3.20 \(\int \frac {\sin ^{-1}(a x)^2}{x^4} \, dx\)

Optimal. Leaf size=116 \[ \frac {1}{3} i a^3 \text {Li}_2\left (-e^{i \sin ^{-1}(a x)}\right )-\frac {1}{3} i a^3 \text {Li}_2\left (e^{i \sin ^{-1}(a x)}\right )-\frac {2}{3} a^3 \sin ^{-1}(a x) \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )-\frac {a \sqrt {1-a^2 x^2} \sin ^{-1}(a x)}{3 x^2}-\frac {a^2}{3 x}-\frac {\sin ^{-1}(a x)^2}{3 x^3} \]

[Out]

-1/3*a^2/x-1/3*arcsin(a*x)^2/x^3-2/3*a^3*arcsin(a*x)*arctanh(I*a*x+(-a^2*x^2+1)^(1/2))+1/3*I*a^3*polylog(2,-I*
a*x-(-a^2*x^2+1)^(1/2))-1/3*I*a^3*polylog(2,I*a*x+(-a^2*x^2+1)^(1/2))-1/3*a*arcsin(a*x)*(-a^2*x^2+1)^(1/2)/x^2

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Rubi [A]  time = 0.17, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {4627, 4701, 4709, 4183, 2279, 2391, 30} \[ \frac {1}{3} i a^3 \text {PolyLog}\left (2,-e^{i \sin ^{-1}(a x)}\right )-\frac {1}{3} i a^3 \text {PolyLog}\left (2,e^{i \sin ^{-1}(a x)}\right )-\frac {a \sqrt {1-a^2 x^2} \sin ^{-1}(a x)}{3 x^2}-\frac {a^2}{3 x}-\frac {2}{3} a^3 \sin ^{-1}(a x) \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )-\frac {\sin ^{-1}(a x)^2}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[a*x]^2/x^4,x]

[Out]

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

Rule 30

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

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 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 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 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]

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]

Rubi steps

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

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Mathematica [A]  time = 0.60, size = 139, normalized size = 1.20 \[ -\frac {-i a^3 x^3 \text {Li}_2\left (-e^{i \sin ^{-1}(a x)}\right )+i a^3 x^3 \text {Li}_2\left (e^{i \sin ^{-1}(a x)}\right )-a^3 x^3 \sin ^{-1}(a x) \log \left (1-e^{i \sin ^{-1}(a x)}\right )+a^3 x^3 \sin ^{-1}(a x) \log \left (1+e^{i \sin ^{-1}(a x)}\right )+a^2 x^2+a x \sqrt {1-a^2 x^2} \sin ^{-1}(a x)+\sin ^{-1}(a x)^2}{3 x^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcSin[a*x]^2/x^4,x]

[Out]

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

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fricas [F]  time = 0.83, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\arcsin \left (a x\right )^{2}}{x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(a*x)^2/x^4,x, algorithm="fricas")

[Out]

integral(arcsin(a*x)^2/x^4, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\arcsin \left (a x\right )^{2}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(a*x)^2/x^4,x, algorithm="giac")

[Out]

integrate(arcsin(a*x)^2/x^4, x)

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maple [A]  time = 0.32, size = 157, normalized size = 1.35 \[ -\frac {a \arcsin \left (a x \right ) \sqrt {-a^{2} x^{2}+1}}{3 x^{2}}-\frac {a^{2}}{3 x}-\frac {\arcsin \left (a x \right )^{2}}{3 x^{3}}-\frac {a^{3} \arcsin \left (a x \right ) \ln \left (1+i a x +\sqrt {-a^{2} x^{2}+1}\right )}{3}+\frac {i a^{3} \polylog \left (2, -i a x -\sqrt {-a^{2} x^{2}+1}\right )}{3}+\frac {a^{3} \arcsin \left (a x \right ) \ln \left (1-i a x -\sqrt {-a^{2} x^{2}+1}\right )}{3}-\frac {i a^{3} \polylog \left (2, i a x +\sqrt {-a^{2} x^{2}+1}\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsin(a*x)^2/x^4,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {2 \, a x^{3} \int \frac {\sqrt {a x + 1} \sqrt {-a x + 1} \arctan \left (a x, \sqrt {a x + 1} \sqrt {-a x + 1}\right )}{a^{2} x^{5} - x^{3}}\,{d x} + \arctan \left (a x, \sqrt {a x + 1} \sqrt {-a x + 1}\right )^{2}}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(a*x)^2/x^4,x, algorithm="maxima")

[Out]

-1/3*(6*a*x^3*integrate(1/3*sqrt(a*x + 1)*sqrt(-a*x + 1)*arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1))/(a^2*x^5 -
 x^3), x) + arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1))^2)/x^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {asin}\left (a\,x\right )}^2}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(a*x)^2/x^4,x)

[Out]

int(asin(a*x)^2/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asin}^{2}{\left (a x \right )}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asin(a*x)**2/x**4,x)

[Out]

Integral(asin(a*x)**2/x**4, x)

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