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