\(\int \frac {(1+2 x)^{7/2}}{1+x+x^2} \, dx\) [1320]

Optimal result

Integrand size = 18, antiderivative size = 183 \[ \int \frac {(1+2 x)^{7/2}}{1+x+x^2} \, dx=-12 \sqrt {1+2 x}+\frac {4}{5} (1+2 x)^{5/2}-3 \sqrt {2} \sqrt [4]{3} \arctan \left (1-\frac {\sqrt {2} \sqrt {1+2 x}}{\sqrt [4]{3}}\right )+3 \sqrt {2} \sqrt [4]{3} \arctan \left (1+\frac {\sqrt {2} \sqrt {1+2 x}}{\sqrt [4]{3}}\right )-\frac {3 \sqrt [4]{3} \log \left (1+\sqrt {3}+2 x-\sqrt {2} \sqrt [4]{3} \sqrt {1+2 x}\right )}{\sqrt {2}}+\frac {3 \sqrt [4]{3} \log \left (1+\sqrt {3}+2 x+\sqrt {2} \sqrt [4]{3} \sqrt {1+2 x}\right )}{\sqrt {2}} \]

[Out]

4/5*(1+2*x)^(5/2)-3/2*3^(1/4)*ln(1+2*x+3^(1/2)-3^(1/4)*2^(1/2)*(1+2*x)^(1/2))*2^(1/2)+3/2*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*3^(1/4)*arctan(-1+1/3*2^(1/2)*(1+2*x)^(1/2)*3^(3/4))*2^(1/2)+3*3
^(1/4)*arctan(1+1/3*2^(1/2)*(1+2*x)^(1/2)*3^(3/4))*2^(1/2)-12*(1+2*x)^(1/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 13, 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} \[ \int \frac {(1+2 x)^{7/2}}{1+x+x^2} \, dx=-3 \sqrt {2} \sqrt [4]{3} \arctan \left (1-\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}\right )+3 \sqrt {2} \sqrt [4]{3} \arctan \left (\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}+1\right )+\frac {4}{5} (2 x+1)^{5/2}-12 \sqrt {2 x+1}-\frac {3 \sqrt [4]{3} \log \left (2 x-\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1\right )}{\sqrt {2}}+\frac {3 \sqrt [4]{3} \log \left (2 x+\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1\right )}{\sqrt {2}} \]

[In]

Int[(1 + 2*x)^(7/2)/(1 + x + x^2),x]

[Out]

-12*Sqrt[1 + 2*x] + (4*(1 + 2*x)^(5/2))/5 - 3*Sqrt[2]*3^(1/4)*ArcTan[1 - (Sqrt[2]*Sqrt[1 + 2*x])/3^(1/4)] + 3*
Sqrt[2]*3^(1/4)*ArcTan[1 + (Sqrt[2]*Sqrt[1 + 2*x])/3^(1/4)] - (3*3^(1/4)*Log[1 + Sqrt[3] + 2*x - Sqrt[2]*3^(1/
4)*Sqrt[1 + 2*x]])/Sqrt[2] + (3*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}& = \frac {4}{5} (1+2 x)^{5/2}-3 \int \frac {(1+2 x)^{3/2}}{1+x+x^2} \, dx \\ & = -12 \sqrt {1+2 x}+\frac {4}{5} (1+2 x)^{5/2}+9 \int \frac {1}{\sqrt {1+2 x} \left (1+x+x^2\right )} \, dx \\ & = -12 \sqrt {1+2 x}+\frac {4}{5} (1+2 x)^{5/2}+\frac {9}{2} \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (\frac {3}{4}+\frac {x^2}{4}\right )} \, dx,x,1+2 x\right ) \\ & = -12 \sqrt {1+2 x}+\frac {4}{5} (1+2 x)^{5/2}+9 \text {Subst}\left (\int \frac {1}{\frac {3}{4}+\frac {x^4}{4}} \, dx,x,\sqrt {1+2 x}\right ) \\ & = -12 \sqrt {1+2 x}+\frac {4}{5} (1+2 x)^{5/2}+\frac {1}{2} \left (3 \sqrt {3}\right ) \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} \left (3 \sqrt {3}\right ) \text {Subst}\left (\int \frac {\sqrt {3}+x^2}{\frac {3}{4}+\frac {x^4}{4}} \, dx,x,\sqrt {1+2 x}\right ) \\ & = -12 \sqrt {1+2 x}+\frac {4}{5} (1+2 x)^{5/2}-\frac {\left (3 \sqrt [4]{3}\right ) \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 {\left (3 \sqrt [4]{3}\right ) \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}}+\left (3 \sqrt {3}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {3}-\sqrt {2} \sqrt [4]{3} x+x^2} \, dx,x,\sqrt {1+2 x}\right )+\left (3 \sqrt {3}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {3}+\sqrt {2} \sqrt [4]{3} x+x^2} \, dx,x,\sqrt {1+2 x}\right ) \\ & = -12 \sqrt {1+2 x}+\frac {4}{5} (1+2 x)^{5/2}-\frac {3 \sqrt [4]{3} \log \left (1+\sqrt {3}+2 x-\sqrt {2} \sqrt [4]{3} \sqrt {1+2 x}\right )}{\sqrt {2}}+\frac {3 \sqrt [4]{3} \log \left (1+\sqrt {3}+2 x+\sqrt {2} \sqrt [4]{3} \sqrt {1+2 x}\right )}{\sqrt {2}}+\left (3 \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 (3 \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 ) \\ & = -12 \sqrt {1+2 x}+\frac {4}{5} (1+2 x)^{5/2}-3 \sqrt {2} \sqrt [4]{3} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {1+2 x}}{\sqrt [4]{3}}\right )+3 \sqrt {2} \sqrt [4]{3} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {1+2 x}}{\sqrt [4]{3}}\right )-\frac {3 \sqrt [4]{3} \log \left (1+\sqrt {3}+2 x-\sqrt {2} \sqrt [4]{3} \sqrt {1+2 x}\right )}{\sqrt {2}}+\frac {3 \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 = 0.23 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.61 \[ \int \frac {(1+2 x)^{7/2}}{1+x+x^2} \, dx=\frac {8}{5} \sqrt {1+2 x} \left (-7+2 x+2 x^2\right )+3 \sqrt {2} \sqrt [4]{3} \arctan \left (\frac {-3+\sqrt {3}+2 \sqrt {3} x}{3^{3/4} \sqrt {2+4 x}}\right )+3 \sqrt {2} \sqrt [4]{3} \text {arctanh}\left (\frac {3^{3/4} \sqrt {2+4 x}}{3+\sqrt {3}+2 \sqrt {3} x}\right ) \]

[In]

Integrate[(1 + 2*x)^(7/2)/(1 + x + x^2),x]

[Out]

(8*Sqrt[1 + 2*x]*(-7 + 2*x + 2*x^2))/5 + 3*Sqrt[2]*3^(1/4)*ArcTan[(-3 + Sqrt[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)]

Maple [A] (verified)

Time = 3.69 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.64

method	result	size
derivativedivides	\(\frac {4 \left (1+2 x \right )^{\frac {5}{2}}}{5}-12 \sqrt {1+2 x}+\frac {3 \,3^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {1+2 x +\sqrt {3}+3^{\frac {1}{4}} \sqrt {2}\, \sqrt {1+2 x}}{1+2 x +\sqrt {3}-3^{\frac {1}{4}} \sqrt {2}\, \sqrt {1+2 x}}\right )+2 \arctan \left (1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )+2 \arctan \left (-1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )\right )}{2}\)	\(118\)
default	\(\frac {4 \left (1+2 x \right )^{\frac {5}{2}}}{5}-12 \sqrt {1+2 x}+\frac {3 \,3^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {1+2 x +\sqrt {3}+3^{\frac {1}{4}} \sqrt {2}\, \sqrt {1+2 x}}{1+2 x +\sqrt {3}-3^{\frac {1}{4}} \sqrt {2}\, \sqrt {1+2 x}}\right )+2 \arctan \left (1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )+2 \arctan \left (-1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )\right )}{2}\)	\(118\)
risch	\(\frac {8 \left (2 x^{2}+2 x -7\right ) \sqrt {1+2 x}}{5}+\frac {3 \,3^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {1+2 x +\sqrt {3}+3^{\frac {1}{4}} \sqrt {2}\, \sqrt {1+2 x}}{1+2 x +\sqrt {3}-3^{\frac {1}{4}} \sqrt {2}\, \sqrt {1+2 x}}\right )+2 \arctan \left (1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )+2 \arctan \left (-1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )\right )}{2}\)	\(119\)
pseudoelliptic	\(\frac {8 \left (2 x^{2}+2 x -7\right ) \sqrt {1+2 x}}{5}+\frac {3 \,3^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {1+2 x +\sqrt {3}+3^{\frac {1}{4}} \sqrt {2}\, \sqrt {1+2 x}}{1+2 x +\sqrt {3}-3^{\frac {1}{4}} \sqrt {2}\, \sqrt {1+2 x}}\right )+2 \arctan \left (1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )+2 \arctan \left (-1+\frac {\sqrt {2}\, \sqrt {1+2 x}\, 3^{\frac {3}{4}}}{3}\right )\right )}{2}\)	\(119\)
trager	\(\left (\frac {16}{5} x^{2}+\frac {16}{5} x -\frac {56}{5}\right ) \sqrt {1+2 x}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{4} x +4 \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}\right ) x +2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}\right )+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}\right ) x +12 \sqrt {1+2 x}+6 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}\right )}{x \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}-x -2}\right )-3 \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{5} x -4 x \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{3}-2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{3}+3 x \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )+6 \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )-12 \sqrt {1+2 x}}{x \operatorname {RootOf}\left (\textit {\_Z}^{4}+3\right )^{2}+x +2}\right )\)	\(249\)

[In]

int((1+2*x)^(7/2)/(x^2+x+1),x,method=_RETURNVERBOSE)

[Out]

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^(1/2)+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 = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.51 \[ \int \frac {(1+2 x)^{7/2}}{1+x+x^2} \, dx=\frac {8}{5} \, {\left (2 \, x^{2} + 2 \, x - 7\right )} \sqrt {2 \, x + 1} + 3 \, \left (-3\right )^{\frac {1}{4}} \log \left (\sqrt {2 \, x + 1} + \left (-3\right )^{\frac {1}{4}}\right ) + 3 i \, \left (-3\right )^{\frac {1}{4}} \log \left (\sqrt {2 \, x + 1} + i \, \left (-3\right )^{\frac {1}{4}}\right ) - 3 i \, \left (-3\right )^{\frac {1}{4}} \log \left (\sqrt {2 \, x + 1} - i \, \left (-3\right )^{\frac {1}{4}}\right ) - 3 \, \left (-3\right )^{\frac {1}{4}} \log \left (\sqrt {2 \, x + 1} - \left (-3\right )^{\frac {1}{4}}\right ) \]

[In]

integrate((1+2*x)^(7/2)/(x^2+x+1),x, algorithm="fricas")

[Out]

8/5*(2*x^2 + 2*x - 7)*sqrt(2*x + 1) + 3*(-3)^(1/4)*log(sqrt(2*x + 1) + (-3)^(1/4)) + 3*I*(-3)^(1/4)*log(sqrt(2
*x + 1) + I*(-3)^(1/4)) - 3*I*(-3)^(1/4)*log(sqrt(2*x + 1) - I*(-3)^(1/4)) - 3*(-3)^(1/4)*log(sqrt(2*x + 1) -
(-3)^(1/4))

Sympy [F]

\[ \int \frac {(1+2 x)^{7/2}}{1+x+x^2} \, dx=\int \frac {\left (2 x + 1\right )^{\frac {7}{2}}}{x^{2} + x + 1}\, dx \]

[In]

integrate((1+2*x)**(7/2)/(x**2+x+1),x)

[Out]

Integral((2*x + 1)**(7/2)/(x**2 + x + 1), x)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.82 \[ \int \frac {(1+2 x)^{7/2}}{1+x+x^2} \, dx=\frac {4}{5} \, {\left (2 \, x + 1\right )}^{\frac {5}{2}} + 3 \cdot 3^{\frac {1}{4}} \sqrt {2} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} + 2 \, \sqrt {2 \, x + 1}\right )}\right ) + 3 \cdot 3^{\frac {1}{4}} \sqrt {2} \arctan \left (-\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} - 2 \, \sqrt {2 \, x + 1}\right )}\right ) + \frac {3}{2} \cdot 3^{\frac {1}{4}} \sqrt {2} \log \left (3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) - \frac {3}{2} \cdot 3^{\frac {1}{4}} \sqrt {2} \log \left (-3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) - 12 \, \sqrt {2 \, x + 1} \]

[In]

integrate((1+2*x)^(7/2)/(x^2+x+1),x, algorithm="maxima")

[Out]

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)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.75 \[ \int \frac {(1+2 x)^{7/2}}{1+x+x^2} \, dx=\frac {4}{5} \, {\left (2 \, x + 1\right )}^{\frac {5}{2}} + 3 \cdot 12^{\frac {1}{4}} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} + 2 \, \sqrt {2 \, x + 1}\right )}\right ) + 3 \cdot 12^{\frac {1}{4}} \arctan \left (-\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} - 2 \, \sqrt {2 \, x + 1}\right )}\right ) + \frac {3}{2} \cdot 12^{\frac {1}{4}} \log \left (3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) - \frac {3}{2} \cdot 12^{\frac {1}{4}} \log \left (-3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) - 12 \, \sqrt {2 \, x + 1} \]

[In]

integrate((1+2*x)^(7/2)/(x^2+x+1),x, algorithm="giac")

[Out]

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)*sqrt(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)

Mupad [B] (verification not implemented)

Time = 9.34 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.41 \[ \int \frac {(1+2 x)^{7/2}}{1+x+x^2} \, dx=\frac {4\,{\left (2\,x+1\right )}^{5/2}}{5}-12\,\sqrt {2\,x+1}+\sqrt {2}\,3^{1/4}\,\mathrm {atan}\left (\sqrt {2}\,3^{3/4}\,\sqrt {2\,x+1}\,\left (\frac {1}{6}-\frac {1}{6}{}\mathrm {i}\right )\right )\,\left (3+3{}\mathrm {i}\right )+\sqrt {2}\,3^{1/4}\,\mathrm {atan}\left (\sqrt {2}\,3^{3/4}\,\sqrt {2\,x+1}\,\left (\frac {1}{6}+\frac {1}{6}{}\mathrm {i}\right )\right )\,\left (3-3{}\mathrm {i}\right ) \]

[In]

int((2*x + 1)^(7/2)/(x + x^2 + 1),x)

[Out]

(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)