Integrand size = 23, antiderivative size = 61 \[ \int \frac {f^{a+4 b x}}{c+d f^{e+2 b x}} \, dx=\frac {f^{a-e+2 b x}}{2 b d \log (f)}-\frac {c f^{a-2 e} \log \left (c+d f^{e+2 b x}\right )}{2 b d^2 \log (f)} \]
[Out]
Time = 0.04 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2280, 45} \[ \int \frac {f^{a+4 b x}}{c+d f^{e+2 b x}} \, dx=\frac {f^{a+2 b x-e}}{2 b d \log (f)}-\frac {c f^{a-2 e} \log \left (d f^{2 b x+e}+c\right )}{2 b d^2 \log (f)} \]
[In]
[Out]
Rule 45
Rule 2280
Rubi steps \begin{align*} \text {integral}& = \frac {f^{a-2 e} \text {Subst}\left (\int \frac {x}{c+d x} \, dx,x,f^{e+2 b x}\right )}{2 b \log (f)} \\ & = \frac {f^{a-2 e} \text {Subst}\left (\int \left (\frac {1}{d}-\frac {c}{d (c+d x)}\right ) \, dx,x,f^{e+2 b x}\right )}{2 b \log (f)} \\ & = \frac {f^{a-e+2 b x}}{2 b d \log (f)}-\frac {c f^{a-2 e} \log \left (c+d f^{e+2 b x}\right )}{2 b d^2 \log (f)} \\ \end{align*}
Time = 0.31 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.20 \[ \int \frac {f^{a+4 b x}}{c+d f^{e+2 b x}} \, dx=\frac {f^{a-2 e} \left (d f^{e+2 b x}-2 b c x \log (f)-c \log \left (c+d f^{e+2 b x}\right )+c \log \left (b d^3 f^{e+2 b x} \log (f)\right )\right )}{2 b d^2 \log (f)} \]
[In]
[Out]
Time = 0.04 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.25
method | result | size |
norman | \(\frac {f^{-2 e} f^{a} {\mathrm e}^{\left (2 b x +e \right ) \ln \left (f \right )}}{2 \ln \left (f \right ) b d}-\frac {c \,f^{-2 e} f^{a} \ln \left (c +d \,{\mathrm e}^{\left (2 b x +e \right ) \ln \left (f \right )}\right )}{2 \ln \left (f \right ) b \,d^{2}}\) | \(76\) |
risch | \(\frac {f^{a} f^{-2 e} f^{2 b x +e}}{2 \ln \left (f \right ) b d}+\frac {f^{a} f^{-2 e} c e}{2 b \,d^{2}}-\frac {f^{a} f^{-2 e} c \ln \left (f^{2 b x +e}+\frac {c}{d}\right )}{2 \ln \left (f \right ) b \,d^{2}}\) | \(96\) |
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.87 \[ \int \frac {f^{a+4 b x}}{c+d f^{e+2 b x}} \, dx=\frac {d f^{2 \, b x + e} f^{a - 2 \, e} - c f^{a - 2 \, e} \log \left (d f^{2 \, b x + e} + c\right )}{2 \, b d^{2} \log \left (f\right )} \]
[In]
[Out]
Time = 0.55 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.51 \[ \int \frac {f^{a+4 b x}}{c+d f^{e+2 b x}} \, dx=\frac {\left (\begin {cases} x & \text {for}\: b = 0 \vee f = 1 \\\frac {e^{2 b x \log {\left (f \right )}}}{2 b \log {\left (f \right )}} & \text {otherwise} \end {cases}\right ) e^{a \log {\left (f \right )}} e^{- e \log {\left (f \right )}}}{d} - \frac {c e^{\left (a - 2 e\right ) \log {\left (f \right )}} \log {\left (\frac {c e^{\frac {a \log {\left (f \right )}}{2}} e^{- e \log {\left (f \right )}}}{d} + \sqrt {e^{\left (a + 4 b x\right ) \log {\left (f \right )}}} \right )}}{2 b d^{2} \log {\left (f \right )}} \]
[In]
[Out]
none
Time = 0.18 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.07 \[ \int \frac {f^{a+4 b x}}{c+d f^{e+2 b x}} \, dx=-\frac {c f^{a - 2 \, e} \log \left (d f^{2 \, b x + e} + c\right )}{2 \, b d^{2} \log \left (f\right )} + \frac {{\left (d f^{2 \, b x + e} + c\right )} f^{a - 2 \, e}}{2 \, b d^{2} \log \left (f\right )} \]
[In]
[Out]
none
Time = 0.29 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.07 \[ \int \frac {f^{a+4 b x}}{c+d f^{e+2 b x}} \, dx=\frac {f^{a} {\left (\frac {f^{2 \, b x} f^{e}}{d f^{2 \, e} \log \left (f\right )} - \frac {c \log \left ({\left | d f^{2 \, b x} f^{e} + c \right |}\right )}{d^{2} f^{2 \, e} \log \left (f\right )}\right )}}{2 \, b} \]
[In]
[Out]
Time = 0.18 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.77 \[ \int \frac {f^{a+4 b x}}{c+d f^{e+2 b x}} \, dx=-\frac {f^{a-2\,e}\,\left (\frac {c\,\ln \left (c+d\,f^{e+2\,b\,x}\right )}{2}-\frac {d\,f^{e+2\,b\,x}}{2}\right )}{b\,d^2\,\ln \left (f\right )} \]
[In]
[Out]