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

Optimal. Leaf size=250 \[ \frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d \sqrt {d-c^2 d x^2}}-\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^3 d \sqrt {d-c^2 d x^2}}-\frac {i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{c^3 d \sqrt {d-c^2 d x^2}}+\frac {2 b \sqrt {1-c^2 x^2} \log \left (1+e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^3 d \sqrt {d-c^2 d x^2}}-\frac {i b^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{c^3 d \sqrt {d-c^2 d x^2}} \]

[Out]

x*(a+b*arcsin(c*x))^2/c^2/d/(-c^2*d*x^2+d)^(1/2)-I*(a+b*arcsin(c*x))^2*(-c^2*x^2+1)^(1/2)/c^3/d/(-c^2*d*x^2+d)
^(1/2)-1/3*(a+b*arcsin(c*x))^3*(-c^2*x^2+1)^(1/2)/b/c^3/d/(-c^2*d*x^2+d)^(1/2)+2*b*(a+b*arcsin(c*x))*ln(1+(I*c
*x+(-c^2*x^2+1)^(1/2))^2)*(-c^2*x^2+1)^(1/2)/c^3/d/(-c^2*d*x^2+d)^(1/2)-I*b^2*polylog(2,-(I*c*x+(-c^2*x^2+1)^(
1/2))^2)*(-c^2*x^2+1)^(1/2)/c^3/d/(-c^2*d*x^2+d)^(1/2)

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Rubi [A]  time = 0.36, antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {4703, 4643, 4641, 4675, 3719, 2190, 2279, 2391} \[ -\frac {i b^2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right )}{c^3 d \sqrt {d-c^2 d x^2}}-\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^3 d \sqrt {d-c^2 d x^2}}+\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d \sqrt {d-c^2 d x^2}}-\frac {i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{c^3 d \sqrt {d-c^2 d x^2}}+\frac {2 b \sqrt {1-c^2 x^2} \log \left (1+e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^3 d \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a + b*ArcSin[c*x])^2)/(c^2*d*Sqrt[d - c^2*d*x^2]) - (I*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2)/(c^3*d*Sqr
t[d - c^2*d*x^2]) - (Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^3)/(3*b*c^3*d*Sqrt[d - c^2*d*x^2]) + (2*b*Sqrt[1 -
c^2*x^2]*(a + b*ArcSin[c*x])*Log[1 + E^((2*I)*ArcSin[c*x])])/(c^3*d*Sqrt[d - c^2*d*x^2]) - (I*b^2*Sqrt[1 - c^2
*x^2]*PolyLog[2, -E^((2*I)*ArcSin[c*x])])/(c^3*d*Sqrt[d - c^2*d*x^2])

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*(m + 1)), x
] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*(e + f*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 4641

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

Rule 4643

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[1 - c^2*x^2]/Sq
rt[d + e*x^2], Int[(a + b*ArcSin[c*x])^n/Sqrt[1 - c^2*x^2], x], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*
d + e, 0] &&  !GtQ[d, 0]

Rule 4675

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Dist[e^(-1), Subst[In
t[(a + b*x)^n*Tan[x], x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[n, 0]

Rule 4703

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

Rubi steps

\begin {align*} \int \frac {x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{\left (d-c^2 d x^2\right )^{3/2}} \, dx &=\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d \sqrt {d-c^2 d x^2}}-\frac {\int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt {d-c^2 d x^2}} \, dx}{c^2 d}-\frac {\left (2 b \sqrt {1-c^2 x^2}\right ) \int \frac {x \left (a+b \sin ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{c d \sqrt {d-c^2 d x^2}}\\ &=\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d \sqrt {d-c^2 d x^2}}-\frac {\left (2 b \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int (a+b x) \tan (x) \, dx,x,\sin ^{-1}(c x)\right )}{c^3 d \sqrt {d-c^2 d x^2}}-\frac {\sqrt {1-c^2 x^2} \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt {1-c^2 x^2}} \, dx}{c^2 d \sqrt {d-c^2 d x^2}}\\ &=\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d \sqrt {d-c^2 d x^2}}-\frac {i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{c^3 d \sqrt {d-c^2 d x^2}}-\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^3 d \sqrt {d-c^2 d x^2}}+\frac {\left (4 i b \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {e^{2 i x} (a+b x)}{1+e^{2 i x}} \, dx,x,\sin ^{-1}(c x)\right )}{c^3 d \sqrt {d-c^2 d x^2}}\\ &=\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d \sqrt {d-c^2 d x^2}}-\frac {i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{c^3 d \sqrt {d-c^2 d x^2}}-\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^3 d \sqrt {d-c^2 d x^2}}+\frac {2 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{c^3 d \sqrt {d-c^2 d x^2}}-\frac {\left (2 b^2 \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^3 d \sqrt {d-c^2 d x^2}}\\ &=\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d \sqrt {d-c^2 d x^2}}-\frac {i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{c^3 d \sqrt {d-c^2 d x^2}}-\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^3 d \sqrt {d-c^2 d x^2}}+\frac {2 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{c^3 d \sqrt {d-c^2 d x^2}}+\frac {\left (i b^2 \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{c^3 d \sqrt {d-c^2 d x^2}}\\ &=\frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d \sqrt {d-c^2 d x^2}}-\frac {i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{c^3 d \sqrt {d-c^2 d x^2}}-\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^3 d \sqrt {d-c^2 d x^2}}+\frac {2 b \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{c^3 d \sqrt {d-c^2 d x^2}}-\frac {i b^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{c^3 d \sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.61, size = 295, normalized size = 1.18 \[ -\frac {a^2 x \sqrt {-d \left (c^2 x^2-1\right )}}{c^2 d^2 \left (c^2 x^2-1\right )}+\frac {a^2 \tan ^{-1}\left (\frac {c x \sqrt {-d \left (c^2 x^2-1\right )}}{\sqrt {d} \left (c^2 x^2-1\right )}\right )}{c^3 d^{3/2}}+\frac {a b \left (2 c x \sin ^{-1}(c x)-\sqrt {1-c^2 x^2} \left (\sin ^{-1}(c x)^2-2 \log \left (\sqrt {1-c^2 x^2}\right )\right )\right )}{c^3 d \sqrt {d \left (1-c^2 x^2\right )}}+\frac {b^2 \left (\sin ^{-1}(c x) \left (-\sqrt {1-c^2 x^2} \left (\sin ^{-1}(c x)+3 i\right ) \sin ^{-1}(c x)+6 \sqrt {1-c^2 x^2} \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )+3 c x \sin ^{-1}(c x)\right )-3 i \sqrt {1-c^2 x^2} \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )\right )}{3 c^3 d \sqrt {d \left (1-c^2 x^2\right )}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((a^2*x*Sqrt[-(d*(-1 + c^2*x^2))])/(c^2*d^2*(-1 + c^2*x^2))) + (a^2*ArcTan[(c*x*Sqrt[-(d*(-1 + c^2*x^2))])/(S
qrt[d]*(-1 + c^2*x^2))])/(c^3*d^(3/2)) + (a*b*(2*c*x*ArcSin[c*x] - Sqrt[1 - c^2*x^2]*(ArcSin[c*x]^2 - 2*Log[Sq
rt[1 - c^2*x^2]])))/(c^3*d*Sqrt[d*(1 - c^2*x^2)]) + (b^2*(ArcSin[c*x]*(3*c*x*ArcSin[c*x] - Sqrt[1 - c^2*x^2]*A
rcSin[c*x]*(3*I + ArcSin[c*x]) + 6*Sqrt[1 - c^2*x^2]*Log[1 + E^((2*I)*ArcSin[c*x])]) - (3*I)*Sqrt[1 - c^2*x^2]
*PolyLog[2, -E^((2*I)*ArcSin[c*x])]))/(3*c^3*d*Sqrt[d*(1 - c^2*x^2)])

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} x^{2} \arcsin \left (c x\right )^{2} + 2 \, a b x^{2} \arcsin \left (c x\right ) + a^{2} x^{2}\right )} \sqrt {-c^{2} d x^{2} + d}}{c^{4} d^{2} x^{4} - 2 \, c^{2} d^{2} x^{2} + d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*x^2*arcsin(c*x)^2 + 2*a*b*x^2*arcsin(c*x) + a^2*x^2)*sqrt(-c^2*d*x^2 + d)/(c^4*d^2*x^4 - 2*c^2*d
^2*x^2 + d^2), x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep^4-1)]index.cc index_m i_lex_is_greater Error: Bad Argument Value

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maple [B]  time = 0.41, size = 581, normalized size = 2.32 \[ \frac {a^{2} x}{c^{2} d \sqrt {-c^{2} d \,x^{2}+d}}-\frac {a^{2} \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{c^{2} d \sqrt {c^{2} d}}+\frac {b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )^{3}}{3 d^{2} c^{3} \left (c^{2} x^{2}-1\right )}+\frac {i b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )^{2} \sqrt {-c^{2} x^{2}+1}}{d^{2} c^{3} \left (c^{2} x^{2}-1\right )}-\frac {b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )^{2} x}{d^{2} c^{2} \left (c^{2} x^{2}-1\right )}-\frac {2 b^{2} \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d^{2} c^{3} \left (c^{2} x^{2}-1\right )}+\frac {i b^{2} \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \polylog \left (2, -\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d^{2} c^{3} \left (c^{2} x^{2}-1\right )}+\frac {a b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )^{2}}{d^{2} c^{3} \left (c^{2} x^{2}-1\right )}+\frac {2 i a b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )}{d^{2} c^{3} \left (c^{2} x^{2}-1\right )}-\frac {2 a b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) x}{d^{2} c^{2} \left (c^{2} x^{2}-1\right )}-\frac {2 a b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d^{2} c^{3} \left (c^{2} x^{2}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*x/c^2/d/(-c^2*d*x^2+d)^(1/2)-a^2/c^2/d/(c^2*d)^(1/2)*arctan((c^2*d)^(1/2)*x/(-c^2*d*x^2+d)^(1/2))+1/3*b^2*
(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/d^2/c^3/(c^2*x^2-1)*arcsin(c*x)^3+I*b^2*(-d*(c^2*x^2-1))^(1/2)*arcsi
n(c*x)^2/d^2/c^3/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)-b^2*(-d*(c^2*x^2-1))^(1/2)*arcsin(c*x)^2/d^2/c^2/(c^2*x^2-1)*x
-2*b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^2/c^3/(c^2*x^2-1)*arcsin(c*x)*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2
))^2)+I*b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^2/c^3/(c^2*x^2-1)*polylog(2,-(I*c*x+(-c^2*x^2+1)^(1/2)
)^2)+a*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/d^2/c^3/(c^2*x^2-1)*arcsin(c*x)^2+2*I*a*b*(-c^2*x^2+1)^(1/2
)*(-d*(c^2*x^2-1))^(1/2)/d^2/c^3/(c^2*x^2-1)*arcsin(c*x)-2*a*b*(-d*(c^2*x^2-1))^(1/2)*arcsin(c*x)/d^2/c^2/(c^2
*x^2-1)*x-2*a*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/d^2/c^3/(c^2*x^2-1)*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^
2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*(x/(sqrt(-c^2*d*x^2 + d)*c^2*d) - arcsin(c*x)/(c^3*d^(3/2))) + sqrt(d)*integrate((b^2*x^2*arctan2(c*x, sqr
t(c*x + 1)*sqrt(-c*x + 1))^2 + 2*a*b*x^2*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)))*sqrt(c*x + 1)*sqrt(-c*x +
1)/(c^4*d^2*x^4 - 2*c^2*d^2*x^2 + d^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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