3.1.49 \(\int \frac {x (a+b \text {ArcSin}(c x))}{(d-c^2 d x^2)^3} \, dx\) [49]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4767, 198, 197} \begin {gather*} \frac {a+b \text {ArcSin}(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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(b*x)/(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 197

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]

Rule 198

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 4767

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/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^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]

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*}

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Mathematica [A]
time = 0.07, size = 62, normalized size = 0.75 \begin {gather*} \frac {\frac {b c x \left (-3+2 c^2 x^2\right )}{3 \left (1-c^2 x^2\right )^{3/2}}+\frac {a+b \text {ArcSin}(c x)}{\left (-1+c^2 x^2\right )^2}}{4 c^2 d^3} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(150\) vs. \(2(73)=146\).
time = 0.08, size = 151, normalized size = 1.82

method	result	size
derivativedivides	\(\frac {\frac {a}{4 d^{3} \left (c^{2} x^{2}-1\right )^{2}}-\frac {b \left (-\frac {\arcsin \left (c x \right )}{4 \left (c^{2} x^{2}-1\right )^{2}}+\frac {\sqrt {-\left (c x -1\right )^{2}-2 c x +2}}{48 \left (c x -1\right )^{2}}-\frac {\sqrt {-\left (c x -1\right )^{2}-2 c x +2}}{12 \left (c x -1\right )}-\frac {\sqrt {-\left (c x +1\right )^{2}+2 c x +2}}{12 \left (c x +1\right )}-\frac {\sqrt {-\left (c x +1\right )^{2}+2 c x +2}}{48 \left (c x +1\right )^{2}}\right )}{d^{3}}}{c^{2}}\)	\(151\)
default	\(\frac {\frac {a}{4 d^{3} \left (c^{2} x^{2}-1\right )^{2}}-\frac {b \left (-\frac {\arcsin \left (c x \right )}{4 \left (c^{2} x^{2}-1\right )^{2}}+\frac {\sqrt {-\left (c x -1\right )^{2}-2 c x +2}}{48 \left (c x -1\right )^{2}}-\frac {\sqrt {-\left (c x -1\right )^{2}-2 c x +2}}{12 \left (c x -1\right )}-\frac {\sqrt {-\left (c x +1\right )^{2}+2 c x +2}}{12 \left (c x +1\right )}-\frac {\sqrt {-\left (c x +1\right )^{2}+2 c x +2}}{48 \left (c x +1\right )^{2}}\right )}{d^{3}}}{c^{2}}\)	\(151\)

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,method=_RETURNVERBOSE)

[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)+1/48/(c*x-1)^2*(-(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)-1/12/(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)))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 1.75, size = 88, normalized size = 1.06 \begin {gather*} -\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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 172 vs. \(2 (72) = 144\).
time = 0.42, size = 172, normalized size = 2.07 \begin {gather*} \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 {gather*}

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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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