3.374 \(\int \frac {a+b \sin ^{-1}(\frac {c}{x})}{x} \, dx\)

Optimal. Leaf size=67 \[ a \log (x)+\frac {1}{2} i b \text {Li}_2\left (e^{2 i \sin ^{-1}\left (\frac {c}{x}\right )}\right )+\frac {1}{2} i b \sin ^{-1}\left (\frac {c}{x}\right )^2-b \sin ^{-1}\left (\frac {c}{x}\right ) \log \left (1-e^{2 i \sin ^{-1}\left (\frac {c}{x}\right )}\right ) \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {6742, 4830, 3717, 2190, 2279, 2391} \[ \frac {1}{2} i b \text {PolyLog}\left (2,e^{2 i \sin ^{-1}\left (\frac {c}{x}\right )}\right )+a \log (x)+\frac {1}{2} i b \sin ^{-1}\left (\frac {c}{x}\right )^2-b \sin ^{-1}\left (\frac {c}{x}\right ) \log \left (1-e^{2 i \sin ^{-1}\left (\frac {c}{x}\right )}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c/x])/x,x]

[Out]

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

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 4830

Int[ArcSin[(a_.)*(x_)^(p_)]^(n_.)/(x_), x_Symbol] :> Dist[1/p, Subst[Int[x^n*Cot[x], x], x, ArcSin[a*x^p]], x]
/; FreeQ[{a, p}, x] && IGtQ[n, 0]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 61, normalized size = 0.91 \[ a \log (x)+\frac {1}{2} i b \left (\sin ^{-1}\left (\frac {c}{x}\right )^2+\text {Li}_2\left (e^{2 i \sin ^{-1}\left (\frac {c}{x}\right )}\right )\right )-b \sin ^{-1}\left (\frac {c}{x}\right ) \log \left (1-e^{2 i \sin ^{-1}\left (\frac {c}{x}\right )}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSin[c/x])/x,x]

[Out]

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

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fricas [F]  time = 1.38, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \arcsin \left (\frac {c}{x}\right ) + a}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c/x))/x,x, algorithm="fricas")

[Out]

integral((b*arcsin(c/x) + a)/x, x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c/x))/x,x, algorithm="giac")

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:Warning, integrat
ion of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [abs(x)]Undef/Un
signed Inf encountered in limitLimit: Max order reached or unable to make series expansion Error: Bad Argument
Value

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maple [A]  time = 0.01, size = 141, normalized size = 2.10 \[ -a \ln \left (\frac {c}{x}\right )+\frac {i b \arcsin \left (\frac {c}{x}\right )^{2}}{2}-b \arcsin \left (\frac {c}{x}\right ) \ln \left (1+\frac {i c}{x}+\sqrt {1-\frac {c^{2}}{x^{2}}}\right )-b \arcsin \left (\frac {c}{x}\right ) \ln \left (1-\frac {i c}{x}-\sqrt {1-\frac {c^{2}}{x^{2}}}\right )+i b \polylog \left (2, -\frac {i c}{x}-\sqrt {1-\frac {c^{2}}{x^{2}}}\right )+i b \polylog \left (2, \frac {i c}{x}+\sqrt {1-\frac {c^{2}}{x^{2}}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(c/x))/x,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c/x))/x,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.44, size = 57, normalized size = 0.85 \[ \frac {b\,{\mathrm {asin}\left (\frac {c}{x}\right )}^2\,1{}\mathrm {i}}{2}+a\,\ln \relax (x)+\frac {b\,\mathrm {polylog}\left (2,{\mathrm {e}}^{\mathrm {asin}\left (\frac {c}{x}\right )\,2{}\mathrm {i}}\right )\,1{}\mathrm {i}}{2}-b\,\ln \left (1-{\mathrm {e}}^{\mathrm {asin}\left (\frac {c}{x}\right )\,2{}\mathrm {i}}\right )\,\mathrm {asin}\left (\frac {c}{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c/x))/x,x)

[Out]

(b*asin(c/x)^2*1i)/2 + a*log(x) + (b*polylog(2, exp(asin(c/x)*2i))*1i)/2 - b*log(1 - exp(asin(c/x)*2i))*asin(c
/x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(c/x))/x,x)

[Out]

Integral((a + b*asin(c/x))/x, x)

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