3.1325 \(\int \frac {1}{(1+2 x)^{3/2} (1+x+x^2)} \, dx\)

Optimal. Leaf size=180 \[ -\frac {4}{3 \sqrt {2 x+1}}-\frac {\log \left (2 x-\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1\right )}{3 \sqrt {2} \sqrt [4]{3}}+\frac {\log \left (2 x+\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1\right )}{3 \sqrt {2} \sqrt [4]{3}}+\frac {\sqrt {2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}\right )}{3 \sqrt [4]{3}}-\frac {\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}+1\right )}{3 \sqrt [4]{3}} \]

[Out]

-1/18*3^(3/4)*ln(1+2*x+3^(1/2)-3^(1/4)*2^(1/2)*(1+2*x)^(1/2))*2^(1/2)+1/18*3^(3/4)*ln(1+2*x+3^(1/2)+3^(1/4)*2^
(1/2)*(1+2*x)^(1/2))*2^(1/2)-1/9*3^(3/4)*arctan(-1+1/3*2^(1/2)*(1+2*x)^(1/2)*3^(3/4))*2^(1/2)-1/9*3^(3/4)*arct
an(1+1/3*2^(1/2)*(1+2*x)^(1/2)*3^(3/4))*2^(1/2)-4/3/(1+2*x)^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {693, 694, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac {4}{3 \sqrt {2 x+1}}-\frac {\log \left (2 x-\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1\right )}{3 \sqrt {2} \sqrt [4]{3}}+\frac {\log \left (2 x+\sqrt {2} \sqrt [4]{3} \sqrt {2 x+1}+\sqrt {3}+1\right )}{3 \sqrt {2} \sqrt [4]{3}}+\frac {\sqrt {2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}\right )}{3 \sqrt [4]{3}}-\frac {\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {2 x+1}}{\sqrt [4]{3}}+1\right )}{3 \sqrt [4]{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

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 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 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 693

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-2*b*d*(d + e*x)^(m
+ 1)*(a + b*x + c*x^2)^(p + 1))/(d^2*(m + 1)*(b^2 - 4*a*c)), x] + Dist[(b^2*(m + 2*p + 3))/(d^2*(m + 1)*(b^2
- 4*a*c)), Int[(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] && LtQ[m, -1] && (IntegerQ[2*p] || (IntegerQ[m] && Rationa
lQ[p]) || IntegerQ[(m + 2*p + 3)/2])

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 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 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]

Rubi steps

\begin {align*} \int \frac {1}{(1+2 x)^{3/2} \left (1+x+x^2\right )} \, dx &=-\frac {4}{3 \sqrt {1+2 x}}-\frac {1}{3} \int \frac {\sqrt {1+2 x}}{1+x+x^2} \, dx\\ &=-\frac {4}{3 \sqrt {1+2 x}}-\frac {1}{6} \operatorname {Subst}\left (\int \frac {\sqrt {x}}{\frac {3}{4}+\frac {x^2}{4}} \, dx,x,1+2 x\right )\\ &=-\frac {4}{3 \sqrt {1+2 x}}-\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^2}{\frac {3}{4}+\frac {x^4}{4}} \, dx,x,\sqrt {1+2 x}\right )\\ &=-\frac {4}{3 \sqrt {1+2 x}}+\frac {1}{6} \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}{6} \operatorname {Subst}\left (\int \frac {\sqrt {3}+x^2}{\frac {3}{4}+\frac {x^4}{4}} \, dx,x,\sqrt {1+2 x}\right )\\ &=-\frac {4}{3 \sqrt {1+2 x}}-\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{\sqrt {3}-\sqrt {2} \sqrt [4]{3} x+x^2} \, dx,x,\sqrt {1+2 x}\right )-\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{\sqrt {3}+\sqrt {2} \sqrt [4]{3} x+x^2} \, dx,x,\sqrt {1+2 x}\right )-\frac {\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 )}{3 \sqrt {2} \sqrt [4]{3}}-\frac {\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 )}{3 \sqrt {2} \sqrt [4]{3}}\\ &=-\frac {4}{3 \sqrt {1+2 x}}-\frac {\log \left (1+\sqrt {3}+2 x-\sqrt {2} \sqrt [4]{3} \sqrt {1+2 x}\right )}{3 \sqrt {2} \sqrt [4]{3}}+\frac {\log \left (1+\sqrt {3}+2 x+\sqrt {2} \sqrt [4]{3} \sqrt {1+2 x}\right )}{3 \sqrt {2} \sqrt [4]{3}}-\frac {\sqrt {2} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2+4 x}}{\sqrt [4]{3}}\right )}{3 \sqrt [4]{3}}+\frac {\sqrt {2} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2+4 x}}{\sqrt [4]{3}}\right )}{3 \sqrt [4]{3}}\\ &=-\frac {4}{3 \sqrt {1+2 x}}+\frac {\sqrt {2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {1+2 x}}{\sqrt [4]{3}}\right )}{3 \sqrt [4]{3}}-\frac {\sqrt {2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {1+2 x}}{\sqrt [4]{3}}\right )}{3 \sqrt [4]{3}}-\frac {\log \left (1+\sqrt {3}+2 x-\sqrt {2} \sqrt [4]{3} \sqrt {1+2 x}\right )}{3 \sqrt {2} \sqrt [4]{3}}+\frac {\log \left (1+\sqrt {3}+2 x+\sqrt {2} \sqrt [4]{3} \sqrt {1+2 x}\right )}{3 \sqrt {2} \sqrt [4]{3}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 32, normalized size = 0.18 \[ -\frac {4 \, _2F_1\left (-\frac {1}{4},1;\frac {3}{4};-\frac {1}{3} (2 x+1)^2\right )}{3 \sqrt {2 x+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Hypergeometric2F1[-1/4, 1, 3/4, -1/3*(1 + 2*x)^2])/(3*Sqrt[1 + 2*x])

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fricas [A]  time = 0.76, size = 225, normalized size = 1.25 \[ \frac {4 \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (2 \, x + 1\right )} \arctan \left (\frac {1}{3} \cdot 3^{\frac {3}{4}} \sqrt {2} \sqrt {3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1} - \frac {1}{3} \cdot 3^{\frac {3}{4}} \sqrt {2} \sqrt {2 \, x + 1} - 1\right ) + 4 \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (2 \, x + 1\right )} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} \sqrt {-4 \cdot 3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 8 \, x + 4 \, \sqrt {3} + 4} - \frac {1}{3} \cdot 3^{\frac {3}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 1\right ) + 3^{\frac {3}{4}} \sqrt {2} {\left (2 \, x + 1\right )} \log \left (4 \cdot 3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 8 \, x + 4 \, \sqrt {3} + 4\right ) - 3^{\frac {3}{4}} \sqrt {2} {\left (2 \, x + 1\right )} \log \left (-4 \cdot 3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 8 \, x + 4 \, \sqrt {3} + 4\right ) - 24 \, \sqrt {2 \, x + 1}}{18 \, {\left (2 \, x + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*(4*3^(3/4)*sqrt(2)*(2*x + 1)*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) + 4*3^(3/4)*sqrt(2)*(2*x + 1)*arctan(1/6*3^(3/4)*sqrt(2)*sqrt(
-4*3^(1/4)*sqrt(2)*sqrt(2*x + 1) + 8*x + 4*sqrt(3) + 4) - 1/3*3^(3/4)*sqrt(2)*sqrt(2*x + 1) + 1) + 3^(3/4)*sqr
t(2)*(2*x + 1)*log(4*3^(1/4)*sqrt(2)*sqrt(2*x + 1) + 8*x + 4*sqrt(3) + 4) - 3^(3/4)*sqrt(2)*(2*x + 1)*log(-4*3
^(1/4)*sqrt(2)*sqrt(2*x + 1) + 8*x + 4*sqrt(3) + 4) - 24*sqrt(2*x + 1))/(2*x + 1)

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giac [A]  time = 0.20, size = 129, normalized size = 0.72 \[ -\frac {1}{9} \cdot 108^{\frac {1}{4}} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} + 2 \, \sqrt {2 \, x + 1}\right )}\right ) - \frac {1}{9} \cdot 108^{\frac {1}{4}} \arctan \left (-\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} - 2 \, \sqrt {2 \, x + 1}\right )}\right ) + \frac {1}{18} \cdot 108^{\frac {1}{4}} \log \left (3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) - \frac {1}{18} \cdot 108^{\frac {1}{4}} \log \left (-3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) - \frac {4}{3 \, \sqrt {2 \, x + 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*108^(1/4)*arctan(1/6*3^(3/4)*sqrt(2)*(3^(1/4)*sqrt(2) + 2*sqrt(2*x + 1))) - 1/9*108^(1/4)*arctan(-1/6*3^(
3/4)*sqrt(2)*(3^(1/4)*sqrt(2) - 2*sqrt(2*x + 1))) + 1/18*108^(1/4)*log(3^(1/4)*sqrt(2)*sqrt(2*x + 1) + 2*x + s
qrt(3) + 1) - 1/18*108^(1/4)*log(-3^(1/4)*sqrt(2)*sqrt(2*x + 1) + 2*x + sqrt(3) + 1) - 4/3/sqrt(2*x + 1)

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maple [A]  time = 0.05, size = 120, normalized size = 0.67 \[ -\frac {3^{\frac {3}{4}} \sqrt {2}\, \arctan \left (-1+\frac {\sqrt {2}\, \sqrt {2 x +1}\, 3^{\frac {3}{4}}}{3}\right )}{9}-\frac {3^{\frac {3}{4}} \sqrt {2}\, \arctan \left (1+\frac {\sqrt {2}\, \sqrt {2 x +1}\, 3^{\frac {3}{4}}}{3}\right )}{9}-\frac {\sqrt {2}\, 3^{\frac {3}{4}} \ln \left (\frac {2 x +1+\sqrt {3}-3^{\frac {1}{4}} \sqrt {2}\, \sqrt {2 x +1}}{2 x +1+\sqrt {3}+3^{\frac {1}{4}} \sqrt {2}\, \sqrt {2 x +1}}\right )}{18}-\frac {4}{3 \sqrt {2 x +1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/3/(2*x+1)^(1/2)-1/9*3^(3/4)*2^(1/2)*arctan(1+1/3*2^(1/2)*(2*x+1)^(1/2)*3^(3/4))-1/9*3^(3/4)*2^(1/2)*arctan(
-1+1/3*2^(1/2)*(2*x+1)^(1/2)*3^(3/4))-1/18*2^(1/2)*3^(3/4)*ln((2*x+1+3^(1/2)-3^(1/4)*2^(1/2)*(2*x+1)^(1/2))/(2
*x+1+3^(1/2)+3^(1/4)*2^(1/2)*(2*x+1)^(1/2)))

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maxima [A]  time = 3.00, size = 141, normalized size = 0.78 \[ -\frac {1}{9} \cdot 3^{\frac {3}{4}} \sqrt {2} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} + 2 \, \sqrt {2 \, x + 1}\right )}\right ) - \frac {1}{9} \cdot 3^{\frac {3}{4}} \sqrt {2} \arctan \left (-\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (3^{\frac {1}{4}} \sqrt {2} - 2 \, \sqrt {2 \, x + 1}\right )}\right ) + \frac {1}{18} \cdot 3^{\frac {3}{4}} \sqrt {2} \log \left (3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) - \frac {1}{18} \cdot 3^{\frac {3}{4}} \sqrt {2} \log \left (-3^{\frac {1}{4}} \sqrt {2} \sqrt {2 \, x + 1} + 2 \, x + \sqrt {3} + 1\right ) - \frac {4}{3 \, \sqrt {2 \, x + 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*3^(3/4)*sqrt(2)*arctan(1/6*3^(3/4)*sqrt(2)*(3^(1/4)*sqrt(2) + 2*sqrt(2*x + 1))) - 1/9*3^(3/4)*sqrt(2)*arc
tan(-1/6*3^(3/4)*sqrt(2)*(3^(1/4)*sqrt(2) - 2*sqrt(2*x + 1))) + 1/18*3^(3/4)*sqrt(2)*log(3^(1/4)*sqrt(2)*sqrt(
2*x + 1) + 2*x + sqrt(3) + 1) - 1/18*3^(3/4)*sqrt(2)*log(-3^(1/4)*sqrt(2)*sqrt(2*x + 1) + 2*x + sqrt(3) + 1) -
4/3/sqrt(2*x + 1)

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mupad [B]  time = 0.53, size = 66, normalized size = 0.37 \[ -\frac {4}{3\,\sqrt {2\,x+1}}+\sqrt {2}\,3^{3/4}\,\mathrm {atan}\left (\sqrt {2}\,3^{3/4}\,\sqrt {2\,x+1}\,\left (\frac {1}{6}-\frac {1}{6}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{9}+\frac {1}{9}{}\mathrm {i}\right )+\sqrt {2}\,3^{3/4}\,\mathrm {atan}\left (\sqrt {2}\,3^{3/4}\,\sqrt {2\,x+1}\,\left (\frac {1}{6}+\frac {1}{6}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{9}-\frac {1}{9}{}\mathrm {i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- 4/(3*(2*x + 1)^(1/2)) - 2^(1/2)*3^(3/4)*atan(2^(1/2)*3^(3/4)*(2*x + 1)^(1/2)*(1/6 - 1i/6))*(1/9 - 1i/9) - 2^
(1/2)*3^(3/4)*atan(2^(1/2)*3^(3/4)*(2*x + 1)^(1/2)*(1/6 + 1i/6))*(1/9 + 1i/9)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (2 x + 1\right )^{\frac {3}{2}} \left (x^{2} + x + 1\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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