\(\int \frac {1+2 x^2+3 x^3}{\sqrt [3]{1+x^3}} \, dx\) [252]

Optimal result

Integrand size = 22, antiderivative size = 21 \[ \int \frac {1+2 x^2+3 x^3}{\sqrt [3]{1+x^3}} \, dx=\left (1+x^3\right )^{2/3}+x \left (1+x^3\right )^{2/3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1907, 245, 267, 327} \[ \int \frac {1+2 x^2+3 x^3}{\sqrt [3]{1+x^3}} \, dx=\left (x^3+1\right )^{2/3} x+\left (x^3+1\right )^{2/3} \]

[In]

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

[Out]

(1 + x^3)^(2/3) + x*(1 + x^3)^(2/3)

Rule 245

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + 2*Rt[b, 3]*(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sq
rt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]

Rule 267

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

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 1907

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n])

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{\sqrt [3]{1+x^3}}+\frac {2 x^2}{\sqrt [3]{1+x^3}}+\frac {3 x^3}{\sqrt [3]{1+x^3}}\right ) \, dx \\ & = 2 \int \frac {x^2}{\sqrt [3]{1+x^3}} \, dx+3 \int \frac {x^3}{\sqrt [3]{1+x^3}} \, dx+\int \frac {1}{\sqrt [3]{1+x^3}} \, dx \\ & = \left (1+x^3\right )^{2/3}+x \left (1+x^3\right )^{2/3}+\frac {\arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (-x+\sqrt [3]{1+x^3}\right )-\int \frac {1}{\sqrt [3]{1+x^3}} \, dx \\ & = \left (1+x^3\right )^{2/3}+x \left (1+x^3\right )^{2/3} \\ \end{align*}

Mathematica [A] (verified)

Time = 10.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \frac {1+2 x^2+3 x^3}{\sqrt [3]{1+x^3}} \, dx=(1+x) \left (1+x^3\right )^{2/3} \]

[In]

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

[Out]

(1 + x)*(1 + x^3)^(2/3)

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.57

method	result	size
trager	\(\left (1+x \right ) \left (x^{3}+1\right )^{\frac {2}{3}}\)	\(12\)
risch	\(\left (1+x \right ) \left (x^{3}+1\right )^{\frac {2}{3}}\)	\(12\)
gosper	\(\frac {\left (1+x \right )^{2} \left (x^{2}-x +1\right )}{\left (x^{3}+1\right )^{\frac {1}{3}}}\)	\(22\)
meijerg	\(x \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {1}{3}\right ], \left [\frac {4}{3}\right ], -x^{3}\right )+\frac {3 x^{4} \operatorname {hypergeom}\left (\left [\frac {1}{3}, \frac {4}{3}\right ], \left [\frac {7}{3}\right ], -x^{3}\right )}{4}+\frac {2 x^{3} \operatorname {hypergeom}\left (\left [\frac {1}{3}, 1\right ], \left [2\right ], -x^{3}\right )}{3}\)	\(47\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.52 \[ \int \frac {1+2 x^2+3 x^3}{\sqrt [3]{1+x^3}} \, dx={\left (x^{3} + 1\right )}^{\frac {2}{3}} {\left (x + 1\right )} \]

[In]

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

[Out]

(x^3 + 1)^(2/3)*(x + 1)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.44 (sec) , antiderivative size = 65, normalized size of antiderivative = 3.10 \[ \int \frac {1+2 x^2+3 x^3}{\sqrt [3]{1+x^3}} \, dx=\frac {x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{\Gamma \left (\frac {7}{3}\right )} + \frac {x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{3} \\ \frac {4}{3} \end {matrix}\middle | {x^{3} e^{i \pi }} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \left (x^{3} + 1\right )^{\frac {2}{3}} \]

[In]

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

[Out]

x**4*gamma(4/3)*hyper((1/3, 4/3), (7/3,), x**3*exp_polar(I*pi))/gamma(7/3) + x*gamma(1/3)*hyper((1/3, 1/3), (4
/3,), x**3*exp_polar(I*pi))/(3*gamma(4/3)) + (x**3 + 1)**(2/3)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.52 \[ \int \frac {1+2 x^2+3 x^3}{\sqrt [3]{1+x^3}} \, dx={\left (x^{3} + 1\right )}^{\frac {2}{3}} + \frac {{\left (x^{3} + 1\right )}^{\frac {2}{3}}}{x^{2} {\left (\frac {x^{3} + 1}{x^{3}} - 1\right )}} \]

[In]

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

[Out]

(x^3 + 1)^(2/3) + (x^3 + 1)^(2/3)/(x^2*((x^3 + 1)/x^3 - 1))

Giac [F]

\[ \int \frac {1+2 x^2+3 x^3}{\sqrt [3]{1+x^3}} \, dx=\int { \frac {3 \, x^{3} + 2 \, x^{2} + 1}{{\left (x^{3} + 1\right )}^{\frac {1}{3}}} \,d x } \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 14.71 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.52 \[ \int \frac {1+2 x^2+3 x^3}{\sqrt [3]{1+x^3}} \, dx={\left (x^3+1\right )}^{2/3}\,\left (x+1\right ) \]

[In]

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

[Out]

(x^3 + 1)^(2/3)*(x + 1)