3.2.84 \(\int \frac {x^3 (a+b \text {ArcSin}(c x))^2}{d-c^2 d x^2} \, dx\) [184]

Optimal. Leaf size=210 \[ \frac {b^2 x^2}{4 c^2 d}-\frac {b x \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))}{2 c^3 d}+\frac {(a+b \text {ArcSin}(c x))^2}{4 c^4 d}-\frac {x^2 (a+b \text {ArcSin}(c x))^2}{2 c^2 d}+\frac {i (a+b \text {ArcSin}(c x))^3}{3 b c^4 d}-\frac {(a+b \text {ArcSin}(c x))^2 \log \left (1+e^{2 i \text {ArcSin}(c x)}\right )}{c^4 d}+\frac {i b (a+b \text {ArcSin}(c x)) \text {PolyLog}\left (2,-e^{2 i \text {ArcSin}(c x)}\right )}{c^4 d}-\frac {b^2 \text {PolyLog}\left (3,-e^{2 i \text {ArcSin}(c x)}\right )}{2 c^4 d} \]

[Out]

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

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Rubi [A]
time = 0.27, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4795, 4765, 3800, 2221, 2611, 2320, 6724, 4737, 30} \begin {gather*} \frac {i b \text {Li}_2\left (-e^{2 i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{c^4 d}+\frac {i (a+b \text {ArcSin}(c x))^3}{3 b c^4 d}+\frac {(a+b \text {ArcSin}(c x))^2}{4 c^4 d}-\frac {\log \left (1+e^{2 i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))^2}{c^4 d}-\frac {x^2 (a+b \text {ArcSin}(c x))^2}{2 c^2 d}-\frac {b x \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))}{2 c^3 d}-\frac {b^2 \text {Li}_3\left (-e^{2 i \text {ArcSin}(c x)}\right )}{2 c^4 d}+\frac {b^2 x^2}{4 c^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*x^2)/(4*c^2*d) - (b*x*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/(2*c^3*d) + (a + b*ArcSin[c*x])^2/(4*c^4*d)
- (x^2*(a + b*ArcSin[c*x])^2)/(2*c^2*d) + ((I/3)*(a + b*ArcSin[c*x])^3)/(b*c^4*d) - ((a + b*ArcSin[c*x])^2*Log
[1 + E^((2*I)*ArcSin[c*x])])/(c^4*d) + (I*b*(a + b*ArcSin[c*x])*PolyLog[2, -E^((2*I)*ArcSin[c*x])])/(c^4*d) -
(b^2*PolyLog[3, -E^((2*I)*ArcSin[c*x])])/(2*c^4*d)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

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 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]

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int \frac {x^3 \left (a+b \sin ^{-1}(c x)\right )^2}{d-c^2 d x^2} \, dx &=-\frac {x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d}+\frac {\int \frac {x \left (a+b \sin ^{-1}(c x)\right )^2}{d-c^2 d x^2} \, dx}{c^2}+\frac {b \int \frac {x^2 \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx}{c d}\\ &=-\frac {b x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c^3 d}-\frac {x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d}+\frac {\text {Subst}\left (\int (a+b x)^2 \tan (x) \, dx,x,\sin ^{-1}(c x)\right )}{c^4 d}+\frac {b \int \frac {a+b \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \, dx}{2 c^3 d}+\frac {b^2 \int x \, dx}{2 c^2 d}\\ &=\frac {b^2 x^2}{4 c^2 d}-\frac {b x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c^3 d}+\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{4 c^4 d}-\frac {x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d}+\frac {i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^4 d}-\frac {(2 i) \text {Subst}\left (\int \frac {e^{2 i x} (a+b x)^2}{1+e^{2 i x}} \, dx,x,\sin ^{-1}(c x)\right )}{c^4 d}\\ &=\frac {b^2 x^2}{4 c^2 d}-\frac {b x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c^3 d}+\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{4 c^4 d}-\frac {x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d}+\frac {i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^4 d}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{c^4 d}+\frac {(2 b) \text {Subst}\left (\int (a+b x) \log \left (1+e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^4 d}\\ &=\frac {b^2 x^2}{4 c^2 d}-\frac {b x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c^3 d}+\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{4 c^4 d}-\frac {x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d}+\frac {i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^4 d}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{c^4 d}+\frac {i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{c^4 d}-\frac {\left (i b^2\right ) \text {Subst}\left (\int \text {Li}_2\left (-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^4 d}\\ &=\frac {b^2 x^2}{4 c^2 d}-\frac {b x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c^3 d}+\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{4 c^4 d}-\frac {x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d}+\frac {i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^4 d}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{c^4 d}+\frac {i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{c^4 d}-\frac {b^2 \text {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{2 c^4 d}\\ &=\frac {b^2 x^2}{4 c^2 d}-\frac {b x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c^3 d}+\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{4 c^4 d}-\frac {x^2 \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 d}+\frac {i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c^4 d}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )}{c^4 d}+\frac {i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{c^4 d}-\frac {b^2 \text {Li}_3\left (-e^{2 i \sin ^{-1}(c x)}\right )}{2 c^4 d}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(459\) vs. \(2(210)=420\).
time = 0.25, size = 459, normalized size = 2.19 \begin {gather*} -\frac {12 a^2 c^2 x^2+12 a b c x \sqrt {1-c^2 x^2}+48 i a b \pi \text {ArcSin}(c x)+24 a b c^2 x^2 \text {ArcSin}(c x)-24 i a b \text {ArcSin}(c x)^2-8 i b^2 \text {ArcSin}(c x)^3-24 a b \text {ArcTan}\left (\frac {c x}{-1+\sqrt {1-c^2 x^2}}\right )+3 b^2 \cos (2 \text {ArcSin}(c x))-6 b^2 \text {ArcSin}(c x)^2 \cos (2 \text {ArcSin}(c x))+96 a b \pi \log \left (1+e^{-i \text {ArcSin}(c x)}\right )+24 a b \pi \log \left (1-i e^{i \text {ArcSin}(c x)}\right )+48 a b \text {ArcSin}(c x) \log \left (1-i e^{i \text {ArcSin}(c x)}\right )-24 a b \pi \log \left (1+i e^{i \text {ArcSin}(c x)}\right )+48 a b \text {ArcSin}(c x) \log \left (1+i e^{i \text {ArcSin}(c x)}\right )+24 b^2 \text {ArcSin}(c x)^2 \log \left (1+e^{2 i \text {ArcSin}(c x)}\right )+12 a^2 \log \left (1-c^2 x^2\right )-96 a b \pi \log \left (\cos \left (\frac {1}{2} \text {ArcSin}(c x)\right )\right )+24 a b \pi \log \left (-\cos \left (\frac {1}{4} (\pi +2 \text {ArcSin}(c x))\right )\right )-24 a b \pi \log \left (\sin \left (\frac {1}{4} (\pi +2 \text {ArcSin}(c x))\right )\right )-48 i a b \text {PolyLog}\left (2,-i e^{i \text {ArcSin}(c x)}\right )-48 i a b \text {PolyLog}\left (2,i e^{i \text {ArcSin}(c x)}\right )-24 i b^2 \text {ArcSin}(c x) \text {PolyLog}\left (2,-e^{2 i \text {ArcSin}(c x)}\right )+12 b^2 \text {PolyLog}\left (3,-e^{2 i \text {ArcSin}(c x)}\right )+6 b^2 \text {ArcSin}(c x) \sin (2 \text {ArcSin}(c x))}{24 c^4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/24*(12*a^2*c^2*x^2 + 12*a*b*c*x*Sqrt[1 - c^2*x^2] + (48*I)*a*b*Pi*ArcSin[c*x] + 24*a*b*c^2*x^2*ArcSin[c*x]
- (24*I)*a*b*ArcSin[c*x]^2 - (8*I)*b^2*ArcSin[c*x]^3 - 24*a*b*ArcTan[(c*x)/(-1 + Sqrt[1 - c^2*x^2])] + 3*b^2*C
os[2*ArcSin[c*x]] - 6*b^2*ArcSin[c*x]^2*Cos[2*ArcSin[c*x]] + 96*a*b*Pi*Log[1 + E^((-I)*ArcSin[c*x])] + 24*a*b*
Pi*Log[1 - I*E^(I*ArcSin[c*x])] + 48*a*b*ArcSin[c*x]*Log[1 - I*E^(I*ArcSin[c*x])] - 24*a*b*Pi*Log[1 + I*E^(I*A
rcSin[c*x])] + 48*a*b*ArcSin[c*x]*Log[1 + I*E^(I*ArcSin[c*x])] + 24*b^2*ArcSin[c*x]^2*Log[1 + E^((2*I)*ArcSin[
c*x])] + 12*a^2*Log[1 - c^2*x^2] - 96*a*b*Pi*Log[Cos[ArcSin[c*x]/2]] + 24*a*b*Pi*Log[-Cos[(Pi + 2*ArcSin[c*x])
/4]] - 24*a*b*Pi*Log[Sin[(Pi + 2*ArcSin[c*x])/4]] - (48*I)*a*b*PolyLog[2, (-I)*E^(I*ArcSin[c*x])] - (48*I)*a*b
*PolyLog[2, I*E^(I*ArcSin[c*x])] - (24*I)*b^2*ArcSin[c*x]*PolyLog[2, -E^((2*I)*ArcSin[c*x])] + 12*b^2*PolyLog[
3, -E^((2*I)*ArcSin[c*x])] + 6*b^2*ArcSin[c*x]*Sin[2*ArcSin[c*x]])/(c^4*d)

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Maple [A]
time = 0.35, size = 380, normalized size = 1.81

method	result	size
derivativedivides	\(\frac {-\frac {a^{2} c^{2} x^{2}}{2 d}-\frac {a^{2} \ln \left (c x -1\right )}{2 d}-\frac {a^{2} \ln \left (c x +1\right )}{2 d}+\frac {i a b \arcsin \left (c x \right )^{2}}{d}-\frac {b^{2} \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, c x}{2 d}-\frac {b^{2} \arcsin \left (c x \right )^{2} c^{2} x^{2}}{2 d}+\frac {b^{2} \arcsin \left (c x \right )^{2}}{4 d}+\frac {b^{2} c^{2} x^{2}}{4 d}-\frac {b^{2}}{8 d}-\frac {b^{2} \arcsin \left (c x \right )^{2} \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d}+\frac {i b^{2} \arcsin \left (c x \right )^{3}}{3 d}-\frac {b^{2} \polylog \left (3, -\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{2 d}+\frac {i b^{2} \arcsin \left (c x \right ) \polylog \left (2, -\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d}-\frac {a b \sqrt {-c^{2} x^{2}+1}\, c x}{2 d}-\frac {a b \arcsin \left (c x \right ) c^{2} x^{2}}{d}+\frac {a b \arcsin \left (c x \right )}{2 d}-\frac {2 a b \arcsin \left (c x \right ) \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d}+\frac {i a b \polylog \left (2, -\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d}}{c^{4}}\)	\(380\)
default	\(\frac {-\frac {a^{2} c^{2} x^{2}}{2 d}-\frac {a^{2} \ln \left (c x -1\right )}{2 d}-\frac {a^{2} \ln \left (c x +1\right )}{2 d}+\frac {i a b \arcsin \left (c x \right )^{2}}{d}-\frac {b^{2} \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}\, c x}{2 d}-\frac {b^{2} \arcsin \left (c x \right )^{2} c^{2} x^{2}}{2 d}+\frac {b^{2} \arcsin \left (c x \right )^{2}}{4 d}+\frac {b^{2} c^{2} x^{2}}{4 d}-\frac {b^{2}}{8 d}-\frac {b^{2} \arcsin \left (c x \right )^{2} \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d}+\frac {i b^{2} \arcsin \left (c x \right )^{3}}{3 d}-\frac {b^{2} \polylog \left (3, -\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{2 d}+\frac {i b^{2} \arcsin \left (c x \right ) \polylog \left (2, -\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d}-\frac {a b \sqrt {-c^{2} x^{2}+1}\, c x}{2 d}-\frac {a b \arcsin \left (c x \right ) c^{2} x^{2}}{d}+\frac {a b \arcsin \left (c x \right )}{2 d}-\frac {2 a b \arcsin \left (c x \right ) \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d}+\frac {i a b \polylog \left (2, -\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d}}{c^{4}}\)	\(380\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^4*(-1/2*a^2/d*c^2*x^2-1/2*a^2/d*ln(c*x-1)-1/2*a^2/d*ln(c*x+1)+I*a*b/d*arcsin(c*x)^2-1/2*b^2/d*arcsin(c*x)*
(-c^2*x^2+1)^(1/2)*c*x-1/2*b^2/d*arcsin(c*x)^2*c^2*x^2+1/4*b^2/d*arcsin(c*x)^2+1/4*b^2/d*c^2*x^2-1/8*b^2/d-b^2
/d*arcsin(c*x)^2*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)+I*b^2/d*arcsin(c*x)*polylog(2,-(I*c*x+(-c^2*x^2+1)^(1/2))^
2)-1/2*b^2*polylog(3,-(I*c*x+(-c^2*x^2+1)^(1/2))^2)/d+1/3*I*b^2/d*arcsin(c*x)^3-1/2*a*b/d*(-c^2*x^2+1)^(1/2)*c
*x-a*b/d*arcsin(c*x)*c^2*x^2+1/2*a*b/d*arcsin(c*x)-2*a*b/d*arcsin(c*x)*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)+I*a*
b/d*polylog(2,-(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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a^2*(x^2/(c^2*d) + log(c^2*x^2 - 1)/(c^4*d)) - 1/2*(b^2*c^2*x^2*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)
)^2 + 2*c^4*d*integrate((2*a*b*c^3*x^3*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + (b^2*c^2*x^2*arctan2(c*x,
sqrt(c*x + 1)*sqrt(-c*x + 1)) + b^2*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(c*x + 1) + b^2*arctan2(c*x,
sqrt(c*x + 1)*sqrt(-c*x + 1))*log(-c*x + 1))*sqrt(c*x + 1)*sqrt(-c*x + 1))/(c^5*d*x^2 - c^3*d), x) + b^2*arct
an2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2*log(c*x + 1) + b^2*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2*log(-
c*x + 1))/(c^4*d)

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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