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

Optimal. Leaf size=83 \[ \frac{a+b \sin ^{-1}(c x)}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}-\frac{b x}{6 c d^3 \sqrt{1-c^2 x^2}}-\frac{b x}{12 c d^3 \left (1-c^2 x^2\right )^{3/2}} \]

[Out]

-(b*x)/(12*c*d^3*(1 - c^2*x^2)^(3/2)) - (b*x)/(6*c*d^3*Sqrt[1 - c^2*x^2]) + (a + b*ArcSin[c*x])/(4*c^2*d^3*(1
- c^2*x^2)^2)

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Rubi [A]  time = 0.0539175, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {4677, 192, 191} \[ \frac{a+b \sin ^{-1}(c x)}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}-\frac{b x}{6 c d^3 \sqrt{1-c^2 x^2}}-\frac{b x}{12 c d^3 \left (1-c^2 x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*x)/(12*c*d^3*(1 - c^2*x^2)^(3/2)) - (b*x)/(6*c*d^3*Sqrt[1 - c^2*x^2]) + (a + b*ArcSin[c*x])/(4*c^2*d^3*(1
- c^2*x^2)^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 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

\begin{align*} \int \frac{x \left (a+b \sin ^{-1}(c x)\right )}{\left (d-c^2 d x^2\right )^3} \, dx &=\frac{a+b \sin ^{-1}(c x)}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}-\frac{b \int \frac{1}{\left (1-c^2 x^2\right )^{5/2}} \, dx}{4 c d^3}\\ &=-\frac{b x}{12 c d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac{a+b \sin ^{-1}(c x)}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}-\frac{b \int \frac{1}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{6 c d^3}\\ &=-\frac{b x}{12 c d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac{b x}{6 c d^3 \sqrt{1-c^2 x^2}}+\frac{a+b \sin ^{-1}(c x)}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}\\ \end{align*}

Mathematica [A]  time = 0.100991, size = 62, normalized size = 0.75 \[ \frac{\frac{a+b \sin ^{-1}(c x)}{\left (c^2 x^2-1\right )^2}+\frac{b c x \left (2 c^2 x^2-3\right )}{3 \left (1-c^2 x^2\right )^{3/2}}}{4 c^2 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c*x*(-3 + 2*c^2*x^2))/(3*(1 - c^2*x^2)^(3/2)) + (a + b*ArcSin[c*x])/(-1 + c^2*x^2)^2)/(4*c^2*d^3)

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Maple [B]  time = 0.01, size = 151, normalized size = 1.8 \begin{align*}{\frac{1}{{c}^{2}} \left ({\frac{a}{4\,{d}^{3} \left ({c}^{2}{x}^{2}-1 \right ) ^{2}}}-{\frac{b}{{d}^{3}} \left ( -{\frac{\arcsin \left ( cx \right ) }{4\, \left ({c}^{2}{x}^{2}-1 \right ) ^{2}}}-{\frac{1}{12\,cx-12}\sqrt{- \left ( cx-1 \right ) ^{2}-2\,cx+2}}-{\frac{1}{12\,cx+12}\sqrt{- \left ( cx+1 \right ) ^{2}+2\,cx+2}}+{\frac{1}{48\, \left ( cx-1 \right ) ^{2}}\sqrt{- \left ( cx-1 \right ) ^{2}-2\,cx+2}}-{\frac{1}{48\, \left ( cx+1 \right ) ^{2}}\sqrt{- \left ( cx+1 \right ) ^{2}+2\,cx+2}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^2*(1/4*a/d^3/(c^2*x^2-1)^2-b/d^3*(-1/4/(c^2*x^2-1)^2*arcsin(c*x)-1/12/(c*x-1)*(-(c*x-1)^2-2*c*x+2)^(1/2)-1
/12/(c*x+1)*(-(c*x+1)^2+2*c*x+2)^(1/2)+1/48/(c*x-1)^2*(-(c*x-1)^2-2*c*x+2)^(1/2)-1/48/(c*x+1)^2*(-(c*x+1)^2+2*
c*x+2)^(1/2)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left ({\left (c^{6} d^{3} x^{4} - 2 \, c^{4} d^{3} x^{2} + c^{2} d^{3}\right )} \int \frac{e^{\left (\frac{1}{2} \, \log \left (c x + 1\right ) + \frac{1}{2} \, \log \left (-c x + 1\right )\right )}}{c^{9} d^{3} x^{8} - 3 \, c^{7} d^{3} x^{6} + 3 \, c^{5} d^{3} x^{4} - c^{3} d^{3} x^{2} -{\left (c^{7} d^{3} x^{6} - 3 \, c^{5} d^{3} x^{4} + 3 \, c^{3} d^{3} x^{2} - c d^{3}\right )}{\left (c x + 1\right )}{\left (c x - 1\right )}}\,{d x} + \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )\right )} b}{4 \,{\left (c^{6} d^{3} x^{4} - 2 \, c^{4} d^{3} x^{2} + c^{2} d^{3}\right )}} + \frac{a}{4 \,{\left (c^{6} d^{3} x^{4} - 2 \, c^{4} d^{3} x^{2} + c^{2} d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*(c^6*d^3*x^4 - 2*c^4*d^3*x^2 + c^2*d^3)*integrate(1/4*e^(1/2*log(c*x + 1) + 1/2*log(-c*x + 1))/(c^9*d^3
*x^8 - 3*c^7*d^3*x^6 + 3*c^5*d^3*x^4 - c^3*d^3*x^2 + (c^7*d^3*x^6 - 3*c^5*d^3*x^4 + 3*c^3*d^3*x^2 - c*d^3)*e^(
log(c*x + 1) + log(-c*x + 1))), x) + arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)))*b/(c^6*d^3*x^4 - 2*c^4*d^3*x^
2 + c^2*d^3) + 1/4*a/(c^6*d^3*x^4 - 2*c^4*d^3*x^2 + c^2*d^3)

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Fricas [A]  time = 2.26994, size = 186, normalized size = 2.24 \begin{align*} -\frac{3 \, a c^{4} x^{4} - 6 \, a c^{2} x^{2} - 3 \, b \arcsin \left (c x\right ) -{\left (2 \, b c^{3} x^{3} - 3 \, b c x\right )} \sqrt{-c^{2} x^{2} + 1}}{12 \,{\left (c^{6} d^{3} x^{4} - 2 \, c^{4} d^{3} x^{2} + c^{2} d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*a*c^4*x^4 - 6*a*c^2*x^2 - 3*b*arcsin(c*x) - (2*b*c^3*x^3 - 3*b*c*x)*sqrt(-c^2*x^2 + 1))/(c^6*d^3*x^4
- 2*c^4*d^3*x^2 + c^2*d^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int \frac{a x}{c^{6} x^{6} - 3 c^{4} x^{4} + 3 c^{2} x^{2} - 1}\, dx + \int \frac{b x \operatorname{asin}{\left (c x \right )}}{c^{6} x^{6} - 3 c^{4} x^{4} + 3 c^{2} x^{2} - 1}\, dx}{d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(Integral(a*x/(c**6*x**6 - 3*c**4*x**4 + 3*c**2*x**2 - 1), x) + Integral(b*x*asin(c*x)/(c**6*x**6 - 3*c**4*x*
*4 + 3*c**2*x**2 - 1), x))/d**3

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Giac [B]  time = 1.32488, size = 232, normalized size = 2.8 \begin{align*} \frac{b c^{2} x^{4} \arcsin \left (c x\right )}{4 \,{\left (c^{2} x^{2} - 1\right )}^{2} d^{3}} + \frac{a c^{2} x^{4}}{4 \,{\left (c^{2} x^{2} - 1\right )}^{2} d^{3}} + \frac{b c x^{3}}{12 \,{\left (c^{2} x^{2} - 1\right )} \sqrt{-c^{2} x^{2} + 1} d^{3}} - \frac{b x^{2} \arcsin \left (c x\right )}{2 \,{\left (c^{2} x^{2} - 1\right )} d^{3}} - \frac{a x^{2}}{2 \,{\left (c^{2} x^{2} - 1\right )} d^{3}} - \frac{b x}{4 \, \sqrt{-c^{2} x^{2} + 1} c d^{3}} + \frac{b \arcsin \left (c x\right )}{4 \, c^{2} d^{3}} + \frac{a}{4 \, c^{2} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b*c^2*x^4*arcsin(c*x)/((c^2*x^2 - 1)^2*d^3) + 1/4*a*c^2*x^4/((c^2*x^2 - 1)^2*d^3) + 1/12*b*c*x^3/((c^2*x^2
- 1)*sqrt(-c^2*x^2 + 1)*d^3) - 1/2*b*x^2*arcsin(c*x)/((c^2*x^2 - 1)*d^3) - 1/2*a*x^2/((c^2*x^2 - 1)*d^3) - 1/
4*b*x/(sqrt(-c^2*x^2 + 1)*c*d^3) + 1/4*b*arcsin(c*x)/(c^2*d^3) + 1/4*a/(c^2*d^3)