3.143 \(\int \frac {g+h x}{\sqrt [3]{\frac {-c^2 g^2+b c g h+2 b^2 h^2}{9 c h^2}+b x+c x^2} (\frac {f (b^2-\frac {-c^2 g^2+b c g h+2 b^2 h^2}{3 h^2})}{c^2}+\frac {b f x}{c}+f x^2)} \, dx\)

Optimal. Leaf size=488 \[ \frac {3^{2/3} h \sqrt [3]{\frac {c h^2 \left (\frac {(c g-2 b h) (b h+c g)}{c h^2}-9 b x-9 c x^2\right )}{(2 c g-b h)^2}} \log \left (\frac {f \left (b^2 h^2-b c g h+c^2 g^2\right )}{3 c^2 h^2}+\frac {b f x}{c}+f x^2\right )}{2 f \sqrt [3]{-\frac {(c g-2 b h) (b h+c g)}{c h^2}+9 b x+9 c x^2}}-\frac {3\ 3^{2/3} h \sqrt [3]{\frac {c h^2 \left (\frac {(c g-2 b h) (b h+c g)}{c h^2}-9 b x-9 c x^2\right )}{(2 c g-b h)^2}} \log \left (\left (1-\frac {3 h (b+2 c x)}{2 c g-b h}\right )^{2/3}+\sqrt [3]{2} \sqrt [3]{\frac {3 h (b+2 c x)}{2 c g-b h}+1}\right )}{2 f \sqrt [3]{-\frac {(c g-2 b h) (b h+c g)}{c h^2}+9 b x+9 c x^2}}+\frac {3 \sqrt [6]{3} h \sqrt [3]{\frac {c h^2 \left (\frac {(c g-2 b h) (b h+c g)}{c h^2}-9 b x-9 c x^2\right )}{(2 c g-b h)^2}} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2^{2/3} \left (1-\frac {3 h (b+2 c x)}{2 c g-b h}\right )^{2/3}}{\sqrt {3} \sqrt [3]{\frac {3 h (b+2 c x)}{2 c g-b h}+1}}\right )}{f \sqrt [3]{-\frac {(c g-2 b h) (b h+c g)}{c h^2}+9 b x+9 c x^2}} \]

[Out]

-3*3^(1/6)*h*(c*h^2*((-2*b*h+c*g)*(b*h+c*g)/c/h^2-9*b*x-9*c*x^2)/(-b*h+2*c*g)^2)^(1/3)*arctan(-1/3*3^(1/2)+1/3
*2^(2/3)*(1-3*h*(2*c*x+b)/(-b*h+2*c*g))^(2/3)/(1+3*h*(2*c*x+b)/(-b*h+2*c*g))^(1/3)*3^(1/2))/f/(-(-2*b*h+c*g)*(
b*h+c*g)/c/h^2+9*b*x+9*c*x^2)^(1/3)+1/2*3^(2/3)*h*(c*h^2*((-2*b*h+c*g)*(b*h+c*g)/c/h^2-9*b*x-9*c*x^2)/(-b*h+2*
c*g)^2)^(1/3)*ln(1/3*f*(b^2*h^2-b*c*g*h+c^2*g^2)/c^2/h^2+b*f*x/c+f*x^2)/f/(-(-2*b*h+c*g)*(b*h+c*g)/c/h^2+9*b*x
+9*c*x^2)^(1/3)-3/2*3^(2/3)*h*(c*h^2*((-2*b*h+c*g)*(b*h+c*g)/c/h^2-9*b*x-9*c*x^2)/(-b*h+2*c*g)^2)^(1/3)*ln((1-
3*h*(2*c*x+b)/(-b*h+2*c*g))^(2/3)+2^(1/3)*(1+3*h*(2*c*x+b)/(-b*h+2*c*g))^(1/3))/f/(-(-2*b*h+c*g)*(b*h+c*g)/c/h
^2+9*b*x+9*c*x^2)^(1/3)

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Rubi [A]  time = 0.36, antiderivative size = 488, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 104, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {1041, 1040} \[ \frac {3^{2/3} h \sqrt [3]{\frac {c h^2 \left (\frac {(c g-2 b h) (b h+c g)}{c h^2}-9 b x-9 c x^2\right )}{(2 c g-b h)^2}} \log \left (\frac {f \left (b^2 h^2-b c g h+c^2 g^2\right )}{3 c^2 h^2}+\frac {b f x}{c}+f x^2\right )}{2 f \sqrt [3]{-\frac {(c g-2 b h) (b h+c g)}{c h^2}+9 b x+9 c x^2}}-\frac {3\ 3^{2/3} h \sqrt [3]{\frac {c h^2 \left (\frac {(c g-2 b h) (b h+c g)}{c h^2}-9 b x-9 c x^2\right )}{(2 c g-b h)^2}} \log \left (\left (1-\frac {3 h (b+2 c x)}{2 c g-b h}\right )^{2/3}+\sqrt [3]{2} \sqrt [3]{\frac {3 h (b+2 c x)}{2 c g-b h}+1}\right )}{2 f \sqrt [3]{-\frac {(c g-2 b h) (b h+c g)}{c h^2}+9 b x+9 c x^2}}+\frac {3 \sqrt [6]{3} h \sqrt [3]{\frac {c h^2 \left (\frac {(c g-2 b h) (b h+c g)}{c h^2}-9 b x-9 c x^2\right )}{(2 c g-b h)^2}} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2^{2/3} \left (1-\frac {3 h (b+2 c x)}{2 c g-b h}\right )^{2/3}}{\sqrt {3} \sqrt [3]{\frac {3 h (b+2 c x)}{2 c g-b h}+1}}\right )}{f \sqrt [3]{-\frac {(c g-2 b h) (b h+c g)}{c h^2}+9 b x+9 c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(g + h*x)/(((-(c^2*g^2) + b*c*g*h + 2*b^2*h^2)/(9*c*h^2) + b*x + c*x^2)^(1/3)*((f*(b^2 - (-(c^2*g^2) + b*c
*g*h + 2*b^2*h^2)/(3*h^2)))/c^2 + (b*f*x)/c + f*x^2)),x]

[Out]

(3*3^(1/6)*h*((c*h^2*(((c*g - 2*b*h)*(c*g + b*h))/(c*h^2) - 9*b*x - 9*c*x^2))/(2*c*g - b*h)^2)^(1/3)*ArcTan[1/
Sqrt[3] - (2^(2/3)*(1 - (3*h*(b + 2*c*x))/(2*c*g - b*h))^(2/3))/(Sqrt[3]*(1 + (3*h*(b + 2*c*x))/(2*c*g - b*h))
^(1/3))])/(f*(-(((c*g - 2*b*h)*(c*g + b*h))/(c*h^2)) + 9*b*x + 9*c*x^2)^(1/3)) + (3^(2/3)*h*((c*h^2*(((c*g - 2
*b*h)*(c*g + b*h))/(c*h^2) - 9*b*x - 9*c*x^2))/(2*c*g - b*h)^2)^(1/3)*Log[(f*(c^2*g^2 - b*c*g*h + b^2*h^2))/(3
*c^2*h^2) + (b*f*x)/c + f*x^2])/(2*f*(-(((c*g - 2*b*h)*(c*g + b*h))/(c*h^2)) + 9*b*x + 9*c*x^2)^(1/3)) - (3*3^
(2/3)*h*((c*h^2*(((c*g - 2*b*h)*(c*g + b*h))/(c*h^2) - 9*b*x - 9*c*x^2))/(2*c*g - b*h)^2)^(1/3)*Log[(1 - (3*h*
(b + 2*c*x))/(2*c*g - b*h))^(2/3) + 2^(1/3)*(1 + (3*h*(b + 2*c*x))/(2*c*g - b*h))^(1/3)])/(2*f*(-(((c*g - 2*b*
h)*(c*g + b*h))/(c*h^2)) + 9*b*x + 9*c*x^2)^(1/3))

Rule 1040

Int[((g_.) + (h_.)*(x_))/(((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(1/3)*((d_.) + (e_.)*(x_) + (f_.)*(x_)^2)), x_Sy
mbol] :> With[{q = ((-9*c*h^2)/(2*c*g - b*h)^2)^(1/3)}, Simp[(Sqrt[3]*h*q*ArcTan[1/Sqrt[3] - (2^(2/3)*(1 - (3*
h*(b + 2*c*x))/(2*c*g - b*h))^(2/3))/(Sqrt[3]*(1 + (3*h*(b + 2*c*x))/(2*c*g - b*h))^(1/3))])/f, x] + (-Simp[(3
*h*q*Log[(1 - (3*h*(b + 2*c*x))/(2*c*g - b*h))^(2/3) + 2^(1/3)*(1 + (3*h*(b + 2*c*x))/(2*c*g - b*h))^(1/3)])/(
2*f), x] + Simp[(h*q*Log[d + e*x + f*x^2])/(2*f), x])] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && EqQ[c*e - b*f,
0] && EqQ[c^2*d - f*(b^2 - 3*a*c), 0] && EqQ[c^2*g^2 - b*c*g*h - 2*b^2*h^2 + 9*a*c*h^2, 0] && GtQ[(-9*c*h^2)/
(2*c*g - b*h)^2, 0]

Rule 1041

Int[((g_.) + (h_.)*(x_))/(((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(1/3)*((d_.) + (e_.)*(x_) + (f_.)*(x_)^2)), x_Sy
mbol] :> With[{q = -(c/(b^2 - 4*a*c))}, Dist[(q*(a + b*x + c*x^2))^(1/3)/(a + b*x + c*x^2)^(1/3), Int[(g + h*x
)/((q*a + b*q*x + c*q*x^2)^(1/3)*(d + e*x + f*x^2)), x], x]] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && EqQ[c*e
- b*f, 0] && EqQ[c^2*d - f*(b^2 - 3*a*c), 0] && EqQ[c^2*g^2 - b*c*g*h - 2*b^2*h^2 + 9*a*c*h^2, 0] &&  !GtQ[4*a
- b^2/c, 0]

Rubi steps

\begin {align*} \int \frac {g+h x}{\sqrt [3]{\frac {-c^2 g^2+b c g h+2 b^2 h^2}{9 c h^2}+b x+c x^2} \left (\frac {f \left (b^2-\frac {-c^2 g^2+b c g h+2 b^2 h^2}{3 h^2}\right )}{c^2}+\frac {b f x}{c}+f x^2\right )} \, dx &=\frac {\sqrt [3]{-\frac {c \left (\frac {-c^2 g^2+b c g h+2 b^2 h^2}{9 c h^2}+b x+c x^2\right )}{b^2-\frac {4 \left (-c^2 g^2+b c g h+2 b^2 h^2\right )}{9 h^2}}} \int \frac {g+h x}{\left (\frac {f \left (b^2-\frac {-c^2 g^2+b c g h+2 b^2 h^2}{3 h^2}\right )}{c^2}+\frac {b f x}{c}+f x^2\right ) \sqrt [3]{-\frac {-c^2 g^2+b c g h+2 b^2 h^2}{9 h^2 \left (b^2-\frac {4 \left (-c^2 g^2+b c g h+2 b^2 h^2\right )}{9 h^2}\right )}-\frac {b c x}{b^2-\frac {4 \left (-c^2 g^2+b c g h+2 b^2 h^2\right )}{9 h^2}}-\frac {c^2 x^2}{b^2-\frac {4 \left (-c^2 g^2+b c g h+2 b^2 h^2\right )}{9 h^2}}}} \, dx}{\sqrt [3]{\frac {-c^2 g^2+b c g h+2 b^2 h^2}{9 c h^2}+b x+c x^2}}\\ &=\frac {3 \sqrt [6]{3} h \sqrt [3]{\frac {c h^2 \left (\frac {(c g-2 b h) (c g+b h)}{c h^2}-9 b x-9 c x^2\right )}{(2 c g-b h)^2}} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2^{2/3} \left (1-\frac {3 h (b+2 c x)}{2 c g-b h}\right )^{2/3}}{\sqrt {3} \sqrt [3]{1+\frac {3 h (b+2 c x)}{2 c g-b h}}}\right )}{f \sqrt [3]{-\frac {(c g-2 b h) (c g+b h)}{c h^2}+9 b x+9 c x^2}}+\frac {3^{2/3} h \sqrt [3]{\frac {c h^2 \left (\frac {(c g-2 b h) (c g+b h)}{c h^2}-9 b x-9 c x^2\right )}{(2 c g-b h)^2}} \log \left (\frac {f \left (c^2 g^2-b c g h+b^2 h^2\right )}{3 c^2 h^2}+\frac {b f x}{c}+f x^2\right )}{2 f \sqrt [3]{-\frac {(c g-2 b h) (c g+b h)}{c h^2}+9 b x+9 c x^2}}-\frac {3\ 3^{2/3} h \sqrt [3]{\frac {c h^2 \left (\frac {(c g-2 b h) (c g+b h)}{c h^2}-9 b x-9 c x^2\right )}{(2 c g-b h)^2}} \log \left (\left (1-\frac {3 h (b+2 c x)}{2 c g-b h}\right )^{2/3}+\sqrt [3]{2} \sqrt [3]{1+\frac {3 h (b+2 c x)}{2 c g-b h}}\right )}{2 f \sqrt [3]{-\frac {(c g-2 b h) (c g+b h)}{c h^2}+9 b x+9 c x^2}}\\ \end {align*}

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Mathematica [F]  time = 0.54, size = 0, normalized size = 0.00 \[ \int \frac {g+h x}{\sqrt [3]{\frac {-c^2 g^2+b c g h+2 b^2 h^2}{9 c h^2}+b x+c x^2} \left (\frac {f \left (b^2-\frac {-c^2 g^2+b c g h+2 b^2 h^2}{3 h^2}\right )}{c^2}+\frac {b f x}{c}+f x^2\right )} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(g + h*x)/(((-(c^2*g^2) + b*c*g*h + 2*b^2*h^2)/(9*c*h^2) + b*x + c*x^2)^(1/3)*((f*(b^2 - (-(c^2*g^2)
+ b*c*g*h + 2*b^2*h^2)/(3*h^2)))/c^2 + (b*f*x)/c + f*x^2)),x]

[Out]

Integrate[(g + h*x)/(((-(c^2*g^2) + b*c*g*h + 2*b^2*h^2)/(9*c*h^2) + b*x + c*x^2)^(1/3)*((f*(b^2 - (-(c^2*g^2)
+ b*c*g*h + 2*b^2*h^2)/(3*h^2)))/c^2 + (b*f*x)/c + f*x^2)), x]

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x+g)/(1/9*(2*b^2*h^2+b*c*g*h-c^2*g^2)/c/h^2+b*x+c*x^2)^(1/3)/(f*(b^2+1/3*(-2*b^2*h^2-b*c*g*h+c^2*
g^2)/h^2)/c^2+b*f*x/c+f*x^2),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {3 \, {\left (h x + g\right )}}{{\left (c x^{2} + b x - \frac {c^{2} g^{2} - b c g h - 2 \, b^{2} h^{2}}{9 \, c h^{2}}\right )}^{\frac {1}{3}} {\left (3 \, f x^{2} + \frac {3 \, b f x}{c} + \frac {{\left (3 \, b^{2} + \frac {c^{2} g^{2} - b c g h - 2 \, b^{2} h^{2}}{h^{2}}\right )} f}{c^{2}}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x+g)/(1/9*(2*b^2*h^2+b*c*g*h-c^2*g^2)/c/h^2+b*x+c*x^2)^(1/3)/(f*(b^2+1/3*(-2*b^2*h^2-b*c*g*h+c^2*
g^2)/h^2)/c^2+b*f*x/c+f*x^2),x, algorithm="giac")

[Out]

integrate(3*(h*x + g)/((c*x^2 + b*x - 1/9*(c^2*g^2 - b*c*g*h - 2*b^2*h^2)/(c*h^2))^(1/3)*(3*f*x^2 + 3*b*f*x/c
+ (3*b^2 + (c^2*g^2 - b*c*g*h - 2*b^2*h^2)/h^2)*f/c^2)), x)

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maple [F]  time = 0.45, size = 0, normalized size = 0.00 \[ \int \frac {h x +g}{\left (c \,x^{2}+b x +\frac {2 b^{2} h^{2}+b c g h -c^{2} g^{2}}{9 c \,h^{2}}\right )^{\frac {1}{3}} \left (f \,x^{2}+\frac {b f x}{c}+\frac {\left (b^{2}+\frac {-2 b^{2} h^{2}-b c g h +c^{2} g^{2}}{3 h^{2}}\right ) f}{c^{2}}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((h*x+g)/(1/9*(2*b^2*h^2+b*c*g*h-c^2*g^2)/c/h^2+b*x+c*x^2)^(1/3)/(f*(b^2+1/3*(-2*b^2*h^2-b*c*g*h+c^2*g^2)/h
^2)/c^2+b*f*x/c+f*x^2),x)

[Out]

int((h*x+g)/(1/9*(2*b^2*h^2+b*c*g*h-c^2*g^2)/c/h^2+b*x+c*x^2)^(1/3)/(f*(b^2+1/3*(-2*b^2*h^2-b*c*g*h+c^2*g^2)/h
^2)/c^2+b*f*x/c+f*x^2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ 3 \, \int \frac {h x + g}{{\left (c x^{2} + b x - \frac {c^{2} g^{2} - b c g h - 2 \, b^{2} h^{2}}{9 \, c h^{2}}\right )}^{\frac {1}{3}} {\left (3 \, f x^{2} + \frac {3 \, b f x}{c} + \frac {{\left (3 \, b^{2} + \frac {c^{2} g^{2} - b c g h - 2 \, b^{2} h^{2}}{h^{2}}\right )} f}{c^{2}}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x+g)/(1/9*(2*b^2*h^2+b*c*g*h-c^2*g^2)/c/h^2+b*x+c*x^2)^(1/3)/(f*(b^2+1/3*(-2*b^2*h^2-b*c*g*h+c^2*
g^2)/h^2)/c^2+b*f*x/c+f*x^2),x, algorithm="maxima")

[Out]

3*integrate((h*x + g)/((c*x^2 + b*x - 1/9*(c^2*g^2 - b*c*g*h - 2*b^2*h^2)/(c*h^2))^(1/3)*(3*f*x^2 + 3*b*f*x/c
+ (3*b^2 + (c^2*g^2 - b*c*g*h - 2*b^2*h^2)/h^2)*f/c^2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {g+h\,x}{{\left (b\,x+c\,x^2+\frac {\frac {2\,b^2\,h^2}{9}+\frac {b\,c\,g\,h}{9}-\frac {c^2\,g^2}{9}}{c\,h^2}\right )}^{1/3}\,\left (f\,x^2-\frac {f\,\left (\frac {\frac {2\,b^2\,h^2}{3}+\frac {b\,c\,g\,h}{3}-\frac {c^2\,g^2}{3}}{h^2}-b^2\right )}{c^2}+\frac {b\,f\,x}{c}\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g + h*x)/((b*x + c*x^2 + ((2*b^2*h^2)/9 - (c^2*g^2)/9 + (b*c*g*h)/9)/(c*h^2))^(1/3)*(f*x^2 - (f*(((2*b^2*
h^2)/3 - (c^2*g^2)/3 + (b*c*g*h)/3)/h^2 - b^2))/c^2 + (b*f*x)/c)),x)

[Out]

int((g + h*x)/((b*x + c*x^2 + ((2*b^2*h^2)/9 - (c^2*g^2)/9 + (b*c*g*h)/9)/(c*h^2))^(1/3)*(f*x^2 - (f*(((2*b^2*
h^2)/3 - (c^2*g^2)/3 + (b*c*g*h)/3)/h^2 - b^2))/c^2 + (b*f*x)/c)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {3 \cdot 3^{\frac {2}{3}} c^{2} h^{2} \left (\int \frac {g}{b^{2} h^{2} \sqrt [3]{\frac {2 b^{2}}{c} + \frac {b g}{h} + 9 b x - \frac {c g^{2}}{h^{2}} + 9 c x^{2}} - b c g h \sqrt [3]{\frac {2 b^{2}}{c} + \frac {b g}{h} + 9 b x - \frac {c g^{2}}{h^{2}} + 9 c x^{2}} + 3 b c h^{2} x \sqrt [3]{\frac {2 b^{2}}{c} + \frac {b g}{h} + 9 b x - \frac {c g^{2}}{h^{2}} + 9 c x^{2}} + c^{2} g^{2} \sqrt [3]{\frac {2 b^{2}}{c} + \frac {b g}{h} + 9 b x - \frac {c g^{2}}{h^{2}} + 9 c x^{2}} + 3 c^{2} h^{2} x^{2} \sqrt [3]{\frac {2 b^{2}}{c} + \frac {b g}{h} + 9 b x - \frac {c g^{2}}{h^{2}} + 9 c x^{2}}}\, dx + \int \frac {h x}{b^{2} h^{2} \sqrt [3]{\frac {2 b^{2}}{c} + \frac {b g}{h} + 9 b x - \frac {c g^{2}}{h^{2}} + 9 c x^{2}} - b c g h \sqrt [3]{\frac {2 b^{2}}{c} + \frac {b g}{h} + 9 b x - \frac {c g^{2}}{h^{2}} + 9 c x^{2}} + 3 b c h^{2} x \sqrt [3]{\frac {2 b^{2}}{c} + \frac {b g}{h} + 9 b x - \frac {c g^{2}}{h^{2}} + 9 c x^{2}} + c^{2} g^{2} \sqrt [3]{\frac {2 b^{2}}{c} + \frac {b g}{h} + 9 b x - \frac {c g^{2}}{h^{2}} + 9 c x^{2}} + 3 c^{2} h^{2} x^{2} \sqrt [3]{\frac {2 b^{2}}{c} + \frac {b g}{h} + 9 b x - \frac {c g^{2}}{h^{2}} + 9 c x^{2}}}\, dx\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((h*x+g)/(1/9*(2*b**2*h**2+b*c*g*h-c**2*g**2)/c/h**2+b*x+c*x**2)**(1/3)/(f*(b**2+1/3*(-2*b**2*h**2-b*
c*g*h+c**2*g**2)/h**2)/c**2+b*f*x/c+f*x**2),x)

[Out]

3*3**(2/3)*c**2*h**2*(Integral(g/(b**2*h**2*(2*b**2/c + b*g/h + 9*b*x - c*g**2/h**2 + 9*c*x**2)**(1/3) - b*c*g
*h*(2*b**2/c + b*g/h + 9*b*x - c*g**2/h**2 + 9*c*x**2)**(1/3) + 3*b*c*h**2*x*(2*b**2/c + b*g/h + 9*b*x - c*g**
2/h**2 + 9*c*x**2)**(1/3) + c**2*g**2*(2*b**2/c + b*g/h + 9*b*x - c*g**2/h**2 + 9*c*x**2)**(1/3) + 3*c**2*h**2
*x**2*(2*b**2/c + b*g/h + 9*b*x - c*g**2/h**2 + 9*c*x**2)**(1/3)), x) + Integral(h*x/(b**2*h**2*(2*b**2/c + b*
g/h + 9*b*x - c*g**2/h**2 + 9*c*x**2)**(1/3) - b*c*g*h*(2*b**2/c + b*g/h + 9*b*x - c*g**2/h**2 + 9*c*x**2)**(1
/3) + 3*b*c*h**2*x*(2*b**2/c + b*g/h + 9*b*x - c*g**2/h**2 + 9*c*x**2)**(1/3) + c**2*g**2*(2*b**2/c + b*g/h +
9*b*x - c*g**2/h**2 + 9*c*x**2)**(1/3) + 3*c**2*h**2*x**2*(2*b**2/c + b*g/h + 9*b*x - c*g**2/h**2 + 9*c*x**2)*
*(1/3)), x))/f

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