\(\int \sqrt {3-x+2 x^2} (2+3 x+5 x^2)^2 \, dx\) [94]

Optimal result

Integrand size = 27, antiderivative size = 124 \[ \int \sqrt {3-x+2 x^2} \left (2+3 x+5 x^2\right )^2 \, dx=\frac {12371 (1-4 x) \sqrt {3-x+2 x^2}}{16384}-\frac {2107 \left (3-x+2 x^2\right )^{3/2}}{3072}+\frac {769}{256} x \left (3-x+2 x^2\right )^{3/2}+\frac {63}{16} x^2 \left (3-x+2 x^2\right )^{3/2}+\frac {25}{12} x^3 \left (3-x+2 x^2\right )^{3/2}+\frac {284533 \text {arcsinh}\left (\frac {1-4 x}{\sqrt {23}}\right )}{32768 \sqrt {2}} \] Output:

12371/16384*(1-4*x)*(2*x^2-x+3)^(1/2)-2107/3072*(2*x^2-x+3)^(3/2)+769/256* 
x*(2*x^2-x+3)^(3/2)+63/16*x^2*(2*x^2-x+3)^(3/2)+25/12*x^3*(2*x^2-x+3)^(3/2 
)+284533/65536*arcsinh(1/23*(1-4*x)*23^(1/2))*2^(1/2)
 

Mathematica [A] (verified)

Time = 0.65 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.60 \[ \int \sqrt {3-x+2 x^2} \left (2+3 x+5 x^2\right )^2 \, dx=\frac {4 \sqrt {3-x+2 x^2} \left (-64023+328204 x+365536 x^2+408960 x^3+284672 x^4+204800 x^5\right )+853599 \sqrt {2} \log \left (1-4 x+2 \sqrt {6-2 x+4 x^2}\right )}{196608} \] Input:

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

Output:

(4*Sqrt[3 - x + 2*x^2]*(-64023 + 328204*x + 365536*x^2 + 408960*x^3 + 2846 
72*x^4 + 204800*x^5) + 853599*Sqrt[2]*Log[1 - 4*x + 2*Sqrt[6 - 2*x + 4*x^2 
]])/196608
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.16, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {2192, 27, 2192, 27, 2192, 27, 1160, 1087, 1090, 222}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \sqrt {2 x^2-x+3} \left (5 x^2+3 x+2\right )^2 \, dx\)

\(\Big \downarrow \) 2192

\(\displaystyle \frac {1}{12} \int \frac {3}{2} \sqrt {2 x^2-x+3} \left (315 x^3+82 x^2+96 x+32\right )dx+\frac {25}{12} \left (2 x^2-x+3\right )^{3/2} x^3\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{8} \int \sqrt {2 x^2-x+3} \left (315 x^3+82 x^2+96 x+32\right )dx+\frac {25}{12} \left (2 x^2-x+3\right )^{3/2} x^3\)

\(\Big \downarrow \) 2192

\(\displaystyle \frac {1}{8} \left (\frac {1}{10} \int \frac {5}{2} \sqrt {2 x^2-x+3} \left (769 x^2-372 x+128\right )dx+\frac {63}{2} \left (2 x^2-x+3\right )^{3/2} x^2\right )+\frac {25}{12} \left (2 x^2-x+3\right )^{3/2} x^3\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{8} \left (\frac {1}{4} \int \sqrt {2 x^2-x+3} \left (769 x^2-372 x+128\right )dx+\frac {63}{2} \left (2 x^2-x+3\right )^{3/2} x^2\right )+\frac {25}{12} \left (2 x^2-x+3\right )^{3/2} x^3\)

\(\Big \downarrow \) 2192

\(\displaystyle \frac {1}{8} \left (\frac {1}{4} \left (\frac {1}{8} \int -\frac {1}{2} (2107 x+2566) \sqrt {2 x^2-x+3}dx+\frac {769}{8} x \left (2 x^2-x+3\right )^{3/2}\right )+\frac {63}{2} \left (2 x^2-x+3\right )^{3/2} x^2\right )+\frac {25}{12} \left (2 x^2-x+3\right )^{3/2} x^3\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{8} \left (\frac {1}{4} \left (\frac {769}{8} x \left (2 x^2-x+3\right )^{3/2}-\frac {1}{16} \int (2107 x+2566) \sqrt {2 x^2-x+3}dx\right )+\frac {63}{2} \left (2 x^2-x+3\right )^{3/2} x^2\right )+\frac {25}{12} \left (2 x^2-x+3\right )^{3/2} x^3\)

\(\Big \downarrow \) 1160

\(\displaystyle \frac {1}{8} \left (\frac {1}{4} \left (\frac {1}{16} \left (-\frac {12371}{4} \int \sqrt {2 x^2-x+3}dx-\frac {2107}{6} \left (2 x^2-x+3\right )^{3/2}\right )+\frac {769}{8} x \left (2 x^2-x+3\right )^{3/2}\right )+\frac {63}{2} \left (2 x^2-x+3\right )^{3/2} x^2\right )+\frac {25}{12} \left (2 x^2-x+3\right )^{3/2} x^3\)

\(\Big \downarrow \) 1087

\(\displaystyle \frac {1}{8} \left (\frac {1}{4} \left (\frac {1}{16} \left (-\frac {12371}{4} \left (\frac {23}{16} \int \frac {1}{\sqrt {2 x^2-x+3}}dx-\frac {1}{8} (1-4 x) \sqrt {2 x^2-x+3}\right )-\frac {2107}{6} \left (2 x^2-x+3\right )^{3/2}\right )+\frac {769}{8} x \left (2 x^2-x+3\right )^{3/2}\right )+\frac {63}{2} \left (2 x^2-x+3\right )^{3/2} x^2\right )+\frac {25}{12} \left (2 x^2-x+3\right )^{3/2} x^3\)

\(\Big \downarrow \) 1090

\(\displaystyle \frac {1}{8} \left (\frac {1}{4} \left (\frac {1}{16} \left (-\frac {12371}{4} \left (\frac {1}{16} \sqrt {\frac {23}{2}} \int \frac {1}{\sqrt {\frac {1}{23} (4 x-1)^2+1}}d(4 x-1)-\frac {1}{8} (1-4 x) \sqrt {2 x^2-x+3}\right )-\frac {2107}{6} \left (2 x^2-x+3\right )^{3/2}\right )+\frac {769}{8} x \left (2 x^2-x+3\right )^{3/2}\right )+\frac {63}{2} \left (2 x^2-x+3\right )^{3/2} x^2\right )+\frac {25}{12} \left (2 x^2-x+3\right )^{3/2} x^3\)

\(\Big \downarrow \) 222

\(\displaystyle \frac {1}{8} \left (\frac {1}{4} \left (\frac {1}{16} \left (-\frac {12371}{4} \left (\frac {23 \text {arcsinh}\left (\frac {4 x-1}{\sqrt {23}}\right )}{16 \sqrt {2}}-\frac {1}{8} (1-4 x) \sqrt {2 x^2-x+3}\right )-\frac {2107}{6} \left (2 x^2-x+3\right )^{3/2}\right )+\frac {769}{8} x \left (2 x^2-x+3\right )^{3/2}\right )+\frac {63}{2} \left (2 x^2-x+3\right )^{3/2} x^2\right )+\frac {25}{12} \left (2 x^2-x+3\right )^{3/2} x^3\)

Input:

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

Output:

(25*x^3*(3 - x + 2*x^2)^(3/2))/12 + ((63*x^2*(3 - x + 2*x^2)^(3/2))/2 + (( 
769*x*(3 - x + 2*x^2)^(3/2))/8 + ((-2107*(3 - x + 2*x^2)^(3/2))/6 - (12371 
*(-1/8*((1 - 4*x)*Sqrt[3 - x + 2*x^2]) + (23*ArcSinh[(-1 + 4*x)/Sqrt[23]]) 
/(16*Sqrt[2])))/4)/16)/4)/8
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt 
[a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
 

rule 1087

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] - Simp[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] && 
GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
 

rule 1090

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

rule 1160

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] + Simp[(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[p, -1]
 

rule 2192

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = 
Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x + 
 c*x^2)^(p + 1)/(c*(q + 2*p + 1))), x] + Simp[1/(c*(q + 2*p + 1))   Int[(a 
+ b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b 
*e*(q + p)*x^(q - 1) - c*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c 
, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]
 

Maple [A] (verified)

Time = 2.38 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.44

method	result	size
risch	\(\frac {\left (204800 x^{5}+284672 x^{4}+408960 x^{3}+365536 x^{2}+328204 x -64023\right ) \sqrt {2 x^{2}-x +3}}{49152}-\frac {284533 \sqrt {2}\, \operatorname {arcsinh}\left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{65536}\)	\(55\)
trager	\(\left (\frac {25}{6} x^{5}+\frac {139}{24} x^{4}+\frac {1065}{128} x^{3}+\frac {11423}{1536} x^{2}+\frac {82051}{12288} x -\frac {21341}{16384}\right ) \sqrt {2 x^{2}-x +3}-\frac {284533 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) \ln \left (4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right ) x +4 \sqrt {2 x^{2}-x +3}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-2\right )\right )}{65536}\)	\(81\)
default	\(-\frac {12371 \left (4 x -1\right ) \sqrt {2 x^{2}-x +3}}{16384}-\frac {284533 \sqrt {2}\, \operatorname {arcsinh}\left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{65536}-\frac {2107 \left (2 x^{2}-x +3\right )^{\frac {3}{2}}}{3072}+\frac {769 x \left (2 x^{2}-x +3\right )^{\frac {3}{2}}}{256}+\frac {63 x^{2} \left (2 x^{2}-x +3\right )^{\frac {3}{2}}}{16}+\frac {25 x^{3} \left (2 x^{2}-x +3\right )^{\frac {3}{2}}}{12}\)	\(98\)

Input:

int((2*x^2-x+3)^(1/2)*(5*x^2+3*x+2)^2,x,method=_RETURNVERBOSE)
 

Output:

1/49152*(204800*x^5+284672*x^4+408960*x^3+365536*x^2+328204*x-64023)*(2*x^ 
2-x+3)^(1/2)-284533/65536*2^(1/2)*arcsinh(4/23*23^(1/2)*(x-1/4))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.63 \[ \int \sqrt {3-x+2 x^2} \left (2+3 x+5 x^2\right )^2 \, dx=\frac {1}{49152} \, {\left (204800 \, x^{5} + 284672 \, x^{4} + 408960 \, x^{3} + 365536 \, x^{2} + 328204 \, x - 64023\right )} \sqrt {2 \, x^{2} - x + 3} + \frac {284533}{131072} \, \sqrt {2} \log \left (4 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) \] Input:

integrate((2*x^2-x+3)^(1/2)*(5*x^2+3*x+2)^2,x, algorithm="fricas")
 

Output:

1/49152*(204800*x^5 + 284672*x^4 + 408960*x^3 + 365536*x^2 + 328204*x - 64 
023)*sqrt(2*x^2 - x + 3) + 284533/131072*sqrt(2)*log(4*sqrt(2)*sqrt(2*x^2 
- x + 3)*(4*x - 1) - 32*x^2 + 16*x - 25)
 

Sympy [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.56 \[ \int \sqrt {3-x+2 x^2} \left (2+3 x+5 x^2\right )^2 \, dx=\sqrt {2 x^{2} - x + 3} \cdot \left (\frac {25 x^{5}}{6} + \frac {139 x^{4}}{24} + \frac {1065 x^{3}}{128} + \frac {11423 x^{2}}{1536} + \frac {82051 x}{12288} - \frac {21341}{16384}\right ) - \frac {284533 \sqrt {2} \operatorname {asinh}{\left (\frac {4 \sqrt {23} \left (x - \frac {1}{4}\right )}{23} \right )}}{65536} \] Input:

integrate((2*x**2-x+3)**(1/2)*(5*x**2+3*x+2)**2,x)
 

Output:

sqrt(2*x**2 - x + 3)*(25*x**5/6 + 139*x**4/24 + 1065*x**3/128 + 11423*x**2 
/1536 + 82051*x/12288 - 21341/16384) - 284533*sqrt(2)*asinh(4*sqrt(23)*(x 
- 1/4)/23)/65536
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.88 \[ \int \sqrt {3-x+2 x^2} \left (2+3 x+5 x^2\right )^2 \, dx=\frac {25}{12} \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}} x^{3} + \frac {63}{16} \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}} x^{2} + \frac {769}{256} \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}} x - \frac {2107}{3072} \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}} - \frac {12371}{4096} \, \sqrt {2 \, x^{2} - x + 3} x - \frac {284533}{65536} \, \sqrt {2} \operatorname {arsinh}\left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) + \frac {12371}{16384} \, \sqrt {2 \, x^{2} - x + 3} \] Input:

integrate((2*x^2-x+3)^(1/2)*(5*x^2+3*x+2)^2,x, algorithm="maxima")
 

Output:

25/12*(2*x^2 - x + 3)^(3/2)*x^3 + 63/16*(2*x^2 - x + 3)^(3/2)*x^2 + 769/25 
6*(2*x^2 - x + 3)^(3/2)*x - 2107/3072*(2*x^2 - x + 3)^(3/2) - 12371/4096*s 
qrt(2*x^2 - x + 3)*x - 284533/65536*sqrt(2)*arcsinh(1/23*sqrt(23)*(4*x - 1 
)) + 12371/16384*sqrt(2*x^2 - x + 3)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.59 \[ \int \sqrt {3-x+2 x^2} \left (2+3 x+5 x^2\right )^2 \, dx=\frac {1}{49152} \, {\left (4 \, {\left (8 \, {\left (4 \, {\left (16 \, {\left (100 \, x + 139\right )} x + 3195\right )} x + 11423\right )} x + 82051\right )} x - 64023\right )} \sqrt {2 \, x^{2} - x + 3} + \frac {284533}{65536} \, \sqrt {2} \log \left (-2 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} + 1\right ) \] Input:

integrate((2*x^2-x+3)^(1/2)*(5*x^2+3*x+2)^2,x, algorithm="giac")
 

Output:

1/49152*(4*(8*(4*(16*(100*x + 139)*x + 3195)*x + 11423)*x + 82051)*x - 640 
23)*sqrt(2*x^2 - x + 3) + 284533/65536*sqrt(2)*log(-2*sqrt(2)*(sqrt(2)*x - 
 sqrt(2*x^2 - x + 3)) + 1)
 

Mupad [B] (verification not implemented)

Time = 16.18 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.23 \[ \int \sqrt {3-x+2 x^2} \left (2+3 x+5 x^2\right )^2 \, dx=\frac {63\,x^2\,{\left (2\,x^2-x+3\right )}^{3/2}}{16}+\frac {25\,x^3\,{\left (2\,x^2-x+3\right )}^{3/2}}{12}-\frac {29509\,\sqrt {2}\,\ln \left (\sqrt {2\,x^2-x+3}+\frac {\sqrt {2}\,\left (2\,x-\frac {1}{2}\right )}{2}\right )}{8192}-\frac {1283\,\left (\frac {x}{2}-\frac {1}{8}\right )\,\sqrt {2\,x^2-x+3}}{256}-\frac {2107\,\sqrt {2\,x^2-x+3}\,\left (32\,x^2-4\,x+45\right )}{49152}+\frac {769\,x\,{\left (2\,x^2-x+3\right )}^{3/2}}{256}-\frac {48461\,\sqrt {2}\,\ln \left (2\,\sqrt {2\,x^2-x+3}+\frac {\sqrt {2}\,\left (4\,x-1\right )}{2}\right )}{65536} \] Input:

int((2*x^2 - x + 3)^(1/2)*(3*x + 5*x^2 + 2)^2,x)
 

Output:

(63*x^2*(2*x^2 - x + 3)^(3/2))/16 + (25*x^3*(2*x^2 - x + 3)^(3/2))/12 - (2 
9509*2^(1/2)*log((2*x^2 - x + 3)^(1/2) + (2^(1/2)*(2*x - 1/2))/2))/8192 - 
(1283*(x/2 - 1/8)*(2*x^2 - x + 3)^(1/2))/256 - (2107*(2*x^2 - x + 3)^(1/2) 
*(32*x^2 - 4*x + 45))/49152 + (769*x*(2*x^2 - x + 3)^(3/2))/256 - (48461*2 
^(1/2)*log(2*(2*x^2 - x + 3)^(1/2) + (2^(1/2)*(4*x - 1))/2))/65536
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.98 \[ \int \sqrt {3-x+2 x^2} \left (2+3 x+5 x^2\right )^2 \, dx=\frac {25 \sqrt {2 x^{2}-x +3}\, x^{5}}{6}+\frac {139 \sqrt {2 x^{2}-x +3}\, x^{4}}{24}+\frac {1065 \sqrt {2 x^{2}-x +3}\, x^{3}}{128}+\frac {11423 \sqrt {2 x^{2}-x +3}\, x^{2}}{1536}+\frac {82051 \sqrt {2 x^{2}-x +3}\, x}{12288}-\frac {21341 \sqrt {2 x^{2}-x +3}}{16384}-\frac {284533 \sqrt {2}\, \mathrm {log}\left (\frac {2 \sqrt {2 x^{2}-x +3}\, \sqrt {2}+4 x -1}{\sqrt {23}}\right )}{65536} \] Input:

int((2*x^2-x+3)^(1/2)*(5*x^2+3*x+2)^2,x)
 

Output:

(819200*sqrt(2*x**2 - x + 3)*x**5 + 1138688*sqrt(2*x**2 - x + 3)*x**4 + 16 
35840*sqrt(2*x**2 - x + 3)*x**3 + 1462144*sqrt(2*x**2 - x + 3)*x**2 + 1312 
816*sqrt(2*x**2 - x + 3)*x - 256092*sqrt(2*x**2 - x + 3) - 853599*sqrt(2)* 
log((2*sqrt(2*x**2 - x + 3)*sqrt(2) + 4*x - 1)/sqrt(23)))/196608