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 = 0.446832, antiderivative size = 546, normalized size of antiderivative = 1., 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 100
Rule 159
Rule 128
Rule 45
Rule 37
Rule 69
Rule 94
Rule 90
Rule 79
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*}
Mathematica [A] time = 10.469, size = 274, normalized size = 0.5 \[ \frac{2^{4-m} (2 x+1)^{1-m} \, _2F_1(1-m,-m;2-m;-6 x-3)}{m-1}-\frac{560 (-6 x-3)^m (3 x+2)^{m+1} (2 x+1)^{-m} \, _2F_1(m+1,m+1;m+2;6 x+4)}{m+1}-\frac{13720\ 3^{m+2} (-2 x-1)^m (3 x+2)^{m+1} (2 x+1)^{-m} \, _2F_1(m+1,m+3;m+2;6 x+4)}{m+1}-\frac{16807\ 3^{m+4} (-2 x-1)^m (3 x+2)^{m+1} (2 x+1)^{-m} \, _2F_1(m+1,m+5;m+2;6 x+4)}{m+1}-\frac{12005\ 3^{m+3} (-2 x-1)^m (4 x+2)^{-m} (6 x+4)^{m+1} \, _2F_1(m+1,m+4;m+2;6 x+4)}{m+1}+\frac{3920 (3 x+2)^{m+1} (2 x+1)^{-m-1}}{m+1} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.057, size = 0, normalized size = 0. \begin{align*} \int \left ( 5-4\,x \right ) ^{5} \left ( 1+2\,x \right ) ^{-5-m} \left ( 2+3\,x \right ) ^{m}\, 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 (3 \, x + 2\right )}^{m}{\left (2 \, x + 1\right )}^{-m - 5}{\left (4 \, x - 5\right )}^{5}\,{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 (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 ) \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 (3 \, x + 2\right )}^{m}{\left (2 \, x + 1\right )}^{-m - 5}{\left (4 \, x - 5\right )}^{5}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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