Integrand size = 19, antiderivative size = 350 \[ \int \frac {(d+e x)^m}{\left (b x+c x^2\right )^3} \, dx=-\frac {(d+e x)^{1+m} (b (c d-b e)+c (2 c d-b e) x)}{2 b^2 d (c d-b e) \left (b x+c x^2\right )^2}+\frac {(d+e x)^{1+m} \left (b (c d-b e) \left (6 c^2 d^2-b^2 e^2 (1-m)-b c d e (4+m)\right )+c (2 c d-b e) \left (6 c^2 d^2-6 b c d e-b^2 e^2 (1-m)\right ) x\right )}{2 b^4 d^2 (c d-b e)^2 \left (b x+c x^2\right )}+\frac {c^3 \left (12 c^2 d^2-6 b c d e (4-m)+b^2 e^2 \left (12-7 m+m^2\right )\right ) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {c (d+e x)}{c d-b e}\right )}{2 b^5 (c d-b e)^3 (1+m)}-\frac {\left (12 c^2 d^2-6 b c d e m-b^2 e^2 (1-m) m\right ) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,1+\frac {e x}{d}\right )}{2 b^5 d^3 (1+m)} \]
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Time = 0.29 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {754, 836, 844, 67, 70} \[ \int \frac {(d+e x)^m}{\left (b x+c x^2\right )^3} \, dx=-\frac {(d+e x)^{m+1} (c x (2 c d-b e)+b (c d-b e))}{2 b^2 d \left (b x+c x^2\right )^2 (c d-b e)}-\frac {(d+e x)^{m+1} \left (-b^2 e^2 (1-m) m-6 b c d e m+12 c^2 d^2\right ) \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {e x}{d}+1\right )}{2 b^5 d^3 (m+1)}+\frac {c^3 (d+e x)^{m+1} \left (b^2 e^2 \left (m^2-7 m+12\right )-6 b c d e (4-m)+12 c^2 d^2\right ) \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {c (d+e x)}{c d-b e}\right )}{2 b^5 (m+1) (c d-b e)^3}+\frac {(d+e x)^{m+1} \left (c x (2 c d-b e) \left (-b^2 e^2 (1-m)-6 b c d e+6 c^2 d^2\right )+b (c d-b e) \left (-b^2 e^2 (1-m)-b c d e (m+4)+6 c^2 d^2\right )\right )}{2 b^4 d^2 \left (b x+c x^2\right ) (c d-b e)^2} \]
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Rule 67
Rule 70
Rule 754
Rule 836
Rule 844
Rubi steps \begin{align*} \text {integral}& = -\frac {(d+e x)^{1+m} (b (c d-b e)+c (2 c d-b e) x)}{2 b^2 d (c d-b e) \left (b x+c x^2\right )^2}-\frac {\int \frac {(d+e x)^m \left (6 c^2 d^2-b^2 e^2 (1-m)-b c d e (4+m)+c e (2 c d-b e) (2-m) x\right )}{\left (b x+c x^2\right )^2} \, dx}{2 b^2 d (c d-b e)} \\ & = -\frac {(d+e x)^{1+m} (b (c d-b e)+c (2 c d-b e) x)}{2 b^2 d (c d-b e) \left (b x+c x^2\right )^2}+\frac {(d+e x)^{1+m} \left (b (c d-b e) \left (6 c^2 d^2-b^2 e^2 (1-m)-b c d e (4+m)\right )+c (2 c d-b e) \left (6 c^2 d^2-6 b c d e-b^2 e^2 (1-m)\right ) x\right )}{2 b^4 d^2 (c d-b e)^2 \left (b x+c x^2\right )}+\frac {\int \frac {(d+e x)^m \left ((c d-b e)^2 \left (12 c^2 d^2-6 b c d e m-b^2 e^2 (1-m) m\right )-c e (2 c d-b e) \left (6 c^2 d^2-6 b c d e-b^2 e^2 (1-m)\right ) m x\right )}{b x+c x^2} \, dx}{2 b^4 d^2 (c d-b e)^2} \\ & = -\frac {(d+e x)^{1+m} (b (c d-b e)+c (2 c d-b e) x)}{2 b^2 d (c d-b e) \left (b x+c x^2\right )^2}+\frac {(d+e x)^{1+m} \left (b (c d-b e) \left (6 c^2 d^2-b^2 e^2 (1-m)-b c d e (4+m)\right )+c (2 c d-b e) \left (6 c^2 d^2-6 b c d e-b^2 e^2 (1-m)\right ) x\right )}{2 b^4 d^2 (c d-b e)^2 \left (b x+c x^2\right )}+\frac {\int \left (\frac {(c d-b e)^2 \left (12 c^2 d^2-6 b c d e m-b^2 e^2 (1-m) m\right ) (d+e x)^m}{b x}+\frac {c^3 d^2 \left (-12 c^2 d^2+6 b c d e (4-m)-b^2 e^2 \left (12-7 m+m^2\right )\right ) (d+e x)^m}{b (b+c x)}\right ) \, dx}{2 b^4 d^2 (c d-b e)^2} \\ & = -\frac {(d+e x)^{1+m} (b (c d-b e)+c (2 c d-b e) x)}{2 b^2 d (c d-b e) \left (b x+c x^2\right )^2}+\frac {(d+e x)^{1+m} \left (b (c d-b e) \left (6 c^2 d^2-b^2 e^2 (1-m)-b c d e (4+m)\right )+c (2 c d-b e) \left (6 c^2 d^2-6 b c d e-b^2 e^2 (1-m)\right ) x\right )}{2 b^4 d^2 (c d-b e)^2 \left (b x+c x^2\right )}+\frac {\left (12 c^2 d^2-6 b c d e m-b^2 e^2 (1-m) m\right ) \int \frac {(d+e x)^m}{x} \, dx}{2 b^5 d^2}-\frac {\left (c^3 \left (12 c^2 d^2-6 b c d e (4-m)+b^2 e^2 \left (12-7 m+m^2\right )\right )\right ) \int \frac {(d+e x)^m}{b+c x} \, dx}{2 b^5 (c d-b e)^2} \\ & = -\frac {(d+e x)^{1+m} (b (c d-b e)+c (2 c d-b e) x)}{2 b^2 d (c d-b e) \left (b x+c x^2\right )^2}+\frac {(d+e x)^{1+m} \left (b (c d-b e) \left (6 c^2 d^2-b^2 e^2 (1-m)-b c d e (4+m)\right )+c (2 c d-b e) \left (6 c^2 d^2-6 b c d e-b^2 e^2 (1-m)\right ) x\right )}{2 b^4 d^2 (c d-b e)^2 \left (b x+c x^2\right )}+\frac {c^3 \left (12 c^2 d^2-6 b c d e (4-m)+b^2 e^2 \left (12-7 m+m^2\right )\right ) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {c (d+e x)}{c d-b e}\right )}{2 b^5 (c d-b e)^3 (1+m)}-\frac {\left (12 c^2 d^2-6 b c d e m-b^2 e^2 (1-m) m\right ) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;1+\frac {e x}{d}\right )}{2 b^5 d^3 (1+m)} \\ \end{align*}
Time = 0.61 (sec) , antiderivative size = 337, normalized size of antiderivative = 0.96 \[ \int \frac {(d+e x)^m}{\left (b x+c x^2\right )^3} \, dx=-\frac {(d+e x)^{1+m} \left (2 b^4 d^2 (c d-b e)^3 (1+m)-2 b^3 d (c d-b e)^3 (4 c d-b e (-1+m)) (1+m) x+x^2 \left (-2 b^2 c d (c d-b e)^2 (1+m) \left (6 c^2 d^2+b^2 e^2 (-1+m)-b c d e (4+m)\right )-(b+c x) \left (-2 b c d (2 c d-b e) (-c d+b e) \left (6 c^2 d^2-6 b c d e+b^2 e^2 (-1+m)\right ) (1+m)+(b+c x) \left (2 c^3 d^3 \left (12 c^2 d^2+6 b c d e (-4+m)+b^2 e^2 \left (12-7 m+m^2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {c (d+e x)}{c d-b e}\right )-2 (c d-b e)^3 \left (12 c^2 d^2-6 b c d e m+b^2 e^2 (-1+m) m\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,1+\frac {e x}{d}\right )\right )\right )\right )\right )}{4 b^5 d^3 (c d-b e)^3 (1+m) x^2 (b+c x)^2} \]
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\[\int \frac {\left (e x +d \right )^{m}}{\left (c \,x^{2}+b x \right )^{3}}d x\]
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\[ \int \frac {(d+e x)^m}{\left (b x+c x^2\right )^3} \, dx=\int { \frac {{\left (e x + d\right )}^{m}}{{\left (c x^{2} + b x\right )}^{3}} \,d x } \]
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\[ \int \frac {(d+e x)^m}{\left (b x+c x^2\right )^3} \, dx=\int \frac {\left (d + e x\right )^{m}}{x^{3} \left (b + c x\right )^{3}}\, dx \]
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\[ \int \frac {(d+e x)^m}{\left (b x+c x^2\right )^3} \, dx=\int { \frac {{\left (e x + d\right )}^{m}}{{\left (c x^{2} + b x\right )}^{3}} \,d x } \]
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\[ \int \frac {(d+e x)^m}{\left (b x+c x^2\right )^3} \, dx=\int { \frac {{\left (e x + d\right )}^{m}}{{\left (c x^{2} + b x\right )}^{3}} \,d x } \]
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Timed out. \[ \int \frac {(d+e x)^m}{\left (b x+c x^2\right )^3} \, dx=\int \frac {{\left (d+e\,x\right )}^m}{{\left (c\,x^2+b\,x\right )}^3} \,d x \]
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