3.1.100 \(\int \frac {a+b \text {ArcTan}(c x^3)}{x} \, dx\) [100]

Optimal. Leaf size=39 \[ a \log (x)+\frac {1}{6} i b \text {PolyLog}\left (2,-i c x^3\right )-\frac {1}{6} i b \text {PolyLog}\left (2,i c x^3\right ) \]

[Out]

a*ln(x)+1/6*I*b*polylog(2,-I*c*x^3)-1/6*I*b*polylog(2,I*c*x^3)

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Rubi [A]
time = 0.03, 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 = {4944, 4940, 2438} \begin {gather*} a \log (x)+\frac {1}{6} i b \text {Li}_2\left (-i c x^3\right )-\frac {1}{6} i b \text {Li}_2\left (i c x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x^3])/x,x]

[Out]

a*Log[x] + (I/6)*b*PolyLog[2, (-I)*c*x^3] - (I/6)*b*PolyLog[2, I*c*x^3]

Rule 2438

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 4940

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 4944

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^3\right )}{x} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {a+b \tan ^{-1}(c x)}{x} \, dx,x,x^3\right )\\ &=a \log (x)+\frac {1}{6} (i b) \text {Subst}\left (\int \frac {\log (1-i c x)}{x} \, dx,x,x^3\right )-\frac {1}{6} (i b) \text {Subst}\left (\int \frac {\log (1+i c x)}{x} \, dx,x,x^3\right )\\ &=a \log (x)+\frac {1}{6} i b \text {Li}_2\left (-i c x^3\right )-\frac {1}{6} i b \text {Li}_2\left (i c x^3\right )\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 39, normalized size = 1.00 \begin {gather*} a \log (x)+\frac {1}{6} i b \text {PolyLog}\left (2,-i c x^3\right )-\frac {1}{6} i b \text {PolyLog}\left (2,i c x^3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c*x^3])/x,x]

[Out]

a*Log[x] + (I/6)*b*PolyLog[2, (-I)*c*x^3] - (I/6)*b*PolyLog[2, I*c*x^3]

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.04, size = 63, normalized size = 1.62

method	result	size
default	\(a \ln \left (x \right )+b \ln \left (x \right ) \arctan \left (c \,x^{3}\right )-\frac {b \left (\munderset {\textit {\_R1} =\RootOf \left (c^{2} \textit {\_Z}^{6}+1\right )}{\sum }\frac {\ln \left (x \right ) \ln \left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )+\dilog \left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )}{\textit {\_R1}^{3}}\right )}{2 c}\)	\(63\)
risch	\(-\frac {i \left (\munderset {\textit {\_R1} =\RootOf \left (c \,\textit {\_Z}^{3}+\RootOf \left (\textit {\_Z}^{2}+1, \mathit {index} =1\right )\right )}{\sum }\left (\ln \left (x \right ) \ln \left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )+\dilog \left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )\right )\right ) b}{2}+\frac {i \ln \left (x \right ) \ln \left (-i c \,x^{3}+1\right ) b}{2}+a \ln \left (x \right )+\frac {i \left (\munderset {\textit {\_R1} =\RootOf \left (c \,\textit {\_Z}^{3}-\RootOf \left (\textit {\_Z}^{2}+1, \mathit {index} =1\right )\right )}{\sum }\left (\ln \left (x \right ) \ln \left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )+\dilog \left (\frac {\textit {\_R1} -x}{\textit {\_R1}}\right )\right )\right ) b}{2}-\frac {i \ln \left (x \right ) \ln \left (i c \,x^{3}+1\right ) b}{2}\)	\(134\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(c*x^3))/x,x,method=_RETURNVERBOSE)

[Out]

a*ln(x)+b*ln(x)*arctan(c*x^3)-1/2*b/c*sum(1/_R1^3*(ln(x)*ln((_R1-x)/_R1)+dilog((_R1-x)/_R1)),_R1=RootOf(_Z^6*c
^2+1))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^3))/x,x, algorithm="maxima")

[Out]

b*integrate(arctan(c*x^3)/x, x) + a*log(x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^3))/x,x, algorithm="fricas")

[Out]

integral((b*arctan(c*x^3) + a)/x, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b \operatorname {atan}{\left (c x^{3} \right )}}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x**3))/x,x)

[Out]

Integral((a + b*atan(c*x**3))/x, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x^3))/x,x, algorithm="giac")

[Out]

integrate((b*arctan(c*x^3) + a)/x, x)

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Mupad [B]
time = 0.35, size = 32, normalized size = 0.82 \begin {gather*} a\,\ln \left (x\right )-\frac {b\,\left ({\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x^3\,1{}\mathrm {i}\right )-{\mathrm {Li}}_{\mathrm {2}}\left (1{}\mathrm {i}\,c\,x^3+1\right )\right )\,1{}\mathrm {i}}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x^3))/x,x)

[Out]

a*log(x) - (b*(dilog(1 - c*x^3*1i) - dilog(c*x^3*1i + 1))*1i)/6

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