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

Optimal. Leaf size=193 \[ i b c^2 d \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )-\frac {d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{2 x^2}-\frac {b c d \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{x}+\frac {i c^2 d \left (a+b \sin ^{-1}(c x)\right )^3}{3 b}-\frac {1}{2} c^2 d \left (a+b \sin ^{-1}(c x)\right )^2-c^2 d \log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2-\frac {1}{2} b^2 c^2 d \text {Li}_3\left (e^{2 i \sin ^{-1}(c x)}\right )+b^2 c^2 d \log (x) \]

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 193, 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 = {4695, 4625, 3717, 2190, 2531, 2282, 6589, 4693, 29, 4641} \[ i b c^2 d \text {PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )-\frac {1}{2} b^2 c^2 d \text {PolyLog}\left (3,e^{2 i \sin ^{-1}(c x)}\right )-\frac {d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{2 x^2}-\frac {b c d \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{x}+\frac {i c^2 d \left (a+b \sin ^{-1}(c x)\right )^3}{3 b}-\frac {1}{2} c^2 d \left (a+b \sin ^{-1}(c x)\right )^2-c^2 d \log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2+b^2 c^2 d \log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

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 4693

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

Rule 4695

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 + 1)), x] + (-Dist[(2*e*p)/(f^2*(m + 1)), Int[(f*x)^
(m + 2)*(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 + 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}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[p, 0] && LtQ[m, -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^3} \, dx &=-\frac {d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{2 x^2}+(b c d) \int \frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{x^2} \, dx-\left (c^2 d\right ) \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{x} \, dx\\ &=-\frac {b c d \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac {d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{2 x^2}-\left (c^2 d\right ) \operatorname {Subst}\left (\int (a+b x)^2 \cot (x) \, dx,x,\sin ^{-1}(c x)\right )+\left (b^2 c^2 d\right ) \int \frac {1}{x} \, dx-\left (b c^3 d\right ) \int \frac {a+b \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \, dx\\ &=-\frac {b c d \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac {1}{2} c^2 d \left (a+b \sin ^{-1}(c x)\right )^2-\frac {d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{2 x^2}+\frac {i c^2 d \left (a+b \sin ^{-1}(c x)\right )^3}{3 b}+b^2 c^2 d \log (x)+\left (2 i c^2 d\right ) \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 {b c d \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac {1}{2} c^2 d \left (a+b \sin ^{-1}(c x)\right )^2-\frac {d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{2 x^2}+\frac {i c^2 d \left (a+b \sin ^{-1}(c x)\right )^3}{3 b}-c^2 d \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )+b^2 c^2 d \log (x)+\left (2 b c^2 d\right ) \operatorname {Subst}\left (\int (a+b x) \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )\\ &=-\frac {b c d \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac {1}{2} c^2 d \left (a+b \sin ^{-1}(c x)\right )^2-\frac {d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{2 x^2}+\frac {i c^2 d \left (a+b \sin ^{-1}(c x)\right )^3}{3 b}-c^2 d \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )+b^2 c^2 d \log (x)+i b c^2 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 c^2 d\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )\\ &=-\frac {b c d \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac {1}{2} c^2 d \left (a+b \sin ^{-1}(c x)\right )^2-\frac {d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{2 x^2}+\frac {i c^2 d \left (a+b \sin ^{-1}(c x)\right )^3}{3 b}-c^2 d \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )+b^2 c^2 d \log (x)+i b c^2 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 c^2 d\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )\\ &=-\frac {b c d \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac {1}{2} c^2 d \left (a+b \sin ^{-1}(c x)\right )^2-\frac {d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{2 x^2}+\frac {i c^2 d \left (a+b \sin ^{-1}(c x)\right )^3}{3 b}-c^2 d \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )+b^2 c^2 d \log (x)+i b c^2 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 c^2 d \text {Li}_3\left (e^{2 i \sin ^{-1}(c x)}\right )\\ \end {align*}

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Mathematica [A]  time = 0.39, size = 236, normalized size = 1.22 \[ \frac {1}{2} d \left (-2 a^2 c^2 \log (x)-\frac {a^2}{x^2}+2 i a b c^2 \left (\text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )+\sin ^{-1}(c x) \left (\sin ^{-1}(c x)+2 i \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )\right )\right )-\frac {2 a b \left (c x \sqrt {1-c^2 x^2}+\sin ^{-1}(c x)\right )}{x^2}+\frac {1}{12} i b^2 c^2 \left (-24 \sin ^{-1}(c x) \text {Li}_2\left (e^{-2 i \sin ^{-1}(c x)}\right )+12 i \text {Li}_3\left (e^{-2 i \sin ^{-1}(c x)}\right )-8 \sin ^{-1}(c x)^3+24 i \sin ^{-1}(c x)^2 \log \left (1-e^{-2 i \sin ^{-1}(c x)}\right )+\pi ^3\right )-\frac {b^2 \left (-2 c^2 x^2 \log (c x)+2 c x \sqrt {1-c^2 x^2} \sin ^{-1}(c x)+\sin ^{-1}(c x)^2\right )}{x^2}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {a^{2} c^{2} d x^{2} - a^{2} d + {\left (b^{2} c^{2} d x^{2} - b^{2} d\right )} \arcsin \left (c x\right )^{2} + 2 \, {\left (a b c^{2} d x^{2} - a b d\right )} \arcsin \left (c x\right )}{x^{3}}, x\right ) \]

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^3,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^3, x)

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

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^3,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.59, size = 564, normalized size = 2.92 \[ -c^{2} d \,a^{2} \ln \left (c x \right )-\frac {d \,a^{2}}{2 x^{2}}+\frac {i c^{2} d \,b^{2} \arcsin \left (c x \right )^{3}}{3}+2 i c^{2} d \,b^{2} \arcsin \left (c x \right ) \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )-\frac {c d \,b^{2} \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}}{x}-\frac {d \,b^{2} \arcsin \left (c x \right )^{2}}{2 x^{2}}-c^{2} d \,b^{2} \arcsin \left (c x \right )^{2} \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )+i c^{2} d a b -2 c^{2} d \,b^{2} \polylog \left (3, -i c x -\sqrt {-c^{2} x^{2}+1}\right )-c^{2} d \,b^{2} \arcsin \left (c x \right )^{2} \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )+i c^{2} d \,b^{2} \arcsin \left (c x \right )-2 c^{2} d \,b^{2} \polylog \left (3, i c x +\sqrt {-c^{2} x^{2}+1}\right )+c^{2} d \,b^{2} \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )-2 c^{2} d \,b^{2} \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )+c^{2} d \,b^{2} \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-1\right )+2 i c^{2} d a b \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )+2 i c^{2} d a b \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )-\frac {c d a b \sqrt {-c^{2} x^{2}+1}}{x}-\frac {d a b \arcsin \left (c x \right )}{x^{2}}-2 c^{2} d a b \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )-2 c^{2} d a b \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )+2 i c^{2} d \,b^{2} \arcsin \left (c x \right ) \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )+i c^{2} d a b \arcsin \left (c x \right )^{2} \]

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^3,x)

[Out]

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

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

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^3,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,\left (d-c^2\,d\,x^2\right )}{x^3} \,d x \]

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^3,x)

[Out]

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

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

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**3,x)

[Out]

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

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