3.1016 \(\int \frac{1}{x (3+3 x+x^2) \sqrt [3]{3+3 x+3 x^2+x^3}} \, dx\)

Optimal. Leaf size=90 \[ -\frac{\log \left (1-(x+1)^3\right )}{6 \sqrt [3]{3}}+\frac{\log \left (\sqrt [3]{3} (x+1)-\sqrt [3]{(x+1)^3+2}\right )}{2 \sqrt [3]{3}}-\frac{\tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{3} (x+1)}{\sqrt [3]{(x+1)^3+2}}+1}{\sqrt{3}}\right )}{3^{5/6}} \]

[Out]

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

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Rubi [A]  time = 0.114392, antiderivative size = 123, normalized size of antiderivative = 1.37, number of steps used = 9, number of rules used = 9, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.29, Rules used = {433, 431, 377, 200, 31, 634, 617, 204, 628} \[ \frac{\log \left (1-\frac{\sqrt [3]{3} (x+1)}{\sqrt [3]{(x+1)^3+2}}\right )}{3 \sqrt [3]{3}}-\frac{\log \left (\frac{3^{2/3} (x+1)^2}{\left ((x+1)^3+2\right )^{2/3}}+\frac{\sqrt [3]{3} (x+1)}{\sqrt [3]{(x+1)^3+2}}+1\right )}{6 \sqrt [3]{3}}-\frac{\tan ^{-1}\left (\frac{2 (x+1)}{\sqrt [6]{3} \sqrt [3]{(x+1)^3+2}}+\frac{1}{\sqrt{3}}\right )}{3^{5/6}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 433

Int[(u_)^(p_.)*(v_)^(q_.)*(x_)^(m_.), x_Symbol] :> Int[NormalizePseudoBinomial[x^(m/p)*u, x]^p*NormalizePseudo
Binomial[v, x]^q, x] /; FreeQ[{p, q}, x] && IntegersQ[p, m/p] && PseudoBinomialPairQ[x^(m/p)*u, v, x]

Rule 431

Int[((a_.) + (b_.)*(u_)^(n_))^(p_.)*((c_.) + (d_.)*(u_)^(n_))^(q_.), x_Symbol] :> Dist[1/Coefficient[u, x, 1],
Subst[Int[(a + b*x^n)^p*(c + d*x^n)^q, x], x, u], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && LinearQ[u, x] && N
eQ[u, x]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, x]

Rule 31

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

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

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])

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]

Rubi steps

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

Mathematica [A]  time = 0.14659, size = 120, normalized size = 1.33 \[ \frac{\sqrt{3} \left (2 \log \left (1-\frac{\sqrt [3]{3} (x+1)}{\sqrt [3]{(x+1)^3+2}}\right )-\log \left (\frac{3^{2/3} (x+1)^2}{\left ((x+1)^3+2\right )^{2/3}}+\frac{\sqrt [3]{3} (x+1)}{\sqrt [3]{(x+1)^3+2}}+1\right )\right )-6 \tan ^{-1}\left (\frac{2 (x+1)}{\sqrt [6]{3} \sqrt [3]{(x+1)^3+2}}+\frac{1}{\sqrt{3}}\right )}{6\ 3^{5/6}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.165, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{x \left ({x}^{2}+3\,x+3 \right ) }{\frac{1}{\sqrt [3]{{x}^{3}+3\,{x}^{2}+3\,x+3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 79.788, size = 1281, normalized size = 14.23 \begin{align*} -\frac{1}{54} \cdot 3^{\frac{2}{3}} \log \left (\frac{3 \cdot 3^{\frac{2}{3}}{\left (7 \, x^{4} + 28 \, x^{3} + 42 \, x^{2} + 30 \, x + 9\right )}{\left (x^{3} + 3 \, x^{2} + 3 \, x + 3\right )}^{\frac{2}{3}} + 3^{\frac{1}{3}}{\left (31 \, x^{6} + 186 \, x^{5} + 465 \, x^{4} + 666 \, x^{3} + 603 \, x^{2} + 324 \, x + 81\right )} + 9 \,{\left (5 \, x^{5} + 25 \, x^{4} + 50 \, x^{3} + 54 \, x^{2} + 33 \, x + 9\right )}{\left (x^{3} + 3 \, x^{2} + 3 \, x + 3\right )}^{\frac{1}{3}}}{x^{6} + 6 \, x^{5} + 15 \, x^{4} + 18 \, x^{3} + 9 \, x^{2}}\right ) + \frac{1}{27} \cdot 3^{\frac{2}{3}} \log \left (\frac{2 \cdot 3^{\frac{2}{3}}{\left (x^{3} + 3 \, x^{2} + 3 \, x\right )} - 9 \cdot 3^{\frac{1}{3}}{\left (x^{3} + 3 \, x^{2} + 3 \, x + 3\right )}^{\frac{1}{3}}{\left (x^{2} + 2 \, x + 1\right )} + 9 \,{\left (x^{3} + 3 \, x^{2} + 3 \, x + 3\right )}^{\frac{2}{3}}{\left (x + 1\right )}}{x^{3} + 3 \, x^{2} + 3 \, x}\right ) - \frac{1}{9} \cdot 3^{\frac{1}{6}} \arctan \left (\frac{3^{\frac{1}{6}}{\left (12 \cdot 3^{\frac{2}{3}}{\left (7 \, x^{7} + 49 \, x^{6} + 147 \, x^{5} + 240 \, x^{4} + 225 \, x^{3} + 117 \, x^{2} + 27 \, x\right )}{\left (x^{3} + 3 \, x^{2} + 3 \, x + 3\right )}^{\frac{2}{3}} - 3^{\frac{1}{3}}{\left (127 \, x^{9} + 1143 \, x^{8} + 4572 \, x^{7} + 11070 \, x^{6} + 18414 \, x^{5} + 22032 \, x^{4} + 18900 \, x^{3} + 11178 \, x^{2} + 4131 \, x + 729\right )} - 18 \,{\left (31 \, x^{8} + 248 \, x^{7} + 868 \, x^{6} + 1782 \, x^{5} + 2400 \, x^{4} + 2196 \, x^{3} + 1332 \, x^{2} + 486 \, x + 81\right )}{\left (x^{3} + 3 \, x^{2} + 3 \, x + 3\right )}^{\frac{1}{3}}\right )}}{3 \,{\left (251 \, x^{9} + 2259 \, x^{8} + 9036 \, x^{7} + 21546 \, x^{6} + 34398 \, x^{5} + 38556 \, x^{4} + 30348 \, x^{3} + 16038 \, x^{2} + 5103 \, x + 729\right )}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/54*3^(2/3)*log((3*3^(2/3)*(7*x^4 + 28*x^3 + 42*x^2 + 30*x + 9)*(x^3 + 3*x^2 + 3*x + 3)^(2/3) + 3^(1/3)*(31*
x^6 + 186*x^5 + 465*x^4 + 666*x^3 + 603*x^2 + 324*x + 81) + 9*(5*x^5 + 25*x^4 + 50*x^3 + 54*x^2 + 33*x + 9)*(x
^3 + 3*x^2 + 3*x + 3)^(1/3))/(x^6 + 6*x^5 + 15*x^4 + 18*x^3 + 9*x^2)) + 1/27*3^(2/3)*log((2*3^(2/3)*(x^3 + 3*x
^2 + 3*x) - 9*3^(1/3)*(x^3 + 3*x^2 + 3*x + 3)^(1/3)*(x^2 + 2*x + 1) + 9*(x^3 + 3*x^2 + 3*x + 3)^(2/3)*(x + 1))
/(x^3 + 3*x^2 + 3*x)) - 1/9*3^(1/6)*arctan(1/3*3^(1/6)*(12*3^(2/3)*(7*x^7 + 49*x^6 + 147*x^5 + 240*x^4 + 225*x
^3 + 117*x^2 + 27*x)*(x^3 + 3*x^2 + 3*x + 3)^(2/3) - 3^(1/3)*(127*x^9 + 1143*x^8 + 4572*x^7 + 11070*x^6 + 1841
4*x^5 + 22032*x^4 + 18900*x^3 + 11178*x^2 + 4131*x + 729) - 18*(31*x^8 + 248*x^7 + 868*x^6 + 1782*x^5 + 2400*x
^4 + 2196*x^3 + 1332*x^2 + 486*x + 81)*(x^3 + 3*x^2 + 3*x + 3)^(1/3))/(251*x^9 + 2259*x^8 + 9036*x^7 + 21546*x
^6 + 34398*x^5 + 38556*x^4 + 30348*x^3 + 16038*x^2 + 5103*x + 729))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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