Integrand size = 21, antiderivative size = 150 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^6} \, dx=-\frac {b c d^2}{20 x^4}+\frac {b c d \left (3 c^2 d-10 e\right )}{30 x^2}-\frac {d^2 (a+b \arctan (c x))}{5 x^5}-\frac {2 d e (a+b \arctan (c x))}{3 x^3}-\frac {e^2 (a+b \arctan (c x))}{x}+\frac {1}{15} b c \left (3 c^4 d^2-10 c^2 d e+15 e^2\right ) \log (x)-\frac {1}{30} b c \left (3 c^4 d^2-10 c^2 d e+15 e^2\right ) \log \left (1+c^2 x^2\right ) \]
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Time = 0.12 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {276, 5096, 12, 1265, 907} \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^6} \, dx=-\frac {d^2 (a+b \arctan (c x))}{5 x^5}-\frac {2 d e (a+b \arctan (c x))}{3 x^3}-\frac {e^2 (a+b \arctan (c x))}{x}+\frac {b c d \left (3 c^2 d-10 e\right )}{30 x^2}-\frac {1}{30} b c \left (3 c^4 d^2-10 c^2 d e+15 e^2\right ) \log \left (c^2 x^2+1\right )+\frac {1}{15} b c \log (x) \left (3 c^4 d^2-10 c^2 d e+15 e^2\right )-\frac {b c d^2}{20 x^4} \]
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Rule 12
Rule 276
Rule 907
Rule 1265
Rule 5096
Rubi steps \begin{align*} \text {integral}& = -\frac {d^2 (a+b \arctan (c x))}{5 x^5}-\frac {2 d e (a+b \arctan (c x))}{3 x^3}-\frac {e^2 (a+b \arctan (c x))}{x}-(b c) \int \frac {-3 d^2-10 d e x^2-15 e^2 x^4}{15 x^5 \left (1+c^2 x^2\right )} \, dx \\ & = -\frac {d^2 (a+b \arctan (c x))}{5 x^5}-\frac {2 d e (a+b \arctan (c x))}{3 x^3}-\frac {e^2 (a+b \arctan (c x))}{x}-\frac {1}{15} (b c) \int \frac {-3 d^2-10 d e x^2-15 e^2 x^4}{x^5 \left (1+c^2 x^2\right )} \, dx \\ & = -\frac {d^2 (a+b \arctan (c x))}{5 x^5}-\frac {2 d e (a+b \arctan (c x))}{3 x^3}-\frac {e^2 (a+b \arctan (c x))}{x}-\frac {1}{30} (b c) \text {Subst}\left (\int \frac {-3 d^2-10 d e x-15 e^2 x^2}{x^3 \left (1+c^2 x\right )} \, dx,x,x^2\right ) \\ & = -\frac {d^2 (a+b \arctan (c x))}{5 x^5}-\frac {2 d e (a+b \arctan (c x))}{3 x^3}-\frac {e^2 (a+b \arctan (c x))}{x}-\frac {1}{30} (b c) \text {Subst}\left (\int \left (-\frac {3 d^2}{x^3}+\frac {d \left (3 c^2 d-10 e\right )}{x^2}+\frac {-3 c^4 d^2+10 c^2 d e-15 e^2}{x}+\frac {3 c^6 d^2-10 c^4 d e+15 c^2 e^2}{1+c^2 x}\right ) \, dx,x,x^2\right ) \\ & = -\frac {b c d^2}{20 x^4}+\frac {b c d \left (3 c^2 d-10 e\right )}{30 x^2}-\frac {d^2 (a+b \arctan (c x))}{5 x^5}-\frac {2 d e (a+b \arctan (c x))}{3 x^3}-\frac {e^2 (a+b \arctan (c x))}{x}+\frac {1}{15} b c \left (3 c^4 d^2-10 c^2 d e+15 e^2\right ) \log (x)-\frac {1}{30} b c \left (3 c^4 d^2-10 c^2 d e+15 e^2\right ) \log \left (1+c^2 x^2\right ) \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.99 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^6} \, dx=\frac {1}{60} \left (-\frac {12 d^2 (a+b \arctan (c x))}{x^5}-\frac {40 d e (a+b \arctan (c x))}{x^3}-\frac {60 e^2 (a+b \arctan (c x))}{x}+30 b c e^2 \left (2 \log (x)-\log \left (1+c^2 x^2\right )\right )-20 b c d e \left (\frac {1}{x^2}+2 c^2 \log (x)-c^2 \log \left (1+c^2 x^2\right )\right )-3 b c d^2 \left (\frac {1}{x^4}-\frac {2 c^2}{x^2}-4 c^4 \log (x)+2 c^4 \log \left (1+c^2 x^2\right )\right )\right ) \]
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Time = 0.28 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.11
method | result | size |
parts | \(a \left (-\frac {e^{2}}{x}-\frac {2 e d}{3 x^{3}}-\frac {d^{2}}{5 x^{5}}\right )+b \,c^{5} \left (-\frac {\arctan \left (c x \right ) e^{2}}{c^{5} x}-\frac {2 \arctan \left (c x \right ) d e}{3 c^{5} x^{3}}-\frac {\arctan \left (c x \right ) d^{2}}{5 c^{5} x^{5}}-\frac {\left (-3 c^{4} d^{2}+10 c^{2} d e -15 e^{2}\right ) \ln \left (c x \right )-\frac {d \left (3 c^{2} d -10 e \right )}{2 x^{2}}+\frac {3 d^{2}}{4 x^{4}}+\frac {\left (3 c^{4} d^{2}-10 c^{2} d e +15 e^{2}\right ) \ln \left (c^{2} x^{2}+1\right )}{2}}{15 c^{4}}\right )\) | \(167\) |
derivativedivides | \(c^{5} \left (\frac {a \left (-\frac {e^{2}}{c x}-\frac {d^{2}}{5 c \,x^{5}}-\frac {2 d e}{3 c \,x^{3}}\right )}{c^{4}}+\frac {b \left (-\frac {\arctan \left (c x \right ) e^{2}}{c x}-\frac {\arctan \left (c x \right ) d^{2}}{5 c \,x^{5}}-\frac {2 \arctan \left (c x \right ) d e}{3 c \,x^{3}}-\frac {\left (3 c^{4} d^{2}-10 c^{2} d e +15 e^{2}\right ) \ln \left (c^{2} x^{2}+1\right )}{30}-\frac {\left (-3 c^{4} d^{2}+10 c^{2} d e -15 e^{2}\right ) \ln \left (c x \right )}{15}+\frac {d \left (3 c^{2} d -10 e \right )}{30 x^{2}}-\frac {d^{2}}{20 x^{4}}\right )}{c^{4}}\right )\) | \(178\) |
default | \(c^{5} \left (\frac {a \left (-\frac {e^{2}}{c x}-\frac {d^{2}}{5 c \,x^{5}}-\frac {2 d e}{3 c \,x^{3}}\right )}{c^{4}}+\frac {b \left (-\frac {\arctan \left (c x \right ) e^{2}}{c x}-\frac {\arctan \left (c x \right ) d^{2}}{5 c \,x^{5}}-\frac {2 \arctan \left (c x \right ) d e}{3 c \,x^{3}}-\frac {\left (3 c^{4} d^{2}-10 c^{2} d e +15 e^{2}\right ) \ln \left (c^{2} x^{2}+1\right )}{30}-\frac {\left (-3 c^{4} d^{2}+10 c^{2} d e -15 e^{2}\right ) \ln \left (c x \right )}{15}+\frac {d \left (3 c^{2} d -10 e \right )}{30 x^{2}}-\frac {d^{2}}{20 x^{4}}\right )}{c^{4}}\right )\) | \(178\) |
parallelrisch | \(\frac {12 \ln \left (x \right ) b \,c^{5} d^{2} x^{5}-6 \ln \left (c^{2} x^{2}+1\right ) x^{5} b \,c^{5} d^{2}-6 b \,c^{5} d^{2} x^{5}-40 \ln \left (x \right ) b \,c^{3} d e \,x^{5}+20 \ln \left (c^{2} x^{2}+1\right ) x^{5} b \,c^{3} d e +20 b \,c^{3} d e \,x^{5}+60 \ln \left (x \right ) b c \,e^{2} x^{5}-30 \ln \left (c^{2} x^{2}+1\right ) x^{5} b c \,e^{2}+6 x^{3} b \,c^{3} d^{2}-60 x^{4} \arctan \left (c x \right ) b \,e^{2}-60 a \,e^{2} x^{4}-20 b c e d \,x^{3}-40 x^{2} \arctan \left (c x \right ) b d e -40 a d e \,x^{2}-3 b c \,d^{2} x -12 b \,d^{2} \arctan \left (c x \right )-12 d^{2} a}{60 x^{5}}\) | \(219\) |
risch | \(\frac {i b \left (15 x^{4} e^{2}+10 x^{2} e d +3 d^{2}\right ) \ln \left (i c x +1\right )}{30 x^{5}}-\frac {-12 \ln \left (x \right ) b \,c^{5} d^{2} x^{5}+6 \ln \left (-c^{2} x^{2}-1\right ) b \,c^{5} d^{2} x^{5}+40 \ln \left (x \right ) b \,c^{3} d e \,x^{5}-20 \ln \left (-c^{2} x^{2}-1\right ) b \,c^{3} d e \,x^{5}-60 \ln \left (x \right ) b c \,e^{2} x^{5}+30 \ln \left (-c^{2} x^{2}-1\right ) b c \,e^{2} x^{5}+30 i b \,e^{2} \ln \left (-i c x +1\right ) x^{4}-6 x^{3} b \,c^{3} d^{2}+20 i b d e \ln \left (-i c x +1\right ) x^{2}+60 a \,e^{2} x^{4}+20 b c e d \,x^{3}+6 i b \,d^{2} \ln \left (-i c x +1\right )+40 a d e \,x^{2}+3 b c \,d^{2} x +12 d^{2} a}{60 x^{5}}\) | \(251\) |
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Time = 0.25 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.07 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^6} \, dx=-\frac {60 \, a e^{2} x^{4} + 2 \, {\left (3 \, b c^{5} d^{2} - 10 \, b c^{3} d e + 15 \, b c e^{2}\right )} x^{5} \log \left (c^{2} x^{2} + 1\right ) - 4 \, {\left (3 \, b c^{5} d^{2} - 10 \, b c^{3} d e + 15 \, b c e^{2}\right )} x^{5} \log \left (x\right ) + 3 \, b c d^{2} x + 40 \, a d e x^{2} - 2 \, {\left (3 \, b c^{3} d^{2} - 10 \, b c d e\right )} x^{3} + 12 \, a d^{2} + 4 \, {\left (15 \, b e^{2} x^{4} + 10 \, b d e x^{2} + 3 \, b d^{2}\right )} \arctan \left (c x\right )}{60 \, x^{5}} \]
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Time = 0.55 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.57 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^6} \, dx=\begin {cases} - \frac {a d^{2}}{5 x^{5}} - \frac {2 a d e}{3 x^{3}} - \frac {a e^{2}}{x} + \frac {b c^{5} d^{2} \log {\left (x \right )}}{5} - \frac {b c^{5} d^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{10} + \frac {b c^{3} d^{2}}{10 x^{2}} - \frac {2 b c^{3} d e \log {\left (x \right )}}{3} + \frac {b c^{3} d e \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{3} - \frac {b c d^{2}}{20 x^{4}} - \frac {b c d e}{3 x^{2}} + b c e^{2} \log {\left (x \right )} - \frac {b c e^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2} - \frac {b d^{2} \operatorname {atan}{\left (c x \right )}}{5 x^{5}} - \frac {2 b d e \operatorname {atan}{\left (c x \right )}}{3 x^{3}} - \frac {b e^{2} \operatorname {atan}{\left (c x \right )}}{x} & \text {for}\: c \neq 0 \\a \left (- \frac {d^{2}}{5 x^{5}} - \frac {2 d e}{3 x^{3}} - \frac {e^{2}}{x}\right ) & \text {otherwise} \end {cases} \]
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Time = 0.20 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.11 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^6} \, dx=-\frac {1}{20} \, {\left ({\left (2 \, c^{4} \log \left (c^{2} x^{2} + 1\right ) - 2 \, c^{4} \log \left (x^{2}\right ) - \frac {2 \, c^{2} x^{2} - 1}{x^{4}}\right )} c + \frac {4 \, \arctan \left (c x\right )}{x^{5}}\right )} b d^{2} + \frac {1}{3} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} + 1\right ) - c^{2} \log \left (x^{2}\right ) - \frac {1}{x^{2}}\right )} c - \frac {2 \, \arctan \left (c x\right )}{x^{3}}\right )} b d e - \frac {1}{2} \, {\left (c {\left (\log \left (c^{2} x^{2} + 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \arctan \left (c x\right )}{x}\right )} b e^{2} - \frac {a e^{2}}{x} - \frac {2 \, a d e}{3 \, x^{3}} - \frac {a d^{2}}{5 \, x^{5}} \]
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\[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^6} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{2} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{6}} \,d x } \]
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Time = 0.52 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.19 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^6} \, dx=\frac {b\,c^3\,d^2}{10\,x^2}-\frac {a\,e^2}{x}-\frac {b\,c^5\,d^2\,\ln \left (c^2\,x^2+1\right )}{10}-\frac {a\,d^2}{5\,x^5}+\frac {b\,c^5\,d^2\,\ln \left (x\right )}{5}-\frac {2\,a\,d\,e}{3\,x^3}-\frac {b\,c\,e^2\,\ln \left (c^2\,x^2+1\right )}{2}-\frac {b\,c\,d^2}{20\,x^4}+b\,c\,e^2\,\ln \left (x\right )-\frac {b\,d^2\,\mathrm {atan}\left (c\,x\right )}{5\,x^5}-\frac {b\,e^2\,\mathrm {atan}\left (c\,x\right )}{x}+\frac {b\,c^3\,d\,e\,\ln \left (c^2\,x^2+1\right )}{3}-\frac {2\,b\,c^3\,d\,e\,\ln \left (x\right )}{3}-\frac {b\,c\,d\,e}{3\,x^2}-\frac {2\,b\,d\,e\,\mathrm {atan}\left (c\,x\right )}{3\,x^3} \]
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