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

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

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.08, 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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fricas [A]  time = 0.53, size = 91, normalized size = 0.91 \[ \frac {3 \, a c^{4} x^{4} + 3 \, {\left (2 \, b c^{2} x^{2} - b\right )} \arcsin \left (c x\right ) - {\left (4 \, b c^{3} x^{3} - 3 \, b c x\right )} \sqrt {-c^{2} x^{2} + 1}}{12 \, {\left (c^{8} d^{3} x^{4} - 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}\right )}} \]

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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giac [A]  time = 0.32, size = 124, normalized size = 1.24 \[ \frac {b x^{4} \arcsin \left (c x\right )}{4 \, {\left (c^{2} x^{2} - 1\right )}^{2} d^{3}} + \frac {a x^{4}}{4 \, {\left (c^{2} x^{2} - 1\right )}^{2} d^{3}} + \frac {b x^{3}}{12 \, {\left (c^{2} x^{2} - 1\right )} \sqrt {-c^{2} x^{2} + 1} c d^{3}} + \frac {b x}{4 \, \sqrt {-c^{2} x^{2} + 1} c^{3} d^{3}} - \frac {b \arcsin \left (c x\right )}{4 \, c^{4} d^{3}} - \frac {a}{4 \, c^{4} d^{3}} \]

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)

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maple [B]  time = 0.02, size = 212, normalized size = 2.12 \[ \frac {-\frac {a \left (-\frac {1}{16 \left (c x +1\right )^{2}}+\frac {3}{16 \left (c x +1\right )}-\frac {1}{16 \left (c x -1\right )^{2}}-\frac {3}{16 \left (c x -1\right )}\right )}{d^{3}}-\frac {b \left (-\frac {\arcsin \left (c x \right )}{16 \left (c x +1\right )^{2}}+\frac {3 \arcsin \left (c x \right )}{16 \left (c x +1\right )}-\frac {\arcsin \left (c x \right )}{16 \left (c x -1\right )^{2}}-\frac {3 \arcsin \left (c x \right )}{16 \left (c x -1\right )}+\frac {\sqrt {-\left (c x +1\right )^{2}+2 c x +2}}{6 c x +6}+\frac {\sqrt {-\left (c x -1\right )^{2}-2 c x +2}}{48 \left (c x -1\right )^{2}}+\frac {\sqrt {-\left (c x -1\right )^{2}-2 c x +2}}{6 c x -6}-\frac {\sqrt {-\left (c x +1\right )^{2}+2 c x +2}}{48 \left (c x +1\right )^{2}}\right )}{d^{3}}}{c^{4}} \]

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)+1/48/(c*x-1)^2*(-(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)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left (2 \, c^{2} x^{2} - 1\right )} a}{4 \, {\left (c^{8} d^{3} x^{4} - 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}\right )}} + \frac {{\left ({\left (2 \, c^{2} x^{2} - 1\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right ) + {\left (c^{8} d^{3} x^{4} - 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}\right )} \int \frac {{\left (2 \, c^{2} x^{2} - 1\right )} e^{\left (\frac {1}{2} \, \log \left (c x + 1\right ) + \frac {1}{2} \, \log \left (-c x + 1\right )\right )}}{c^{11} d^{3} x^{8} - 3 \, c^{9} d^{3} x^{6} + 3 \, c^{7} d^{3} x^{4} - c^{5} d^{3} x^{2} - {\left (c^{9} d^{3} x^{6} - 3 \, c^{7} d^{3} x^{4} + 3 \, c^{5} d^{3} x^{2} - c^{3} d^{3}\right )} {\left (c x + 1\right )} {\left (c x - 1\right )}}\,{d x}\right )} b}{4 \, {\left (c^{8} d^{3} x^{4} - 2 \, c^{6} d^{3} x^{2} + c^{4} d^{3}\right )}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{{\left (d-c^2\,d\,x^2\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {a x^{3}}{c^{6} x^{6} - 3 c^{4} x^{4} + 3 c^{2} x^{2} - 1}\, dx + \int \frac {b x^{3} \operatorname {asin}{\left (c x \right )}}{c^{6} x^{6} - 3 c^{4} x^{4} + 3 c^{2} x^{2} - 1}\, dx}{d^{3}} \]

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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