\(\int \frac {2^{2/3} \sqrt [3]{a}+2 \sqrt [3]{b} x}{(2^{2/3} \sqrt [3]{a}-\sqrt [3]{b} x) \sqrt {a-b x^3}} \, dx\) [48]

Optimal result

Integrand size = 55, antiderivative size = 65 \[ \int \frac {2^{2/3} \sqrt [3]{a}+2 \sqrt [3]{b} x}{\left (2^{2/3} \sqrt [3]{a}-\sqrt [3]{b} x\right ) \sqrt {a-b x^3}} \, dx=-\frac {2\ 2^{2/3} \arctan \left (\frac {\sqrt {3} \sqrt [6]{a} \left (\sqrt [3]{a}-\sqrt [3]{2} \sqrt [3]{b} x\right )}{\sqrt {a-b x^3}}\right )}{\sqrt {3} \sqrt [6]{a} \sqrt [3]{b}} \]

[Out]

-2/3*2^(2/3)*arctan(a^(1/6)*(a^(1/3)-2^(1/3)*b^(1/3)*x)*3^(1/2)/(-b*x^3+a)^(1/2))/a^(1/6)/b^(1/3)*3^(1/2)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {2162, 209} \[ \int \frac {2^{2/3} \sqrt [3]{a}+2 \sqrt [3]{b} x}{\left (2^{2/3} \sqrt [3]{a}-\sqrt [3]{b} x\right ) \sqrt {a-b x^3}} \, dx=-\frac {2\ 2^{2/3} \arctan \left (\frac {\sqrt {3} \sqrt [6]{a} \left (\sqrt [3]{a}-\sqrt [3]{2} \sqrt [3]{b} x\right )}{\sqrt {a-b x^3}}\right )}{\sqrt {3} \sqrt [6]{a} \sqrt [3]{b}} \]

[In]

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

[Out]

(-2*2^(2/3)*ArcTan[(Sqrt[3]*a^(1/6)*(a^(1/3) - 2^(1/3)*b^(1/3)*x))/Sqrt[a - b*x^3]])/(Sqrt[3]*a^(1/6)*b^(1/3))

Rule 209

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

Rule 2162

Int[((e_) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_Symbol] :> Dist[2*(e/d), Subst[Int[
1/(1 + 3*a*x^2), x], x, (1 + 2*d*(x/c))/Sqrt[a + b*x^3]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[d*e - c*f,
0] && EqQ[b*c^3 - 4*a*d^3, 0] && EqQ[2*d*e + c*f, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (2\ 2^{2/3} \sqrt [3]{a}\right ) \text {Subst}\left (\int \frac {1}{1+3 a x^2} \, dx,x,\frac {1-\frac {\sqrt [3]{2} \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {a-b x^3}}\right )}{\sqrt [3]{b}} \\ & = -\frac {2\ 2^{2/3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [6]{a} \left (\sqrt [3]{a}-\sqrt [3]{2} \sqrt [3]{b} x\right )}{\sqrt {a-b x^3}}\right )}{\sqrt {3} \sqrt [6]{a} \sqrt [3]{b}} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.39 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.03 \[ \int \frac {2^{2/3} \sqrt [3]{a}+2 \sqrt [3]{b} x}{\left (2^{2/3} \sqrt [3]{a}-\sqrt [3]{b} x\right ) \sqrt {a-b x^3}} \, dx=\frac {2\ 2^{2/3} \arctan \left (\frac {\sqrt {a-b x^3}}{\sqrt {3} \left (\sqrt {a}-\sqrt [3]{2} \sqrt [6]{a} \sqrt [3]{b} x\right )}\right )}{\sqrt {3} \sqrt [6]{a} \sqrt [3]{b}} \]

[In]

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

[Out]

(2*2^(2/3)*ArcTan[Sqrt[a - b*x^3]/(Sqrt[3]*(Sqrt[a] - 2^(1/3)*a^(1/6)*b^(1/3)*x))])/(Sqrt[3]*a^(1/6)*b^(1/3))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F(-1)]

Timed out. \[ \int \frac {2^{2/3} \sqrt [3]{a}+2 \sqrt [3]{b} x}{\left (2^{2/3} \sqrt [3]{a}-\sqrt [3]{b} x\right ) \sqrt {a-b x^3}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-1)]

Timed out. \[ \int \frac {2^{2/3} \sqrt [3]{a}+2 \sqrt [3]{b} x}{\left (2^{2/3} \sqrt [3]{a}-\sqrt [3]{b} x\right ) \sqrt {a-b x^3}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 10.97 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.65 \[ \int \frac {2^{2/3} \sqrt [3]{a}+2 \sqrt [3]{b} x}{\left (2^{2/3} \sqrt [3]{a}-\sqrt [3]{b} x\right ) \sqrt {a-b x^3}} \, dx=\frac {2^{2/3}\,\sqrt {3}\,\ln \left (\frac {\left (\sqrt {a-b\,x^3}-\sqrt {3}\,\sqrt {a}\,1{}\mathrm {i}+2^{1/3}\,\sqrt {3}\,a^{1/6}\,b^{1/3}\,x\,1{}\mathrm {i}\right )\,{\left (\sqrt {3}\,\sqrt {a}\,1{}\mathrm {i}+\sqrt {a-b\,x^3}-2^{1/3}\,\sqrt {3}\,a^{1/6}\,b^{1/3}\,x\,1{}\mathrm {i}\right )}^3}{{\left (2^{2/3}\,a^{1/3}-b^{1/3}\,x\right )}^6}\right )\,1{}\mathrm {i}}{3\,a^{1/6}\,b^{1/3}} \]

[In]

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

[Out]

(2^(2/3)*3^(1/2)*log((((a - b*x^3)^(1/2) - 3^(1/2)*a^(1/2)*1i + 2^(1/3)*3^(1/2)*a^(1/6)*b^(1/3)*x*1i)*(3^(1/2)
*a^(1/2)*1i + (a - b*x^3)^(1/2) - 2^(1/3)*3^(1/2)*a^(1/6)*b^(1/3)*x*1i)^3)/(2^(2/3)*a^(1/3) - b^(1/3)*x)^6)*1i
)/(3*a^(1/6)*b^(1/3))