3.14.26 \(\int \frac {1}{(1+2 x)^{5/2} (1+x+x^2)} \, dx\) [1326]

Optimal. Leaf size=180 \[ -\frac {4}{9 (1+2 x)^{3/2}}+\frac {\sqrt {2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {1+2 x}}{\sqrt [4]{3}}\right )}{3\ 3^{3/4}}-\frac {\sqrt {2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {1+2 x}}{\sqrt [4]{3}}\right )}{3\ 3^{3/4}}+\frac {\log \left (1+\sqrt {3}+2 x-\sqrt {2} \sqrt [4]{3} \sqrt {1+2 x}\right )}{3 \sqrt {2} 3^{3/4}}-\frac {\log \left (1+\sqrt {3}+2 x+\sqrt {2} \sqrt [4]{3} \sqrt {1+2 x}\right )}{3 \sqrt {2} 3^{3/4}} \]

[Out]

-4/9/(1+2*x)^(3/2)+1/18*3^(1/4)*ln(1+2*x+3^(1/2)-3^(1/4)*2^(1/2)*(1+2*x)^(1/2))*2^(1/2)-1/18*3^(1/4)*ln(1+2*x+
3^(1/2)+3^(1/4)*2^(1/2)*(1+2*x)^(1/2))*2^(1/2)-1/9*3^(1/4)*arctan(-1+1/3*2^(1/2)*(1+2*x)^(1/2)*3^(3/4))*2^(1/2
)-1/9*3^(1/4)*arctan(1+1/3*2^(1/2)*(1+2*x)^(1/2)*3^(3/4))*2^(1/2)

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Rubi [A]
time = 0.09, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {707, 708, 335, 217, 1179, 642, 1176, 631, 210} \begin {gather*} \frac {\sqrt {2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}\right )}{3\ 3^{3/4}}-\frac {\sqrt {2} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}+1\right )}{3\ 3^{3/4}}-\frac {4}{9 (2 x+1)^{3/2}}+\frac {\log \left (2 x-\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1\right )}{3 \sqrt {2} 3^{3/4}}-\frac {\log \left (2 x+\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1\right )}{3 \sqrt {2} 3^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((1 + 2*x)^(5/2)*(1 + x + x^2)),x]

[Out]

-4/(9*(1 + 2*x)^(3/2)) + (Sqrt[2]*ArcTan[1 - (Sqrt[2]*Sqrt[1 + 2*x])/3^(1/4)])/(3*3^(3/4)) - (Sqrt[2]*ArcTan[1
+ (Sqrt[2]*Sqrt[1 + 2*x])/3^(1/4)])/(3*3^(3/4)) + Log[1 + Sqrt[3] + 2*x - Sqrt[2]*3^(1/4)*Sqrt[1 + 2*x]]/(3*S
qrt[2]*3^(3/4)) - Log[1 + Sqrt[3] + 2*x + Sqrt[2]*3^(1/4)*Sqrt[1 + 2*x]]/(3*Sqrt[2]*3^(3/4))

Rule 210

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

Rule 217

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

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

Rule 707

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[-2*b*d*(d + e*x)^(m
+ 1)*((a + b*x + c*x^2)^(p + 1)/(d^2*(m + 1)*(b^2 - 4*a*c))), x] + Dist[b^2*((m + 2*p + 3)/(d^2*(m + 1)*(b^2 -
4*a*c))), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*
c, 0] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[m, -1] && (IntegerQ[2*p] || (IntegerQ[m] && Rationa
lQ[p]) || IntegerQ[(m + 2*p + 3)/2])

Rule 708

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[x^m*(
a - b^2/(4*c) + (c*x^2)/e^2)^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps

\begin {align*} \int \frac {1}{(1+2 x)^{5/2} \left (1+x+x^2\right )} \, dx &=-\frac {4}{9 (1+2 x)^{3/2}}-\frac {1}{3} \int \frac {1}{\sqrt {1+2 x} \left (1+x+x^2\right )} \, dx\\ &=-\frac {4}{9 (1+2 x)^{3/2}}-\frac {1}{6} \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (\frac {3}{4}+\frac {x^2}{4}\right )} \, dx,x,1+2 x\right )\\ &=-\frac {4}{9 (1+2 x)^{3/2}}-\frac {1}{3} \text {Subst}\left (\int \frac {1}{\frac {3}{4}+\frac {x^4}{4}} \, dx,x,\sqrt {1+2 x}\right )\\ &=-\frac {4}{9 (1+2 x)^{3/2}}-\frac {\text {Subst}\left (\int \frac {\sqrt {3}-x^2}{\frac {3}{4}+\frac {x^4}{4}} \, dx,x,\sqrt {1+2 x}\right )}{6 \sqrt {3}}-\frac {\text {Subst}\left (\int \frac {\sqrt {3}+x^2}{\frac {3}{4}+\frac {x^4}{4}} \, dx,x,\sqrt {1+2 x}\right )}{6 \sqrt {3}}\\ &=-\frac {4}{9 (1+2 x)^{3/2}}+\frac {\text {Subst}\left (\int \frac {\sqrt {2} \sqrt [4]{3}+2 x}{-\sqrt {3}-\sqrt {2} \sqrt [4]{3} x-x^2} \, dx,x,\sqrt {1+2 x}\right )}{3 \sqrt {2} 3^{3/4}}+\frac {\text {Subst}\left (\int \frac {\sqrt {2} \sqrt [4]{3}-2 x}{-\sqrt {3}+\sqrt {2} \sqrt [4]{3} x-x^2} \, dx,x,\sqrt {1+2 x}\right )}{3 \sqrt {2} 3^{3/4}}-\frac {\text {Subst}\left (\int \frac {1}{\sqrt {3}-\sqrt {2} \sqrt [4]{3} x+x^2} \, dx,x,\sqrt {1+2 x}\right )}{3 \sqrt {3}}-\frac {\text {Subst}\left (\int \frac {1}{\sqrt {3}+\sqrt {2} \sqrt [4]{3} x+x^2} \, dx,x,\sqrt {1+2 x}\right )}{3 \sqrt {3}}\\ &=-\frac {4}{9 (1+2 x)^{3/2}}+\frac {\log \left (1+\sqrt {3}+2 x-\sqrt {2} \sqrt [4]{3} \sqrt {1+2 x}\right )}{3 \sqrt {2} 3^{3/4}}-\frac {\log \left (1+\sqrt {3}+2 x+\sqrt {2} \sqrt [4]{3} \sqrt {1+2 x}\right )}{3 \sqrt {2} 3^{3/4}}-\frac {\sqrt {2} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2+4 x}}{\sqrt [4]{3}}\right )}{3\ 3^{3/4}}+\frac {\sqrt {2} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2+4 x}}{\sqrt [4]{3}}\right )}{3\ 3^{3/4}}\\ &=-\frac {4}{9 (1+2 x)^{3/2}}+\frac {\sqrt {2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {1+2 x}}{\sqrt [4]{3}}\right )}{3\ 3^{3/4}}-\frac {\sqrt {2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {1+2 x}}{\sqrt [4]{3}}\right )}{3\ 3^{3/4}}+\frac {\log \left (1+\sqrt {3}+2 x-\sqrt {2} \sqrt [4]{3} \sqrt {1+2 x}\right )}{3 \sqrt {2} 3^{3/4}}-\frac {\log \left (1+\sqrt {3}+2 x+\sqrt {2} \sqrt [4]{3} \sqrt {1+2 x}\right )}{3 \sqrt {2} 3^{3/4}}\\ \end {align*}

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Mathematica [A]
time = 0.19, size = 104, normalized size = 0.58 \begin {gather*} \frac {1}{9} \left (-\frac {4}{(1+2 x)^{3/2}}-\sqrt {2} \sqrt [4]{3} \tan ^{-1}\left (\frac {-3+\sqrt {3}+2 \sqrt {3} x}{3^{3/4} \sqrt {2+4 x}}\right )-\sqrt {2} \sqrt [4]{3} \tanh ^{-1}\left (\frac {3^{3/4} \sqrt {2+4 x}}{3+\sqrt {3}+2 \sqrt {3} x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((1 + 2*x)^(5/2)*(1 + x + x^2)),x]

[Out]

(-4/(1 + 2*x)^(3/2) - Sqrt[2]*3^(1/4)*ArcTan[(-3 + Sqrt[3] + 2*Sqrt[3]*x)/(3^(3/4)*Sqrt[2 + 4*x])] - Sqrt[2]*3
^(1/4)*ArcTanh[(3^(3/4)*Sqrt[2 + 4*x])/(3 + Sqrt[3] + 2*Sqrt[3]*x)])/9

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Maple [A]
time = 1.25, size = 109, normalized size = 0.61

method	result	size
derivativedivides	\(-\frac {4}{9 \left (2 x +1\right )^{\frac {3}{2}}}-\frac {3^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {1+2 x +\sqrt {3}+3^{\frac {1}{4}} \sqrt {2}\, \sqrt {2 x +1}}{1+2 x +\sqrt {3}-3^{\frac {1}{4}} \sqrt {2}\, \sqrt {2 x +1}}\right )+2 \arctan \left (1+\frac {\sqrt {2}\, \sqrt {2 x +1}\, 3^{\frac {3}{4}}}{3}\right )+2 \arctan \left (-1+\frac {\sqrt {2}\, \sqrt {2 x +1}\, 3^{\frac {3}{4}}}{3}\right )\right )}{18}\)	\(109\)
default	\(-\frac {4}{9 \left (2 x +1\right )^{\frac {3}{2}}}-\frac {3^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {1+2 x +\sqrt {3}+3^{\frac {1}{4}} \sqrt {2}\, \sqrt {2 x +1}}{1+2 x +\sqrt {3}-3^{\frac {1}{4}} \sqrt {2}\, \sqrt {2 x +1}}\right )+2 \arctan \left (1+\frac {\sqrt {2}\, \sqrt {2 x +1}\, 3^{\frac {3}{4}}}{3}\right )+2 \arctan \left (-1+\frac {\sqrt {2}\, \sqrt {2 x +1}\, 3^{\frac {3}{4}}}{3}\right )\right )}{18}\)	\(109\)
trager	\(-\frac {4}{9 \left (2 x +1\right )^{\frac {3}{2}}}-\frac {\RootOf \left (\textit {\_Z}^{4}+3\right ) \ln \left (-\frac {\RootOf \left (\textit {\_Z}^{4}+3\right )^{5} x -4 \RootOf \left (\textit {\_Z}^{4}+3\right )^{3} x -2 \RootOf \left (\textit {\_Z}^{4}+3\right )^{3}+3 \RootOf \left (\textit {\_Z}^{4}+3\right ) x +6 \RootOf \left (\textit {\_Z}^{4}+3\right )+12 \sqrt {2 x +1}}{\RootOf \left (\textit {\_Z}^{4}+3\right )^{2} x +x +2}\right )}{9}-\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+3\right )^{2}\right ) \ln \left (-\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+3\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}+3\right )^{4} x +4 \RootOf \left (\textit {\_Z}^{4}+3\right )^{2} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+3\right )^{2}\right ) x +2 \RootOf \left (\textit {\_Z}^{4}+3\right )^{2} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+3\right )^{2}\right )+3 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+3\right )^{2}\right ) x +12 \sqrt {2 x +1}+6 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}+3\right )^{2}\right )}{\RootOf \left (\textit {\_Z}^{4}+3\right )^{2} x -x -2}\right )}{9}\)	\(241\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/9/(2*x+1)^(3/2)-1/18*3^(1/4)*2^(1/2)*(ln((1+2*x+3^(1/2)+3^(1/4)*2^(1/2)*(2*x+1)^(1/2))/(1+2*x+3^(1/2)-3^(1/
4)*2^(1/2)*(2*x+1)^(1/2)))+2*arctan(1+1/3*2^(1/2)*(2*x+1)^(1/2)*3^(3/4))+2*arctan(-1+1/3*2^(1/2)*(2*x+1)^(1/2)
*3^(3/4)))

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Maxima [A]
time = 0.53, size = 141, normalized size = 0.78 \begin {gather*} -\frac {1}{9} \cdot 3^{\frac {1}{4}} \sqrt {2} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} + 2 \, \sqrt {2 \, x + 1}\right )}\right ) - \frac {1}{9} \cdot 3^{\frac {1}{4}} \sqrt {2} \arctan \left (-\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} - 2 \, \sqrt {2 \, x + 1}\right )}\right ) - \frac {1}{18} \cdot 3^{\frac {1}{4}} \sqrt {2} \log \left (3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) + \frac {1}{18} \cdot 3^{\frac {1}{4}} \sqrt {2} \log \left (-3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) - \frac {4}{9 \, {\left (2 \, x + 1\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*3^(1/4)*sqrt(2)*arctan(1/6*3^(3/4)*sqrt(2)*(3^(1/4)*sqrt(2) + 2*sqrt(2*x + 1))) - 1/9*3^(1/4)*sqrt(2)*arc
tan(-1/6*3^(3/4)*sqrt(2)*(3^(1/4)*sqrt(2) - 2*sqrt(2*x + 1))) - 1/18*3^(1/4)*sqrt(2)*log(3^(1/4)*sqrt(2)*sqrt(
2*x + 1) + 2*x + sqrt(3) + 1) + 1/18*3^(1/4)*sqrt(2)*log(-3^(1/4)*sqrt(2)*sqrt(2*x + 1) + 2*x + sqrt(3) + 1) -
4/9/(2*x + 1)^(3/2)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 252 vs. \(2 (125) = 250\).
time = 2.57, size = 252, normalized size = 1.40 \begin {gather*} \frac {4 \cdot 27^{\frac {3}{4}} \sqrt {2} {\left (4 \, x^{2} + 4 \, x + 1\right )} \arctan \left (\frac {1}{9} \cdot 27^{\frac {1}{4}} \sqrt {2} \sqrt {27^{\frac {3}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 18 \, x + 9 \, \sqrt {3} + 9} - \frac {1}{3} \cdot 27^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} - 1\right ) + 4 \cdot 27^{\frac {3}{4}} \sqrt {2} {\left (4 \, x^{2} + 4 \, x + 1\right )} \arctan \left (\frac {1}{54} \cdot 27^{\frac {1}{4}} \sqrt {2} \sqrt {-36 \cdot 27^{\frac {3}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 648 \, x + 324 \, \sqrt {3} + 324} - \frac {1}{3} \cdot 27^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 1\right ) - 27^{\frac {3}{4}} \sqrt {2} {\left (4 \, x^{2} + 4 \, x + 1\right )} \log \left (36 \cdot 27^{\frac {3}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 648 \, x + 324 \, \sqrt {3} + 324\right ) + 27^{\frac {3}{4}} \sqrt {2} {\left (4 \, x^{2} + 4 \, x + 1\right )} \log \left (-36 \cdot 27^{\frac {3}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 648 \, x + 324 \, \sqrt {3} + 324\right ) - 72 \, \sqrt {2 \, x + 1}}{162 \, {\left (4 \, x^{2} + 4 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/162*(4*27^(3/4)*sqrt(2)*(4*x^2 + 4*x + 1)*arctan(1/9*27^(1/4)*sqrt(2)*sqrt(27^(3/4)*sqrt(2)*sqrt(2*x + 1) +
18*x + 9*sqrt(3) + 9) - 1/3*27^(1/4)*sqrt(2)*sqrt(2*x + 1) - 1) + 4*27^(3/4)*sqrt(2)*(4*x^2 + 4*x + 1)*arctan(
1/54*27^(1/4)*sqrt(2)*sqrt(-36*27^(3/4)*sqrt(2)*sqrt(2*x + 1) + 648*x + 324*sqrt(3) + 324) - 1/3*27^(1/4)*sqrt
(2)*sqrt(2*x + 1) + 1) - 27^(3/4)*sqrt(2)*(4*x^2 + 4*x + 1)*log(36*27^(3/4)*sqrt(2)*sqrt(2*x + 1) + 648*x + 32
4*sqrt(3) + 324) + 27^(3/4)*sqrt(2)*(4*x^2 + 4*x + 1)*log(-36*27^(3/4)*sqrt(2)*sqrt(2*x + 1) + 648*x + 324*sqr
t(3) + 324) - 72*sqrt(2*x + 1))/(4*x^2 + 4*x + 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 2.58, size = 129, normalized size = 0.72 \begin {gather*} -\frac {1}{9} \cdot 12^{\frac {1}{4}} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} + 2 \, \sqrt {2 \, x + 1}\right )}\right ) - \frac {1}{9} \cdot 12^{\frac {1}{4}} \arctan \left (-\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} - 2 \, \sqrt {2 \, x + 1}\right )}\right ) - \frac {1}{18} \cdot 12^{\frac {1}{4}} \log \left (3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) + \frac {1}{18} \cdot 12^{\frac {1}{4}} \log \left (-3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) - \frac {4}{9 \, {\left (2 \, x + 1\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*12^(1/4)*arctan(1/6*3^(3/4)*sqrt(2)*(3^(1/4)*sqrt(2) + 2*sqrt(2*x + 1))) - 1/9*12^(1/4)*arctan(-1/6*3^(3/
4)*sqrt(2)*(3^(1/4)*sqrt(2) - 2*sqrt(2*x + 1))) - 1/18*12^(1/4)*log(3^(1/4)*sqrt(2)*sqrt(2*x + 1) + 2*x + sqrt
(3) + 1) + 1/18*12^(1/4)*log(-3^(1/4)*sqrt(2)*sqrt(2*x + 1) + 2*x + sqrt(3) + 1) - 4/9/(2*x + 1)^(3/2)

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Mupad [B]
time = 0.10, size = 66, normalized size = 0.37 \begin {gather*} -\frac {4}{9\,{\left (2\,x+1\right )}^{3/2}}+\sqrt {2}\,3^{1/4}\,\mathrm {atan}\left (\sqrt {2}\,3^{3/4}\,\sqrt {2\,x+1}\,\left (\frac {1}{6}-\frac {1}{6}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{9}-\frac {1}{9}{}\mathrm {i}\right )+\sqrt {2}\,3^{1/4}\,\mathrm {atan}\left (\sqrt {2}\,3^{3/4}\,\sqrt {2\,x+1}\,\left (\frac {1}{6}+\frac {1}{6}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{9}+\frac {1}{9}{}\mathrm {i}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- 4/(9*(2*x + 1)^(3/2)) - 2^(1/2)*3^(1/4)*atan(2^(1/2)*3^(3/4)*(2*x + 1)^(1/2)*(1/6 - 1i/6))*(1/9 + 1i/9) - 2^
(1/2)*3^(1/4)*atan(2^(1/2)*3^(3/4)*(2*x + 1)^(1/2)*(1/6 + 1i/6))*(1/9 - 1i/9)

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