(Sqrt[e]*Sqrt[1 - c^2*x^2])/Sqrt[c^2*d + e]])/(2*d*Sqrt[
e]*Sqrt[c^2*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 86

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), In
t[(e + f*x)^p/(a + b*x), x], x] - Dist[d/(b*c - a*d), Int[(e + f*x)^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d,
e, f, p}, x] &&  !IntegerQ[p]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{x \left (a+b \text{sech}^{-1}(c x)\right )}{\left (d+e x^2\right )^2} \, dx &=-\frac{a+b \text{sech}^{-1}(c x)}{2 e \left (d+e x^2\right )}-\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} \left (d+e x^2\right )} \, dx}{2 e}\\ &=-\frac{a+b \text{sech}^{-1}(c x)}{2 e \left (d+e x^2\right )}-\frac{\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{1}{x \sqrt{1-c^2 x^2} \left (d+e x^2\right )} \, dx}{2 e}\\ &=-\frac{a+b \text{sech}^{-1}(c x)}{2 e \left (d+e x^2\right )}-\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} (d+e x)} \, dx,x,x^2\right )}{4 e}\\ &=-\frac{a+b \text{sech}^{-1}(c x)}{2 e \left (d+e x^2\right )}+\frac{\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-c^2 x} (d+e x)} \, dx,x,x^2\right )}{4 d}-\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}} \, dx,x,x^2\right )}{4 d e}\\ &=-\frac{a+b \text{sech}^{-1}(c x)}{2 e \left (d+e x^2\right )}-\frac{\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \operatorname{Subst}\left (\int \frac{1}{d+\frac{e}{c^2}-\frac{e x^2}{c^2}} \, dx,x,\sqrt{1-c^2 x^2}\right )}{2 c^2 d}+\frac{\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{c^2}-\frac{x^2}{c^2}} \, dx,x,\sqrt{1-c^2 x^2}\right )}{2 c^2 d e}\\ &=-\frac{a+b \text{sech}^{-1}(c x)}{2 e \left (d+e x^2\right )}+\frac{b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )}{2 d e}-\frac{b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{1-c^2 x^2}}{\sqrt{c^2 d+e}}\right )}{2 d \sqrt{e} \sqrt{c^2 d+e}}\\ \end{align*}

Mathematica [C]  time = 0.963418, size = 345, normalized size = 2.35 \[ -\frac{\frac{2 a}{d+e x^2}+\frac{b \sqrt{e} \log \left (\frac{4 \left (\frac{c^2 d^{3/2} \sqrt{e} x+i d e}{\sqrt{c^2 d+e} \left (\sqrt{d}+i \sqrt{e} x\right )}+\frac{d e \sqrt{\frac{1-c x}{c x+1}} (c x+1)}{e x-i \sqrt{d} \sqrt{e}}\right )}{b}\right )}{d \sqrt{c^2 d+e}}+\frac{b \sqrt{e} \log \left (\frac{4 \left (\frac{d e+i c^2 d^{3/2} \sqrt{e} x}{\sqrt{c^2 d+e} \left (\sqrt{e} x+i \sqrt{d}\right )}+\frac{d e \sqrt{\frac{1-c x}{c x+1}} (c x+1)}{e x+i \sqrt{d} \sqrt{e}}\right )}{b}\right )}{d \sqrt{c^2 d+e}}+\frac{2 b \text{sech}^{-1}(c x)}{d+e x^2}-\frac{2 b \log \left (c x \sqrt{\frac{1-c x}{c x+1}}+\sqrt{\frac{1-c x}{c x+1}}+1\right )}{d}+\frac{2 b \log (x)}{d}}{4 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2*a)/(d + e*x^2) + (2*b*ArcSech[c*x])/(d + e*x^2) + (2*b*Log[x])/d - (2*b*Log[1 + Sqrt[(1 - c*x)/(1 + c*x)]
+ c*x*Sqrt[(1 - c*x)/(1 + c*x)]])/d + (b*Sqrt[e]*Log[(4*((I*d*e + c^2*d^(3/2)*Sqrt[e]*x)/(Sqrt[c^2*d + e]*(Sq
rt[d] + I*Sqrt[e]*x)) + (d*e*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x))/((-I)*Sqrt[d]*Sqrt[e] + e*x)))/b])/(d*Sqrt[c
^2*d + e]) + (b*Sqrt[e]*Log[(4*((d*e + I*c^2*d^(3/2)*Sqrt[e]*x)/(Sqrt[c^2*d + e]*(I*Sqrt[d] + Sqrt[e]*x)) + (d
*e*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x))/(I*Sqrt[d]*Sqrt[e] + e*x)))/b])/(d*Sqrt[c^2*d + e]))/(4*e)

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Maple [B]  time = 0.283, size = 844, normalized size = 5.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*c^2*a/e/(c^2*e*x^2+c^2*d)-1/2*c^2*b/e/(c^2*e*x^2+c^2*d)*arcsech(c*x)-1/2*c^3*b*(-(c*x-1)/c/x)^(1/2)*x*((c
*x+1)/c/x)^(1/2)/(-c^2*x^2+1)^(1/2)/((-c^2*d*e)^(1/2)+e)/((-c^2*d*e)^(1/2)-e)*arctanh(1/(-c^2*x^2+1)^(1/2))+1/
4*c^3*b*(-(c*x-1)/c/x)^(1/2)*x*((c*x+1)/c/x)^(1/2)/(-c^2*x^2+1)^(1/2)/((-c^2*d*e)^(1/2)+e)/((-c^2*d*e)^(1/2)-e
)/((c^2*d+e)/e)^(1/2)*ln(2*((-c^2*x^2+1)^(1/2)*((c^2*d+e)/e)^(1/2)*e+(-c^2*d*e)^(1/2)*c*x+e)/(c*x*e+(-c^2*d*e)
^(1/2)))+1/4*c^3*b*(-(c*x-1)/c/x)^(1/2)*x*((c*x+1)/c/x)^(1/2)/(-c^2*x^2+1)^(1/2)/((-c^2*d*e)^(1/2)+e)/((-c^2*d
*e)^(1/2)-e)/((c^2*d+e)/e)^(1/2)*ln(2*(-(-c^2*x^2+1)^(1/2)*((c^2*d+e)/e)^(1/2)*e+(-c^2*d*e)^(1/2)*c*x-e)/(-c*x
*e+(-c^2*d*e)^(1/2)))-1/2*c*b*(-(c*x-1)/c/x)^(1/2)*x*((c*x+1)/c/x)^(1/2)/(-c^2*x^2+1)^(1/2)/d/((-c^2*d*e)^(1/2
)+e)/((-c^2*d*e)^(1/2)-e)*arctanh(1/(-c^2*x^2+1)^(1/2))*e+1/4*c*b*(-(c*x-1)/c/x)^(1/2)*x*((c*x+1)/c/x)^(1/2)/(
-c^2*x^2+1)^(1/2)/d/((-c^2*d*e)^(1/2)+e)/((-c^2*d*e)^(1/2)-e)/((c^2*d+e)/e)^(1/2)*ln(2*((-c^2*x^2+1)^(1/2)*((c
^2*d+e)/e)^(1/2)*e+(-c^2*d*e)^(1/2)*c*x+e)/(c*x*e+(-c^2*d*e)^(1/2)))*e+1/4*c*b*(-(c*x-1)/c/x)^(1/2)*x*((c*x+1)
/c/x)^(1/2)/(-c^2*x^2+1)^(1/2)/d/((-c^2*d*e)^(1/2)+e)/((-c^2*d*e)^(1/2)-e)/((c^2*d+e)/e)^(1/2)*ln(2*(-(-c^2*x^
2+1)^(1/2)*((c^2*d+e)/e)^(1/2)*e+(-c^2*d*e)^(1/2)*c*x-e)/(-c*x*e+(-c^2*d*e)^(1/2)))*e

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

[Out]

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

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Fricas [B]  time = 1.97457, size = 1265, normalized size = 8.61 \begin{align*} \left [-\frac{2 \, a c^{2} d^{2} + 2 \, a d e - \sqrt{c^{2} d e + e^{2}}{\left (b e x^{2} + b d\right )} \log \left (\frac{c^{4} d^{2} + 4 \, c^{2} d e -{\left (c^{4} d e + 2 \, c^{2} e^{2}\right )} x^{2} + 4 \,{\left (c^{3} d e + c e^{2}\right )} x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} + 4 \, e^{2} + 2 \,{\left (c^{2} e x^{2} - c^{2} d -{\left (c^{3} d + 2 \, c e\right )} x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} - 2 \, e\right )} \sqrt{c^{2} d e + e^{2}}}{e x^{2} + d}\right ) + 2 \,{\left (b c^{2} d^{2} + b d e +{\left (b c^{2} d e + b e^{2}\right )} x^{2}\right )} \log \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{x}\right ) + 2 \,{\left (b c^{2} d^{2} + b d e\right )} \log \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right )}{4 \,{\left (c^{2} d^{3} e + d^{2} e^{2} +{\left (c^{2} d^{2} e^{2} + d e^{3}\right )} x^{2}\right )}}, -\frac{a c^{2} d^{2} + a d e + \sqrt{-c^{2} d e - e^{2}}{\left (b e x^{2} + b d\right )} \arctan \left (\frac{\sqrt{-c^{2} d e - e^{2}} c d x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} - \sqrt{-c^{2} d e - e^{2}}{\left (e x^{2} + d\right )}}{{\left (c^{2} d e + e^{2}\right )} x^{2}}\right ) +{\left (b c^{2} d^{2} + b d e +{\left (b c^{2} d e + b e^{2}\right )} x^{2}\right )} \log \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{x}\right ) +{\left (b c^{2} d^{2} + b d e\right )} \log \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right )}{2 \,{\left (c^{2} d^{3} e + d^{2} e^{2} +{\left (c^{2} d^{2} e^{2} + d e^{3}\right )} x^{2}\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)^2,x, algorithm="fricas")

[Out]

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

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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*asech(c*x))/(e*x**2+d)**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{arsech}\left (c x\right ) + a\right )} x}{{\left (e x^{2} + d\right )}^{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)^2,x, algorithm="giac")

[Out]

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