3.31 \(\int \frac {\sin ^{-1}(a x)^3}{x^5} \, dx\)

Optimal. Leaf size=169 \[ -\frac {1}{2} i a^4 \text {Li}_2\left (e^{2 i \sin ^{-1}(a x)}\right )-\frac {1}{2} i a^4 \sin ^{-1}(a x)^2+a^4 \sin ^{-1}(a x) \log \left (1-e^{2 i \sin ^{-1}(a x)}\right )-\frac {a^2 \sin ^{-1}(a x)}{4 x^2}-\frac {a \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{4 x^3}-\frac {a^3 \sqrt {1-a^2 x^2}}{4 x}-\frac {a^3 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{2 x}-\frac {\sin ^{-1}(a x)^3}{4 x^4} \]

[Out]

-1/4*a^2*arcsin(a*x)/x^2-1/2*I*a^4*arcsin(a*x)^2-1/4*arcsin(a*x)^3/x^4+a^4*arcsin(a*x)*ln(1-(I*a*x+(-a^2*x^2+1
)^(1/2))^2)-1/2*I*a^4*polylog(2,(I*a*x+(-a^2*x^2+1)^(1/2))^2)-1/4*a^3*(-a^2*x^2+1)^(1/2)/x-1/4*a*arcsin(a*x)^2
*(-a^2*x^2+1)^(1/2)/x^3-1/2*a^3*arcsin(a*x)^2*(-a^2*x^2+1)^(1/2)/x

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Rubi [A]  time = 0.29, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.900, Rules used = {4627, 4701, 4681, 4625, 3717, 2190, 2279, 2391, 264} \[ -\frac {1}{2} i a^4 \text {PolyLog}\left (2,e^{2 i \sin ^{-1}(a x)}\right )-\frac {a^3 \sqrt {1-a^2 x^2}}{4 x}-\frac {a^3 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{2 x}-\frac {a^2 \sin ^{-1}(a x)}{4 x^2}-\frac {a \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{4 x^3}-\frac {1}{2} i a^4 \sin ^{-1}(a x)^2+a^4 \sin ^{-1}(a x) \log \left (1-e^{2 i \sin ^{-1}(a x)}\right )-\frac {\sin ^{-1}(a x)^3}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[a*x]^3/x^5,x]

[Out]

-(a^3*Sqrt[1 - a^2*x^2])/(4*x) - (a^2*ArcSin[a*x])/(4*x^2) - (I/2)*a^4*ArcSin[a*x]^2 - (a*Sqrt[1 - a^2*x^2]*Ar
cSin[a*x]^2)/(4*x^3) - (a^3*Sqrt[1 - a^2*x^2]*ArcSin[a*x]^2)/(2*x) - ArcSin[a*x]^3/(4*x^4) + a^4*ArcSin[a*x]*L
og[1 - E^((2*I)*ArcSin[a*x])] - (I/2)*a^4*PolyLog[2, E^((2*I)*ArcSin[a*x])]

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

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 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 4681

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[(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*ArcSi
n[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && EqQ[m + 2*p
+ 3, 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]

Rubi steps

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

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Mathematica [A]  time = 0.63, size = 116, normalized size = 0.69 \[ \frac {1}{4} \left (-\frac {\sin ^{-1}(a x)^3}{x^4}+a^4 \left (-\frac {\sqrt {1-a^2 x^2} \left (\left (\frac {1}{a^2 x^2}+2\right ) \sin ^{-1}(a x)^2+1\right )}{a x}-\sin ^{-1}(a x) \left (\frac {1}{a^2 x^2}+2 i \sin ^{-1}(a x)-4 \log \left (1-e^{2 i \sin ^{-1}(a x)}\right )\right )-2 i \text {Li}_2\left (e^{2 i \sin ^{-1}(a x)}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[a*x]^3/x^5,x]

[Out]

(-(ArcSin[a*x]^3/x^4) + a^4*(-((Sqrt[1 - a^2*x^2]*(1 + (2 + 1/(a^2*x^2))*ArcSin[a*x]^2))/(a*x)) - ArcSin[a*x]*
(1/(a^2*x^2) + (2*I)*ArcSin[a*x] - 4*Log[1 - E^((2*I)*ArcSin[a*x])]) - (2*I)*PolyLog[2, E^((2*I)*ArcSin[a*x])]
))/4

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fricas [F]  time = 0.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\arcsin \left (a x\right )^{3}}{x^{5}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(a*x)^3/x^5,x, algorithm="fricas")

[Out]

integral(arcsin(a*x)^3/x^5, 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(arcsin(a*x)^3/x^5,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Eval
uation time: 0.75sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.22, size = 225, normalized size = 1.33 \[ -\frac {i a^{4} \arcsin \left (a x \right )^{2}}{2}-\frac {a^{3} \arcsin \left (a x \right )^{2} \sqrt {-a^{2} x^{2}+1}}{2 x}+\frac {i a^{4}}{4}-\frac {a^{3} \sqrt {-a^{2} x^{2}+1}}{4 x}-\frac {a \arcsin \left (a x \right )^{2} \sqrt {-a^{2} x^{2}+1}}{4 x^{3}}-\frac {a^{2} \arcsin \left (a x \right )}{4 x^{2}}-\frac {\arcsin \left (a x \right )^{3}}{4 x^{4}}+a^{4} \arcsin \left (a x \right ) \ln \left (1+i a x +\sqrt {-a^{2} x^{2}+1}\right )+a^{4} \arcsin \left (a x \right ) \ln \left (1-i a x -\sqrt {-a^{2} x^{2}+1}\right )-i a^{4} \polylog \left (2, i a x +\sqrt {-a^{2} x^{2}+1}\right )-i a^{4} \polylog \left (2, -i a x -\sqrt {-a^{2} x^{2}+1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsin(a*x)^3/x^5,x)

[Out]

-1/2*I*a^4*arcsin(a*x)^2-1/2*a^3*arcsin(a*x)^2*(-a^2*x^2+1)^(1/2)/x+1/4*I*a^4-1/4*a^3*(-a^2*x^2+1)^(1/2)/x-1/4
*a*arcsin(a*x)^2*(-a^2*x^2+1)^(1/2)/x^3-1/4*a^2*arcsin(a*x)/x^2-1/4*arcsin(a*x)^3/x^4+a^4*arcsin(a*x)*ln(1+I*a
*x+(-a^2*x^2+1)^(1/2))+a^4*arcsin(a*x)*ln(1-I*a*x-(-a^2*x^2+1)^(1/2))-I*a^4*polylog(2,I*a*x+(-a^2*x^2+1)^(1/2)
)-I*a^4*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 \[ -\frac {\frac {1}{4} \, {\left ({\left (2 \, a^{2} x^{2} + 1\right )} \sqrt {a x + 1} \sqrt {-a x + 1} \arctan \left (a x, \sqrt {a x + 1} \sqrt {-a x + 1}\right )^{2} + 12 \, x^{3} \int \frac {9 \, \sqrt {a x + 1} \sqrt {-a x + 1} \arctan \left (a x, \sqrt {a x + 1} \sqrt {-a x + 1}\right )^{2} - 2 \, {\left (2 \, a^{5} x^{5} - a^{3} x^{3} - a x\right )} \arctan \left (a x, \sqrt {a x + 1} \sqrt {-a x + 1}\right )}{12 \, {\left (a^{2} x^{6} - x^{4}\right )}}\,{d x}\right )} a x + \arctan \left (a x, \sqrt {a x + 1} \sqrt {-a x + 1}\right )^{3}}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(a*x)^3/x^5,x, algorithm="maxima")

[Out]

-1/4*(12*a*x^4*integrate(1/4*sqrt(a*x + 1)*sqrt(-a*x + 1)*arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1))^2/(a^2*x^
6 - x^4), x) + arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1))^3)/x^4

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {asin}\left (a\,x\right )}^3}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(a*x)^3/x^5,x)

[Out]

int(asin(a*x)^3/x^5, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asin}^{3}{\left (a x \right )}}{x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asin(a*x)**3/x**5,x)

[Out]

Integral(asin(a*x)**3/x**5, x)

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