Optimal. Leaf size=197 \[ \frac {i c^3 \log \left (2-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}+\frac {c^2 \left (a+b \tan ^{-1}(c x)\right )}{d x}-\frac {a+b \tan ^{-1}(c x)}{3 d x^3}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )}{2 d x^2}-\frac {b c^3 \text {Li}_2\left (\frac {2}{i c x+1}-1\right )}{2 d}-\frac {4 b c^3 \log (x)}{3 d}+\frac {i b c^3 \tan ^{-1}(c x)}{2 d}+\frac {i b c^2}{2 d x}+\frac {2 b c^3 \log \left (c^2 x^2+1\right )}{3 d}-\frac {b c}{6 d x^2} \]
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Rubi [A] time = 0.34, antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 11, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {4870, 4852, 266, 44, 325, 203, 36, 29, 31, 4868, 2447} \[ -\frac {b c^3 \text {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )}{2 d}+\frac {c^2 \left (a+b \tan ^{-1}(c x)\right )}{d x}+\frac {i c^3 \log \left (2-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )}{2 d x^2}-\frac {a+b \tan ^{-1}(c x)}{3 d x^3}+\frac {2 b c^3 \log \left (c^2 x^2+1\right )}{3 d}+\frac {i b c^2}{2 d x}-\frac {4 b c^3 \log (x)}{3 d}+\frac {i b c^3 \tan ^{-1}(c x)}{2 d}-\frac {b c}{6 d x^2} \]
Antiderivative was successfully verified.
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Rule 29
Rule 31
Rule 36
Rule 44
Rule 203
Rule 266
Rule 325
Rule 2447
Rule 4852
Rule 4868
Rule 4870
Rubi steps
\begin {align*} \int \frac {a+b \tan ^{-1}(c x)}{x^4 (d+i c d x)} \, dx &=-\left ((i c) \int \frac {a+b \tan ^{-1}(c x)}{x^3 (d+i c d x)} \, dx\right )+\frac {\int \frac {a+b \tan ^{-1}(c x)}{x^4} \, dx}{d}\\ &=-\frac {a+b \tan ^{-1}(c x)}{3 d x^3}-c^2 \int \frac {a+b \tan ^{-1}(c x)}{x^2 (d+i c d x)} \, dx-\frac {(i c) \int \frac {a+b \tan ^{-1}(c x)}{x^3} \, dx}{d}+\frac {(b c) \int \frac {1}{x^3 \left (1+c^2 x^2\right )} \, dx}{3 d}\\ &=-\frac {a+b \tan ^{-1}(c x)}{3 d x^3}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )}{2 d x^2}+\left (i c^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{x (d+i c d x)} \, dx+\frac {(b c) \operatorname {Subst}\left (\int \frac {1}{x^2 \left (1+c^2 x\right )} \, dx,x,x^2\right )}{6 d}-\frac {c^2 \int \frac {a+b \tan ^{-1}(c x)}{x^2} \, dx}{d}-\frac {\left (i b c^2\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx}{2 d}\\ &=\frac {i b c^2}{2 d x}-\frac {a+b \tan ^{-1}(c x)}{3 d x^3}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )}{2 d x^2}+\frac {c^2 \left (a+b \tan ^{-1}(c x)\right )}{d x}+\frac {i c^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+i c x}\right )}{d}+\frac {(b c) \operatorname {Subst}\left (\int \left (\frac {1}{x^2}-\frac {c^2}{x}+\frac {c^4}{1+c^2 x}\right ) \, dx,x,x^2\right )}{6 d}-\frac {\left (b c^3\right ) \int \frac {1}{x \left (1+c^2 x^2\right )} \, dx}{d}+\frac {\left (i b c^4\right ) \int \frac {1}{1+c^2 x^2} \, dx}{2 d}-\frac {\left (i b c^4\right ) \int \frac {\log \left (2-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d}\\ &=-\frac {b c}{6 d x^2}+\frac {i b c^2}{2 d x}+\frac {i b c^3 \tan ^{-1}(c x)}{2 d}-\frac {a+b \tan ^{-1}(c x)}{3 d x^3}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )}{2 d x^2}+\frac {c^2 \left (a+b \tan ^{-1}(c x)\right )}{d x}-\frac {b c^3 \log (x)}{3 d}+\frac {b c^3 \log \left (1+c^2 x^2\right )}{6 d}+\frac {i c^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+i c x}\right )}{d}-\frac {b c^3 \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{2 d}-\frac {\left (b c^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )}{2 d}\\ &=-\frac {b c}{6 d x^2}+\frac {i b c^2}{2 d x}+\frac {i b c^3 \tan ^{-1}(c x)}{2 d}-\frac {a+b \tan ^{-1}(c x)}{3 d x^3}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )}{2 d x^2}+\frac {c^2 \left (a+b \tan ^{-1}(c x)\right )}{d x}-\frac {b c^3 \log (x)}{3 d}+\frac {b c^3 \log \left (1+c^2 x^2\right )}{6 d}+\frac {i c^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+i c x}\right )}{d}-\frac {b c^3 \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{2 d}-\frac {\left (b c^3\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )}{2 d}+\frac {\left (b c^5\right ) \operatorname {Subst}\left (\int \frac {1}{1+c^2 x} \, dx,x,x^2\right )}{2 d}\\ &=-\frac {b c}{6 d x^2}+\frac {i b c^2}{2 d x}+\frac {i b c^3 \tan ^{-1}(c x)}{2 d}-\frac {a+b \tan ^{-1}(c x)}{3 d x^3}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )}{2 d x^2}+\frac {c^2 \left (a+b \tan ^{-1}(c x)\right )}{d x}-\frac {4 b c^3 \log (x)}{3 d}+\frac {2 b c^3 \log \left (1+c^2 x^2\right )}{3 d}+\frac {i c^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+i c x}\right )}{d}-\frac {b c^3 \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{2 d}\\ \end {align*}
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Mathematica [C] time = 0.14, size = 254, normalized size = 1.29 \[ \frac {6 i a c^3 x^3 \log (x)+6 i a c^3 x^3 \log \left (\frac {2 i}{-c x+i}\right )+6 a c^2 x^2+3 i a c x-2 a-3 b c^3 x^3 \text {Li}_2(-i c x)+3 b c^3 x^3 \text {Li}_2(i c x)-3 b c^3 x^3 \text {Li}_2\left (\frac {c x+i}{c x-i}\right )-8 b c^3 x^3 \log (x)+6 i b c^3 x^3 \log \left (\frac {2 i}{-c x+i}\right ) \tan ^{-1}(c x)+3 i b c^2 x^2 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-c^2 x^2\right )+6 b c^2 x^2 \tan ^{-1}(c x)+4 b c^3 x^3 \log \left (c^2 x^2+1\right )-b c x+3 i b c x \tan ^{-1}(c x)-2 b \tan ^{-1}(c x)}{6 d x^3} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \log \left (-\frac {c x + i}{c x - i}\right ) - 2 i \, a}{2 \, c d x^{5} - 2 i \, d x^{4}}, x\right ) \]
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 [B] time = 0.08, size = 369, normalized size = 1.87 \[ -\frac {a}{3 d \,x^{3}}+\frac {i c a}{2 d \,x^{2}}+\frac {i c b \arctan \left (c x \right )}{2 d \,x^{2}}+\frac {c^{2} a}{d x}+\frac {i c^{3} b \arctan \left (c x \right ) \ln \left (c x \right )}{d}+\frac {c^{3} a \arctan \left (c x \right )}{d}-\frac {b \arctan \left (c x \right )}{3 d \,x^{3}}+\frac {i b \,c^{2}}{2 d x}+\frac {i b \,c^{3} \arctan \left (c x \right )}{2 d}+\frac {c^{2} b \arctan \left (c x \right )}{d x}+\frac {i c^{3} a \ln \left (c x \right )}{d}-\frac {c^{3} b \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{2 d}-\frac {c^{3} b \dilog \left (-\frac {i \left (c x +i\right )}{2}\right )}{2 d}+\frac {c^{3} b \ln \left (c x -i\right )^{2}}{4 d}-\frac {c^{3} b \ln \left (c x \right ) \ln \left (i c x +1\right )}{2 d}+\frac {c^{3} b \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2 d}-\frac {c^{3} b \dilog \left (i c x +1\right )}{2 d}+\frac {c^{3} b \dilog \left (-i c x +1\right )}{2 d}-\frac {i c^{3} b \arctan \left (c x \right ) \ln \left (c x -i\right )}{d}-\frac {b c}{6 d \,x^{2}}-\frac {4 c^{3} b \ln \left (c x \right )}{3 d}+\frac {2 b \,c^{3} \ln \left (c^{2} x^{2}+1\right )}{3 d}-\frac {i c^{3} a \ln \left (c^{2} x^{2}+1\right )}{2 d} \]
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 \[ -\frac {1}{6} \, {\left (\frac {6 i \, c^{3} \log \left (i \, c x + 1\right )}{d} - \frac {6 i \, c^{3} \log \relax (x)}{d} - \frac {6 \, c^{2} x^{2} + 3 i \, c x - 2}{d x^{3}}\right )} a + {\left (-i \, c \int \frac {\arctan \left (c x\right )}{c^{2} d x^{5} + d x^{3}}\,{d x} + \int \frac {\arctan \left (c x\right )}{c^{2} d x^{6} + d x^{4}}\,{d x}\right )} b \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {atan}\left (c\,x\right )}{x^4\,\left (d+c\,d\,x\,1{}\mathrm {i}\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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