Integrand size = 26, antiderivative size = 142 \[ \int (5-4 x)^3 (1+2 x)^{-1-m} (2+3 x)^m \, dx=-\frac {2}{9} (5-4 x)^2 (1+2 x)^{-m} (2+3 x)^{1+m}-\frac {(1+2 x)^{-m} (2+3 x)^{1+m} \left (9261-512 m+4 m^2-4 (109-2 m) m x\right )}{27 m}+\frac {2^{-1-m} \left (27783-8324 m+390 m^2-4 m^3\right ) (1+2 x)^{1-m} \operatorname {Hypergeometric2F1}(1-m,-m,2-m,-3 (1+2 x))}{27 (1-m) m} \]
[Out]
Time = 0.12 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {102, 151, 71} \[ \int (5-4 x)^3 (1+2 x)^{-1-m} (2+3 x)^m \, dx=\frac {2^{-m-1} \left (-4 m^3+390 m^2-8324 m+27783\right ) (2 x+1)^{1-m} \operatorname {Hypergeometric2F1}(1-m,-m,2-m,-3 (2 x+1))}{27 (1-m) m}-\frac {(3 x+2)^{m+1} \left (4 m^2-4 (109-2 m) m x-512 m+9261\right ) (2 x+1)^{-m}}{27 m}-\frac {2}{9} (5-4 x)^2 (3 x+2)^{m+1} (2 x+1)^{-m} \]
[In]
[Out]
Rule 71
Rule 102
Rule 151
Rubi steps \begin{align*} \text {integral}& = -\frac {2}{9} (5-4 x)^2 (1+2 x)^{-m} (2+3 x)^{1+m}+\frac {1}{18} \int (5-4 x) (1+2 x)^{-1-m} (2+3 x)^m (2 (223-10 m)-8 (109-2 m) x) \, dx \\ & = -\frac {2}{9} (5-4 x)^2 (1+2 x)^{-m} (2+3 x)^{1+m}-\frac {(1+2 x)^{-m} (2+3 x)^{1+m} \left (9261-512 m+4 m^2-4 (109-2 m) m x\right )}{27 m}+\frac {(288 (109-2 m) (1+m) (2+m)+6 (1+m) (-6 (-8 (223-10 m)-40 (109-2 m))-128 (109-2 m) m)+4 (180 (223-10 m)+12 (-8 (223-10 m)-40 (109-2 m)) m-128 (109-2 m) (1-m) m)) \int (1+2 x)^{-m} (2+3 x)^m \, dx}{432 m} \\ & = -\frac {2}{9} (5-4 x)^2 (1+2 x)^{-m} (2+3 x)^{1+m}-\frac {(1+2 x)^{-m} (2+3 x)^{1+m} \left (9261-512 m+4 m^2-4 (109-2 m) m x\right )}{27 m}+\frac {2^{-1-m} \left (27783-8324 m+390 m^2-4 m^3\right ) (1+2 x)^{1-m} \, _2F_1(1-m,-m;2-m;-3 (1+2 x))}{27 (1-m) m} \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.75 \[ \int (5-4 x)^3 (1+2 x)^{-1-m} (2+3 x)^m \, dx=\frac {2^{-1-m} (1+2 x)^{-m} \left (-\left ((-1+m) (4+6 x)^{1+m} \left (9261+m^2 (4+8 x)+m \left (-362-676 x+96 x^2\right )\right )\right )+\left (-27783+8324 m-390 m^2+4 m^3\right ) (1+2 x) \operatorname {Hypergeometric2F1}(1-m,-m,2-m,-3-6 x)\right )}{27 (-1+m) m} \]
[In]
[Out]
\[\int \left (5-4 x \right )^{3} \left (1+2 x \right )^{-1-m} \left (2+3 x \right )^{m}d x\]
[In]
[Out]
\[ \int (5-4 x)^3 (1+2 x)^{-1-m} (2+3 x)^m \, dx=\int { -{\left (3 \, x + 2\right )}^{m} {\left (2 \, x + 1\right )}^{-m - 1} {\left (4 \, x - 5\right )}^{3} \,d x } \]
[In]
[Out]
\[ \int (5-4 x)^3 (1+2 x)^{-1-m} (2+3 x)^m \, dx=- \int \left (- 125 \left (2 x + 1\right )^{- m - 1} \left (3 x + 2\right )^{m}\right )\, dx - \int 300 x \left (2 x + 1\right )^{- m - 1} \left (3 x + 2\right )^{m}\, dx - \int \left (- 240 x^{2} \left (2 x + 1\right )^{- m - 1} \left (3 x + 2\right )^{m}\right )\, dx - \int 64 x^{3} \left (2 x + 1\right )^{- m - 1} \left (3 x + 2\right )^{m}\, dx \]
[In]
[Out]
\[ \int (5-4 x)^3 (1+2 x)^{-1-m} (2+3 x)^m \, dx=\int { -{\left (3 \, x + 2\right )}^{m} {\left (2 \, x + 1\right )}^{-m - 1} {\left (4 \, x - 5\right )}^{3} \,d x } \]
[In]
[Out]
\[ \int (5-4 x)^3 (1+2 x)^{-1-m} (2+3 x)^m \, dx=\int { -{\left (3 \, x + 2\right )}^{m} {\left (2 \, x + 1\right )}^{-m - 1} {\left (4 \, x - 5\right )}^{3} \,d x } \]
[In]
[Out]
Timed out. \[ \int (5-4 x)^3 (1+2 x)^{-1-m} (2+3 x)^m \, dx=-\int \frac {{\left (3\,x+2\right )}^m\,{\left (4\,x-5\right )}^3}{{\left (2\,x+1\right )}^{m+1}} \,d x \]
[In]
[Out]