Optimal. Leaf size=152 \[ -\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac {2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-a c^2 d^2 \log (x)-\frac {1}{2} i b c^2 d^2 \text {Li}_2(-i c x)+\frac {1}{2} i b c^2 d^2 \text {Li}_2(i c x)-i b c^2 d^2 \log \left (c^2 x^2+1\right )+2 i b c^2 d^2 \log (x)-\frac {1}{2} b c^2 d^2 \tan ^{-1}(c x)-\frac {b c d^2}{2 x} \]
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Rubi [A] time = 0.15, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {4876, 4852, 325, 203, 266, 36, 29, 31, 4848, 2391} \[ -\frac {1}{2} i b c^2 d^2 \text {PolyLog}(2,-i c x)+\frac {1}{2} i b c^2 d^2 \text {PolyLog}(2,i c x)-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac {2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-a c^2 d^2 \log (x)-i b c^2 d^2 \log \left (c^2 x^2+1\right )+2 i b c^2 d^2 \log (x)-\frac {1}{2} b c^2 d^2 \tan ^{-1}(c x)-\frac {b c d^2}{2 x} \]
Antiderivative was successfully verified.
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Rule 29
Rule 31
Rule 36
Rule 203
Rule 266
Rule 325
Rule 2391
Rule 4848
Rule 4852
Rule 4876
Rubi steps
\begin {align*} \int \frac {(d+i c d x)^2 \left (a+b \tan ^{-1}(c x)\right )}{x^3} \, dx &=\int \left (\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{x^3}+\frac {2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{x^2}-\frac {c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}\right ) \, dx\\ &=d^2 \int \frac {a+b \tan ^{-1}(c x)}{x^3} \, dx+\left (2 i c d^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{x^2} \, dx-\left (c^2 d^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{x} \, dx\\ &=-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac {2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-a c^2 d^2 \log (x)+\frac {1}{2} \left (b c d^2\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx-\frac {1}{2} \left (i b c^2 d^2\right ) \int \frac {\log (1-i c x)}{x} \, dx+\frac {1}{2} \left (i b c^2 d^2\right ) \int \frac {\log (1+i c x)}{x} \, dx+\left (2 i b c^2 d^2\right ) \int \frac {1}{x \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {b c d^2}{2 x}-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac {2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-a c^2 d^2 \log (x)-\frac {1}{2} i b c^2 d^2 \text {Li}_2(-i c x)+\frac {1}{2} i b c^2 d^2 \text {Li}_2(i c x)+\left (i b c^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )-\frac {1}{2} \left (b c^3 d^2\right ) \int \frac {1}{1+c^2 x^2} \, dx\\ &=-\frac {b c d^2}{2 x}-\frac {1}{2} b c^2 d^2 \tan ^{-1}(c x)-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac {2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-a c^2 d^2 \log (x)-\frac {1}{2} i b c^2 d^2 \text {Li}_2(-i c x)+\frac {1}{2} i b c^2 d^2 \text {Li}_2(i c x)+\left (i b c^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-\left (i b c^4 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{1+c^2 x} \, dx,x,x^2\right )\\ &=-\frac {b c d^2}{2 x}-\frac {1}{2} b c^2 d^2 \tan ^{-1}(c x)-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac {2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-a c^2 d^2 \log (x)+2 i b c^2 d^2 \log (x)-i b c^2 d^2 \log \left (1+c^2 x^2\right )-\frac {1}{2} i b c^2 d^2 \text {Li}_2(-i c x)+\frac {1}{2} i b c^2 d^2 \text {Li}_2(i c x)\\ \end {align*}
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Mathematica [C] time = 0.08, size = 139, normalized size = 0.91 \[ -\frac {d^2 \left (2 a c^2 x^2 \log (x)+4 i a c x+a+b c x \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-c^2 x^2\right )+i b c^2 x^2 \text {Li}_2(-i c x)-i b c^2 x^2 \text {Li}_2(i c x)-4 i b c^2 x^2 \log (x)+2 i b c^2 x^2 \log \left (c^2 x^2+1\right )+4 i b c x \tan ^{-1}(c x)+b \tan ^{-1}(c x)\right )}{2 x^2} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {2 \, a c^{2} d^{2} x^{2} - 4 i \, a c d^{2} x - 2 \, a d^{2} - {\left (-i \, b c^{2} d^{2} x^{2} - 2 \, b c d^{2} x + i \, b d^{2}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{2 \, x^{3}}, x\right ) \]
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 \[ \mathit {sage}_{0} x \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 217, normalized size = 1.43 \[ -c^{2} d^{2} a \ln \left (c x \right )-\frac {2 i c \,d^{2} a}{x}-\frac {d^{2} a}{2 x^{2}}-c^{2} d^{2} b \ln \left (c x \right ) \arctan \left (c x \right )-\frac {2 i c \,d^{2} b \arctan \left (c x \right )}{x}-\frac {d^{2} b \arctan \left (c x \right )}{2 x^{2}}-\frac {i c^{2} d^{2} b \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {i c^{2} d^{2} b \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {i c^{2} d^{2} b \dilog \left (i c x +1\right )}{2}+\frac {i c^{2} d^{2} b \dilog \left (-i c x +1\right )}{2}-\frac {b c \,d^{2}}{2 x}+2 i c^{2} d^{2} b \ln \left (c x \right )-i b \,c^{2} d^{2} \ln \left (c^{2} x^{2}+1\right )-\frac {b \,c^{2} d^{2} \arctan \left (c x \right )}{2} \]
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 \[ -b c^{2} d^{2} \int \frac {\arctan \left (c x\right )}{x}\,{d x} - a c^{2} d^{2} \log \relax (x) - i \, {\left (c {\left (\log \left (c^{2} x^{2} + 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \arctan \left (c x\right )}{x}\right )} b c d^{2} - \frac {1}{2} \, {\left ({\left (c \arctan \left (c x\right ) + \frac {1}{x}\right )} c + \frac {\arctan \left (c x\right )}{x^{2}}\right )} b d^{2} - \frac {2 i \, a c d^{2}}{x} - \frac {a d^{2}}{2 \, x^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.74, size = 161, normalized size = 1.06 \[ \left \{\begin {array}{cl} -\frac {a\,d^2}{2\,x^2} & \text {\ if\ \ }c=0\\ b\,d^2\,\left (c^2\,\ln \relax (x)-\frac {c^2\,\ln \left (c^2\,x^2+1\right )}{2}\right )\,2{}\mathrm {i}+\frac {b\,c^2\,d^2\,{\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}-\frac {b\,c^2\,d^2\,{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}-\frac {b\,d^2\,\left (c^3\,\mathrm {atan}\left (c\,x\right )+\frac {c^2}{x}\right )}{2\,c}-\frac {a\,d^2\,\left (2\,c^2\,x^2\,\ln \relax (x)+1+c\,x\,4{}\mathrm {i}\right )}{2\,x^2}-\frac {b\,d^2\,\mathrm {atan}\left (c\,x\right )}{2\,x^2}-\frac {b\,c\,d^2\,\mathrm {atan}\left (c\,x\right )\,2{}\mathrm {i}}{x} & \text {\ if\ \ }c\neq 0 \end {array}\right . \]
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 \[ - d^{2} \left (\int \left (- \frac {a}{x^{3}}\right )\, dx + \int \frac {a c^{2}}{x}\, dx + \int \left (- \frac {b \operatorname {atan}{\left (c x \right )}}{x^{3}}\right )\, dx + \int \left (- \frac {2 i a c}{x^{2}}\right )\, dx + \int \frac {b c^{2} \operatorname {atan}{\left (c x \right )}}{x}\, dx + \int \left (- \frac {2 i b c \operatorname {atan}{\left (c x \right )}}{x^{2}}\right )\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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