\(\int (5-x) (2+5 x+3 x^2)^{5/2} \, dx\) [2436]

Optimal result

Integrand size = 20, antiderivative size = 126 \[ \int (5-x) \left (2+5 x+3 x^2\right )^{5/2} \, dx=\frac {175 (5+6 x) \sqrt {2+5 x+3 x^2}}{82944}-\frac {175 (5+6 x) \left (2+5 x+3 x^2\right )^{3/2}}{10368}+\frac {35}{216} (5+6 x) \left (2+5 x+3 x^2\right )^{5/2}-\frac {1}{21} \left (2+5 x+3 x^2\right )^{7/2}-\frac {175 \text {arctanh}\left (\frac {5+6 x}{2 \sqrt {3} \sqrt {2+5 x+3 x^2}}\right )}{165888 \sqrt {3}} \]

[Out]

-175/10368*(5+6*x)*(3*x^2+5*x+2)^(3/2)+35/216*(5+6*x)*(3*x^2+5*x+2)^(5/2)-1/21*(3*x^2+5*x+2)^(7/2)-175/497664*
arctanh(1/6*(5+6*x)*3^(1/2)/(3*x^2+5*x+2)^(1/2))*3^(1/2)+175/82944*(5+6*x)*(3*x^2+5*x+2)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {654, 626, 635, 212} \[ \int (5-x) \left (2+5 x+3 x^2\right )^{5/2} \, dx=-\frac {175 \text {arctanh}\left (\frac {6 x+5}{2 \sqrt {3} \sqrt {3 x^2+5 x+2}}\right )}{165888 \sqrt {3}}-\frac {1}{21} \left (3 x^2+5 x+2\right )^{7/2}+\frac {35}{216} (6 x+5) \left (3 x^2+5 x+2\right )^{5/2}-\frac {175 (6 x+5) \left (3 x^2+5 x+2\right )^{3/2}}{10368}+\frac {175 (6 x+5) \sqrt {3 x^2+5 x+2}}{82944} \]

[In]

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

[Out]

(175*(5 + 6*x)*Sqrt[2 + 5*x + 3*x^2])/82944 - (175*(5 + 6*x)*(2 + 5*x + 3*x^2)^(3/2))/10368 + (35*(5 + 6*x)*(2
+ 5*x + 3*x^2)^(5/2))/216 - (2 + 5*x + 3*x^2)^(7/2)/21 - (175*ArcTanh[(5 + 6*x)/(2*Sqrt[3]*Sqrt[2 + 5*x + 3*x
^2])])/(165888*Sqrt[3])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 626

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^p/(2*c*(2*p + 1
))), x] - Dist[p*((b^2 - 4*a*c)/(2*c*(2*p + 1))), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 635

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 654

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*((a + b*x + c*x^2)^(p +
1)/(2*c*(p + 1))), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{21} \left (2+5 x+3 x^2\right )^{7/2}+\frac {35}{6} \int \left (2+5 x+3 x^2\right )^{5/2} \, dx \\ & = \frac {35}{216} (5+6 x) \left (2+5 x+3 x^2\right )^{5/2}-\frac {1}{21} \left (2+5 x+3 x^2\right )^{7/2}-\frac {175}{432} \int \left (2+5 x+3 x^2\right )^{3/2} \, dx \\ & = -\frac {175 (5+6 x) \left (2+5 x+3 x^2\right )^{3/2}}{10368}+\frac {35}{216} (5+6 x) \left (2+5 x+3 x^2\right )^{5/2}-\frac {1}{21} \left (2+5 x+3 x^2\right )^{7/2}+\frac {175 \int \sqrt {2+5 x+3 x^2} \, dx}{6912} \\ & = \frac {175 (5+6 x) \sqrt {2+5 x+3 x^2}}{82944}-\frac {175 (5+6 x) \left (2+5 x+3 x^2\right )^{3/2}}{10368}+\frac {35}{216} (5+6 x) \left (2+5 x+3 x^2\right )^{5/2}-\frac {1}{21} \left (2+5 x+3 x^2\right )^{7/2}-\frac {175 \int \frac {1}{\sqrt {2+5 x+3 x^2}} \, dx}{165888} \\ & = \frac {175 (5+6 x) \sqrt {2+5 x+3 x^2}}{82944}-\frac {175 (5+6 x) \left (2+5 x+3 x^2\right )^{3/2}}{10368}+\frac {35}{216} (5+6 x) \left (2+5 x+3 x^2\right )^{5/2}-\frac {1}{21} \left (2+5 x+3 x^2\right )^{7/2}-\frac {175 \text {Subst}\left (\int \frac {1}{12-x^2} \, dx,x,\frac {5+6 x}{\sqrt {2+5 x+3 x^2}}\right )}{82944} \\ & = \frac {175 (5+6 x) \sqrt {2+5 x+3 x^2}}{82944}-\frac {175 (5+6 x) \left (2+5 x+3 x^2\right )^{3/2}}{10368}+\frac {35}{216} (5+6 x) \left (2+5 x+3 x^2\right )^{5/2}-\frac {1}{21} \left (2+5 x+3 x^2\right )^{7/2}-\frac {175 \tanh ^{-1}\left (\frac {5+6 x}{2 \sqrt {3} \sqrt {2+5 x+3 x^2}}\right )}{165888 \sqrt {3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.64 \[ \int (5-x) \left (2+5 x+3 x^2\right )^{5/2} \, dx=\frac {-3 \sqrt {2+5 x+3 x^2} \left (-1568541-9651790 x-23110872 x^2-26388720 x^3-13454208 x^4-1347840 x^5+746496 x^6\right )-1225 \sqrt {3} \text {arctanh}\left (\frac {\sqrt {\frac {2}{3}+\frac {5 x}{3}+x^2}}{1+x}\right )}{1741824} \]

[In]

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

[Out]

(-3*Sqrt[2 + 5*x + 3*x^2]*(-1568541 - 9651790*x - 23110872*x^2 - 26388720*x^3 - 13454208*x^4 - 1347840*x^5 + 7
46496*x^6) - 1225*Sqrt[3]*ArcTanh[Sqrt[2/3 + (5*x)/3 + x^2]/(1 + x)])/1741824

Maple [A] (verified)

Time = 0.33 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.60

method	result	size
risch	\(-\frac {\left (746496 x^{6}-1347840 x^{5}-13454208 x^{4}-26388720 x^{3}-23110872 x^{2}-9651790 x -1568541\right ) \sqrt {3 x^{2}+5 x +2}}{580608}-\frac {175 \ln \left (\frac {\left (\frac {5}{2}+3 x \right ) \sqrt {3}}{3}+\sqrt {3 x^{2}+5 x +2}\right ) \sqrt {3}}{497664}\)	\(75\)
trager	\(\left (-\frac {9}{7} x^{6}+\frac {65}{28} x^{5}+\frac {3893}{168} x^{4}+\frac {61085}{1344} x^{3}+\frac {962953}{24192} x^{2}+\frac {4825895}{290304} x +\frac {522847}{193536}\right ) \sqrt {3 x^{2}+5 x +2}+\frac {175 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (-6 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) x +6 \sqrt {3 x^{2}+5 x +2}-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right )\right )}{497664}\)	\(86\)
default	\(\frac {35 \left (5+6 x \right ) \left (3 x^{2}+5 x +2\right )^{\frac {5}{2}}}{216}-\frac {175 \left (5+6 x \right ) \left (3 x^{2}+5 x +2\right )^{\frac {3}{2}}}{10368}+\frac {175 \left (5+6 x \right ) \sqrt {3 x^{2}+5 x +2}}{82944}-\frac {175 \ln \left (\frac {\left (\frac {5}{2}+3 x \right ) \sqrt {3}}{3}+\sqrt {3 x^{2}+5 x +2}\right ) \sqrt {3}}{497664}-\frac {\left (3 x^{2}+5 x +2\right )^{\frac {7}{2}}}{21}\)	\(102\)

[In]

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

[Out]

-1/580608*(746496*x^6-1347840*x^5-13454208*x^4-26388720*x^3-23110872*x^2-9651790*x-1568541)*(3*x^2+5*x+2)^(1/2
)-175/497664*ln(1/3*(5/2+3*x)*3^(1/2)+(3*x^2+5*x+2)^(1/2))*3^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.66 \[ \int (5-x) \left (2+5 x+3 x^2\right )^{5/2} \, dx=-\frac {1}{580608} \, {\left (746496 \, x^{6} - 1347840 \, x^{5} - 13454208 \, x^{4} - 26388720 \, x^{3} - 23110872 \, x^{2} - 9651790 \, x - 1568541\right )} \sqrt {3 \, x^{2} + 5 \, x + 2} + \frac {175}{995328} \, \sqrt {3} \log \left (-4 \, \sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (6 \, x + 5\right )} + 72 \, x^{2} + 120 \, x + 49\right ) \]

[In]

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

[Out]

-1/580608*(746496*x^6 - 1347840*x^5 - 13454208*x^4 - 26388720*x^3 - 23110872*x^2 - 9651790*x - 1568541)*sqrt(3
*x^2 + 5*x + 2) + 175/995328*sqrt(3)*log(-4*sqrt(3)*sqrt(3*x^2 + 5*x + 2)*(6*x + 5) + 72*x^2 + 120*x + 49)

Sympy [A] (verification not implemented)

Time = 0.59 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.71 \[ \int (5-x) \left (2+5 x+3 x^2\right )^{5/2} \, dx=\sqrt {3 x^{2} + 5 x + 2} \left (- \frac {9 x^{6}}{7} + \frac {65 x^{5}}{28} + \frac {3893 x^{4}}{168} + \frac {61085 x^{3}}{1344} + \frac {962953 x^{2}}{24192} + \frac {4825895 x}{290304} + \frac {522847}{193536}\right ) - \frac {175 \sqrt {3} \log {\left (6 x + 2 \sqrt {3} \sqrt {3 x^{2} + 5 x + 2} + 5 \right )}}{497664} \]

[In]

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

[Out]

sqrt(3*x**2 + 5*x + 2)*(-9*x**6/7 + 65*x**5/28 + 3893*x**4/168 + 61085*x**3/1344 + 962953*x**2/24192 + 4825895
*x/290304 + 522847/193536) - 175*sqrt(3)*log(6*x + 2*sqrt(3)*sqrt(3*x**2 + 5*x + 2) + 5)/497664

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.03 \[ \int (5-x) \left (2+5 x+3 x^2\right )^{5/2} \, dx=-\frac {1}{21} \, {\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {7}{2}} + \frac {35}{36} \, {\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {5}{2}} x + \frac {175}{216} \, {\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {5}{2}} - \frac {175}{1728} \, {\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}} x - \frac {875}{10368} \, {\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}} + \frac {175}{13824} \, \sqrt {3 \, x^{2} + 5 \, x + 2} x - \frac {175}{497664} \, \sqrt {3} \log \left (2 \, \sqrt {3} \sqrt {3 \, x^{2} + 5 \, x + 2} + 6 \, x + 5\right ) + \frac {875}{82944} \, \sqrt {3 \, x^{2} + 5 \, x + 2} \]

[In]

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

[Out]

-1/21*(3*x^2 + 5*x + 2)^(7/2) + 35/36*(3*x^2 + 5*x + 2)^(5/2)*x + 175/216*(3*x^2 + 5*x + 2)^(5/2) - 175/1728*(
3*x^2 + 5*x + 2)^(3/2)*x - 875/10368*(3*x^2 + 5*x + 2)^(3/2) + 175/13824*sqrt(3*x^2 + 5*x + 2)*x - 175/497664*
sqrt(3)*log(2*sqrt(3)*sqrt(3*x^2 + 5*x + 2) + 6*x + 5) + 875/82944*sqrt(3*x^2 + 5*x + 2)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.63 \[ \int (5-x) \left (2+5 x+3 x^2\right )^{5/2} \, dx=-\frac {1}{580608} \, {\left (2 \, {\left (12 \, {\left (18 \, {\left (8 \, {\left (6 \, {\left (36 \, x - 65\right )} x - 3893\right )} x - 61085\right )} x - 962953\right )} x - 4825895\right )} x - 1568541\right )} \sqrt {3 \, x^{2} + 5 \, x + 2} + \frac {175}{497664} \, \sqrt {3} \log \left ({\left | -2 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 5 \, x + 2}\right )} - 5 \right |}\right ) \]

[In]

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

[Out]

-1/580608*(2*(12*(18*(8*(6*(36*x - 65)*x - 3893)*x - 61085)*x - 962953)*x - 4825895)*x - 1568541)*sqrt(3*x^2 +
5*x + 2) + 175/497664*sqrt(3)*log(abs(-2*sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 + 5*x + 2)) - 5))

Mupad [F(-1)]

Timed out. \[ \int (5-x) \left (2+5 x+3 x^2\right )^{5/2} \, dx=\int -\left (x-5\right )\,{\left (3\,x^2+5\,x+2\right )}^{5/2} \,d x \]

[In]

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

[Out]

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