3.4.53 \(\int \frac {x}{\sqrt {-a-b x^3} (-2 (5+3 \sqrt {3}) a-b x^3)} \, dx\) [353]

Optimal. Leaf size=330 \[ \frac {\left (2-\sqrt {3}\right ) \tan ^{-1}\left (\frac {\sqrt [4]{3} \sqrt [6]{a} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{\sqrt {2} \sqrt {-a-b x^3}}\right )}{3 \sqrt {2} \sqrt [4]{3} a^{5/6} b^{2/3}}+\frac {\left (2-\sqrt {3}\right ) \tan ^{-1}\left (\frac {\sqrt [4]{3} \left (1-\sqrt {3}\right ) \sqrt [6]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt {2} \sqrt {-a-b x^3}}\right )}{6 \sqrt {2} \sqrt [4]{3} a^{5/6} b^{2/3}}+\frac {\left (2-\sqrt {3}\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{3} \left (1+\sqrt {3}\right ) \sqrt [6]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt {2} \sqrt {-a-b x^3}}\right )}{2 \sqrt {2} 3^{3/4} a^{5/6} b^{2/3}}-\frac {\left (2-\sqrt {3}\right ) \tanh ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt {-a-b x^3}}{\sqrt {2} 3^{3/4} \sqrt {a}}\right )}{3 \sqrt {2} 3^{3/4} a^{5/6} b^{2/3}} \]

[Out]

1/36*arctan(1/2*3^(1/4)*a^(1/6)*(a^(1/3)+b^(1/3)*x)*(1-3^(1/2))*2^(1/2)/(-b*x^3-a)^(1/2))*(2-3^(1/2))*3^(3/4)/
a^(5/6)/b^(2/3)*2^(1/2)+1/18*arctan(1/2*3^(1/4)*a^(1/6)*(-2*b^(1/3)*x+a^(1/3)*(1+3^(1/2)))*2^(1/2)/(-b*x^3-a)^
(1/2))*(2-3^(1/2))*3^(3/4)/a^(5/6)/b^(2/3)*2^(1/2)+1/12*arctanh(1/2*3^(1/4)*a^(1/6)*(a^(1/3)+b^(1/3)*x)*(1+3^(
1/2))*2^(1/2)/(-b*x^3-a)^(1/2))*(2-3^(1/2))*3^(1/4)/a^(5/6)/b^(2/3)*2^(1/2)-1/18*arctanh(1/6*(1-3^(1/2))*(-b*x
^3-a)^(1/2)*3^(1/4)*2^(1/2)/a^(1/2))*(2-3^(1/2))*3^(1/4)/a^(5/6)/b^(2/3)*2^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 330, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {501} \begin {gather*} \frac {\left (2-\sqrt {3}\right ) \text {ArcTan}\left (\frac {\sqrt [4]{3} \sqrt [6]{a} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{\sqrt {2} \sqrt {-a-b x^3}}\right )}{3 \sqrt {2} \sqrt [4]{3} a^{5/6} b^{2/3}}+\frac {\left (2-\sqrt {3}\right ) \text {ArcTan}\left (\frac {\sqrt [4]{3} \left (1-\sqrt {3}\right ) \sqrt [6]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt {2} \sqrt {-a-b x^3}}\right )}{6 \sqrt {2} \sqrt [4]{3} a^{5/6} b^{2/3}}+\frac {\left (2-\sqrt {3}\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{3} \left (1+\sqrt {3}\right ) \sqrt [6]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt {2} \sqrt {-a-b x^3}}\right )}{2 \sqrt {2} 3^{3/4} a^{5/6} b^{2/3}}-\frac {\left (2-\sqrt {3}\right ) \tanh ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt {-a-b x^3}}{\sqrt {2} 3^{3/4} \sqrt {a}}\right )}{3 \sqrt {2} 3^{3/4} a^{5/6} b^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(Sqrt[-a - b*x^3]*(-2*(5 + 3*Sqrt[3])*a - b*x^3)),x]

[Out]

((2 - Sqrt[3])*ArcTan[(3^(1/4)*a^(1/6)*((1 + Sqrt[3])*a^(1/3) - 2*b^(1/3)*x))/(Sqrt[2]*Sqrt[-a - b*x^3])])/(3*
Sqrt[2]*3^(1/4)*a^(5/6)*b^(2/3)) + ((2 - Sqrt[3])*ArcTan[(3^(1/4)*(1 - Sqrt[3])*a^(1/6)*(a^(1/3) + b^(1/3)*x))
/(Sqrt[2]*Sqrt[-a - b*x^3])])/(6*Sqrt[2]*3^(1/4)*a^(5/6)*b^(2/3)) + ((2 - Sqrt[3])*ArcTanh[(3^(1/4)*(1 + Sqrt[
3])*a^(1/6)*(a^(1/3) + b^(1/3)*x))/(Sqrt[2]*Sqrt[-a - b*x^3])])/(2*Sqrt[2]*3^(3/4)*a^(5/6)*b^(2/3)) - ((2 - Sq
rt[3])*ArcTanh[((1 - Sqrt[3])*Sqrt[-a - b*x^3])/(Sqrt[2]*3^(3/4)*Sqrt[a])])/(3*Sqrt[2]*3^(3/4)*a^(5/6)*b^(2/3)
)

Rule 501

Int[(x_)/(Sqrt[(a_) + (b_.)*(x_)^3]*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[b/a, 3], r = Simplify[(b
*c - 10*a*d)/(6*a*d)]}, Simp[q*(2 - r)*(ArcTanh[(1 - r)*(Sqrt[a + b*x^3]/(Sqrt[2]*Rt[-a, 2]*r^(3/2)))]/(3*Sqrt
[2]*Rt[-a, 2]*d*r^(3/2))), x] + (-Simp[q*(2 - r)*(ArcTanh[Rt[-a, 2]*Sqrt[r]*(1 + r)*((1 + q*x)/(Sqrt[2]*Sqrt[a
+ b*x^3]))]/(2*Sqrt[2]*Rt[-a, 2]*d*r^(3/2))), x] - Simp[q*(2 - r)*(ArcTan[Rt[-a, 2]*Sqrt[r]*((1 + r - 2*q*x)/
(Sqrt[2]*Sqrt[a + b*x^3]))]/(3*Sqrt[2]*Rt[-a, 2]*d*Sqrt[r])), x] - Simp[q*(2 - r)*(ArcTan[Rt[-a, 2]*(1 - r)*Sq
rt[r]*((1 + q*x)/(Sqrt[2]*Sqrt[a + b*x^3]))]/(6*Sqrt[2]*Rt[-a, 2]*d*Sqrt[r])), x])] /; FreeQ[{a, b, c, d}, x]
&& NeQ[b*c - a*d, 0] && EqQ[b^2*c^2 - 20*a*b*c*d - 8*a^2*d^2, 0] && NegQ[a]

Rubi steps

\begin {align*} \int \frac {x}{\sqrt {-a-b x^3} \left (-2 \left (5+3 \sqrt {3}\right ) a-b x^3\right )} \, dx &=\frac {\left (2-\sqrt {3}\right ) \tan ^{-1}\left (\frac {\sqrt [4]{3} \sqrt [6]{a} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{\sqrt {2} \sqrt {-a-b x^3}}\right )}{3 \sqrt {2} \sqrt [4]{3} a^{5/6} b^{2/3}}+\frac {\left (2-\sqrt {3}\right ) \tan ^{-1}\left (\frac {\sqrt [4]{3} \left (1-\sqrt {3}\right ) \sqrt [6]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt {2} \sqrt {-a-b x^3}}\right )}{6 \sqrt {2} \sqrt [4]{3} a^{5/6} b^{2/3}}+\frac {\left (2-\sqrt {3}\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{3} \left (1+\sqrt {3}\right ) \sqrt [6]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt {2} \sqrt {-a-b x^3}}\right )}{2 \sqrt {2} 3^{3/4} a^{5/6} b^{2/3}}-\frac {\left (2-\sqrt {3}\right ) \tanh ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt {-a-b x^3}}{\sqrt {2} 3^{3/4} \sqrt {a}}\right )}{3 \sqrt {2} 3^{3/4} a^{5/6} b^{2/3}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
time = 10.05, size = 87, normalized size = 0.26 \begin {gather*} -\frac {x^2 \sqrt {1+\frac {b x^3}{a}} F_1\left (\frac {2}{3};\frac {1}{2},1;\frac {5}{3};-\frac {b x^3}{a},-\frac {b x^3}{10 a+6 \sqrt {3} a}\right )}{\left (20 a+12 \sqrt {3} a\right ) \sqrt {-a-b x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/(Sqrt[-a - b*x^3]*(-2*(5 + 3*Sqrt[3])*a - b*x^3)),x]

[Out]

-((x^2*Sqrt[1 + (b*x^3)/a]*AppellF1[2/3, 1/2, 1, 5/3, -((b*x^3)/a), -((b*x^3)/(10*a + 6*Sqrt[3]*a))])/((20*a +
12*Sqrt[3]*a)*Sqrt[-a - b*x^3]))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.38, size = 541, normalized size = 1.64 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(-b*x^3-2*a*(5+3*3^(1/2)))/(-b*x^3-a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/27*I/b^3/a*2^(1/2)*sum(1/_alpha*(-a*b^2)^(1/3)*(1/2*I*b*(2*x+1/b*((-a*b^2)^(1/3)-I*3^(1/2)*(-a*b^2)^(1/3)))/
(-a*b^2)^(1/3))^(1/2)*(b*(x-1/b*(-a*b^2)^(1/3))/(-3*(-a*b^2)^(1/3)+I*3^(1/2)*(-a*b^2)^(1/3)))^(1/2)*(-1/2*I*b*
(2*x+1/b*((-a*b^2)^(1/3)+I*3^(1/2)*(-a*b^2)^(1/3)))/(-a*b^2)^(1/3))^(1/2)/(-b*x^3-a)^(1/2)*(-3*I*(-a*b^2)^(1/3
)*_alpha*3^(1/2)*b+4*b^2*_alpha^2*3^(1/2)+3*I*(-a*b^2)^(2/3)*3^(1/2)+6*I*(-a*b^2)^(1/3)*_alpha*b-2*(-a*b^2)^(1
/3)*_alpha*3^(1/2)*b-6*b^2*_alpha^2-6*I*(-a*b^2)^(2/3)-2*(-a*b^2)^(2/3)*3^(1/2)+3*(-a*b^2)^(1/3)*_alpha*b+3*(-
a*b^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*
b^2)^(1/3))^(1/2),-1/6/b*(2*I*(-a*b^2)^(1/3)*_alpha^2*3^(1/2)*b-I*(-a*b^2)^(2/3)*_alpha*3^(1/2)-4*I*(-a*b^2)^(
1/3)*_alpha^2*b+2*I*(-a*b^2)^(2/3)*_alpha+2*(-a*b^2)^(2/3)*_alpha*3^(1/2)+I*3^(1/2)*a*b-3*(-a*b^2)^(2/3)*_alph
a-2*I*a*b+2*3^(1/2)*a*b-3*a*b)/a,(I*3^(1/2)/b*(-a*b^2)^(1/3)/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(
1/3)))^(1/2)),_alpha=RootOf(b*_Z^3+6*3^(1/2)*a+10*a))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(-b*x^3-2*a*(5+3*3^(1/2)))/(-b*x^3-a)^(1/2),x, algorithm="maxima")

[Out]

-integrate(x/((b*x^3 + 2*a*(3*sqrt(3) + 5))*sqrt(-b*x^3 - a)), x)

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(-b*x^3-2*a*(5+3*3^(1/2)))/(-b*x^3-a)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {x}{10 a \sqrt {- a - b x^{3}} + 6 \sqrt {3} a \sqrt {- a - b x^{3}} + b x^{3} \sqrt {- a - b x^{3}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(-b*x**3-2*a*(5+3*3**(1/2)))/(-b*x**3-a)**(1/2),x)

[Out]

-Integral(x/(10*a*sqrt(-a - b*x**3) + 6*sqrt(3)*a*sqrt(-a - b*x**3) + b*x**3*sqrt(-a - b*x**3)), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(-b*x^3-2*a*(5+3*3^(1/2)))/(-b*x^3-a)^(1/2),x, algorithm="giac")

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:index.cc index_m
operator + Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int -\frac {x}{\sqrt {-b\,x^3-a}\,\left (b\,x^3+2\,a\,\left (3\,\sqrt {3}+5\right )\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-x/((- a - b*x^3)^(1/2)*(b*x^3 + 2*a*(3*3^(1/2) + 5))),x)

[Out]

int(-x/((- a - b*x^3)^(1/2)*(b*x^3 + 2*a*(3*3^(1/2) + 5))), x)

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