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