\(\int (1+x+x^2)^{5/2} \, dx\) [273]

Optimal result

Integrand size = 10, antiderivative size = 74 \[ \int \left (1+x+x^2\right )^{5/2} \, dx=\frac {45}{512} (1+2 x) \sqrt {1+x+x^2}+\frac {5}{64} (1+2 x) \left (1+x+x^2\right )^{3/2}+\frac {1}{12} (1+2 x) \left (1+x+x^2\right )^{5/2}+\frac {135 \text {arcsinh}\left (\frac {1+2 x}{\sqrt {3}}\right )}{1024} \]

[Out]

5/64*(1+2*x)*(x^2+x+1)^(3/2)+1/12*(1+2*x)*(x^2+x+1)^(5/2)+135/1024*arcsinh(1/3*(1+2*x)*3^(1/2))+45/512*(1+2*x)
*(x^2+x+1)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {626, 633, 221} \[ \int \left (1+x+x^2\right )^{5/2} \, dx=\frac {135 \text {arcsinh}\left (\frac {2 x+1}{\sqrt {3}}\right )}{1024}+\frac {1}{12} (2 x+1) \left (x^2+x+1\right )^{5/2}+\frac {5}{64} (2 x+1) \left (x^2+x+1\right )^{3/2}+\frac {45}{512} (2 x+1) \sqrt {x^2+x+1} \]

[In]

Int[(1 + x + x^2)^(5/2),x]

[Out]

(45*(1 + 2*x)*Sqrt[1 + x + x^2])/512 + (5*(1 + 2*x)*(1 + x + x^2)^(3/2))/64 + ((1 + 2*x)*(1 + x + x^2)^(5/2))/
12 + (135*ArcSinh[(1 + 2*x)/Sqrt[3]])/1024

Rule 221

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 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 633

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*(-4*(c/(b^2 - 4*a*c)))^p), Subst[Int[Si
mp[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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{12} (1+2 x) \left (1+x+x^2\right )^{5/2}+\frac {5}{8} \int \left (1+x+x^2\right )^{3/2} \, dx \\ & = \frac {5}{64} (1+2 x) \left (1+x+x^2\right )^{3/2}+\frac {1}{12} (1+2 x) \left (1+x+x^2\right )^{5/2}+\frac {45}{128} \int \sqrt {1+x+x^2} \, dx \\ & = \frac {45}{512} (1+2 x) \sqrt {1+x+x^2}+\frac {5}{64} (1+2 x) \left (1+x+x^2\right )^{3/2}+\frac {1}{12} (1+2 x) \left (1+x+x^2\right )^{5/2}+\frac {135 \int \frac {1}{\sqrt {1+x+x^2}} \, dx}{1024} \\ & = \frac {45}{512} (1+2 x) \sqrt {1+x+x^2}+\frac {5}{64} (1+2 x) \left (1+x+x^2\right )^{3/2}+\frac {1}{12} (1+2 x) \left (1+x+x^2\right )^{5/2}+\frac {\left (45 \sqrt {3}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{3}}} \, dx,x,1+2 x\right )}{1024} \\ & = \frac {45}{512} (1+2 x) \sqrt {1+x+x^2}+\frac {5}{64} (1+2 x) \left (1+x+x^2\right )^{3/2}+\frac {1}{12} (1+2 x) \left (1+x+x^2\right )^{5/2}+\frac {135 \text {arcsinh}\left (\frac {1+2 x}{\sqrt {3}}\right )}{1024} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.84 \[ \int \left (1+x+x^2\right )^{5/2} \, dx=\frac {\sqrt {1+x+x^2} \left (383+1142 x+1256 x^2+1264 x^3+640 x^4+256 x^5\right )}{1536}-\frac {135 \log \left (-1-2 x+2 \sqrt {1+x+x^2}\right )}{1024} \]

[In]

Integrate[(1 + x + x^2)^(5/2),x]

[Out]

(Sqrt[1 + x + x^2]*(383 + 1142*x + 1256*x^2 + 1264*x^3 + 640*x^4 + 256*x^5))/1536 - (135*Log[-1 - 2*x + 2*Sqrt
[1 + x + x^2]])/1024

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.65

method	result	size
risch	\(\frac {\left (256 x^{5}+640 x^{4}+1264 x^{3}+1256 x^{2}+1142 x +383\right ) \sqrt {x^{2}+x +1}}{1536}+\frac {135 \,\operatorname {arcsinh}\left (\frac {2 \sqrt {3}\, \left (x +\frac {1}{2}\right )}{3}\right )}{1024}\)	\(48\)
trager	\(\left (\frac {1}{6} x^{5}+\frac {5}{12} x^{4}+\frac {79}{96} x^{3}+\frac {157}{192} x^{2}+\frac {571}{768} x +\frac {383}{1536}\right ) \sqrt {x^{2}+x +1}+\frac {135 \ln \left (1+2 x +2 \sqrt {x^{2}+x +1}\right )}{1024}\)	\(54\)
default	\(\frac {\left (1+2 x \right ) \left (x^{2}+x +1\right )^{\frac {5}{2}}}{12}+\frac {5 \left (1+2 x \right ) \left (x^{2}+x +1\right )^{\frac {3}{2}}}{64}+\frac {45 \left (1+2 x \right ) \sqrt {x^{2}+x +1}}{512}+\frac {135 \,\operatorname {arcsinh}\left (\frac {2 \sqrt {3}\, \left (x +\frac {1}{2}\right )}{3}\right )}{1024}\)	\(58\)

[In]

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

[Out]

1/1536*(256*x^5+640*x^4+1264*x^3+1256*x^2+1142*x+383)*(x^2+x+1)^(1/2)+135/1024*arcsinh(2/3*3^(1/2)*(x+1/2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.73 \[ \int \left (1+x+x^2\right )^{5/2} \, dx=\frac {1}{1536} \, {\left (256 \, x^{5} + 640 \, x^{4} + 1264 \, x^{3} + 1256 \, x^{2} + 1142 \, x + 383\right )} \sqrt {x^{2} + x + 1} - \frac {135}{1024} \, \log \left (-2 \, x + 2 \, \sqrt {x^{2} + x + 1} - 1\right ) \]

[In]

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

[Out]

1/1536*(256*x^5 + 640*x^4 + 1264*x^3 + 1256*x^2 + 1142*x + 383)*sqrt(x^2 + x + 1) - 135/1024*log(-2*x + 2*sqrt
(x^2 + x + 1) - 1)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (71) = 142\).

Time = 0.45 (sec) , antiderivative size = 168, normalized size of antiderivative = 2.27 \[ \int \left (1+x+x^2\right )^{5/2} \, dx=\left (\frac {x}{2} + \frac {1}{4}\right ) \sqrt {x^{2} + x + 1} + 2 \left (\frac {x^{2}}{3} + \frac {x}{12} + \frac {5}{24}\right ) \sqrt {x^{2} + x + 1} + 3 \sqrt {x^{2} + x + 1} \left (\frac {x^{3}}{4} + \frac {x^{2}}{24} + \frac {7 x}{96} - \frac {37}{192}\right ) + 2 \sqrt {x^{2} + x + 1} \left (\frac {x^{4}}{5} + \frac {x^{3}}{40} + \frac {3 x^{2}}{80} - \frac {27 x}{320} + \frac {33}{640}\right ) + \sqrt {x^{2} + x + 1} \left (\frac {x^{5}}{6} + \frac {x^{4}}{60} + \frac {11 x^{3}}{480} - \frac {47 x^{2}}{960} + \frac {103 x}{3840} + \frac {443}{7680}\right ) + \frac {135 \operatorname {asinh}{\left (\frac {2 \sqrt {3} \left (x + \frac {1}{2}\right )}{3} \right )}}{1024} \]

[In]

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

[Out]

(x/2 + 1/4)*sqrt(x**2 + x + 1) + 2*(x**2/3 + x/12 + 5/24)*sqrt(x**2 + x + 1) + 3*sqrt(x**2 + x + 1)*(x**3/4 +
x**2/24 + 7*x/96 - 37/192) + 2*sqrt(x**2 + x + 1)*(x**4/5 + x**3/40 + 3*x**2/80 - 27*x/320 + 33/640) + sqrt(x*
*2 + x + 1)*(x**5/6 + x**4/60 + 11*x**3/480 - 47*x**2/960 + 103*x/3840 + 443/7680) + 135*asinh(2*sqrt(3)*(x +
1/2)/3)/1024

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.04 \[ \int \left (1+x+x^2\right )^{5/2} \, dx=\frac {1}{6} \, {\left (x^{2} + x + 1\right )}^{\frac {5}{2}} x + \frac {1}{12} \, {\left (x^{2} + x + 1\right )}^{\frac {5}{2}} + \frac {5}{32} \, {\left (x^{2} + x + 1\right )}^{\frac {3}{2}} x + \frac {5}{64} \, {\left (x^{2} + x + 1\right )}^{\frac {3}{2}} + \frac {45}{256} \, \sqrt {x^{2} + x + 1} x + \frac {45}{512} \, \sqrt {x^{2} + x + 1} + \frac {135}{1024} \, \operatorname {arsinh}\left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) \]

[In]

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

[Out]

1/6*(x^2 + x + 1)^(5/2)*x + 1/12*(x^2 + x + 1)^(5/2) + 5/32*(x^2 + x + 1)^(3/2)*x + 5/64*(x^2 + x + 1)^(3/2) +
45/256*sqrt(x^2 + x + 1)*x + 45/512*sqrt(x^2 + x + 1) + 135/1024*arcsinh(1/3*sqrt(3)*(2*x + 1))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.73 \[ \int \left (1+x+x^2\right )^{5/2} \, dx=\frac {1}{1536} \, {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (2 \, x + 5\right )} x + 79\right )} x + 157\right )} x + 571\right )} x + 383\right )} \sqrt {x^{2} + x + 1} - \frac {135}{1024} \, \log \left (-2 \, x + 2 \, \sqrt {x^{2} + x + 1} - 1\right ) \]

[In]

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

[Out]

1/1536*(2*(4*(2*(8*(2*x + 5)*x + 79)*x + 157)*x + 571)*x + 383)*sqrt(x^2 + x + 1) - 135/1024*log(-2*x + 2*sqrt
(x^2 + x + 1) - 1)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.76 \[ \int \left (1+x+x^2\right )^{5/2} \, dx=\frac {135\,\ln \left (x+\sqrt {x^2+x+1}+\frac {1}{2}\right )}{1024}+\frac {5\,\left (x+\frac {1}{2}\right )\,{\left (x^2+x+1\right )}^{3/2}}{32}+\frac {\left (x+\frac {1}{2}\right )\,{\left (x^2+x+1\right )}^{5/2}}{6}+\frac {45\,\left (\frac {x}{2}+\frac {1}{4}\right )\,\sqrt {x^2+x+1}}{128} \]

[In]

int((x + x^2 + 1)^(5/2),x)

[Out]

(135*log(x + (x + x^2 + 1)^(1/2) + 1/2))/1024 + (5*(x + 1/2)*(x + x^2 + 1)^(3/2))/32 + ((x + 1/2)*(x + x^2 + 1
)^(5/2))/6 + (45*(x/2 + 1/4)*(x + x^2 + 1)^(1/2))/128