3.73 \(\int \frac{x}{\sqrt{1-x^3} (4-x^3)} \, dx\)
Optimal. Leaf size=127 \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{3} \left (1-\sqrt [3]{2} x\right )}{\sqrt{1-x^3}}\right )}{3\ 2^{2/3} \sqrt{3}}+\frac{\tan ^{-1}\left (\frac{\sqrt{1-x^3}}{\sqrt{3}}\right )}{3\ 2^{2/3} \sqrt{3}}-\frac{\tanh ^{-1}\left (\frac{\sqrt [3]{2} x+1}{\sqrt{1-x^3}}\right )}{3\ 2^{2/3}}+\frac{\tanh ^{-1}\left (\sqrt{1-x^3}\right )}{9\ 2^{2/3}} \]
[Out]
-ArcTan[(Sqrt[3]*(1 - 2^(1/3)*x))/Sqrt[1 - x^3]]/(3*2^(2/3)*Sqrt[3]) + ArcTan[Sqrt[1 - x^3]/Sqrt[3]]/(3*2^(2/3
)*Sqrt[3]) - ArcTanh[(1 + 2^(1/3)*x)/Sqrt[1 - x^3]]/(3*2^(2/3)) + ArcTanh[Sqrt[1 - x^3]]/(9*2^(2/3))
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Rubi [A] time = 0.0192024, antiderivative size = 127, normalized size of antiderivative = 1.,
number of steps used = 1, number of rules used = 1, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.045, Rules used =
{484} \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{3} \left (1-\sqrt [3]{2} x\right )}{\sqrt{1-x^3}}\right )}{3\ 2^{2/3} \sqrt{3}}+\frac{\tan ^{-1}\left (\frac{\sqrt{1-x^3}}{\sqrt{3}}\right )}{3\ 2^{2/3} \sqrt{3}}-\frac{\tanh ^{-1}\left (\frac{\sqrt [3]{2} x+1}{\sqrt{1-x^3}}\right )}{3\ 2^{2/3}}+\frac{\tanh ^{-1}\left (\sqrt{1-x^3}\right )}{9\ 2^{2/3}} \]
Antiderivative was successfully verified.
[In]
Int[x/(Sqrt[1 - x^3]*(4 - x^3)),x]
[Out]
-ArcTan[(Sqrt[3]*(1 - 2^(1/3)*x))/Sqrt[1 - x^3]]/(3*2^(2/3)*Sqrt[3]) + ArcTan[Sqrt[1 - x^3]/Sqrt[3]]/(3*2^(2/3
)*Sqrt[3]) - ArcTanh[(1 + 2^(1/3)*x)/Sqrt[1 - x^3]]/(3*2^(2/3)) + ArcTanh[Sqrt[1 - x^3]]/(9*2^(2/3))
Rule 484
Int[(x_)/(((a_) + (b_.)*(x_)^3)*Sqrt[(c_) + (d_.)*(x_)^3]), x_Symbol] :> With[{q = Rt[d/c, 3]}, Simp[(q*ArcTan
h[Sqrt[c + d*x^3]/Rt[c, 2]])/(9*2^(2/3)*b*Rt[c, 2]), x] + (-Simp[(q*ArcTanh[(Rt[c, 2]*(1 - 2^(1/3)*q*x))/Sqrt[
c + d*x^3]])/(3*2^(2/3)*b*Rt[c, 2]), x] + Simp[(q*ArcTan[Sqrt[c + d*x^3]/(Sqrt[3]*Rt[c, 2])])/(3*2^(2/3)*Sqrt[
3]*b*Rt[c, 2]), x] - Simp[(q*ArcTan[(Sqrt[3]*Rt[c, 2]*(1 + 2^(1/3)*q*x))/Sqrt[c + d*x^3]])/(3*2^(2/3)*Sqrt[3]*
b*Rt[c, 2]), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[4*b*c - a*d, 0] && PosQ[c]
Rubi steps
\begin{align*} \int \frac{x}{\sqrt{1-x^3} \left (4-x^3\right )} \, dx &=-\frac{\tan ^{-1}\left (\frac{\sqrt{3} \left (1-\sqrt [3]{2} x\right )}{\sqrt{1-x^3}}\right )}{3\ 2^{2/3} \sqrt{3}}+\frac{\tan ^{-1}\left (\frac{\sqrt{1-x^3}}{\sqrt{3}}\right )}{3\ 2^{2/3} \sqrt{3}}-\frac{\tanh ^{-1}\left (\frac{1+\sqrt [3]{2} x}{\sqrt{1-x^3}}\right )}{3\ 2^{2/3}}+\frac{\tanh ^{-1}\left (\sqrt{1-x^3}\right )}{9\ 2^{2/3}}\\ \end{align*}
Mathematica [C] time = 0.0217669, size = 28, normalized size = 0.22 \[ \frac{1}{8} x^2 F_1\left (\frac{2}{3};\frac{1}{2},1;\frac{5}{3};x^3,\frac{x^3}{4}\right ) \]
Warning: Unable to verify antiderivative.
[In]
Integrate[x/(Sqrt[1 - x^3]*(4 - x^3)),x]
[Out]
(x^2*AppellF1[2/3, 1/2, 1, 5/3, x^3, x^3/4])/8
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Maple [C] time = 0.099, size = 164, normalized size = 1.3 \begin{align*}{\frac{i}{36}}\sqrt{2}\sum _{{\it \_alpha}={\it RootOf} \left ({{\it \_Z}}^{3}-4 \right ) }{{{\it \_alpha}}^{2} \left ( -2\,{{\it \_alpha}}^{2}+{\it \_alpha}+1+i\sqrt{3} \left ( 1-{\it \_alpha} \right ) \right ) \sqrt{{\frac{i}{2}} \left ( 2\,x+1-i\sqrt{3} \right ) }\sqrt{{\frac{-1+x}{i\sqrt{3}-3}}}\sqrt{-{\frac{i}{2}} \left ( 2\,x+1+i\sqrt{3} \right ) }{\it EllipticPi} \left ({\frac{\sqrt{3}}{3}\sqrt{i \left ( x+{\frac{1}{2}}-{\frac{i}{2}}\sqrt{3} \right ) \sqrt{3}}},{\frac{{\it \_alpha}}{2}}-{\frac{i}{3}}{{\it \_alpha}}^{2}\sqrt{3}-{\frac{1}{2}}+{\frac{i}{6}}{\it \_alpha}\,\sqrt{3}+{\frac{i}{6}}\sqrt{3},\sqrt{{\frac{i\sqrt{3}}{-{\frac{3}{2}}+{\frac{i}{2}}\sqrt{3}}}} \right ){\frac{1}{\sqrt{-{x}^{3}+1}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x/(-x^3+4)/(-x^3+1)^(1/2),x)
[Out]
1/36*I*2^(1/2)*sum(_alpha^2*(1/2*I*(2*x+1-I*3^(1/2)))^(1/2)*((-1+x)/(I*3^(1/2)-3))^(1/2)*(-1/2*I*(2*x+1+I*3^(1
/2)))^(1/2)/(-x^3+1)^(1/2)*(-2*_alpha^2+_alpha+1+I*3^(1/2)*(1-_alpha))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2-1/2*I*
3^(1/2))*3^(1/2))^(1/2),1/2*_alpha-1/3*I*_alpha^2*3^(1/2)-1/2+1/6*I*_alpha*3^(1/2)+1/6*I*3^(1/2),(I*3^(1/2)/(-
3/2+1/2*I*3^(1/2)))^(1/2)),_alpha=RootOf(_Z^3-4))
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{x}{{\left (x^{3} - 4\right )} \sqrt{-x^{3} + 1}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x/(-x^3+4)/(-x^3+1)^(1/2),x, algorithm="maxima")
[Out]
-integrate(x/((x^3 - 4)*sqrt(-x^3 + 1)), x)
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Fricas [B] time = 5.56681, size = 3299, normalized size = 25.98 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x/(-x^3+4)/(-x^3+1)^(1/2),x, algorithm="fricas")
[Out]
-1/31104*432^(5/6)*sqrt(3)*log(144*(36*x^9 - 8208*x^6 + 9504*x^3 - 648*2^(2/3)*(x^8 - 5*x^5 + 4*x^2) + (2592*x
^6 - 2592*x^3 - 432^(5/6)*sqrt(3)*(x^7 - 26*x^4 + 16*x) - 216*432^(1/6)*sqrt(3)*(7*x^5 - 4*x^2))*sqrt(-x^3 + 1
) + 3888*2^(1/3)*(x^7 - x^4) - 2304)/(x^9 - 12*x^6 + 48*x^3 - 64)) - 1/31104*432^(5/6)*sqrt(3)*log(36*(36*x^9
- 8208*x^6 + 9504*x^3 - 648*2^(2/3)*(x^8 - 5*x^5 + 4*x^2) + (2592*x^6 - 2592*x^3 - 432^(5/6)*sqrt(3)*(x^7 - 26
*x^4 + 16*x) - 216*432^(1/6)*sqrt(3)*(7*x^5 - 4*x^2))*sqrt(-x^3 + 1) + 3888*2^(1/3)*(x^7 - x^4) - 2304)/(x^9 -
12*x^6 + 48*x^3 - 64)) + 1/31104*432^(5/6)*sqrt(3)*log(144*(36*x^9 - 8208*x^6 + 9504*x^3 - 648*2^(2/3)*(x^8 -
5*x^5 + 4*x^2) - (2592*x^6 - 2592*x^3 - 432^(5/6)*sqrt(3)*(x^7 - 26*x^4 + 16*x) - 216*432^(1/6)*sqrt(3)*(7*x^
5 - 4*x^2))*sqrt(-x^3 + 1) + 3888*2^(1/3)*(x^7 - x^4) - 2304)/(x^9 - 12*x^6 + 48*x^3 - 64)) + 1/31104*432^(5/6
)*sqrt(3)*log(36*(36*x^9 - 8208*x^6 + 9504*x^3 - 648*2^(2/3)*(x^8 - 5*x^5 + 4*x^2) - (2592*x^6 - 2592*x^3 - 43
2^(5/6)*sqrt(3)*(x^7 - 26*x^4 + 16*x) - 216*432^(1/6)*sqrt(3)*(7*x^5 - 4*x^2))*sqrt(-x^3 + 1) + 3888*2^(1/3)*(
x^7 - x^4) - 2304)/(x^9 - 12*x^6 + 48*x^3 - 64)) - 1/1944*432^(5/6)*arctan(1/216*sqrt(-x^3 + 1)*(72*432^(1/6)*
x^2 + 432^(5/6)*x + 72*sqrt(3))/(2*x^3 - 1)) + 1/3888*432^(5/6)*arctan(-1/648*(6*sqrt(-x^3 + 1)*(432^(5/6)*(x^
4 + 2*x) - 36*sqrt(3)*(x^3 - 4) + 18*432^(1/6)*(x^5 + 8*x^2)) + (108*sqrt(3)*2^(2/3)*(x^5 - x^2) - 216*sqrt(3)
*2^(1/3)*(x^4 - x) - 108*sqrt(3)*(x^6 - x^3) - sqrt(-x^3 + 1)*(432^(5/6)*(2*x^4 + x) - 36*sqrt(3)*(5*x^3 - 8)
- 18*432^(1/6)*(x^5 - 10*x^2)))*sqrt((36*x^9 - 8208*x^6 + 9504*x^3 - 648*2^(2/3)*(x^8 - 5*x^5 + 4*x^2) + (2592
*x^6 - 2592*x^3 - 432^(5/6)*sqrt(3)*(x^7 - 26*x^4 + 16*x) - 216*432^(1/6)*sqrt(3)*(7*x^5 - 4*x^2))*sqrt(-x^3 +
1) + 3888*2^(1/3)*(x^7 - x^4) - 2304)/(x^9 - 12*x^6 + 48*x^3 - 64)))/(x^6 + 3*x^3 - 4)) + 1/3888*432^(5/6)*ar
ctan(-1/648*(6*sqrt(-x^3 + 1)*(432^(5/6)*(x^4 + 2*x) - 36*sqrt(3)*(x^3 - 4) + 18*432^(1/6)*(x^5 + 8*x^2)) - (1
08*sqrt(3)*2^(2/3)*(x^5 - x^2) - 216*sqrt(3)*2^(1/3)*(x^4 - x) - 108*sqrt(3)*(x^6 - x^3) + sqrt(-x^3 + 1)*(432
^(5/6)*(2*x^4 + x) - 36*sqrt(3)*(5*x^3 - 8) - 18*432^(1/6)*(x^5 - 10*x^2)))*sqrt((36*x^9 - 8208*x^6 + 9504*x^3
- 648*2^(2/3)*(x^8 - 5*x^5 + 4*x^2) - (2592*x^6 - 2592*x^3 - 432^(5/6)*sqrt(3)*(x^7 - 26*x^4 + 16*x) - 216*43
2^(1/6)*sqrt(3)*(7*x^5 - 4*x^2))*sqrt(-x^3 + 1) + 3888*2^(1/3)*(x^7 - x^4) - 2304)/(x^9 - 12*x^6 + 48*x^3 - 64
)))/(x^6 + 3*x^3 - 4))
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{x}{x^{3} \sqrt{1 - x^{3}} - 4 \sqrt{1 - x^{3}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x/(-x**3+4)/(-x**3+1)**(1/2),x)
[Out]
-Integral(x/(x**3*sqrt(1 - x**3) - 4*sqrt(1 - x**3)), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{x}{{\left (x^{3} - 4\right )} \sqrt{-x^{3} + 1}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x/(-x^3+4)/(-x^3+1)^(1/2),x, algorithm="giac")
[Out]
integrate(-x/((x^3 - 4)*sqrt(-x^3 + 1)), x)