\(\int \frac {(1+2 x)^{3/2}}{1+x+x^2} \, dx\) [1322]

Optimal result

Integrand size = 18, antiderivative size = 168 \[ \int \frac {(1+2 x)^{3/2}}{1+x+x^2} \, dx=4 \sqrt {1+2 x}+\sqrt {2} \sqrt [4]{3} \arctan \left (1-\frac {\sqrt {2} \sqrt {1+2 x}}{\sqrt [4]{3}}\right )-\sqrt {2} \sqrt [4]{3} \arctan \left (1+\frac {\sqrt {2} \sqrt {1+2 x}}{\sqrt [4]{3}}\right )+\frac {\sqrt [4]{3} \log \left (1+\sqrt {3}+2 x-\sqrt {2} \sqrt [4]{3} \sqrt {1+2 x}\right )}{\sqrt {2}}-\frac {\sqrt [4]{3} \log \left (1+\sqrt {3}+2 x+\sqrt {2} \sqrt [4]{3} \sqrt {1+2 x}\right )}{\sqrt {2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {706, 708, 335, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {(1+2 x)^{3/2}}{1+x+x^2} \, dx=\sqrt {2} \sqrt [4]{3} \arctan \left (1-\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}\right )-\sqrt {2} \sqrt [4]{3} \arctan \left (\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}+1\right )+4 \sqrt {2 x+1}+\frac {\sqrt [4]{3} \log \left (2 x-\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1\right )}{\sqrt {2}}-\frac {\sqrt [4]{3} \log \left (2 x+\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1\right )}{\sqrt {2}} \]

[In]

Int[(1 + 2*x)^(3/2)/(1 + x + x^2),x]

[Out]

4*Sqrt[1 + 2*x] + Sqrt[2]*3^(1/4)*ArcTan[1 - (Sqrt[2]*Sqrt[1 + 2*x])/3^(1/4)] - Sqrt[2]*3^(1/4)*ArcTan[1 + (Sq
rt[2]*Sqrt[1 + 2*x])/3^(1/4)] + (3^(1/4)*Log[1 + Sqrt[3] + 2*x - Sqrt[2]*3^(1/4)*Sqrt[1 + 2*x]])/Sqrt[2] - (3^
(1/4)*Log[1 + Sqrt[3] + 2*x + Sqrt[2]*3^(1/4)*Sqrt[1 + 2*x]])/Sqrt[2]

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 706

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2*d*(d + e*x)^(m - 1
)*((a + b*x + c*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] + Dist[d^2*(m - 1)*((b^2 - 4*a*c)/(b^2*(m + 2*p + 1))), In
t[(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] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] &
& RationalQ[p]) || OddQ[m])

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.60 \[ \int \frac {(1+2 x)^{3/2}}{1+x+x^2} \, dx=4 \sqrt {1+2 x}-\sqrt {2} \sqrt [4]{3} \arctan \left (\frac {-3+\sqrt {3}+2 \sqrt {3} x}{3^{3/4} \sqrt {2+4 x}}\right )-\sqrt {2} \sqrt [4]{3} \text {arctanh}\left (\frac {3^{3/4} \sqrt {2+4 x}}{3+\sqrt {3}+2 \sqrt {3} x}\right ) \]

[In]

Integrate[(1 + 2*x)^(3/2)/(1 + x + x^2),x]

[Out]

4*Sqrt[1 + 2*x] - 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)]

Maple [A] (verified)

Time = 2.57 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.65

method	result	size
derivativedivides	\(4 \sqrt {1+2 x}-\frac {3^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {1+2 x +\sqrt {3}+3^{\frac {1}{4}} \sqrt {2}\, \sqrt {1+2 x}}{1+2 x +\sqrt {3}-3^{\frac {1}{4}} \sqrt {2}\, \sqrt {1+2 x}}\right )+2 \arctan \left (1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )+2 \arctan \left (-1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )\right )}{2}\)	\(109\)
default	\(4 \sqrt {1+2 x}-\frac {3^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {1+2 x +\sqrt {3}+3^{\frac {1}{4}} \sqrt {2}\, \sqrt {1+2 x}}{1+2 x +\sqrt {3}-3^{\frac {1}{4}} \sqrt {2}\, \sqrt {1+2 x}}\right )+2 \arctan \left (1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )+2 \arctan \left (-1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )\right )}{2}\)	\(109\)
risch	\(4 \sqrt {1+2 x}-\frac {3^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {1+2 x +\sqrt {3}+3^{\frac {1}{4}} \sqrt {2}\, \sqrt {1+2 x}}{1+2 x +\sqrt {3}-3^{\frac {1}{4}} \sqrt {2}\, \sqrt {1+2 x}}\right )+2 \arctan \left (1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )+2 \arctan \left (-1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )\right )}{2}\)	\(109\)
pseudoelliptic	\(4 \sqrt {1+2 x}-\frac {\ln \left (\frac {1+2 x +\sqrt {3}+3^{\frac {1}{4}} \sqrt {2}\, \sqrt {1+2 x}}{1+2 x +\sqrt {3}-3^{\frac {1}{4}} \sqrt {2}\, \sqrt {1+2 x}}\right ) 3^{\frac {1}{4}} \sqrt {2}}{2}-3^{\frac {1}{4}} \arctan \left (1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right ) \sqrt {2}-3^{\frac {1}{4}} \arctan \left (-1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right ) \sqrt {2}\)	\(120\)
trager	\(4 \sqrt {1+2 x}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{5} x -4 x \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{3}-2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{3}+3 x \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )+6 \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )-12 \sqrt {1+2 x}}{x \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}+x +2}\right )+\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{4} x +4 \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}\right ) x +2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}\right )+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}\right ) x -12 \sqrt {1+2 x}+6 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}\right )}{x \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}-x -2}\right )\)	\(237\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.39 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.49 \[ \int \frac {(1+2 x)^{3/2}}{1+x+x^2} \, dx=-\left (-3\right )^{\frac {1}{4}} \log \left (\sqrt {2 \, x + 1} + \left (-3\right )^{\frac {1}{4}}\right ) - i \, \left (-3\right )^{\frac {1}{4}} \log \left (\sqrt {2 \, x + 1} + i \, \left (-3\right )^{\frac {1}{4}}\right ) + i \, \left (-3\right )^{\frac {1}{4}} \log \left (\sqrt {2 \, x + 1} - i \, \left (-3\right )^{\frac {1}{4}}\right ) + \left (-3\right )^{\frac {1}{4}} \log \left (\sqrt {2 \, x + 1} - \left (-3\right )^{\frac {1}{4}}\right ) + 4 \, \sqrt {2 \, x + 1} \]

[In]

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

[Out]

-(-3)^(1/4)*log(sqrt(2*x + 1) + (-3)^(1/4)) - I*(-3)^(1/4)*log(sqrt(2*x + 1) + I*(-3)^(1/4)) + I*(-3)^(1/4)*lo
g(sqrt(2*x + 1) - I*(-3)^(1/4)) + (-3)^(1/4)*log(sqrt(2*x + 1) - (-3)^(1/4)) + 4*sqrt(2*x + 1)

Sympy [F]

\[ \int \frac {(1+2 x)^{3/2}}{1+x+x^2} \, dx=\int \frac {\left (2 x + 1\right )^{\frac {3}{2}}}{x^{2} + x + 1}\, dx \]

[In]

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

[Out]

Integral((2*x + 1)**(3/2)/(x**2 + x + 1), x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.84 \[ \int \frac {(1+2 x)^{3/2}}{1+x+x^2} \, dx=-3^{\frac {1}{4}} \sqrt {2} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} + 2 \, \sqrt {2 \, x + 1}\right )}\right ) - 3^{\frac {1}{4}} \sqrt {2} \arctan \left (-\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} - 2 \, \sqrt {2 \, x + 1}\right )}\right ) - \frac {1}{2} \cdot 3^{\frac {1}{4}} \sqrt {2} \log \left (3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) + \frac {1}{2} \cdot 3^{\frac {1}{4}} \sqrt {2} \log \left (-3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) + 4 \, \sqrt {2 \, x + 1} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.77 \[ \int \frac {(1+2 x)^{3/2}}{1+x+x^2} \, dx=-12^{\frac {1}{4}} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} + 2 \, \sqrt {2 \, x + 1}\right )}\right ) - 12^{\frac {1}{4}} \arctan \left (-\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} - 2 \, \sqrt {2 \, x + 1}\right )}\right ) - \frac {1}{2} \cdot 12^{\frac {1}{4}} \log \left (3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) + \frac {1}{2} \cdot 12^{\frac {1}{4}} \log \left (-3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) + 4 \, \sqrt {2 \, x + 1} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.36 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.39 \[ \int \frac {(1+2 x)^{3/2}}{1+x+x^2} \, dx=4\,\sqrt {2\,x+1}+\sqrt {2}\,3^{1/4}\,\mathrm {atan}\left (\sqrt {2}\,3^{3/4}\,\sqrt {2\,x+1}\,\left (\frac {1}{6}-\frac {1}{6}{}\mathrm {i}\right )\right )\,\left (-1-\mathrm {i}\right )+\sqrt {2}\,3^{1/4}\,\mathrm {atan}\left (\sqrt {2}\,3^{3/4}\,\sqrt {2\,x+1}\,\left (\frac {1}{6}+\frac {1}{6}{}\mathrm {i}\right )\right )\,\left (-1+1{}\mathrm {i}\right ) \]

[In]

int((2*x + 1)^(3/2)/(x + x^2 + 1),x)

[Out]

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