Integrand size = 21, antiderivative size = 72 \[ \int (c e+d e x)^3 (a+b \arctan (c+d x)) \, dx=\frac {1}{4} b e^3 x-\frac {b e^3 (c+d x)^3}{12 d}-\frac {b e^3 \arctan (c+d x)}{4 d}+\frac {e^3 (c+d x)^4 (a+b \arctan (c+d x))}{4 d} \]
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Time = 0.04 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {5151, 12, 4946, 308, 209} \[ \int (c e+d e x)^3 (a+b \arctan (c+d x)) \, dx=\frac {e^3 (c+d x)^4 (a+b \arctan (c+d x))}{4 d}-\frac {b e^3 \arctan (c+d x)}{4 d}-\frac {b e^3 (c+d x)^3}{12 d}+\frac {1}{4} b e^3 x \]
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Rule 12
Rule 209
Rule 308
Rule 4946
Rule 5151
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int e^3 x^3 (a+b \arctan (x)) \, dx,x,c+d x\right )}{d} \\ & = \frac {e^3 \text {Subst}\left (\int x^3 (a+b \arctan (x)) \, dx,x,c+d x\right )}{d} \\ & = \frac {e^3 (c+d x)^4 (a+b \arctan (c+d x))}{4 d}-\frac {\left (b e^3\right ) \text {Subst}\left (\int \frac {x^4}{1+x^2} \, dx,x,c+d x\right )}{4 d} \\ & = \frac {e^3 (c+d x)^4 (a+b \arctan (c+d x))}{4 d}-\frac {\left (b e^3\right ) \text {Subst}\left (\int \left (-1+x^2+\frac {1}{1+x^2}\right ) \, dx,x,c+d x\right )}{4 d} \\ & = \frac {1}{4} b e^3 x-\frac {b e^3 (c+d x)^3}{12 d}+\frac {e^3 (c+d x)^4 (a+b \arctan (c+d x))}{4 d}-\frac {\left (b e^3\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,c+d x\right )}{4 d} \\ & = \frac {1}{4} b e^3 x-\frac {b e^3 (c+d x)^3}{12 d}-\frac {b e^3 \arctan (c+d x)}{4 d}+\frac {e^3 (c+d x)^4 (a+b \arctan (c+d x))}{4 d} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.78 \[ \int (c e+d e x)^3 (a+b \arctan (c+d x)) \, dx=\frac {e^3 \left (-\frac {1}{4} b \left (-d x+\frac {1}{3} (c+d x)^3+\arctan (c+d x)\right )+\frac {1}{4} (c+d x)^4 (a+b \arctan (c+d x))\right )}{d} \]
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Time = 5.70 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {\frac {e^{3} a \left (d x +c \right )^{4}}{4}+e^{3} b \left (\frac {\left (d x +c \right )^{4} \arctan \left (d x +c \right )}{4}-\frac {\left (d x +c \right )^{3}}{12}+\frac {d x}{4}+\frac {c}{4}-\frac {\arctan \left (d x +c \right )}{4}\right )}{d}\) | \(64\) |
default | \(\frac {\frac {e^{3} a \left (d x +c \right )^{4}}{4}+e^{3} b \left (\frac {\left (d x +c \right )^{4} \arctan \left (d x +c \right )}{4}-\frac {\left (d x +c \right )^{3}}{12}+\frac {d x}{4}+\frac {c}{4}-\frac {\arctan \left (d x +c \right )}{4}\right )}{d}\) | \(64\) |
parts | \(\frac {e^{3} a \left (d x +c \right )^{4}}{4 d}+\frac {e^{3} b \left (\frac {\left (d x +c \right )^{4} \arctan \left (d x +c \right )}{4}-\frac {\left (d x +c \right )^{3}}{12}+\frac {d x}{4}+\frac {c}{4}-\frac {\arctan \left (d x +c \right )}{4}\right )}{d}\) | \(66\) |
parallelrisch | \(\frac {3 d^{5} e^{3} b \arctan \left (d x +c \right ) x^{4}+3 x^{4} a \,d^{5} e^{3}+12 b c \,d^{4} e^{3} \arctan \left (d x +c \right ) x^{3}+12 x^{3} a c \,d^{4} e^{3}+18 x^{2} \arctan \left (d x +c \right ) b \,c^{2} d^{3} e^{3}-x^{3} b \,d^{4} e^{3}+18 x^{2} a \,c^{2} d^{3} e^{3}+12 x \arctan \left (d x +c \right ) b \,c^{3} d^{2} e^{3}-3 x^{2} b c \,d^{3} e^{3}+12 x a \,c^{3} d^{2} e^{3}+3 \arctan \left (d x +c \right ) b \,c^{4} d \,e^{3}-3 x b \,c^{2} d^{2} e^{3}-42 a \,c^{4} d \,e^{3}+9 b \,c^{3} d \,e^{3}+3 x b \,d^{2} e^{3}-18 d \,e^{3} c^{2} a -3 e^{3} b \arctan \left (d x +c \right ) d -3 b c d \,e^{3}}{12 d^{2}}\) | \(252\) |
risch | \(\frac {i e^{3} d^{2} b c \,x^{3} \ln \left (1-i \left (d x +c \right )\right )}{2}+\frac {i e^{3} d^{3} b \,x^{4} \ln \left (1-i \left (d x +c \right )\right )}{8}+\frac {i e^{3} b \,c^{3} x \ln \left (1-i \left (d x +c \right )\right )}{2}+\frac {3 i e^{3} d b \,c^{2} x^{2} \ln \left (1-i \left (d x +c \right )\right )}{4}+\frac {e^{3} a \,d^{3} x^{4}}{4}-\frac {i e^{3} \left (d x +c \right )^{4} b \ln \left (1+i \left (d x +c \right )\right )}{8 d}+\frac {3 e^{3} a \,c^{2} d \,x^{2}}{2}+\frac {i e^{3} b \,c^{4} \ln \left (d^{2} x^{2}+2 c d x +c^{2}+1\right )}{16 d}-\frac {b \,e^{3} \arctan \left (d x +c \right )}{4 d}-\frac {e^{3} b \,d^{2} x^{3}}{12}+e^{3} c^{3} a x +e^{3} a c \,d^{2} x^{3}-\frac {e^{3} b \,c^{2} x}{4}-\frac {e^{3} b c d \,x^{2}}{4}+\frac {e^{3} b \,c^{4} \arctan \left (d x +c \right )}{8 d}+\frac {b \,e^{3} x}{4}\) | \(276\) |
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Leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (64) = 128\).
Time = 0.26 (sec) , antiderivative size = 151, normalized size of antiderivative = 2.10 \[ \int (c e+d e x)^3 (a+b \arctan (c+d x)) \, dx=\frac {3 \, a d^{4} e^{3} x^{4} + {\left (12 \, a c - b\right )} d^{3} e^{3} x^{3} + 3 \, {\left (6 \, a c^{2} - b c\right )} d^{2} e^{3} x^{2} + 3 \, {\left (4 \, a c^{3} - b c^{2} + b\right )} d e^{3} x + 3 \, {\left (b d^{4} e^{3} x^{4} + 4 \, b c d^{3} e^{3} x^{3} + 6 \, b c^{2} d^{2} e^{3} x^{2} + 4 \, b c^{3} d e^{3} x + {\left (b c^{4} - b\right )} e^{3}\right )} \arctan \left (d x + c\right )}{12 \, d} \]
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Leaf count of result is larger than twice the leaf count of optimal. 231 vs. \(2 (61) = 122\).
Time = 23.08 (sec) , antiderivative size = 231, normalized size of antiderivative = 3.21 \[ \int (c e+d e x)^3 (a+b \arctan (c+d x)) \, dx=\begin {cases} a c^{3} e^{3} x + \frac {3 a c^{2} d e^{3} x^{2}}{2} + a c d^{2} e^{3} x^{3} + \frac {a d^{3} e^{3} x^{4}}{4} + \frac {b c^{4} e^{3} \operatorname {atan}{\left (c + d x \right )}}{4 d} + b c^{3} e^{3} x \operatorname {atan}{\left (c + d x \right )} + \frac {3 b c^{2} d e^{3} x^{2} \operatorname {atan}{\left (c + d x \right )}}{2} - \frac {b c^{2} e^{3} x}{4} + b c d^{2} e^{3} x^{3} \operatorname {atan}{\left (c + d x \right )} - \frac {b c d e^{3} x^{2}}{4} + \frac {b d^{3} e^{3} x^{4} \operatorname {atan}{\left (c + d x \right )}}{4} - \frac {b d^{2} e^{3} x^{3}}{12} + \frac {b e^{3} x}{4} - \frac {b e^{3} \operatorname {atan}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\c^{3} e^{3} x \left (a + b \operatorname {atan}{\left (c \right )}\right ) & \text {otherwise} \end {cases} \]
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Leaf count of result is larger than twice the leaf count of optimal. 370 vs. \(2 (64) = 128\).
Time = 0.29 (sec) , antiderivative size = 370, normalized size of antiderivative = 5.14 \[ \int (c e+d e x)^3 (a+b \arctan (c+d x)) \, dx=\frac {1}{4} \, a d^{3} e^{3} x^{4} + a c d^{2} e^{3} x^{3} + \frac {3}{2} \, a c^{2} d e^{3} x^{2} + \frac {3}{2} \, {\left (x^{2} \arctan \left (d x + c\right ) - d {\left (\frac {x}{d^{2}} + \frac {{\left (c^{2} - 1\right )} \arctan \left (\frac {d^{2} x + c d}{d}\right )}{d^{3}} - \frac {c \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{d^{3}}\right )}\right )} b c^{2} d e^{3} + \frac {1}{2} \, {\left (2 \, x^{3} \arctan \left (d x + c\right ) - d {\left (\frac {d x^{2} - 4 \, c x}{d^{3}} - \frac {2 \, {\left (c^{3} - 3 \, c\right )} \arctan \left (\frac {d^{2} x + c d}{d}\right )}{d^{4}} + \frac {{\left (3 \, c^{2} - 1\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{d^{4}}\right )}\right )} b c d^{2} e^{3} + \frac {1}{12} \, {\left (3 \, x^{4} \arctan \left (d x + c\right ) - d {\left (\frac {d^{2} x^{3} - 3 \, c d x^{2} + 3 \, {\left (3 \, c^{2} - 1\right )} x}{d^{4}} + \frac {3 \, {\left (c^{4} - 6 \, c^{2} + 1\right )} \arctan \left (\frac {d^{2} x + c d}{d}\right )}{d^{5}} - \frac {6 \, {\left (c^{3} - c\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{d^{5}}\right )}\right )} b d^{3} e^{3} + a c^{3} e^{3} x + \frac {{\left (2 \, {\left (d x + c\right )} \arctan \left (d x + c\right ) - \log \left ({\left (d x + c\right )}^{2} + 1\right )\right )} b c^{3} e^{3}}{2 \, d} \]
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\[ \int (c e+d e x)^3 (a+b \arctan (c+d x)) \, dx=\int { {\left (d e x + c e\right )}^{3} {\left (b \arctan \left (d x + c\right ) + a\right )} \,d x } \]
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Time = 0.75 (sec) , antiderivative size = 371, normalized size of antiderivative = 5.15 \[ \int (c e+d e x)^3 (a+b \arctan (c+d x)) \, dx=\mathrm {atan}\left (c+d\,x\right )\,\left (b\,c^3\,e^3\,x+\frac {3\,b\,c^2\,d\,e^3\,x^2}{2}+b\,c\,d^2\,e^3\,x^3+\frac {b\,d^3\,e^3\,x^4}{4}\right )-x^3\,\left (\frac {d^2\,e^3\,\left (b-20\,a\,c\right )}{12}+\frac {2\,a\,c\,d^2\,e^3}{3}\right )+x^2\,\left (\frac {c\,\left (\frac {d^2\,e^3\,\left (b-20\,a\,c\right )}{4}+2\,a\,c\,d^2\,e^3\right )}{d}+\frac {d\,e^3\,\left (10\,a\,c^2-b\,c+a\right )}{2}-\frac {a\,d\,e^3\,\left (4\,c^2+4\right )}{8}\right )+x\,\left (\frac {c\,e^3\,\left (20\,a\,c^2-3\,b\,c+6\,a\right )}{2}+\frac {\left (4\,c^2+4\right )\,\left (\frac {d^2\,e^3\,\left (b-20\,a\,c\right )}{4}+2\,a\,c\,d^2\,e^3\right )}{4\,d^2}-\frac {2\,c\,\left (\frac {2\,c\,\left (\frac {d^2\,e^3\,\left (b-20\,a\,c\right )}{4}+2\,a\,c\,d^2\,e^3\right )}{d}+d\,e^3\,\left (10\,a\,c^2-b\,c+a\right )-\frac {a\,d\,e^3\,\left (4\,c^2+4\right )}{4}\right )}{d}\right )+\frac {a\,d^3\,e^3\,x^4}{4}-\frac {b\,e^3\,\mathrm {atan}\left (\frac {\frac {b\,c\,e^3\,\left (c^2+1\right )\,\left (c-1\right )\,\left (c+1\right )}{4}+\frac {b\,d\,e^3\,x\,\left (c^2+1\right )\,\left (c-1\right )\,\left (c+1\right )}{4}}{\frac {b\,e^3}{4}-\frac {b\,c^4\,e^3}{4}}\right )\,\left (c^2+1\right )\,\left (c-1\right )\,\left (c+1\right )}{4\,d} \]
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