Integrand size = 16, antiderivative size = 184 \[ \int (d+e x)^4 (a+b \arctan (c x)) \, dx=-\frac {b d e \left (2 c^2 d^2-e^2\right ) x}{c^3}-\frac {b e^2 \left (10 c^2 d^2-e^2\right ) x^2}{10 c^3}-\frac {b d e^3 x^3}{3 c}-\frac {b e^4 x^4}{20 c}-\frac {b d \left (c^4 d^4-10 c^2 d^2 e^2+5 e^4\right ) \arctan (c x)}{5 c^4 e}+\frac {(d+e x)^5 (a+b \arctan (c x))}{5 e}-\frac {b \left (5 c^4 d^4-10 c^2 d^2 e^2+e^4\right ) \log \left (1+c^2 x^2\right )}{10 c^5} \]
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Time = 0.13 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {4972, 716, 649, 209, 266} \[ \int (d+e x)^4 (a+b \arctan (c x)) \, dx=\frac {(d+e x)^5 (a+b \arctan (c x))}{5 e}-\frac {b d \arctan (c x) \left (c^4 d^4-10 c^2 d^2 e^2+5 e^4\right )}{5 c^4 e}-\frac {b e^2 x^2 \left (10 c^2 d^2-e^2\right )}{10 c^3}-\frac {b d e x \left (2 c^2 d^2-e^2\right )}{c^3}-\frac {b \left (5 c^4 d^4-10 c^2 d^2 e^2+e^4\right ) \log \left (c^2 x^2+1\right )}{10 c^5}-\frac {b d e^3 x^3}{3 c}-\frac {b e^4 x^4}{20 c} \]
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Rule 209
Rule 266
Rule 649
Rule 716
Rule 4972
Rubi steps \begin{align*} \text {integral}& = \frac {(d+e x)^5 (a+b \arctan (c x))}{5 e}-\frac {(b c) \int \frac {(d+e x)^5}{1+c^2 x^2} \, dx}{5 e} \\ & = \frac {(d+e x)^5 (a+b \arctan (c x))}{5 e}-\frac {(b c) \int \left (\frac {5 d e^2 \left (2 c^2 d^2-e^2\right )}{c^4}+\frac {e^3 \left (10 c^2 d^2-e^2\right ) x}{c^4}+\frac {5 d e^4 x^2}{c^2}+\frac {e^5 x^3}{c^2}+\frac {c^4 d^5-10 c^2 d^3 e^2+5 d e^4+e \left (5 c^4 d^4-10 c^2 d^2 e^2+e^4\right ) x}{c^4 \left (1+c^2 x^2\right )}\right ) \, dx}{5 e} \\ & = -\frac {b d e \left (2 c^2 d^2-e^2\right ) x}{c^3}-\frac {b e^2 \left (10 c^2 d^2-e^2\right ) x^2}{10 c^3}-\frac {b d e^3 x^3}{3 c}-\frac {b e^4 x^4}{20 c}+\frac {(d+e x)^5 (a+b \arctan (c x))}{5 e}-\frac {b \int \frac {c^4 d^5-10 c^2 d^3 e^2+5 d e^4+e \left (5 c^4 d^4-10 c^2 d^2 e^2+e^4\right ) x}{1+c^2 x^2} \, dx}{5 c^3 e} \\ & = -\frac {b d e \left (2 c^2 d^2-e^2\right ) x}{c^3}-\frac {b e^2 \left (10 c^2 d^2-e^2\right ) x^2}{10 c^3}-\frac {b d e^3 x^3}{3 c}-\frac {b e^4 x^4}{20 c}+\frac {(d+e x)^5 (a+b \arctan (c x))}{5 e}-\frac {\left (b \left (5 c^4 d^4-10 c^2 d^2 e^2+e^4\right )\right ) \int \frac {x}{1+c^2 x^2} \, dx}{5 c^3}-\frac {\left (b d \left (c^4 d^4-10 c^2 d^2 e^2+5 e^4\right )\right ) \int \frac {1}{1+c^2 x^2} \, dx}{5 c^3 e} \\ & = -\frac {b d e \left (2 c^2 d^2-e^2\right ) x}{c^3}-\frac {b e^2 \left (10 c^2 d^2-e^2\right ) x^2}{10 c^3}-\frac {b d e^3 x^3}{3 c}-\frac {b e^4 x^4}{20 c}-\frac {b d \left (c^4 d^4-10 c^2 d^2 e^2+5 e^4\right ) \arctan (c x)}{5 c^4 e}+\frac {(d+e x)^5 (a+b \arctan (c x))}{5 e}-\frac {b \left (5 c^4 d^4-10 c^2 d^2 e^2+e^4\right ) \log \left (1+c^2 x^2\right )}{10 c^5} \\ \end{align*}
Time = 0.52 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.39 \[ \int (d+e x)^4 (a+b \arctan (c x)) \, dx=\frac {(d+e x)^5 (a+b \arctan (c x))-\frac {b \left (c^2 e^2 x \left (-6 e^2 (10 d+e x)+c^2 \left (120 d^3+60 d^2 e x+20 d e^2 x^2+3 e^3 x^3\right )\right )+6 \left (-10 c^2 d^2 e^2 \left (\sqrt {-c^2} d+e\right )+e^4 \left (5 \sqrt {-c^2} d+e\right )+c^4 d^4 \left (\sqrt {-c^2} d+5 e\right )\right ) \log \left (1-\sqrt {-c^2} x\right )-6 \left (c^4 d^4 \left (\sqrt {-c^2} d-5 e\right )-10 c^2 d^2 \left (\sqrt {-c^2} d-e\right ) e^2+\left (5 \sqrt {-c^2} d-e\right ) e^4\right ) \log \left (1+\sqrt {-c^2} x\right )\right )}{12 c^5}}{5 e} \]
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Time = 2.57 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.34
method | result | size |
parts | \(\frac {a \left (e x +d \right )^{5}}{5 e}+\frac {b \left (\frac {c \,e^{4} \arctan \left (c x \right ) x^{5}}{5}+c \,e^{3} \arctan \left (c x \right ) x^{4} d +2 c \,e^{2} \arctan \left (c x \right ) x^{3} d^{2}+2 c e \arctan \left (c x \right ) x^{2} d^{3}+\arctan \left (c x \right ) c x \,d^{4}+\frac {c \arctan \left (c x \right ) d^{5}}{5 e}-\frac {10 c^{4} d^{3} e^{2} x +5 c^{4} d^{2} e^{3} x^{2}+\frac {5 c^{4} d \,e^{4} x^{3}}{3}+\frac {e^{5} c^{4} x^{4}}{4}-5 c^{2} d \,e^{4} x -\frac {e^{5} c^{2} x^{2}}{2}+\frac {\left (5 c^{4} d^{4} e -10 c^{2} d^{2} e^{3}+e^{5}\right ) \ln \left (c^{2} x^{2}+1\right )}{2}+\left (c^{5} d^{5}-10 c^{3} d^{3} e^{2}+5 c d \,e^{4}\right ) \arctan \left (c x \right )}{5 c^{4} e}\right )}{c}\) | \(246\) |
derivativedivides | \(\frac {\frac {a \left (c e x +c d \right )^{5}}{5 c^{4} e}+\frac {b \left (\frac {\arctan \left (c x \right ) c^{5} d^{5}}{5 e}+\arctan \left (c x \right ) c^{5} d^{4} x +2 e \arctan \left (c x \right ) c^{5} d^{3} x^{2}+2 e^{2} \arctan \left (c x \right ) c^{5} d^{2} x^{3}+e^{3} \arctan \left (c x \right ) c^{5} d \,x^{4}+\frac {e^{4} \arctan \left (c x \right ) c^{5} x^{5}}{5}-\frac {10 c^{4} d^{3} e^{2} x +5 c^{4} d^{2} e^{3} x^{2}+\frac {5 c^{4} d \,e^{4} x^{3}}{3}+\frac {e^{5} c^{4} x^{4}}{4}-5 c^{2} d \,e^{4} x -\frac {e^{5} c^{2} x^{2}}{2}+\frac {\left (5 c^{4} d^{4} e -10 c^{2} d^{2} e^{3}+e^{5}\right ) \ln \left (c^{2} x^{2}+1\right )}{2}+\left (c^{5} d^{5}-10 c^{3} d^{3} e^{2}+5 c d \,e^{4}\right ) \arctan \left (c x \right )}{5 e}\right )}{c^{4}}}{c}\) | \(265\) |
default | \(\frac {\frac {a \left (c e x +c d \right )^{5}}{5 c^{4} e}+\frac {b \left (\frac {\arctan \left (c x \right ) c^{5} d^{5}}{5 e}+\arctan \left (c x \right ) c^{5} d^{4} x +2 e \arctan \left (c x \right ) c^{5} d^{3} x^{2}+2 e^{2} \arctan \left (c x \right ) c^{5} d^{2} x^{3}+e^{3} \arctan \left (c x \right ) c^{5} d \,x^{4}+\frac {e^{4} \arctan \left (c x \right ) c^{5} x^{5}}{5}-\frac {10 c^{4} d^{3} e^{2} x +5 c^{4} d^{2} e^{3} x^{2}+\frac {5 c^{4} d \,e^{4} x^{3}}{3}+\frac {e^{5} c^{4} x^{4}}{4}-5 c^{2} d \,e^{4} x -\frac {e^{5} c^{2} x^{2}}{2}+\frac {\left (5 c^{4} d^{4} e -10 c^{2} d^{2} e^{3}+e^{5}\right ) \ln \left (c^{2} x^{2}+1\right )}{2}+\left (c^{5} d^{5}-10 c^{3} d^{3} e^{2}+5 c d \,e^{4}\right ) \arctan \left (c x \right )}{5 e}\right )}{c^{4}}}{c}\) | \(265\) |
parallelrisch | \(-\frac {-12 x^{5} \arctan \left (c x \right ) b \,c^{5} e^{4}-12 x^{5} a \,c^{5} e^{4}-60 x^{4} \arctan \left (c x \right ) b \,c^{5} d \,e^{3}-60 x^{4} a \,c^{5} d \,e^{3}-120 x^{3} \arctan \left (c x \right ) b \,c^{5} d^{2} e^{2}+3 x^{4} b \,c^{4} e^{4}-120 x^{3} a \,c^{5} d^{2} e^{2}-120 x^{2} \arctan \left (c x \right ) b \,c^{5} d^{3} e +20 x^{3} b \,c^{4} d \,e^{3}-120 x^{2} a \,c^{5} d^{3} e -60 x \arctan \left (c x \right ) b \,c^{5} d^{4}+60 x^{2} b \,c^{4} d^{2} e^{2}-60 x a \,c^{5} d^{4}+30 \ln \left (c^{2} x^{2}+1\right ) b \,c^{4} d^{4}+120 x b \,c^{4} d^{3} e -6 x^{2} b \,c^{2} e^{4}-120 \arctan \left (c x \right ) b \,c^{3} d^{3} e -60 \ln \left (c^{2} x^{2}+1\right ) b \,c^{2} d^{2} e^{2}-60 x b \,c^{2} d \,e^{3}+60 \arctan \left (c x \right ) b c d \,e^{3}+6 \ln \left (c^{2} x^{2}+1\right ) b \,e^{4}}{60 c^{5}}\) | \(310\) |
risch | \(\frac {i b \,d^{5} \ln \left (c^{2} x^{2}+1\right )}{20 e}+i e b \,d^{3} x^{2} \ln \left (-i c x +1\right )-\frac {i \left (e x +d \right )^{5} b \ln \left (i c x +1\right )}{10 e}+\frac {i b \,d^{4} x \ln \left (-i c x +1\right )}{2}+\frac {x^{5} e^{4} a}{5}+\frac {i e^{3} b d \,x^{4} \ln \left (-i c x +1\right )}{2}+x^{4} e^{3} d a +i e^{2} b \,d^{2} x^{3} \ln \left (-i c x +1\right )+2 x^{3} e^{2} d^{2} a -\frac {b \,e^{4} x^{4}}{20 c}+\frac {i e^{4} b \,x^{5} \ln \left (-i c x +1\right )}{10}+2 x^{2} e \,d^{3} a -\frac {b d \,e^{3} x^{3}}{3 c}-\frac {b \,d^{5} \arctan \left (c x \right )}{10 e}+x a \,d^{4}-\frac {e^{2} b \,d^{2} x^{2}}{c}-\frac {b \,d^{4} \ln \left (c^{2} x^{2}+1\right )}{2 c}-\frac {2 e b \,d^{3} x}{c}+\frac {2 e b \,d^{3} \arctan \left (c x \right )}{c^{2}}+\frac {e^{4} b \,x^{2}}{10 c^{3}}+\frac {e^{2} b \,d^{2} \ln \left (c^{2} x^{2}+1\right )}{c^{3}}+\frac {e^{3} b d x}{c^{3}}-\frac {e^{3} b d \arctan \left (c x \right )}{c^{4}}-\frac {e^{4} b \ln \left (c^{2} x^{2}+1\right )}{10 c^{5}}\) | \(356\) |
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Time = 0.26 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.43 \[ \int (d+e x)^4 (a+b \arctan (c x)) \, dx=\frac {12 \, a c^{5} e^{4} x^{5} + 3 \, {\left (20 \, a c^{5} d e^{3} - b c^{4} e^{4}\right )} x^{4} + 20 \, {\left (6 \, a c^{5} d^{2} e^{2} - b c^{4} d e^{3}\right )} x^{3} + 6 \, {\left (20 \, a c^{5} d^{3} e - 10 \, b c^{4} d^{2} e^{2} + b c^{2} e^{4}\right )} x^{2} + 60 \, {\left (a c^{5} d^{4} - 2 \, b c^{4} d^{3} e + b c^{2} d e^{3}\right )} x + 12 \, {\left (b c^{5} e^{4} x^{5} + 5 \, b c^{5} d e^{3} x^{4} + 10 \, b c^{5} d^{2} e^{2} x^{3} + 10 \, b c^{5} d^{3} e x^{2} + 5 \, b c^{5} d^{4} x + 10 \, b c^{3} d^{3} e - 5 \, b c d e^{3}\right )} \arctan \left (c x\right ) - 6 \, {\left (5 \, b c^{4} d^{4} - 10 \, b c^{2} d^{2} e^{2} + b e^{4}\right )} \log \left (c^{2} x^{2} + 1\right )}{60 \, c^{5}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 345 vs. \(2 (170) = 340\).
Time = 0.45 (sec) , antiderivative size = 345, normalized size of antiderivative = 1.88 \[ \int (d+e x)^4 (a+b \arctan (c x)) \, dx=\begin {cases} a d^{4} x + 2 a d^{3} e x^{2} + 2 a d^{2} e^{2} x^{3} + a d e^{3} x^{4} + \frac {a e^{4} x^{5}}{5} + b d^{4} x \operatorname {atan}{\left (c x \right )} + 2 b d^{3} e x^{2} \operatorname {atan}{\left (c x \right )} + 2 b d^{2} e^{2} x^{3} \operatorname {atan}{\left (c x \right )} + b d e^{3} x^{4} \operatorname {atan}{\left (c x \right )} + \frac {b e^{4} x^{5} \operatorname {atan}{\left (c x \right )}}{5} - \frac {b d^{4} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2 c} - \frac {2 b d^{3} e x}{c} - \frac {b d^{2} e^{2} x^{2}}{c} - \frac {b d e^{3} x^{3}}{3 c} - \frac {b e^{4} x^{4}}{20 c} + \frac {2 b d^{3} e \operatorname {atan}{\left (c x \right )}}{c^{2}} + \frac {b d^{2} e^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{c^{3}} + \frac {b d e^{3} x}{c^{3}} + \frac {b e^{4} x^{2}}{10 c^{3}} - \frac {b d e^{3} \operatorname {atan}{\left (c x \right )}}{c^{4}} - \frac {b e^{4} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{10 c^{5}} & \text {for}\: c \neq 0 \\a \left (d^{4} x + 2 d^{3} e x^{2} + 2 d^{2} e^{2} x^{3} + d e^{3} x^{4} + \frac {e^{4} x^{5}}{5}\right ) & \text {otherwise} \end {cases} \]
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Time = 0.30 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.37 \[ \int (d+e x)^4 (a+b \arctan (c x)) \, dx=\frac {1}{5} \, a e^{4} x^{5} + a d e^{3} x^{4} + 2 \, a d^{2} e^{2} x^{3} + 2 \, a d^{3} e x^{2} + 2 \, {\left (x^{2} \arctan \left (c x\right ) - c {\left (\frac {x}{c^{2}} - \frac {\arctan \left (c x\right )}{c^{3}}\right )}\right )} b d^{3} e + {\left (2 \, x^{3} \arctan \left (c x\right ) - c {\left (\frac {x^{2}}{c^{2}} - \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} b d^{2} e^{2} + \frac {1}{3} \, {\left (3 \, x^{4} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{3} - 3 \, x}{c^{4}} + \frac {3 \, \arctan \left (c x\right )}{c^{5}}\right )}\right )} b d e^{3} + \frac {1}{20} \, {\left (4 \, x^{5} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b e^{4} + a d^{4} x + \frac {{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b d^{4}}{2 \, c} \]
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\[ \int (d+e x)^4 (a+b \arctan (c x)) \, dx=\int { {\left (e x + d\right )}^{4} {\left (b \arctan \left (c x\right ) + a\right )} \,d x } \]
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Time = 0.92 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.48 \[ \int (d+e x)^4 (a+b \arctan (c x)) \, dx=\frac {a\,e^4\,x^5}{5}+a\,d^4\,x-\frac {b\,d^4\,\ln \left (c^2\,x^2+1\right )}{2\,c}-\frac {b\,e^4\,\ln \left (c^2\,x^2+1\right )}{10\,c^5}+2\,a\,d^2\,e^2\,x^3-\frac {b\,e^4\,x^4}{20\,c}+\frac {b\,e^4\,x^2}{10\,c^3}+b\,d^4\,x\,\mathrm {atan}\left (c\,x\right )+2\,a\,d^3\,e\,x^2+a\,d\,e^3\,x^4+\frac {b\,e^4\,x^5\,\mathrm {atan}\left (c\,x\right )}{5}-\frac {2\,b\,d^3\,e\,x}{c}+\frac {b\,d\,e^3\,x}{c^3}+\frac {2\,b\,d^3\,e\,\mathrm {atan}\left (c\,x\right )}{c^2}-\frac {b\,d\,e^3\,\mathrm {atan}\left (c\,x\right )}{c^4}+2\,b\,d^3\,e\,x^2\,\mathrm {atan}\left (c\,x\right )+b\,d\,e^3\,x^4\,\mathrm {atan}\left (c\,x\right )-\frac {b\,d\,e^3\,x^3}{3\,c}+2\,b\,d^2\,e^2\,x^3\,\mathrm {atan}\left (c\,x\right )+\frac {b\,d^2\,e^2\,\ln \left (c^2\,x^2+1\right )}{c^3}-\frac {b\,d^2\,e^2\,x^2}{c} \]
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