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

Optimal. Leaf size=150 \[ -\frac {b x \left (a+b \sin ^{-1}(c x)\right )}{3 c d^3 \sqrt {1-c^2 x^2}}-\frac {b x \left (a+b \sin ^{-1}(c x)\right )}{6 c d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}+\frac {b^2}{12 c^2 d^3 \left (1-c^2 x^2\right )}-\frac {b^2 \log \left (1-c^2 x^2\right )}{6 c^2 d^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4677, 4655, 4651, 260, 261} \[ -\frac {b x \left (a+b \sin ^{-1}(c x)\right )}{3 c d^3 \sqrt {1-c^2 x^2}}-\frac {b x \left (a+b \sin ^{-1}(c x)\right )}{6 c d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}+\frac {b^2}{12 c^2 d^3 \left (1-c^2 x^2\right )}-\frac {b^2 \log \left (1-c^2 x^2\right )}{6 c^2 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && 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 4655

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

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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.21, size = 162, normalized size = 1.08 \[ \frac {3 a^2-6 a b c x \sqrt {1-c^2 x^2}+2 b \sin ^{-1}(c x) \left (3 a+b c x \sqrt {1-c^2 x^2} \left (2 c^2 x^2-3\right )\right )+4 a b c^3 x^3 \sqrt {1-c^2 x^2}-b^2 c^2 x^2-2 b^2 \left (c^2 x^2-1\right )^2 \log \left (1-c^2 x^2\right )+3 b^2 \sin ^{-1}(c x)^2+b^2}{12 c^2 d^3 \left (c^2 x^2-1\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 3.23, size = 165, normalized size = 1.10 \[ -\frac {b^{2} c^{2} x^{2} - 3 \, b^{2} \arcsin \left (c x\right )^{2} - 6 \, a b \arcsin \left (c x\right ) - 3 \, a^{2} - b^{2} + 2 \, {\left (b^{2} c^{4} x^{4} - 2 \, b^{2} c^{2} x^{2} + b^{2}\right )} \log \left (c^{2} x^{2} - 1\right ) - 2 \, {\left (2 \, a b c^{3} x^{3} - 3 \, a b c x + {\left (2 \, b^{2} c^{3} x^{3} - 3 \, b^{2} c x\right )} \arcsin \left (c x\right )\right )} \sqrt {-c^{2} x^{2} + 1}}{12 \, {\left (c^{6} d^{3} x^{4} - 2 \, c^{4} d^{3} x^{2} + c^{2} d^{3}\right )}} \]

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)^3,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [B]  time = 1.26, size = 395, normalized size = 2.63 \[ \frac {b^{2} c^{2} x^{4} \arcsin \left (c x\right )^{2}}{4 \, {\left (c^{2} x^{2} - 1\right )}^{2} d^{3}} + \frac {a b c^{2} x^{4} \arcsin \left (c x\right )}{2 \, {\left (c^{2} x^{2} - 1\right )}^{2} d^{3}} + \frac {a^{2} c^{2} x^{4}}{4 \, {\left (c^{2} x^{2} - 1\right )}^{2} d^{3}} + \frac {b^{2} c x^{3} \arcsin \left (c x\right )}{6 \, {\left (c^{2} x^{2} - 1\right )} \sqrt {-c^{2} x^{2} + 1} d^{3}} - \frac {b^{2} x^{2} \arcsin \left (c x\right )^{2}}{2 \, {\left (c^{2} x^{2} - 1\right )} d^{3}} + \frac {a b c x^{3}}{6 \, {\left (c^{2} x^{2} - 1\right )} \sqrt {-c^{2} x^{2} + 1} d^{3}} - \frac {a b x^{2} \arcsin \left (c x\right )}{{\left (c^{2} x^{2} - 1\right )} d^{3}} - \frac {a^{2} x^{2}}{2 \, {\left (c^{2} x^{2} - 1\right )} d^{3}} - \frac {b^{2} x^{2}}{12 \, {\left (c^{2} x^{2} - 1\right )} d^{3}} - \frac {b^{2} x \arcsin \left (c x\right )}{2 \, \sqrt {-c^{2} x^{2} + 1} c d^{3}} + \frac {b^{2} \arcsin \left (c x\right )^{2}}{4 \, c^{2} d^{3}} - \frac {a b x}{2 \, \sqrt {-c^{2} x^{2} + 1} c d^{3}} + \frac {a b \arcsin \left (c x\right )}{2 \, c^{2} d^{3}} - \frac {b^{2} \log \relax (2)}{3 \, c^{2} d^{3}} - \frac {b^{2} \log \left ({\left | -c^{2} x^{2} + 1 \right |}\right )}{6 \, c^{2} d^{3}} + \frac {a^{2}}{4 \, c^{2} d^{3}} + \frac {b^{2}}{12 \, c^{2} d^{3}} \]

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)^3,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.05, size = 335, normalized size = 2.23 \[ \frac {a^{2}}{4 c^{2} d^{3} \left (c^{2} x^{2}-1\right )^{2}}+\frac {b^{2} \arcsin \left (c x \right )^{2}}{4 c^{2} d^{3} \left (c^{2} x^{2}-1\right )^{2}}-\frac {b^{2} \arcsin \left (c x \right ) x \sqrt {-c^{2} x^{2}+1}}{6 c \,d^{3} \left (c^{2} x^{2}-1\right )^{2}}-\frac {b^{2}}{12 c^{2} d^{3} \left (c^{2} x^{2}-1\right )}+\frac {b^{2} \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right ) x}{3 c \,d^{3} \left (c^{2} x^{2}-1\right )}-\frac {b^{2} \ln \left (-c^{2} x^{2}+1\right )}{6 c^{2} d^{3}}+\frac {a b \arcsin \left (c x \right )}{2 c^{2} d^{3} \left (c^{2} x^{2}-1\right )^{2}}+\frac {a b \sqrt {-\left (c x +1\right )^{2}+2 c x +2}}{6 c^{2} d^{3} \left (c x +1\right )}-\frac {a b \sqrt {-\left (c x -1\right )^{2}-2 c x +2}}{24 c^{2} d^{3} \left (c x -1\right )^{2}}+\frac {a b \sqrt {-\left (c x -1\right )^{2}-2 c x +2}}{6 c^{2} d^{3} \left (c x -1\right )}+\frac {a b \sqrt {-\left (c x +1\right )^{2}+2 c x +2}}{24 c^{2} d^{3} \left (c x +1\right )^{2}} \]

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)^3,x)

[Out]

1/4/c^2*a^2/d^3/(c^2*x^2-1)^2+1/4/c^2*b^2/d^3*arcsin(c*x)^2/(c^2*x^2-1)^2-1/6/c*b^2/d^3*arcsin(c*x)*x*(-c^2*x^
2+1)^(1/2)/(c^2*x^2-1)^2-1/12/c^2*b^2/d^3/(c^2*x^2-1)+1/3/c*b^2/d^3*(-c^2*x^2+1)^(1/2)/(c^2*x^2-1)*arcsin(c*x)
*x-1/6*b^2*ln(-c^2*x^2+1)/c^2/d^3+1/2/c^2*a*b/d^3/(c^2*x^2-1)^2*arcsin(c*x)+1/6/c^2*a*b/d^3/(c*x+1)*(-(c*x+1)^
2+2*c*x+2)^(1/2)-1/24/c^2*a*b/d^3/(c*x-1)^2*(-(c*x-1)^2-2*c*x+2)^(1/2)+1/6/c^2*a*b/d^3/(c*x-1)*(-(c*x-1)^2-2*c
*x+2)^(1/2)+1/24/c^2*a*b/d^3/(c*x+1)^2*(-(c*x+1)^2+2*c*x+2)^(1/2)

________________________________________________________________________________________

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

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)^3,x, algorithm="maxima")

[Out]

1/4*a^2/(c^6*d^3*x^4 - 2*c^4*d^3*x^2 + c^2*d^3) + 1/4*(b^2*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2 + 4*(c
^6*d^3*x^4 - 2*c^4*d^3*x^2 + c^2*d^3)*integrate(-1/2*(4*a*b*c*x*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) - s
qrt(c*x + 1)*sqrt(-c*x + 1)*b^2*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)))/(c^7*d^3*x^6 - 3*c^5*d^3*x^4 + 3*c
^3*d^3*x^2 - c*d^3), x))/(c^6*d^3*x^4 - 2*c^4*d^3*x^2 + c^2*d^3)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

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)**3,x)

[Out]

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

________________________________________________________________________________________