\(\int (5-x) (3+2 x)^{3/2} (2+5 x+3 x^2)^2 \, dx\) [2536]

Optimal result

Integrand size = 27, antiderivative size = 79 \[ \int (5-x) (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )^2 \, dx=\frac {65}{32} (3+2 x)^{5/2}-\frac {1065}{224} (3+2 x)^{7/2}+\frac {217}{48} (3+2 x)^{9/2}-\frac {359}{176} (3+2 x)^{11/2}+\frac {165}{416} (3+2 x)^{13/2}-\frac {3}{160} (3+2 x)^{15/2} \]

[Out]

65/32*(3+2*x)^(5/2)-1065/224*(3+2*x)^(7/2)+217/48*(3+2*x)^(9/2)-359/176*(3+2*x)^(11/2)+165/416*(3+2*x)^(13/2)-
3/160*(3+2*x)^(15/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {785} \[ \int (5-x) (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )^2 \, dx=-\frac {3}{160} (2 x+3)^{15/2}+\frac {165}{416} (2 x+3)^{13/2}-\frac {359}{176} (2 x+3)^{11/2}+\frac {217}{48} (2 x+3)^{9/2}-\frac {1065}{224} (2 x+3)^{7/2}+\frac {65}{32} (2 x+3)^{5/2} \]

[In]

Int[(5 - x)*(3 + 2*x)^(3/2)*(2 + 5*x + 3*x^2)^2,x]

[Out]

(65*(3 + 2*x)^(5/2))/32 - (1065*(3 + 2*x)^(7/2))/224 + (217*(3 + 2*x)^(9/2))/48 - (359*(3 + 2*x)^(11/2))/176 +
(165*(3 + 2*x)^(13/2))/416 - (3*(3 + 2*x)^(15/2))/160

Rule 785

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(d + e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && N
eQ[b^2 - 4*a*c, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {325}{32} (3+2 x)^{3/2}-\frac {1065}{32} (3+2 x)^{5/2}+\frac {651}{16} (3+2 x)^{7/2}-\frac {359}{16} (3+2 x)^{9/2}+\frac {165}{32} (3+2 x)^{11/2}-\frac {9}{32} (3+2 x)^{13/2}\right ) \, dx \\ & = \frac {65}{32} (3+2 x)^{5/2}-\frac {1065}{224} (3+2 x)^{7/2}+\frac {217}{48} (3+2 x)^{9/2}-\frac {359}{176} (3+2 x)^{11/2}+\frac {165}{416} (3+2 x)^{13/2}-\frac {3}{160} (3+2 x)^{15/2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.48 \[ \int (5-x) (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )^2 \, dx=-\frac {(3+2 x)^{5/2} \left (-14304-76260 x-151270 x^2-124005 x^3-27720 x^4+9009 x^5\right )}{15015} \]

[In]

Integrate[(5 - x)*(3 + 2*x)^(3/2)*(2 + 5*x + 3*x^2)^2,x]

[Out]

-1/15015*((3 + 2*x)^(5/2)*(-14304 - 76260*x - 151270*x^2 - 124005*x^3 - 27720*x^4 + 9009*x^5))

Maple [A] (verified)

Time = 0.34 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.44

method	result	size
gosper	\(-\frac {\left (9009 x^{5}-27720 x^{4}-124005 x^{3}-151270 x^{2}-76260 x -14304\right ) \left (3+2 x \right )^{\frac {5}{2}}}{15015}\)	\(35\)
pseudoelliptic	\(-\frac {12 \left (x^{5}-\frac {40}{13} x^{4}-\frac {5905}{429} x^{3}-\frac {21610}{1287} x^{2}-\frac {25420}{3003} x -\frac {4768}{3003}\right ) \sqrt {3+2 x}\, \left (x +\frac {3}{2}\right )^{2}}{5}\)	\(38\)
trager	\(\left (-\frac {12}{5} x^{7}+\frac {12}{65} x^{6}+\frac {35599}{715} x^{5}+\frac {66932}{429} x^{4}+\frac {215755}{1001} x^{3}+\frac {777922}{5005} x^{2}+\frac {285996}{5005} x +\frac {42912}{5005}\right ) \sqrt {3+2 x}\)	\(44\)
risch	\(-\frac {\left (36036 x^{7}-2772 x^{6}-747579 x^{5}-2342620 x^{4}-3236325 x^{3}-2333766 x^{2}-857988 x -128736\right ) \sqrt {3+2 x}}{15015}\)	\(45\)
derivativedivides	\(\frac {65 \left (3+2 x \right )^{\frac {5}{2}}}{32}-\frac {1065 \left (3+2 x \right )^{\frac {7}{2}}}{224}+\frac {217 \left (3+2 x \right )^{\frac {9}{2}}}{48}-\frac {359 \left (3+2 x \right )^{\frac {11}{2}}}{176}+\frac {165 \left (3+2 x \right )^{\frac {13}{2}}}{416}-\frac {3 \left (3+2 x \right )^{\frac {15}{2}}}{160}\)	\(56\)
default	\(\frac {65 \left (3+2 x \right )^{\frac {5}{2}}}{32}-\frac {1065 \left (3+2 x \right )^{\frac {7}{2}}}{224}+\frac {217 \left (3+2 x \right )^{\frac {9}{2}}}{48}-\frac {359 \left (3+2 x \right )^{\frac {11}{2}}}{176}+\frac {165 \left (3+2 x \right )^{\frac {13}{2}}}{416}-\frac {3 \left (3+2 x \right )^{\frac {15}{2}}}{160}\)	\(56\)
meijerg	\(\frac {40095 \sqrt {3}\, \left (-\frac {64 \sqrt {\pi }}{945}+\frac {4 \sqrt {\pi }\, \left (\frac {1120}{81} x^{4}+\frac {800}{27} x^{3}+\frac {8}{3} x^{2}-\frac {16}{3} x +16\right ) \sqrt {1+\frac {2 x}{3}}}{945}\right )}{32 \sqrt {\pi }}+\frac {82377 \sqrt {3}\, \left (\frac {128 \sqrt {\pi }}{3465}-\frac {\sqrt {\pi }\, \left (-\frac {8960}{81} x^{5}-\frac {17920}{81} x^{4}-\frac {320}{27} x^{3}+\frac {64}{3} x^{2}-\frac {128}{3} x +128\right ) \sqrt {1+\frac {2 x}{3}}}{3465}\right )}{64 \sqrt {\pi }}+\frac {32805 \sqrt {3}\, \left (-\frac {1024 \sqrt {\pi }}{45045}+\frac {4 \sqrt {\pi }\, \left (\frac {49280}{243} x^{6}+\frac {31360}{81} x^{5}+\frac {1120}{81} x^{4}-\frac {640}{27} x^{3}+\frac {128}{3} x^{2}-\frac {256}{3} x +256\right ) \sqrt {1+\frac {2 x}{3}}}{45045}\right )}{128 \sqrt {\pi }}+\frac {486 \sqrt {3}\, \left (\frac {16 \sqrt {\pi }}{105}-\frac {2 \sqrt {\pi }\, \left (-\frac {160}{27} x^{3}-\frac {128}{9} x^{2}-\frac {8}{3} x +8\right ) \sqrt {1+\frac {2 x}{3}}}{105}\right )}{\sqrt {\pi }}+\frac {135 \sqrt {3}\, \left (-\frac {8 \sqrt {\pi }}{15}+\frac {4 \sqrt {\pi }\, \left (\frac {8}{9} x^{2}+\frac {8}{3} x +2\right ) \sqrt {1+\frac {2 x}{3}}}{15}\right )}{2 \sqrt {\pi }}-\frac {59049 \sqrt {3}\, \left (\frac {2048 \sqrt {\pi }}{135135}-\frac {2 \sqrt {\pi }\, \left (-\frac {512512}{729} x^{7}-\frac {315392}{243} x^{6}-\frac {896}{27} x^{5}+\frac {4480}{81} x^{4}-\frac {2560}{27} x^{3}+\frac {512}{3} x^{2}-\frac {1024}{3} x +1024\right ) \sqrt {1+\frac {2 x}{3}}}{135135}\right )}{256 \sqrt {\pi }}\)	\(293\)

[In]

int((5-x)*(3+2*x)^(3/2)*(3*x^2+5*x+2)^2,x,method=_RETURNVERBOSE)

[Out]

-1/15015*(9009*x^5-27720*x^4-124005*x^3-151270*x^2-76260*x-14304)*(3+2*x)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.56 \[ \int (5-x) (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )^2 \, dx=-\frac {1}{15015} \, {\left (36036 \, x^{7} - 2772 \, x^{6} - 747579 \, x^{5} - 2342620 \, x^{4} - 3236325 \, x^{3} - 2333766 \, x^{2} - 857988 \, x - 128736\right )} \sqrt {2 \, x + 3} \]

[In]

integrate((5-x)*(3+2*x)^(3/2)*(3*x^2+5*x+2)^2,x, algorithm="fricas")

[Out]

-1/15015*(36036*x^7 - 2772*x^6 - 747579*x^5 - 2342620*x^4 - 3236325*x^3 - 2333766*x^2 - 857988*x - 128736)*sqr
t(2*x + 3)

Sympy [A] (verification not implemented)

Time = 0.99 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.89 \[ \int (5-x) (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )^2 \, dx=- \frac {3 \left (2 x + 3\right )^{\frac {15}{2}}}{160} + \frac {165 \left (2 x + 3\right )^{\frac {13}{2}}}{416} - \frac {359 \left (2 x + 3\right )^{\frac {11}{2}}}{176} + \frac {217 \left (2 x + 3\right )^{\frac {9}{2}}}{48} - \frac {1065 \left (2 x + 3\right )^{\frac {7}{2}}}{224} + \frac {65 \left (2 x + 3\right )^{\frac {5}{2}}}{32} \]

[In]

integrate((5-x)*(3+2*x)**(3/2)*(3*x**2+5*x+2)**2,x)

[Out]

-3*(2*x + 3)**(15/2)/160 + 165*(2*x + 3)**(13/2)/416 - 359*(2*x + 3)**(11/2)/176 + 217*(2*x + 3)**(9/2)/48 - 1
065*(2*x + 3)**(7/2)/224 + 65*(2*x + 3)**(5/2)/32

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.70 \[ \int (5-x) (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )^2 \, dx=-\frac {3}{160} \, {\left (2 \, x + 3\right )}^{\frac {15}{2}} + \frac {165}{416} \, {\left (2 \, x + 3\right )}^{\frac {13}{2}} - \frac {359}{176} \, {\left (2 \, x + 3\right )}^{\frac {11}{2}} + \frac {217}{48} \, {\left (2 \, x + 3\right )}^{\frac {9}{2}} - \frac {1065}{224} \, {\left (2 \, x + 3\right )}^{\frac {7}{2}} + \frac {65}{32} \, {\left (2 \, x + 3\right )}^{\frac {5}{2}} \]

[In]

integrate((5-x)*(3+2*x)^(3/2)*(3*x^2+5*x+2)^2,x, algorithm="maxima")

[Out]

-3/160*(2*x + 3)^(15/2) + 165/416*(2*x + 3)^(13/2) - 359/176*(2*x + 3)^(11/2) + 217/48*(2*x + 3)^(9/2) - 1065/
224*(2*x + 3)^(7/2) + 65/32*(2*x + 3)^(5/2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.70 \[ \int (5-x) (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )^2 \, dx=-\frac {3}{160} \, {\left (2 \, x + 3\right )}^{\frac {15}{2}} + \frac {165}{416} \, {\left (2 \, x + 3\right )}^{\frac {13}{2}} - \frac {359}{176} \, {\left (2 \, x + 3\right )}^{\frac {11}{2}} + \frac {217}{48} \, {\left (2 \, x + 3\right )}^{\frac {9}{2}} - \frac {1065}{224} \, {\left (2 \, x + 3\right )}^{\frac {7}{2}} + \frac {65}{32} \, {\left (2 \, x + 3\right )}^{\frac {5}{2}} \]

[In]

integrate((5-x)*(3+2*x)^(3/2)*(3*x^2+5*x+2)^2,x, algorithm="giac")

[Out]

-3/160*(2*x + 3)^(15/2) + 165/416*(2*x + 3)^(13/2) - 359/176*(2*x + 3)^(11/2) + 217/48*(2*x + 3)^(9/2) - 1065/
224*(2*x + 3)^(7/2) + 65/32*(2*x + 3)^(5/2)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.70 \[ \int (5-x) (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )^2 \, dx=\frac {65\,{\left (2\,x+3\right )}^{5/2}}{32}-\frac {1065\,{\left (2\,x+3\right )}^{7/2}}{224}+\frac {217\,{\left (2\,x+3\right )}^{9/2}}{48}-\frac {359\,{\left (2\,x+3\right )}^{11/2}}{176}+\frac {165\,{\left (2\,x+3\right )}^{13/2}}{416}-\frac {3\,{\left (2\,x+3\right )}^{15/2}}{160} \]

[In]

int(-(2*x + 3)^(3/2)*(x - 5)*(5*x + 3*x^2 + 2)^2,x)

[Out]

(65*(2*x + 3)^(5/2))/32 - (1065*(2*x + 3)^(7/2))/224 + (217*(2*x + 3)^(9/2))/48 - (359*(2*x + 3)^(11/2))/176 +
(165*(2*x + 3)^(13/2))/416 - (3*(2*x + 3)^(15/2))/160