Optimal. Leaf size=445 \[ -\frac{(b c-a d) (a+b x)^{m+1} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \left (-3 a^2 b d^2 f^2 \left (m^2-5 m+6\right ) (5 d e-c f (m+1))+a^3 d^3 f^3 \left (-m^3+9 m^2-26 m+24\right )+3 a b^2 d f (2-m) \left (c^2 f^2 \left (m^2+3 m+2\right )-10 c d e f (m+1)+20 d^2 e^2\right )+b^3 \left (-\left (15 c^2 d e f^2 \left (m^2+3 m+2\right )-c^3 f^3 \left (m^3+6 m^2+11 m+6\right )-60 c d^2 e^2 f (m+1)+60 d^3 e^3\right )\right )\right ) \, _2F_1\left (m-1,m+1;m+2;-\frac{d (a+b x)}{b c-a d}\right )}{60 b^5 d^3 (m+1)}+\frac{f (a+b x)^{m+1} (c+d x)^{2-m} \left (a^2 d^2 f^2 \left (m^2-7 m+12\right )-a b d f \left (15 d e (3-m)-c f \left (-2 m^2+2 m+9\right )\right )-3 b d f x (a d f (4-m)-b (7 d e-c f (m+3)))+b^2 \left (c^2 f^2 \left (m^2+5 m+6\right )-15 c d e f (m+2)+48 d^2 e^2\right )\right )}{60 b^3 d^3}+\frac{f (e+f x)^2 (a+b x)^{m+1} (c+d x)^{2-m}}{5 b d} \]
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Rubi [A] time = 0.515915, antiderivative size = 444, 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 = {100, 147, 70, 69} \[ -\frac{(b c-a d) (a+b x)^{m+1} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \left (-3 a^2 b d^2 f^2 \left (m^2-5 m+6\right ) (5 d e-c f (m+1))+a^3 d^3 f^3 \left (-m^3+9 m^2-26 m+24\right )+3 a b^2 d f (2-m) \left (c^2 f^2 \left (m^2+3 m+2\right )-10 c d e f (m+1)+20 d^2 e^2\right )+b^3 \left (-\left (15 c^2 d e f^2 \left (m^2+3 m+2\right )-c^3 f^3 \left (m^3+6 m^2+11 m+6\right )-60 c d^2 e^2 f (m+1)+60 d^3 e^3\right )\right )\right ) \, _2F_1\left (m-1,m+1;m+2;-\frac{d (a+b x)}{b c-a d}\right )}{60 b^5 d^3 (m+1)}+\frac{f (a+b x)^{m+1} (c+d x)^{2-m} \left (a^2 d^2 f^2 \left (m^2-7 m+12\right )-a b d f \left (15 d e (3-m)-c f \left (-2 m^2+2 m+9\right )\right )+3 b d f x (-a d f (4-m)-b c f (m+3)+7 b d e)+b^2 \left (c^2 f^2 \left (m^2+5 m+6\right )-15 c d e f (m+2)+48 d^2 e^2\right )\right )}{60 b^3 d^3}+\frac{f (e+f x)^2 (a+b x)^{m+1} (c+d x)^{2-m}}{5 b d} \]
Antiderivative was successfully verified.
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Rule 100
Rule 147
Rule 70
Rule 69
Rubi steps
\begin{align*} \int (a+b x)^m (c+d x)^{1-m} (e+f x)^3 \, dx &=\frac{f (a+b x)^{1+m} (c+d x)^{2-m} (e+f x)^2}{5 b d}+\frac{\int (a+b x)^m (c+d x)^{1-m} (e+f x) (-a f (2 c f+d e (2-m))+b e (5 d e-c f (1+m))+f (7 b d e-a d f (4-m)-b c f (3+m)) x) \, dx}{5 b d}\\ &=\frac{f (a+b x)^{1+m} (c+d x)^{2-m} (e+f x)^2}{5 b d}+\frac{f (a+b x)^{1+m} (c+d x)^{2-m} \left (a^2 d^2 f^2 \left (12-7 m+m^2\right )-a b d f \left (15 d e (3-m)-c f \left (9+2 m-2 m^2\right )\right )+b^2 \left (48 d^2 e^2-15 c d e f (2+m)+c^2 f^2 \left (6+5 m+m^2\right )\right )+3 b d f (7 b d e-a d f (4-m)-b c f (3+m)) x\right )}{60 b^3 d^3}-\frac{\left (a^3 d^3 f^3 \left (24-26 m+9 m^2-m^3\right )-3 a^2 b d^2 f^2 \left (6-5 m+m^2\right ) (5 d e-c f (1+m))+3 a b^2 d f (2-m) \left (20 d^2 e^2-10 c d e f (1+m)+c^2 f^2 \left (2+3 m+m^2\right )\right )-b^3 \left (60 d^3 e^3-60 c d^2 e^2 f (1+m)+15 c^2 d e f^2 \left (2+3 m+m^2\right )-c^3 f^3 \left (6+11 m+6 m^2+m^3\right )\right )\right ) \int (a+b x)^m (c+d x)^{1-m} \, dx}{60 b^3 d^3}\\ &=\frac{f (a+b x)^{1+m} (c+d x)^{2-m} (e+f x)^2}{5 b d}+\frac{f (a+b x)^{1+m} (c+d x)^{2-m} \left (a^2 d^2 f^2 \left (12-7 m+m^2\right )-a b d f \left (15 d e (3-m)-c f \left (9+2 m-2 m^2\right )\right )+b^2 \left (48 d^2 e^2-15 c d e f (2+m)+c^2 f^2 \left (6+5 m+m^2\right )\right )+3 b d f (7 b d e-a d f (4-m)-b c f (3+m)) x\right )}{60 b^3 d^3}-\frac{\left ((b c-a d) \left (a^3 d^3 f^3 \left (24-26 m+9 m^2-m^3\right )-3 a^2 b d^2 f^2 \left (6-5 m+m^2\right ) (5 d e-c f (1+m))+3 a b^2 d f (2-m) \left (20 d^2 e^2-10 c d e f (1+m)+c^2 f^2 \left (2+3 m+m^2\right )\right )-b^3 \left (60 d^3 e^3-60 c d^2 e^2 f (1+m)+15 c^2 d e f^2 \left (2+3 m+m^2\right )-c^3 f^3 \left (6+11 m+6 m^2+m^3\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}{60 b^4 d^3}\\ &=\frac{f (a+b x)^{1+m} (c+d x)^{2-m} (e+f x)^2}{5 b d}+\frac{f (a+b x)^{1+m} (c+d x)^{2-m} \left (a^2 d^2 f^2 \left (12-7 m+m^2\right )-a b d f \left (15 d e (3-m)-c f \left (9+2 m-2 m^2\right )\right )+b^2 \left (48 d^2 e^2-15 c d e f (2+m)+c^2 f^2 \left (6+5 m+m^2\right )\right )+3 b d f (7 b d e-a d f (4-m)-b c f (3+m)) x\right )}{60 b^3 d^3}-\frac{(b c-a d) \left (a^3 d^3 f^3 \left (24-26 m+9 m^2-m^3\right )-3 a^2 b d^2 f^2 \left (6-5 m+m^2\right ) (5 d e-c f (1+m))+3 a b^2 d f (2-m) \left (20 d^2 e^2-10 c d e f (1+m)+c^2 f^2 \left (2+3 m+m^2\right )\right )-b^3 \left (60 d^3 e^3-60 c d^2 e^2 f (1+m)+15 c^2 d e f^2 \left (2+3 m+m^2\right )-c^3 f^3 \left (6+11 m+6 m^2+m^3\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 )}{60 b^5 d^3 (1+m)}\\ \end{align*}
Mathematica [A] time = 0.834999, size = 317, normalized size = 0.71 \[ \frac{(a+b x)^{m+1} (c+d x)^{-m} \left (b^3 (c+d x) (d e-c f)^2 \left (\frac{b (c+d x)}{b c-a d}\right )^{m-1} (a d f (m-2)-b c f (m+3)+5 b d e) \, _2F_1\left (m-1,m+1;m+2;\frac{d (a+b x)}{a d-b c}\right )+f^2 (b c-a d)^3 \left (\frac{b (c+d x)}{b c-a d}\right )^m (a d f (m-4)-b c f (m+3)+7 b d e) \, _2F_1\left (m-3,m+1;m+2;\frac{d (a+b x)}{a d-b c}\right )+2 b f (b c-a d)^2 (d e-c f) \left (\frac{b (c+d x)}{b c-a d}\right )^m (a d f (m-3)-b c f (m+3)+6 b d e) \, _2F_1\left (m-2,m+1;m+2;\frac{d (a+b x)}{a d-b c}\right )+b^4 d^2 f (m+1) (c+d x)^2 (e+f x)^2\right )}{5 b^5 d^3 (m+1)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.068, size = 0, normalized size = 0. \begin{align*} \int \left ( bx+a \right ) ^{m} \left ( dx+c \right ) ^{1-m} \left ( fx+e \right ) ^{3}\, 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 )}^{3}{\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^{3} x^{3} + 3 \, e f^{2} x^{2} + 3 \, e^{2} f x + e^{3}\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 )}^{3}{\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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