3.6.91 \(\int \frac {x^2 (a+b \text {ArcSin}(c x))^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}} \, dx\) [591]

Optimal. Leaf size=295 \[ \frac {x (a+b \text {ArcSin}(c x))^2}{c^2 d e \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {i \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))^2}{c^3 d e \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {\sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))^3}{3 b c^3 d e \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {2 b \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x)) \log \left (1+e^{2 i \text {ArcSin}(c x)}\right )}{c^3 d e \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {i b^2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,-e^{2 i \text {ArcSin}(c x)}\right )}{c^3 d e \sqrt {d+c d x} \sqrt {e-c e x}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.53, antiderivative size = 295, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {4823, 4791, 4737, 4765, 3800, 2221, 2317, 2438} \begin {gather*} \frac {x (a+b \text {ArcSin}(c x))^2}{c^2 d e \sqrt {c d x+d} \sqrt {e-c e x}}-\frac {\sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))^3}{3 b c^3 d e \sqrt {c d x+d} \sqrt {e-c e x}}-\frac {i \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))^2}{c^3 d e \sqrt {c d x+d} \sqrt {e-c e x}}+\frac {2 b \sqrt {1-c^2 x^2} \log \left (1+e^{2 i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{c^3 d e \sqrt {c d x+d} \sqrt {e-c e x}}-\frac {i b^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (-e^{2 i \text {ArcSin}(c x)}\right )}{c^3 d e \sqrt {c d x+d} \sqrt {e-c e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2317

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 2438

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 3800

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 4737

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

Rule 4765

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 4791

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/(2*c*(p + 1)))*Simp[(d
+ e*x^2)^p/(1 - c^2*x^2)^p], Int[(f*x)^(m - 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[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] && IGtQ[m, 1]

Rule 4823

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((h_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(p_)*((f_) + (g_.)*(x_))^(
q_), x_Symbol] :> Dist[((-d^2)*(g/e))^IntPart[q]*(d + e*x)^FracPart[q]*((f + g*x)^FracPart[q]/(1 - c^2*x^2)^Fr
acPart[q]), Int[(h*x)^m*(d + e*x)^(p - q)*(1 - c^2*x^2)^q*(a + b*ArcSin[c*x])^n, x], x] /; FreeQ[{a, b, c, d,
e, f, g, h, m, n}, x] && EqQ[e*f + d*g, 0] && EqQ[c^2*d^2 - e^2, 0] && HalfIntegerQ[p, q] && GeQ[p - q, 0]

Rubi steps

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

________________________________________________________________________________________

Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(636\) vs. \(2(295)=590\).
time = 1.51, size = 636, normalized size = 2.16 \begin {gather*} \frac {3 a^2 c \sqrt {d} e x+3 a^2 \sqrt {e} \sqrt {d+c d x} \sqrt {e-c e x} \text {ArcTan}\left (\frac {c x \sqrt {d+c d x} \sqrt {e-c e x}}{\sqrt {d} \sqrt {e} \left (-1+c^2 x^2\right )}\right )+3 a b \sqrt {d} e \left (2 c x \text {ArcSin}(c x)+\sqrt {1-c^2 x^2} \left (-\text {ArcSin}(c x)^2+2 \left (\log \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )-\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )+\log \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )+\sin \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )\right )\right )\right )+b^2 \sqrt {d} e \left (6 i \pi \sqrt {1-c^2 x^2} \text {ArcSin}(c x)+3 c x \text {ArcSin}(c x)^2-3 i \sqrt {1-c^2 x^2} \text {ArcSin}(c x)^2-\sqrt {1-c^2 x^2} \text {ArcSin}(c x)^3+12 \pi \sqrt {1-c^2 x^2} \log \left (1+e^{-i \text {ArcSin}(c x)}\right )+3 \pi \sqrt {1-c^2 x^2} \log \left (1-i e^{i \text {ArcSin}(c x)}\right )+6 \sqrt {1-c^2 x^2} \text {ArcSin}(c x) \log \left (1-i e^{i \text {ArcSin}(c x)}\right )-3 \pi \sqrt {1-c^2 x^2} \log \left (1+i e^{i \text {ArcSin}(c x)}\right )+6 \sqrt {1-c^2 x^2} \text {ArcSin}(c x) \log \left (1+i e^{i \text {ArcSin}(c x)}\right )-12 \pi \sqrt {1-c^2 x^2} \log \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )+3 \pi \sqrt {1-c^2 x^2} \log \left (-\cos \left (\frac {1}{4} (\pi +2 \text {ArcSin}(c x))\right )\right )-3 \pi \sqrt {1-c^2 x^2} \log \left (\sin \left (\frac {1}{4} (\pi +2 \text {ArcSin}(c x))\right )\right )-6 i \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,-i e^{i \text {ArcSin}(c x)}\right )-6 i \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,i e^{i \text {ArcSin}(c x)}\right )\right )}{3 c^3 d^{3/2} e^2 \sqrt {d+c d x} \sqrt {e-c e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a^2*c*Sqrt[d]*e*x + 3*a^2*Sqrt[e]*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*ArcTan[(c*x*Sqrt[d + c*d*x]*Sqrt[e - c*e*
x])/(Sqrt[d]*Sqrt[e]*(-1 + c^2*x^2))] + 3*a*b*Sqrt[d]*e*(2*c*x*ArcSin[c*x] + Sqrt[1 - c^2*x^2]*(-ArcSin[c*x]^2
+ 2*(Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]] + Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]]))) + b^2*Sqr
t[d]*e*((6*I)*Pi*Sqrt[1 - c^2*x^2]*ArcSin[c*x] + 3*c*x*ArcSin[c*x]^2 - (3*I)*Sqrt[1 - c^2*x^2]*ArcSin[c*x]^2 -
Sqrt[1 - c^2*x^2]*ArcSin[c*x]^3 + 12*Pi*Sqrt[1 - c^2*x^2]*Log[1 + E^((-I)*ArcSin[c*x])] + 3*Pi*Sqrt[1 - c^2*x
^2]*Log[1 - I*E^(I*ArcSin[c*x])] + 6*Sqrt[1 - c^2*x^2]*ArcSin[c*x]*Log[1 - I*E^(I*ArcSin[c*x])] - 3*Pi*Sqrt[1
- c^2*x^2]*Log[1 + I*E^(I*ArcSin[c*x])] + 6*Sqrt[1 - c^2*x^2]*ArcSin[c*x]*Log[1 + I*E^(I*ArcSin[c*x])] - 12*Pi
*Sqrt[1 - c^2*x^2]*Log[Cos[ArcSin[c*x]/2]] + 3*Pi*Sqrt[1 - c^2*x^2]*Log[-Cos[(Pi + 2*ArcSin[c*x])/4]] - 3*Pi*S
qrt[1 - c^2*x^2]*Log[Sin[(Pi + 2*ArcSin[c*x])/4]] - (6*I)*Sqrt[1 - c^2*x^2]*PolyLog[2, (-I)*E^(I*ArcSin[c*x])]
- (6*I)*Sqrt[1 - c^2*x^2]*PolyLog[2, I*E^(I*ArcSin[c*x])]))/(3*c^3*d^(3/2)*e^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x
])

________________________________________________________________________________________

Maple [F]
time = 0.52, size = 0, normalized size = 0.00 \[\int \frac {x^{2} \left (a +b \arcsin \left (c x \right )\right )^{2}}{\left (c d x +d \right )^{\frac {3}{2}} \left (-c e x +e \right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

a^2*(x*e^(-1)/(sqrt(-c^2*d*x^2*e + d*e)*c^2*d) - arcsin(c*x)*e^(-3/2)/(c^3*d^(3/2))) + sqrt(d)*e^(1/2)*integra
te((b^2*x^2*arctan2(c*x, sqrt(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*e^2 - 2*c^2*d^2*x^2*e^2 + d^2*e^2), x)

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________