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

Optimal. Leaf size=333 \[ \frac {2 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{d \sqrt {d-c^2 d x^2}}-\frac {2 i c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{d \sqrt {d-c^2 d x^2}}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d x \sqrt {d-c^2 d x^2}}+\frac {4 b c \sqrt {1-c^2 x^2} \log \left (1+e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d \sqrt {d-c^2 d x^2}}-\frac {4 b c \sqrt {1-c^2 x^2} \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d \sqrt {d-c^2 d x^2}}-\frac {i b^2 c \sqrt {1-c^2 x^2} \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}-\frac {i b^2 c \sqrt {1-c^2 x^2} \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{d \sqrt {d-c^2 d x^2}} \]

[Out]

-(a+b*arcsin(c*x))^2/d/x/(-c^2*d*x^2+d)^(1/2)+2*c^2*x*(a+b*arcsin(c*x))^2/d/(-c^2*d*x^2+d)^(1/2)-2*I*c*(a+b*ar
csin(c*x))^2*(-c^2*x^2+1)^(1/2)/d/(-c^2*d*x^2+d)^(1/2)-4*b*c*(a+b*arcsin(c*x))*arctanh((I*c*x+(-c^2*x^2+1)^(1/
2))^2)*(-c^2*x^2+1)^(1/2)/d/(-c^2*d*x^2+d)^(1/2)+4*b*c*(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)/d/(-c^2*d*x^2+d)^(1/2)-I*b^2*c*polylog(2,-(I*c*x+(-c^2*x^2+1)^(1/2))^2)*(-c^2*x^2+1)^(1/2)/d/
(-c^2*d*x^2+d)^(1/2)-I*b^2*c*polylog(2,(I*c*x+(-c^2*x^2+1)^(1/2))^2)*(-c^2*x^2+1)^(1/2)/d/(-c^2*d*x^2+d)^(1/2)

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Rubi [A]  time = 0.44, antiderivative size = 333, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {4701, 4653, 4675, 3719, 2190, 2279, 2391, 4679, 4419, 4183} \[ -\frac {i b^2 c \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}-\frac {i b^2 c \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}+\frac {2 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{d \sqrt {d-c^2 d x^2}}-\frac {2 i c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{d \sqrt {d-c^2 d x^2}}-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d x \sqrt {d-c^2 d x^2}}+\frac {4 b c \sqrt {1-c^2 x^2} \log \left (1+e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d \sqrt {d-c^2 d x^2}}-\frac {4 b c \sqrt {1-c^2 x^2} \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b*ArcSin[c*x])^2/(d*x*Sqrt[d - c^2*d*x^2])) + (2*c^2*x*(a + b*ArcSin[c*x])^2)/(d*Sqrt[d - c^2*d*x^2]) -
((2*I)*c*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2)/(d*Sqrt[d - c^2*d*x^2]) - (4*b*c*Sqrt[1 - c^2*x^2]*(a + b*A
rcSin[c*x])*ArcTanh[E^((2*I)*ArcSin[c*x])])/(d*Sqrt[d - c^2*d*x^2]) + (4*b*c*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c
*x])*Log[1 + E^((2*I)*ArcSin[c*x])])/(d*Sqrt[d - c^2*d*x^2]) - (I*b^2*c*Sqrt[1 - c^2*x^2]*PolyLog[2, -E^((2*I)
*ArcSin[c*x])])/(d*Sqrt[d - c^2*d*x^2]) - (I*b^2*c*Sqrt[1 - c^2*x^2]*PolyLog[2, E^((2*I)*ArcSin[c*x])])/(d*Sqr
t[d - c^2*d*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 3719

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 4183

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E^(I*(e + f*
x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[(d*m)/f, Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IGtQ[m, 0]

Rule 4419

Int[Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dist[
2^n, Int[(c + d*x)^m*Csc[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d, m}, x] && IntegerQ[n] && RationalQ[m]

Rule 4653

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(x*(a + b*ArcSin[c
*x])^n)/(d*Sqrt[d + e*x^2]), x] - Dist[(b*c*n*Sqrt[1 - c^2*x^2])/(d*Sqrt[d + e*x^2]), Int[(x*(a + b*ArcSin[c*x
])^(n - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0]

Rule 4675

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 4679

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Dist[1/d, Subst[Int[(a
+ b*x)^n/(Cos[x]*Sin[x]), x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[n
, 0]

Rule 4701

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 + 1)*(a + b*ArcSin[c*x])^n)/(d*f*(m + 1)), x] + (Dist[(c^2*(m + 2*p + 3))/(f^2*(m
+ 1)), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^F
racPart[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, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[m, -1] && Inte
gerQ[m]

Rubi steps

\begin {align*} \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{x^2 \left (d-c^2 d x^2\right )^{3/2}} \, dx &=-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d x \sqrt {d-c^2 d x^2}}+\left (2 c^2\right ) \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{\left (d-c^2 d x^2\right )^{3/2}} \, dx+\frac {\left (2 b c \sqrt {1-c^2 x^2}\right ) \int \frac {a+b \sin ^{-1}(c x)}{x \left (1-c^2 x^2\right )} \, dx}{d \sqrt {d-c^2 d x^2}}\\ &=-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d x \sqrt {d-c^2 d x^2}}+\frac {2 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{d \sqrt {d-c^2 d x^2}}+\frac {\left (2 b c \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int (a+b x) \csc (x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{d \sqrt {d-c^2 d x^2}}-\frac {\left (4 b c^3 \sqrt {1-c^2 x^2}\right ) \int \frac {x \left (a+b \sin ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{d \sqrt {d-c^2 d x^2}}\\ &=-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d x \sqrt {d-c^2 d x^2}}+\frac {2 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{d \sqrt {d-c^2 d x^2}}+\frac {\left (4 b c \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int (a+b x) \csc (2 x) \, dx,x,\sin ^{-1}(c x)\right )}{d \sqrt {d-c^2 d x^2}}-\frac {\left (4 b c \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int (a+b x) \tan (x) \, dx,x,\sin ^{-1}(c x)\right )}{d \sqrt {d-c^2 d x^2}}\\ &=-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d x \sqrt {d-c^2 d x^2}}+\frac {2 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{d \sqrt {d-c^2 d x^2}}-\frac {2 i c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{d \sqrt {d-c^2 d x^2}}-\frac {4 b c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}+\frac {\left (8 i b c \sqrt {1-c^2 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 )}{d \sqrt {d-c^2 d x^2}}-\frac {\left (2 b^2 c \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d \sqrt {d-c^2 d x^2}}+\frac {\left (2 b^2 c \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d \sqrt {d-c^2 d x^2}}\\ &=-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d x \sqrt {d-c^2 d x^2}}+\frac {2 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{d \sqrt {d-c^2 d x^2}}-\frac {2 i c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{d \sqrt {d-c^2 d x^2}}-\frac {4 b c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}+\frac {4 b c \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 )}{d \sqrt {d-c^2 d x^2}}+\frac {\left (i b^2 c \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}-\frac {\left (i b^2 c \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}-\frac {\left (4 b^2 c \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d \sqrt {d-c^2 d x^2}}\\ &=-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d x \sqrt {d-c^2 d x^2}}+\frac {2 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{d \sqrt {d-c^2 d x^2}}-\frac {2 i c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{d \sqrt {d-c^2 d x^2}}-\frac {4 b c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}+\frac {4 b c \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 )}{d \sqrt {d-c^2 d x^2}}+\frac {i b^2 c \sqrt {1-c^2 x^2} \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}-\frac {i b^2 c \sqrt {1-c^2 x^2} \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}+\frac {\left (2 i b^2 c \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}\\ &=-\frac {\left (a+b \sin ^{-1}(c x)\right )^2}{d x \sqrt {d-c^2 d x^2}}+\frac {2 c^2 x \left (a+b \sin ^{-1}(c x)\right )^2}{d \sqrt {d-c^2 d x^2}}-\frac {2 i c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{d \sqrt {d-c^2 d x^2}}-\frac {4 b c \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}+\frac {4 b c \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 )}{d \sqrt {d-c^2 d x^2}}-\frac {i b^2 c \sqrt {1-c^2 x^2} \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}-\frac {i b^2 c \sqrt {1-c^2 x^2} \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{d \sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.78, size = 322, normalized size = 0.97 \[ \frac {2 a^2 c^2 x^2-a^2+2 a b c x \sqrt {1-c^2 x^2} \log (c x)+a b c x \sqrt {1-c^2 x^2} \log \left (1-c^2 x^2\right )+4 a b c^2 x^2 \sin ^{-1}(c x)-2 a b \sin ^{-1}(c x)-i b^2 c x \sqrt {1-c^2 x^2} \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )-i b^2 c x \sqrt {1-c^2 x^2} \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )+2 b^2 c^2 x^2 \sin ^{-1}(c x)^2-2 i b^2 c x \sqrt {1-c^2 x^2} \sin ^{-1}(c x)^2+2 b^2 c x \sqrt {1-c^2 x^2} \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )+2 b^2 c x \sqrt {1-c^2 x^2} \sin ^{-1}(c x) \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )-b^2 \sin ^{-1}(c x)^2}{d x \sqrt {d-c^2 d x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-c^{2} d x^{2} + d} {\left (b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}\right )}}{c^{4} d^{2} x^{6} - 2 \, c^{2} d^{2} x^{4} + d^{2} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))^2/x^2/(-c^2*d*x^2+d)^(3/2),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)/(c^4*d^2*x^6 - 2*c^2*d^2*x^4 + d^2
*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.38, size = 807, normalized size = 2.42 \[ -\frac {a^{2}}{d x \sqrt {-c^{2} d \,x^{2}+d}}+\frac {2 a^{2} c^{2} x}{d \sqrt {-c^{2} d \,x^{2}+d}}+\frac {2 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}{\left (c^{2} x^{2}-1\right ) d^{2}}-\frac {2 b^{2} \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right )^{2} x \,c^{2}}{\left (c^{2} x^{2}-1\right ) d^{2}}+\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 ) d^{2} x}-\frac {2 b^{2} \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )}{\left (c^{2} x^{2}-1\right ) d^{2}}-\frac {2 b^{2} \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c \arcsin \left (c x \right ) \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{\left (c^{2} x^{2}-1\right ) d^{2}}-\frac {2 b^{2} \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{\left (c^{2} x^{2}-1\right ) d^{2}}+\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 )}{\left (c^{2} x^{2}-1\right ) d^{2}}+\frac {i b^{2} \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, c \polylog \left (2, -\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{\left (c^{2} x^{2}-1\right ) d^{2}}+\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 )}{\left (c^{2} x^{2}-1\right ) d^{2}}+\frac {4 i a b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) c}{\left (c^{2} x^{2}-1\right ) d^{2}}-\frac {4 a b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \arcsin \left (c x \right ) x \,c^{2}}{\left (c^{2} x^{2}-1\right ) d^{2}}+\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 ) d^{2} 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 )^{4}-1\right ) c}{\left (c^{2} x^{2}-1\right ) d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a b c {\left (\frac {\log \left (c x + 1\right )}{d^{\frac {3}{2}}} + \frac {\log \left (c x - 1\right )}{d^{\frac {3}{2}}} + \frac {2 \, \log \relax (x)}{d^{\frac {3}{2}}}\right )} + 2 \, {\left (\frac {2 \, c^{2} x}{\sqrt {-c^{2} d x^{2} + d} d} - \frac {1}{\sqrt {-c^{2} d x^{2} + d} d x}\right )} a b \arcsin \left (c x\right ) + {\left (\frac {2 \, c^{2} x}{\sqrt {-c^{2} d x^{2} + d} d} - \frac {1}{\sqrt {-c^{2} d x^{2} + d} d x}\right )} a^{2} - \frac {\frac {\frac {1}{4} \, {\left (7 \, {\left (2 \, c^{2} x^{2} - 1\right )} \sqrt {c x + 1} \sqrt {-c x + 1} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} - 4 \, {\left (c^{2} d x^{3} - d x\right )} \int \frac {9 \, \sqrt {c x + 1} \sqrt {-c x + 1} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} + 14 \, {\left (2 \, c^{5} x^{5} - 3 \, c^{3} x^{3} + c x\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{4 \, {\left (c^{4} d x^{6} - 2 \, c^{2} d x^{4} + d x^{2}\right )}}\,{d x}\right )} b^{2}}{4 \, {\left (c^{2} d x^{3} - d x\right )}}}{\sqrt {d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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