3.26.88 \(\int \frac {1}{\sqrt [3]{b e-c e x} \sqrt [3]{b^2+b c x+c^2 x^2}} \, dx\) [2588]

Optimal. Leaf size=196 \[ -\frac {\sqrt [3]{b^3 e-c^3 e x^3} \tan ^{-1}\left (\frac {1-\frac {2 c \sqrt [3]{e} x}{\sqrt [3]{b^3 e-c^3 e x^3}}}{\sqrt {3}}\right )}{\sqrt {3} c \sqrt [3]{e} \sqrt [3]{b e-c e x} \sqrt [3]{b^2+b c x+c^2 x^2}}+\frac {\sqrt [3]{b^3 e-c^3 e x^3} \log \left (c \sqrt [3]{e} x+\sqrt [3]{b^3 e-c^3 e x^3}\right )}{2 c \sqrt [3]{e} \sqrt [3]{b e-c e x} \sqrt [3]{b^2+b c x+c^2 x^2}} \]

[Out]

1/2*(-c^3*e*x^3+b^3*e)^(1/3)*ln(c*e^(1/3)*x+(-c^3*e*x^3+b^3*e)^(1/3))/c/e^(1/3)/(-c*e*x+b*e)^(1/3)/(c^2*x^2+b*
c*x+b^2)^(1/3)-1/3*(-c^3*e*x^3+b^3*e)^(1/3)*arctan(1/3*(1-2*c*e^(1/3)*x/(-c^3*e*x^3+b^3*e)^(1/3))*3^(1/2))/c/e
^(1/3)/(-c*e*x+b*e)^(1/3)/(c^2*x^2+b*c*x+b^2)^(1/3)*3^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {727, 245} \begin {gather*} \frac {\sqrt [3]{b^3 e-c^3 e x^3} \log \left (\sqrt [3]{b^3 e-c^3 e x^3}+c \sqrt [3]{e} x\right )}{2 c \sqrt [3]{e} \sqrt [3]{b^2+b c x+c^2 x^2} \sqrt [3]{b e-c e x}}-\frac {\sqrt [3]{b^3 e-c^3 e x^3} \text {ArcTan}\left (\frac {1-\frac {2 c \sqrt [3]{e} x}{\sqrt [3]{b^3 e-c^3 e x^3}}}{\sqrt {3}}\right )}{\sqrt {3} c \sqrt [3]{e} \sqrt [3]{b^2+b c x+c^2 x^2} \sqrt [3]{b e-c e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((b^3*e - c^3*e*x^3)^(1/3)*ArcTan[(1 - (2*c*e^(1/3)*x)/(b^3*e - c^3*e*x^3)^(1/3))/Sqrt[3]])/(Sqrt[3]*c*e^(1/
3)*(b*e - c*e*x)^(1/3)*(b^2 + b*c*x + c^2*x^2)^(1/3))) + ((b^3*e - c^3*e*x^3)^(1/3)*Log[c*e^(1/3)*x + (b^3*e -
c^3*e*x^3)^(1/3)])/(2*c*e^(1/3)*(b*e - c*e*x)^(1/3)*(b^2 + b*c*x + c^2*x^2)^(1/3))

Rule 245

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + 2*Rt[b, 3]*(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sq
rt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]

Rule 727

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(d + e*x)^FracPart[p]
*((a + b*x + c*x^2)^FracPart[p]/(a*d + c*e*x^3)^FracPart[p]), Int[(d + e*x)^(m - p)*(a*d + c*e*x^3)^p, x], x]
/; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b*d + a*e, 0] && EqQ[c*d + b*e, 0] && IGtQ[m - p + 1, 0] &&  !Intege
rQ[p]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt [3]{b e-c e x} \sqrt [3]{b^2+b c x+c^2 x^2}} \, dx &=\frac {\sqrt [3]{b^3 e-c^3 e x^3} \int \frac {1}{\sqrt [3]{b^3 e-c^3 e x^3}} \, dx}{\sqrt [3]{b e-c e x} \sqrt [3]{b^2+b c x+c^2 x^2}}\\ &=-\frac {\sqrt [3]{b^3 e-c^3 e x^3} \tan ^{-1}\left (\frac {1-\frac {2 c \sqrt [3]{e} x}{\sqrt [3]{b^3 e-c^3 e x^3}}}{\sqrt {3}}\right )}{\sqrt {3} c \sqrt [3]{e} \sqrt [3]{b e-c e x} \sqrt [3]{b^2+b c x+c^2 x^2}}+\frac {\sqrt [3]{b^3 e-c^3 e x^3} \log \left (c \sqrt [3]{e} x+\sqrt [3]{b^3 e-c^3 e x^3}\right )}{2 c \sqrt [3]{e} \sqrt [3]{b e-c e x} \sqrt [3]{b^2+b c x+c^2 x^2}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
time = 10.14, size = 241, normalized size = 1.23 \begin {gather*} -\frac {3 (e (b-c x))^{2/3} \sqrt [3]{\frac {b c-\sqrt {3} \sqrt {-b^2 c^2}+2 c^2 x}{3 b c-\sqrt {3} \sqrt {-b^2 c^2}}} \sqrt [3]{\frac {b c+\sqrt {3} \sqrt {-b^2 c^2}+2 c^2 x}{3 b c+\sqrt {3} \sqrt {-b^2 c^2}}} F_1\left (\frac {2}{3};\frac {1}{3},\frac {1}{3};\frac {5}{3};\frac {2 c (b-c x)}{3 b c+\sqrt {3} \sqrt {-b^2 c^2}},\frac {2 c (b-c x)}{3 b c-\sqrt {3} \sqrt {-b^2 c^2}}\right )}{2 c e \sqrt [3]{b^2+b c x+c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(e*(b - c*x))^(2/3)*((b*c - Sqrt[3]*Sqrt[-(b^2*c^2)] + 2*c^2*x)/(3*b*c - Sqrt[3]*Sqrt[-(b^2*c^2)]))^(1/3)*
((b*c + Sqrt[3]*Sqrt[-(b^2*c^2)] + 2*c^2*x)/(3*b*c + Sqrt[3]*Sqrt[-(b^2*c^2)]))^(1/3)*AppellF1[2/3, 1/3, 1/3,
5/3, (2*c*(b - c*x))/(3*b*c + Sqrt[3]*Sqrt[-(b^2*c^2)]), (2*c*(b - c*x))/(3*b*c - Sqrt[3]*Sqrt[-(b^2*c^2)])])/
(2*c*e*(b^2 + b*c*x + c^2*x^2)^(1/3))

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Maple [F]
time = 0.48, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (-c e x +b e \right )^{\frac {1}{3}} \left (c^{2} x^{2}+b c x +b^{2}\right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate(1/((c^2*x^2 + b*c*x + b^2)^(1/3)*(-c*x*e + b*e)^(1/3)), x)

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Fricas [A]
time = 143.07, size = 218, normalized size = 1.11 \begin {gather*} -\frac {{\left (2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (c^{2} x^{2} + b c x + b^{2}\right )}^{\frac {2}{3}} \left (-{\left (c x - b\right )} e\right )^{\frac {2}{3}} c x e^{\frac {2}{3}} + {\left (c^{3} x^{3} - b^{3}\right )} e^{\frac {4}{3}}\right )} e^{\left (-\frac {4}{3}\right )}}{3 \, {\left (c^{3} x^{3} - b^{3}\right )}}\right ) e^{\frac {2}{3}} + e^{\frac {2}{3}} \log \left (c^{2} x^{2} e - {\left (c^{2} x^{2} + b c x + b^{2}\right )}^{\frac {1}{3}} \left (-{\left (c x - b\right )} e\right )^{\frac {1}{3}} c x e^{\frac {2}{3}} + {\left (c^{2} x^{2} + b c x + b^{2}\right )}^{\frac {2}{3}} \left (-{\left (c x - b\right )} e\right )^{\frac {2}{3}} e^{\frac {1}{3}}\right ) - 2 \, e^{\frac {2}{3}} \log \left (c x e + {\left (c^{2} x^{2} + b c x + b^{2}\right )}^{\frac {1}{3}} \left (-{\left (c x - b\right )} e\right )^{\frac {1}{3}} e^{\frac {2}{3}}\right )\right )} e^{\left (-1\right )}}{6 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*sqrt(3)*arctan(1/3*sqrt(3)*(2*(c^2*x^2 + b*c*x + b^2)^(2/3)*(-(c*x - b)*e)^(2/3)*c*x*e^(2/3) + (c^3*x^
3 - b^3)*e^(4/3))*e^(-4/3)/(c^3*x^3 - b^3))*e^(2/3) + e^(2/3)*log(c^2*x^2*e - (c^2*x^2 + b*c*x + b^2)^(1/3)*(-
(c*x - b)*e)^(1/3)*c*x*e^(2/3) + (c^2*x^2 + b*c*x + b^2)^(2/3)*(-(c*x - b)*e)^(2/3)*e^(1/3)) - 2*e^(2/3)*log(c
*x*e + (c^2*x^2 + b*c*x + b^2)^(1/3)*(-(c*x - b)*e)^(1/3)*e^(2/3)))*e^(-1)/c

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt [3]{- e \left (- b + c x\right )} \sqrt [3]{b^{2} + b c x + c^{2} x^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/((-e*(-b + c*x))**(1/3)*(b**2 + b*c*x + c**2*x**2)**(1/3)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((c^2*x^2 + b*c*x + b^2)^(1/3)*(-c*x*e + b*e)^(1/3)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (b\,e-c\,e\,x\right )}^{1/3}\,{\left (b^2+b\,c\,x+c^2\,x^2\right )}^{1/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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