Integrand size = 32, antiderivative size = 119 \[ \int (f x)^{-1-(1+q) r} \left (d+e x^r\right )^q \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {b n (f x)^{-((1+q) r)} \left (d+e x^r\right )^q \left (1+\frac {e x^r}{d}\right )^{-q} \operatorname {Hypergeometric2F1}\left (-1-q,-1-q,-q,-\frac {e x^r}{d}\right )}{f (1+q)^2 r^2}-\frac {(f x)^{-((1+q) r)} \left (d+e x^r\right )^{1+q} \left (a+b \log \left (c x^n\right )\right )}{d f (1+q) r} \]
[Out]
Time = 0.09 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {2373, 372, 371} \[ \int (f x)^{-1-(1+q) r} \left (d+e x^r\right )^q \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {(f x)^{-((q+1) r)} \left (d+e x^r\right )^{q+1} \left (a+b \log \left (c x^n\right )\right )}{d f (q+1) r}-\frac {b n (f x)^{-((q+1) r)} \left (d+e x^r\right )^q \left (\frac {e x^r}{d}+1\right )^{-q} \operatorname {Hypergeometric2F1}\left (-q-1,-q-1,-q,-\frac {e x^r}{d}\right )}{f (q+1)^2 r^2} \]
[In]
[Out]
Rule 371
Rule 372
Rule 2373
Rubi steps \begin{align*} \text {integral}& = -\frac {(f x)^{-((1+q) r)} \left (d+e x^r\right )^{1+q} \left (a+b \log \left (c x^n\right )\right )}{d f (1+q) r}+\frac {(b n) \int (f x)^{-1-(1+q) r} \left (d+e x^r\right )^{1+q} \, dx}{d (1+q) r} \\ & = -\frac {(f x)^{-((1+q) r)} \left (d+e x^r\right )^{1+q} \left (a+b \log \left (c x^n\right )\right )}{d f (1+q) r}+\frac {\left (b n \left (d+e x^r\right )^q \left (1+\frac {e x^r}{d}\right )^{-q}\right ) \int (f x)^{-1-(1+q) r} \left (1+\frac {e x^r}{d}\right )^{1+q} \, dx}{(1+q) r} \\ & = -\frac {b n (f x)^{-((1+q) r)} \left (d+e x^r\right )^q \left (1+\frac {e x^r}{d}\right )^{-q} \, _2F_1\left (-1-q,-1-q;-q;-\frac {e x^r}{d}\right )}{f (1+q)^2 r^2}-\frac {(f x)^{-((1+q) r)} \left (d+e x^r\right )^{1+q} \left (a+b \log \left (c x^n\right )\right )}{d f (1+q) r} \\ \end{align*}
Time = 0.46 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.82 \[ \int (f x)^{-1-(1+q) r} \left (d+e x^r\right )^q \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {(f x)^{-((1+q) r)} \left (d+e x^r\right )^q \left (b n \left (1+\frac {e x^r}{d}\right )^{-q} \operatorname {Hypergeometric2F1}\left (-1-q,-1-q,-q,-\frac {e x^r}{d}\right )+\frac {(1+q) r \left (d+e x^r\right ) \left (a+b \log \left (c x^n\right )\right )}{d}\right )}{f (1+q)^2 r^2} \]
[In]
[Out]
\[\int \left (f x \right )^{-1-\left (1+q \right ) r} \left (d +e \,x^{r}\right )^{q} \left (a +b \ln \left (c \,x^{n}\right )\right )d x\]
[In]
[Out]
\[ \int (f x)^{-1-(1+q) r} \left (d+e x^r\right )^q \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} {\left (e x^{r} + d\right )}^{q} \left (f x\right )^{-{\left (q + 1\right )} r - 1} \,d x } \]
[In]
[Out]
Timed out. \[ \int (f x)^{-1-(1+q) r} \left (d+e x^r\right )^q \left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Timed out} \]
[In]
[Out]
\[ \int (f x)^{-1-(1+q) r} \left (d+e x^r\right )^q \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} {\left (e x^{r} + d\right )}^{q} \left (f x\right )^{-{\left (q + 1\right )} r - 1} \,d x } \]
[In]
[Out]
\[ \int (f x)^{-1-(1+q) r} \left (d+e x^r\right )^q \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { {\left (b \log \left (c x^{n}\right ) + a\right )} {\left (e x^{r} + d\right )}^{q} \left (f x\right )^{-{\left (q + 1\right )} r - 1} \,d x } \]
[In]
[Out]
Timed out. \[ \int (f x)^{-1-(1+q) r} \left (d+e x^r\right )^q \left (a+b \log \left (c x^n\right )\right ) \, dx=\int \frac {{\left (d+e\,x^r\right )}^q\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (f\,x\right )}^{r\,\left (q+1\right )+1}} \,d x \]
[In]
[Out]