Integrand size = 18, antiderivative size = 141 \[ \int \frac {(1+2 x)^{7/2}}{1+x+x^2} \, dx=-12 \sqrt {1+2 x}+\frac {4}{5} (1+2 x)^{5/2}-3 \sqrt {2} \sqrt [4]{3} \arctan \left (1-\frac {\sqrt {2} \sqrt {1+2 x}}{\sqrt [4]{3}}\right )+3 \sqrt {2} \sqrt [4]{3} \arctan \left (1+\frac {\sqrt {2} \sqrt {1+2 x}}{\sqrt [4]{3}}\right )+3 \sqrt {2} \sqrt [4]{3} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{3} \sqrt {1+2 x}}{1+\sqrt {3}+2 x}\right ) \] Output:
-12*(1+2*x)^(1/2)+4/5*(1+2*x)^(5/2)+3*2^(1/2)*3^(1/4)*arctan(-1+1/3*2^(1/2 )*(1+2*x)^(1/2)*3^(3/4))+3*2^(1/2)*3^(1/4)*arctan(1+1/3*2^(1/2)*(1+2*x)^(1 /2)*3^(3/4))+3*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 = 0.27 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.79 \[ \int \frac {(1+2 x)^{7/2}}{1+x+x^2} \, dx=\frac {8}{5} \sqrt {1+2 x} \left (-7+2 x+2 x^2\right )+3 \sqrt {2} \sqrt [4]{3} \arctan \left (\frac {-3+\sqrt {3}+2 \sqrt {3} x}{3^{3/4} \sqrt {2+4 x}}\right )+3 \sqrt {2} \sqrt [4]{3} \text {arctanh}\left (\frac {3^{3/4} \sqrt {2+4 x}}{3+\sqrt {3}+2 \sqrt {3} x}\right ) \] Input:
Integrate[(1 + 2*x)^(7/2)/(1 + x + x^2),x]
Output:
(8*Sqrt[1 + 2*x]*(-7 + 2*x + 2*x^2))/5 + 3*Sqrt[2]*3^(1/4)*ArcTan[(-3 + Sq rt[3] + 2*Sqrt[3]*x)/(3^(3/4)*Sqrt[2 + 4*x])] + 3*Sqrt[2]*3^(1/4)*ArcTanh[ (3^(3/4)*Sqrt[2 + 4*x])/(3 + Sqrt[3] + 2*Sqrt[3]*x)]
Time = 0.44 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.50, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.722, Rules used = {1116, 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.
| \(\displaystyle \int \frac {(2 x+1)^{7/2}}{x^2+x+1} \, dx\) |
| \(\Big \downarrow \) 1116 |
| \(\displaystyle \frac {4}{5} (2 x+1)^{5/2}-3 \int \frac {(2 x+1)^{3/2}}{x^2+x+1}dx\) |
| \(\Big \downarrow \) 1116 |
| \(\displaystyle \frac {4}{5} (2 x+1)^{5/2}-3 \left (4 \sqrt {2 x+1}-3 \int \frac {1}{\sqrt {2 x+1} \left (x^2+x+1\right )}dx\right )\) |
| \(\Big \downarrow \) 1118 |
| \(\displaystyle \frac {4}{5} (2 x+1)^{5/2}-3 \left (4 \sqrt {2 x+1}-\frac {3}{2} \int \frac {4}{\sqrt {2 x+1} \left ((2 x+1)^2+3\right )}d(2 x+1)\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {4}{5} (2 x+1)^{5/2}-3 \left (4 \sqrt {2 x+1}-6 \int \frac {1}{\sqrt {2 x+1} \left ((2 x+1)^2+3\right )}d(2 x+1)\right )\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {4}{5} (2 x+1)^{5/2}-3 \left (4 \sqrt {2 x+1}-12 \int \frac {1}{(2 x+1)^2+3}d\sqrt {2 x+1}\right )\) |
| \(\Big \downarrow \) 755 |
| \(\displaystyle \frac {4}{5} (2 x+1)^{5/2}-3 \left (4 \sqrt {2 x+1}-12 \left (\frac {\int \frac {-2 x+\sqrt {3}-1}{(2 x+1)^2+3}d\sqrt {2 x+1}}{2 \sqrt {3}}+\frac {\int \frac {2 x+\sqrt {3}+1}{(2 x+1)^2+3}d\sqrt {2 x+1}}{2 \sqrt {3}}\right )\right )\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {4}{5} (2 x+1)^{5/2}-3 \left (4 \sqrt {2 x+1}-12 \left (\frac {\frac {1}{2} \int \frac {1}{2 x-\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1}d\sqrt {2 x+1}+\frac {1}{2} \int \frac {1}{2 x+\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1}d\sqrt {2 x+1}}{2 \sqrt {3}}+\frac {\int \frac {-2 x+\sqrt {3}-1}{(2 x+1)^2+3}d\sqrt {2 x+1}}{2 \sqrt {3}}\right )\right )\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {4}{5} (2 x+1)^{5/2}-3 \left (4 \sqrt {2 x+1}-12 \left (\frac {\frac {\int \frac {1}{-2 x-2}d\left (1-\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}\right )}{\sqrt {2} \sqrt [4]{3}}-\frac {\int \frac {1}{-2 x-2}d\left (\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}+1\right )}{\sqrt {2} \sqrt [4]{3}}}{2 \sqrt {3}}+\frac {\int \frac {-2 x+\sqrt {3}-1}{(2 x+1)^2+3}d\sqrt {2 x+1}}{2 \sqrt {3}}\right )\right )\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {4}{5} (2 x+1)^{5/2}-3 \left (4 \sqrt {2 x+1}-12 \left (\frac {\int \frac {-2 x+\sqrt {3}-1}{(2 x+1)^2+3}d\sqrt {2 x+1}}{2 \sqrt {3}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}+1\right )}{\sqrt {2} \sqrt [4]{3}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}\right )}{\sqrt {2} \sqrt [4]{3}}}{2 \sqrt {3}}\right )\right )\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {4}{5} (2 x+1)^{5/2}-3 \left (4 \sqrt {2 x+1}-12 \left (\frac {-\frac {\int -\frac {\sqrt {2} \sqrt [4]{3}-2 \sqrt {2 x+1}}{2 x-\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1}d\sqrt {2 x+1}}{2 \sqrt {2} \sqrt [4]{3}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {2 x+1}+\sqrt [4]{3}\right )}{2 x+\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1}d\sqrt {2 x+1}}{2 \sqrt {2} \sqrt [4]{3}}}{2 \sqrt {3}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}+1\right )}{\sqrt {2} \sqrt [4]{3}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}\right )}{\sqrt {2} \sqrt [4]{3}}}{2 \sqrt {3}}\right )\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {4}{5} (2 x+1)^{5/2}-3 \left (4 \sqrt {2 x+1}-12 \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{3}-2 \sqrt {2 x+1}}{2 x-\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1}d\sqrt {2 x+1}}{2 \sqrt {2} \sqrt [4]{3}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {2 x+1}+\sqrt [4]{3}\right )}{2 x+\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1}d\sqrt {2 x+1}}{2 \sqrt {2} \sqrt [4]{3}}}{2 \sqrt {3}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}+1\right )}{\sqrt {2} \sqrt [4]{3}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}\right )}{\sqrt {2} \sqrt [4]{3}}}{2 \sqrt {3}}\right )\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {4}{5} (2 x+1)^{5/2}-3 \left (4 \sqrt {2 x+1}-12 \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{3}-2 \sqrt {2 x+1}}{2 x-\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1}d\sqrt {2 x+1}}{2 \sqrt {2} \sqrt [4]{3}}+\frac {\int \frac {\sqrt {2} \sqrt {2 x+1}+\sqrt [4]{3}}{2 x+\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1}d\sqrt {2 x+1}}{2 \sqrt [4]{3}}}{2 \sqrt {3}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}+1\right )}{\sqrt {2} \sqrt [4]{3}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}\right )}{\sqrt {2} \sqrt [4]{3}}}{2 \sqrt {3}}\right )\right )\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {4}{5} (2 x+1)^{5/2}-3 \left (4 \sqrt {2 x+1}-12 \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}+1\right )}{\sqrt {2} \sqrt [4]{3}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}\right )}{\sqrt {2} \sqrt [4]{3}}}{2 \sqrt {3}}+\frac {\frac {\log \left (2 x+\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1\right )}{2 \sqrt {2} \sqrt [4]{3}}-\frac {\log \left (2 x-\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1\right )}{2 \sqrt {2} \sqrt [4]{3}}}{2 \sqrt {3}}\right )\right )\) |
Input:
Int[(1 + 2*x)^(7/2)/(1 + x + x^2),x]
Output:
(4*(1 + 2*x)^(5/2))/5 - 3*(4*Sqrt[1 + 2*x] - 12*((-(ArcTan[1 - (Sqrt[2]*Sq rt[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 = 2.40 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.84
| method | result | size |
| derivativedivides | \(\frac {4 \left (1+2 x \right )^{\frac {5}{2}}}{5}-12 \sqrt {1+2 x}+\frac {3 \,3^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {1+2 x +3^{\frac {1}{4}} \sqrt {1+2 x}\, \sqrt {2}+\sqrt {3}}{1+2 x -3^{\frac {1}{4}} \sqrt {1+2 x}\, \sqrt {2}+\sqrt {3}}\right )+2 \arctan \left (1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )+2 \arctan \left (-1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )\right )}{2}\) | \(118\) |
| default | \(\frac {4 \left (1+2 x \right )^{\frac {5}{2}}}{5}-12 \sqrt {1+2 x}+\frac {3 \,3^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {1+2 x +3^{\frac {1}{4}} \sqrt {1+2 x}\, \sqrt {2}+\sqrt {3}}{1+2 x -3^{\frac {1}{4}} \sqrt {1+2 x}\, \sqrt {2}+\sqrt {3}}\right )+2 \arctan \left (1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )+2 \arctan \left (-1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )\right )}{2}\) | \(118\) |
| risch | \(\frac {8 \left (2 x^{2}+2 x -7\right ) \sqrt {1+2 x}}{5}+\frac {3 \,3^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {1+2 x +3^{\frac {1}{4}} \sqrt {1+2 x}\, \sqrt {2}+\sqrt {3}}{1+2 x -3^{\frac {1}{4}} \sqrt {1+2 x}\, \sqrt {2}+\sqrt {3}}\right )+2 \arctan \left (1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )+2 \arctan \left (-1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )\right )}{2}\) | \(119\) |
| pseudoelliptic | \(\frac {8 \left (2 x^{2}+2 x -7\right ) \sqrt {1+2 x}}{5}+\frac {3 \,3^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {1+2 x +3^{\frac {1}{4}} \sqrt {1+2 x}\, \sqrt {2}+\sqrt {3}}{1+2 x -3^{\frac {1}{4}} \sqrt {1+2 x}\, \sqrt {2}+\sqrt {3}}\right )+2 \arctan \left (1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )+2 \arctan \left (-1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )\right )}{2}\) | \(119\) |
| trager | \(\left (\frac {16}{5} x^{2}+\frac {16}{5} x -\frac {56}{5}\right ) \sqrt {1+2 x}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{4} x +4 \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}\right ) x +2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}\right )+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}\right ) x +12 \sqrt {1+2 x}+6 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}\right )}{x \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}-x -2}\right )+3 \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{5} x -4 x \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{3}-2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{3}+3 x \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )+12 \sqrt {1+2 x}+6 \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )}{x \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}+x +2}\right )\) | \(250\) |
Input:
int((1+2*x)^(7/2)/(x^2+x+1),x,method=_RETURNVERBOSE)
Output:
4/5*(1+2*x)^(5/2)-12*(1+2*x)^(1/2)+3/2*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 = 112, normalized size of antiderivative = 0.79 \[ \int \frac {(1+2 x)^{7/2}}{1+x+x^2} \, dx=\frac {8}{5} \, {\left (2 \, x^{2} + 2 \, x - 7\right )} \sqrt {2 \, x + 1} + 6 \, \left (\frac {3}{4}\right )^{\frac {1}{4}} \arctan \left (\frac {4}{3} \, \left (\frac {3}{4}\right )^{\frac {3}{4}} \sqrt {2 \, x + 1} + 1\right ) + 6 \, \left (\frac {3}{4}\right )^{\frac {1}{4}} \arctan \left (\frac {4}{3} \, \left (\frac {3}{4}\right )^{\frac {3}{4}} \sqrt {2 \, x + 1} - 1\right ) + 3 \, \left (\frac {3}{4}\right )^{\frac {1}{4}} \log \left (2 \, x + 2 \, \left (\frac {3}{4}\right )^{\frac {1}{4}} \sqrt {2 \, x + 1} + \sqrt {3} + 1\right ) - 3 \, \left (\frac {3}{4}\right )^{\frac {1}{4}} \log \left (2 \, x - 2 \, \left (\frac {3}{4}\right )^{\frac {1}{4}} \sqrt {2 \, x + 1} + \sqrt {3} + 1\right ) \] Input:
integrate((1+2*x)^(7/2)/(x^2+x+1),x, algorithm="fricas")
Output:
8/5*(2*x^2 + 2*x - 7)*sqrt(2*x + 1) + 6*(3/4)^(1/4)*arctan(4/3*(3/4)^(3/4) *sqrt(2*x + 1) + 1) + 6*(3/4)^(1/4)*arctan(4/3*(3/4)^(3/4)*sqrt(2*x + 1) - 1) + 3*(3/4)^(1/4)*log(2*x + 2*(3/4)^(1/4)*sqrt(2*x + 1) + sqrt(3) + 1) - 3*(3/4)^(1/4)*log(2*x - 2*(3/4)^(1/4)*sqrt(2*x + 1) + sqrt(3) + 1)
\[ \int \frac {(1+2 x)^{7/2}}{1+x+x^2} \, dx=\int \frac {\left (2 x + 1\right )^{\frac {7}{2}}}{x^{2} + x + 1}\, dx \] Input:
integrate((1+2*x)**(7/2)/(x**2+x+1),x)
Output:
Integral((2*x + 1)**(7/2)/(x**2 + x + 1), x)
Time = 0.11 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.06 \[ \int \frac {(1+2 x)^{7/2}}{1+x+x^2} \, dx=\frac {4}{5} \, {\left (2 \, x + 1\right )}^{\frac {5}{2}} + 3 \cdot 3^{\frac {1}{4}} \sqrt {2} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} + 2 \, \sqrt {2 \, x + 1}\right )}\right ) + 3 \cdot 3^{\frac {1}{4}} \sqrt {2} \arctan \left (-\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} - 2 \, \sqrt {2 \, x + 1}\right )}\right ) + \frac {3}{2} \cdot 3^{\frac {1}{4}} \sqrt {2} \log \left (3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) - \frac {3}{2} \cdot 3^{\frac {1}{4}} \sqrt {2} \log \left (-3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) - 12 \, \sqrt {2 \, x + 1} \] Input:
integrate((1+2*x)^(7/2)/(x^2+x+1),x, algorithm="maxima")
Output:
4/5*(2*x + 1)^(5/2) + 3*3^(1/4)*sqrt(2)*arctan(1/6*3^(3/4)*sqrt(2)*(3^(1/4 )*sqrt(2) + 2*sqrt(2*x + 1))) + 3*3^(1/4)*sqrt(2)*arctan(-1/6*3^(3/4)*sqrt (2)*(3^(1/4)*sqrt(2) - 2*sqrt(2*x + 1))) + 3/2*3^(1/4)*sqrt(2)*log(3^(1/4) *sqrt(2)*sqrt(2*x + 1) + 2*x + sqrt(3) + 1) - 3/2*3^(1/4)*sqrt(2)*log(-3^( 1/4)*sqrt(2)*sqrt(2*x + 1) + 2*x + sqrt(3) + 1) - 12*sqrt(2*x + 1)
Time = 0.15 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.98 \[ \int \frac {(1+2 x)^{7/2}}{1+x+x^2} \, dx=\frac {4}{5} \, {\left (2 \, x + 1\right )}^{\frac {5}{2}} + 3 \cdot 12^{\frac {1}{4}} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} + 2 \, \sqrt {2 \, x + 1}\right )}\right ) + 3 \cdot 12^{\frac {1}{4}} \arctan \left (-\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} - 2 \, \sqrt {2 \, x + 1}\right )}\right ) + \frac {3}{2} \cdot 12^{\frac {1}{4}} \log \left (3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) - \frac {3}{2} \cdot 12^{\frac {1}{4}} \log \left (-3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) - 12 \, \sqrt {2 \, x + 1} \] Input:
integrate((1+2*x)^(7/2)/(x^2+x+1),x, algorithm="giac")
Output:
4/5*(2*x + 1)^(5/2) + 3*12^(1/4)*arctan(1/6*3^(3/4)*sqrt(2)*(3^(1/4)*sqrt( 2) + 2*sqrt(2*x + 1))) + 3*12^(1/4)*arctan(-1/6*3^(3/4)*sqrt(2)*(3^(1/4)*s qrt(2) - 2*sqrt(2*x + 1))) + 3/2*12^(1/4)*log(3^(1/4)*sqrt(2)*sqrt(2*x + 1 ) + 2*x + sqrt(3) + 1) - 3/2*12^(1/4)*log(-3^(1/4)*sqrt(2)*sqrt(2*x + 1) + 2*x + sqrt(3) + 1) - 12*sqrt(2*x + 1)
Time = 5.36 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.53 \[ \int \frac {(1+2 x)^{7/2}}{1+x+x^2} \, dx=\frac {4\,{\left (2\,x+1\right )}^{5/2}}{5}-12\,\sqrt {2\,x+1}+\sqrt {2}\,3^{1/4}\,\mathrm {atan}\left (\sqrt {2}\,3^{3/4}\,\sqrt {2\,x+1}\,\left (\frac {1}{6}-\frac {1}{6}{}\mathrm {i}\right )\right )\,\left (3+3{}\mathrm {i}\right )+\sqrt {2}\,3^{1/4}\,\mathrm {atan}\left (\sqrt {2}\,3^{3/4}\,\sqrt {2\,x+1}\,\left (\frac {1}{6}+\frac {1}{6}{}\mathrm {i}\right )\right )\,\left (3-3{}\mathrm {i}\right ) \] Input:
int((2*x + 1)^(7/2)/(x + x^2 + 1),x)
Output:
(4*(2*x + 1)^(5/2))/5 - 12*(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))*(3 + 3i) + 2^(1/2)*3^(1/4)*atan(2^(1 /2)*3^(3/4)*(2*x + 1)^(1/2)*(1/6 + 1i/6))*(3 - 3i)
Time = 0.22 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.06 \[ \int \frac {(1+2 x)^{7/2}}{1+x+x^2} \, dx=3 \sqrt {2}\, 3^{\frac {1}{4}} \mathit {atan} \left (\frac {\left (2 \sqrt {2 x +1}-\sqrt {2}\, 3^{\frac {1}{4}}\right ) 3^{\frac {3}{4}}}{3 \sqrt {2}}\right )+3 \sqrt {2}\, 3^{\frac {1}{4}} \mathit {atan} \left (\frac {\left (2 \sqrt {2 x +1}+\sqrt {2}\, 3^{\frac {1}{4}}\right ) 3^{\frac {3}{4}}}{3 \sqrt {2}}\right )+\frac {16 \sqrt {2 x +1}\, x^{2}}{5}+\frac {16 \sqrt {2 x +1}\, x}{5}-\frac {56 \sqrt {2 x +1}}{5}-\frac {3 \sqrt {2}\, 3^{\frac {1}{4}} \mathrm {log}\left (-\sqrt {2 x +1}\, \sqrt {2}\, 3^{\frac {1}{4}}+\sqrt {3}+2 x +1\right )}{2}+\frac {3 \sqrt {2}\, 3^{\frac {1}{4}} \mathrm {log}\left (\sqrt {2 x +1}\, \sqrt {2}\, 3^{\frac {1}{4}}+\sqrt {3}+2 x +1\right )}{2} \] Input:
int((1+2*x)^(7/2)/(x^2+x+1),x)
Output:
(30*sqrt(2)*3**(1/4)*atan((2*sqrt(2*x + 1) - sqrt(2)*3**(1/4))/(sqrt(2)*3* *(1/4))) + 30*sqrt(2)*3**(1/4)*atan((2*sqrt(2*x + 1) + sqrt(2)*3**(1/4))/( sqrt(2)*3**(1/4))) + 32*sqrt(2*x + 1)*x**2 + 32*sqrt(2*x + 1)*x - 112*sqrt (2*x + 1) - 15*sqrt(2)*3**(1/4)*log( - sqrt(2*x + 1)*sqrt(2)*3**(1/4) + sq rt(3) + 2*x + 1) + 15*sqrt(2)*3**(1/4)*log(sqrt(2*x + 1)*sqrt(2)*3**(1/4) + sqrt(3) + 2*x + 1))/10