3.64 \(\int \frac {a+b \tan ^{-1}(c x^2)}{x} \, dx\)

Optimal. Leaf size=39 \[ a \log (x)+\frac {1}{4} i b \text {Li}_2\left (-i c x^2\right )-\frac {1}{4} i b \text {Li}_2\left (i c x^2\right ) \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {5031, 4848, 2391} \[ \frac {1}{4} i b \text {PolyLog}\left (2,-i c x^2\right )-\frac {1}{4} i b \text {PolyLog}\left (2,i c x^2\right )+a \log (x) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x^2])/x,x]

[Out]

a*Log[x] + (I/4)*b*PolyLog[2, (-I)*c*x^2] - (I/4)*b*PolyLog[2, I*c*x^2]

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 4848

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[(I*b)/2, Int[Log[1 - I*c*x
]/x, x], x] - Dist[(I*b)/2, Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]

Rule 5031

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/n, Subst[Int[(a + b*ArcTan[c*x])^p
/x, x], x, x^n], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int \frac {a+b \tan ^{-1}\left (c x^2\right )}{x} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {a+b \tan ^{-1}(c x)}{x} \, dx,x,x^2\right )\\ &=a \log (x)+\frac {1}{4} (i b) \operatorname {Subst}\left (\int \frac {\log (1-i c x)}{x} \, dx,x,x^2\right )-\frac {1}{4} (i b) \operatorname {Subst}\left (\int \frac {\log (1+i c x)}{x} \, dx,x,x^2\right )\\ &=a \log (x)+\frac {1}{4} i b \text {Li}_2\left (-i c x^2\right )-\frac {1}{4} i b \text {Li}_2\left (i c x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 39, normalized size = 1.00 \[ a \log (x)+\frac {1}{4} i b \text {Li}_2\left (-i c x^2\right )-\frac {1}{4} i b \text {Li}_2\left (i c x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c*x^2])/x,x]

[Out]

a*Log[x] + (I/4)*b*PolyLog[2, (-I)*c*x^2] - (I/4)*b*PolyLog[2, I*c*x^2]

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \arctan \left (c x^{2}\right ) + a}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^2))/x,x, algorithm="fricas")

[Out]

integral((b*arctan(c*x^2) + a)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^2))/x,x, algorithm="giac")

[Out]

integrate((b*arctan(c*x^2) + a)/x, x)

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maple [C]  time = 0.11, size = 63, normalized size = 1.62 \[ a \ln \relax (x )+b \ln \relax (x ) \arctan \left (c \,x^{2}\right )-\frac {b \left (\munderset {\textit {\_R1} =\RootOf \left (c^{2} \textit {\_Z}^{4}+1\right )}{\sum }\frac {\ln \relax (x ) \ln \left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )+\dilog \left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )}{\textit {\_R1}^{2}}\right )}{2 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(c*x^2))/x,x)

[Out]

a*ln(x)+b*ln(x)*arctan(c*x^2)-1/2*b/c*sum(1/_R1^2*(ln(x)*ln((_R1-x)/_R1)+dilog((_R1-x)/_R1)),_R1=RootOf(_Z^4*c
^2+1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^2))/x,x, algorithm="maxima")

[Out]

b*integrate(arctan(c*x^2)/x, x) + a*log(x)

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mupad [B]  time = 0.33, size = 32, normalized size = 0.82 \[ a\,\ln \relax (x)-\frac {b\,\left ({\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x^2\,1{}\mathrm {i}\right )-{\mathrm {Li}}_{\mathrm {2}}\left (1{}\mathrm {i}\,c\,x^2+1\right )\right )\,1{}\mathrm {i}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x^2))/x,x)

[Out]

a*log(x) - (b*(dilog(1 - c*x^2*1i) - dilog(c*x^2*1i + 1))*1i)/4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x**2))/x,x)

[Out]

Integral((a + b*atan(c*x**2))/x, x)

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