3.1320 \(\int \frac{(1+2 x)^{7/2}}{1+x+x^2} \, dx\)

Optimal. Leaf size=183 \[ \frac{4}{5} (2 x+1)^{5/2}-12 \sqrt{2 x+1}-\frac{3 \sqrt [4]{3} \log \left (2 x-\sqrt{2} \sqrt [4]{3} \sqrt{2 x+1}+\sqrt{3}+1\right )}{\sqrt{2}}+\frac{3 \sqrt [4]{3} \log \left (2 x+\sqrt{2} \sqrt [4]{3} \sqrt{2 x+1}+\sqrt{3}+1\right )}{\sqrt{2}}-3 \sqrt{2} \sqrt [4]{3} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{2 x+1}}{\sqrt [4]{3}}\right )+3 \sqrt{2} \sqrt [4]{3} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{2 x+1}}{\sqrt [4]{3}}+1\right ) \]

[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]

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Rubi [A]  time = 0.17134, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {692, 694, 329, 211, 1165, 628, 1162, 617, 204} \[ \frac{4}{5} (2 x+1)^{5/2}-12 \sqrt{2 x+1}-\frac{3 \sqrt [4]{3} \log \left (2 x-\sqrt{2} \sqrt [4]{3} \sqrt{2 x+1}+\sqrt{3}+1\right )}{\sqrt{2}}+\frac{3 \sqrt [4]{3} \log \left (2 x+\sqrt{2} \sqrt [4]{3} \sqrt{2 x+1}+\sqrt{3}+1\right )}{\sqrt{2}}-3 \sqrt{2} \sqrt [4]{3} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{2 x+1}}{\sqrt [4]{3}}\right )+3 \sqrt{2} \sqrt [4]{3} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{2 x+1}}{\sqrt [4]{3}}+1\right ) \]

Antiderivative was successfully verified.

[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 692

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 694

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 329

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 211

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 1165

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]

Rule 628

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 1162

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 617

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 204

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

Rubi steps

\begin{align*} \int \frac{(1+2 x)^{7/2}}{1+x+x^2} \, dx &=\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} \operatorname{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 \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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]  time = 0.0929894, size = 182, normalized size = 0.99 \[ \frac{16}{5} \sqrt{2 x+1} x^2+\frac{16}{5} \sqrt{2 x+1} x-\frac{56}{5} \sqrt{2 x+1}-\frac{3 \sqrt [4]{3} \log \left (2 x-\sqrt [4]{3} \sqrt{4 x+2}+\sqrt{3}+1\right )}{\sqrt{2}}+\frac{3 \sqrt [4]{3} \log \left (2 x+\sqrt [4]{3} \sqrt{4 x+2}+\sqrt{3}+1\right )}{\sqrt{2}}-3 \sqrt{2} \sqrt [4]{3} \tan ^{-1}\left (1-\frac{\sqrt{4 x+2}}{\sqrt [4]{3}}\right )+3 \sqrt{2} \sqrt [4]{3} \tan ^{-1}\left (\frac{\sqrt{4 x+2}}{\sqrt [4]{3}}+1\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-56*Sqrt[1 + 2*x])/5 + (16*x*Sqrt[1 + 2*x])/5 + (16*x^2*Sqrt[1 + 2*x])/5 - 3*Sqrt[2]*3^(1/4)*ArcTan[1 - Sqrt[
2 + 4*x]/3^(1/4)] + 3*Sqrt[2]*3^(1/4)*ArcTan[1 + Sqrt[2 + 4*x]/3^(1/4)] - (3*3^(1/4)*Log[1 + Sqrt[3] + 2*x - 3
^(1/4)*Sqrt[2 + 4*x]])/Sqrt[2] + (3*3^(1/4)*Log[1 + Sqrt[3] + 2*x + 3^(1/4)*Sqrt[2 + 4*x]])/Sqrt[2]

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Maple [A]  time = 0.046, size = 129, normalized size = 0.7 \begin{align*}{\frac{4}{5} \left ( 1+2\,x \right ) ^{{\frac{5}{2}}}}-12\,\sqrt{1+2\,x}+3\,\sqrt [4]{3}\arctan \left ( 1+1/3\,\sqrt{2}\sqrt{1+2\,x}{3}^{3/4} \right ) \sqrt{2}+3\,\sqrt [4]{3}\arctan \left ( -1+1/3\,\sqrt{2}\sqrt{1+2\,x}{3}^{3/4} \right ) \sqrt{2}+{\frac{3\,\sqrt [4]{3}\sqrt{2}}{2}\ln \left ({ \left ( 1+2\,x+\sqrt{3}+\sqrt [4]{3}\sqrt{2}\sqrt{1+2\,x} \right ) \left ( 1+2\,x+\sqrt{3}-\sqrt [4]{3}\sqrt{2}\sqrt{1+2\,x} \right ) ^{-1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/5*(1+2*x)^(5/2)-12*(1+2*x)^(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)*arc
tan(-1+1/3*2^(1/2)*(1+2*x)^(1/2)*3^(3/4))*2^(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)))

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Maxima [A]  time = 1.80035, size = 203, normalized size = 1.11 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.70997, size = 651, normalized size = 3.56 \begin{align*} -6 \cdot 3^{\frac{1}{4}} \sqrt{2} \arctan \left (\frac{1}{3} \cdot 3^{\frac{3}{4}} \sqrt{2} \sqrt{3^{\frac{1}{4}} \sqrt{2} \sqrt{2 \, x + 1} + 2 \, x + \sqrt{3} + 1} - \frac{1}{3} \cdot 3^{\frac{3}{4}} \sqrt{2} \sqrt{2 \, x + 1} - 1\right ) - 6 \cdot 3^{\frac{1}{4}} \sqrt{2} \arctan \left (\frac{1}{3} \cdot 3^{\frac{3}{4}} \sqrt{2} \sqrt{-3^{\frac{1}{4}} \sqrt{2} \sqrt{2 \, x + 1} + 2 \, x + \sqrt{3} + 1} - \frac{1}{3} \cdot 3^{\frac{3}{4}} \sqrt{2} \sqrt{2 \, x + 1} + 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 ) - \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{8}{5} \,{\left (2 \, x^{2} + 2 \, x - 7\right )} \sqrt{2 \, x + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6*3^(1/4)*sqrt(2)*arctan(1/3*3^(3/4)*sqrt(2)*sqrt(3^(1/4)*sqrt(2)*sqrt(2*x + 1) + 2*x + sqrt(3) + 1) - 1/3*3^
(3/4)*sqrt(2)*sqrt(2*x + 1) - 1) - 6*3^(1/4)*sqrt(2)*arctan(1/3*3^(3/4)*sqrt(2)*sqrt(-3^(1/4)*sqrt(2)*sqrt(2*x
+ 1) + 2*x + sqrt(3) + 1) - 1/3*3^(3/4)*sqrt(2)*sqrt(2*x + 1) + 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) + 8/5*(2*x^2 + 2*x - 7)*sqrt(2*x + 1)

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Sympy [A]  time = 54.7249, size = 180, normalized size = 0.98 \begin{align*} \frac{4 \left (2 x + 1\right )^{\frac{5}{2}}}{5} - 12 \sqrt{2 x + 1} - \frac{3 \sqrt{2} \sqrt [4]{3} \log{\left (2 x - \sqrt{2} \sqrt [4]{3} \sqrt{2 x + 1} + 1 + \sqrt{3} \right )}}{2} + \frac{3 \sqrt{2} \sqrt [4]{3} \log{\left (2 x + \sqrt{2} \sqrt [4]{3} \sqrt{2 x + 1} + 1 + \sqrt{3} \right )}}{2} + 3 \sqrt{2} \sqrt [4]{3} \operatorname{atan}{\left (\frac{\sqrt{2} \cdot 3^{\frac{3}{4}} \sqrt{2 x + 1}}{3} - 1 \right )} + 3 \sqrt{2} \sqrt [4]{3} \operatorname{atan}{\left (\frac{\sqrt{2} \cdot 3^{\frac{3}{4}} \sqrt{2 x + 1}}{3} + 1 \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*(2*x + 1)**(5/2)/5 - 12*sqrt(2*x + 1) - 3*sqrt(2)*3**(1/4)*log(2*x - sqrt(2)*3**(1/4)*sqrt(2*x + 1) + 1 + sq
rt(3))/2 + 3*sqrt(2)*3**(1/4)*log(2*x + sqrt(2)*3**(1/4)*sqrt(2*x + 1) + 1 + sqrt(3))/2 + 3*sqrt(2)*3**(1/4)*a
tan(sqrt(2)*3**(3/4)*sqrt(2*x + 1)/3 - 1) + 3*sqrt(2)*3**(1/4)*atan(sqrt(2)*3**(3/4)*sqrt(2*x + 1)/3 + 1)

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Giac [A]  time = 1.18237, size = 186, normalized size = 1.02 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)