Optimal. Leaf size=228 \[ -i c d (a+b \text {ArcTan}(c x))^2-\frac {d (a+b \text {ArcTan}(c x))^2}{x}+2 i c d (a+b \text {ArcTan}(c x))^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+2 b c d (a+b \text {ArcTan}(c x)) \log \left (2-\frac {2}{1-i c x}\right )-i b^2 c d \text {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )+b c d (a+b \text {ArcTan}(c x)) \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )-b c d (a+b \text {ArcTan}(c x)) \text {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )-\frac {1}{2} i b^2 c d \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )+\frac {1}{2} i b^2 c d \text {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right ) \]
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Rubi [A]
time = 0.34, antiderivative size = 228, normalized size of antiderivative = 1.00, number of steps
used = 12, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {4996, 4946,
5044, 4988, 2497, 4942, 5108, 5004, 5114, 6745} \begin {gather*} b c d \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) (a+b \text {ArcTan}(c x))-b c d \text {Li}_2\left (\frac {2}{i c x+1}-1\right ) (a+b \text {ArcTan}(c x))-i c d (a+b \text {ArcTan}(c x))^2-\frac {d (a+b \text {ArcTan}(c x))^2}{x}+2 b c d \log \left (2-\frac {2}{1-i c x}\right ) (a+b \text {ArcTan}(c x))+2 i c d \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) (a+b \text {ArcTan}(c x))^2-i b^2 c d \text {Li}_2\left (\frac {2}{1-i c x}-1\right )-\frac {1}{2} i b^2 c d \text {Li}_3\left (1-\frac {2}{i c x+1}\right )+\frac {1}{2} i b^2 c d \text {Li}_3\left (\frac {2}{i c x+1}-1\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 2497
Rule 4942
Rule 4946
Rule 4988
Rule 4996
Rule 5004
Rule 5044
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^2} \, dx &=\int \left (\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{x^2}+\frac {i c 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^2} \, dx+(i c d) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x} \, dx\\ &=-\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{x}+2 i c d \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+(2 b c d) \int \frac {a+b \tan ^{-1}(c x)}{x \left (1+c^2 x^2\right )} \, dx-\left (4 i b c^2 d\right ) \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\\ &=-i c d \left (a+b \tan ^{-1}(c x)\right )^2-\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{x}+2 i c 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)}{x (i+c x)} \, dx+\left (2 i b c^2 d\right ) \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-\left (2 i b c^2 d\right ) \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\\ &=-i c d \left (a+b \tan ^{-1}(c x)\right )^2-\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{x}+2 i c d \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+2 b c d \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )+b c d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )-b c d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )-\left (b^2 c^2 d\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx+\left (b^2 c^2 d\right ) \int \frac {\text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (2 b^2 c^2 d\right ) \int \frac {\log \left (2-\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx\\ &=-i c d \left (a+b \tan ^{-1}(c x)\right )^2-\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{x}+2 i c d \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+2 b c d \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )-i b^2 c d \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )+b c d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )-b c d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )-\frac {1}{2} i b^2 c d \text {Li}_3\left (1-\frac {2}{1+i c x}\right )+\frac {1}{2} i b^2 c d \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )\\ \end {align*}
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Mathematica [A]
time = 0.24, size = 289, normalized size = 1.27 \begin {gather*} \frac {i d \left (i a^2+a^2 c x \log (x)+i a b \left (2 \text {ArcTan}(c x)+c x \left (-2 \log (c x)+\log \left (1+c^2 x^2\right )\right )\right )+i b^2 \left (\text {ArcTan}(c x)^2-2 c x \text {ArcTan}(c x) \log \left (1-e^{2 i \text {ArcTan}(c x)}\right )+i c x \left (\text {ArcTan}(c x)^2+\text {PolyLog}\left (2,e^{2 i \text {ArcTan}(c x)}\right )\right )\right )+i a b c x (\text {PolyLog}(2,-i c x)-\text {PolyLog}(2,i c x))+\frac {1}{24} b^2 c x \left (-i \pi ^3+16 i \text {ArcTan}(c x)^3+24 \text {ArcTan}(c x)^2 \log \left (1-e^{-2 i \text {ArcTan}(c x)}\right )-24 \text {ArcTan}(c x)^2 \log \left (1+e^{2 i \text {ArcTan}(c x)}\right )+24 i \text {ArcTan}(c x) \text {PolyLog}\left (2,e^{-2 i \text {ArcTan}(c x)}\right )+24 i \text {ArcTan}(c x) \text {PolyLog}\left (2,-e^{2 i \text {ArcTan}(c x)}\right )+12 \text {PolyLog}\left (3,e^{-2 i \text {ArcTan}(c x)}\right )-12 \text {PolyLog}\left (3,-e^{2 i \text {ArcTan}(c x)}\right )\right )\right )}{x} \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 = 2.64, size = 5899, normalized size = 25.87
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(5899\) |
default | \(\text {Expression too large to display}\) | \(5899\) |
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 \left (- \frac {i a^{2}}{x^{2}}\right )\, dx + \int \frac {a^{2} c}{x}\, dx + \int \left (- \frac {i b^{2} \operatorname {atan}^{2}{\left (c x \right )}}{x^{2}}\right )\, dx + \int \frac {b^{2} c \operatorname {atan}^{2}{\left (c x \right )}}{x}\, dx + \int \left (- \frac {2 i a b \operatorname {atan}{\left (c x \right )}}{x^{2}}\right )\, dx + \int \frac {2 a b c \operatorname {atan}{\left (c x \right )}}{x}\, dx\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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