Integrand size = 18, antiderivative size = 99 \[ \int \frac {x^m (c+d x)^2}{a+b x} \, dx=\frac {c d x^{1+m}}{b (1+m)}+\frac {d (b c-a d) x^{1+m}}{b^2 (1+m)}+\frac {d^2 x^{2+m}}{b (2+m)}+\frac {(b c-a d)^2 x^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {b x}{a}\right )}{a b^2 (1+m)} \]
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Time = 0.04 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {90, 66, 45} \[ \int \frac {x^m (c+d x)^2}{a+b x} \, dx=\frac {x^{m+1} (b c-a d)^2 \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {b x}{a}\right )}{a b^2 (m+1)}+\frac {d x^{m+1} (b c-a d)}{b^2 (m+1)}+\frac {c d x^{m+1}}{b (m+1)}+\frac {d^2 x^{m+2}}{b (m+2)} \]
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Rule 45
Rule 66
Rule 90
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {d (b c-a d) x^m}{b^2}+\frac {(b c-a d)^2 x^m}{b^2 (a+b x)}+\frac {d x^m (c+d x)}{b}\right ) \, dx \\ & = \frac {d (b c-a d) x^{1+m}}{b^2 (1+m)}+\frac {d \int x^m (c+d x) \, dx}{b}+\frac {(b c-a d)^2 \int \frac {x^m}{a+b x} \, dx}{b^2} \\ & = \frac {d (b c-a d) x^{1+m}}{b^2 (1+m)}+\frac {(b c-a d)^2 x^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {b x}{a}\right )}{a b^2 (1+m)}+\frac {d \int \left (c x^m+d x^{1+m}\right ) \, dx}{b} \\ & = \frac {c d x^{1+m}}{b (1+m)}+\frac {d (b c-a d) x^{1+m}}{b^2 (1+m)}+\frac {d^2 x^{2+m}}{b (2+m)}+\frac {(b c-a d)^2 x^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {b x}{a}\right )}{a b^2 (1+m)} \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.78 \[ \int \frac {x^m (c+d x)^2}{a+b x} \, dx=\frac {x^{1+m} \left (a d (2 b c (2+m)-a d (2+m)+b d (1+m) x)+(b c-a d)^2 (2+m) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {b x}{a}\right )\right )}{a b^2 (1+m) (2+m)} \]
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\[\int \frac {x^{m} \left (d x +c \right )^{2}}{b x +a}d x\]
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\[ \int \frac {x^m (c+d x)^2}{a+b x} \, dx=\int { \frac {{\left (d x + c\right )}^{2} x^{m}}{b x + a} \,d x } \]
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Result contains complex when optimal does not.
Time = 1.62 (sec) , antiderivative size = 212, normalized size of antiderivative = 2.14 \[ \int \frac {x^m (c+d x)^2}{a+b x} \, dx=\frac {c^{2} m x^{m + 1} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{a \Gamma \left (m + 2\right )} + \frac {c^{2} x^{m + 1} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{a \Gamma \left (m + 2\right )} + \frac {2 c d m x^{m + 2} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{a \Gamma \left (m + 3\right )} + \frac {4 c d x^{m + 2} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{a \Gamma \left (m + 3\right )} + \frac {d^{2} m x^{m + 3} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 3\right ) \Gamma \left (m + 3\right )}{a \Gamma \left (m + 4\right )} + \frac {3 d^{2} x^{m + 3} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 3\right ) \Gamma \left (m + 3\right )}{a \Gamma \left (m + 4\right )} \]
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\[ \int \frac {x^m (c+d x)^2}{a+b x} \, dx=\int { \frac {{\left (d x + c\right )}^{2} x^{m}}{b x + a} \,d x } \]
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\[ \int \frac {x^m (c+d x)^2}{a+b x} \, dx=\int { \frac {{\left (d x + c\right )}^{2} x^{m}}{b x + a} \,d x } \]
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Timed out. \[ \int \frac {x^m (c+d x)^2}{a+b x} \, dx=\int \frac {x^m\,{\left (c+d\,x\right )}^2}{a+b\,x} \,d x \]
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