Optimal. Leaf size=63 \[ -\frac {a c}{6 x^2}-\frac {c \text {ArcTan}(a x)}{3 x^3}-\frac {a^2 c \text {ArcTan}(a x)}{x}+\frac {2}{3} a^3 c \log (x)-\frac {1}{3} a^3 c \log \left (1+a^2 x^2\right ) \]
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Rubi [A]
time = 0.06, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps
used = 10, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {5070, 4946,
272, 46, 36, 29, 31} \begin {gather*} \frac {2}{3} a^3 c \log (x)-\frac {a^2 c \text {ArcTan}(a x)}{x}-\frac {1}{3} a^3 c \log \left (a^2 x^2+1\right )-\frac {c \text {ArcTan}(a x)}{3 x^3}-\frac {a c}{6 x^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 29
Rule 31
Rule 36
Rule 46
Rule 272
Rule 4946
Rule 5070
Rubi steps
\begin {align*} \int \frac {\left (c+a^2 c x^2\right ) \tan ^{-1}(a x)}{x^4} \, dx &=c \int \frac {\tan ^{-1}(a x)}{x^4} \, dx+\left (a^2 c\right ) \int \frac {\tan ^{-1}(a x)}{x^2} \, dx\\ &=-\frac {c \tan ^{-1}(a x)}{3 x^3}-\frac {a^2 c \tan ^{-1}(a x)}{x}+\frac {1}{3} (a c) \int \frac {1}{x^3 \left (1+a^2 x^2\right )} \, dx+\left (a^3 c\right ) \int \frac {1}{x \left (1+a^2 x^2\right )} \, dx\\ &=-\frac {c \tan ^{-1}(a x)}{3 x^3}-\frac {a^2 c \tan ^{-1}(a x)}{x}+\frac {1}{6} (a c) \text {Subst}\left (\int \frac {1}{x^2 \left (1+a^2 x\right )} \, dx,x,x^2\right )+\frac {1}{2} \left (a^3 c\right ) \text {Subst}\left (\int \frac {1}{x \left (1+a^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac {c \tan ^{-1}(a x)}{3 x^3}-\frac {a^2 c \tan ^{-1}(a x)}{x}+\frac {1}{6} (a c) \text {Subst}\left (\int \left (\frac {1}{x^2}-\frac {a^2}{x}+\frac {a^4}{1+a^2 x}\right ) \, dx,x,x^2\right )+\frac {1}{2} \left (a^3 c\right ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-\frac {1}{2} \left (a^5 c\right ) \text {Subst}\left (\int \frac {1}{1+a^2 x} \, dx,x,x^2\right )\\ &=-\frac {a c}{6 x^2}-\frac {c \tan ^{-1}(a x)}{3 x^3}-\frac {a^2 c \tan ^{-1}(a x)}{x}+\frac {2}{3} a^3 c \log (x)-\frac {1}{3} a^3 c \log \left (1+a^2 x^2\right )\\ \end {align*}
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Mathematica [A]
time = 0.02, size = 58, normalized size = 0.92 \begin {gather*} \frac {c \left (-2 \left (1+3 a^2 x^2\right ) \text {ArcTan}(a x)+a x \left (-1+4 a^2 x^2 \log (x)-2 a^2 x^2 \log \left (1+a^2 x^2\right )\right )\right )}{6 x^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.12, size = 60, normalized size = 0.95
method | result | size |
derivativedivides | \(a^{3} \left (-\frac {c \arctan \left (a x \right )}{a x}-\frac {c \arctan \left (a x \right )}{3 a^{3} x^{3}}-\frac {c \left (\ln \left (a^{2} x^{2}+1\right )+\frac {1}{2 a^{2} x^{2}}-2 \ln \left (a x \right )\right )}{3}\right )\) | \(60\) |
default | \(a^{3} \left (-\frac {c \arctan \left (a x \right )}{a x}-\frac {c \arctan \left (a x \right )}{3 a^{3} x^{3}}-\frac {c \left (\ln \left (a^{2} x^{2}+1\right )+\frac {1}{2 a^{2} x^{2}}-2 \ln \left (a x \right )\right )}{3}\right )\) | \(60\) |
risch | \(\frac {i c \left (3 a^{2} x^{2}+1\right ) \ln \left (i a x +1\right )}{6 x^{3}}+\frac {c \left (4 \ln \left (x \right ) a^{3} x^{3}-2 \ln \left (-3 a^{2} x^{2}-3\right ) a^{3} x^{3}-3 i a^{2} x^{2} \ln \left (-i a x +1\right )-a x -i \ln \left (-i a x +1\right )\right )}{6 x^{3}}\) | \(95\) |
meijerg | \(\frac {a^{3} c \left (-\frac {4 \arctan \left (\sqrt {a^{2} x^{2}}\right )}{\sqrt {a^{2} x^{2}}}-2 \ln \left (a^{2} x^{2}+1\right )+4 \ln \left (x \right )+4 \ln \left (a \right )\right )}{4}+\frac {a^{3} c \left (\frac {-\frac {4 a^{2} x^{2}}{9}+\frac {4}{3}}{a^{2} x^{2}}-\frac {4 \arctan \left (\sqrt {a^{2} x^{2}}\right )}{3 a^{2} x^{2} \sqrt {a^{2} x^{2}}}+\frac {2 \ln \left (a^{2} x^{2}+1\right )}{3}+\frac {4}{9}-\frac {4 \ln \left (x \right )}{3}-\frac {4 \ln \left (a \right )}{3}-\frac {2}{a^{2} x^{2}}\right )}{4}\) | \(131\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.26, size = 56, normalized size = 0.89 \begin {gather*} -\frac {1}{6} \, {\left (2 \, a^{2} c \log \left (a^{2} x^{2} + 1\right ) - 2 \, a^{2} c \log \left (x^{2}\right ) + \frac {c}{x^{2}}\right )} a - \frac {{\left (3 \, a^{2} c x^{2} + c\right )} \arctan \left (a x\right )}{3 \, x^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 1.49, size = 57, normalized size = 0.90 \begin {gather*} -\frac {2 \, a^{3} c x^{3} \log \left (a^{2} x^{2} + 1\right ) - 4 \, a^{3} c x^{3} \log \left (x\right ) + a c x + 2 \, {\left (3 \, a^{2} c x^{2} + c\right )} \arctan \left (a x\right )}{6 \, x^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.37, size = 61, normalized size = 0.97 \begin {gather*} \begin {cases} \frac {2 a^{3} c \log {\left (x \right )}}{3} - \frac {a^{3} c \log {\left (x^{2} + \frac {1}{a^{2}} \right )}}{3} - \frac {a^{2} c \operatorname {atan}{\left (a x \right )}}{x} - \frac {a c}{6 x^{2}} - \frac {c \operatorname {atan}{\left (a x \right )}}{3 x^{3}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \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 [B]
time = 0.15, size = 57, normalized size = 0.90 \begin {gather*} \frac {c\,\left (4\,a^3\,\ln \left (x\right )-2\,a^3\,\ln \left (a^2\,x^2+1\right )\right )}{6}-\frac {\frac {c\,\mathrm {atan}\left (a\,x\right )}{3}+\frac {a\,c\,x}{6}+a^2\,c\,x^2\,\mathrm {atan}\left (a\,x\right )}{x^3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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