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

Optimal. Leaf size=106 \[ \text{Unintegrable}\left (\frac{1}{x^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )},x\right )-\frac{c \cos \left (\frac{2 a}{b}\right ) \text{CosIntegral}\left (\frac{2 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{2 b}-\frac{c \sin \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{2 b}-\frac{3 c \log \left (a+b \sin ^{-1}(c x)\right )}{2 b} \]

[Out]

-(c*Cos[(2*a)/b]*CosIntegral[(2*(a + b*ArcSin[c*x]))/b])/(2*b) - (3*c*Log[a + b*ArcSin[c*x]])/(2*b) - (c*Sin[(
2*a)/b]*SinIntegral[(2*(a + b*ArcSin[c*x]))/b])/(2*b) + Unintegrable[1/(x^2*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*
x])), x]

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Rubi [A]  time = 0.576431, 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 = {} \[ \int \frac{\left (1-c^2 x^2\right )^{3/2}}{x^2 \left (a+b \sin ^{-1}(c x)\right )} \, dx \]

Verification is Not applicable to the result.

[In]

Int[(1 - c^2*x^2)^(3/2)/(x^2*(a + b*ArcSin[c*x])),x]

[Out]

-(c*Cos[(2*a)/b]*CosIntegral[(2*a)/b + 2*ArcSin[c*x]])/(2*b) - (3*c*Log[a + b*ArcSin[c*x]])/(2*b) - (c*Sin[(2*
a)/b]*SinIntegral[(2*a)/b + 2*ArcSin[c*x]])/(2*b) + Defer[Int][1/(x^2*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])),
x]

Rubi steps

\begin{align*} \int \frac{\left (1-c^2 x^2\right )^{3/2}}{x^2 \left (a+b \sin ^{-1}(c x)\right )} \, dx &=\int \left (-\frac{2 c^2}{\sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}+\frac{1}{x^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}+\frac{c^4 x^2}{\sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}\right ) \, dx\\ &=-\left (\left (2 c^2\right ) \int \frac{1}{\sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )} \, dx\right )+c^4 \int \frac{x^2}{\sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )} \, dx+\int \frac{1}{x^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )} \, dx\\ &=-\frac{2 c \log \left (a+b \sin ^{-1}(c x)\right )}{b}+c \operatorname{Subst}\left (\int \frac{\sin ^2(x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )+\int \frac{1}{x^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )} \, dx\\ &=-\frac{2 c \log \left (a+b \sin ^{-1}(c x)\right )}{b}+c \operatorname{Subst}\left (\int \left (\frac{1}{2 (a+b x)}-\frac{\cos (2 x)}{2 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c x)\right )+\int \frac{1}{x^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )} \, dx\\ &=-\frac{3 c \log \left (a+b \sin ^{-1}(c x)\right )}{2 b}-\frac{1}{2} c \operatorname{Subst}\left (\int \frac{\cos (2 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )+\int \frac{1}{x^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )} \, dx\\ &=-\frac{3 c \log \left (a+b \sin ^{-1}(c x)\right )}{2 b}-\frac{1}{2} \left (c \cos \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )-\frac{1}{2} \left (c \sin \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )+\int \frac{1}{x^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )} \, dx\\ &=-\frac{c \cos \left (\frac{2 a}{b}\right ) \text{Ci}\left (\frac{2 a}{b}+2 \sin ^{-1}(c x)\right )}{2 b}-\frac{3 c \log \left (a+b \sin ^{-1}(c x)\right )}{2 b}-\frac{c \sin \left (\frac{2 a}{b}\right ) \text{Si}\left (\frac{2 a}{b}+2 \sin ^{-1}(c x)\right )}{2 b}+\int \frac{1}{x^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )} \, dx\\ \end{align*}

Mathematica [A]  time = 1.15088, size = 0, normalized size = 0. \[ \int \frac{\left (1-c^2 x^2\right )^{3/2}}{x^2 \left (a+b \sin ^{-1}(c x)\right )} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(1 - c^2*x^2)^(3/2)/(x^2*(a + b*ArcSin[c*x])),x]

[Out]

Integrate[(1 - c^2*x^2)^(3/2)/(x^2*(a + b*ArcSin[c*x])), x]

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Maple [A]  time = 0.273, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{2} \left ( a+b\arcsin \left ( cx \right ) \right ) } \left ( -{c}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-c^2*x^2+1)^(3/2)/x^2/(a+b*arcsin(c*x)),x)

[Out]

int((-c^2*x^2+1)^(3/2)/x^2/(a+b*arcsin(c*x)),x)

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Maxima [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-c^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{{\left (b \arcsin \left (c x\right ) + a\right )} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*x^2+1)^(3/2)/x^2/(a+b*arcsin(c*x)),x, algorithm="maxima")

[Out]

integrate((-c^2*x^2 + 1)^(3/2)/((b*arcsin(c*x) + a)*x^2), x)

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Fricas [A]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (-c^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{b x^{2} \arcsin \left (c x\right ) + a x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*x^2+1)^(3/2)/x^2/(a+b*arcsin(c*x)),x, algorithm="fricas")

[Out]

integral((-c^2*x^2 + 1)^(3/2)/(b*x^2*arcsin(c*x) + a*x^2), x)

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Sympy [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (- \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac{3}{2}}}{x^{2} \left (a + b \operatorname{asin}{\left (c x \right )}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c**2*x**2+1)**(3/2)/x**2/(a+b*asin(c*x)),x)

[Out]

Integral((-(c*x - 1)*(c*x + 1))**(3/2)/(x**2*(a + b*asin(c*x))), x)

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Giac [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-c^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{{\left (b \arcsin \left (c x\right ) + a\right )} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*x^2+1)^(3/2)/x^2/(a+b*arcsin(c*x)),x, algorithm="giac")

[Out]

integrate((-c^2*x^2 + 1)^(3/2)/((b*arcsin(c*x) + a)*x^2), x)