\(\int \frac {x (a+b \sec ^{-1}(c x))}{(d+e x^2)^{3/2}} \, dx\) [143]

Optimal result

Integrand size = 21, antiderivative size = 80 \[ \int \frac {x \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=-\frac {a+b \sec ^{-1}(c x)}{e \sqrt {d+e x^2}}-\frac {b c x \arctan \left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {-1+c^2 x^2}}\right )}{\sqrt {d} e \sqrt {c^2 x^2}} \]

[Out]

-b*c*x*arctan((e*x^2+d)^(1/2)/d^(1/2)/(c^2*x^2-1)^(1/2))/e/d^(1/2)/(c^2*x^2)^(1/2)+(-a-b*arcsec(c*x))/e/(e*x^2
+d)^(1/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {5344, 457, 95, 210} \[ \int \frac {x \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=-\frac {a+b \sec ^{-1}(c x)}{e \sqrt {d+e x^2}}-\frac {b c x \arctan \left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {c^2 x^2-1}}\right )}{\sqrt {d} e \sqrt {c^2 x^2}} \]

[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 95

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 210

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

Rule 457

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 5344

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]

Rubi steps \begin{align*} \text {integral}& = -\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) \text {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) \text {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 \arctan \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 [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 0.24 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.99 \[ \int \frac {x \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {-b \sqrt {1+\frac {d}{e x^2}} \operatorname {AppellF1}\left (1,\frac {1}{2},\frac {1}{2},2,\frac {1}{c^2 x^2},-\frac {d}{e x^2}\right )-2 c x \left (a+b \sec ^{-1}(c x)\right )}{2 c e x \sqrt {d+e x^2}} \]

[In]

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

[Out]

(-(b*Sqrt[1 + d/(e*x^2)]*AppellF1[1, 1/2, 1/2, 2, 1/(c^2*x^2), -(d/(e*x^2))]) - 2*c*x*(a + b*ArcSec[c*x]))/(2*
c*e*x*Sqrt[d + e*x^2])

Maple [F]

\[\int \frac {x \left (a +b \,\operatorname {arcsec}\left (c x \right )\right )}{\left (e \,x^{2}+d \right )^{\frac {3}{2}}}d x\]

[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)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 283, normalized size of antiderivative = 3.54 \[ \int \frac {x \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\left [-\frac {{\left (b e x^{2} + b d\right )} \sqrt {-d} \log \left (\frac {{\left (c^{4} d^{2} - 6 \, c^{2} d e + e^{2}\right )} x^{4} - 8 \, {\left (c^{2} d^{2} - d e\right )} x^{2} - 4 \, \sqrt {c^{2} x^{2} - 1} {\left ({\left (c^{2} d - e\right )} x^{2} - 2 \, d\right )} \sqrt {e x^{2} + d} \sqrt {-d} + 8 \, d^{2}}{x^{4}}\right ) + 4 \, \sqrt {e x^{2} + d} {\left (b d \operatorname {arcsec}\left (c x\right ) + a d\right )}}{4 \, {\left (d e^{2} x^{2} + d^{2} e\right )}}, -\frac {{\left (b e x^{2} + b d\right )} \sqrt {d} \arctan \left (-\frac {\sqrt {c^{2} x^{2} - 1} {\left ({\left (c^{2} d - e\right )} x^{2} - 2 \, d\right )} \sqrt {e x^{2} + d} \sqrt {d}}{2 \, {\left (c^{2} d e x^{4} + {\left (c^{2} d^{2} - d e\right )} x^{2} - d^{2}\right )}}\right ) + 2 \, \sqrt {e x^{2} + d} {\left (b d \operatorname {arcsec}\left (c x\right ) + a d\right )}}{2 \, {\left (d e^{2} x^{2} + d^{2} e\right )}}\right ] \]

[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)]

Sympy [F]

\[ \int \frac {x \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\int \frac {x \left (a + b \operatorname {asec}{\left (c x \right )}\right )}{\left (d + e x^{2}\right )^{\frac {3}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {x \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\int { \frac {{\left (b \operatorname {arcsec}\left (c x\right ) + a\right )} x}{{\left (e x^{2} + d\right )}^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

-(sqrt(e*x^2 + d)*e*integrate((c^2*e*x^3*log(c) - e*x*log(c) + ((c^2*log(c) - c^2)*e*x^3 - (c^2*d + e*log(c))*
x)*e^(log(c*x + 1) + log(c*x - 1)) + (c^2*e*x^3 - e*x + (c^2*e*x^3 - e*x)*e^(log(c*x + 1) + log(c*x - 1)))*log
(x))/((c^2*e^2*x^4 + (c^2*d*e - e^2)*x^2 - d*e + (c^2*e^2*x^4 + (c^2*d*e - e^2)*x^2 - d*e)*e^(log(c*x + 1) + l
og(c*x - 1)))*sqrt(e*x^2 + d)), x) + arctan(sqrt(c*x + 1)*sqrt(c*x - 1)))*b/(sqrt(e*x^2 + d)*e) - a/(sqrt(e*x^
2 + d)*e)

Giac [F]

\[ \int \frac {x \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\int { \frac {{\left (b \operatorname {arcsec}\left (c x\right ) + a\right )} x}{{\left (e x^{2} + d\right )}^{\frac {3}{2}}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {x \left (a+b \sec ^{-1}(c x)\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\int \frac {x\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}{{\left (e\,x^2+d\right )}^{3/2}} \,d x \]

[In]

int((x*(a + b*acos(1/(c*x))))/(d + e*x^2)^(3/2),x)

[Out]

int((x*(a + b*acos(1/(c*x))))/(d + e*x^2)^(3/2), x)