3.1.95 \(\int \frac {1+x}{(1+x+x^2) \sqrt [3]{a+b x^3}} \, dx\) [95]

Optimal. Leaf size=154 \[ \frac {\tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{a+b}}+\frac {\tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a+b}}}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{a+b}}+\frac {\log \left (\sqrt [3]{a+b}-\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{a+b}}-\frac {\log \left (\sqrt [3]{a+b} x-\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{a+b}} \]

[Out]

1/2*ln((a+b)^(1/3)-(b*x^3+a)^(1/3))/(a+b)^(1/3)-1/2*ln((a+b)^(1/3)*x-(b*x^3+a)^(1/3))/(a+b)^(1/3)+1/3*arctan(1
/3*(1+2*(a+b)^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))/(a+b)^(1/3)*3^(1/2)+1/3*arctan(1/3*(1+2*(b*x^3+a)^(1/3)/(a+b)^
(1/3))*3^(1/2))/(a+b)^(1/3)*3^(1/2)

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Rubi [A]
time = 0.12, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2183, 384, 455, 57, 631, 210, 31} \begin {gather*} \frac {\log \left (\sqrt [3]{a+b}-\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{a+b}}-\frac {\log \left (x \sqrt [3]{a+b}-\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{a+b}}+\frac {\tan ^{-1}\left (\frac {\frac {2 x \sqrt [3]{a+b}}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{a+b}}+\frac {\tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{a+b x^3}}{\sqrt [3]{a+b}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{a+b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[(1 + (2*(a + b)^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*(a + b)^(1/3)) + ArcTan[(1 + (2*(a + b*x^
3)^(1/3))/(a + b)^(1/3))/Sqrt[3]]/(Sqrt[3]*(a + b)^(1/3)) + Log[(a + b)^(1/3) - (a + b*x^3)^(1/3)]/(2*(a + b)^
(1/3)) - Log[(a + b)^(1/3)*x - (a + b*x^3)^(1/3)]/(2*(a + b)^(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 57

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 210

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

Rule 384

Int[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/c, 3]}, Simp[
ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*c*q), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*c*q
), x] + Simp[Log[c + d*x^3]/(6*c*q), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 455

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

Rule 631

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 2183

Int[(Px_.)*((c_) + (d_.)*(x_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^3)^(p_.), x_Symbol] :> Dist[1/c^q, Int[E
xpandIntegrand[(c^3 - d^3*x^3)^q*(a + b*x^3)^p, Px/(c - d*x)^q, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] &&
PolyQ[Px, x] && EqQ[d^2 - c*e, 0] && ILtQ[q, 0] && RationalQ[p] && EqQ[Denominator[p], 3]

Rubi steps

\begin {align*} \int \frac {1+x}{\left (1+x+x^2\right ) \sqrt [3]{a+b x^3}} \, dx &=\int \left (\frac {1-\frac {i}{\sqrt {3}}}{\left (1-i \sqrt {3}+2 x\right ) \sqrt [3]{a+b x^3}}+\frac {1+\frac {i}{\sqrt {3}}}{\left (1+i \sqrt {3}+2 x\right ) \sqrt [3]{a+b x^3}}\right ) \, dx\\ &=\frac {1}{3} \left (3-i \sqrt {3}\right ) \int \frac {1}{\left (1-i \sqrt {3}+2 x\right ) \sqrt [3]{a+b x^3}} \, dx+\frac {1}{3} \left (3+i \sqrt {3}\right ) \int \frac {1}{\left (1+i \sqrt {3}+2 x\right ) \sqrt [3]{a+b x^3}} \, dx\\ \end {align*}

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Mathematica [F]
time = 10.13, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1+x}{\left (1+x+x^2\right ) \sqrt [3]{a+b x^3}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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Mathics [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {cought exception: maximum recursion depth exceeded while calling a Python object} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[(1 + x)/((1 + x + x^2)*(a + b*x^3)^(1/3)),x]')

[Out]

cought exception: maximum recursion depth exceeded while calling a Python object

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Maple [F]
time = 0.09, size = 0, normalized size = 0.00 \[\int \frac {1+x}{\left (x^{2}+x +1\right ) \left (b \,x^{3}+a \right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F] N/A
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Could not integrate

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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