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

Optimal. Leaf size=100 \[ \frac{x^4 \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac{b x^3}{12 c d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac{b x}{4 c^3 d^3 \sqrt{1-c^2 x^2}}-\frac{b \sin ^{-1}(c x)}{4 c^4 d^3} \]

[Out]

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

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Rubi [A]  time = 0.0844497, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {4681, 288, 216} \[ \frac{x^4 \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac{b x^3}{12 c d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac{b x}{4 c^3 d^3 \sqrt{1-c^2 x^2}}-\frac{b \sin ^{-1}(c x)}{4 c^4 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4681

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(
(f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSin[c*x])^n)/(d*f*(m + 1)), x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x
^2)^FracPart[p])/(f*(m + 1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSi
n[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && EqQ[m + 2*p
+ 3, 0] && NeQ[m, -1]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

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

Mathematica [A]  time = 0.0726101, size = 79, normalized size = 0.79 \[ \frac{a \left (6 c^2 x^2-3\right )+b c x \sqrt{1-c^2 x^2} \left (3-4 c^2 x^2\right )+3 b \left (2 c^2 x^2-1\right ) \sin ^{-1}(c x)}{12 c^4 d^3 \left (c^2 x^2-1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.02, size = 212, normalized size = 2.1 \begin{align*}{\frac{1}{{c}^{4}} \left ( -{\frac{a}{{d}^{3}} \left ( -{\frac{1}{16\, \left ( cx-1 \right ) ^{2}}}-{\frac{3}{16\,cx-16}}-{\frac{1}{16\, \left ( cx+1 \right ) ^{2}}}+{\frac{3}{16\,cx+16}} \right ) }-{\frac{b}{{d}^{3}} \left ( -{\frac{\arcsin \left ( cx \right ) }{16\, \left ( cx-1 \right ) ^{2}}}-{\frac{3\,\arcsin \left ( cx \right ) }{16\,cx-16}}-{\frac{\arcsin \left ( cx \right ) }{16\, \left ( cx+1 \right ) ^{2}}}+{\frac{3\,\arcsin \left ( cx \right ) }{16\,cx+16}}+{\frac{1}{6\,cx-6}\sqrt{- \left ( cx-1 \right ) ^{2}-2\,cx+2}}+{\frac{1}{6\,cx+6}\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^3*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^3,x)

[Out]

1/c^4*(-a/d^3*(-1/16/(c*x-1)^2-3/16/(c*x-1)-1/16/(c*x+1)^2+3/16/(c*x+1))-b/d^3*(-1/16*arcsin(c*x)/(c*x-1)^2-3/
16*arcsin(c*x)/(c*x-1)-1/16*arcsin(c*x)/(c*x+1)^2+3/16*arcsin(c*x)/(c*x+1)+1/6/(c*x-1)*(-(c*x-1)^2-2*c*x+2)^(1
/2)+1/6/(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 (2 \, c^{2} x^{2} - 1\right )} a}{4 \,{\left (c^{8} d^{3} x^{4} - 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}\right )}} + \frac{{\left ({\left (2 \, c^{2} x^{2} - 1\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) +{\left (c^{8} d^{3} x^{4} - 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}\right )} \int \frac{{\left (2 \, c^{2} x^{2} - 1\right )} e^{\left (\frac{1}{2} \, \log \left (c x + 1\right ) + \frac{1}{2} \, \log \left (-c x + 1\right )\right )}}{c^{11} d^{3} x^{8} - 3 \, c^{9} d^{3} x^{6} + 3 \, c^{7} d^{3} x^{4} - c^{5} d^{3} x^{2} -{\left (c^{9} d^{3} x^{6} - 3 \, c^{7} d^{3} x^{4} + 3 \, c^{5} d^{3} x^{2} - c^{3} d^{3}\right )}{\left (c x + 1\right )}{\left (c x - 1\right )}}\,{d x}\right )} b}{4 \,{\left (c^{8} d^{3} x^{4} - 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*c^2*x^2 - 1)*a/(c^8*d^3*x^4 - 2*c^6*d^3*x^2 + c^4*d^3) + 1/4*((2*c^2*x^2 - 1)*arctan2(c*x, sqrt(c*x + 1
)*sqrt(-c*x + 1)) + 4*(c^8*d^3*x^4 - 2*c^6*d^3*x^2 + c^4*d^3)*integrate(1/4*(2*c^2*x^2 - 1)*e^(1/2*log(c*x + 1
) + 1/2*log(-c*x + 1))/(c^11*d^3*x^8 - 3*c^9*d^3*x^6 + 3*c^7*d^3*x^4 - c^5*d^3*x^2 + (c^9*d^3*x^6 - 3*c^7*d^3*
x^4 + 3*c^5*d^3*x^2 - c^3*d^3)*e^(log(c*x + 1) + log(-c*x + 1))), x))*b/(c^8*d^3*x^4 - 2*c^6*d^3*x^2 + c^4*d^3
)

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Fricas [A]  time = 2.09677, size = 188, normalized size = 1.88 \begin{align*} \frac{3 \, a c^{4} x^{4} + 3 \,{\left (2 \, b c^{2} x^{2} - b\right )} \arcsin \left (c x\right ) -{\left (4 \, b c^{3} x^{3} - 3 \, b c x\right )} \sqrt{-c^{2} x^{2} + 1}}{12 \,{\left (c^{8} d^{3} x^{4} - 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*a*c^4*x^4 + 3*(2*b*c^2*x^2 - b)*arcsin(c*x) - (4*b*c^3*x^3 - 3*b*c*x)*sqrt(-c^2*x^2 + 1))/(c^8*d^3*x^4
- 2*c^6*d^3*x^2 + c^4*d^3)

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

[Out]

-(Integral(a*x**3/(c**6*x**6 - 3*c**4*x**4 + 3*c**2*x**2 - 1), x) + Integral(b*x**3*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 [A]  time = 1.368, size = 167, normalized size = 1.67 \begin{align*} \frac{b x^{4} \arcsin \left (c x\right )}{4 \,{\left (c^{2} x^{2} - 1\right )}^{2} d^{3}} + \frac{a x^{4}}{4 \,{\left (c^{2} x^{2} - 1\right )}^{2} d^{3}} + \frac{b x^{3}}{12 \,{\left (c^{2} x^{2} - 1\right )} \sqrt{-c^{2} x^{2} + 1} c d^{3}} + \frac{b x}{4 \, \sqrt{-c^{2} x^{2} + 1} c^{3} d^{3}} - \frac{b \arcsin \left (c x\right )}{4 \, c^{4} d^{3}} - \frac{a}{4 \, c^{4} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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