Optimal. Leaf size=287 \[ \frac {i b^2 \text {Li}_2\left (\frac {2}{1-i (c+d x)}-1\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{d e^4}-\frac {b^2 \left (a+b \tan ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {i \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4}-\frac {b \log \left (2-\frac {2}{1-i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )^2}{d e^4}-\frac {b^3 \text {Li}_3\left (\frac {2}{1-i (c+d x)}-1\right )}{2 d e^4}+\frac {b^3 \log (c+d x)}{d e^4}-\frac {b^3 \log \left ((c+d x)^2+1\right )}{2 d e^4} \]
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Rubi [A] time = 0.50, antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {5043, 12, 4852, 4918, 266, 36, 29, 31, 4884, 4924, 4868, 4992, 6610} \[ \frac {i b^2 \text {PolyLog}\left (2,-1+\frac {2}{1-i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )}{d e^4}-\frac {b^3 \text {PolyLog}\left (3,-1+\frac {2}{1-i (c+d x)}\right )}{2 d e^4}-\frac {b^2 \left (a+b \tan ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {i \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4}-\frac {b \log \left (2-\frac {2}{1-i (c+d x)}\right ) \left (a+b \tan ^{-1}(c+d x)\right )^2}{d e^4}+\frac {b^3 \log (c+d x)}{d e^4}-\frac {b^3 \log \left ((c+d x)^2+1\right )}{2 d e^4} \]
Antiderivative was successfully verified.
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Rule 12
Rule 29
Rule 31
Rule 36
Rule 266
Rule 4852
Rule 4868
Rule 4884
Rule 4918
Rule 4924
Rule 4992
Rule 5043
Rule 6610
Rubi steps
\begin {align*} \int \frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{(c e+d e x)^4} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^3}{e^4 x^4} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^3}{x^4} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^2}{x^3 \left (1+x^2\right )} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^2}{x^3} \, dx,x,c+d x\right )}{d e^4}-\frac {b \operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^2}{x \left (1+x^2\right )} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}+\frac {i \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}-\frac {(i b) \operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right )^2}{x (i+x)} \, dx,x,c+d x\right )}{d e^4}+\frac {b^2 \operatorname {Subst}\left (\int \frac {a+b \tan ^{-1}(x)}{x^2 \left (1+x^2\right )} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}+\frac {i \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (2-\frac {2}{1-i (c+d x)}\right )}{d e^4}+\frac {b^2 \operatorname {Subst}\left (\int \frac {a+b \tan ^{-1}(x)}{x^2} \, dx,x,c+d x\right )}{d e^4}-\frac {b^2 \operatorname {Subst}\left (\int \frac {a+b \tan ^{-1}(x)}{1+x^2} \, dx,x,c+d x\right )}{d e^4}+\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(x)\right ) \log \left (2-\frac {2}{1-i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {b^2 \left (a+b \tan ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}+\frac {i \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (2-\frac {2}{1-i (c+d x)}\right )}{d e^4}+\frac {i b^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text {Li}_2\left (-1+\frac {2}{1-i (c+d x)}\right )}{d e^4}-\frac {\left (i b^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-1+\frac {2}{1-i x}\right )}{1+x^2} \, dx,x,c+d x\right )}{d e^4}+\frac {b^3 \operatorname {Subst}\left (\int \frac {1}{x \left (1+x^2\right )} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {b^2 \left (a+b \tan ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}+\frac {i \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (2-\frac {2}{1-i (c+d x)}\right )}{d e^4}+\frac {i b^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text {Li}_2\left (-1+\frac {2}{1-i (c+d x)}\right )}{d e^4}-\frac {b^3 \text {Li}_3\left (-1+\frac {2}{1-i (c+d x)}\right )}{2 d e^4}+\frac {b^3 \operatorname {Subst}\left (\int \frac {1}{x (1+x)} \, dx,x,(c+d x)^2\right )}{2 d e^4}\\ &=-\frac {b^2 \left (a+b \tan ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}+\frac {i \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (2-\frac {2}{1-i (c+d x)}\right )}{d e^4}+\frac {i b^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text {Li}_2\left (-1+\frac {2}{1-i (c+d x)}\right )}{d e^4}-\frac {b^3 \text {Li}_3\left (-1+\frac {2}{1-i (c+d x)}\right )}{2 d e^4}+\frac {b^3 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,(c+d x)^2\right )}{2 d e^4}-\frac {b^3 \operatorname {Subst}\left (\int \frac {1}{1+x} \, dx,x,(c+d x)^2\right )}{2 d e^4}\\ &=-\frac {b^2 \left (a+b \tan ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}+\frac {i \left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4}-\frac {\left (a+b \tan ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b^3 \log (c+d x)}{d e^4}-\frac {b^3 \log \left (1+(c+d x)^2\right )}{2 d e^4}-\frac {b \left (a+b \tan ^{-1}(c+d x)\right )^2 \log \left (2-\frac {2}{1-i (c+d x)}\right )}{d e^4}+\frac {i b^2 \left (a+b \tan ^{-1}(c+d x)\right ) \text {Li}_2\left (-1+\frac {2}{1-i (c+d x)}\right )}{d e^4}-\frac {b^3 \text {Li}_3\left (-1+\frac {2}{1-i (c+d x)}\right )}{2 d e^4}\\ \end {align*}
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Mathematica [A] time = 1.22, size = 360, normalized size = 1.25 \[ \frac {-\frac {8 a^3}{(c+d x)^3}+12 a^2 b \log \left (c^2+2 c d x+d^2 x^2+1\right )-\frac {12 a^2 b}{(c+d x)^2}-24 a^2 b \log (c+d x)-\frac {24 a^2 b \tan ^{-1}(c+d x)}{(c+d x)^3}+24 a b^2 \left (i \text {Li}_2\left (e^{2 i \tan ^{-1}(c+d x)}\right )-\frac {(c+d x)^2+\tan ^{-1}(c+d x)^2}{(c+d x)^3}+\tan ^{-1}(c+d x) \left (-\frac {1}{(c+d x)^2}+i \tan ^{-1}(c+d x)-2 \log \left (1-e^{2 i \tan ^{-1}(c+d x)}\right )-1\right )\right )+b^3 \left (-24 i \tan ^{-1}(c+d x) \text {Li}_2\left (e^{-2 i \tan ^{-1}(c+d x)}\right )-12 \text {Li}_3\left (e^{-2 i \tan ^{-1}(c+d x)}\right )+24 \log \left (\frac {c+d x}{\sqrt {(c+d x)^2+1}}\right )-\frac {8 \tan ^{-1}(c+d x)^3}{(c+d x)^3}-8 i \tan ^{-1}(c+d x)^3-\frac {12 \tan ^{-1}(c+d x)^2}{(c+d x)^2}-12 \tan ^{-1}(c+d x)^2-\frac {24 \tan ^{-1}(c+d x)}{c+d x}-24 \tan ^{-1}(c+d x)^2 \log \left (1-e^{-2 i \tan ^{-1}(c+d x)}\right )+i \pi ^3\right )}{24 d e^4} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \arctan \left (d x + c\right )^{3} + 3 \, a b^{2} \arctan \left (d x + c\right )^{2} + 3 \, a^{2} b \arctan \left (d x + c\right ) + a^{3}}{d^{4} e^{4} x^{4} + 4 \, c d^{3} e^{4} x^{3} + 6 \, c^{2} d^{2} e^{4} x^{2} + 4 \, c^{3} d e^{4} x + c^{4} e^{4}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [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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maple [C] time = 1.63, size = 7083, normalized size = 24.68 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [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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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {atan}\left (c+d\,x\right )\right )}^3}{{\left (c\,e+d\,e\,x\right )}^4} \,d x \]
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 \[ \frac {\int \frac {a^{3}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {b^{3} \operatorname {atan}^{3}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {3 a b^{2} \operatorname {atan}^{2}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {3 a^{2} b \operatorname {atan}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx}{e^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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