Optimal. Leaf size=281 \[ \frac{d^2 (a+b x)^{m+1} (2 a d f-b (c f (2-m)+d e m)) \, _2F_1\left (1,m+1;m+2;-\frac{d (a+b x)}{b c-a d}\right )}{(m+1) (b c-a d)^2 (d e-c f)^3}-\frac{f^2 (a+b x)^{m+1} (2 a d f-b (c f m+d e (2-m))) \, _2F_1\left (1,m+1;m+2;-\frac{f (a+b x)}{b e-a f}\right )}{(m+1) (b e-a f)^2 (d e-c f)^3}+\frac{f (a+b x)^{m+1} (-2 a d f+b c f+b d e)}{(e+f x) (b c-a d) (b e-a f) (d e-c f)^2}+\frac{d (a+b x)^{m+1}}{(c+d x) (e+f x) (b c-a d) (d e-c f)} \]
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Rubi [A] time = 0.418147, antiderivative size = 281, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {103, 151, 156, 68} \[ \frac{d^2 (a+b x)^{m+1} (2 a d f-b c f (2-m)-b d e m) \, _2F_1\left (1,m+1;m+2;-\frac{d (a+b x)}{b c-a d}\right )}{(m+1) (b c-a d)^2 (d e-c f)^3}-\frac{f^2 (a+b x)^{m+1} (2 a d f-b c f m-b d e (2-m)) \, _2F_1\left (1,m+1;m+2;-\frac{f (a+b x)}{b e-a f}\right )}{(m+1) (b e-a f)^2 (d e-c f)^3}+\frac{f (a+b x)^{m+1} (-2 a d f+b c f+b d e)}{(e+f x) (b c-a d) (b e-a f) (d e-c f)^2}+\frac{d (a+b x)^{m+1}}{(c+d x) (e+f x) (b c-a d) (d e-c f)} \]
Antiderivative was successfully verified.
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Rule 103
Rule 151
Rule 156
Rule 68
Rubi steps
\begin{align*} \int \frac{(a+b x)^m}{(c+d x)^2 (e+f x)^2} \, dx &=\frac{d (a+b x)^{1+m}}{(b c-a d) (d e-c f) (c+d x) (e+f x)}+\frac{\int \frac{(a+b x)^m (2 a d f-b (c f+d e m)+b d f (1-m) x)}{(c+d x) (e+f x)^2} \, dx}{(b c-a d) (d e-c f)}\\ &=\frac{f (b d e+b c f-2 a d f) (a+b x)^{1+m}}{(b c-a d) (b e-a f) (d e-c f)^2 (e+f x)}+\frac{d (a+b x)^{1+m}}{(b c-a d) (d e-c f) (c+d x) (e+f x)}-\frac{\int \frac{(a+b x)^m \left (2 a^2 d^2 f^2-a b d f (d e+c f) (2+m)+b^2 \left (2 c d e f+d^2 e^2 m+c^2 f^2 m\right )+b d f (b d e+b c f-2 a d f) m x\right )}{(c+d x) (e+f x)} \, dx}{(b c-a d) (b e-a f) (d e-c f)^2}\\ &=\frac{f (b d e+b c f-2 a d f) (a+b x)^{1+m}}{(b c-a d) (b e-a f) (d e-c f)^2 (e+f x)}+\frac{d (a+b x)^{1+m}}{(b c-a d) (d e-c f) (c+d x) (e+f x)}+\frac{\left (d^2 (2 a d f-b c f (2-m)-b d e m)\right ) \int \frac{(a+b x)^m}{c+d x} \, dx}{(b c-a d) (d e-c f)^3}-\frac{\left (f^2 (2 a d f-b d e (2-m)-b c f m)\right ) \int \frac{(a+b x)^m}{e+f x} \, dx}{(b e-a f) (d e-c f)^3}\\ &=\frac{f (b d e+b c f-2 a d f) (a+b x)^{1+m}}{(b c-a d) (b e-a f) (d e-c f)^2 (e+f x)}+\frac{d (a+b x)^{1+m}}{(b c-a d) (d e-c f) (c+d x) (e+f x)}+\frac{d^2 (2 a d f-b c f (2-m)-b d e m) (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)^2 (d e-c f)^3 (1+m)}-\frac{f^2 (2 a d f-b d e (2-m)-b c f m) (a+b x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac{f (a+b x)}{b e-a f}\right )}{(b e-a f)^2 (d e-c f)^3 (1+m)}\\ \end{align*}
Mathematica [A] time = 0.545703, size = 247, normalized size = 0.88 \[ \frac{(a+b x)^{m+1} \left (-\frac{f^2 (b c-a d)^2 (-2 a d f+b c f m-b d e (m-2)) \, _2F_1\left (1,m+1;m+2;\frac{f (a+b x)}{a f-b e}\right )-d^2 (b e-a f)^2 (-2 a d f-b c f (m-2)+b d e m) \, _2F_1\left (1,m+1;m+2;\frac{d (a+b x)}{a d-b c}\right )}{(m+1) (b c-a d) (b e-a f)^2 (d e-c f)^2}-\frac{f (-2 a d f+b c f+b d e)}{(e+f x) (b e-a f) (d e-c f)}-\frac{d}{(c+d x) (e+f x)}\right )}{(b c-a d) (c f-d e)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.087, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) ^{m}}{ \left ( dx+c \right ) ^{2} \left ( fx+e \right ) ^{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{m}}{{\left (d x + c\right )}^{2}{\left (f x + e\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x + a\right )}^{m}}{d^{2} f^{2} x^{4} + c^{2} e^{2} + 2 \,{\left (d^{2} e f + c d f^{2}\right )} x^{3} +{\left (d^{2} e^{2} + 4 \, c d e f + c^{2} f^{2}\right )} x^{2} + 2 \,{\left (c d e^{2} + c^{2} e f\right )} x}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{m}}{{\left (d x + c\right )}^{2}{\left (f x + e\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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