Integrand size = 15, antiderivative size = 51 \[ \int \frac {(c+d x)^n}{(a+b x)^2} \, dx=\frac {d (c+d x)^{1+n} \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,\frac {b (c+d x)}{b c-a d}\right )}{(b c-a d)^2 (1+n)} \]
[Out]
Time = 0.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {70} \[ \int \frac {(c+d x)^n}{(a+b x)^2} \, dx=\frac {d (c+d x)^{n+1} \operatorname {Hypergeometric2F1}\left (2,n+1,n+2,\frac {b (c+d x)}{b c-a d}\right )}{(n+1) (b c-a d)^2} \]
[In]
[Out]
Rule 70
Rubi steps \begin{align*} \text {integral}& = \frac {d (c+d x)^{1+n} \, _2F_1\left (2,1+n;2+n;\frac {b (c+d x)}{b c-a d}\right )}{(b c-a d)^2 (1+n)} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.02 \[ \int \frac {(c+d x)^n}{(a+b x)^2} \, dx=\frac {d (c+d x)^{1+n} \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,-\frac {b (c+d x)}{-b c+a d}\right )}{(-b c+a d)^2 (1+n)} \]
[In]
[Out]
\[\int \frac {\left (d x +c \right )^{n}}{\left (b x +a \right )^{2}}d x\]
[In]
[Out]
\[ \int \frac {(c+d x)^n}{(a+b x)^2} \, dx=\int { \frac {{\left (d x + c\right )}^{n}}{{\left (b x + a\right )}^{2}} \,d x } \]
[In]
[Out]
\[ \int \frac {(c+d x)^n}{(a+b x)^2} \, dx=\int \frac {\left (c + d x\right )^{n}}{\left (a + b x\right )^{2}}\, dx \]
[In]
[Out]
\[ \int \frac {(c+d x)^n}{(a+b x)^2} \, dx=\int { \frac {{\left (d x + c\right )}^{n}}{{\left (b x + a\right )}^{2}} \,d x } \]
[In]
[Out]
\[ \int \frac {(c+d x)^n}{(a+b x)^2} \, dx=\int { \frac {{\left (d x + c\right )}^{n}}{{\left (b x + a\right )}^{2}} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {(c+d x)^n}{(a+b x)^2} \, dx=\int \frac {{\left (c+d\,x\right )}^n}{{\left (a+b\,x\right )}^2} \,d x \]
[In]
[Out]