3.666 \(\int \frac {(a+b \sin ^{-1}(c x))^2}{(d+e x^2)^{3/2}} \, dx\)

Optimal. Leaf size=25 \[ \text {Int}\left (\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{\left (d+e x^2\right )^{3/2}},x\right ) \]

[Out]

Unintegrable((a+b*arcsin(c*x))^2/(e*x^2+d)^(3/2),x)

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Rubi [A]  time = 0.04, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{\left (d+e x^2\right )^{3/2}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

Defer[Int][(a + b*ArcSin[c*x])^2/(d + e*x^2)^(3/2), x]

Rubi steps

\begin {align*} \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{\left (d+e x^2\right )^{3/2}} \, dx &=\int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{\left (d+e x^2\right )^{3/2}} \, dx\\ \end {align*}

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Mathematica [A]  time = 4.33, size = 0, normalized size = 0.00 \[ \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{\left (d+e x^2\right )^{3/2}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}\right )} \sqrt {e x^{2} + d}}{e^{2} x^{4} + 2 \, d e x^{2} + d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*arcsin(c*x)^2 + 2*a*b*arcsin(c*x) + a^2)*sqrt(e*x^2 + d)/(e^2*x^4 + 2*d*e*x^2 + d^2), x)

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giac [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (e x^{2} + d\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.68, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arcsin \left (c x \right )\right )^{2}}{\left (e \,x^{2}+d \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))^2/(e*x^2+d)^(3/2),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-c^2*d>0)', see `assume?` for
more details)Is e-c^2*d zero or nonzero?

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mupad [A]  time = 0.00, size = -1, normalized size = -0.04 \[ \int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{{\left (e\,x^2+d\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{\left (d + e x^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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