Integrand size = 22, antiderivative size = 187 \[ \int \frac {(a+b x)^m}{(c+d x) (e+f x)^2} \, dx=-\frac {f (a+b x)^{1+m}}{(b e-a f) (d e-c f) (e+f x)}+\frac {d^2 (a+b x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {d (a+b x)}{b c-a d}\right )}{(b c-a d) (d e-c f)^2 (1+m)}+\frac {f (a d f-b (d e (1-m)+c f m)) (a+b x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {f (a+b x)}{b e-a f}\right )}{(b e-a f)^2 (d e-c f)^2 (1+m)} \]
[Out]
Time = 0.13 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {105, 162, 70} \[ \int \frac {(a+b x)^m}{(c+d x) (e+f x)^2} \, dx=\frac {d^2 (a+b x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {d (a+b x)}{b c-a d}\right )}{(m+1) (b c-a d) (d e-c f)^2}+\frac {f (a+b x)^{m+1} (a d f-b c f m-b d e (1-m)) \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {f (a+b x)}{b e-a f}\right )}{(m+1) (b e-a f)^2 (d e-c f)^2}-\frac {f (a+b x)^{m+1}}{(e+f x) (b e-a f) (d e-c f)} \]
[In]
[Out]
Rule 70
Rule 105
Rule 162
Rubi steps \begin{align*} \text {integral}& = -\frac {f (a+b x)^{1+m}}{(b e-a f) (d e-c f) (e+f x)}-\frac {\int \frac {(a+b x)^m (a d f-b (d e+c f m)-b d f m x)}{(c+d x) (e+f x)} \, dx}{(b e-a f) (d e-c f)} \\ & = -\frac {f (a+b x)^{1+m}}{(b e-a f) (d e-c f) (e+f x)}+\frac {d^2 \int \frac {(a+b x)^m}{c+d x} \, dx}{(d e-c f)^2}+\frac {(f (a d f-b d e (1-m)-b c f m)) \int \frac {(a+b x)^m}{e+f x} \, dx}{(b e-a f) (d e-c f)^2} \\ & = -\frac {f (a+b x)^{1+m}}{(b e-a f) (d e-c f) (e+f x)}+\frac {d^2 (a+b x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {d (a+b x)}{b c-a d}\right )}{(b c-a d) (d e-c f)^2 (1+m)}+\frac {f (a d f-b d e (1-m)-b c f m) (a+b x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {f (a+b x)}{b e-a f}\right )}{(b e-a f)^2 (d e-c f)^2 (1+m)} \\ \end{align*}
Time = 0.22 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.93 \[ \int \frac {(a+b x)^m}{(c+d x) (e+f x)^2} \, dx=\frac {(a+b x)^{1+m} \left (-\frac {f}{e+f x}-\frac {d^2 (b e-a f) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {d (a+b x)}{-b c+a d}\right )}{(b c-a d) (-d e+c f) (1+m)}+\frac {f (a d f+b d e (-1+m)-b c f m) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {f (a+b x)}{-b e+a f}\right )}{(b e-a f) (d e-c f) (1+m)}\right )}{(b e-a f) (d e-c f)} \]
[In]
[Out]
\[\int \frac {\left (b x +a \right )^{m}}{\left (d x +c \right ) \left (f x +e \right )^{2}}d x\]
[In]
[Out]
\[ \int \frac {(a+b x)^m}{(c+d x) (e+f x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{m}}{{\left (d x + c\right )} {\left (f x + e\right )}^{2}} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {(a+b x)^m}{(c+d x) (e+f x)^2} \, dx=\text {Timed out} \]
[In]
[Out]
\[ \int \frac {(a+b x)^m}{(c+d x) (e+f x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{m}}{{\left (d x + c\right )} {\left (f x + e\right )}^{2}} \,d x } \]
[In]
[Out]
\[ \int \frac {(a+b x)^m}{(c+d x) (e+f x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{m}}{{\left (d x + c\right )} {\left (f x + e\right )}^{2}} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {(a+b x)^m}{(c+d x) (e+f x)^2} \, dx=\int \frac {{\left (a+b\,x\right )}^m}{{\left (e+f\,x\right )}^2\,\left (c+d\,x\right )} \,d x \]
[In]
[Out]