Optimal. Leaf size=260 \[ \frac{(b c-a d) (a+b x)^{m+1} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \left (a^2 d^2 f^2 \left (m^2-5 m+6\right )-2 a b d f (2-m) (4 d e-c f (m+1))+b^2 \left (c^2 f^2 \left (m^2+3 m+2\right )-8 c d e f (m+1)+12 d^2 e^2\right )\right ) \, _2F_1\left (m-1,m+1;m+2;-\frac{d (a+b x)}{b c-a d}\right )}{12 b^4 d^2 (m+1)}-\frac{f (a+b x)^{m+1} (c+d x)^{2-m} (a d f (3-m)-b (5 d e-c f (m+2)))}{12 b^2 d^2}+\frac{f (e+f x) (a+b x)^{m+1} (c+d x)^{2-m}}{4 b d} \]
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Rubi [A] time = 0.228239, antiderivative size = 259, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {90, 80, 70, 69} \[ \frac{(b c-a d) (a+b x)^{m+1} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \left (a^2 d^2 f^2 \left (m^2-5 m+6\right )-2 a b d f (2-m) (4 d e-c f (m+1))+b^2 \left (c^2 f^2 \left (m^2+3 m+2\right )-8 c d e f (m+1)+12 d^2 e^2\right )\right ) \, _2F_1\left (m-1,m+1;m+2;-\frac{d (a+b x)}{b c-a d}\right )}{12 b^4 d^2 (m+1)}+\frac{f (a+b x)^{m+1} (c+d x)^{2-m} (-a d f (3-m)-b c f (m+2)+5 b d e)}{12 b^2 d^2}+\frac{f (e+f x) (a+b x)^{m+1} (c+d x)^{2-m}}{4 b d} \]
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)^{1-m} (e+f x)^2 \, dx &=\frac{f (a+b x)^{1+m} (c+d x)^{2-m} (e+f x)}{4 b d}+\frac{\int (a+b x)^m (c+d x)^{1-m} (-a f (c f+d e (2-m))+b e (4 d e-c f (1+m))+f (5 b d e-a d f (3-m)-b c f (2+m)) x) \, dx}{4 b d}\\ &=\frac{f (5 b d e-a d f (3-m)-b c f (2+m)) (a+b x)^{1+m} (c+d x)^{2-m}}{12 b^2 d^2}+\frac{f (a+b x)^{1+m} (c+d x)^{2-m} (e+f x)}{4 b d}+\frac{\left (a^2 d^2 f^2 \left (6-5 m+m^2\right )-2 a b d f (2-m) (4 d e-c f (1+m))+b^2 \left (12 d^2 e^2-8 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)^{1-m} \, dx}{12 b^2 d^2}\\ &=\frac{f (5 b d e-a d f (3-m)-b c f (2+m)) (a+b x)^{1+m} (c+d x)^{2-m}}{12 b^2 d^2}+\frac{f (a+b x)^{1+m} (c+d x)^{2-m} (e+f x)}{4 b d}+\frac{\left ((b c-a d) \left (a^2 d^2 f^2 \left (6-5 m+m^2\right )-2 a b d f (2-m) (4 d e-c f (1+m))+b^2 \left (12 d^2 e^2-8 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 )^{1-m} \, dx}{12 b^3 d^2}\\ &=\frac{f (5 b d e-a d f (3-m)-b c f (2+m)) (a+b x)^{1+m} (c+d x)^{2-m}}{12 b^2 d^2}+\frac{f (a+b x)^{1+m} (c+d x)^{2-m} (e+f x)}{4 b d}+\frac{(b c-a d) \left (a^2 d^2 f^2 \left (6-5 m+m^2\right )-2 a b d f (2-m) (4 d e-c f (1+m))+b^2 \left (12 d^2 e^2-8 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 (-1+m,1+m;2+m;-\frac{d (a+b x)}{b c-a d}\right )}{12 b^4 d^2 (1+m)}\\ \end{align*}
Mathematica [A] time = 0.353413, size = 223, normalized size = 0.86 \[ \frac{(a+b x)^{m+1} (c+d x)^{-m} \left ((b c-a d) \left (\frac{b (c+d x)}{b c-a d}\right )^m \left (a^2 d^2 f^2 \left (m^2-5 m+6\right )-2 a b d f (m-2) (c f (m+1)-4 d e)+b^2 \left (c^2 f^2 \left (m^2+3 m+2\right )-8 c d e f (m+1)+12 d^2 e^2\right )\right ) \, _2F_1\left (m-1,m+1;m+2;\frac{d (a+b x)}{a d-b c}\right )+b^2 f (m+1) (c+d x)^2 (a d f (m-3)-b c f (m+2)+5 b d e)+3 b^3 d f (m+1) (c+d x)^2 (e+f x)\right )}{12 b^4 d^2 (m+1)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.067, size = 0, normalized size = 0. \begin{align*} \int \left ( bx+a \right ) ^{m} \left ( dx+c \right ) ^{1-m} \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{\left (f x + e\right )}^{2}{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m + 1}\,{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 ({\left (f^{2} x^{2} + 2 \, e f x + e^{2}\right )}{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m + 1}, 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{\left (f x + e\right )}^{2}{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m + 1}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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