Optimal. Leaf size=216 \[ -d (a+b \text {ArcTan}(c x))^2+i c d x (a+b \text {ArcTan}(c x))^2+2 d (a+b \text {ArcTan}(c x))^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+2 i b d (a+b \text {ArcTan}(c x)) \log \left (\frac {2}{1+i c x}\right )-b^2 d \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )-i b d (a+b \text {ArcTan}(c x)) \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )+i b d (a+b \text {ArcTan}(c x)) \text {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )-\frac {1}{2} b^2 d \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 d \text {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right ) \]
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Rubi [A]
time = 0.30, antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps
used = 13, number of rules used = 11, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {4996, 4930,
5040, 4964, 2449, 2352, 4942, 5108, 5004, 5114, 6745} \begin {gather*} -i b d \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) (a+b \text {ArcTan}(c x))+i b d \text {Li}_2\left (\frac {2}{i c x+1}-1\right ) (a+b \text {ArcTan}(c x))-d (a+b \text {ArcTan}(c x))^2+i c d x (a+b \text {ArcTan}(c x))^2+2 i b d \log \left (\frac {2}{1+i c x}\right ) (a+b \text {ArcTan}(c x))+2 d \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) (a+b \text {ArcTan}(c x))^2+b^2 (-d) \text {Li}_2\left (1-\frac {2}{i c x+1}\right )-\frac {1}{2} b^2 d \text {Li}_3\left (1-\frac {2}{i c x+1}\right )+\frac {1}{2} b^2 d \text {Li}_3\left (\frac {2}{i c x+1}-1\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 2352
Rule 2449
Rule 4930
Rule 4942
Rule 4964
Rule 4996
Rule 5004
Rule 5040
Rule 5108
Rule 5114
Rule 6745
Rubi steps
\begin {align*} \int \frac {(d+i c d x) \left (a+b \tan ^{-1}(c x)\right )^2}{x} \, dx &=\int \left (i c d \left (a+b \tan ^{-1}(c x)\right )^2+\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{x}\right ) \, dx\\ &=d \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x} \, dx+(i c d) \int \left (a+b \tan ^{-1}(c x)\right )^2 \, dx\\ &=i c d x \left (a+b \tan ^{-1}(c x)\right )^2+2 d \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )-(4 b c d) \int \frac {\left (a+b \tan ^{-1}(c x)\right ) \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (2 i b c^2 d\right ) \int \frac {x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx\\ &=-d \left (a+b \tan ^{-1}(c x)\right )^2+i c d x \left (a+b \tan ^{-1}(c x)\right )^2+2 d \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+(2 i b c d) \int \frac {a+b \tan ^{-1}(c x)}{i-c x} \, dx+(2 b c d) \int \frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-(2 b c d) \int \frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx\\ &=-d \left (a+b \tan ^{-1}(c x)\right )^2+i c d x \left (a+b \tan ^{-1}(c x)\right )^2+2 d \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+2 i b d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )-i b d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )+i b d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )+\left (i b^2 c d\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (i b^2 c d\right ) \int \frac {\text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (2 i b^2 c d\right ) \int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx\\ &=-d \left (a+b \tan ^{-1}(c x)\right )^2+i c d x \left (a+b \tan ^{-1}(c x)\right )^2+2 d \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+2 i b d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )-i b d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )+i b d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )-\frac {1}{2} b^2 d \text {Li}_3\left (1-\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 d \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )-\left (2 b^2 d\right ) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i c x}\right )\\ &=-d \left (a+b \tan ^{-1}(c x)\right )^2+i c d x \left (a+b \tan ^{-1}(c x)\right )^2+2 d \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+2 i b d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )-b^2 d \text {Li}_2\left (1-\frac {2}{1+i c x}\right )-i b d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )+i b d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )-\frac {1}{2} b^2 d \text {Li}_3\left (1-\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 d \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )\\ \end {align*}
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Mathematica [A]
time = 0.26, size = 272, normalized size = 1.26 \begin {gather*} d \left (i a^2 c x+a^2 \log (c x)+i a b \left (2 c x \text {ArcTan}(c x)-\log \left (1+c^2 x^2\right )\right )+b^2 \left (\text {ArcTan}(c x) \left ((1+i c x) \text {ArcTan}(c x)+2 i \log \left (1+e^{2 i \text {ArcTan}(c x)}\right )\right )+\text {PolyLog}\left (2,-e^{2 i \text {ArcTan}(c x)}\right )\right )+i a b (\text {PolyLog}(2,-i c x)-\text {PolyLog}(2,i c x))+b^2 \left (-\frac {i \pi ^3}{24}+\frac {2}{3} i \text {ArcTan}(c x)^3+\text {ArcTan}(c x)^2 \log \left (1-e^{-2 i \text {ArcTan}(c x)}\right )-\text {ArcTan}(c x)^2 \log \left (1+e^{2 i \text {ArcTan}(c x)}\right )+i \text {ArcTan}(c x) \text {PolyLog}\left (2,e^{-2 i \text {ArcTan}(c x)}\right )+i \text {ArcTan}(c x) \text {PolyLog}\left (2,-e^{2 i \text {ArcTan}(c x)}\right )+\frac {1}{2} \text {PolyLog}\left (3,e^{-2 i \text {ArcTan}(c x)}\right )-\frac {1}{2} \text {PolyLog}\left (3,-e^{2 i \text {ArcTan}(c x)}\right )\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
4.
time = 2.57, size = 7034, normalized size = 32.56
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(7034\) |
default | \(\text {Expression too large to display}\) | \(7034\) |
Verification of antiderivative is not currently implemented for this CAS.
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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.
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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.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} i d \left (\int a^{2} c\, dx + \int \left (- \frac {i a^{2}}{x}\right )\, dx + \int b^{2} c \operatorname {atan}^{2}{\left (c x \right )}\, dx + \int \left (- \frac {i b^{2} \operatorname {atan}^{2}{\left (c x \right )}}{x}\right )\, dx + \int 2 a b c \operatorname {atan}{\left (c x \right )}\, dx + \int \left (- \frac {2 i a b \operatorname {atan}{\left (c x \right )}}{x}\right )\, dx\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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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.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2\,\left (d+c\,d\,x\,1{}\mathrm {i}\right )}{x} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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