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

Optimal. Leaf size=227 \[ -\frac {c \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{3 b \sqrt {1-c^2 x^2}}-\frac {i c \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt {1-c^2 x^2}}-\frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{x}+\frac {2 b c \sqrt {d-c^2 d x^2} \log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}-\frac {i b^2 c \sqrt {d-c^2 d x^2} \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.32, antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {4693, 4625, 3717, 2190, 2279, 2391, 4641} \[ -\frac {i b^2 c \sqrt {d-c^2 d x^2} \text {PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )}{\sqrt {1-c^2 x^2}}-\frac {c \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{3 b \sqrt {1-c^2 x^2}}-\frac {i c \sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt {1-c^2 x^2}}-\frac {\sqrt {d-c^2 d x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{x}+\frac {2 b c \sqrt {d-c^2 d x^2} \log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2279

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 2391

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 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]

Rubi steps

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

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Mathematica [A]  time = 1.05, size = 257, normalized size = 1.13 \[ -\frac {a^2 \sqrt {d-c^2 d x^2}}{x}+a^2 c \sqrt {d} \tan ^{-1}\left (\frac {c x \sqrt {d-c^2 d x^2}}{\sqrt {d} \left (c^2 x^2-1\right )}\right )-\frac {a b \sqrt {d-c^2 d x^2} \left (2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x)-2 c x \log (c x)+c x \sin ^{-1}(c x)^2\right )}{x \sqrt {1-c^2 x^2}}-\frac {b^2 c \sqrt {d-c^2 d x^2} \left (\sin ^{-1}(c x) \left (\left (\frac {3 \sqrt {1-c^2 x^2}}{c x}+3 i\right ) \sin ^{-1}(c x)+\sin ^{-1}(c x)^2-6 \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )\right )+3 i \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )\right )}{3 \sqrt {1-c^2 x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-c^{2} d x^{2} + d} {\left (b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}\right )}}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))^2/x^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [B]  time = 0.42, size = 762, normalized size = 3.36 \[ -\frac {a^{2} \left (-c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}{d x}-a^{2} c^{2} x \sqrt {-c^{2} d \,x^{2}+d}-\frac {a^{2} c^{2} d \arctan \left (\frac {\sqrt {c^{2} d}\, x}{\sqrt {-c^{2} d \,x^{2}+d}}\right )}{\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} c}{3 c^{2} x^{2}-3}+\frac {2 i b^{2} \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )}{c^{2} x^{2}-1}-\frac {b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )^{2} x \,c^{2}}{c^{2} x^{2}-1}+\frac {b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )^{2}}{\left (c^{2} x^{2}-1\right ) x}-\frac {2 b^{2} \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right ) \arcsin \left (c x \right )}{c^{2} x^{2}-1}-\frac {2 b^{2} \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right ) \arcsin \left (c x \right )}{c^{2} x^{2}-1}+\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}{c^{2} x^{2}-1}+\frac {i b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )^{2} \sqrt {-c^{2} x^{2}+1}\, c}{c^{2} x^{2}-1}+\frac {a b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \arcsin \left (c x \right )^{2} c}{c^{2} x^{2}-1}+\frac {2 i b^{2} \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )}{c^{2} x^{2}-1}-\frac {2 a b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) x \,c^{2}}{c^{2} x^{2}-1}+\frac {2 a b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )}{\left (c^{2} x^{2}-1\right ) x}-\frac {2 a b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} x^{2}+1}\, \ln \left (\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}-1\right ) c}{c^{2} x^{2}-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))^2/x^2,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-c^2*d*x^2+d)^(1/2)*(a+b*arcsin(c*x))^2/x^2,x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*asin(c*x))^2*(d - c^2*d*x^2)^(1/2))/x^2,x)

[Out]

int(((a + b*asin(c*x))^2*(d - c^2*d*x^2)^(1/2))/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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