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

Optimal. Leaf size=89 \[ -\frac{b x \left (a+b \sin ^{-1}(c x)\right )}{c d^2 \sqrt{1-c^2 x^2}}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac{b^2 \log \left (1-c^2 x^2\right )}{2 c^2 d^2} \]

[Out]

-((b*x*(a + b*ArcSin[c*x]))/(c*d^2*Sqrt[1 - c^2*x^2])) + (a + b*ArcSin[c*x])^2/(2*c^2*d^2*(1 - c^2*x^2)) - (b^
2*Log[1 - c^2*x^2])/(2*c^2*d^2)

________________________________________________________________________________________

Rubi [A]  time = 0.0985197, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {4677, 4651, 260} \[ -\frac{b x \left (a+b \sin ^{-1}(c x)\right )}{c d^2 \sqrt{1-c^2 x^2}}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (1-c^2 x^2\right )}-\frac{b^2 \log \left (1-c^2 x^2\right )}{2 c^2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*x*(a + b*ArcSin[c*x]))/(c*d^2*Sqrt[1 - c^2*x^2])) + (a + b*ArcSin[c*x])^2/(2*c^2*d^2*(1 - c^2*x^2)) - (b^
2*Log[1 - c^2*x^2])/(2*c^2*d^2)

Rule 4677

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)^
(p + 1)*(a + b*ArcSin[c*x])^n)/(2*e*(p + 1)), x] + Dist[(b*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p + 1
)*(1 - c^2*x^2)^FracPart[p]), Int[(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x] /; FreeQ[{a, b,
c, d, e, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && NeQ[p, -1]

Rule 4651

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(x*(a + b*ArcSin[c
*x])^n)/(d*Sqrt[d + e*x^2]), x] - Dist[(b*c*n)/Sqrt[d], Int[(x*(a + b*ArcSin[c*x])^(n - 1))/(d + e*x^2), x], x
] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[d, 0]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

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

Mathematica [A]  time = 0.189782, size = 75, normalized size = 0.84 \[ -\frac{\frac{2 b c x \left (a+b \sin ^{-1}(c x)\right )}{\sqrt{1-c^2 x^2}}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{c^2 x^2-1}+b^2 \log \left (1-c^2 x^2\right )}{2 c^2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2*b*c*x*(a + b*ArcSin[c*x]))/Sqrt[1 - c^2*x^2] + (a + b*ArcSin[c*x])^2/(-1 + c^2*x^2) + b^2*Log[1 - c^2*x^2
])/(2*c^2*d^2)

________________________________________________________________________________________

Maple [B]  time = 0.03, size = 205, normalized size = 2.3 \begin{align*} -{\frac{{a}^{2}}{2\,{c}^{2}{d}^{2} \left ({c}^{2}{x}^{2}-1 \right ) }}-{\frac{{b}^{2} \left ( \arcsin \left ( cx \right ) \right ) ^{2}}{2\,{c}^{2}{d}^{2} \left ({c}^{2}{x}^{2}-1 \right ) }}+{\frac{{b}^{2}\arcsin \left ( cx \right ) x}{c{d}^{2} \left ({c}^{2}{x}^{2}-1 \right ) }\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{{b}^{2}\ln \left ( -{c}^{2}{x}^{2}+1 \right ) }{2\,{c}^{2}{d}^{2}}}-{\frac{ab\arcsin \left ( cx \right ) }{{c}^{2}{d}^{2} \left ({c}^{2}{x}^{2}-1 \right ) }}+{\frac{ab}{2\,{c}^{2}{d}^{2} \left ( cx-1 \right ) }\sqrt{- \left ( cx-1 \right ) ^{2}-2\,cx+2}}+{\frac{ab}{2\,{c}^{2}{d}^{2} \left ( cx+1 \right ) }\sqrt{- \left ( cx+1 \right ) ^{2}+2\,cx+2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/c^2*a^2/d^2/(c^2*x^2-1)-1/2/c^2*b^2/d^2*arcsin(c*x)^2/(c^2*x^2-1)+1/c*b^2/d^2*arcsin(c*x)/(c^2*x^2-1)*(-c
^2*x^2+1)^(1/2)*x-1/2*b^2*ln(-c^2*x^2+1)/c^2/d^2-1/c^2*a*b/d^2/(c^2*x^2-1)*arcsin(c*x)+1/2/c^2*a*b/d^2/(c*x-1)
*(-(c*x-1)^2-2*c*x+2)^(1/2)+1/2/c^2*a*b/d^2/(c*x+1)*(-(c*x+1)^2+2*c*x+2)^(1/2)

________________________________________________________________________________________

Maxima [B]  time = 1.67815, size = 495, normalized size = 5.56 \begin{align*} \frac{1}{2} \,{\left (\frac{{\left (\frac{\sqrt{-c^{2} x^{2} + 1} c^{2} d^{2}}{c^{6} d^{4} + \sqrt{c^{6} d^{4}} c^{4} d^{2} x} - \frac{\sqrt{-c^{2} x^{2} + 1} c^{2} d^{2}}{c^{6} d^{4} - \sqrt{c^{6} d^{4}} c^{4} d^{2} x}\right )} c^{5} d^{2}}{\sqrt{c^{6} d^{4}}} - \frac{2 \, \arcsin \left (c x\right )}{c^{4} d^{2} x^{2} - c^{2} d^{2}}\right )} a b - \frac{1}{2} \,{\left (\frac{c^{6} d^{2}{\left (\frac{\log \left (c x + 1\right )}{c^{5} d^{2}} + \frac{\log \left (c x - 1\right )}{c^{5} d^{2}}\right )}}{\sqrt{c^{6} d^{4}}} - \frac{{\left (\frac{\sqrt{-c^{2} x^{2} + 1} c^{2} d^{2}}{c^{6} d^{4} + \sqrt{c^{6} d^{4}} c^{4} d^{2} x} - \frac{\sqrt{-c^{2} x^{2} + 1} c^{2} d^{2}}{c^{6} d^{4} - \sqrt{c^{6} d^{4}} c^{4} d^{2} x}\right )} c^{5} d^{2} \arcsin \left (c x\right )}{\sqrt{c^{6} d^{4}}}\right )} b^{2} - \frac{b^{2} \arcsin \left (c x\right )^{2}}{2 \,{\left (c^{4} d^{2} x^{2} - c^{2} d^{2}\right )}} - \frac{a^{2}}{2 \,{\left (c^{4} d^{2} x^{2} - c^{2} d^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((sqrt(-c^2*x^2 + 1)*c^2*d^2/(c^6*d^4 + sqrt(c^6*d^4)*c^4*d^2*x) - sqrt(-c^2*x^2 + 1)*c^2*d^2/(c^6*d^4 - s
qrt(c^6*d^4)*c^4*d^2*x))*c^5*d^2/sqrt(c^6*d^4) - 2*arcsin(c*x)/(c^4*d^2*x^2 - c^2*d^2))*a*b - 1/2*(c^6*d^2*(lo
g(c*x + 1)/(c^5*d^2) + log(c*x - 1)/(c^5*d^2))/sqrt(c^6*d^4) - (sqrt(-c^2*x^2 + 1)*c^2*d^2/(c^6*d^4 + sqrt(c^6
*d^4)*c^4*d^2*x) - sqrt(-c^2*x^2 + 1)*c^2*d^2/(c^6*d^4 - sqrt(c^6*d^4)*c^4*d^2*x))*c^5*d^2*arcsin(c*x)/sqrt(c^
6*d^4))*b^2 - 1/2*b^2*arcsin(c*x)^2/(c^4*d^2*x^2 - c^2*d^2) - 1/2*a^2/(c^4*d^2*x^2 - c^2*d^2)

________________________________________________________________________________________

Fricas [A]  time = 2.65029, size = 230, normalized size = 2.58 \begin{align*} -\frac{b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2} +{\left (b^{2} c^{2} x^{2} - b^{2}\right )} \log \left (c^{2} x^{2} - 1\right ) - 2 \,{\left (b^{2} c x \arcsin \left (c x\right ) + a b c x\right )} \sqrt{-c^{2} x^{2} + 1}}{2 \,{\left (c^{4} d^{2} x^{2} - c^{2} d^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a^{2} x}{c^{4} x^{4} - 2 c^{2} x^{2} + 1}\, dx + \int \frac{b^{2} x \operatorname{asin}^{2}{\left (c x \right )}}{c^{4} x^{4} - 2 c^{2} x^{2} + 1}\, dx + \int \frac{2 a b x \operatorname{asin}{\left (c x \right )}}{c^{4} x^{4} - 2 c^{2} x^{2} + 1}\, dx}{d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a**2*x/(c**4*x**4 - 2*c**2*x**2 + 1), x) + Integral(b**2*x*asin(c*x)**2/(c**4*x**4 - 2*c**2*x**2 + 1
), x) + Integral(2*a*b*x*asin(c*x)/(c**4*x**4 - 2*c**2*x**2 + 1), x))/d**2

________________________________________________________________________________________

Giac [B]  time = 1.55962, size = 275, normalized size = 3.09 \begin{align*} -\frac{b^{2} x^{2} \arcsin \left (c x\right )^{2}}{2 \,{\left (c^{2} x^{2} - 1\right )} d^{2}} - \frac{a b x^{2} \arcsin \left (c x\right )}{{\left (c^{2} x^{2} - 1\right )} d^{2}} - \frac{a^{2} x^{2}}{2 \,{\left (c^{2} x^{2} - 1\right )} d^{2}} - \frac{b^{2} x \arcsin \left (c x\right )}{\sqrt{-c^{2} x^{2} + 1} c d^{2}} + \frac{b^{2} \arcsin \left (c x\right )^{2}}{2 \, c^{2} d^{2}} - \frac{a b x}{\sqrt{-c^{2} x^{2} + 1} c d^{2}} + \frac{a b \arcsin \left (c x\right )}{c^{2} d^{2}} - \frac{b^{2} \log \left (2\right )}{c^{2} d^{2}} - \frac{b^{2} \log \left ({\left | -c^{2} x^{2} + 1 \right |}\right )}{2 \, c^{2} d^{2}} + \frac{a^{2}}{2 \, c^{2} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b^2*x^2*arcsin(c*x)^2/((c^2*x^2 - 1)*d^2) - a*b*x^2*arcsin(c*x)/((c^2*x^2 - 1)*d^2) - 1/2*a^2*x^2/((c^2*x
^2 - 1)*d^2) - b^2*x*arcsin(c*x)/(sqrt(-c^2*x^2 + 1)*c*d^2) + 1/2*b^2*arcsin(c*x)^2/(c^2*d^2) - a*b*x/(sqrt(-c
^2*x^2 + 1)*c*d^2) + a*b*arcsin(c*x)/(c^2*d^2) - b^2*log(2)/(c^2*d^2) - 1/2*b^2*log(abs(-c^2*x^2 + 1))/(c^2*d^
2) + 1/2*a^2/(c^2*d^2)