Integrand size = 21, antiderivative size = 61 \[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^2} \, dx=-\frac {a+b \arctan (c+d x)}{d e^2 (c+d x)}+\frac {b \log (c+d x)}{d e^2}-\frac {b \log \left (1+(c+d x)^2\right )}{2 d e^2} \]
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Time = 0.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5151, 12, 4946, 272, 36, 29, 31} \[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^2} \, dx=-\frac {a+b \arctan (c+d x)}{d e^2 (c+d x)}+\frac {b \log (c+d x)}{d e^2}-\frac {b \log \left ((c+d x)^2+1\right )}{2 d e^2} \]
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Rule 12
Rule 29
Rule 31
Rule 36
Rule 272
Rule 4946
Rule 5151
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {a+b \arctan (x)}{e^2 x^2} \, dx,x,c+d x\right )}{d} \\ & = \frac {\text {Subst}\left (\int \frac {a+b \arctan (x)}{x^2} \, dx,x,c+d x\right )}{d e^2} \\ & = -\frac {a+b \arctan (c+d x)}{d e^2 (c+d x)}+\frac {b \text {Subst}\left (\int \frac {1}{x \left (1+x^2\right )} \, dx,x,c+d x\right )}{d e^2} \\ & = -\frac {a+b \arctan (c+d x)}{d e^2 (c+d x)}+\frac {b \text {Subst}\left (\int \frac {1}{x (1+x)} \, dx,x,(c+d x)^2\right )}{2 d e^2} \\ & = -\frac {a+b \arctan (c+d x)}{d e^2 (c+d x)}+\frac {b \text {Subst}\left (\int \frac {1}{x} \, dx,x,(c+d x)^2\right )}{2 d e^2}-\frac {b \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,(c+d x)^2\right )}{2 d e^2} \\ & = -\frac {a+b \arctan (c+d x)}{d e^2 (c+d x)}+\frac {b \log (c+d x)}{d e^2}-\frac {b \log \left (1+(c+d x)^2\right )}{2 d e^2} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.85 \[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^2} \, dx=\frac {\frac {-a-b \arctan (c+d x)}{c+d x}+b \left (\log (c+d x)-\frac {1}{2} \log \left (1+(c+d x)^2\right )\right )}{d e^2} \]
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Time = 0.28 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.95
method | result | size |
derivativedivides | \(\frac {-\frac {a}{e^{2} \left (d x +c \right )}+\frac {b \left (-\frac {\arctan \left (d x +c \right )}{d x +c}+\ln \left (d x +c \right )-\frac {\ln \left (1+\left (d x +c \right )^{2}\right )}{2}\right )}{e^{2}}}{d}\) | \(58\) |
default | \(\frac {-\frac {a}{e^{2} \left (d x +c \right )}+\frac {b \left (-\frac {\arctan \left (d x +c \right )}{d x +c}+\ln \left (d x +c \right )-\frac {\ln \left (1+\left (d x +c \right )^{2}\right )}{2}\right )}{e^{2}}}{d}\) | \(58\) |
parts | \(-\frac {a}{d \,e^{2} \left (d x +c \right )}+\frac {b \left (-\frac {\arctan \left (d x +c \right )}{d x +c}+\ln \left (d x +c \right )-\frac {\ln \left (1+\left (d x +c \right )^{2}\right )}{2}\right )}{e^{2} d}\) | \(60\) |
parallelrisch | \(\frac {6 \ln \left (d x +c \right ) x b c \,d^{2}-3 \ln \left (d^{2} x^{2}+2 c d x +c^{2}+1\right ) x b c \,d^{2}+6 \ln \left (d x +c \right ) b \,c^{2} d -3 \ln \left (d^{2} x^{2}+2 c d x +c^{2}+1\right ) b \,c^{2} d +2 x a \,d^{2}-6 b \arctan \left (d x +c \right ) c d -4 a c d}{6 \left (d x +c \right ) c \,d^{2} e^{2}}\) | \(121\) |
risch | \(\frac {i b \ln \left (1+i \left (d x +c \right )\right )}{2 d \,e^{2} \left (d x +c \right )}-\frac {-2 \ln \left (-d x -c \right ) b d x +\ln \left (-d^{2} x^{2}-2 c d x -c^{2}-1\right ) b d x -2 \ln \left (-d x -c \right ) b c +\ln \left (-d^{2} x^{2}-2 c d x -c^{2}-1\right ) b c +i b \ln \left (1-i \left (d x +c \right )\right )+2 a}{2 e^{2} \left (d x +c \right ) d}\) | \(140\) |
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Time = 0.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.23 \[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^2} \, dx=-\frac {2 \, b \arctan \left (d x + c\right ) + {\left (b d x + b c\right )} \log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right ) - 2 \, {\left (b d x + b c\right )} \log \left (d x + c\right ) + 2 \, a}{2 \, {\left (d^{2} e^{2} x + c d e^{2}\right )}} \]
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Result contains complex when optimal does not.
Time = 12.22 (sec) , antiderivative size = 223, normalized size of antiderivative = 3.66 \[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^2} \, dx=\begin {cases} - \frac {a}{c d e^{2} + d^{2} e^{2} x} + \frac {b c \log {\left (\frac {c}{d} + x \right )}}{c d e^{2} + d^{2} e^{2} x} - \frac {b c \log {\left (\frac {c}{d} + x - \frac {i}{d} \right )}}{c d e^{2} + d^{2} e^{2} x} + \frac {i b c \operatorname {atan}{\left (c + d x \right )}}{c d e^{2} + d^{2} e^{2} x} + \frac {b d x \log {\left (\frac {c}{d} + x \right )}}{c d e^{2} + d^{2} e^{2} x} - \frac {b d x \log {\left (\frac {c}{d} + x - \frac {i}{d} \right )}}{c d e^{2} + d^{2} e^{2} x} + \frac {i b d x \operatorname {atan}{\left (c + d x \right )}}{c d e^{2} + d^{2} e^{2} x} - \frac {b \operatorname {atan}{\left (c + d x \right )}}{c d e^{2} + d^{2} e^{2} x} & \text {for}\: d \neq 0 \\\frac {x \left (a + b \operatorname {atan}{\left (c \right )}\right )}{c^{2} e^{2}} & \text {otherwise} \end {cases} \]
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Time = 0.21 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.51 \[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^2} \, dx=-\frac {1}{2} \, {\left (d {\left (\frac {\log \left (d^{2} x^{2} + 2 \, c d x + c^{2} + 1\right )}{d^{2} e^{2}} - \frac {2 \, \log \left (d x + c\right )}{d^{2} e^{2}}\right )} + \frac {2 \, \arctan \left (d x + c\right )}{d^{2} e^{2} x + c d e^{2}}\right )} b - \frac {a}{d^{2} e^{2} x + c d e^{2}} \]
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\[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^2} \, dx=\int { \frac {b \arctan \left (d x + c\right ) + a}{{\left (d e x + c e\right )}^{2}} \,d x } \]
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Time = 0.75 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.44 \[ \int \frac {a+b \arctan (c+d x)}{(c e+d e x)^2} \, dx=\frac {b\,\ln \left (c+d\,x\right )}{d\,e^2}-\frac {b\,\mathrm {atan}\left (c+d\,x\right )}{x\,d^2\,e^2+c\,d\,e^2}-\frac {b\,\ln \left (c^2+2\,c\,d\,x+d^2\,x^2+1\right )}{2\,d\,e^2}-\frac {a}{x\,d^2\,e^2+c\,d\,e^2} \]
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