Integrand size = 30, antiderivative size = 237 \[ \int (a+b x)^m (c+d x)^{-m-n} (e+f x)^{-3+n} \, dx=-\frac {f (a+b x)^{1+m} (c+d x)^{1-m-n} (e+f x)^{-2+n}}{(b e-a f) (d e-c f) (2-n)}-\frac {(a d f (1+m)-b (d e (2-n)-c f (1-m-n))) (a+b x)^{1+m} (c+d x)^{-m-n} \left (\frac {(b e-a f) (c+d x)}{(b c-a d) (e+f x)}\right )^{m+n} (e+f x)^{-1+n} \operatorname {Hypergeometric2F1}\left (1+m,m+n,2+m,-\frac {(d e-c f) (a+b x)}{(b c-a d) (e+f x)}\right )}{(b e-a f)^2 (d e-c f) (1+m) (2-n)} \]
[Out]
Time = 0.06 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.99, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {98, 134} \[ \int (a+b x)^m (c+d x)^{-m-n} (e+f x)^{-3+n} \, dx=-\frac {(a+b x)^{m+1} (e+f x)^{n-1} (c+d x)^{-m-n} (a d f (m+1)+b c f (-m-n+1)-b d e (2-n)) \left (\frac {(c+d x) (b e-a f)}{(e+f x) (b c-a d)}\right )^{m+n} \operatorname {Hypergeometric2F1}\left (m+1,m+n,m+2,-\frac {(d e-c f) (a+b x)}{(b c-a d) (e+f x)}\right )}{(m+1) (2-n) (b e-a f)^2 (d e-c f)}-\frac {f (a+b x)^{m+1} (e+f x)^{n-2} (c+d x)^{-m-n+1}}{(2-n) (b e-a f) (d e-c f)} \]
[In]
[Out]
Rule 98
Rule 134
Rubi steps \begin{align*} \text {integral}& = -\frac {f (a+b x)^{1+m} (c+d x)^{1-m-n} (e+f x)^{-2+n}}{(b e-a f) (d e-c f) (2-n)}-\frac {(a d f (1+m)-b d e (2-n)+b c f (1-m-n)) \int (a+b x)^m (c+d x)^{-m-n} (e+f x)^{-2+n} \, dx}{(b e-a f) (d e-c f) (2-n)} \\ & = -\frac {f (a+b x)^{1+m} (c+d x)^{1-m-n} (e+f x)^{-2+n}}{(b e-a f) (d e-c f) (2-n)}-\frac {(a d f (1+m)-b d e (2-n)+b c f (1-m-n)) (a+b x)^{1+m} (c+d x)^{-m-n} \left (\frac {(b e-a f) (c+d x)}{(b c-a d) (e+f x)}\right )^{m+n} (e+f x)^{-1+n} \, _2F_1\left (1+m,m+n;2+m;-\frac {(d e-c f) (a+b x)}{(b c-a d) (e+f x)}\right )}{(b e-a f)^2 (d e-c f) (1+m) (2-n)} \\ \end{align*}
Time = 0.19 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.78 \[ \int (a+b x)^m (c+d x)^{-m-n} (e+f x)^{-3+n} \, dx=\frac {(a+b x)^{1+m} (c+d x)^{-m-n} (e+f x)^{-2+n} \left (f (c+d x)+\frac {(a d f (1+m)+b d e (-2+n)-b c f (-1+m+n)) \left (\frac {(b e-a f) (c+d x)}{(b c-a d) (e+f x)}\right )^{m+n} (e+f x) \operatorname {Hypergeometric2F1}\left (1+m,m+n,2+m,\frac {(-d e+c f) (a+b x)}{(b c-a d) (e+f x)}\right )}{(b e-a f) (1+m)}\right )}{(b e-a f) (d e-c f) (-2+n)} \]
[In]
[Out]
\[\int \left (b x +a \right )^{m} \left (d x +c \right )^{-n -m} \left (f x +e \right )^{-3+n}d x\]
[In]
[Out]
\[ \int (a+b x)^m (c+d x)^{-m-n} (e+f x)^{-3+n} \, dx=\int { {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - n} {\left (f x + e\right )}^{n - 3} \,d x } \]
[In]
[Out]
Timed out. \[ \int (a+b x)^m (c+d x)^{-m-n} (e+f x)^{-3+n} \, dx=\text {Timed out} \]
[In]
[Out]
\[ \int (a+b x)^m (c+d x)^{-m-n} (e+f x)^{-3+n} \, dx=\int { {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - n} {\left (f x + e\right )}^{n - 3} \,d x } \]
[In]
[Out]
\[ \int (a+b x)^m (c+d x)^{-m-n} (e+f x)^{-3+n} \, dx=\int { {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - n} {\left (f x + e\right )}^{n - 3} \,d x } \]
[In]
[Out]
Timed out. \[ \int (a+b x)^m (c+d x)^{-m-n} (e+f x)^{-3+n} \, dx=\int \frac {{\left (e+f\,x\right )}^{n-3}\,{\left (a+b\,x\right )}^m}{{\left (c+d\,x\right )}^{m+n}} \,d x \]
[In]
[Out]