3.657 \(\int \frac {(1-x^2)^{3/2} \cos ^{-1}(x)}{x} \, dx\)

Optimal. Leaf size=95 \[ -i \operatorname {PolyLog}\left (2,-i e^{i \cos ^{-1}(x)}\right )+i \operatorname {PolyLog}\left (2,i e^{i \cos ^{-1}(x)}\right )-\frac {x^3}{9}+\frac {1}{3} \left (1-x^2\right )^{3/2} \cos ^{-1}(x)+\sqrt {1-x^2} \cos ^{-1}(x)+\frac {4 x}{3}+2 i \cos ^{-1}(x) \tan ^{-1}\left (e^{i \cos ^{-1}(x)}\right ) \]

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Rubi [A]  time = 0.16, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {4700, 4698, 4710, 4181, 2279, 2391, 8} \[ -i \text {PolyLog}\left (2,-i e^{i \cos ^{-1}(x)}\right )+i \text {PolyLog}\left (2,i e^{i \cos ^{-1}(x)}\right )-\frac {x^3}{9}+\frac {1}{3} \left (1-x^2\right )^{3/2} \cos ^{-1}(x)+\sqrt {1-x^2} \cos ^{-1}(x)+\frac {4 x}{3}+2 i \cos ^{-1}(x) \tan ^{-1}\left (e^{i \cos ^{-1}(x)}\right ) \]

Antiderivative was successfully verified.

[In]

Int[((1 - x^2)^(3/2)*ArcCos[x])/x,x]

[Out]

(4*x)/3 - x^3/9 + Sqrt[1 - x^2]*ArcCos[x] + ((1 - x^2)^(3/2)*ArcCos[x])/3 + (2*I)*ArcCos[x]*ArcTan[E^(I*ArcCos
[x])] - I*PolyLog[2, (-I)*E^(I*ArcCos[x])] + I*PolyLog[2, I*E^(I*ArcCos[x])]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 4181

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

Rule 4698

Int[((a_.) + ArcCos[(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*ArcCos[c*x])^n)/(f*(m + 2)), x] + (Dist[Sqrt[d + e*x^2]/((m + 2)*Sqrt[1 -
c^2*x^2]), Int[((f*x)^m*(a + b*ArcCos[c*x])^n)/Sqrt[1 - c^2*x^2], x], x] + Dist[(b*c*n*Sqrt[d + e*x^2])/(f*(m
+ 2)*Sqrt[1 - c^2*x^2]), Int[(f*x)^(m + 1)*(a + b*ArcCos[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, m}
, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] &&  !LtQ[m, -1] && (RationalQ[m] || EqQ[n, 1])

Rule 4700

Int[((a_.) + ArcCos[(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*ArcCos[c*x])^n)/(f*(m + 2*p + 1)), x] + (Dist[(2*d*p)/(m + 2*p + 1), Int[(
f*x)^m*(d + e*x^2)^(p - 1)*(a + b*ArcCos[c*x])^n, x], x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(
f*(m + 2*p + 1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 - c^2*x^2)^(p - 1/2)*(a + b*ArcCos[c*x])^(n -
1), x], x]) /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[p, 0] &&  !LtQ[m, -1]
&& (RationalQ[m] || EqQ[n, 1])

Rule 4710

Int[(((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> -Dist[(c^(m +
1)*Sqrt[d])^(-1), Subst[Int[(a + b*x)^n*Cos[x]^m, x], x, ArcCos[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 {\left (1-x^2\right )^{3/2} \cos ^{-1}(x)}{x} \, dx &=\frac {1}{3} \left (1-x^2\right )^{3/2} \cos ^{-1}(x)+\frac {1}{3} \int \left (1-x^2\right ) \, dx+\int \frac {\sqrt {1-x^2} \cos ^{-1}(x)}{x} \, dx\\ &=\frac {x}{3}-\frac {x^3}{9}+\sqrt {1-x^2} \cos ^{-1}(x)+\frac {1}{3} \left (1-x^2\right )^{3/2} \cos ^{-1}(x)+\int 1 \, dx+\int \frac {\cos ^{-1}(x)}{x \sqrt {1-x^2}} \, dx\\ &=\frac {4 x}{3}-\frac {x^3}{9}+\sqrt {1-x^2} \cos ^{-1}(x)+\frac {1}{3} \left (1-x^2\right )^{3/2} \cos ^{-1}(x)-\operatorname {Subst}\left (\int x \sec (x) \, dx,x,\cos ^{-1}(x)\right )\\ &=\frac {4 x}{3}-\frac {x^3}{9}+\sqrt {1-x^2} \cos ^{-1}(x)+\frac {1}{3} \left (1-x^2\right )^{3/2} \cos ^{-1}(x)+2 i \cos ^{-1}(x) \tan ^{-1}\left (e^{i \cos ^{-1}(x)}\right )+\operatorname {Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\cos ^{-1}(x)\right )-\operatorname {Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\cos ^{-1}(x)\right )\\ &=\frac {4 x}{3}-\frac {x^3}{9}+\sqrt {1-x^2} \cos ^{-1}(x)+\frac {1}{3} \left (1-x^2\right )^{3/2} \cos ^{-1}(x)+2 i \cos ^{-1}(x) \tan ^{-1}\left (e^{i \cos ^{-1}(x)}\right )-i \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i \cos ^{-1}(x)}\right )+i \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i \cos ^{-1}(x)}\right )\\ &=\frac {4 x}{3}-\frac {x^3}{9}+\sqrt {1-x^2} \cos ^{-1}(x)+\frac {1}{3} \left (1-x^2\right )^{3/2} \cos ^{-1}(x)+2 i \cos ^{-1}(x) \tan ^{-1}\left (e^{i \cos ^{-1}(x)}\right )-i \text {Li}_2\left (-i e^{i \cos ^{-1}(x)}\right )+i \text {Li}_2\left (i e^{i \cos ^{-1}(x)}\right )\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 119, normalized size = 1.25 \[ -i \operatorname {PolyLog}\left (2,-i e^{i \cos ^{-1}(x)}\right )+i \operatorname {PolyLog}\left (2,i e^{i \cos ^{-1}(x)}\right )+\sqrt {1-x^2} \cos ^{-1}(x)+\frac {1}{36} \left (12 \left (1-x^2\right )^{3/2} \cos ^{-1}(x)+9 x-\cos \left (3 \cos ^{-1}(x)\right )\right )+x-\cos ^{-1}(x) \log \left (1-i e^{i \cos ^{-1}(x)}\right )+\cos ^{-1}(x) \log \left (1+i e^{i \cos ^{-1}(x)}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((1 - x^2)^(3/2)*ArcCos[x])/x,x]

[Out]

x + Sqrt[1 - x^2]*ArcCos[x] + (9*x + 12*(1 - x^2)^(3/2)*ArcCos[x] - Cos[3*ArcCos[x]])/36 - ArcCos[x]*Log[1 - I
*E^(I*ArcCos[x])] + ArcCos[x]*Log[1 + I*E^(I*ArcCos[x])] - I*PolyLog[2, (-I)*E^(I*ArcCos[x])] + I*PolyLog[2, I
*E^(I*ArcCos[x])]

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (1-x^2\right )^{3/2} \cos ^{-1}(x)}{x} \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[((1 - x^2)^(3/2)*ArcCos[x])/x,x]

[Out]

Could not integrate

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fricas [F]  time = 1.07, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (x^{2} - 1\right )} \sqrt {-x^{2} + 1} \arccos \relax (x)}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2+1)^(3/2)*arccos(x)/x,x, algorithm="fricas")

[Out]

integral(-(x^2 - 1)*sqrt(-x^2 + 1)*arccos(x)/x, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-x^{2} + 1\right )}^{\frac {3}{2}} \arccos \relax (x)}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2+1)^(3/2)*arccos(x)/x,x, algorithm="giac")

[Out]

integrate((-x^2 + 1)^(3/2)*arccos(x)/x, x)

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maple [B]  time = 0.44, size = 230, normalized size = 2.42

method	result	size
default	\(\frac {\left (i+3 \arccos \relax (x )\right ) \left (4 i x^{3}-4 \sqrt {-x^{2}+1}\, x^{2}-3 i x +\sqrt {-x^{2}+1}\right )}{72}-\frac {5 \left (\arccos \relax (x )+i\right ) \left (i x -\sqrt {-x^{2}+1}\right )}{8}+\frac {5 \left (\arccos \relax (x )-i\right ) \left (i x +\sqrt {-x^{2}+1}\right )}{8}-\frac {\left (-i+3 \arccos \relax (x )\right ) \left (4 i x^{3}+4 \sqrt {-x^{2}+1}\, x^{2}-3 i x -\sqrt {-x^{2}+1}\right )}{72}-i \left (i \arccos \relax (x ) \ln \left (1+i \left (x +i \sqrt {-x^{2}+1}\right )\right )-i \arccos \relax (x ) \ln \left (1-i \left (x +i \sqrt {-x^{2}+1}\right )\right )+\dilog \left (1+i \left (x +i \sqrt {-x^{2}+1}\right )\right )-\dilog \left (1-i \left (x +i \sqrt {-x^{2}+1}\right )\right )\right )\)	\(230\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^2+1)^(3/2)*arccos(x)/x,x,method=_RETURNVERBOSE)

[Out]

1/72*(I+3*arccos(x))*(4*I*x^3-4*(-x^2+1)^(1/2)*x^2-3*I*x+(-x^2+1)^(1/2))-5/8*(arccos(x)+I)*(I*x-(-x^2+1)^(1/2)
)+5/8*(arccos(x)-I)*(I*x+(-x^2+1)^(1/2))-1/72*(-I+3*arccos(x))*(4*I*x^3+4*(-x^2+1)^(1/2)*x^2-3*I*x-(-x^2+1)^(1
/2))-I*(I*arccos(x)*ln(1+I*(x+I*(-x^2+1)^(1/2)))-I*arccos(x)*ln(1-I*(x+I*(-x^2+1)^(1/2)))+dilog(1+I*(x+I*(-x^2
+1)^(1/2)))-dilog(1-I*(x+I*(-x^2+1)^(1/2))))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2+1)^(3/2)*arccos(x)/x,x, algorithm="maxima")

[Out]

integrate((-x^2 + 1)^(3/2)*arccos(x)/x, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {acos}\relax (x)\,{\left (1-x^2\right )}^{3/2}}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((acos(x)*(1 - x^2)^(3/2))/x,x)

[Out]

int((acos(x)*(1 - x^2)^(3/2))/x, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**2+1)**(3/2)*acos(x)/x,x)

[Out]

Timed out

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