Integrand size = 20, antiderivative size = 52 \[ \int \frac {(1-2 x) (2+3 x)^m}{3+5 x} \, dx=-\frac {2 (2+3 x)^{1+m}}{15 (1+m)}-\frac {11 (2+3 x)^{1+m} \operatorname {Hypergeometric2F1}(1,1+m,2+m,5 (2+3 x))}{5 (1+m)} \]
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Time = 0.01 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {81, 70} \[ \int \frac {(1-2 x) (2+3 x)^m}{3+5 x} \, dx=-\frac {11 (3 x+2)^{m+1} \operatorname {Hypergeometric2F1}(1,m+1,m+2,5 (3 x+2))}{5 (m+1)}-\frac {2 (3 x+2)^{m+1}}{15 (m+1)} \]
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Rule 70
Rule 81
Rubi steps \begin{align*} \text {integral}& = -\frac {2 (2+3 x)^{1+m}}{15 (1+m)}+\frac {11}{5} \int \frac {(2+3 x)^m}{3+5 x} \, dx \\ & = -\frac {2 (2+3 x)^{1+m}}{15 (1+m)}-\frac {11 (2+3 x)^{1+m} \, _2F_1(1,1+m;2+m;5 (2+3 x))}{5 (1+m)} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.71 \[ \int \frac {(1-2 x) (2+3 x)^m}{3+5 x} \, dx=-\frac {(2+3 x)^{1+m} (2+33 \operatorname {Hypergeometric2F1}(1,1+m,2+m,5 (2+3 x)))}{15 (1+m)} \]
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\[\int \frac {\left (1-2 x \right ) \left (2+3 x \right )^{m}}{3+5 x}d x\]
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\[ \int \frac {(1-2 x) (2+3 x)^m}{3+5 x} \, dx=\int { -\frac {{\left (3 \, x + 2\right )}^{m} {\left (2 \, x - 1\right )}}{5 \, x + 3} \,d x } \]
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\[ \int \frac {(1-2 x) (2+3 x)^m}{3+5 x} \, dx=- \int \left (- \frac {\left (3 x + 2\right )^{m}}{5 x + 3}\right )\, dx - \int \frac {2 x \left (3 x + 2\right )^{m}}{5 x + 3}\, dx \]
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\[ \int \frac {(1-2 x) (2+3 x)^m}{3+5 x} \, dx=\int { -\frac {{\left (3 \, x + 2\right )}^{m} {\left (2 \, x - 1\right )}}{5 \, x + 3} \,d x } \]
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\[ \int \frac {(1-2 x) (2+3 x)^m}{3+5 x} \, dx=\int { -\frac {{\left (3 \, x + 2\right )}^{m} {\left (2 \, x - 1\right )}}{5 \, x + 3} \,d x } \]
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Timed out. \[ \int \frac {(1-2 x) (2+3 x)^m}{3+5 x} \, dx=-\int \frac {\left (2\,x-1\right )\,{\left (3\,x+2\right )}^m}{5\,x+3} \,d x \]
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