3.74 \(\int \frac{x}{(4-d x^3) \sqrt{-1+d x^3}} \, dx\)

Optimal. Leaf size=157 \[ -\frac{\tan ^{-1}\left (\frac{\sqrt [3]{2} \sqrt [3]{d} x+1}{\sqrt{d x^3-1}}\right )}{3\ 2^{2/3} d^{2/3}}-\frac{\tan ^{-1}\left (\sqrt{d x^3-1}\right )}{9\ 2^{2/3} d^{2/3}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{3} \left (1-\sqrt [3]{2} \sqrt [3]{d} x\right )}{\sqrt{d x^3-1}}\right )}{3\ 2^{2/3} \sqrt{3} d^{2/3}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d x^3-1}}{\sqrt{3}}\right )}{3\ 2^{2/3} \sqrt{3} d^{2/3}} \]

[Out]

-ArcTan[(1 + 2^(1/3)*d^(1/3)*x)/Sqrt[-1 + d*x^3]]/(3*2^(2/3)*d^(2/3)) - ArcTan[Sqrt[-1 + d*x^3]]/(9*2^(2/3)*d^
(2/3)) - ArcTanh[(Sqrt[3]*(1 - 2^(1/3)*d^(1/3)*x))/Sqrt[-1 + d*x^3]]/(3*2^(2/3)*Sqrt[3]*d^(2/3)) - ArcTanh[Sqr
t[-1 + d*x^3]/Sqrt[3]]/(3*2^(2/3)*Sqrt[3]*d^(2/3))

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Rubi [A]  time = 0.0329981, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.043, Rules used = {485} \[ -\frac{\tan ^{-1}\left (\frac{\sqrt [3]{2} \sqrt [3]{d} x+1}{\sqrt{d x^3-1}}\right )}{3\ 2^{2/3} d^{2/3}}-\frac{\tan ^{-1}\left (\sqrt{d x^3-1}\right )}{9\ 2^{2/3} d^{2/3}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{3} \left (1-\sqrt [3]{2} \sqrt [3]{d} x\right )}{\sqrt{d x^3-1}}\right )}{3\ 2^{2/3} \sqrt{3} d^{2/3}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d x^3-1}}{\sqrt{3}}\right )}{3\ 2^{2/3} \sqrt{3} d^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[x/((4 - d*x^3)*Sqrt[-1 + d*x^3]),x]

[Out]

-ArcTan[(1 + 2^(1/3)*d^(1/3)*x)/Sqrt[-1 + d*x^3]]/(3*2^(2/3)*d^(2/3)) - ArcTan[Sqrt[-1 + d*x^3]]/(9*2^(2/3)*d^
(2/3)) - ArcTanh[(Sqrt[3]*(1 - 2^(1/3)*d^(1/3)*x))/Sqrt[-1 + d*x^3]]/(3*2^(2/3)*Sqrt[3]*d^(2/3)) - ArcTanh[Sqr
t[-1 + d*x^3]/Sqrt[3]]/(3*2^(2/3)*Sqrt[3]*d^(2/3))

Rule 485

Int[(x_)/(((a_) + (b_.)*(x_)^3)*Sqrt[(c_) + (d_.)*(x_)^3]), x_Symbol] :> With[{q = Rt[d/c, 3]}, -Simp[(q*ArcTa
n[Sqrt[c + d*x^3]/Rt[-c, 2]])/(9*2^(2/3)*b*Rt[-c, 2]), x] + (-Simp[(q*ArcTan[(Rt[-c, 2]*(1 - 2^(1/3)*q*x))/Sqr
t[c + d*x^3]])/(3*2^(2/3)*b*Rt[-c, 2]), x] - Simp[(q*ArcTanh[Sqrt[c + d*x^3]/(Sqrt[3]*Rt[-c, 2])])/(3*2^(2/3)*
Sqrt[3]*b*Rt[-c, 2]), x] - Simp[(q*ArcTanh[(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] && NegQ[c]

Rubi steps

\begin{align*} \int \frac{x}{\left (4-d x^3\right ) \sqrt{-1+d x^3}} \, dx &=-\frac{\tan ^{-1}\left (\frac{1+\sqrt [3]{2} \sqrt [3]{d} x}{\sqrt{-1+d x^3}}\right )}{3\ 2^{2/3} d^{2/3}}-\frac{\tan ^{-1}\left (\sqrt{-1+d x^3}\right )}{9\ 2^{2/3} d^{2/3}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{3} \left (1-\sqrt [3]{2} \sqrt [3]{d} x\right )}{\sqrt{-1+d x^3}}\right )}{3\ 2^{2/3} \sqrt{3} d^{2/3}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{-1+d x^3}}{\sqrt{3}}\right )}{3\ 2^{2/3} \sqrt{3} d^{2/3}}\\ \end{align*}

Mathematica [C]  time = 0.0290417, size = 54, normalized size = 0.34 \[ \frac{x^2 \sqrt{1-d x^3} F_1\left (\frac{2}{3};\frac{1}{2},1;\frac{5}{3};d x^3,\frac{d x^3}{4}\right )}{8 \sqrt{d x^3-1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x/((4 - d*x^3)*Sqrt[-1 + d*x^3]),x]

[Out]

(x^2*Sqrt[1 - d*x^3]*AppellF1[2/3, 1/2, 1, 5/3, d*x^3, (d*x^3)/4])/(8*Sqrt[-1 + d*x^3])

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Maple [C]  time = 0.106, size = 240, normalized size = 1.5 \begin{align*} -{\frac{i}{9}}\sqrt{2}\sum _{{\it \_alpha}={\it RootOf} \left ( d{{\it \_Z}}^{3}-4 \right ) }{\frac{1}{{\it \_alpha}}\sqrt{-{\frac{i}{2}} \left ( 2\,x+{\frac{1}{\sqrt [3]{d}}}+{i\sqrt{3}{\frac{1}{\sqrt [3]{d}}}} \right ) \sqrt [3]{d}}\sqrt{{ \left ( x-{\frac{1}{\sqrt [3]{d}}} \right ) \left ( -3\,{\frac{1}{\sqrt [3]{d}}}-{i\sqrt{3}{\frac{1}{\sqrt [3]{d}}}} \right ) ^{-1}}}\sqrt{{\frac{i}{2}} \left ( 2\,x+{\frac{1}{\sqrt [3]{d}}}-{i\sqrt{3}{\frac{1}{\sqrt [3]{d}}}} \right ) \sqrt [3]{d}} \left ( -2\,{{\it \_alpha}}^{2}d+i\sqrt{3}{\it \_alpha}\,{d}^{{\frac{2}{3}}}-i\sqrt{3}\sqrt [3]{d}+{\it \_alpha}\,{d}^{{\frac{2}{3}}}+\sqrt [3]{d} \right ){\it EllipticPi} \left ({\frac{\sqrt{3}}{3}\sqrt{-i \left ( x+{\frac{1}{2}{\frac{1}{\sqrt [3]{d}}}}+{{\frac{i}{2}}\sqrt{3}{\frac{1}{\sqrt [3]{d}}}} \right ) \sqrt{3}\sqrt [3]{d}}},{\frac{i}{3}}\sqrt{3}{{\it \_alpha}}^{2}{d}^{{\frac{2}{3}}}-{\frac{i}{6}}\sqrt{3}{\it \_alpha}\,\sqrt [3]{d}-{\frac{i}{6}}\sqrt{3}+{\frac{{\it \_alpha}}{2}\sqrt [3]{d}}-{\frac{1}{2}},\sqrt{{-i\sqrt{3}{\frac{1}{\sqrt [3]{d}}} \left ( -{\frac{3}{2}{\frac{1}{\sqrt [3]{d}}}}-{{\frac{i}{2}}\sqrt{3}{\frac{1}{\sqrt [3]{d}}}} \right ) ^{-1}}} \right ){d}^{-{\frac{4}{3}}}{\frac{1}{\sqrt{d{x}^{3}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(-d*x^3+4)/(d*x^3-1)^(1/2),x)

[Out]

-1/9*I*2^(1/2)*sum(1/_alpha/d^(4/3)*(-1/2*I*(2*x+1/d^(1/3)+I*3^(1/2)/d^(1/3))*d^(1/3))^(1/2)*((x-1/d^(1/3))/(-
3/d^(1/3)-I*3^(1/2)/d^(1/3)))^(1/2)*(1/2*I*(2*x+1/d^(1/3)-I*3^(1/2)/d^(1/3))*d^(1/3))^(1/2)/(d*x^3-1)^(1/2)*(-
2*_alpha^2*d+I*3^(1/2)*_alpha*d^(2/3)-I*3^(1/2)*d^(1/3)+_alpha*d^(2/3)+d^(1/3))*EllipticPi(1/3*3^(1/2)*(-I*(x+
1/2/d^(1/3)+1/2*I*3^(1/2)/d^(1/3))*3^(1/2)*d^(1/3))^(1/2),1/3*I*3^(1/2)*_alpha^2*d^(2/3)-1/6*I*3^(1/2)*_alpha*
d^(1/3)-1/6*I*3^(1/2)+1/2*_alpha*d^(1/3)-1/2,(-I*3^(1/2)/d^(1/3)/(-3/2/d^(1/3)-1/2*I*3^(1/2)/d^(1/3)))^(1/2)),
_alpha=RootOf(_Z^3*d-4))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{x}{\sqrt{d x^{3} - 1}{\left (d x^{3} - 4\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(-d*x^3+4)/(d*x^3-1)^(1/2),x, algorithm="maxima")

[Out]

-integrate(x/(sqrt(d*x^3 - 1)*(d*x^3 - 4)), x)

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Fricas [B]  time = 5.7955, size = 4625, normalized size = 29.46 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(-d*x^3+4)/(d*x^3-1)^(1/2),x, algorithm="fricas")

[Out]

-1/9*sqrt(3)*(1/432)^(1/6)*(d^(-4))^(1/6)*arctan(1/3*(3*(sqrt(3)*sqrt(1/3)*d^2*sqrt(d^(-4))*x + 2*sqrt(3)*(1/4
32)^(1/6)*d*(d^(-4))^(1/6)*x^2 - 24*sqrt(3)*(1/432)^(5/6)*(d^4*x^3 - 4*d^3)*(d^(-4))^(5/6))*sqrt(d*x^3 - 1) +
(2*sqrt(3)*(1/2)^(1/3)*(d^2*x^3 - d)*(d^(-4))^(1/3) + sqrt(3)*(d*x^4 - x) + 3*(sqrt(3)*sqrt(1/3)*d^2*sqrt(d^(-
4))*x + 2*sqrt(3)*(1/432)^(1/6)*d*(d^(-4))^(1/6)*x^2 + 24*sqrt(3)*(1/432)^(5/6)*(d^4*x^3 + 2*d^3)*(d^(-4))^(5/
6))*sqrt(d*x^3 - 1))*sqrt((d^3*x^9 - 60*d^2*x^6 - 24*(1/2)^(2/3)*(d^5*x^7 - 5*d^4*x^4 + 4*d^3*x)*(d^(-4))^(2/3
) + 12*(1/2)^(1/3)*(d^4*x^8 + 7*d^3*x^5 - 8*d^2*x^2)*(d^(-4))^(1/3) + 12*(648*(1/432)^(5/6)*d^5*(d^(-4))^(5/6)
*x^5 - sqrt(1/3)*(d^4*x^6 + 16*d^3*x^3 - 8*d^2)*sqrt(d^(-4)) - (1/432)^(1/6)*(d^3*x^7 - 2*d^2*x^4 - 8*d*x)*(d^
(-4))^(1/6))*sqrt(d*x^3 - 1) + 32)/(d^3*x^9 - 12*d^2*x^6 + 48*d*x^3 - 64)))/(d*x^4 - x)) - 1/9*sqrt(3)*(1/432)
^(1/6)*(d^(-4))^(1/6)*arctan(1/3*(3*(sqrt(3)*sqrt(1/3)*d^2*sqrt(d^(-4))*x + 2*sqrt(3)*(1/432)^(1/6)*d*(d^(-4))
^(1/6)*x^2 - 24*sqrt(3)*(1/432)^(5/6)*(d^4*x^3 - 4*d^3)*(d^(-4))^(5/6))*sqrt(d*x^3 - 1) - (2*sqrt(3)*(1/2)^(1/
3)*(d^2*x^3 - d)*(d^(-4))^(1/3) + sqrt(3)*(d*x^4 - x) - 3*(sqrt(3)*sqrt(1/3)*d^2*sqrt(d^(-4))*x + 2*sqrt(3)*(1
/432)^(1/6)*d*(d^(-4))^(1/6)*x^2 + 24*sqrt(3)*(1/432)^(5/6)*(d^4*x^3 + 2*d^3)*(d^(-4))^(5/6))*sqrt(d*x^3 - 1))
*sqrt((d^3*x^9 - 60*d^2*x^6 - 24*(1/2)^(2/3)*(d^5*x^7 - 5*d^4*x^4 + 4*d^3*x)*(d^(-4))^(2/3) + 12*(1/2)^(1/3)*(
d^4*x^8 + 7*d^3*x^5 - 8*d^2*x^2)*(d^(-4))^(1/3) - 12*(648*(1/432)^(5/6)*d^5*(d^(-4))^(5/6)*x^5 - sqrt(1/3)*(d^
4*x^6 + 16*d^3*x^3 - 8*d^2)*sqrt(d^(-4)) - (1/432)^(1/6)*(d^3*x^7 - 2*d^2*x^4 - 8*d*x)*(d^(-4))^(1/6))*sqrt(d*
x^3 - 1) + 32)/(d^3*x^9 - 12*d^2*x^6 + 48*d*x^3 - 64)))/(d*x^4 - x)) + 1/18*(1/432)^(1/6)*(d^(-4))^(1/6)*log((
d^3*x^9 + 66*d^2*x^6 - 72*d*x^3 + 48*(1/2)^(2/3)*(d^5*x^7 + d^4*x^4 - 2*d^3*x)*(d^(-4))^(2/3) + 12*(1/2)^(1/3)
*(d^4*x^8 + 7*d^3*x^5 - 8*d^2*x^2)*(d^(-4))^(1/3) + 6*(1296*(1/432)^(5/6)*d^5*(d^(-4))^(5/6)*x^5 + sqrt(1/3)*(
5*d^4*x^6 + 20*d^3*x^3 - 16*d^2)*sqrt(d^(-4)) + 2*(1/432)^(1/6)*(d^3*x^7 + 16*d^2*x^4 - 8*d*x)*(d^(-4))^(1/6))
*sqrt(d*x^3 - 1) + 32)/(d^3*x^9 - 12*d^2*x^6 + 48*d*x^3 - 64)) - 1/18*(1/432)^(1/6)*(d^(-4))^(1/6)*log((d^3*x^
9 + 66*d^2*x^6 - 72*d*x^3 + 48*(1/2)^(2/3)*(d^5*x^7 + d^4*x^4 - 2*d^3*x)*(d^(-4))^(2/3) + 12*(1/2)^(1/3)*(d^4*
x^8 + 7*d^3*x^5 - 8*d^2*x^2)*(d^(-4))^(1/3) - 6*(1296*(1/432)^(5/6)*d^5*(d^(-4))^(5/6)*x^5 + sqrt(1/3)*(5*d^4*
x^6 + 20*d^3*x^3 - 16*d^2)*sqrt(d^(-4)) + 2*(1/432)^(1/6)*(d^3*x^7 + 16*d^2*x^4 - 8*d*x)*(d^(-4))^(1/6))*sqrt(
d*x^3 - 1) + 32)/(d^3*x^9 - 12*d^2*x^6 + 48*d*x^3 - 64)) - 1/36*(1/432)^(1/6)*(d^(-4))^(1/6)*log((d^3*x^9 - 60
*d^2*x^6 - 24*(1/2)^(2/3)*(d^5*x^7 - 5*d^4*x^4 + 4*d^3*x)*(d^(-4))^(2/3) + 12*(1/2)^(1/3)*(d^4*x^8 + 7*d^3*x^5
- 8*d^2*x^2)*(d^(-4))^(1/3) + 12*(648*(1/432)^(5/6)*d^5*(d^(-4))^(5/6)*x^5 - sqrt(1/3)*(d^4*x^6 + 16*d^3*x^3
- 8*d^2)*sqrt(d^(-4)) - (1/432)^(1/6)*(d^3*x^7 - 2*d^2*x^4 - 8*d*x)*(d^(-4))^(1/6))*sqrt(d*x^3 - 1) + 32)/(d^3
*x^9 - 12*d^2*x^6 + 48*d*x^3 - 64)) + 1/36*(1/432)^(1/6)*(d^(-4))^(1/6)*log((d^3*x^9 - 60*d^2*x^6 - 24*(1/2)^(
2/3)*(d^5*x^7 - 5*d^4*x^4 + 4*d^3*x)*(d^(-4))^(2/3) + 12*(1/2)^(1/3)*(d^4*x^8 + 7*d^3*x^5 - 8*d^2*x^2)*(d^(-4)
)^(1/3) - 12*(648*(1/432)^(5/6)*d^5*(d^(-4))^(5/6)*x^5 - sqrt(1/3)*(d^4*x^6 + 16*d^3*x^3 - 8*d^2)*sqrt(d^(-4))
- (1/432)^(1/6)*(d^3*x^7 - 2*d^2*x^4 - 8*d*x)*(d^(-4))^(1/6))*sqrt(d*x^3 - 1) + 32)/(d^3*x^9 - 12*d^2*x^6 + 4
8*d*x^3 - 64))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{x}{d x^{3} \sqrt{d x^{3} - 1} - 4 \sqrt{d x^{3} - 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(-d*x**3+4)/(d*x**3-1)**(1/2),x)

[Out]

-Integral(x/(d*x**3*sqrt(d*x**3 - 1) - 4*sqrt(d*x**3 - 1)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{x}{\sqrt{d x^{3} - 1}{\left (d x^{3} - 4\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(-d*x^3+4)/(d*x^3-1)^(1/2),x, algorithm="giac")

[Out]

integrate(-x/(sqrt(d*x^3 - 1)*(d*x^3 - 4)), x)