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

Optimal. Leaf size=74 \[ \frac {45}{512} (1+2 x) \sqrt {1+x+x^2}+\frac {5}{64} (1+2 x) \left (1+x+x^2\right )^{3/2}+\frac {1}{12} (1+2 x) \left (1+x+x^2\right )^{5/2}+\frac {135 \sinh ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )}{1024} \]

[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)

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Rubi [A]
time = 0.01, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {626, 633, 221} \begin {gather*} \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}+\frac {135 \sinh ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{1024} \end {gather*}

Antiderivative was successfully verified.

[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*} \int \left (1+x+x^2\right )^{5/2} \, dx &=\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 \sinh ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )}{1024}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 62, normalized size = 0.84 \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[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

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Maple [A]
time = 0.12, size = 58, normalized size = 0.78

method result size
risch \(\frac {\left (256 x^{5}+640 x^{4}+1264 x^{3}+1256 x^{2}+1142 x +383\right ) \sqrt {x^{2}+x +1}}{1536}+\frac {135 \arcsinh \left (\frac {2 \sqrt {3}\, \left (x +\frac {1}{2}\right )}{3}\right )}{1024}\) \(48\)
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 \(\frac {\left (1+2 x \right ) \left (x^{2}+x +1\right )^{\frac {5}{2}}}{12}+\frac {5 \left (1+2 x \right ) \left (x^{2}+x +1\right )^{\frac {3}{2}}}{64}+\frac {45 \left (1+2 x \right ) \sqrt {x^{2}+x +1}}{512}+\frac {135 \arcsinh \left (\frac {2 \sqrt {3}\, \left (x +\frac {1}{2}\right )}{3}\right )}{1024}\) \(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 2.64, size = 77, normalized size = 1.04 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Fricas [A]
time = 0.57, size = 54, normalized size = 0.73 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (x^{2} + x + 1\right )^{\frac {5}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x**2 + x + 1)**(5/2), x)

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Giac [A]
time = 0.75, size = 54, normalized size = 0.73 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Mupad [B]
time = 0.07, size = 56, normalized size = 0.76 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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