3.105 \(\int \frac{x^3 (a+b \sec ^{-1}(c x))}{(d+e x^2)^3} \, dx\)
Optimal. Leaf size=157 \[ \frac{x^4 \left (a+b \sec ^{-1}(c x)\right )}{4 d \left (d+e x^2\right )^2}-\frac{b c x \left (c^2 d+2 e\right ) \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{c^2 x^2-1}}{\sqrt{c^2 d+e}}\right )}{8 d e^{3/2} \sqrt{c^2 x^2} \left (c^2 d+e\right )^{3/2}}+\frac{b c x \sqrt{c^2 x^2-1}}{8 e \sqrt{c^2 x^2} \left (c^2 d+e\right ) \left (d+e x^2\right )} \]
[Out]
(b*c*x*Sqrt[-1 + c^2*x^2])/(8*e*(c^2*d + e)*Sqrt[c^2*x^2]*(d + e*x^2)) + (x^4*(a + b*ArcSec[c*x]))/(4*d*(d + e
*x^2)^2) - (b*c*(c^2*d + 2*e)*x*ArcTan[(Sqrt[e]*Sqrt[-1 + c^2*x^2])/Sqrt[c^2*d + e]])/(8*d*e^(3/2)*(c^2*d + e)
^(3/2)*Sqrt[c^2*x^2])
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Rubi [A] time = 0.174512, antiderivative size = 157, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used =
{264, 5238, 12, 446, 78, 63, 205} \[ \frac{x^4 \left (a+b \sec ^{-1}(c x)\right )}{4 d \left (d+e x^2\right )^2}-\frac{b c x \left (c^2 d+2 e\right ) \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{c^2 x^2-1}}{\sqrt{c^2 d+e}}\right )}{8 d e^{3/2} \sqrt{c^2 x^2} \left (c^2 d+e\right )^{3/2}}+\frac{b c x \sqrt{c^2 x^2-1}}{8 e \sqrt{c^2 x^2} \left (c^2 d+e\right ) \left (d+e x^2\right )} \]
Antiderivative was successfully verified.
[In]
Int[(x^3*(a + b*ArcSec[c*x]))/(d + e*x^2)^3,x]
[Out]
(b*c*x*Sqrt[-1 + c^2*x^2])/(8*e*(c^2*d + e)*Sqrt[c^2*x^2]*(d + e*x^2)) + (x^4*(a + b*ArcSec[c*x]))/(4*d*(d + e
*x^2)^2) - (b*c*(c^2*d + 2*e)*x*ArcTan[(Sqrt[e]*Sqrt[-1 + c^2*x^2])/Sqrt[c^2*d + e]])/(8*d*e^(3/2)*(c^2*d + e)
^(3/2)*Sqrt[c^2*x^2])
Rule 264
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]
Rule 5238
Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSec[c*x], u, x] - Dist[(b*c*x)/Sqrt[c^2*x^2], Int[SimplifyI
ntegrand[u/(x*Sqrt[c^2*x^2 - 1]), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && ((IGtQ[p, 0] && !(ILtQ
[(m - 1)/2, 0] && GtQ[m + 2*p + 3, 0])) || (IGtQ[(m + 1)/2, 0] && !(ILtQ[p, 0] && GtQ[m + 2*p + 3, 0])) || (I
LtQ[(m + 2*p + 1)/2, 0] && !ILtQ[(m - 1)/2, 0]))
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
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 78
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ
[p, n]))))
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^3 \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx &=\frac{x^4 \left (a+b \sec ^{-1}(c x)\right )}{4 d \left (d+e x^2\right )^2}-\frac{(b c x) \int \frac{x^3}{4 d \sqrt{-1+c^2 x^2} \left (d+e x^2\right )^2} \, dx}{\sqrt{c^2 x^2}}\\ &=\frac{x^4 \left (a+b \sec ^{-1}(c x)\right )}{4 d \left (d+e x^2\right )^2}-\frac{(b c x) \int \frac{x^3}{\sqrt{-1+c^2 x^2} \left (d+e x^2\right )^2} \, dx}{4 d \sqrt{c^2 x^2}}\\ &=\frac{x^4 \left (a+b \sec ^{-1}(c x)\right )}{4 d \left (d+e x^2\right )^2}-\frac{(b c x) \operatorname{Subst}\left (\int \frac{x}{\sqrt{-1+c^2 x} (d+e x)^2} \, dx,x,x^2\right )}{8 d \sqrt{c^2 x^2}}\\ &=\frac{b c x \sqrt{-1+c^2 x^2}}{8 e \left (c^2 d+e\right ) \sqrt{c^2 x^2} \left (d+e x^2\right )}+\frac{x^4 \left (a+b \sec ^{-1}(c x)\right )}{4 d \left (d+e x^2\right )^2}-\frac{\left (b c \left (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 e \left (c^2 d+e\right ) \sqrt{c^2 x^2}}\\ &=\frac{b c x \sqrt{-1+c^2 x^2}}{8 e \left (c^2 d+e\right ) \sqrt{c^2 x^2} \left (d+e x^2\right )}+\frac{x^4 \left (a+b \sec ^{-1}(c x)\right )}{4 d \left (d+e x^2\right )^2}-\frac{\left (b \left (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 e \left (c^2 d+e\right ) \sqrt{c^2 x^2}}\\ &=\frac{b c x \sqrt{-1+c^2 x^2}}{8 e \left (c^2 d+e\right ) \sqrt{c^2 x^2} \left (d+e x^2\right )}+\frac{x^4 \left (a+b \sec ^{-1}(c x)\right )}{4 d \left (d+e x^2\right )^2}-\frac{b c \left (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 e^{3/2} \left (c^2 d+e\right )^{3/2} \sqrt{c^2 x^2}}\\ \end{align*}
Mathematica [C] time = 1.31066, size = 389, normalized size = 2.48 \[ -\frac{\frac{8 a}{d+e x^2}-\frac{4 a d}{\left (d+e x^2\right )^2}+\frac{b \sqrt{e} \left (c^2 d+2 e\right ) \log \left (-\frac{16 d e^{3/2} \sqrt{c^2 (-d)-e} \left (\sqrt{e}+c x \left (-\sqrt{1-\frac{1}{c^2 x^2}} \sqrt{c^2 (-d)-e}+i c \sqrt{d}\right )\right )}{b \left (c^2 d+2 e\right ) \left (\sqrt{e} x+i \sqrt{d}\right )}\right )}{d \left (c^2 (-d)-e\right )^{3/2}}+\frac{b \sqrt{e} \left (c^2 d+2 e\right ) \log \left (\frac{16 i d e^{3/2} \sqrt{c^2 (-d)-e} \left (-\sqrt{e}+c x \left (\sqrt{1-\frac{1}{c^2 x^2}} \sqrt{c^2 (-d)-e}+i c \sqrt{d}\right )\right )}{b \left (c^2 d+2 e\right ) \left (\sqrt{d}+i \sqrt{e} x\right )}\right )}{d \left (c^2 (-d)-e\right )^{3/2}}-\frac{2 b c e x \sqrt{1-\frac{1}{c^2 x^2}}}{\left (c^2 d+e\right ) \left (d+e x^2\right )}+\frac{4 b \sec ^{-1}(c x) \left (d+2 e x^2\right )}{\left (d+e x^2\right )^2}+\frac{4 b \sin ^{-1}\left (\frac{1}{c x}\right )}{d}}{16 e^2} \]
Antiderivative was successfully verified.
[In]
Integrate[(x^3*(a + b*ArcSec[c*x]))/(d + e*x^2)^3,x]
[Out]
-((-4*a*d)/(d + e*x^2)^2 + (8*a)/(d + e*x^2) - (2*b*c*e*Sqrt[1 - 1/(c^2*x^2)]*x)/((c^2*d + e)*(d + e*x^2)) + (
4*b*(d + 2*e*x^2)*ArcSec[c*x])/(d + e*x^2)^2 + (4*b*ArcSin[1/(c*x)])/d + (b*Sqrt[e]*(c^2*d + 2*e)*Log[(-16*d*S
qrt[-(c^2*d) - e]*e^(3/2)*(Sqrt[e] + c*(I*c*Sqrt[d] - Sqrt[-(c^2*d) - e]*Sqrt[1 - 1/(c^2*x^2)])*x))/(b*(c^2*d
+ 2*e)*(I*Sqrt[d] + Sqrt[e]*x))])/(d*(-(c^2*d) - e)^(3/2)) + (b*Sqrt[e]*(c^2*d + 2*e)*Log[((16*I)*d*Sqrt[-(c^2
*d) - e]*e^(3/2)*(-Sqrt[e] + c*(I*c*Sqrt[d] + Sqrt[-(c^2*d) - e]*Sqrt[1 - 1/(c^2*x^2)])*x))/(b*(c^2*d + 2*e)*(
Sqrt[d] + I*Sqrt[e]*x))])/(d*(-(c^2*d) - e)^(3/2)))/(16*e^2)
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Maple [B] time = 0.277, size = 1870, normalized size = 11.9 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^3*(a+b*arcsec(c*x))/(e*x^2+d)^3,x)
[Out]
-1/2*c^2*a/e^2/(c^2*e*x^2+c^2*d)+1/4*c^4*a/e^2*d/(c^2*e*x^2+c^2*d)^2-1/2*c^2*b*arcsec(c*x)/e^2/(c^2*e*x^2+c^2*
d)+1/4*c^4*b*arcsec(c*x)/e^2*d/(c^2*e*x^2+c^2*d)^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))+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))-1/16*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)))-1/16*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)))-1/16*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)))-1/16*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)))+1/4*c*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*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))-1/8*c^3*b/((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))+1/8*c*b/((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))-1/8*c*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)))-1/8*c*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)))-1/8*c*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)))-1/8*c*
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)))
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{{\left (2 \, e x^{2} + d\right )} a}{4 \,{\left (e^{4} x^{4} + 2 \, d e^{3} x^{2} + d^{2} e^{2}\right )}} - \frac{{\left ({\left (2 \, e x^{2} + d\right )} \arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right ) -{\left (e^{4} x^{4} + 2 \, d e^{3} x^{2} + d^{2} e^{2}\right )} \int \frac{{\left (2 \, c^{2} e x^{3} + c^{2} d x\right )} e^{\left (\frac{1}{2} \, \log \left (c x + 1\right ) + \frac{1}{2} \, \log \left (c x - 1\right )\right )}}{c^{2} e^{4} x^{6} +{\left (2 \, c^{2} d e^{3} - e^{4}\right )} x^{4} - d^{2} e^{2} +{\left (c^{2} e^{4} x^{6} +{\left (2 \, c^{2} d e^{3} - e^{4}\right )} x^{4} - d^{2} e^{2} +{\left (c^{2} d^{2} e^{2} - 2 \, d e^{3}\right )} x^{2}\right )}{\left (c x + 1\right )}{\left (c x - 1\right )} +{\left (c^{2} d^{2} e^{2} - 2 \, d e^{3}\right )} x^{2}}\,{d x}\right )} b}{4 \,{\left (e^{4} x^{4} + 2 \, d e^{3} x^{2} + d^{2} e^{2}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(a+b*arcsec(c*x))/(e*x^2+d)^3,x, algorithm="maxima")
[Out]
-1/4*(2*e*x^2 + d)*a/(e^4*x^4 + 2*d*e^3*x^2 + d^2*e^2) - 1/4*((2*e*x^2 + d)*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)
) - 4*(e^4*x^4 + 2*d*e^3*x^2 + d^2*e^2)*integrate(1/4*(2*c^2*e*x^3 + c^2*d*x)*e^(1/2*log(c*x + 1) + 1/2*log(c*
x - 1))/(c^2*e^4*x^6 + (2*c^2*d*e^3 - e^4)*x^4 - d^2*e^2 + (c^2*d^2*e^2 - 2*d*e^3)*x^2 + (c^2*e^4*x^6 + (2*c^2
*d*e^3 - e^4)*x^4 - d^2*e^2 + (c^2*d^2*e^2 - 2*d*e^3)*x^2)*e^(log(c*x + 1) + log(c*x - 1))), x))*b/(e^4*x^4 +
2*d*e^3*x^2 + d^2*e^2)
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Fricas [B] time = 4.35592, size = 2093, normalized size = 13.33 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(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 + 8*(a*c^4*d^3*e + 2*a*c^2*d^2*e^2 + a*d*e^3)*x^2 + (b*c^2*d
^3 + (b*c^2*d*e^2 + 2*b*e^3)*x^4 + 2*b*d^2*e + 2*(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 + 2*(b*c^4*d^3*e + 2*b*c^2*d^2*e^2 + b*d*e^3)*x^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^5*e^2 + 2*c^2*d^4*e^3 + d^3*e^4 + (c^4*d^3*e^4 + 2*c^2*d^2*e^5 + d*e^6)*x^4 + 2*(c^4*d^4*e^3 + 2*c^2*d^3*e
^4 + d^2*e^5)*x^2), -1/8*(2*a*c^4*d^4 + 4*a*c^2*d^3*e + 2*a*d^2*e^2 + 4*(a*c^4*d^3*e + 2*a*c^2*d^2*e^2 + a*d*e
^3)*x^2 + (b*c^2*d^3 + (b*c^2*d*e^2 + 2*b*e^3)*x^4 + 2*b*d^2*e + 2*(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 +
2*(b*c^4*d^3*e + 2*b*c^2*d^2*e^2 + b*d*e^3)*x^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)*sqrt(c^2*x^2 - 1))/(c^4*d^5*e^2 + 2
*c^2*d^4*e^3 + d^3*e^4 + (c^4*d^3*e^4 + 2*c^2*d^2*e^5 + d*e^6)*x^4 + 2*(c^4*d^4*e^3 + 2*c^2*d^3*e^4 + d^2*e^5)
*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**3*(a+b*asec(c*x))/(e*x**2+d)**3,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^{3}}{{\left (e x^{2} + d\right )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^3*(a+b*arcsec(c*x))/(e*x^2+d)^3,x, algorithm="giac")
[Out]
integrate((b*arcsec(c*x) + a)*x^3/(e*x^2 + d)^3, x)