3.106 \(\int \frac{x (a+b \sec ^{-1}(c x))}{(d+e x^2)^3} \, dx\)
Optimal. Leaf size=193 \[ -\frac{a+b \sec ^{-1}(c x)}{4 e \left (d+e x^2\right )^2}+\frac{b c x \tan ^{-1}\left (\sqrt{c^2 x^2-1}\right )}{4 d^2 e \sqrt{c^2 x^2}}-\frac{b c x \left (3 c^2 d+2 e\right ) \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{c^2 x^2-1}}{\sqrt{c^2 d+e}}\right )}{8 d^2 \sqrt{e} \sqrt{c^2 x^2} \left (c^2 d+e\right )^{3/2}}-\frac{b c x \sqrt{c^2 x^2-1}}{8 d \sqrt{c^2 x^2} \left (c^2 d+e\right ) \left (d+e x^2\right )} \]
[Out]
-(b*c*x*Sqrt[-1 + c^2*x^2])/(8*d*(c^2*d + e)*Sqrt[c^2*x^2]*(d + e*x^2)) - (a + b*ArcSec[c*x])/(4*e*(d + e*x^2)
^2) + (b*c*x*ArcTan[Sqrt[-1 + c^2*x^2]])/(4*d^2*e*Sqrt[c^2*x^2]) - (b*c*(3*c^2*d + 2*e)*x*ArcTan[(Sqrt[e]*Sqrt
[-1 + c^2*x^2])/Sqrt[c^2*d + e]])/(8*d^2*Sqrt[e]*(c^2*d + e)^(3/2)*Sqrt[c^2*x^2])
________________________________________________________________________________________
Rubi [A] time = 0.18969, antiderivative size = 193, 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 =
{5236, 446, 103, 156, 63, 205} \[ -\frac{a+b \sec ^{-1}(c x)}{4 e \left (d+e x^2\right )^2}+\frac{b c x \tan ^{-1}\left (\sqrt{c^2 x^2-1}\right )}{4 d^2 e \sqrt{c^2 x^2}}-\frac{b c x \left (3 c^2 d+2 e\right ) \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{c^2 x^2-1}}{\sqrt{c^2 d+e}}\right )}{8 d^2 \sqrt{e} \sqrt{c^2 x^2} \left (c^2 d+e\right )^{3/2}}-\frac{b c x \sqrt{c^2 x^2-1}}{8 d \sqrt{c^2 x^2} \left (c^2 d+e\right ) \left (d+e x^2\right )} \]
Antiderivative was successfully verified.
[In]
Int[(x*(a + b*ArcSec[c*x]))/(d + e*x^2)^3,x]
[Out]
-(b*c*x*Sqrt[-1 + c^2*x^2])/(8*d*(c^2*d + e)*Sqrt[c^2*x^2]*(d + e*x^2)) - (a + b*ArcSec[c*x])/(4*e*(d + e*x^2)
^2) + (b*c*x*ArcTan[Sqrt[-1 + c^2*x^2]])/(4*d^2*e*Sqrt[c^2*x^2]) - (b*c*(3*c^2*d + 2*e)*x*ArcTan[(Sqrt[e]*Sqrt
[-1 + c^2*x^2])/Sqrt[c^2*d + e]])/(8*d^2*Sqrt[e]*(c^2*d + e)^(3/2)*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 103
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])
Rule 156
Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
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 205
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rubi steps
\begin{align*} \int \frac{x \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx &=-\frac{a+b \sec ^{-1}(c x)}{4 e \left (d+e x^2\right )^2}+\frac{(b c x) \int \frac{1}{x \sqrt{-1+c^2 x^2} \left (d+e x^2\right )^2} \, dx}{4 e \sqrt{c^2 x^2}}\\ &=-\frac{a+b \sec ^{-1}(c x)}{4 e \left (d+e x^2\right )^2}+\frac{(b c x) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{-1+c^2 x} (d+e x)^2} \, dx,x,x^2\right )}{8 e \sqrt{c^2 x^2}}\\ &=-\frac{b c x \sqrt{-1+c^2 x^2}}{8 d \left (c^2 d+e\right ) \sqrt{c^2 x^2} \left (d+e x^2\right )}-\frac{a+b \sec ^{-1}(c x)}{4 e \left (d+e x^2\right )^2}-\frac{(b c x) \operatorname{Subst}\left (\int \frac{-c^2 d-e+\frac{1}{2} c^2 e x}{x \sqrt{-1+c^2 x} (d+e x)} \, dx,x,x^2\right )}{8 d e \left (c^2 d+e\right ) \sqrt{c^2 x^2}}\\ &=-\frac{b c x \sqrt{-1+c^2 x^2}}{8 d \left (c^2 d+e\right ) \sqrt{c^2 x^2} \left (d+e x^2\right )}-\frac{a+b \sec ^{-1}(c x)}{4 e \left (d+e x^2\right )^2}+\frac{(b c x) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{-1+c^2 x}} \, dx,x,x^2\right )}{8 d^2 e \sqrt{c^2 x^2}}-\frac{\left (b c \left (3 c^2 d+2 e\right ) x\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-1+c^2 x} (d+e x)} \, dx,x,x^2\right )}{16 d^2 \left (c^2 d+e\right ) \sqrt{c^2 x^2}}\\ &=-\frac{b c x \sqrt{-1+c^2 x^2}}{8 d \left (c^2 d+e\right ) \sqrt{c^2 x^2} \left (d+e x^2\right )}-\frac{a+b \sec ^{-1}(c x)}{4 e \left (d+e x^2\right )^2}+\frac{(b x) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{c^2}+\frac{x^2}{c^2}} \, dx,x,\sqrt{-1+c^2 x^2}\right )}{4 c d^2 e \sqrt{c^2 x^2}}-\frac{\left (b \left (3 c^2 d+2 e\right ) 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 )}{8 c d^2 \left (c^2 d+e\right ) \sqrt{c^2 x^2}}\\ &=-\frac{b c x \sqrt{-1+c^2 x^2}}{8 d \left (c^2 d+e\right ) \sqrt{c^2 x^2} \left (d+e x^2\right )}-\frac{a+b \sec ^{-1}(c x)}{4 e \left (d+e x^2\right )^2}+\frac{b c x \tan ^{-1}\left (\sqrt{-1+c^2 x^2}\right )}{4 d^2 e \sqrt{c^2 x^2}}-\frac{b c \left (3 c^2 d+2 e\right ) x \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{-1+c^2 x^2}}{\sqrt{c^2 d+e}}\right )}{8 d^2 \sqrt{e} \left (c^2 d+e\right )^{3/2} \sqrt{c^2 x^2}}\\ \end{align*}
Mathematica [C] time = 0.877326, size = 386, normalized size = 2. \[ \frac{1}{16} \left (-\frac{4 a}{e \left (d+e x^2\right )^2}-\frac{b \left (3 c^2 d+2 e\right ) \log \left (-\frac{16 d^2 \sqrt{e} \sqrt{c^2 (-d)-e} \left (\sqrt{e}+c x \left (-\sqrt{1-\frac{1}{c^2 x^2}} \sqrt{c^2 (-d)-e}+i c \sqrt{d}\right )\right )}{b \left (3 c^2 d+2 e\right ) \left (\sqrt{e} x+i \sqrt{d}\right )}\right )}{d^2 \sqrt{e} \left (c^2 (-d)-e\right )^{3/2}}-\frac{b \left (3 c^2 d+2 e\right ) \log \left (\frac{16 i d^2 \sqrt{e} \sqrt{c^2 (-d)-e} \left (-\sqrt{e}+c x \left (\sqrt{1-\frac{1}{c^2 x^2}} \sqrt{c^2 (-d)-e}+i c \sqrt{d}\right )\right )}{b \left (3 c^2 d+2 e\right ) \left (\sqrt{d}+i \sqrt{e} x\right )}\right )}{d^2 \sqrt{e} \left (c^2 (-d)-e\right )^{3/2}}-\frac{2 b c x \sqrt{1-\frac{1}{c^2 x^2}}}{d \left (c^2 d+e\right ) \left (d+e x^2\right )}-\frac{4 b \sin ^{-1}\left (\frac{1}{c x}\right )}{d^2 e}-\frac{4 b \sec ^{-1}(c x)}{e \left (d+e x^2\right )^2}\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[(x*(a + b*ArcSec[c*x]))/(d + e*x^2)^3,x]
[Out]
((-4*a)/(e*(d + e*x^2)^2) - (2*b*c*Sqrt[1 - 1/(c^2*x^2)]*x)/(d*(c^2*d + e)*(d + e*x^2)) - (4*b*ArcSec[c*x])/(e
*(d + e*x^2)^2) - (4*b*ArcSin[1/(c*x)])/(d^2*e) - (b*(3*c^2*d + 2*e)*Log[(-16*d^2*Sqrt[-(c^2*d) - e]*Sqrt[e]*(
Sqrt[e] + c*(I*c*Sqrt[d] - Sqrt[-(c^2*d) - e]*Sqrt[1 - 1/(c^2*x^2)])*x))/(b*(3*c^2*d + 2*e)*(I*Sqrt[d] + Sqrt[
e]*x))])/(d^2*(-(c^2*d) - e)^(3/2)*Sqrt[e]) - (b*(3*c^2*d + 2*e)*Log[((16*I)*d^2*Sqrt[-(c^2*d) - e]*Sqrt[e]*(-
Sqrt[e] + c*(I*c*Sqrt[d] + Sqrt[-(c^2*d) - e]*Sqrt[1 - 1/(c^2*x^2)])*x))/(b*(3*c^2*d + 2*e)*(Sqrt[d] + I*Sqrt[
e]*x))])/(d^2*(-(c^2*d) - e)^(3/2)*Sqrt[e]))/16
________________________________________________________________________________________
Maple [B] time = 0.262, size = 1840, normalized size = 9.5 \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*arcsec(c*x))/(e*x^2+d)^3,x)
[Out]
-1/4*c^4*a/e/(c^2*e*x^2+c^2*d)^2-1/4*c^4*b/e/(c^2*e*x^2+c^2*d)^2*arcsec(c*x)+1/4*c^3*b*(c^2*x^2-1)^(1/2)*e/((c
^2*x^2-1)/c^2/x^2)^(1/2)*x/d/(c^2*d+e)/(e*c*x+(-c^2*e*d)^(1/2))/(-e*c*x+(-c^2*e*d)^(1/2))*arctan(1/(c^2*x^2-1)
^(1/2))+1/4*c^3*b*(c^2*x^2-1)^(1/2)/((c^2*x^2-1)/c^2/x^2)^(1/2)/x/(c^2*d+e)/(e*c*x+(-c^2*e*d)^(1/2))/(-e*c*x+(
-c^2*e*d)^(1/2))*arctan(1/(c^2*x^2-1)^(1/2))-3/16*c^3*b*(c^2*x^2-1)^(1/2)*e/((c^2*x^2-1)/c^2/x^2)^(1/2)*x/d/(-
(c^2*d+e)/e)^(1/2)/(c^2*d+e)/(e*c*x+(-c^2*e*d)^(1/2))/(-e*c*x+(-c^2*e*d)^(1/2))*ln(-2*((-(c^2*d+e)/e)^(1/2)*(c
^2*x^2-1)^(1/2)*e+(-c^2*e*d)^(1/2)*c*x-e)/(-e*c*x+(-c^2*e*d)^(1/2)))-3/16*c^3*b*(c^2*x^2-1)^(1/2)/((c^2*x^2-1)
/c^2/x^2)^(1/2)/x/(-(c^2*d+e)/e)^(1/2)/(c^2*d+e)/(e*c*x+(-c^2*e*d)^(1/2))/(-e*c*x+(-c^2*e*d)^(1/2))*ln(-2*((-(
c^2*d+e)/e)^(1/2)*(c^2*x^2-1)^(1/2)*e+(-c^2*e*d)^(1/2)*c*x-e)/(-e*c*x+(-c^2*e*d)^(1/2)))-3/16*c^3*b*(c^2*x^2-1
)^(1/2)*e/((c^2*x^2-1)/c^2/x^2)^(1/2)*x/d/(-(c^2*d+e)/e)^(1/2)/(c^2*d+e)/(e*c*x+(-c^2*e*d)^(1/2))/(-e*c*x+(-c^
2*e*d)^(1/2))*ln(2*((-(c^2*d+e)/e)^(1/2)*(c^2*x^2-1)^(1/2)*e-(-c^2*e*d)^(1/2)*c*x-e)/(e*c*x+(-c^2*e*d)^(1/2)))
-3/16*c^3*b*(c^2*x^2-1)^(1/2)/((c^2*x^2-1)/c^2/x^2)^(1/2)/x/(-(c^2*d+e)/e)^(1/2)/(c^2*d+e)/(e*c*x+(-c^2*e*d)^(
1/2))/(-e*c*x+(-c^2*e*d)^(1/2))*ln(2*((-(c^2*d+e)/e)^(1/2)*(c^2*x^2-1)^(1/2)*e-(-c^2*e*d)^(1/2)*c*x-e)/(e*c*x+
(-c^2*e*d)^(1/2)))+1/4*c*b*(c^2*x^2-1)^(1/2)/((c^2*x^2-1)/c^2/x^2)^(1/2)*x/d^2/(c^2*d+e)/(-e*c*x+(-c^2*e*d)^(1
/2))/(e*c*x+(-c^2*e*d)^(1/2))*arctan(1/(c^2*x^2-1)^(1/2))*e^2+1/4*c*b*(c^2*x^2-1)^(1/2)/((c^2*x^2-1)/c^2/x^2)^
(1/2)/x/d/(c^2*d+e)/(-e*c*x+(-c^2*e*d)^(1/2))/(e*c*x+(-c^2*e*d)^(1/2))*arctan(1/(c^2*x^2-1)^(1/2))*e+1/8*c^3*b
/((c^2*x^2-1)/c^2/x^2)^(1/2)*x/d/(c^2*d+e)/(-e*c*x+(-c^2*e*d)^(1/2))/(e*c*x+(-c^2*e*d)^(1/2))*e-1/8*c*b/((c^2*
x^2-1)/c^2/x^2)^(1/2)/x/d/(c^2*d+e)/(-e*c*x+(-c^2*e*d)^(1/2))/(e*c*x+(-c^2*e*d)^(1/2))*e-1/8*c*b*(c^2*x^2-1)^(
1/2)/((c^2*x^2-1)/c^2/x^2)^(1/2)*x/d^2/(c^2*d+e)/(-e*c*x+(-c^2*e*d)^(1/2))/(-(c^2*d+e)/e)^(1/2)/(e*c*x+(-c^2*e
*d)^(1/2))*ln(-2*((-(c^2*d+e)/e)^(1/2)*(c^2*x^2-1)^(1/2)*e+(-c^2*e*d)^(1/2)*c*x-e)/(-e*c*x+(-c^2*e*d)^(1/2)))*
e^2-1/8*c*b*(c^2*x^2-1)^(1/2)/((c^2*x^2-1)/c^2/x^2)^(1/2)/x/d/(c^2*d+e)/(-e*c*x+(-c^2*e*d)^(1/2))/(-(c^2*d+e)/
e)^(1/2)/(e*c*x+(-c^2*e*d)^(1/2))*ln(-2*((-(c^2*d+e)/e)^(1/2)*(c^2*x^2-1)^(1/2)*e+(-c^2*e*d)^(1/2)*c*x-e)/(-e*
c*x+(-c^2*e*d)^(1/2)))*e-1/8*c*b*(c^2*x^2-1)^(1/2)/((c^2*x^2-1)/c^2/x^2)^(1/2)*x/d^2/(c^2*d+e)/(-e*c*x+(-c^2*e
*d)^(1/2))/(-(c^2*d+e)/e)^(1/2)/(e*c*x+(-c^2*e*d)^(1/2))*ln(2*((-(c^2*d+e)/e)^(1/2)*(c^2*x^2-1)^(1/2)*e-(-c^2*
e*d)^(1/2)*c*x-e)/(e*c*x+(-c^2*e*d)^(1/2)))*e^2-1/8*c*b*(c^2*x^2-1)^(1/2)/((c^2*x^2-1)/c^2/x^2)^(1/2)/x/d/(c^2
*d+e)/(-e*c*x+(-c^2*e*d)^(1/2))/(-(c^2*d+e)/e)^(1/2)/(e*c*x+(-c^2*e*d)^(1/2))*ln(2*((-(c^2*d+e)/e)^(1/2)*(c^2*
x^2-1)^(1/2)*e-(-c^2*e*d)^(1/2)*c*x-e)/(e*c*x+(-c^2*e*d)^(1/2)))*e
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left ({\left (c^{2} e^{3} x^{4} + 2 \, c^{2} d e^{2} x^{2} + c^{2} d^{2} e\right )} \int \frac{x e^{\left (\frac{1}{2} \, \log \left (c x + 1\right ) + \frac{1}{2} \, \log \left (c x - 1\right )\right )}}{c^{2} e^{3} x^{6} +{\left (2 \, c^{2} d e^{2} - e^{3}\right )} x^{4} +{\left (c^{2} e^{3} x^{6} +{\left (2 \, c^{2} d e^{2} - e^{3}\right )} x^{4} - d^{2} e +{\left (c^{2} d^{2} e - 2 \, d e^{2}\right )} x^{2}\right )}{\left (c x + 1\right )}{\left (c x - 1\right )} - d^{2} e +{\left (c^{2} d^{2} e - 2 \, d e^{2}\right )} x^{2}}\,{d x} - \arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right )\right )} b}{4 \,{\left (e^{3} x^{4} + 2 \, d e^{2} x^{2} + d^{2} e\right )}} - \frac{a}{4 \,{\left (e^{3} x^{4} + 2 \, d e^{2} x^{2} + d^{2} e\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,x, algorithm="maxima")
[Out]
1/4*(4*(c^2*e^3*x^4 + 2*c^2*d*e^2*x^2 + c^2*d^2*e)*integrate(1/4*x*e^(1/2*log(c*x + 1) + 1/2*log(c*x - 1))/(c^
2*e^3*x^6 + (2*c^2*d*e^2 - e^3)*x^4 - d^2*e + (c^2*d^2*e - 2*d*e^2)*x^2 + (c^2*e^3*x^6 + (2*c^2*d*e^2 - e^3)*x
^4 - d^2*e + (c^2*d^2*e - 2*d*e^2)*x^2)*e^(log(c*x + 1) + log(c*x - 1))), x) - arctan(sqrt(c*x + 1)*sqrt(c*x -
1)))*b/(e^3*x^4 + 2*d*e^2*x^2 + d^2*e) - 1/4*a/(e^3*x^4 + 2*d*e^2*x^2 + d^2*e)
________________________________________________________________________________________
Fricas [B] time = 5.00913, size = 1839, normalized size = 9.53 \begin{align*} \text{result too large to display} \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,x, algorithm="fricas")
[Out]
[-1/16*(4*a*c^4*d^4 + 8*a*c^2*d^3*e + 4*a*d^2*e^2 + (3*b*c^2*d^3 + (3*b*c^2*d*e^2 + 2*b*e^3)*x^4 + 2*b*d^2*e +
2*(3*b*c^2*d^2*e + 2*b*d*e^2)*x^2)*sqrt(-c^2*d*e - e^2)*log((c^2*e*x^2 - c^2*d + 2*sqrt(-c^2*d*e - e^2)*sqrt(
c^2*x^2 - 1) - 2*e)/(e*x^2 + d)) + 4*(b*c^4*d^4 + 2*b*c^2*d^3*e + b*d^2*e^2)*arcsec(c*x) - 8*(b*c^4*d^4 + 2*b*
c^2*d^3*e + b*d^2*e^2 + (b*c^4*d^2*e^2 + 2*b*c^2*d*e^3 + b*e^4)*x^4 + 2*(b*c^4*d^3*e + 2*b*c^2*d^2*e^2 + b*d*e
^3)*x^2)*arctan(-c*x + sqrt(c^2*x^2 - 1)) + 2*(b*c^2*d^3*e + b*d^2*e^2 + (b*c^2*d^2*e^2 + b*d*e^3)*x^2)*sqrt(c
^2*x^2 - 1))/(c^4*d^6*e + 2*c^2*d^5*e^2 + d^4*e^3 + (c^4*d^4*e^3 + 2*c^2*d^3*e^4 + d^2*e^5)*x^4 + 2*(c^4*d^5*e
^2 + 2*c^2*d^4*e^3 + d^3*e^4)*x^2), -1/8*(2*a*c^4*d^4 + 4*a*c^2*d^3*e + 2*a*d^2*e^2 + (3*b*c^2*d^3 + (3*b*c^2*
d*e^2 + 2*b*e^3)*x^4 + 2*b*d^2*e + 2*(3*b*c^2*d^2*e + 2*b*d*e^2)*x^2)*sqrt(c^2*d*e + e^2)*arctan(sqrt(c^2*d*e
+ e^2)*sqrt(c^2*x^2 - 1)/(c^2*d + e)) + 2*(b*c^4*d^4 + 2*b*c^2*d^3*e + b*d^2*e^2)*arcsec(c*x) - 4*(b*c^4*d^4 +
2*b*c^2*d^3*e + b*d^2*e^2 + (b*c^4*d^2*e^2 + 2*b*c^2*d*e^3 + b*e^4)*x^4 + 2*(b*c^4*d^3*e + 2*b*c^2*d^2*e^2 +
b*d*e^3)*x^2)*arctan(-c*x + sqrt(c^2*x^2 - 1)) + (b*c^2*d^3*e + b*d^2*e^2 + (b*c^2*d^2*e^2 + b*d*e^3)*x^2)*sqr
t(c^2*x^2 - 1))/(c^4*d^6*e + 2*c^2*d^5*e^2 + d^4*e^3 + (c^4*d^4*e^3 + 2*c^2*d^3*e^4 + d^2*e^5)*x^4 + 2*(c^4*d^
5*e^2 + 2*c^2*d^4*e^3 + d^3*e^4)*x^2)]
________________________________________________________________________________________
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,x)
[Out]
Timed out
________________________________________________________________________________________
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 )}^{3}}\,{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,x, algorithm="giac")
[Out]
integrate((b*arcsec(c*x) + a)*x/(e*x^2 + d)^3, x)