Integrand size = 18, antiderivative size = 127 \[ \int \frac {(1+2 x)^{3/2}}{1+x+x^2} \, dx=4 \sqrt {1+2 x}+\sqrt {2} \sqrt [4]{3} \arctan \left (1-\frac {\sqrt {2} \sqrt {1+2 x}}{\sqrt [4]{3}}\right )-\sqrt {2} \sqrt [4]{3} \arctan \left (1+\frac {\sqrt {2} \sqrt {1+2 x}}{\sqrt [4]{3}}\right )-\sqrt {2} \sqrt [4]{3} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{3} \sqrt {1+2 x}}{1+\sqrt {3}+2 x}\right ) \] Output:
4*(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)*3^( 1/4)*arctanh(2^(1/2)*3^(1/4)*(1+2*x)^(1/2)/(1+3^(1/2)+2*x))
Time = 0.18 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.79 \[ \int \frac {(1+2 x)^{3/2}}{1+x+x^2} \, dx=4 \sqrt {1+2 x}-\sqrt {2} \sqrt [4]{3} \arctan \left (\frac {-3+\sqrt {3}+2 \sqrt {3} x}{3^{3/4} \sqrt {2+4 x}}\right )-\sqrt {2} \sqrt [4]{3} \text {arctanh}\left (\frac {3^{3/4} \sqrt {2+4 x}}{3+\sqrt {3}+2 \sqrt {3} x}\right ) \] Input:
Integrate[(1 + 2*x)^(3/2)/(1 + x + x^2),x]
Output:
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)]
Time = 0.39 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.54, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {1116, 1118, 27, 266, 755, 1476, 1082, 217, 1479, 25, 27, 1103}
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.
\(\displaystyle \int \frac {(2 x+1)^{3/2}}{x^2+x+1} \, dx\) |
\(\Big \downarrow \) 1116 |
\(\displaystyle 4 \sqrt {2 x+1}-3 \int \frac {1}{\sqrt {2 x+1} \left (x^2+x+1\right )}dx\) |
\(\Big \downarrow \) 1118 |
\(\displaystyle 4 \sqrt {2 x+1}-\frac {3}{2} \int \frac {4}{\sqrt {2 x+1} \left ((2 x+1)^2+3\right )}d(2 x+1)\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \sqrt {2 x+1}-6 \int \frac {1}{\sqrt {2 x+1} \left ((2 x+1)^2+3\right )}d(2 x+1)\) |
\(\Big \downarrow \) 266 |
\(\displaystyle 4 \sqrt {2 x+1}-12 \int \frac {1}{(2 x+1)^2+3}d\sqrt {2 x+1}\) |
\(\Big \downarrow \) 755 |
\(\displaystyle 4 \sqrt {2 x+1}-12 \left (\frac {\int \frac {-2 x+\sqrt {3}-1}{(2 x+1)^2+3}d\sqrt {2 x+1}}{2 \sqrt {3}}+\frac {\int \frac {2 x+\sqrt {3}+1}{(2 x+1)^2+3}d\sqrt {2 x+1}}{2 \sqrt {3}}\right )\) |
\(\Big \downarrow \) 1476 |
\(\displaystyle 4 \sqrt {2 x+1}-12 \left (\frac {\frac {1}{2} \int \frac {1}{2 x-\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1}d\sqrt {2 x+1}+\frac {1}{2} \int \frac {1}{2 x+\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1}d\sqrt {2 x+1}}{2 \sqrt {3}}+\frac {\int \frac {-2 x+\sqrt {3}-1}{(2 x+1)^2+3}d\sqrt {2 x+1}}{2 \sqrt {3}}\right )\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle 4 \sqrt {2 x+1}-12 \left (\frac {\frac {\int \frac {1}{-2 x-2}d\left (1-\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}\right )}{\sqrt {2} \sqrt [4]{3}}-\frac {\int \frac {1}{-2 x-2}d\left (\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}+1\right )}{\sqrt {2} \sqrt [4]{3}}}{2 \sqrt {3}}+\frac {\int \frac {-2 x+\sqrt {3}-1}{(2 x+1)^2+3}d\sqrt {2 x+1}}{2 \sqrt {3}}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle 4 \sqrt {2 x+1}-12 \left (\frac {\int \frac {-2 x+\sqrt {3}-1}{(2 x+1)^2+3}d\sqrt {2 x+1}}{2 \sqrt {3}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}+1\right )}{\sqrt {2} \sqrt [4]{3}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}\right )}{\sqrt {2} \sqrt [4]{3}}}{2 \sqrt {3}}\right )\) |
\(\Big \downarrow \) 1479 |
\(\displaystyle 4 \sqrt {2 x+1}-12 \left (\frac {-\frac {\int -\frac {\sqrt {2} \sqrt [4]{3}-2 \sqrt {2 x+1}}{2 x-\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1}d\sqrt {2 x+1}}{2 \sqrt {2} \sqrt [4]{3}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {2 x+1}+\sqrt [4]{3}\right )}{2 x+\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1}d\sqrt {2 x+1}}{2 \sqrt {2} \sqrt [4]{3}}}{2 \sqrt {3}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}+1\right )}{\sqrt {2} \sqrt [4]{3}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}\right )}{\sqrt {2} \sqrt [4]{3}}}{2 \sqrt {3}}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle 4 \sqrt {2 x+1}-12 \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{3}-2 \sqrt {2 x+1}}{2 x-\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1}d\sqrt {2 x+1}}{2 \sqrt {2} \sqrt [4]{3}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {2 x+1}+\sqrt [4]{3}\right )}{2 x+\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1}d\sqrt {2 x+1}}{2 \sqrt {2} \sqrt [4]{3}}}{2 \sqrt {3}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}+1\right )}{\sqrt {2} \sqrt [4]{3}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}\right )}{\sqrt {2} \sqrt [4]{3}}}{2 \sqrt {3}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \sqrt {2 x+1}-12 \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{3}-2 \sqrt {2 x+1}}{2 x-\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1}d\sqrt {2 x+1}}{2 \sqrt {2} \sqrt [4]{3}}+\frac {\int \frac {\sqrt {2} \sqrt {2 x+1}+\sqrt [4]{3}}{2 x+\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1}d\sqrt {2 x+1}}{2 \sqrt [4]{3}}}{2 \sqrt {3}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}+1\right )}{\sqrt {2} \sqrt [4]{3}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}\right )}{\sqrt {2} \sqrt [4]{3}}}{2 \sqrt {3}}\right )\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle 4 \sqrt {2 x+1}-12 \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}+1\right )}{\sqrt {2} \sqrt [4]{3}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}\right )}{\sqrt {2} \sqrt [4]{3}}}{2 \sqrt {3}}+\frac {\frac {\log \left (2 x+\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1\right )}{2 \sqrt {2} \sqrt [4]{3}}-\frac {\log \left (2 x-\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1\right )}{2 \sqrt {2} \sqrt [4]{3}}}{2 \sqrt {3}}\right )\) |
Input:
Int[(1 + 2*x)^(3/2)/(1 + x + x^2),x]
Output:
4*Sqrt[1 + 2*x] - 12*((-(ArcTan[1 - (Sqrt[2]*Sqrt[1 + 2*x])/3^(1/4)]/(Sqrt [2]*3^(1/4))) + ArcTan[1 + (Sqrt[2]*Sqrt[1 + 2*x])/3^(1/4)]/(Sqrt[2]*3^(1/ 4)))/(2*Sqrt[3]) + (-1/2*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]]/(2*Sqrt[2]*3^(1/4)))/(2*Sqrt[3]))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] ], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r) Int[(r - s*x^2)/(a + b*x^4) , x], x] + Simp[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]]))
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[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])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[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]
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S ymbol] :> Simp[2*d*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] + Simp[d^2*(m - 1)*((b^2 - 4*a*c)/(b^2*(m + 2*p + 1))) Int[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && 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])
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S ymbol] :> Simp[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] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 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]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Time = 1.40 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.86
method | result | size |
derivativedivides | \(4 \sqrt {1+2 x}-\frac {3^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {1+2 x +3^{\frac {1}{4}} \sqrt {1+2 x}\, \sqrt {2}+\sqrt {3}}{1+2 x -3^{\frac {1}{4}} \sqrt {1+2 x}\, \sqrt {2}+\sqrt {3}}\right )+2 \arctan \left (1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )+2 \arctan \left (-1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )\right )}{2}\) | \(109\) |
default | \(4 \sqrt {1+2 x}-\frac {3^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {1+2 x +3^{\frac {1}{4}} \sqrt {1+2 x}\, \sqrt {2}+\sqrt {3}}{1+2 x -3^{\frac {1}{4}} \sqrt {1+2 x}\, \sqrt {2}+\sqrt {3}}\right )+2 \arctan \left (1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )+2 \arctan \left (-1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )\right )}{2}\) | \(109\) |
risch | \(4 \sqrt {1+2 x}-\frac {3^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {1+2 x +3^{\frac {1}{4}} \sqrt {1+2 x}\, \sqrt {2}+\sqrt {3}}{1+2 x -3^{\frac {1}{4}} \sqrt {1+2 x}\, \sqrt {2}+\sqrt {3}}\right )+2 \arctan \left (1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )+2 \arctan \left (-1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )\right )}{2}\) | \(109\) |
pseudoelliptic | \(4 \sqrt {1+2 x}-\frac {\ln \left (\frac {1+2 x +3^{\frac {1}{4}} \sqrt {1+2 x}\, \sqrt {2}+\sqrt {3}}{1+2 x -3^{\frac {1}{4}} \sqrt {1+2 x}\, \sqrt {2}+\sqrt {3}}\right ) 3^{\frac {1}{4}} \sqrt {2}}{2}-\sqrt {2}\, 3^{\frac {1}{4}} \arctan \left (1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )-\sqrt {2}\, 3^{\frac {1}{4}} \arctan \left (-1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )\) | \(120\) |
trager | \(4 \sqrt {1+2 x}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{4} x +4 \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}\right ) x +2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}\right )+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}\right ) x +6 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}\right )-12 \sqrt {1+2 x}}{x \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}-x -2}\right )+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right ) \ln \left (-\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{5} x +4 x \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{3}+2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{3}-3 x \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )+12 \sqrt {1+2 x}-6 \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )}{x \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}+x +2}\right )\) | \(239\) |
Input:
int((1+2*x)^(3/2)/(x^2+x+1),x,method=_RETURNVERBOSE)
Output:
4*(1+2*x)^(1/2)-1/2*3^(1/4)*2^(1/2)*(ln((1+2*x+3^(1/4)*(1+2*x)^(1/2)*2^(1/ 2)+3^(1/2))/(1+2*x-3^(1/4)*(1+2*x)^(1/2)*2^(1/2)+3^(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)))
Time = 0.08 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.80 \[ \int \frac {(1+2 x)^{3/2}}{1+x+x^2} \, dx=-2 \, \left (\frac {3}{4}\right )^{\frac {1}{4}} \arctan \left (\frac {4}{3} \, \left (\frac {3}{4}\right )^{\frac {3}{4}} \sqrt {2 \, x + 1} + 1\right ) - 2 \, \left (\frac {3}{4}\right )^{\frac {1}{4}} \arctan \left (\frac {4}{3} \, \left (\frac {3}{4}\right )^{\frac {3}{4}} \sqrt {2 \, x + 1} - 1\right ) - \left (\frac {3}{4}\right )^{\frac {1}{4}} \log \left (2 \, x + 2 \, \left (\frac {3}{4}\right )^{\frac {1}{4}} \sqrt {2 \, x + 1} + \sqrt {3} + 1\right ) + \left (\frac {3}{4}\right )^{\frac {1}{4}} \log \left (2 \, x - 2 \, \left (\frac {3}{4}\right )^{\frac {1}{4}} \sqrt {2 \, x + 1} + \sqrt {3} + 1\right ) + 4 \, \sqrt {2 \, x + 1} \] Input:
integrate((1+2*x)^(3/2)/(x^2+x+1),x, algorithm="fricas")
Output:
-2*(3/4)^(1/4)*arctan(4/3*(3/4)^(3/4)*sqrt(2*x + 1) + 1) - 2*(3/4)^(1/4)*a rctan(4/3*(3/4)^(3/4)*sqrt(2*x + 1) - 1) - (3/4)^(1/4)*log(2*x + 2*(3/4)^( 1/4)*sqrt(2*x + 1) + sqrt(3) + 1) + (3/4)^(1/4)*log(2*x - 2*(3/4)^(1/4)*sq rt(2*x + 1) + sqrt(3) + 1) + 4*sqrt(2*x + 1)
\[ \int \frac {(1+2 x)^{3/2}}{1+x+x^2} \, dx=\int \frac {\left (2 x + 1\right )^{\frac {3}{2}}}{x^{2} + x + 1}\, dx \] Input:
integrate((1+2*x)**(3/2)/(x**2+x+1),x)
Output:
Integral((2*x + 1)**(3/2)/(x**2 + x + 1), x)
Time = 0.11 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.11 \[ \int \frac {(1+2 x)^{3/2}}{1+x+x^2} \, dx=-3^{\frac {1}{4}} \sqrt {2} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} + 2 \, \sqrt {2 \, x + 1}\right )}\right ) - 3^{\frac {1}{4}} \sqrt {2} \arctan \left (-\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} - 2 \, \sqrt {2 \, x + 1}\right )}\right ) - \frac {1}{2} \cdot 3^{\frac {1}{4}} \sqrt {2} \log \left (3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) + \frac {1}{2} \cdot 3^{\frac {1}{4}} \sqrt {2} \log \left (-3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) + 4 \, \sqrt {2 \, x + 1} \] Input:
integrate((1+2*x)^(3/2)/(x^2+x+1),x, algorithm="maxima")
Output:
-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)
Time = 0.15 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.02 \[ \int \frac {(1+2 x)^{3/2}}{1+x+x^2} \, dx=-12^{\frac {1}{4}} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} + 2 \, \sqrt {2 \, x + 1}\right )}\right ) - 12^{\frac {1}{4}} \arctan \left (-\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} - 2 \, \sqrt {2 \, x + 1}\right )}\right ) - \frac {1}{2} \cdot 12^{\frac {1}{4}} \log \left (3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) + \frac {1}{2} \cdot 12^{\frac {1}{4}} \log \left (-3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) + 4 \, \sqrt {2 \, x + 1} \] Input:
integrate((1+2*x)^(3/2)/(x^2+x+1),x, algorithm="giac")
Output:
-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*s qrt(2*x + 1)
Time = 0.05 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.52 \[ \int \frac {(1+2 x)^{3/2}}{1+x+x^2} \, dx=4\,\sqrt {2\,x+1}+\sqrt {2}\,3^{1/4}\,\mathrm {atan}\left (\sqrt {2}\,3^{3/4}\,\sqrt {2\,x+1}\,\left (\frac {1}{6}-\frac {1}{6}{}\mathrm {i}\right )\right )\,\left (-1-\mathrm {i}\right )+\sqrt {2}\,3^{1/4}\,\mathrm {atan}\left (\sqrt {2}\,3^{3/4}\,\sqrt {2\,x+1}\,\left (\frac {1}{6}+\frac {1}{6}{}\mathrm {i}\right )\right )\,\left (-1+1{}\mathrm {i}\right ) \] Input:
int((2*x + 1)^(3/2)/(x + x^2 + 1),x)
Output:
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)
Time = 0.20 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.02 \[ \int \frac {(1+2 x)^{3/2}}{1+x+x^2} \, dx=-\sqrt {2}\, 3^{\frac {1}{4}} \mathit {atan} \left (\frac {\left (2 \sqrt {2 x +1}-\sqrt {2}\, 3^{\frac {1}{4}}\right ) 3^{\frac {3}{4}}}{3 \sqrt {2}}\right )-\sqrt {2}\, 3^{\frac {1}{4}} \mathit {atan} \left (\frac {\left (2 \sqrt {2 x +1}+\sqrt {2}\, 3^{\frac {1}{4}}\right ) 3^{\frac {3}{4}}}{3 \sqrt {2}}\right )+4 \sqrt {2 x +1}+\frac {\sqrt {2}\, 3^{\frac {1}{4}} \mathrm {log}\left (-\sqrt {2 x +1}\, \sqrt {2}\, 3^{\frac {1}{4}}+\sqrt {3}+2 x +1\right )}{2}-\frac {\sqrt {2}\, 3^{\frac {1}{4}} \mathrm {log}\left (\sqrt {2 x +1}\, \sqrt {2}\, 3^{\frac {1}{4}}+\sqrt {3}+2 x +1\right )}{2} \] Input:
int((1+2*x)^(3/2)/(x^2+x+1),x)
Output:
( - 2*sqrt(2)*3**(1/4)*atan((2*sqrt(2*x + 1) - sqrt(2)*3**(1/4))/(sqrt(2)* 3**(1/4))) - 2*sqrt(2)*3**(1/4)*atan((2*sqrt(2*x + 1) + sqrt(2)*3**(1/4))/ (sqrt(2)*3**(1/4))) + 8*sqrt(2*x + 1) + sqrt(2)*3**(1/4)*log( - sqrt(2*x + 1)*sqrt(2)*3**(1/4) + sqrt(3) + 2*x + 1) - sqrt(2)*3**(1/4)*log(sqrt(2*x + 1)*sqrt(2)*3**(1/4) + sqrt(3) + 2*x + 1))/2