Optimal. Leaf size=546 \[ -\frac {7 (13-2 m) (5-4 x)^3 (1+2 x)^{-4-m} (2+3 x)^{1+m}}{3 (4+m)}-\frac {2}{3} (5-4 x)^4 (1+2 x)^{-4-m} (2+3 x)^{1+m}+\frac {24334 (105-2 m) (1+2 x)^{-3-m} (2+3 x)^{1+m}}{81 (3+m)}+\frac {1127 (13-2 m) (27+2 m) (1+2 x)^{-3-m} (2+3 x)^{1+m}}{3 (3+m) (4+m)}-\frac {322 (13-2 m) (5-4 x) (1+2 x)^{-3-m} (2+3 x)^{1+m}}{3 (4+m)}-\frac {48668 (105-2 m) (1+2 x)^{-2-m} (2+3 x)^{1+m}}{27 \left (6+5 m+m^2\right )}-\frac {322 (13-2 m) \left (579+52 m+2 m^2\right ) (1+2 x)^{-2-m} (2+3 x)^{1+m}}{3 (2+m) (3+m) (4+m)}+\frac {322 (13-2 m) \left (579+52 m+2 m^2\right ) (1+2 x)^{-1-m} (2+3 x)^{1+m}}{(3+m) (4+m) \left (2+3 m+m^2\right )}+\frac {48668 (105-2 m) (1+2 x)^{-1-m} (2+3 x)^{1+m}}{9 \left (6+11 m+6 m^2+m^3\right )}-\frac {4232 (105-2 m) (1+2 x)^{-3-m} (2+3 x)^{2+m}}{27 (3+m)}+\frac {4232 (105-2 m) (1+2 x)^{-2-m} (2+3 x)^{2+m}}{9 \left (6+5 m+m^2\right )}+\frac {736 (105-2 m) (1+2 x)^{-3-m} (2+3 x)^{3+m}}{27 (3+m)}-\frac {2^{3-m} (105-2 m) (1+2 x)^{-3-m} \, _2F_1(-3-m,-3-m;-2-m;-3 (1+2 x))}{81 (3+m)} \]
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Rubi [A]
time = 0.29, antiderivative size = 546, normalized size of antiderivative = 1.00, number of steps
used = 16, number of rules used = 9, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {102, 165, 128,
47, 37, 71, 96, 92, 80} \begin {gather*} -\frac {2^{3-m} (105-2 m) (2 x+1)^{-m-3} \, _2F_1(-m-3,-m-3;-m-2;-3 (2 x+1))}{81 (m+3)}-\frac {322 (13-2 m) \left (2 m^2+52 m+579\right ) (3 x+2)^{m+1} (2 x+1)^{-m-2}}{3 (m+2) (m+3) (m+4)}-\frac {48668 (105-2 m) (3 x+2)^{m+1} (2 x+1)^{-m-2}}{27 \left (m^2+5 m+6\right )}+\frac {4232 (105-2 m) (3 x+2)^{m+2} (2 x+1)^{-m-2}}{9 \left (m^2+5 m+6\right )}+\frac {322 (13-2 m) \left (2 m^2+52 m+579\right ) (3 x+2)^{m+1} (2 x+1)^{-m-1}}{(m+3) (m+4) \left (m^2+3 m+2\right )}+\frac {48668 (105-2 m) (3 x+2)^{m+1} (2 x+1)^{-m-1}}{9 \left (m^3+6 m^2+11 m+6\right )}-\frac {2}{3} (5-4 x)^4 (3 x+2)^{m+1} (2 x+1)^{-m-4}-\frac {7 (13-2 m) (5-4 x)^3 (3 x+2)^{m+1} (2 x+1)^{-m-4}}{3 (m+4)}+\frac {1127 (13-2 m) (2 m+27) (3 x+2)^{m+1} (2 x+1)^{-m-3}}{3 (m+3) (m+4)}-\frac {322 (13-2 m) (5-4 x) (3 x+2)^{m+1} (2 x+1)^{-m-3}}{3 (m+4)}+\frac {24334 (105-2 m) (3 x+2)^{m+1} (2 x+1)^{-m-3}}{81 (m+3)}-\frac {4232 (105-2 m) (3 x+2)^{m+2} (2 x+1)^{-m-3}}{27 (m+3)}+\frac {736 (105-2 m) (3 x+2)^{m+3} (2 x+1)^{-m-3}}{27 (m+3)} \end {gather*}
Antiderivative was successfully verified.
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Rule 37
Rule 47
Rule 71
Rule 80
Rule 92
Rule 96
Rule 102
Rule 128
Rule 165
Rubi steps
\begin {align*} \int (5-4 x)^5 (1+2 x)^{-5-m} (2+3 x)^m \, dx &=-\frac {2}{3} (5-4 x)^4 (1+2 x)^{-4-m} (2+3 x)^{1+m}+\frac {1}{6} \int (5-4 x)^3 (1+2 x)^{-5-m} (2+3 x)^m (-2 (119+10 m)-8 (105-2 m) x) \, dx\\ &=-\frac {2}{3} (5-4 x)^4 (1+2 x)^{-4-m} (2+3 x)^{1+m}+\frac {1}{3} (7 (13-2 m)) \int (5-4 x)^3 (1+2 x)^{-5-m} (2+3 x)^m \, dx-\frac {1}{3} (2 (105-2 m)) \int (5-4 x)^3 (1+2 x)^{-4-m} (2+3 x)^m \, dx\\ &=-\frac {7 (13-2 m) (5-4 x)^3 (1+2 x)^{-4-m} (2+3 x)^{1+m}}{3 (4+m)}-\frac {2}{3} (5-4 x)^4 (1+2 x)^{-4-m} (2+3 x)^{1+m}-\frac {1}{3} (2 (105-2 m)) \int \left (\frac {12167}{27} (1+2 x)^{-4-m} (2+3 x)^m-\frac {2116}{9} (1+2 x)^{-4-m} (2+3 x)^{1+m}+\frac {368}{9} (1+2 x)^{-4-m} (2+3 x)^{2+m}-\frac {64}{27} (1+2 x)^{-4-m} (2+3 x)^{3+m}\right ) \, dx-\frac {(161 (13-2 m)) \int (5-4 x)^2 (1+2 x)^{-4-m} (2+3 x)^m \, dx}{4+m}\\ &=-\frac {7 (13-2 m) (5-4 x)^3 (1+2 x)^{-4-m} (2+3 x)^{1+m}}{3 (4+m)}-\frac {2}{3} (5-4 x)^4 (1+2 x)^{-4-m} (2+3 x)^{1+m}-\frac {322 (13-2 m) (5-4 x) (1+2 x)^{-3-m} (2+3 x)^{1+m}}{3 (4+m)}+\frac {1}{81} (128 (105-2 m)) \int (1+2 x)^{-4-m} (2+3 x)^{3+m} \, dx-\frac {1}{27} (736 (105-2 m)) \int (1+2 x)^{-4-m} (2+3 x)^{2+m} \, dx+\frac {1}{27} (4232 (105-2 m)) \int (1+2 x)^{-4-m} (2+3 x)^{1+m} \, dx-\frac {1}{81} (24334 (105-2 m)) \int (1+2 x)^{-4-m} (2+3 x)^m \, dx+\frac {(161 (13-2 m)) \int (1+2 x)^{-4-m} (2+3 x)^m (-2 (181+10 m)+16 (2+m) x) \, dx}{6 (4+m)}\\ &=-\frac {7 (13-2 m) (5-4 x)^3 (1+2 x)^{-4-m} (2+3 x)^{1+m}}{3 (4+m)}-\frac {2}{3} (5-4 x)^4 (1+2 x)^{-4-m} (2+3 x)^{1+m}+\frac {24334 (105-2 m) (1+2 x)^{-3-m} (2+3 x)^{1+m}}{81 (3+m)}+\frac {1127 (13-2 m) (27+2 m) (1+2 x)^{-3-m} (2+3 x)^{1+m}}{3 (3+m) (4+m)}-\frac {322 (13-2 m) (5-4 x) (1+2 x)^{-3-m} (2+3 x)^{1+m}}{3 (4+m)}-\frac {4232 (105-2 m) (1+2 x)^{-3-m} (2+3 x)^{2+m}}{27 (3+m)}+\frac {736 (105-2 m) (1+2 x)^{-3-m} (2+3 x)^{3+m}}{27 (3+m)}-\frac {2^{3-m} (105-2 m) (1+2 x)^{-3-m} \, _2F_1(-3-m,-3-m;-2-m;-3 (1+2 x))}{81 (3+m)}-\frac {(4232 (105-2 m)) \int (1+2 x)^{-3-m} (2+3 x)^{1+m} \, dx}{9 (3+m)}+\frac {(48668 (105-2 m)) \int (1+2 x)^{-3-m} (2+3 x)^m \, dx}{27 (3+m)}+\frac {\left (322 (13-2 m) \left (579+52 m+2 m^2\right )\right ) \int (1+2 x)^{-3-m} (2+3 x)^m \, dx}{3 (3+m) (4+m)}\\ &=-\frac {7 (13-2 m) (5-4 x)^3 (1+2 x)^{-4-m} (2+3 x)^{1+m}}{3 (4+m)}-\frac {2}{3} (5-4 x)^4 (1+2 x)^{-4-m} (2+3 x)^{1+m}+\frac {24334 (105-2 m) (1+2 x)^{-3-m} (2+3 x)^{1+m}}{81 (3+m)}+\frac {1127 (13-2 m) (27+2 m) (1+2 x)^{-3-m} (2+3 x)^{1+m}}{3 (3+m) (4+m)}-\frac {322 (13-2 m) (5-4 x) (1+2 x)^{-3-m} (2+3 x)^{1+m}}{3 (4+m)}-\frac {48668 (105-2 m) (1+2 x)^{-2-m} (2+3 x)^{1+m}}{27 (2+m) (3+m)}-\frac {322 (13-2 m) \left (579+52 m+2 m^2\right ) (1+2 x)^{-2-m} (2+3 x)^{1+m}}{3 (2+m) (3+m) (4+m)}-\frac {4232 (105-2 m) (1+2 x)^{-3-m} (2+3 x)^{2+m}}{27 (3+m)}+\frac {4232 (105-2 m) (1+2 x)^{-2-m} (2+3 x)^{2+m}}{9 (2+m) (3+m)}+\frac {736 (105-2 m) (1+2 x)^{-3-m} (2+3 x)^{3+m}}{27 (3+m)}-\frac {2^{3-m} (105-2 m) (1+2 x)^{-3-m} \, _2F_1(-3-m,-3-m;-2-m;-3 (1+2 x))}{81 (3+m)}-\frac {(48668 (105-2 m)) \int (1+2 x)^{-2-m} (2+3 x)^m \, dx}{9 (2+m) (3+m)}-\frac {\left (322 (13-2 m) \left (579+52 m+2 m^2\right )\right ) \int (1+2 x)^{-2-m} (2+3 x)^m \, dx}{(2+m) (3+m) (4+m)}\\ &=-\frac {7 (13-2 m) (5-4 x)^3 (1+2 x)^{-4-m} (2+3 x)^{1+m}}{3 (4+m)}-\frac {2}{3} (5-4 x)^4 (1+2 x)^{-4-m} (2+3 x)^{1+m}+\frac {24334 (105-2 m) (1+2 x)^{-3-m} (2+3 x)^{1+m}}{81 (3+m)}+\frac {1127 (13-2 m) (27+2 m) (1+2 x)^{-3-m} (2+3 x)^{1+m}}{3 (3+m) (4+m)}-\frac {322 (13-2 m) (5-4 x) (1+2 x)^{-3-m} (2+3 x)^{1+m}}{3 (4+m)}-\frac {48668 (105-2 m) (1+2 x)^{-2-m} (2+3 x)^{1+m}}{27 (2+m) (3+m)}-\frac {322 (13-2 m) \left (579+52 m+2 m^2\right ) (1+2 x)^{-2-m} (2+3 x)^{1+m}}{3 (2+m) (3+m) (4+m)}+\frac {48668 (105-2 m) (1+2 x)^{-1-m} (2+3 x)^{1+m}}{9 (1+m) (2+m) (3+m)}+\frac {322 (13-2 m) \left (579+52 m+2 m^2\right ) (1+2 x)^{-1-m} (2+3 x)^{1+m}}{(1+m) (2+m) (3+m) (4+m)}-\frac {4232 (105-2 m) (1+2 x)^{-3-m} (2+3 x)^{2+m}}{27 (3+m)}+\frac {4232 (105-2 m) (1+2 x)^{-2-m} (2+3 x)^{2+m}}{9 (2+m) (3+m)}+\frac {736 (105-2 m) (1+2 x)^{-3-m} (2+3 x)^{3+m}}{27 (3+m)}-\frac {2^{3-m} (105-2 m) (1+2 x)^{-3-m} \, _2F_1(-3-m,-3-m;-2-m;-3 (1+2 x))}{81 (3+m)}\\ \end {align*}
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Mathematica [A]
time = 11.44, size = 274, normalized size = 0.50 \begin {gather*} \frac {3920 (1+2 x)^{-1-m} (2+3 x)^{1+m}}{1+m}+\frac {2^{4-m} (1+2 x)^{1-m} \, _2F_1(1-m,-m;2-m;-3-6 x)}{-1+m}-\frac {560 (-3-6 x)^m (1+2 x)^{-m} (2+3 x)^{1+m} \, _2F_1(1+m,1+m;2+m;4+6 x)}{1+m}-\frac {13720\ 3^{2+m} (-1-2 x)^m (1+2 x)^{-m} (2+3 x)^{1+m} \, _2F_1(1+m,3+m;2+m;4+6 x)}{1+m}-\frac {12005\ 3^{3+m} (-1-2 x)^m (2+4 x)^{-m} (4+6 x)^{1+m} \, _2F_1(1+m,4+m;2+m;4+6 x)}{1+m}-\frac {16807\ 3^{4+m} (-1-2 x)^m (1+2 x)^{-m} (2+3 x)^{1+m} \, _2F_1(1+m,5+m;2+m;4+6 x)}{1+m} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \left (5-4 x \right )^{5} \left (1+2 x \right )^{-5-m} \left (2+3 x \right )^{m}\, dx\]
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: SystemError} \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 \frac {{\left (3\,x+2\right )}^m\,{\left (4\,x-5\right )}^5}{{\left (2\,x+1\right )}^{m+5}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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