\(\int \frac {1+x}{(-2+x) \sqrt {1+x^3}} \, dx\) [377]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [B] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 18, antiderivative size = 31 \[ \int \frac {1+x}{(-2+x) \sqrt {1+x^3}} \, dx=-\frac {2}{3} \text {arctanh}\left (\frac {\frac {1}{3}+\frac {2 x}{3}+\frac {x^2}{3}}{\sqrt {1+x^3}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.74, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2163, 212} \[ \int \frac {1+x}{(-2+x) \sqrt {1+x^3}} \, dx=-\frac {2}{3} \text {arctanh}\left (\frac {(x+1)^2}{3 \sqrt {x^3+1}}\right ) \]

[In]

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

[Out]

(-2*ArcTanh[(1 + x)^2/(3*Sqrt[1 + x^3])])/3

Rule 212

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

Rule 2163

Int[((e_) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_Symbol] :> Dist[-2*(e/d), Subst[Int
[1/(9 - a*x^2), x], x, (1 + f*(x/e))^2/Sqrt[a + b*x^3]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[d*e - c*f,
0] && EqQ[b*c^3 + 8*a*d^3, 0] && EqQ[2*d*e + c*f, 0]

Rubi steps \begin{align*} \text {integral}& = -\left (2 \text {Subst}\left (\int \frac {1}{9-x^2} \, dx,x,\frac {(1+x)^2}{\sqrt {1+x^3}}\right )\right ) \\ & = -\frac {2}{3} \text {arctanh}\left (\frac {(1+x)^2}{3 \sqrt {1+x^3}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.94 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {1+x}{(-2+x) \sqrt {1+x^3}} \, dx=-\frac {2}{3} \text {arctanh}\left (\frac {\frac {1}{3}+\frac {2 x}{3}+\frac {x^2}{3}}{\sqrt {1+x^3}}\right ) \]

[In]

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

[Out]

(-2*ArcTanh[(1/3 + (2*x)/3 + x^2/3)/Sqrt[1 + x^3]])/3

Maple [A] (verified)

Time = 2.49 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35

method result size
trager \(-\frac {\ln \left (\frac {x^{3}+6 x \sqrt {x^{3}+1}+12 x^{2}+6 \sqrt {x^{3}+1}-6 x +10}{\left (x -2\right )^{3}}\right )}{3}\) \(42\)
default \(\frac {2 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}}-\frac {2 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \operatorname {EllipticPi}\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \frac {1}{2}-\frac {i \sqrt {3}}{6}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}}\) \(240\)
elliptic \(\frac {2 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}}-\frac {2 \left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \operatorname {EllipticPi}\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \frac {1}{2}-\frac {i \sqrt {3}}{6}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}}\) \(240\)

[In]

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

[Out]

-1/3*ln((x^3+6*x*(x^3+1)^(1/2)+12*x^2+6*(x^3+1)^(1/2)-6*x+10)/(x-2)^3)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (20) = 40\).

Time = 0.25 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \[ \int \frac {1+x}{(-2+x) \sqrt {1+x^3}} \, dx=\frac {1}{3} \, \log \left (\frac {x^{3} + 12 \, x^{2} - 6 \, \sqrt {x^{3} + 1} {\left (x + 1\right )} - 6 \, x + 10}{x^{3} - 6 \, x^{2} + 12 \, x - 8}\right ) \]

[In]

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

[Out]

1/3*log((x^3 + 12*x^2 - 6*sqrt(x^3 + 1)*(x + 1) - 6*x + 10)/(x^3 - 6*x^2 + 12*x - 8))

Sympy [F]

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

[In]

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

[Out]

Integral((x + 1)/(sqrt((x + 1)*(x**2 - x + 1))*(x - 2)), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.16 (sec) , antiderivative size = 204, normalized size of antiderivative = 6.58 \[ \int \frac {1+x}{(-2+x) \sqrt {1+x^3}} \, dx=\frac {\left (3+\sqrt {3}\,1{}\mathrm {i}\right )\,\sqrt {\frac {x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\left (\mathrm {F}\left (\mathrm {asin}\left (\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )-\Pi \left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6};\mathrm {asin}\left (\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )\right )\,\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {\frac {1}{2}-x+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}}{\sqrt {x^3+\left (-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-1\right )\,x-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \]

[In]

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

[Out]

((3^(1/2)*1i + 3)*((x + (3^(1/2)*1i)/2 - 1/2)/((3^(1/2)*1i)/2 - 3/2))^(1/2)*(ellipticF(asin(((x + 1)/((3^(1/2)
*1i)/2 + 3/2))^(1/2)), -((3^(1/2)*1i)/2 + 3/2)/((3^(1/2)*1i)/2 - 3/2)) - ellipticPi((3^(1/2)*1i)/6 + 1/2, asin
(((x + 1)/((3^(1/2)*1i)/2 + 3/2))^(1/2)), -((3^(1/2)*1i)/2 + 3/2)/((3^(1/2)*1i)/2 - 3/2)))*((x + 1)/((3^(1/2)*
1i)/2 + 3/2))^(1/2)*(((3^(1/2)*1i)/2 - x + 1/2)/((3^(1/2)*1i)/2 + 3/2))^(1/2))/(x^3 - x*(((3^(1/2)*1i)/2 - 1/2
)*((3^(1/2)*1i)/2 + 1/2) + 1) - ((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2))^(1/2)