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

Optimal. Leaf size=178 \[ \frac {1}{2} d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2-\frac {1}{2} b c d x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )-i b d \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )-\frac {i d \left (a+b \sin ^{-1}(c x)\right )^3}{3 b}-\frac {1}{4} d \left (a+b \sin ^{-1}(c x)\right )^2+d \log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2+\frac {1}{4} b^2 c^2 d x^2+\frac {1}{2} b^2 d \text {Li}_3\left (e^{2 i \sin ^{-1}(c x)}\right ) \]

[Out]

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

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Rubi [A]  time = 0.24, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4699, 4625, 3717, 2190, 2531, 2282, 6589, 4647, 4641, 30} \[ -i b d \text {PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{2} b^2 d \text {PolyLog}\left (3,e^{2 i \sin ^{-1}(c x)}\right )+\frac {1}{2} d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2-\frac {1}{2} b c d x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )-\frac {i d \left (a+b \sin ^{-1}(c x)\right )^3}{3 b}-\frac {1}{4} d \left (a+b \sin ^{-1}(c x)\right )^2+d \log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2+\frac {1}{4} b^2 c^2 d x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 30

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

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 2282

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 2531

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 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 4625

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[(a + b*x)^n/Tan[x], x], x, ArcSin[c*
x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 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 4647

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(x*Sqrt[d + e*x^2]*(
a + b*ArcSin[c*x])^n)/2, x] + (Dist[Sqrt[d + e*x^2]/(2*Sqrt[1 - c^2*x^2]), Int[(a + b*ArcSin[c*x])^n/Sqrt[1 -
c^2*x^2], x], x] - Dist[(b*c*n*Sqrt[d + e*x^2])/(2*Sqrt[1 - c^2*x^2]), Int[x*(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]

Rule 4699

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

Rule 6589

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

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Mathematica [A]  time = 0.46, size = 236, normalized size = 1.33 \[ \frac {1}{2} d \left (a^2 \left (-c^2\right ) x^2+2 a^2 \log (x)-2 a b c^2 x^2 \sin ^{-1}(c x)+a b \left (\sin ^{-1}(c x)-c x \sqrt {1-c^2 x^2}\right )-2 i a b \left (\sin ^{-1}(c x)^2+\text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )\right )+4 a b \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )+\frac {1}{12} b^2 \left (24 i \sin ^{-1}(c x) \text {Li}_2\left (e^{-2 i \sin ^{-1}(c x)}\right )+12 \text {Li}_3\left (e^{-2 i \sin ^{-1}(c x)}\right )+8 i \sin ^{-1}(c x)^3+24 \sin ^{-1}(c x)^2 \log \left (1-e^{-2 i \sin ^{-1}(c x)}\right )-i \pi ^3\right )-\frac {1}{2} b^2 \sin ^{-1}(c x) \sin \left (2 \sin ^{-1}(c x)\right )+\frac {1}{4} b^2 \left (2 \sin ^{-1}(c x)^2-1\right ) \cos \left (2 \sin ^{-1}(c x)\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {a^{2} c^{2} d x^{2} - a^{2} d + {\left (b^{2} c^{2} d x^{2} - b^{2} d\right )} \arcsin \left (c x\right )^{2} + 2 \, {\left (a b c^{2} d x^{2} - a b d\right )} \arcsin \left (c x\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.35, size = 421, normalized size = 2.37 \[ -\frac {d \,a^{2} c^{2} x^{2}}{2}+d \,a^{2} \ln \left (c x \right )-2 i d a b \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )+d \,b^{2} \arcsin \left (c x \right )^{2} \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )-2 i d a b \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )+2 d \,b^{2} \polylog \left (3, -i c x -\sqrt {-c^{2} x^{2}+1}\right )+d \,b^{2} \arcsin \left (c x \right )^{2} \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )-2 i d \,b^{2} \arcsin \left (c x \right ) \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )+2 d \,b^{2} \polylog \left (3, i c x +\sqrt {-c^{2} x^{2}+1}\right )+\frac {d \,b^{2} \cos \left (2 \arcsin \left (c x \right )\right ) \arcsin \left (c x \right )^{2}}{4}-\frac {d \,b^{2} \cos \left (2 \arcsin \left (c x \right )\right )}{8}-\frac {d \,b^{2} \arcsin \left (c x \right ) \sin \left (2 \arcsin \left (c x \right )\right )}{4}-2 i d \,b^{2} \arcsin \left (c x \right ) \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )+2 d a b \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )+2 d a b \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )-\frac {i d \,b^{2} \arcsin \left (c x \right )^{3}}{3}-i d a b \arcsin \left (c x \right )^{2}+\frac {d a b \cos \left (2 \arcsin \left (c x \right )\right ) \arcsin \left (c x \right )}{2}-\frac {d a b \sin \left (2 \arcsin \left (c x \right )\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a^2*c^2*d*x^2 + a^2*d*log(x) - integrate(((b^2*c^2*d*x^2 - b^2*d)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x +
1))^2 + 2*(a*b*c^2*d*x^2 - a*b*d)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)))/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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