Optimal. Leaf size=445 \[ \frac {f (a+b x)^{1+m} (c+d x)^{2-m} (e+f x)^2}{5 b d}+\frac {f (a+b x)^{1+m} (c+d x)^{2-m} \left (a^2 d^2 f^2 \left (12-7 m+m^2\right )-a b d f \left (15 d e (3-m)-c f \left (9+2 m-2 m^2\right )\right )+b^2 \left (48 d^2 e^2-15 c d e f (2+m)+c^2 f^2 \left (6+5 m+m^2\right )\right )-3 b d f (a d f (4-m)-b (7 d e-c f (3+m))) x\right )}{60 b^3 d^3}-\frac {(b c-a d) \left (a^3 d^3 f^3 \left (24-26 m+9 m^2-m^3\right )-3 a^2 b d^2 f^2 \left (6-5 m+m^2\right ) (5 d e-c f (1+m))+3 a b^2 d f (2-m) \left (20 d^2 e^2-10 c d e f (1+m)+c^2 f^2 \left (2+3 m+m^2\right )\right )-b^3 \left (60 d^3 e^3-60 c d^2 e^2 f (1+m)+15 c^2 d e f^2 \left (2+3 m+m^2\right )-c^3 f^3 \left (6+11 m+6 m^2+m^3\right )\right )\right ) (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (-1+m,1+m;2+m;-\frac {d (a+b x)}{b c-a d}\right )}{60 b^5 d^3 (1+m)} \]
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Rubi [A]
time = 0.34, antiderivative size = 444, normalized size of antiderivative = 1.00, 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 = {102, 152, 72,
71} \begin {gather*} \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 {(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^3 d^3 f^3 \left (-m^3+9 m^2-26 m+24\right )-3 a^2 b d^2 f^2 \left (m^2-5 m+6\right ) (5 d e-c f (m+1))+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 )-\left (b^3 \left (-c^3 f^3 \left (m^3+6 m^2+11 m+6\right )+15 c^2 d e f^2 \left (m^2+3 m+2\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 (e+f x)^2 (a+b x)^{m+1} (c+d x)^{2-m}}{5 b d} \end {gather*}
Antiderivative was successfully verified.
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Rule 71
Rule 72
Rule 102
Rule 152
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*}
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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in
optimal.
time = 0.53, size = 256, normalized size = 0.58 \begin {gather*} \frac {1}{4} (a+b x)^m (c+d x)^{-m} \left (6 c e^2 f x^2 \left (1+\frac {b x}{a}\right )^{-m} \left (1+\frac {d x}{c}\right )^m F_1\left (2;-m,-1+m;3;-\frac {b x}{a},-\frac {d x}{c}\right )+4 c e f^2 x^3 \left (1+\frac {b x}{a}\right )^{-m} \left (1+\frac {d x}{c}\right )^m F_1\left (3;-m,-1+m;4;-\frac {b x}{a},-\frac {d x}{c}\right )+c f^3 x^4 \left (1+\frac {b x}{a}\right )^{-m} \left (1+\frac {d x}{c}\right )^m F_1\left (4;-m,-1+m;5;-\frac {b x}{a},-\frac {d x}{c}\right )-\frac {4 e^3 \left (\frac {d (a+b x)}{-b c+a d}\right )^{-m} (c+d x)^2 \, _2F_1\left (2-m,-m;3-m;\frac {b (c+d x)}{b c-a d}\right )}{d (-2+m)}\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \left (b x +a \right )^{m} \left (d x +c \right )^{1-m} \left (f x +e \right )^{3}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: HeuristicGCDFailed} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (e+f\,x\right )}^3\,{\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^{1-m} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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