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

Optimal. Leaf size=123 \[ \frac {\log \left (2^{2/3} \sqrt [3]{x^2+2 x+3}-2\right )}{2 \sqrt [3]{2}}-\frac {\log \left (\sqrt [3]{2} \left (x^2+2 x+3\right )^{2/3}+2^{2/3} \sqrt [3]{x^2+2 x+3}+2\right )}{4 \sqrt [3]{2}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {2^{2/3} \sqrt [3]{x^2+2 x+3}}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right )}{2 \sqrt [3]{2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 83, normalized size of antiderivative = 0.67, number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {694, 266, 55, 617, 204, 31} \begin {gather*} -\frac {\log (x+1)}{2 \sqrt [3]{2}}+\frac {3 \log \left (\sqrt [3]{2}-\sqrt [3]{(x+1)^2+2}\right )}{4 \sqrt [3]{2}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {2^{2/3} \sqrt [3]{(x+1)^2+2}+1}{\sqrt {3}}\right )}{2 \sqrt [3]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

Rule 55

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), 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 {1}{(1+x) \sqrt [3]{3+2 x+x^2}} \, dx &=\operatorname {Subst}\left (\int \frac {1}{x \sqrt [3]{2+x^2}} \, dx,x,1+x\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x \sqrt [3]{2+x}} \, dx,x,(1+x)^2\right )\\ &=-\frac {\log (1+x)}{2 \sqrt [3]{2}}+\frac {3}{4} \operatorname {Subst}\left (\int \frac {1}{2^{2/3}+\sqrt [3]{2} x+x^2} \, dx,x,\sqrt [3]{2+(1+x)^2}\right )-\frac {3 \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{2}-x} \, dx,x,\sqrt [3]{2+(1+x)^2}\right )}{4 \sqrt [3]{2}}\\ &=-\frac {\log (1+x)}{2 \sqrt [3]{2}}+\frac {3 \log \left (\sqrt [3]{2}-\sqrt [3]{2+(1+x)^2}\right )}{4 \sqrt [3]{2}}-\frac {3 \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2^{2/3} \sqrt [3]{2+(1+x)^2}\right )}{2 \sqrt [3]{2}}\\ &=\frac {\sqrt {3} \tan ^{-1}\left (\frac {1+2^{2/3} \sqrt [3]{2+(1+x)^2}}{\sqrt {3}}\right )}{2 \sqrt [3]{2}}-\frac {\log (1+x)}{2 \sqrt [3]{2}}+\frac {3 \log \left (\sqrt [3]{2}-\sqrt [3]{2+(1+x)^2}\right )}{4 \sqrt [3]{2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 72, normalized size = 0.59 \begin {gather*} \frac {2 \sqrt {3} \tan ^{-1}\left (\frac {2^{2/3} \sqrt [3]{x^2+2 x+3}+1}{\sqrt {3}}\right )-2 \log (x+1)+3 \log \left (\sqrt [3]{2}-\sqrt [3]{(x+1)^2+2}\right )}{4 \sqrt [3]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.24, size = 123, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2^{2/3} \sqrt [3]{3+2 x+x^2}}{\sqrt {3}}\right )}{2 \sqrt [3]{2}}+\frac {\log \left (-2+2^{2/3} \sqrt [3]{3+2 x+x^2}\right )}{2 \sqrt [3]{2}}-\frac {\log \left (2+2^{2/3} \sqrt [3]{3+2 x+x^2}+\sqrt [3]{2} \left (3+2 x+x^2\right )^{2/3}\right )}{4 \sqrt [3]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.56, size = 93, normalized size = 0.76 \begin {gather*} \frac {1}{4} \, \sqrt {3} 2^{\frac {2}{3}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {1}{6}} {\left (2^{\frac {5}{6}} + 2 \, \sqrt {2} {\left (x^{2} + 2 \, x + 3\right )}^{\frac {1}{3}}\right )}\right ) - \frac {1}{8} \cdot 2^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (x^{2} + 2 \, x + 3\right )}^{\frac {1}{3}} + {\left (x^{2} + 2 \, x + 3\right )}^{\frac {2}{3}}\right ) + \frac {1}{4} \cdot 2^{\frac {2}{3}} \log \left (-2^{\frac {1}{3}} + {\left (x^{2} + 2 \, x + 3\right )}^{\frac {1}{3}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 8.28, size = 1196, normalized size = 9.72

method	result	size
trager	\(\text {Expression too large to display}\)	\(1196\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*RootOf(_Z^3-4)*ln((7*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*RootOf(_Z^3-4)^3*x^2-2*RootOf(Roo
tOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)^2*RootOf(_Z^3-4)^2*x^2+14*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4
)+4*_Z^2)*RootOf(_Z^3-4)^3*x-4*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)^2*RootOf(_Z^3-4)^2*x-48*(x^
2+2*x+3)^(2/3)*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*RootOf(_Z^3-4)^2+15*(x^2+2*x+3)^(1/3)*RootO
f(_Z^3-4)^2+126*(x^2+2*x+3)^(1/3)*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*RootOf(_Z^3-4)+35*RootOf
(_Z^3-4)*x^2-10*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*x^2+70*RootOf(_Z^3-4)*x-20*RootOf(RootOf(_
Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*x-126*(x^2+2*x+3)^(2/3)+133*RootOf(_Z^3-4)-38*RootOf(RootOf(_Z^3-4)^2+2*_
Z*RootOf(_Z^3-4)+4*_Z^2))/(1+x)^2)-1/4*ln((7*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*RootOf(_Z^3-4
)^3*x^2+16*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)^2*RootOf(_Z^3-4)^2*x^2+14*RootOf(RootOf(_Z^3-4)
^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*RootOf(_Z^3-4)^3*x+32*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)^2*Roo
tOf(_Z^3-4)^2*x+48*(x^2+2*x+3)^(2/3)*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*RootOf(_Z^3-4)^2+63*(
x^2+2*x+3)^(1/3)*RootOf(_Z^3-4)^2+30*(x^2+2*x+3)^(1/3)*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*Roo
tOf(_Z^3-4)-21*RootOf(_Z^3-4)*x^2-48*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*x^2-42*RootOf(_Z^3-4)
*x-96*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*x-30*(x^2+2*x+3)^(2/3)-133*RootOf(_Z^3-4)-304*RootOf
(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2))/(1+x)^2)*RootOf(_Z^3-4)-1/2*ln((7*RootOf(RootOf(_Z^3-4)^2+2*_Z*
RootOf(_Z^3-4)+4*_Z^2)*RootOf(_Z^3-4)^3*x^2+16*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)^2*RootOf(_Z
^3-4)^2*x^2+14*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*RootOf(_Z^3-4)^3*x+32*RootOf(RootOf(_Z^3-4)
^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)^2*RootOf(_Z^3-4)^2*x+48*(x^2+2*x+3)^(2/3)*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_
Z^3-4)+4*_Z^2)*RootOf(_Z^3-4)^2+63*(x^2+2*x+3)^(1/3)*RootOf(_Z^3-4)^2+30*(x^2+2*x+3)^(1/3)*RootOf(RootOf(_Z^3-
4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*RootOf(_Z^3-4)-21*RootOf(_Z^3-4)*x^2-48*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_
Z^3-4)+4*_Z^2)*x^2-42*RootOf(_Z^3-4)*x-96*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)*x-30*(x^2+2*x+3)
^(2/3)-133*RootOf(_Z^3-4)-304*RootOf(RootOf(_Z^3-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2))/(1+x)^2)*RootOf(RootOf(_Z^3
-4)^2+2*_Z*RootOf(_Z^3-4)+4*_Z^2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________