\(\int e^{\sec ^{-1}(a x)} x^2 \, dx\) [43]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 10, antiderivative size = 99 \[ \int e^{\sec ^{-1}(a x)} x^2 \, dx=-\frac {\left (\frac {12}{5}+\frac {4 i}{5}\right ) e^{(1+3 i) \sec ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{2}-\frac {i}{2},3,\frac {5}{2}-\frac {i}{2},-e^{2 i \sec ^{-1}(a x)}\right )}{a^3}+\frac {\left (\frac {24}{5}+\frac {8 i}{5}\right ) e^{(1+3 i) \sec ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{2}-\frac {i}{2},4,\frac {5}{2}-\frac {i}{2},-e^{2 i \sec ^{-1}(a x)}\right )}{a^3} \] Output:

(-12/5-4/5*I)*exp((1+3*I)*arcsec(a*x))*hypergeom([3, 3/2-1/2*I],[5/2-1/2*I 
],-(1/a/x+I*(1-1/a^2/x^2)^(1/2))^2)/a^3+(24/5+8/5*I)*exp((1+3*I)*arcsec(a* 
x))*hypergeom([4, 3/2-1/2*I],[5/2-1/2*I],-(1/a/x+I*(1-1/a^2/x^2)^(1/2))^2) 
/a^3
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.96 \[ \int e^{\sec ^{-1}(a x)} x^2 \, dx=\frac {e^{\sec ^{-1}(a x)} \left ((-4-4 i) \left (-i+a \sqrt {1-\frac {1}{a^2 x^2}} x\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-\frac {i}{2},1,\frac {3}{2}-\frac {i}{2},-e^{2 i \sec ^{-1}(a x)}\right )+a^4 x^4 \left (5+\cos \left (2 \sec ^{-1}(a x)\right )-\sin \left (2 \sec ^{-1}(a x)\right )\right )\right )}{12 a^4 x} \] Input:

Integrate[E^ArcSec[a*x]*x^2,x]
 

Output:

(E^ArcSec[a*x]*((-4 - 4*I)*(-I + a*Sqrt[1 - 1/(a^2*x^2)]*x)*Hypergeometric 
2F1[1/2 - I/2, 1, 3/2 - I/2, -E^((2*I)*ArcSec[a*x])] + a^4*x^4*(5 + Cos[2* 
ArcSec[a*x]] - Sin[2*ArcSec[a*x]])))/(12*a^4*x)
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5789, 27, 4974, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int x^2 e^{\sec ^{-1}(a x)} \, dx\)

\(\Big \downarrow \) 5789

\(\displaystyle \frac {\int a^2 e^{\sec ^{-1}(a x)} \sqrt {1-\frac {1}{a^2 x^2}} x^4d\sec ^{-1}(a x)}{a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int a^4 e^{\sec ^{-1}(a x)} \sqrt {1-\frac {1}{a^2 x^2}} x^4d\sec ^{-1}(a x)}{a^3}\)

\(\Big \downarrow \) 4974

\(\displaystyle \frac {\int \left (\frac {16 i e^{(1+3 i) \sec ^{-1}(a x)}}{\left (1+e^{2 i \sec ^{-1}(a x)}\right )^4}-\frac {8 i e^{(1+3 i) \sec ^{-1}(a x)}}{\left (1+e^{2 i \sec ^{-1}(a x)}\right )^3}\right )d\sec ^{-1}(a x)}{a^3}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\left (\frac {24}{5}+\frac {8 i}{5}\right ) e^{(1+3 i) \sec ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{2}-\frac {i}{2},4,\frac {5}{2}-\frac {i}{2},-e^{2 i \sec ^{-1}(a x)}\right )-\left (\frac {12}{5}+\frac {4 i}{5}\right ) e^{(1+3 i) \sec ^{-1}(a x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{2}-\frac {i}{2},3,\frac {5}{2}-\frac {i}{2},-e^{2 i \sec ^{-1}(a x)}\right )}{a^3}\)

Input:

Int[E^ArcSec[a*x]*x^2,x]
 

Output:

((-12/5 - (4*I)/5)*E^((1 + 3*I)*ArcSec[a*x])*Hypergeometric2F1[3/2 - I/2, 
3, 5/2 - I/2, -E^((2*I)*ArcSec[a*x])] + (24/5 + (8*I)/5)*E^((1 + 3*I)*ArcS 
ec[a*x])*Hypergeometric2F1[3/2 - I/2, 4, 5/2 - I/2, -E^((2*I)*ArcSec[a*x]) 
])/a^3
 

Defintions of rubi rules used

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 4974
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*(G_)[(d_.) + (e_.)*(x_)]^(m_.)*(H_)[( 
d_.) + (e_.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigToExp[F^(c*(a + b*x)), 
 G[d + e*x]^m*H[d + e*x]^n, x], x] /; FreeQ[{F, a, b, c, d, e}, x] && IGtQ[ 
m, 0] && IGtQ[n, 0] && TrigQ[G] && TrigQ[H]
 

rule 5789
Int[(u_.)*(f_)^(ArcSec[(a_.) + (b_.)*(x_)]^(n_.)*(c_.)), x_Symbol] :> Simp[ 
1/b   Subst[Int[(u /. x -> -a/b + Sec[x]/b)*f^(c*x^n)*Sec[x]*Tan[x], x], x, 
 ArcSec[a + b*x]], x] /; FreeQ[{a, b, c, f}, x] && IGtQ[n, 0]
 
Maple [F]

\[\int {\mathrm e}^{\operatorname {arcsec}\left (a x \right )} x^{2}d x\]

Input:

int(exp(arcsec(a*x))*x^2,x)
 

Output:

int(exp(arcsec(a*x))*x^2,x)
 

Fricas [F]

\[ \int e^{\sec ^{-1}(a x)} x^2 \, dx=\int { x^{2} e^{\left (\operatorname {arcsec}\left (a x\right )\right )} \,d x } \] Input:

integrate(exp(arcsec(a*x))*x^2,x, algorithm="fricas")
 

Output:

integral(x^2*e^(arcsec(a*x)), x)
 

Sympy [F]

\[ \int e^{\sec ^{-1}(a x)} x^2 \, dx=\int x^{2} e^{\operatorname {asec}{\left (a x \right )}}\, dx \] Input:

integrate(exp(asec(a*x))*x**2,x)
 

Output:

Integral(x**2*exp(asec(a*x)), x)
 

Maxima [F]

\[ \int e^{\sec ^{-1}(a x)} x^2 \, dx=\int { x^{2} e^{\left (\operatorname {arcsec}\left (a x\right )\right )} \,d x } \] Input:

integrate(exp(arcsec(a*x))*x^2,x, algorithm="maxima")
 

Output:

integrate(x^2*e^(arcsec(a*x)), x)
 

Giac [F]

\[ \int e^{\sec ^{-1}(a x)} x^2 \, dx=\int { x^{2} e^{\left (\operatorname {arcsec}\left (a x\right )\right )} \,d x } \] Input:

integrate(exp(arcsec(a*x))*x^2,x, algorithm="giac")
 

Output:

integrate(x^2*e^(arcsec(a*x)), x)
 

Mupad [F(-1)]

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

int(x^2*exp(acos(1/(a*x))),x)
 

Output:

int(x^2*exp(acos(1/(a*x))), x)
 

Reduce [F]

\[ \int e^{\sec ^{-1}(a x)} x^2 \, dx=\int e^{\mathit {asec} \left (a x \right )} x^{2}d x \] Input:

int(exp(asec(a*x))*x^2,x)
 

Output:

int(e**asec(a*x)*x**2,x)