3.403 \(\int \frac{(1-c^2 x^2)^{5/2}}{x (a+b \sin ^{-1}(c x))^2} \, dx\)

Optimal. Leaf size=234 \[ -\frac{\text{Unintegrable}\left (\frac{\left (1-c^2 x^2\right )^2}{x^2 \left (a+b \sin ^{-1}(c x)\right )},x\right )}{b c}-\frac{25 \cos \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a+b \sin ^{-1}(c x)}{b}\right )}{8 b^2}-\frac{25 \cos \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (\frac{3 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2}-\frac{5 \cos \left (\frac{5 a}{b}\right ) \text{CosIntegral}\left (\frac{5 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2}-\frac{25 \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a+b \sin ^{-1}(c x)}{b}\right )}{8 b^2}-\frac{25 \sin \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2}-\frac{5 \sin \left (\frac{5 a}{b}\right ) \text{Si}\left (\frac{5 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{16 b^2}-\frac{\left (1-c^2 x^2\right )^3}{b c x \left (a+b \sin ^{-1}(c x)\right )} \]

[Out]

-((1 - c^2*x^2)^3/(b*c*x*(a + b*ArcSin[c*x]))) - (25*Cos[a/b]*CosIntegral[(a + b*ArcSin[c*x])/b])/(8*b^2) - (2
5*Cos[(3*a)/b]*CosIntegral[(3*(a + b*ArcSin[c*x]))/b])/(16*b^2) - (5*Cos[(5*a)/b]*CosIntegral[(5*(a + b*ArcSin
[c*x]))/b])/(16*b^2) - (25*Sin[a/b]*SinIntegral[(a + b*ArcSin[c*x])/b])/(8*b^2) - (25*Sin[(3*a)/b]*SinIntegral
[(3*(a + b*ArcSin[c*x]))/b])/(16*b^2) - (5*Sin[(5*a)/b]*SinIntegral[(5*(a + b*ArcSin[c*x]))/b])/(16*b^2) - Uni
ntegrable[(1 - c^2*x^2)^2/(x^2*(a + b*ArcSin[c*x])), x]/(b*c)

________________________________________________________________________________________

Rubi [A]  time = 0.53435, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \frac{\left (1-c^2 x^2\right )^{5/2}}{x \left (a+b \sin ^{-1}(c x)\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

Int[(1 - c^2*x^2)^(5/2)/(x*(a + b*ArcSin[c*x])^2),x]

[Out]

-((1 - c^2*x^2)^3/(b*c*x*(a + b*ArcSin[c*x]))) - (25*Cos[a/b]*CosIntegral[a/b + ArcSin[c*x]])/(8*b^2) - (25*Co
s[(3*a)/b]*CosIntegral[(3*a)/b + 3*ArcSin[c*x]])/(16*b^2) - (5*Cos[(5*a)/b]*CosIntegral[(5*a)/b + 5*ArcSin[c*x
]])/(16*b^2) - (25*Sin[a/b]*SinIntegral[a/b + ArcSin[c*x]])/(8*b^2) - (25*Sin[(3*a)/b]*SinIntegral[(3*a)/b + 3
*ArcSin[c*x]])/(16*b^2) - (5*Sin[(5*a)/b]*SinIntegral[(5*a)/b + 5*ArcSin[c*x]])/(16*b^2) - Defer[Int][(1 - c^2
*x^2)^2/(x^2*(a + b*ArcSin[c*x])), x]/(b*c)

Rubi steps

\begin{align*} \int \frac{\left (1-c^2 x^2\right )^{5/2}}{x \left (a+b \sin ^{-1}(c x)\right )^2} \, dx &=-\frac{\left (1-c^2 x^2\right )^3}{b c x \left (a+b \sin ^{-1}(c x)\right )}-\frac{\int \frac{\left (1-c^2 x^2\right )^2}{x^2 \left (a+b \sin ^{-1}(c x)\right )} \, dx}{b c}-\frac{(5 c) \int \frac{\left (1-c^2 x^2\right )^2}{a+b \sin ^{-1}(c x)} \, dx}{b}\\ &=-\frac{\left (1-c^2 x^2\right )^3}{b c x \left (a+b \sin ^{-1}(c x)\right )}-\frac{5 \operatorname{Subst}\left (\int \frac{\cos ^5(x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{b}-\frac{\int \frac{\left (1-c^2 x^2\right )^2}{x^2 \left (a+b \sin ^{-1}(c x)\right )} \, dx}{b c}\\ &=-\frac{\left (1-c^2 x^2\right )^3}{b c x \left (a+b \sin ^{-1}(c x)\right )}-\frac{5 \operatorname{Subst}\left (\int \left (\frac{5 \cos (x)}{8 (a+b x)}+\frac{5 \cos (3 x)}{16 (a+b x)}+\frac{\cos (5 x)}{16 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{b}-\frac{\int \frac{\left (1-c^2 x^2\right )^2}{x^2 \left (a+b \sin ^{-1}(c x)\right )} \, dx}{b c}\\ &=-\frac{\left (1-c^2 x^2\right )^3}{b c x \left (a+b \sin ^{-1}(c x)\right )}-\frac{5 \operatorname{Subst}\left (\int \frac{\cos (5 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b}-\frac{25 \operatorname{Subst}\left (\int \frac{\cos (3 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b}-\frac{25 \operatorname{Subst}\left (\int \frac{\cos (x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 b}-\frac{\int \frac{\left (1-c^2 x^2\right )^2}{x^2 \left (a+b \sin ^{-1}(c x)\right )} \, dx}{b c}\\ &=-\frac{\left (1-c^2 x^2\right )^3}{b c x \left (a+b \sin ^{-1}(c x)\right )}-\frac{\int \frac{\left (1-c^2 x^2\right )^2}{x^2 \left (a+b \sin ^{-1}(c x)\right )} \, dx}{b c}-\frac{\left (25 \cos \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 b}-\frac{\left (25 \cos \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b}-\frac{\left (5 \cos \left (\frac{5 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b}-\frac{\left (25 \sin \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{8 b}-\frac{\left (25 \sin \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b}-\frac{\left (5 \sin \left (\frac{5 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{16 b}\\ &=-\frac{\left (1-c^2 x^2\right )^3}{b c x \left (a+b \sin ^{-1}(c x)\right )}-\frac{25 \cos \left (\frac{a}{b}\right ) \text{Ci}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )}{8 b^2}-\frac{25 \cos \left (\frac{3 a}{b}\right ) \text{Ci}\left (\frac{3 a}{b}+3 \sin ^{-1}(c x)\right )}{16 b^2}-\frac{5 \cos \left (\frac{5 a}{b}\right ) \text{Ci}\left (\frac{5 a}{b}+5 \sin ^{-1}(c x)\right )}{16 b^2}-\frac{25 \sin \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sin ^{-1}(c x)\right )}{8 b^2}-\frac{25 \sin \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 a}{b}+3 \sin ^{-1}(c x)\right )}{16 b^2}-\frac{5 \sin \left (\frac{5 a}{b}\right ) \text{Si}\left (\frac{5 a}{b}+5 \sin ^{-1}(c x)\right )}{16 b^2}-\frac{\int \frac{\left (1-c^2 x^2\right )^2}{x^2 \left (a+b \sin ^{-1}(c x)\right )} \, dx}{b c}\\ \end{align*}

Mathematica [A]  time = 12.7212, size = 0, normalized size = 0. \[ \int \frac{\left (1-c^2 x^2\right )^{5/2}}{x \left (a+b \sin ^{-1}(c x)\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(1 - c^2*x^2)^(5/2)/(x*(a + b*ArcSin[c*x])^2),x]

[Out]

Integrate[(1 - c^2*x^2)^(5/2)/(x*(a + b*ArcSin[c*x])^2), x]

________________________________________________________________________________________

Maple [A]  time = 0.424, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{x \left ( a+b\arcsin \left ( cx \right ) \right ) ^{2}} \left ( -{c}^{2}{x}^{2}+1 \right ) ^{{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-c^2*x^2+1)^(5/2)/x/(a+b*arcsin(c*x))^2,x)

[Out]

int((-c^2*x^2+1)^(5/2)/x/(a+b*arcsin(c*x))^2,x)

________________________________________________________________________________________

Maxima [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((-c^2*x^2+1)^(5/2)/x/(a+b*arcsin(c*x))^2,x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [A]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (c^{4} x^{4} - 2 \, c^{2} x^{2} + 1\right )} \sqrt{-c^{2} x^{2} + 1}}{b^{2} x \arcsin \left (c x\right )^{2} + 2 \, a b x \arcsin \left (c x\right ) + a^{2} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*x^2+1)^(5/2)/x/(a+b*arcsin(c*x))^2,x, algorithm="fricas")

[Out]

integral((c^4*x^4 - 2*c^2*x^2 + 1)*sqrt(-c^2*x^2 + 1)/(b^2*x*arcsin(c*x)^2 + 2*a*b*x*arcsin(c*x) + a^2*x), x)

________________________________________________________________________________________

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((-c**2*x**2+1)**(5/2)/x/(a+b*asin(c*x))**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-c^{2} x^{2} + 1\right )}^{\frac{5}{2}}}{{\left (b \arcsin \left (c x\right ) + a\right )}^{2} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*x^2+1)^(5/2)/x/(a+b*arcsin(c*x))^2,x, algorithm="giac")

[Out]

integrate((-c^2*x^2 + 1)^(5/2)/((b*arcsin(c*x) + a)^2*x), x)