Optimal. Leaf size=250 \[ \frac{(a+b x)^{m+1} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \left (a^2 d^2 f^2 \left (m^2-3 m+2\right )-2 a b d f (1-m) (3 d e-c f (m+1))+b^2 \left (c^2 f^2 \left (m^2+3 m+2\right )-6 c d e f (m+1)+6 d^2 e^2\right )\right ) \, _2F_1\left (m,m+1;m+2;-\frac{d (a+b x)}{b c-a d}\right )}{6 b^3 d^2 (m+1)}-\frac{f (a+b x)^{m+1} (c+d x)^{1-m} (a d f (2-m)-b (4 d e-c f (m+2)))}{6 b^2 d^2}+\frac{f (e+f x) (a+b x)^{m+1} (c+d x)^{1-m}}{3 b d} \]
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Rubi [A] time = 0.205017, antiderivative size = 249, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {90, 80, 70, 69} \[ \frac{(a+b x)^{m+1} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \left (a^2 d^2 f^2 \left (m^2-3 m+2\right )-2 a b d f (1-m) (3 d e-c f (m+1))+b^2 \left (c^2 f^2 \left (m^2+3 m+2\right )-6 c d e f (m+1)+6 d^2 e^2\right )\right ) \, _2F_1\left (m,m+1;m+2;-\frac{d (a+b x)}{b c-a d}\right )}{6 b^3 d^2 (m+1)}+\frac{f (a+b x)^{m+1} (c+d x)^{1-m} (-a d f (2-m)-b c f (m+2)+4 b d e)}{6 b^2 d^2}+\frac{f (e+f x) (a+b x)^{m+1} (c+d x)^{1-m}}{3 b d} \]
Antiderivative was successfully verified.
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Rule 90
Rule 80
Rule 70
Rule 69
Rubi steps
\begin{align*} \int (a+b x)^m (c+d x)^{-m} (e+f x)^2 \, dx &=\frac{f (a+b x)^{1+m} (c+d x)^{1-m} (e+f x)}{3 b d}+\frac{\int (a+b x)^m (c+d x)^{-m} \left (3 b d e^2-f (a c f+a d e (1-m)+b c e (1+m))+f (4 b d e-a d f (2-m)-b c f (2+m)) x\right ) \, dx}{3 b d}\\ &=\frac{f (4 b d e-a d f (2-m)-b c f (2+m)) (a+b x)^{1+m} (c+d x)^{1-m}}{6 b^2 d^2}+\frac{f (a+b x)^{1+m} (c+d x)^{1-m} (e+f x)}{3 b d}+\frac{\left (a^2 d^2 f^2 \left (2-3 m+m^2\right )-2 a b d f (1-m) (3 d e-c f (1+m))+b^2 \left (6 d^2 e^2-6 c d e f (1+m)+c^2 f^2 \left (2+3 m+m^2\right )\right )\right ) \int (a+b x)^m (c+d x)^{-m} \, dx}{6 b^2 d^2}\\ &=\frac{f (4 b d e-a d f (2-m)-b c f (2+m)) (a+b x)^{1+m} (c+d x)^{1-m}}{6 b^2 d^2}+\frac{f (a+b x)^{1+m} (c+d x)^{1-m} (e+f x)}{3 b d}+\frac{\left (\left (a^2 d^2 f^2 \left (2-3 m+m^2\right )-2 a b d f (1-m) (3 d e-c f (1+m))+b^2 \left (6 d^2 e^2-6 c d e f (1+m)+c^2 f^2 \left (2+3 m+m^2\right )\right )\right ) (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m\right ) \int (a+b x)^m \left (\frac{b c}{b c-a d}+\frac{b d x}{b c-a d}\right )^{-m} \, dx}{6 b^2 d^2}\\ &=\frac{f (4 b d e-a d f (2-m)-b c f (2+m)) (a+b x)^{1+m} (c+d x)^{1-m}}{6 b^2 d^2}+\frac{f (a+b x)^{1+m} (c+d x)^{1-m} (e+f x)}{3 b d}+\frac{\left (a^2 d^2 f^2 \left (2-3 m+m^2\right )-2 a b d f (1-m) (3 d e-c f (1+m))+b^2 \left (6 d^2 e^2-6 c d e f (1+m)+c^2 f^2 \left (2+3 m+m^2\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,1+m;2+m;-\frac{d (a+b x)}{b c-a d}\right )}{6 b^3 d^2 (1+m)}\\ \end{align*}
Mathematica [A] time = 0.25395, size = 201, normalized size = 0.8 \[ \frac{(a+b x)^{m+1} (c+d x)^{-m} \left (\frac{\left (\frac{b (c+d x)}{b c-a d}\right )^m \left (a^2 d^2 f^2 \left (m^2-3 m+2\right )-2 a b d f (m-1) (c f (m+1)-3 d e)+b^2 \left (c^2 f^2 \left (m^2+3 m+2\right )-6 c d e f (m+1)+6 d^2 e^2\right )\right ) \, _2F_1\left (m,m+1;m+2;\frac{d (a+b x)}{a d-b c}\right )}{m+1}+b f (c+d x) (a d f (m-2)-b c f (m+2)+4 b d e)+2 b^2 d f (c+d x) (e+f x)\right )}{6 b^3 d^2} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.065, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{2} \left ( bx+a \right ) ^{m}}{ \left ( dx+c \right ) ^{m}}}\, 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 (f x + e\right )}^{2}{\left (b x + a\right )}^{m}}{{\left (d x + c\right )}^{m}}\,{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 (f^{2} x^{2} + 2 \, e f x + e^{2}\right )}{\left (b x + a\right )}^{m}}{{\left (d x + c\right )}^{m}}, 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 (f x + e\right )}^{2}{\left (b x + a\right )}^{m}}{{\left (d x + c\right )}^{m}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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