3.20.71 \(\int \frac {\sqrt [3]{6+2 x+x^2}}{1+x} \, dx\)

Optimal. Leaf size=140 \[ \frac {3}{2} \sqrt [3]{x^2+2 x+6}+\frac {1}{2} \sqrt [3]{5} \log \left (5^{2/3} \sqrt [3]{x^2+2 x+6}-5\right )-\frac {1}{4} \sqrt [3]{5} \log \left (\sqrt [3]{5} \left (x^2+2 x+6\right )^{2/3}+5^{2/3} \sqrt [3]{x^2+2 x+6}+5\right )-\frac {1}{2} \sqrt {3} \sqrt [3]{5} \tan ^{-1}\left (\frac {2 \sqrt [3]{x^2+2 x+6}}{\sqrt {3} \sqrt [3]{5}}+\frac {1}{\sqrt {3}}\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 103, normalized size of antiderivative = 0.74, number of steps used = 7, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {694, 266, 50, 57, 617, 204, 31} \begin {gather*} \frac {3}{2} \sqrt [3]{(x+1)^2+5}-\frac {1}{2} \sqrt [3]{5} \log (x+1)+\frac {3}{4} \sqrt [3]{5} \log \left (\sqrt [3]{5}-\sqrt [3]{(x+1)^2+5}\right )-\frac {1}{2} \sqrt {3} \sqrt [3]{5} \tan ^{-1}\left (\frac {2 \sqrt [3]{(x+1)^2+5}+\sqrt [3]{5}}{\sqrt {3} \sqrt [3]{5}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(6 + 2*x + x^2)^(1/3)/(1 + x),x]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 57

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Dist[3/(2*b*q), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x
)^(1/3)], x] - Dist[3/(2*b*q^2), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x]
&& PosQ[(b*c - a*d)/b]

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 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]

Rubi steps

\begin {align*} \int \frac {\sqrt [3]{6+2 x+x^2}}{1+x} \, dx &=\operatorname {Subst}\left (\int \frac {\sqrt [3]{5+x^2}}{x} \, dx,x,1+x\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\sqrt [3]{5+x}}{x} \, dx,x,(1+x)^2\right )\\ &=\frac {3}{2} \sqrt [3]{5+(1+x)^2}+\frac {5}{2} \operatorname {Subst}\left (\int \frac {1}{x (5+x)^{2/3}} \, dx,x,(1+x)^2\right )\\ &=\frac {3}{2} \sqrt [3]{5+(1+x)^2}-\frac {1}{2} \sqrt [3]{5} \log (1+x)-\frac {1}{4} \left (3 \sqrt [3]{5}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{5}-x} \, dx,x,\sqrt [3]{5+(1+x)^2}\right )-\frac {1}{4} \left (3\ 5^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1}{5^{2/3}+\sqrt [3]{5} x+x^2} \, dx,x,\sqrt [3]{5+(1+x)^2}\right )\\ &=\frac {3}{2} \sqrt [3]{5+(1+x)^2}-\frac {1}{2} \sqrt [3]{5} \log (1+x)+\frac {3}{4} \sqrt [3]{5} \log \left (\sqrt [3]{5}-\sqrt [3]{5+(1+x)^2}\right )+\frac {1}{2} \left (3 \sqrt [3]{5}\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{5+(1+x)^2}}{\sqrt [3]{5}}\right )\\ &=\frac {3}{2} \sqrt [3]{5+(1+x)^2}-\frac {1}{2} \sqrt {3} \sqrt [3]{5} \tan ^{-1}\left (\frac {5+2\ 5^{2/3} \sqrt [3]{5+(1+x)^2}}{5 \sqrt {3}}\right )-\frac {1}{2} \sqrt [3]{5} \log (1+x)+\frac {3}{4} \sqrt [3]{5} \log \left (\sqrt [3]{5}-\sqrt [3]{5+(1+x)^2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 131, normalized size = 0.94 \begin {gather*} \frac {1}{4} \left (6 \sqrt [3]{x^2+2 x+6}+2 \sqrt [3]{5} \log \left (\sqrt [3]{5}-\sqrt [3]{(x+1)^2+5}\right )-\sqrt [3]{5} \log \left (\left ((x+1)^2+5\right )^{2/3}+\sqrt [3]{5} \sqrt [3]{(x+1)^2+5}+5^{2/3}\right )-2 \sqrt {3} \sqrt [3]{5} \tan ^{-1}\left (\frac {2 \sqrt [3]{(x+1)^2+5}+\sqrt [3]{5}}{\sqrt {3} \sqrt [3]{5}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6 + 2*x + x^2)^(1/3)/(1 + x),x]

[Out]

(6*(6 + 2*x + x^2)^(1/3) - 2*Sqrt[3]*5^(1/3)*ArcTan[(5^(1/3) + 2*(5 + (1 + x)^2)^(1/3))/(Sqrt[3]*5^(1/3))] + 2
*5^(1/3)*Log[5^(1/3) - (5 + (1 + x)^2)^(1/3)] - 5^(1/3)*Log[5^(2/3) + 5^(1/3)*(5 + (1 + x)^2)^(1/3) + (5 + (1
+ x)^2)^(2/3)])/4

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IntegrateAlgebraic [A]  time = 0.23, size = 140, normalized size = 1.00 \begin {gather*} \frac {3}{2} \sqrt [3]{6+2 x+x^2}-\frac {1}{2} \sqrt {3} \sqrt [3]{5} \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{6+2 x+x^2}}{\sqrt {3} \sqrt [3]{5}}\right )+\frac {1}{2} \sqrt [3]{5} \log \left (-5+5^{2/3} \sqrt [3]{6+2 x+x^2}\right )-\frac {1}{4} \sqrt [3]{5} \log \left (5+5^{2/3} \sqrt [3]{6+2 x+x^2}+\sqrt [3]{5} \left (6+2 x+x^2\right )^{2/3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(6 + 2*x + x^2)^(1/3)/(1 + x),x]

[Out]

(3*(6 + 2*x + x^2)^(1/3))/2 - (Sqrt[3]*5^(1/3)*ArcTan[1/Sqrt[3] + (2*(6 + 2*x + x^2)^(1/3))/(Sqrt[3]*5^(1/3))]
)/2 + (5^(1/3)*Log[-5 + 5^(2/3)*(6 + 2*x + x^2)^(1/3)])/2 - (5^(1/3)*Log[5 + 5^(2/3)*(6 + 2*x + x^2)^(1/3) + 5
^(1/3)*(6 + 2*x + x^2)^(2/3)])/4

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fricas [A]  time = 0.57, size = 102, normalized size = 0.73 \begin {gather*} -\frac {1}{2} \cdot 5^{\frac {1}{3}} \sqrt {3} \arctan \left (\frac {2}{15} \cdot 5^{\frac {2}{3}} \sqrt {3} {\left (x^{2} + 2 \, x + 6\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) - \frac {1}{4} \cdot 5^{\frac {1}{3}} \log \left (5^{\frac {2}{3}} + 5^{\frac {1}{3}} {\left (x^{2} + 2 \, x + 6\right )}^{\frac {1}{3}} + {\left (x^{2} + 2 \, x + 6\right )}^{\frac {2}{3}}\right ) + \frac {1}{2} \cdot 5^{\frac {1}{3}} \log \left (-5^{\frac {1}{3}} + {\left (x^{2} + 2 \, x + 6\right )}^{\frac {1}{3}}\right ) + \frac {3}{2} \, {\left (x^{2} + 2 \, x + 6\right )}^{\frac {1}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+2*x+6)^(1/3)/(1+x),x, algorithm="fricas")

[Out]

-1/2*5^(1/3)*sqrt(3)*arctan(2/15*5^(2/3)*sqrt(3)*(x^2 + 2*x + 6)^(1/3) + 1/3*sqrt(3)) - 1/4*5^(1/3)*log(5^(2/3
) + 5^(1/3)*(x^2 + 2*x + 6)^(1/3) + (x^2 + 2*x + 6)^(2/3)) + 1/2*5^(1/3)*log(-5^(1/3) + (x^2 + 2*x + 6)^(1/3))
+ 3/2*(x^2 + 2*x + 6)^(1/3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{2} + 2 \, x + 6\right )}^{\frac {1}{3}}}{x + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+2*x+6)^(1/3)/(1+x),x, algorithm="giac")

[Out]

integrate((x^2 + 2*x + 6)^(1/3)/(x + 1), x)

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maple [C]  time = 17.38, size = 1277, normalized size = 9.12

method	result	size
trager	\(\text {Expression too large to display}\)	\(1277\)
risch	\(\text {Expression too large to display}\)	\(2698\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+2*x+6)^(1/3)/(1+x),x,method=_RETURNVERBOSE)

[Out]

3/2*(x^2+2*x+6)^(1/3)+15/2*RootOf(RootOf(_Z^3-5)^2+15*_Z*RootOf(_Z^3-5)+225*_Z^2)*ln(-(3600*RootOf(RootOf(_Z^3
-5)^2+15*_Z*RootOf(_Z^3-5)+225*_Z^2)^2*RootOf(_Z^3-5)^3*x^2+255*RootOf(RootOf(_Z^3-5)^2+15*_Z*RootOf(_Z^3-5)+2
25*_Z^2)*RootOf(_Z^3-5)^4*x^2+7200*RootOf(RootOf(_Z^3-5)^2+15*_Z*RootOf(_Z^3-5)+225*_Z^2)^2*RootOf(_Z^3-5)^3*x
+510*RootOf(RootOf(_Z^3-5)^2+15*_Z*RootOf(_Z^3-5)+225*_Z^2)*RootOf(_Z^3-5)^4*x+3825*(x^2+2*x+6)^(2/3)*RootOf(R
ootOf(_Z^3-5)^2+15*_Z*RootOf(_Z^3-5)+225*_Z^2)*RootOf(_Z^3-5)^2+2640*RootOf(RootOf(_Z^3-5)^2+15*_Z*RootOf(_Z^3
-5)+225*_Z^2)*RootOf(_Z^3-5)*x^2+187*RootOf(_Z^3-5)^2*x^2+5280*RootOf(RootOf(_Z^3-5)^2+15*_Z*RootOf(_Z^3-5)+22
5*_Z^2)*RootOf(_Z^3-5)*x+374*RootOf(_Z^3-5)^2*x+19125*(x^2+2*x+6)^(1/3)*RootOf(RootOf(_Z^3-5)^2+15*_Z*RootOf(_
Z^3-5)+225*_Z^2)+21840*RootOf(RootOf(_Z^3-5)^2+15*_Z*RootOf(_Z^3-5)+225*_Z^2)*RootOf(_Z^3-5)+1440*(x^2+2*x+6)^
(2/3)+1440*(x^2+2*x+6)^(1/3)*RootOf(_Z^3-5)+1547*RootOf(_Z^3-5)^2)/(1+x)^2)-1/2*ln(-(3600*RootOf(RootOf(_Z^3-5
)^2+15*_Z*RootOf(_Z^3-5)+225*_Z^2)^2*RootOf(_Z^3-5)^3*x^2-15*RootOf(RootOf(_Z^3-5)^2+15*_Z*RootOf(_Z^3-5)+225*
_Z^2)*RootOf(_Z^3-5)^4*x^2+7200*RootOf(RootOf(_Z^3-5)^2+15*_Z*RootOf(_Z^3-5)+225*_Z^2)^2*RootOf(_Z^3-5)^3*x-30
*RootOf(RootOf(_Z^3-5)^2+15*_Z*RootOf(_Z^3-5)+225*_Z^2)*RootOf(_Z^3-5)^4*x-3825*(x^2+2*x+6)^(2/3)*RootOf(RootO
f(_Z^3-5)^2+15*_Z*RootOf(_Z^3-5)+225*_Z^2)*RootOf(_Z^3-5)^2-1440*RootOf(RootOf(_Z^3-5)^2+15*_Z*RootOf(_Z^3-5)+
225*_Z^2)*RootOf(_Z^3-5)*x^2+6*RootOf(_Z^3-5)^2*x^2-2880*RootOf(RootOf(_Z^3-5)^2+15*_Z*RootOf(_Z^3-5)+225*_Z^2
)*RootOf(_Z^3-5)*x+12*RootOf(_Z^3-5)^2*x-19125*(x^2+2*x+6)^(1/3)*RootOf(RootOf(_Z^3-5)^2+15*_Z*RootOf(_Z^3-5)+
225*_Z^2)-21840*RootOf(RootOf(_Z^3-5)^2+15*_Z*RootOf(_Z^3-5)+225*_Z^2)*RootOf(_Z^3-5)+165*(x^2+2*x+6)^(2/3)+16
5*(x^2+2*x+6)^(1/3)*RootOf(_Z^3-5)+91*RootOf(_Z^3-5)^2)/(1+x)^2)*RootOf(_Z^3-5)-15/2*ln(-(3600*RootOf(RootOf(_
Z^3-5)^2+15*_Z*RootOf(_Z^3-5)+225*_Z^2)^2*RootOf(_Z^3-5)^3*x^2-15*RootOf(RootOf(_Z^3-5)^2+15*_Z*RootOf(_Z^3-5)
+225*_Z^2)*RootOf(_Z^3-5)^4*x^2+7200*RootOf(RootOf(_Z^3-5)^2+15*_Z*RootOf(_Z^3-5)+225*_Z^2)^2*RootOf(_Z^3-5)^3
*x-30*RootOf(RootOf(_Z^3-5)^2+15*_Z*RootOf(_Z^3-5)+225*_Z^2)*RootOf(_Z^3-5)^4*x-3825*(x^2+2*x+6)^(2/3)*RootOf(
RootOf(_Z^3-5)^2+15*_Z*RootOf(_Z^3-5)+225*_Z^2)*RootOf(_Z^3-5)^2-1440*RootOf(RootOf(_Z^3-5)^2+15*_Z*RootOf(_Z^
3-5)+225*_Z^2)*RootOf(_Z^3-5)*x^2+6*RootOf(_Z^3-5)^2*x^2-2880*RootOf(RootOf(_Z^3-5)^2+15*_Z*RootOf(_Z^3-5)+225
*_Z^2)*RootOf(_Z^3-5)*x+12*RootOf(_Z^3-5)^2*x-19125*(x^2+2*x+6)^(1/3)*RootOf(RootOf(_Z^3-5)^2+15*_Z*RootOf(_Z^
3-5)+225*_Z^2)-21840*RootOf(RootOf(_Z^3-5)^2+15*_Z*RootOf(_Z^3-5)+225*_Z^2)*RootOf(_Z^3-5)+165*(x^2+2*x+6)^(2/
3)+165*(x^2+2*x+6)^(1/3)*RootOf(_Z^3-5)+91*RootOf(_Z^3-5)^2)/(1+x)^2)*RootOf(RootOf(_Z^3-5)^2+15*_Z*RootOf(_Z^
3-5)+225*_Z^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{2} + 2 \, x + 6\right )}^{\frac {1}{3}}}{x + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+2*x+6)^(1/3)/(1+x),x, algorithm="maxima")

[Out]

integrate((x^2 + 2*x + 6)^(1/3)/(x + 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (x^2+2\,x+6\right )}^{1/3}}{x+1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + x^2 + 6)^(1/3)/(x + 1),x)

[Out]

int((2*x + x^2 + 6)^(1/3)/(x + 1), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt [3]{x^{2} + 2 x + 6}}{x + 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2+2*x+6)**(1/3)/(1+x),x)

[Out]

Integral((x**2 + 2*x + 6)**(1/3)/(x + 1), x)

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