Integrand size = 24, antiderivative size = 51 \[ \int (a+b x)^m (c+d x)^{1+2 n-2 (1+n)} \, dx=\frac {(a+b x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {d (a+b x)}{b c-a d}\right )}{(b c-a d) (1+m)} \]
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Time = 0.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {7, 70} \[ \int (a+b x)^m (c+d x)^{1+2 n-2 (1+n)} \, dx=\frac {(a+b x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {d (a+b x)}{b c-a d}\right )}{(m+1) (b c-a d)} \]
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Rule 7
Rule 70
Rubi steps \begin{align*} \text {integral}& = \int \frac {(a+b x)^m}{c+d x} \, dx \\ & = \frac {(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) (1+m)} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00 \[ \int (a+b x)^m (c+d x)^{1+2 n-2 (1+n)} \, dx=-\frac {(a+b x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {d (a+b x)}{-b c+a d}\right )}{(-b c+a d) (1+m)} \]
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\[\int \frac {\left (b x +a \right )^{m}}{d x +c}d x\]
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\[ \int (a+b x)^m (c+d x)^{1+2 n-2 (1+n)} \, dx=\int { \frac {{\left (b x + a\right )}^{m}}{d x + c} \,d x } \]
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\[ \int (a+b x)^m (c+d x)^{1+2 n-2 (1+n)} \, dx=\int \frac {\left (a + b x\right )^{m}}{c + d x}\, dx \]
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\[ \int (a+b x)^m (c+d x)^{1+2 n-2 (1+n)} \, dx=\int { \frac {{\left (b x + a\right )}^{m}}{d x + c} \,d x } \]
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\[ \int (a+b x)^m (c+d x)^{1+2 n-2 (1+n)} \, dx=\int { \frac {{\left (b x + a\right )}^{m}}{d x + c} \,d x } \]
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Timed out. \[ \int (a+b x)^m (c+d x)^{1+2 n-2 (1+n)} \, dx=\int \frac {{\left (a+b\,x\right )}^m}{c+d\,x} \,d x \]
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