3.3.45 \(\int \frac {x^4 (a+b \text {ArcSin}(c x))^2}{(d-c^2 d x^2)^{3/2}} \, dx\) [245]

Optimal. Leaf size=424 \[ -\frac {b^2 x \left (1-c^2 x^2\right )}{4 c^4 d \sqrt {d-c^2 d x^2}}+\frac {b^2 \sqrt {1-c^2 x^2} \text {ArcSin}(c x)}{4 c^5 d \sqrt {d-c^2 d x^2}}-\frac {b x^2 \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))}{2 c^3 d \sqrt {d-c^2 d x^2}}+\frac {x^3 (a+b \text {ArcSin}(c x))^2}{c^2 d \sqrt {d-c^2 d x^2}}-\frac {i \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))^2}{c^5 d \sqrt {d-c^2 d x^2}}+\frac {3 x \sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x))^2}{2 c^4 d^2}-\frac {\sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))^3}{2 b c^5 d \sqrt {d-c^2 d x^2}}+\frac {2 b \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x)) \log \left (1+e^{2 i \text {ArcSin}(c x)}\right )}{c^5 d \sqrt {d-c^2 d x^2}}-\frac {i b^2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,-e^{2 i \text {ArcSin}(c x)}\right )}{c^5 d \sqrt {d-c^2 d x^2}} \]

[Out]

-1/4*b^2*x*(-c^2*x^2+1)/c^4/d/(-c^2*d*x^2+d)^(1/2)+x^3*(a+b*arcsin(c*x))^2/c^2/d/(-c^2*d*x^2+d)^(1/2)+1/4*b^2*
arcsin(c*x)*(-c^2*x^2+1)^(1/2)/c^5/d/(-c^2*d*x^2+d)^(1/2)-1/2*b*x^2*(a+b*arcsin(c*x))*(-c^2*x^2+1)^(1/2)/c^3/d
/(-c^2*d*x^2+d)^(1/2)-I*(a+b*arcsin(c*x))^2*(-c^2*x^2+1)^(1/2)/c^5/d/(-c^2*d*x^2+d)^(1/2)-1/2*(a+b*arcsin(c*x)
)^3*(-c^2*x^2+1)^(1/2)/b/c^5/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^5/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^5/d/(-c^2*d*x^2+d)^(1/2)+3/2*x*(a+b*arcsin(c*x))^2*(-c^2*d*x^2+d)^(1/2)/c^4/d^2

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Rubi [A]
time = 0.43, antiderivative size = 424, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.379, Rules used = {4791, 4795, 4737, 4723, 327, 222, 4765, 3800, 2221, 2317, 2438} \begin {gather*} \frac {x^3 (a+b \text {ArcSin}(c x))^2}{c^2 d \sqrt {d-c^2 d x^2}}-\frac {\sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))^3}{2 b c^5 d \sqrt {d-c^2 d x^2}}-\frac {i \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))^2}{c^5 d \sqrt {d-c^2 d x^2}}+\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^5 d \sqrt {d-c^2 d x^2}}+\frac {3 x \sqrt {d-c^2 d x^2} (a+b \text {ArcSin}(c x))^2}{2 c^4 d^2}-\frac {b x^2 \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))}{2 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 \text {ArcSin}(c x)}\right )}{c^5 d \sqrt {d-c^2 d x^2}}+\frac {b^2 \sqrt {1-c^2 x^2} \text {ArcSin}(c x)}{4 c^5 d \sqrt {d-c^2 d x^2}}-\frac {b^2 x \left (1-c^2 x^2\right )}{4 c^4 d \sqrt {d-c^2 d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*ArcSin[c*x])^2)/(d - c^2*d*x^2)^(3/2),x]

[Out]

-1/4*(b^2*x*(1 - c^2*x^2))/(c^4*d*Sqrt[d - c^2*d*x^2]) + (b^2*Sqrt[1 - c^2*x^2]*ArcSin[c*x])/(4*c^5*d*Sqrt[d -
c^2*d*x^2]) - (b*x^2*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/(2*c^3*d*Sqrt[d - c^2*d*x^2]) + (x^3*(a + b*ArcSi
n[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^5*d*Sqrt[d - c^2*d*x^2
]) + (3*x*Sqrt[d - c^2*d*x^2]*(a + b*ArcSin[c*x])^2)/(2*c^4*d^2) - (Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^3)/(
2*b*c^5*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
^5*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^5*d*Sqrt[d - c^2*d
*x^2])

Rule 222

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 327

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*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, 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 4723

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

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 4795

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/(e*(m + 2*p + 1))), x] + (Dist[f^2*((m - 1)/(c^2*(m
+ 2*p + 1))), Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, x], x] + Dist[b*f*(n/(c*(m + 2*p + 1)))*S
imp[(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, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && IGtQ[m, 1] && NeQ[m + 2*p + 1, 0]

Rubi steps

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

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Mathematica [A]
time = 1.46, size = 312, normalized size = 0.74 \begin {gather*} \frac {-4 a^2 c \sqrt {d} x \left (-3+c^2 x^2\right )+12 a^2 \sqrt {d-c^2 d x^2} \text {ArcTan}\left (\frac {c x \sqrt {d-c^2 d x^2}}{\sqrt {d} \left (-1+c^2 x^2\right )}\right )+2 a b \sqrt {d} \left (8 c x \text {ArcSin}(c x)+\sqrt {1-c^2 x^2} \left (-6 \text {ArcSin}(c x)^2+\cos (2 \text {ArcSin}(c x))+4 \log \left (1-c^2 x^2\right )+2 \text {ArcSin}(c x) \sin (2 \text {ArcSin}(c x))\right )\right )+b^2 \sqrt {d} \left (8 c x \text {ArcSin}(c x)^2-8 i \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,-e^{2 i \text {ArcSin}(c x)}\right )+\sqrt {1-c^2 x^2} \left (-4 \text {ArcSin}(c x)^3+2 \text {ArcSin}(c x) \left (\cos (2 \text {ArcSin}(c x))+8 \log \left (1+e^{2 i \text {ArcSin}(c x)}\right )\right )-\sin (2 \text {ArcSin}(c x))+2 \text {ArcSin}(c x)^2 (-4 i+\sin (2 \text {ArcSin}(c x)))\right )\right )}{8 c^5 d^{3/2} \sqrt {d-c^2 d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(a + b*ArcSin[c*x])^2)/(d - c^2*d*x^2)^(3/2),x]

[Out]

(-4*a^2*c*Sqrt[d]*x*(-3 + c^2*x^2) + 12*a^2*Sqrt[d - c^2*d*x^2]*ArcTan[(c*x*Sqrt[d - c^2*d*x^2])/(Sqrt[d]*(-1
+ c^2*x^2))] + 2*a*b*Sqrt[d]*(8*c*x*ArcSin[c*x] + Sqrt[1 - c^2*x^2]*(-6*ArcSin[c*x]^2 + Cos[2*ArcSin[c*x]] + 4
*Log[1 - c^2*x^2] + 2*ArcSin[c*x]*Sin[2*ArcSin[c*x]])) + b^2*Sqrt[d]*(8*c*x*ArcSin[c*x]^2 - (8*I)*Sqrt[1 - c^2
*x^2]*PolyLog[2, -E^((2*I)*ArcSin[c*x])] + Sqrt[1 - c^2*x^2]*(-4*ArcSin[c*x]^3 + 2*ArcSin[c*x]*(Cos[2*ArcSin[c
*x]] + 8*Log[1 + E^((2*I)*ArcSin[c*x])]) - Sin[2*ArcSin[c*x]] + 2*ArcSin[c*x]^2*(-4*I + Sin[2*ArcSin[c*x]]))))
/(8*c^5*d^(3/2)*Sqrt[d - c^2*d*x^2])

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 972 vs. \(2 (402 ) = 804\).
time = 0.59, size = 973, normalized size = 2.29

method	result	size
default	\(-\frac {a^{2} x^{3}}{2 c^{2} d \sqrt {-c^{2} d \,x^{2}+d}}+\frac {3 a^{2} x}{2 c^{4} d \sqrt {-c^{2} d \,x^{2}+d}}-\frac {3 a^{2} \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{2 c^{4} 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}}{2 c^{5} d^{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 )}{c^{5} d^{2} \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 )}{c^{5} d^{2} \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 )}\, \arcsin \left (c x \right )^{2}}{c^{5} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b^{2} \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )}{8 c^{5} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {9 b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x \arcsin \left (c x \right )^{2}}{8 c^{4} d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, x}{16 c^{4} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \cos \left (3 \arcsin \left (c x \right )\right )}{8 c^{5} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sin \left (3 \arcsin \left (c x \right )\right ) \arcsin \left (c x \right )^{2}}{8 c^{5} d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sin \left (3 \arcsin \left (c x \right )\right )}{16 c^{5} d^{2} \left (c^{2} x^{2}-1\right )}+\frac {3 a b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )^{2}}{2 c^{5} d^{2} \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 )}{c^{5} d^{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 )}{c^{5} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {a b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}}{8 c^{5} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {9 a b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) x}{4 c^{4} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {a b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \cos \left (3 \arcsin \left (c x \right )\right )}{8 c^{5} d^{2} \left (c^{2} x^{2}-1\right )}-\frac {a b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) \sin \left (3 \arcsin \left (c x \right )\right )}{4 c^{5} d^{2} \left (c^{2} x^{2}-1\right )}\)	\(973\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*arcsin(c*x))^2/(-c^2*d*x^2+d)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*a^2*x^3/c^2/d/(-c^2*d*x^2+d)^(1/2)+3/2*a^2/c^4*x/d/(-c^2*d*x^2+d)^(1/2)-3/2*a^2/c^4/d/(c^2*d)^(1/2)*arcta
n((c^2*d)^(1/2)*x/(-c^2*d*x^2+d)^(1/2))+1/2*b^2*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c^5/d^2/(c^2*x^2-1)*
arcsin(c*x)^3-2*b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/c^5/d^2/(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)/c^5/d^2/(c^2*x^2-1)*polylog(2,-(I*c*x+(-c^2
*x^2+1)^(1/2))^2)+2*I*a*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/c^5/d^2/(c^2*x^2-1)*arcsin(c*x)-1/8*b^2*(-
c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/c^5/d^2/(c^2*x^2-1)*arcsin(c*x)-9/8*b^2*(-d*(c^2*x^2-1))^(1/2)/c^4/d^2
/(c^2*x^2-1)*x*arcsin(c*x)^2+1/16*b^2*(-d*(c^2*x^2-1))^(1/2)/c^4/d^2/(c^2*x^2-1)*x-1/8*b^2*(-d*(c^2*x^2-1))^(1
/2)/c^5/d^2/(c^2*x^2-1)*arcsin(c*x)*cos(3*arcsin(c*x))-1/8*b^2*(-d*(c^2*x^2-1))^(1/2)/c^5/d^2/(c^2*x^2-1)*sin(
3*arcsin(c*x))*arcsin(c*x)^2+1/16*b^2*(-d*(c^2*x^2-1))^(1/2)/c^5/d^2/(c^2*x^2-1)*sin(3*arcsin(c*x))+3/2*a*b*(-
d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c^5/d^2/(c^2*x^2-1)*arcsin(c*x)^2+I*b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^
2-1))^(1/2)/c^5/d^2/(c^2*x^2-1)*arcsin(c*x)^2-2*a*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c^5/d^2/(c^2*x^2
-1)*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)-1/8*a*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/c^5/d^2/(c^2*x^2-1)-9
/4*a*b*(-d*(c^2*x^2-1))^(1/2)/c^4/d^2/(c^2*x^2-1)*arcsin(c*x)*x-1/8*a*b*(-d*(c^2*x^2-1))^(1/2)/c^5/d^2/(c^2*x^
2-1)*cos(3*arcsin(c*x))-1/4*a*b*(-d*(c^2*x^2-1))^(1/2)/c^5/d^2/(c^2*x^2-1)*arcsin(c*x)*sin(3*arcsin(c*x))

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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^4*(a+b*arcsin(c*x))^2/(-c^2*d*x^2+d)^(3/2),x, algorithm="maxima")

[Out]

-1/2*a^2*(x^3/(sqrt(-c^2*d*x^2 + d)*c^2*d) - 3*x/(sqrt(-c^2*d*x^2 + d)*c^4*d) + 3*arcsin(c*x)/(c^5*d^(3/2))) +
sqrt(d)*integrate((b^2*x^4*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2 + 2*a*b*x^4*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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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^4*(a+b*arcsin(c*x))^2/(-c^2*d*x^2+d)^(3/2),x, algorithm="fricas")

[Out]

integral((b^2*x^4*arcsin(c*x)^2 + 2*a*b*x^4*arcsin(c*x) + a^2*x^4)*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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(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,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(t_

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^4\,{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{{\left (d-c^2\,d\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(a + b*asin(c*x))^2)/(d - c^2*d*x^2)^(3/2),x)

[Out]

int((x^4*(a + b*asin(c*x))^2)/(d - c^2*d*x^2)^(3/2), x)

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