Integrand size = 21, antiderivative size = 186 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^8} \, dx=-\frac {b c d^2}{42 x^6}+\frac {b c d \left (5 c^2 d-14 e\right )}{140 x^4}-\frac {b c \left (15 c^4 d^2-42 c^2 d e+35 e^2\right )}{210 x^2}-\frac {d^2 (a+b \arctan (c x))}{7 x^7}-\frac {2 d e (a+b \arctan (c x))}{5 x^5}-\frac {e^2 (a+b \arctan (c x))}{3 x^3}-\frac {1}{105} b c^3 \left (15 c^4 d^2-42 c^2 d e+35 e^2\right ) \log (x)+\frac {1}{210} b c^3 \left (15 c^4 d^2-42 c^2 d e+35 e^2\right ) \log \left (1+c^2 x^2\right ) \]
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Time = 0.15 (sec) , antiderivative size = 186, 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} \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^8} \, dx=-\frac {d^2 (a+b \arctan (c x))}{7 x^7}-\frac {2 d e (a+b \arctan (c x))}{5 x^5}-\frac {e^2 (a+b \arctan (c x))}{3 x^3}+\frac {b c d \left (5 c^2 d-14 e\right )}{140 x^4}-\frac {b c \left (15 c^4 d^2-42 c^2 d e+35 e^2\right )}{210 x^2}+\frac {1}{210} b c^3 \left (15 c^4 d^2-42 c^2 d e+35 e^2\right ) \log \left (c^2 x^2+1\right )-\frac {1}{105} b c^3 \log (x) \left (15 c^4 d^2-42 c^2 d e+35 e^2\right )-\frac {b c d^2}{42 x^6} \]
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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))}{7 x^7}-\frac {2 d e (a+b \arctan (c x))}{5 x^5}-\frac {e^2 (a+b \arctan (c x))}{3 x^3}-(b c) \int \frac {-15 d^2-42 d e x^2-35 e^2 x^4}{105 x^7 \left (1+c^2 x^2\right )} \, dx \\ & = -\frac {d^2 (a+b \arctan (c x))}{7 x^7}-\frac {2 d e (a+b \arctan (c x))}{5 x^5}-\frac {e^2 (a+b \arctan (c x))}{3 x^3}-\frac {1}{105} (b c) \int \frac {-15 d^2-42 d e x^2-35 e^2 x^4}{x^7 \left (1+c^2 x^2\right )} \, dx \\ & = -\frac {d^2 (a+b \arctan (c x))}{7 x^7}-\frac {2 d e (a+b \arctan (c x))}{5 x^5}-\frac {e^2 (a+b \arctan (c x))}{3 x^3}-\frac {1}{210} (b c) \text {Subst}\left (\int \frac {-15 d^2-42 d e x-35 e^2 x^2}{x^4 \left (1+c^2 x\right )} \, dx,x,x^2\right ) \\ & = -\frac {d^2 (a+b \arctan (c x))}{7 x^7}-\frac {2 d e (a+b \arctan (c x))}{5 x^5}-\frac {e^2 (a+b \arctan (c x))}{3 x^3}-\frac {1}{210} (b c) \text {Subst}\left (\int \left (-\frac {15 d^2}{x^4}+\frac {3 d \left (5 c^2 d-14 e\right )}{x^3}+\frac {-15 c^4 d^2+42 c^2 d e-35 e^2}{x^2}+\frac {15 c^6 d^2-42 c^4 d e+35 c^2 e^2}{x}+\frac {-15 c^8 d^2+42 c^6 d e-35 c^4 e^2}{1+c^2 x}\right ) \, dx,x,x^2\right ) \\ & = -\frac {b c d^2}{42 x^6}+\frac {b c d \left (5 c^2 d-14 e\right )}{140 x^4}-\frac {b c \left (15 c^4 d^2-42 c^2 d e+35 e^2\right )}{210 x^2}-\frac {d^2 (a+b \arctan (c x))}{7 x^7}-\frac {2 d e (a+b \arctan (c x))}{5 x^5}-\frac {e^2 (a+b \arctan (c x))}{3 x^3}-\frac {1}{105} b c^3 \left (15 c^4 d^2-42 c^2 d e+35 e^2\right ) \log (x)+\frac {1}{210} b c^3 \left (15 c^4 d^2-42 c^2 d e+35 e^2\right ) \log \left (1+c^2 x^2\right ) \\ \end{align*}
Time = 0.14 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.95 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^8} \, dx=\frac {1}{420} \left (-\frac {60 d^2 (a+b \arctan (c x))}{x^7}-\frac {168 d e (a+b \arctan (c x))}{x^5}-\frac {140 e^2 (a+b \arctan (c x))}{x^3}-70 b c e^2 \left (\frac {1}{x^2}+2 c^2 \log (x)-c^2 \log \left (1+c^2 x^2\right )\right )-42 b c d e \left (\frac {1}{x^4}-\frac {2 c^2}{x^2}-4 c^4 \log (x)+2 c^4 \log \left (1+c^2 x^2\right )\right )-5 b c d^2 \left (\frac {2-3 c^2 x^2+6 c^4 x^4}{x^6}+12 c^6 \log (x)-6 c^6 \log \left (1+c^2 x^2\right )\right )\right ) \]
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Time = 0.26 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.09
method | result | size |
parts | \(a \left (-\frac {d^{2}}{7 x^{7}}-\frac {e^{2}}{3 x^{3}}-\frac {2 e d}{5 x^{5}}\right )+b \,c^{7} \left (-\frac {\arctan \left (c x \right ) d^{2}}{7 c^{7} x^{7}}-\frac {\arctan \left (c x \right ) e^{2}}{3 c^{7} x^{3}}-\frac {2 \arctan \left (c x \right ) d e}{5 c^{7} x^{5}}-\frac {-\frac {-15 c^{4} d^{2}+42 c^{2} d e -35 e^{2}}{2 c^{2} x^{2}}+\left (15 c^{4} d^{2}-42 c^{2} d e +35 e^{2}\right ) \ln \left (c x \right )+\frac {5 d^{2}}{2 c^{2} x^{6}}-\frac {3 d \left (5 c^{2} d -14 e \right )}{4 c^{2} x^{4}}+\frac {\left (-15 c^{4} d^{2}+42 c^{2} d e -35 e^{2}\right ) \ln \left (c^{2} x^{2}+1\right )}{2}}{105 c^{4}}\right )\) | \(202\) |
derivativedivides | \(c^{7} \left (\frac {a \left (-\frac {2 d e}{5 c^{3} x^{5}}-\frac {d^{2}}{7 c^{3} x^{7}}-\frac {e^{2}}{3 c^{3} x^{3}}\right )}{c^{4}}+\frac {b \left (-\frac {2 \arctan \left (c x \right ) d e}{5 c^{3} x^{5}}-\frac {\arctan \left (c x \right ) d^{2}}{7 c^{3} x^{7}}-\frac {\arctan \left (c x \right ) e^{2}}{3 c^{3} x^{3}}-\frac {\left (-15 c^{4} d^{2}+42 c^{2} d e -35 e^{2}\right ) \ln \left (c^{2} x^{2}+1\right )}{210}+\frac {-15 c^{4} d^{2}+42 c^{2} d e -35 e^{2}}{210 c^{2} x^{2}}-\frac {\left (15 c^{4} d^{2}-42 c^{2} d e +35 e^{2}\right ) \ln \left (c x \right )}{105}-\frac {d^{2}}{42 c^{2} x^{6}}+\frac {d \left (5 c^{2} d -14 e \right )}{140 c^{2} x^{4}}\right )}{c^{4}}\right )\) | \(213\) |
default | \(c^{7} \left (\frac {a \left (-\frac {2 d e}{5 c^{3} x^{5}}-\frac {d^{2}}{7 c^{3} x^{7}}-\frac {e^{2}}{3 c^{3} x^{3}}\right )}{c^{4}}+\frac {b \left (-\frac {2 \arctan \left (c x \right ) d e}{5 c^{3} x^{5}}-\frac {\arctan \left (c x \right ) d^{2}}{7 c^{3} x^{7}}-\frac {\arctan \left (c x \right ) e^{2}}{3 c^{3} x^{3}}-\frac {\left (-15 c^{4} d^{2}+42 c^{2} d e -35 e^{2}\right ) \ln \left (c^{2} x^{2}+1\right )}{210}+\frac {-15 c^{4} d^{2}+42 c^{2} d e -35 e^{2}}{210 c^{2} x^{2}}-\frac {\left (15 c^{4} d^{2}-42 c^{2} d e +35 e^{2}\right ) \ln \left (c x \right )}{105}-\frac {d^{2}}{42 c^{2} x^{6}}+\frac {d \left (5 c^{2} d -14 e \right )}{140 c^{2} x^{4}}\right )}{c^{4}}\right )\) | \(213\) |
parallelrisch | \(-\frac {60 \ln \left (x \right ) b \,c^{7} d^{2} x^{7}-30 \ln \left (c^{2} x^{2}+1\right ) b \,c^{7} d^{2} x^{7}-30 b \,c^{7} d^{2} x^{7}-168 \ln \left (x \right ) b \,c^{5} d e \,x^{7}+84 \ln \left (c^{2} x^{2}+1\right ) b \,c^{5} d e \,x^{7}+84 b \,c^{5} d e \,x^{7}+140 \ln \left (x \right ) b \,c^{3} e^{2} x^{7}-70 \ln \left (c^{2} x^{2}+1\right ) b \,c^{3} e^{2} x^{7}-70 b \,c^{3} e^{2} x^{7}+30 b \,c^{5} d^{2} x^{5}-84 b \,c^{3} d e \,x^{5}+70 b c \,e^{2} x^{5}-15 x^{3} b \,c^{3} d^{2}+140 x^{4} \arctan \left (c x \right ) b \,e^{2}+140 a \,e^{2} x^{4}+42 b c e d \,x^{3}+168 x^{2} \arctan \left (c x \right ) b d e +168 a d e \,x^{2}+10 b c \,d^{2} x +60 b \,d^{2} \arctan \left (c x \right )+60 d^{2} a}{420 x^{7}}\) | \(268\) |
risch | \(\frac {i b \left (35 x^{4} e^{2}+42 x^{2} e d +15 d^{2}\right ) \ln \left (i c x +1\right )}{210 x^{7}}-\frac {60 \ln \left (x \right ) b \,c^{7} d^{2} x^{7}-30 \ln \left (c^{2} x^{2}+1\right ) b \,c^{7} d^{2} x^{7}-168 \ln \left (x \right ) b \,c^{5} d e \,x^{7}+84 \ln \left (c^{2} x^{2}+1\right ) b \,c^{5} d e \,x^{7}+140 \ln \left (x \right ) b \,c^{3} e^{2} x^{7}-70 \ln \left (c^{2} x^{2}+1\right ) b \,c^{3} e^{2} x^{7}+30 b \,c^{5} d^{2} x^{5}-84 b \,c^{3} d e \,x^{5}+70 i b \,e^{2} \ln \left (-i c x +1\right ) x^{4}-15 x^{3} b \,c^{3} d^{2}+70 b c \,e^{2} x^{5}+30 i b \,d^{2} \ln \left (-i c x +1\right )+140 a \,e^{2} x^{4}+42 b c e d \,x^{3}+84 i b d e \ln \left (-i c x +1\right ) x^{2}+168 a d e \,x^{2}+10 b c \,d^{2} x +60 d^{2} a}{420 x^{7}}\) | \(285\) |
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Time = 0.27 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.04 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^8} \, dx=\frac {2 \, {\left (15 \, b c^{7} d^{2} - 42 \, b c^{5} d e + 35 \, b c^{3} e^{2}\right )} x^{7} \log \left (c^{2} x^{2} + 1\right ) - 4 \, {\left (15 \, b c^{7} d^{2} - 42 \, b c^{5} d e + 35 \, b c^{3} e^{2}\right )} x^{7} \log \left (x\right ) - 140 \, a e^{2} x^{4} - 2 \, {\left (15 \, b c^{5} d^{2} - 42 \, b c^{3} d e + 35 \, b c e^{2}\right )} x^{5} - 10 \, b c d^{2} x - 168 \, a d e x^{2} + 3 \, {\left (5 \, b c^{3} d^{2} - 14 \, b c d e\right )} x^{3} - 60 \, a d^{2} - 4 \, {\left (35 \, b e^{2} x^{4} + 42 \, b d e x^{2} + 15 \, b d^{2}\right )} \arctan \left (c x\right )}{420 \, x^{7}} \]
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Time = 0.72 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.55 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^8} \, dx=\begin {cases} - \frac {a d^{2}}{7 x^{7}} - \frac {2 a d e}{5 x^{5}} - \frac {a e^{2}}{3 x^{3}} - \frac {b c^{7} d^{2} \log {\left (x \right )}}{7} + \frac {b c^{7} d^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{14} - \frac {b c^{5} d^{2}}{14 x^{2}} + \frac {2 b c^{5} d e \log {\left (x \right )}}{5} - \frac {b c^{5} d e \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{5} + \frac {b c^{3} d^{2}}{28 x^{4}} + \frac {b c^{3} d e}{5 x^{2}} - \frac {b c^{3} e^{2} \log {\left (x \right )}}{3} + \frac {b c^{3} e^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{6} - \frac {b c d^{2}}{42 x^{6}} - \frac {b c d e}{10 x^{4}} - \frac {b c e^{2}}{6 x^{2}} - \frac {b d^{2} \operatorname {atan}{\left (c x \right )}}{7 x^{7}} - \frac {2 b d e \operatorname {atan}{\left (c x \right )}}{5 x^{5}} - \frac {b e^{2} \operatorname {atan}{\left (c x \right )}}{3 x^{3}} & \text {for}\: c \neq 0 \\a \left (- \frac {d^{2}}{7 x^{7}} - \frac {2 d e}{5 x^{5}} - \frac {e^{2}}{3 x^{3}}\right ) & \text {otherwise} \end {cases} \]
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Time = 0.21 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.06 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^8} \, dx=\frac {1}{84} \, {\left ({\left (6 \, c^{6} \log \left (c^{2} x^{2} + 1\right ) - 6 \, c^{6} \log \left (x^{2}\right ) - \frac {6 \, c^{4} x^{4} - 3 \, c^{2} x^{2} + 2}{x^{6}}\right )} c - \frac {12 \, \arctan \left (c x\right )}{x^{7}}\right )} b d^{2} - \frac {1}{10} \, {\left ({\left (2 \, c^{4} \log \left (c^{2} x^{2} + 1\right ) - 2 \, c^{4} \log \left (x^{2}\right ) - \frac {2 \, c^{2} x^{2} - 1}{x^{4}}\right )} c + \frac {4 \, \arctan \left (c x\right )}{x^{5}}\right )} b d e + \frac {1}{6} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} + 1\right ) - c^{2} \log \left (x^{2}\right ) - \frac {1}{x^{2}}\right )} c - \frac {2 \, \arctan \left (c x\right )}{x^{3}}\right )} b e^{2} - \frac {a e^{2}}{3 \, x^{3}} - \frac {2 \, a d e}{5 \, x^{5}} - \frac {a d^{2}}{7 \, x^{7}} \]
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\[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^8} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{2} {\left (b \arctan \left (c x\right ) + a\right )}}{x^{8}} \,d x } \]
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Time = 0.71 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.25 \[ \int \frac {\left (d+e x^2\right )^2 (a+b \arctan (c x))}{x^8} \, dx=-\frac {60\,a\,d^2+60\,b\,d^2\,\mathrm {atan}\left (c\,x\right )+140\,a\,e^2\,x^4-15\,b\,c^3\,d^2\,x^3+30\,b\,c^5\,d^2\,x^5+10\,b\,c\,d^2\,x+168\,a\,d\,e\,x^2+70\,b\,c\,e^2\,x^5+140\,b\,e^2\,x^4\,\mathrm {atan}\left (c\,x\right )+60\,b\,c^7\,d^2\,x^7\,\ln \left (x\right )+140\,b\,c^3\,e^2\,x^7\,\ln \left (x\right )-84\,b\,c^3\,d\,e\,x^5+42\,b\,c\,d\,e\,x^3-30\,b\,c^7\,d^2\,x^7\,\ln \left (c^2\,x^2+1\right )-70\,b\,c^3\,e^2\,x^7\,\ln \left (c^2\,x^2+1\right )+168\,b\,d\,e\,x^2\,\mathrm {atan}\left (c\,x\right )-168\,b\,c^5\,d\,e\,x^7\,\ln \left (x\right )+84\,b\,c^5\,d\,e\,x^7\,\ln \left (c^2\,x^2+1\right )}{420\,x^7} \]
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