Optimal. Leaf size=546 \[ -\frac {2^{3-m} (105-2 m) (2 x+1)^{-m-3} \, _2F_1(-m-3,-m-3;-m-2;-3 (2 x+1))}{81 (m+3)}-\frac {322 (13-2 m) \left (2 m^2+52 m+579\right ) (3 x+2)^{m+1} (2 x+1)^{-m-2}}{3 (m+2) (m+3) (m+4)}-\frac {48668 (105-2 m) (3 x+2)^{m+1} (2 x+1)^{-m-2}}{27 \left (m^2+5 m+6\right )}+\frac {4232 (105-2 m) (3 x+2)^{m+2} (2 x+1)^{-m-2}}{9 \left (m^2+5 m+6\right )}+\frac {322 (13-2 m) \left (2 m^2+52 m+579\right ) (3 x+2)^{m+1} (2 x+1)^{-m-1}}{(m+3) (m+4) \left (m^2+3 m+2\right )}+\frac {48668 (105-2 m) (3 x+2)^{m+1} (2 x+1)^{-m-1}}{9 \left (m^3+6 m^2+11 m+6\right )}-\frac {2}{3} (5-4 x)^4 (3 x+2)^{m+1} (2 x+1)^{-m-4}-\frac {7 (13-2 m) (5-4 x)^3 (3 x+2)^{m+1} (2 x+1)^{-m-4}}{3 (m+4)}+\frac {1127 (13-2 m) (2 m+27) (3 x+2)^{m+1} (2 x+1)^{-m-3}}{3 (m+3) (m+4)}-\frac {322 (13-2 m) (5-4 x) (3 x+2)^{m+1} (2 x+1)^{-m-3}}{3 (m+4)}+\frac {24334 (105-2 m) (3 x+2)^{m+1} (2 x+1)^{-m-3}}{81 (m+3)}-\frac {4232 (105-2 m) (3 x+2)^{m+2} (2 x+1)^{-m-3}}{27 (m+3)}+\frac {736 (105-2 m) (3 x+2)^{m+3} (2 x+1)^{-m-3}}{27 (m+3)} \]
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Rubi [A] time = 0.45, 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 = {100, 159, 128, 45, 37, 69, 94, 90, 79} \[ -\frac {2^{3-m} (105-2 m) (2 x+1)^{-m-3} \, _2F_1(-m-3,-m-3;-m-2;-3 (2 x+1))}{81 (m+3)}-\frac {322 (13-2 m) \left (2 m^2+52 m+579\right ) (3 x+2)^{m+1} (2 x+1)^{-m-2}}{3 (m+2) (m+3) (m+4)}-\frac {48668 (105-2 m) (3 x+2)^{m+1} (2 x+1)^{-m-2}}{27 \left (m^2+5 m+6\right )}+\frac {4232 (105-2 m) (3 x+2)^{m+2} (2 x+1)^{-m-2}}{9 \left (m^2+5 m+6\right )}+\frac {322 (13-2 m) \left (2 m^2+52 m+579\right ) (3 x+2)^{m+1} (2 x+1)^{-m-1}}{(m+3) (m+4) \left (m^2+3 m+2\right )}+\frac {48668 (105-2 m) (3 x+2)^{m+1} (2 x+1)^{-m-1}}{9 \left (m^3+6 m^2+11 m+6\right )}-\frac {2}{3} (5-4 x)^4 (3 x+2)^{m+1} (2 x+1)^{-m-4}-\frac {7 (13-2 m) (5-4 x)^3 (3 x+2)^{m+1} (2 x+1)^{-m-4}}{3 (m+4)}+\frac {1127 (13-2 m) (2 m+27) (3 x+2)^{m+1} (2 x+1)^{-m-3}}{3 (m+3) (m+4)}-\frac {322 (13-2 m) (5-4 x) (3 x+2)^{m+1} (2 x+1)^{-m-3}}{3 (m+4)}+\frac {24334 (105-2 m) (3 x+2)^{m+1} (2 x+1)^{-m-3}}{81 (m+3)}-\frac {4232 (105-2 m) (3 x+2)^{m+2} (2 x+1)^{-m-3}}{27 (m+3)}+\frac {736 (105-2 m) (3 x+2)^{m+3} (2 x+1)^{-m-3}}{27 (m+3)} \]
Antiderivative was successfully verified.
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Rule 37
Rule 45
Rule 69
Rule 79
Rule 90
Rule 94
Rule 100
Rule 128
Rule 159
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 = 10.49, size = 274, normalized size = 0.50 \[ \frac {2^{4-m} (2 x+1)^{1-m} \, _2F_1(1-m,-m;2-m;-6 x-3)}{m-1}-\frac {560 (-6 x-3)^m (3 x+2)^{m+1} (2 x+1)^{-m} \, _2F_1(m+1,m+1;m+2;6 x+4)}{m+1}-\frac {13720\ 3^{m+2} (-2 x-1)^m (3 x+2)^{m+1} (2 x+1)^{-m} \, _2F_1(m+1,m+3;m+2;6 x+4)}{m+1}-\frac {16807\ 3^{m+4} (-2 x-1)^m (3 x+2)^{m+1} (2 x+1)^{-m} \, _2F_1(m+1,m+5;m+2;6 x+4)}{m+1}-\frac {12005\ 3^{m+3} (-2 x-1)^m (4 x+2)^{-m} (6 x+4)^{m+1} \, _2F_1(m+1,m+4;m+2;6 x+4)}{m+1}+\frac {3920 (3 x+2)^{m+1} (2 x+1)^{-m-1}}{m+1} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.89, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (1024 \, x^{5} - 6400 \, x^{4} + 16000 \, x^{3} - 20000 \, x^{2} + 12500 \, x - 3125\right )} {\left (3 \, x + 2\right )}^{m} {\left (2 \, x + 1\right )}^{-m - 5}, x\right ) \]
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 \[ \int -{\left (3 \, x + 2\right )}^{m} {\left (2 \, x + 1\right )}^{-m - 5} {\left (4 \, x - 5\right )}^{5}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.24, size = 0, normalized size = 0.00 \[ \int \left (-4 x +5\right )^{5} \left (2 x +1\right )^{-m -5} \left (3 x +2\right )^{m}\, dx \]
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 \[ -\int {\left (3 \, x + 2\right )}^{m} {\left (2 \, x + 1\right )}^{-m - 5} {\left (4 \, x - 5\right )}^{5}\,{d x} \]
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 \[ -\int \frac {{\left (3\,x+2\right )}^m\,{\left (4\,x-5\right )}^5}{{\left (2\,x+1\right )}^{m+5}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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