3.164 \(\int \frac{(d-c^2 d x^2) (a+b \sin ^{-1}(c x))^2}{x^4} \, dx\)

Optimal. Leaf size=176 \[ -\frac{5}{3} i b^2 c^3 d \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )+\frac{5}{3} i b^2 c^3 d \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )-\frac{b c d \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 x^2}-\frac{d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{3 x^3}+\frac{2 c^2 d \left (a+b \sin ^{-1}(c x)\right )^2}{3 x}+\frac{10}{3} b c^3 d \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )-\frac{b^2 c^2 d}{3 x} \]

[Out]

-(b^2*c^2*d)/(3*x) - (b*c*d*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/(3*x^2) + (2*c^2*d*(a + b*ArcSin[c*x])^2)/(
3*x) - (d*(1 - c^2*x^2)*(a + b*ArcSin[c*x])^2)/(3*x^3) + (10*b*c^3*d*(a + b*ArcSin[c*x])*ArcTanh[E^(I*ArcSin[c
*x])])/3 - ((5*I)/3)*b^2*c^3*d*PolyLog[2, -E^(I*ArcSin[c*x])] + ((5*I)/3)*b^2*c^3*d*PolyLog[2, E^(I*ArcSin[c*x
])]

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Rubi [A]  time = 0.376732, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {4695, 4627, 4709, 4183, 2279, 2391, 4693, 30} \[ -\frac{5}{3} i b^2 c^3 d \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )+\frac{5}{3} i b^2 c^3 d \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )-\frac{b c d \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 x^2}-\frac{d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{3 x^3}+\frac{2 c^2 d \left (a+b \sin ^{-1}(c x)\right )^2}{3 x}+\frac{10}{3} b c^3 d \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )-\frac{b^2 c^2 d}{3 x} \]

Antiderivative was successfully verified.

[In]

Int[((d - c^2*d*x^2)*(a + b*ArcSin[c*x])^2)/x^4,x]

[Out]

-(b^2*c^2*d)/(3*x) - (b*c*d*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/(3*x^2) + (2*c^2*d*(a + b*ArcSin[c*x])^2)/(
3*x) - (d*(1 - c^2*x^2)*(a + b*ArcSin[c*x])^2)/(3*x^3) + (10*b*c^3*d*(a + b*ArcSin[c*x])*ArcTanh[E^(I*ArcSin[c
*x])])/3 - ((5*I)/3)*b^2*c^3*d*PolyLog[2, -E^(I*ArcSin[c*x])] + ((5*I)/3)*b^2*c^3*d*PolyLog[2, E^(I*ArcSin[c*x
])]

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 4627

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

Rule 4709

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

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 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 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 30

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

Rubi steps

\begin{align*} \int \frac{\left (d-c^2 d x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{x^4} \, dx &=-\frac{d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{3 x^3}+\frac{1}{3} (2 b c d) \int \frac{\sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{x^3} \, dx-\frac{1}{3} \left (2 c^2 d\right ) \int \frac{\left (a+b \sin ^{-1}(c x)\right )^2}{x^2} \, dx\\ &=-\frac{b c d \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 x^2}+\frac{2 c^2 d \left (a+b \sin ^{-1}(c x)\right )^2}{3 x}-\frac{d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{3 x^3}+\frac{1}{3} \left (b^2 c^2 d\right ) \int \frac{1}{x^2} \, dx-\frac{1}{3} \left (b c^3 d\right ) \int \frac{a+b \sin ^{-1}(c x)}{x \sqrt{1-c^2 x^2}} \, dx-\frac{1}{3} \left (4 b c^3 d\right ) \int \frac{a+b \sin ^{-1}(c x)}{x \sqrt{1-c^2 x^2}} \, dx\\ &=-\frac{b^2 c^2 d}{3 x}-\frac{b c d \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 x^2}+\frac{2 c^2 d \left (a+b \sin ^{-1}(c x)\right )^2}{3 x}-\frac{d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{3 x^3}-\frac{1}{3} \left (b c^3 d\right ) \operatorname{Subst}\left (\int (a+b x) \csc (x) \, dx,x,\sin ^{-1}(c x)\right )-\frac{1}{3} \left (4 b c^3 d\right ) \operatorname{Subst}\left (\int (a+b x) \csc (x) \, dx,x,\sin ^{-1}(c x)\right )\\ &=-\frac{b^2 c^2 d}{3 x}-\frac{b c d \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 x^2}+\frac{2 c^2 d \left (a+b \sin ^{-1}(c x)\right )^2}{3 x}-\frac{d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{3 x^3}+\frac{10}{3} b c^3 d \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )+\frac{1}{3} \left (b^2 c^3 d\right ) \operatorname{Subst}\left (\int \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )-\frac{1}{3} \left (b^2 c^3 d\right ) \operatorname{Subst}\left (\int \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )+\frac{1}{3} \left (4 b^2 c^3 d\right ) \operatorname{Subst}\left (\int \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )-\frac{1}{3} \left (4 b^2 c^3 d\right ) \operatorname{Subst}\left (\int \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )\\ &=-\frac{b^2 c^2 d}{3 x}-\frac{b c d \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 x^2}+\frac{2 c^2 d \left (a+b \sin ^{-1}(c x)\right )^2}{3 x}-\frac{d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{3 x^3}+\frac{10}{3} b c^3 d \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )-\frac{1}{3} \left (i b^2 c^3 d\right ) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )+\frac{1}{3} \left (i b^2 c^3 d\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )-\frac{1}{3} \left (4 i b^2 c^3 d\right ) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )+\frac{1}{3} \left (4 i b^2 c^3 d\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )\\ &=-\frac{b^2 c^2 d}{3 x}-\frac{b c d \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{3 x^2}+\frac{2 c^2 d \left (a+b \sin ^{-1}(c x)\right )^2}{3 x}-\frac{d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{3 x^3}+\frac{10}{3} b c^3 d \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )-\frac{5}{3} i b^2 c^3 d \text{Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )+\frac{5}{3} i b^2 c^3 d \text{Li}_2\left (e^{i \sin ^{-1}(c x)}\right )\\ \end{align*}

Mathematica [A]  time = 0.743432, size = 266, normalized size = 1.51 \[ \frac{d \left (-5 i b^2 c^3 x^3 \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )+5 i b^2 c^3 x^3 \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )+3 a^2 c^2 x^2-a^2-a b c x \sqrt{1-c^2 x^2}+6 a b c^2 x^2 \sin ^{-1}(c x)+5 a b c^3 x^3 \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )-2 a b \sin ^{-1}(c x)-b^2 c^2 x^2+3 b^2 c^2 x^2 \sin ^{-1}(c x)^2-b^2 c x \sqrt{1-c^2 x^2} \sin ^{-1}(c x)-5 b^2 c^3 x^3 \sin ^{-1}(c x) \log \left (1-e^{i \sin ^{-1}(c x)}\right )+5 b^2 c^3 x^3 \sin ^{-1}(c x) \log \left (1+e^{i \sin ^{-1}(c x)}\right )-b^2 \sin ^{-1}(c x)^2\right )}{3 x^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((d - c^2*d*x^2)*(a + b*ArcSin[c*x])^2)/x^4,x]

[Out]

(d*(-a^2 + 3*a^2*c^2*x^2 - b^2*c^2*x^2 - a*b*c*x*Sqrt[1 - c^2*x^2] - 2*a*b*ArcSin[c*x] + 6*a*b*c^2*x^2*ArcSin[
c*x] - b^2*c*x*Sqrt[1 - c^2*x^2]*ArcSin[c*x] - b^2*ArcSin[c*x]^2 + 3*b^2*c^2*x^2*ArcSin[c*x]^2 + 5*a*b*c^3*x^3
*ArcTanh[Sqrt[1 - c^2*x^2]] - 5*b^2*c^3*x^3*ArcSin[c*x]*Log[1 - E^(I*ArcSin[c*x])] + 5*b^2*c^3*x^3*ArcSin[c*x]
*Log[1 + E^(I*ArcSin[c*x])] - (5*I)*b^2*c^3*x^3*PolyLog[2, -E^(I*ArcSin[c*x])] + (5*I)*b^2*c^3*x^3*PolyLog[2,
E^(I*ArcSin[c*x])]))/(3*x^3)

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Maple [A]  time = 0.361, size = 291, normalized size = 1.7 \begin{align*}{\frac{{c}^{2}d{a}^{2}}{x}}-{\frac{d{a}^{2}}{3\,{x}^{3}}}+{\frac{{c}^{2}d{b}^{2} \left ( \arcsin \left ( cx \right ) \right ) ^{2}}{x}}-{\frac{dc{b}^{2}\arcsin \left ( cx \right ) }{3\,{x}^{2}}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{d{b}^{2} \left ( \arcsin \left ( cx \right ) \right ) ^{2}}{3\,{x}^{3}}}-{\frac{{c}^{2}d{b}^{2}}{3\,x}}+{\frac{5\,d{c}^{3}{b}^{2}\arcsin \left ( cx \right ) }{3}\ln \left ( 1+icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) }-{\frac{5\,i}{3}}{b}^{2}{c}^{3}d{\it polylog} \left ( 2,-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) -{\frac{5\,d{c}^{3}{b}^{2}\arcsin \left ( cx \right ) }{3}\ln \left ( 1-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) }+{\frac{5\,i}{3}}{b}^{2}{c}^{3}d{\it polylog} \left ( 2,icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) +2\,{\frac{{c}^{2}dab\arcsin \left ( cx \right ) }{x}}-{\frac{2\,dab\arcsin \left ( cx \right ) }{3\,{x}^{3}}}-{\frac{dcab}{3\,{x}^{2}}\sqrt{-{c}^{2}{x}^{2}+1}}+{\frac{5\,d{c}^{3}ab}{3}{\it Artanh} \left ({\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}} \right ) } \end{align*}

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^4,x)

[Out]

c^2*d*a^2/x-1/3*d*a^2/x^3+c^2*d*b^2/x*arcsin(c*x)^2-1/3*c*d*b^2/x^2*arcsin(c*x)*(-c^2*x^2+1)^(1/2)-1/3*d*b^2/x
^3*arcsin(c*x)^2-1/3*b^2*c^2*d/x+5/3*c^3*d*b^2*arcsin(c*x)*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))-5/3*I*b^2*c^3*d*poly
log(2,-I*c*x-(-c^2*x^2+1)^(1/2))-5/3*c^3*d*b^2*arcsin(c*x)*ln(1-I*c*x-(-c^2*x^2+1)^(1/2))+5/3*I*b^2*c^3*d*poly
log(2,I*c*x+(-c^2*x^2+1)^(1/2))+2*c^2*d*a*b/x*arcsin(c*x)-2/3*d*a*b*arcsin(c*x)/x^3-1/3*c*d*a*b/x^2*(-c^2*x^2+
1)^(1/2)+5/3*c^3*d*a*b*arctanh(1/(-c^2*x^2+1)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 2 \,{\left (c \log \left (\frac{2 \, \sqrt{-c^{2} x^{2} + 1}}{{\left | x \right |}} + \frac{2}{{\left | x \right |}}\right ) + \frac{\arcsin \left (c x\right )}{x}\right )} a b c^{2} d - \frac{1}{3} \,{\left ({\left (c^{2} \log \left (\frac{2 \, \sqrt{-c^{2} x^{2} + 1}}{{\left | x \right |}} + \frac{2}{{\left | x \right |}}\right ) + \frac{\sqrt{-c^{2} x^{2} + 1}}{x^{2}}\right )} c + \frac{2 \, \arcsin \left (c x\right )}{x^{3}}\right )} a b d + \frac{a^{2} c^{2} d}{x} - \frac{a^{2} d}{3 \, x^{3}} + \frac{2 \, x^{3} \int \frac{{\left (3 \, b^{2} c^{3} d x^{2} - b^{2} c d\right )} \sqrt{c x + 1} \sqrt{-c x + 1} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )}{c^{2} x^{5} - x^{3}}\,{d x} +{\left (3 \, 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}}{3 \, x^{3}} \end{align*}

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^4,x, algorithm="maxima")

[Out]

2*(c*log(2*sqrt(-c^2*x^2 + 1)/abs(x) + 2/abs(x)) + arcsin(c*x)/x)*a*b*c^2*d - 1/3*((c^2*log(2*sqrt(-c^2*x^2 +
1)/abs(x) + 2/abs(x)) + sqrt(-c^2*x^2 + 1)/x^2)*c + 2*arcsin(c*x)/x^3)*a*b*d + a^2*c^2*d/x - 1/3*a^2*d/x^3 + 1
/3*(3*x^3*integrate(2/3*(3*b^2*c^3*d*x^2 - b^2*c*d)*sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1)*sq
rt(-c*x + 1))/(c^2*x^5 - x^3), x) + (3*b^2*c^2*d*x^2 - b^2*d)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2)/x^
3

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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^{4}}, x\right ) \end{align*}

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^4,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^4, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - d \left (\int - \frac{a^{2}}{x^{4}}\, dx + \int \frac{a^{2} c^{2}}{x^{2}}\, dx + \int - \frac{b^{2} \operatorname{asin}^{2}{\left (c x \right )}}{x^{4}}\, dx + \int - \frac{2 a b \operatorname{asin}{\left (c x \right )}}{x^{4}}\, dx + \int \frac{b^{2} c^{2} \operatorname{asin}^{2}{\left (c x \right )}}{x^{2}}\, dx + \int \frac{2 a b c^{2} \operatorname{asin}{\left (c x \right )}}{x^{2}}\, dx\right ) \end{align*}

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**4,x)

[Out]

-d*(Integral(-a**2/x**4, x) + Integral(a**2*c**2/x**2, x) + Integral(-b**2*asin(c*x)**2/x**4, x) + Integral(-2
*a*b*asin(c*x)/x**4, x) + Integral(b**2*c**2*asin(c*x)**2/x**2, x) + Integral(2*a*b*c**2*asin(c*x)/x**2, x))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

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^4,x, algorithm="giac")

[Out]

sage0*x