3.4.52 \(\int \frac {x}{\sqrt {-a+b x^3} (-2 (5+3 \sqrt {3}) a+b x^3)} \, dx\) [352]

Optimal. Leaf size=328 \[ \frac {\left (2-\sqrt {3}\right ) \tan ^{-1}\left (\frac {\sqrt [4]{3} \left (1-\sqrt {3}\right ) \sqrt [6]{a} \left (\sqrt [3]{a}-\sqrt [3]{b} x\right )}{\sqrt {2} \sqrt {-a+b x^3}}\right )}{6 \sqrt {2} \sqrt [4]{3} a^{5/6} b^{2/3}}+\frac {\left (2-\sqrt {3}\right ) \tan ^{-1}\left (\frac {\sqrt [4]{3} \sqrt [6]{a} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+2 \sqrt [3]{b} x\right )}{\sqrt {2} \sqrt {-a+b x^3}}\right )}{3 \sqrt {2} \sqrt [4]{3} a^{5/6} b^{2/3}}+\frac {\left (2-\sqrt {3}\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{3} \left (1+\sqrt {3}\right ) \sqrt [6]{a} \left (\sqrt [3]{a}-\sqrt [3]{b} x\right )}{\sqrt {2} \sqrt {-a+b x^3}}\right )}{2 \sqrt {2} 3^{3/4} a^{5/6} b^{2/3}}-\frac {\left (2-\sqrt {3}\right ) \tanh ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt {-a+b x^3}}{\sqrt {2} 3^{3/4} \sqrt {a}}\right )}{3 \sqrt {2} 3^{3/4} a^{5/6} b^{2/3}} \]

[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 = 328, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {501} \begin {gather*} \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 {b x^3-a}}\right )}{6 \sqrt {2} \sqrt [4]{3} a^{5/6} b^{2/3}}+\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 {b x^3-a}}\right )}{3 \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 {b x^3-a}}\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 {b x^3-a}}{\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)*(1 - Sqrt[3])*a^(1/6)*(a^(1/3) - b^(1/3)*x))/(Sqrt[2]*Sqrt[-a + b*x^3])])/(6*Sq
rt[2]*3^(1/4)*a^(5/6)*b^(2/3)) + ((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])*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} \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 ) \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 ) \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 = 10.06, size = 85, 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*Sq
rt[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 = 0.36, size = 510, normalized size = 1.55

method	result	size
default	\(-\frac {i \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (b \,\textit {\_Z}^{3}-6 \sqrt {3}\, a -10 a \right )}{\sum }\frac {\left (a \,b^{2}\right )^{\frac {1}{3}} \sqrt {-\frac {i b \left (2 x +\frac {i \sqrt {3}\, \left (a \,b^{2}\right )^{\frac {1}{3}}+\left (a \,b^{2}\right )^{\frac {1}{3}}}{b}\right )}{2 \left (a \,b^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {b \left (x -\frac {\left (a \,b^{2}\right )^{\frac {1}{3}}}{b}\right )}{-3 \left (a \,b^{2}\right )^{\frac {1}{3}}-i \sqrt {3}\, \left (a \,b^{2}\right )^{\frac {1}{3}}}}\, \sqrt {2}\, \sqrt {\frac {i b \left (2 x +\frac {-i \sqrt {3}\, \left (a \,b^{2}\right )^{\frac {1}{3}}+\left (a \,b^{2}\right )^{\frac {1}{3}}}{b}\right )}{\left (a \,b^{2}\right )^{\frac {1}{3}}}}\, \left (3 i \left (a \,b^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}\, b +4 b^{2} \underline {\hspace {1.25 ex}}\alpha ^{2} \sqrt {3}-3 i \left (a \,b^{2}\right )^{\frac {2}{3}} \sqrt {3}-6 i \left (a \,b^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha b -2 \left (a \,b^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}\, b -6 b^{2} \underline {\hspace {1.25 ex}}\alpha ^{2}+6 i \left (a \,b^{2}\right )^{\frac {2}{3}}-2 \left (a \,b^{2}\right )^{\frac {2}{3}} \sqrt {3}+3 \left (a \,b^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha b +3 \left (a \,b^{2}\right )^{\frac {2}{3}}\right ) \EllipticPi \left (\frac {\sqrt {3}\, \sqrt {-\frac {i \left (x +\frac {\left (a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (a \,b^{2}\right )^{\frac {1}{3}}}}}{3}, \frac {-2 i \left (a \,b^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha ^{2} \sqrt {3}\, b +i \left (a \,b^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}+4 i \left (a \,b^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha ^{2} b -2 i \left (a \,b^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha +2 \left (a \,b^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}+i \sqrt {3}\, a b -3 \left (a \,b^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha -2 i a b -2 \sqrt {3}\, a b +3 a b}{6 b a}, \sqrt {-\frac {i \sqrt {3}\, \left (a \,b^{2}\right )^{\frac {1}{3}}}{b \left (-\frac {3 \left (a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right )}}\right )}{2 \underline {\hspace {1.25 ex}}\alpha \sqrt {b \,x^{3}-a}}\right )}{27 b^{3} a}\)	\(510\)
elliptic	\(-\frac {i \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (b \,\textit {\_Z}^{3}-6 \sqrt {3}\, a -10 a \right )}{\sum }\frac {\left (a \,b^{2}\right )^{\frac {1}{3}} \sqrt {-\frac {i b \left (2 x +\frac {i \sqrt {3}\, \left (a \,b^{2}\right )^{\frac {1}{3}}+\left (a \,b^{2}\right )^{\frac {1}{3}}}{b}\right )}{2 \left (a \,b^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {b \left (x -\frac {\left (a \,b^{2}\right )^{\frac {1}{3}}}{b}\right )}{-3 \left (a \,b^{2}\right )^{\frac {1}{3}}-i \sqrt {3}\, \left (a \,b^{2}\right )^{\frac {1}{3}}}}\, \sqrt {2}\, \sqrt {\frac {i b \left (2 x +\frac {-i \sqrt {3}\, \left (a \,b^{2}\right )^{\frac {1}{3}}+\left (a \,b^{2}\right )^{\frac {1}{3}}}{b}\right )}{\left (a \,b^{2}\right )^{\frac {1}{3}}}}\, \left (3 i \left (a \,b^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}\, b +4 b^{2} \underline {\hspace {1.25 ex}}\alpha ^{2} \sqrt {3}-3 i \left (a \,b^{2}\right )^{\frac {2}{3}} \sqrt {3}-6 i \left (a \,b^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha b -2 \left (a \,b^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}\, b -6 b^{2} \underline {\hspace {1.25 ex}}\alpha ^{2}+6 i \left (a \,b^{2}\right )^{\frac {2}{3}}-2 \left (a \,b^{2}\right )^{\frac {2}{3}} \sqrt {3}+3 \left (a \,b^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha b +3 \left (a \,b^{2}\right )^{\frac {2}{3}}\right ) \EllipticPi \left (\frac {\sqrt {3}\, \sqrt {-\frac {i \left (x +\frac {\left (a \,b^{2}\right )^{\frac {1}{3}}}{2 b}+\frac {i \sqrt {3}\, \left (a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right ) \sqrt {3}\, b}{\left (a \,b^{2}\right )^{\frac {1}{3}}}}}{3}, \frac {-2 i \left (a \,b^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha ^{2} \sqrt {3}\, b +i \left (a \,b^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}+4 i \left (a \,b^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha ^{2} b -2 i \left (a \,b^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha +2 \left (a \,b^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}+i \sqrt {3}\, a b -3 \left (a \,b^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha -2 i a b -2 \sqrt {3}\, a b +3 a b}{6 b a}, \sqrt {-\frac {i \sqrt {3}\, \left (a \,b^{2}\right )^{\frac {1}{3}}}{b \left (-\frac {3 \left (a \,b^{2}\right )^{\frac {1}{3}}}{2 b}-\frac {i \sqrt {3}\, \left (a \,b^{2}\right )^{\frac {1}{3}}}{2 b}\right )}}\right )}{2 \underline {\hspace {1.25 ex}}\alpha \sqrt {b \,x^{3}-a}}\right )}{27 b^{3} a}\)	\(510\)

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*(I*3^(1/2)*(a*b^2)^(1/3)+(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*(-I*3^(1/2)*(a*b^2)^(1/3)+(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))*Elli
pticPi(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)*_alpha-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}{\sqrt {- a + b x^{3}} \left (- 6 \sqrt {3} a - 10 a + b x^{3}\right )}\, 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/(sqrt(-a + b*x**3)*(-6*sqrt(3)*a - 10*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/((b*x^3 - a)^(1/2)*(b*x^3 - 2*a*(3*3^(1/2) + 5))),x)

[Out]

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

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