\(\int \frac {a+b \sec ^{-1}(c x)}{x (d+e x^2)^{5/2}} \, dx\) [154]

   Optimal result
   Rubi [N/A]
   Mathematica [N/A]
   Maple [N/A] (verified)
   Fricas [N/A]
   Sympy [F(-1)]
   Maxima [F(-2)]
   Giac [N/A]
   Mupad [N/A]

Optimal result

Integrand size = 23, antiderivative size = 23 \[ \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^{5/2}} \, dx=\text {Int}\left (\frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^{5/2}},x\right ) \]

[Out]

Unintegrable((a+b*arcsec(c*x))/x/(e*x^2+d)^(5/2),x)

Rubi [N/A]

Not integrable

Time = 0.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 0, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^{5/2}} \, dx=\int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^{5/2}} \, dx \]

[In]

Int[(a + b*ArcSec[c*x])/(x*(d + e*x^2)^(5/2)),x]

[Out]

Defer[Int][(a + b*ArcSec[c*x])/(x*(d + e*x^2)^(5/2)), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^{5/2}} \, dx \\ \end{align*}

Mathematica [N/A]

Not integrable

Time = 14.07 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^{5/2}} \, dx=\int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^{5/2}} \, dx \]

[In]

Integrate[(a + b*ArcSec[c*x])/(x*(d + e*x^2)^(5/2)),x]

[Out]

Integrate[(a + b*ArcSec[c*x])/(x*(d + e*x^2)^(5/2)), x]

Maple [N/A] (verified)

Not integrable

Time = 1.82 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91

\[\int \frac {a +b \,\operatorname {arcsec}\left (c x \right )}{x \left (e \,x^{2}+d \right )^{\frac {5}{2}}}d x\]

[In]

int((a+b*arcsec(c*x))/x/(e*x^2+d)^(5/2),x)

[Out]

int((a+b*arcsec(c*x))/x/(e*x^2+d)^(5/2),x)

Fricas [N/A]

Not integrable

Time = 0.27 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.30 \[ \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^{5/2}} \, dx=\int { \frac {b \operatorname {arcsec}\left (c x\right ) + a}{{\left (e x^{2} + d\right )}^{\frac {5}{2}} x} \,d x } \]

[In]

integrate((a+b*arcsec(c*x))/x/(e*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(e*x^2 + d)*(b*arcsec(c*x) + a)/(e^3*x^7 + 3*d*e^2*x^5 + 3*d^2*e*x^3 + d^3*x), x)

Sympy [F(-1)]

Timed out. \[ \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+b*asec(c*x))/x/(e*x**2+d)**(5/2),x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((a+b*arcsec(c*x))/x/(e*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [N/A]

Not integrable

Time = 0.38 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^{5/2}} \, dx=\int { \frac {b \operatorname {arcsec}\left (c x\right ) + a}{{\left (e x^{2} + d\right )}^{\frac {5}{2}} x} \,d x } \]

[In]

integrate((a+b*arcsec(c*x))/x/(e*x^2+d)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsec(c*x) + a)/((e*x^2 + d)^(5/2)*x), x)

Mupad [N/A]

Not integrable

Time = 1.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {a+b \sec ^{-1}(c x)}{x \left (d+e x^2\right )^{5/2}} \, dx=\int \frac {a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )}{x\,{\left (e\,x^2+d\right )}^{5/2}} \,d x \]

[In]

int((a + b*acos(1/(c*x)))/(x*(d + e*x^2)^(5/2)),x)

[Out]

int((a + b*acos(1/(c*x)))/(x*(d + e*x^2)^(5/2)), x)