3.6.97 \(\int \frac {(d+e x)^2 (a+c x^2)}{(f+g x)^{3/2}} \, dx\) [597]

Optimal. Leaf size=173 \[ -\frac {2 (e f-d g)^2 \left (c f^2+a g^2\right )}{g^5 \sqrt {f+g x}}-\frac {4 (e f-d g) \left (a e g^2+c f (2 e f-d g)\right ) \sqrt {f+g x}}{g^5}+\frac {2 \left (a e^2 g^2+c \left (6 e^2 f^2-6 d e f g+d^2 g^2\right )\right ) (f+g x)^{3/2}}{3 g^5}-\frac {4 c e (2 e f-d g) (f+g x)^{5/2}}{5 g^5}+\frac {2 c e^2 (f+g x)^{7/2}}{7 g^5} \]

[Out]

2/3*(a*e^2*g^2+c*(d^2*g^2-6*d*e*f*g+6*e^2*f^2))*(g*x+f)^(3/2)/g^5-4/5*c*e*(-d*g+2*e*f)*(g*x+f)^(5/2)/g^5+2/7*c
*e^2*(g*x+f)^(7/2)/g^5-2*(-d*g+e*f)^2*(a*g^2+c*f^2)/g^5/(g*x+f)^(1/2)-4*(-d*g+e*f)*(a*e*g^2+c*f*(-d*g+2*e*f))*
(g*x+f)^(1/2)/g^5

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Rubi [A]
time = 0.13, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {912, 1275} \begin {gather*} \frac {2 (f+g x)^{3/2} \left (a e^2 g^2+c \left (d^2 g^2-6 d e f g+6 e^2 f^2\right )\right )}{3 g^5}-\frac {2 \left (a g^2+c f^2\right ) (e f-d g)^2}{g^5 \sqrt {f+g x}}-\frac {4 \sqrt {f+g x} (e f-d g) \left (a e g^2+c f (2 e f-d g)\right )}{g^5}-\frac {4 c e (f+g x)^{5/2} (2 e f-d g)}{5 g^5}+\frac {2 c e^2 (f+g x)^{7/2}}{7 g^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((d + e*x)^2*(a + c*x^2))/(f + g*x)^(3/2),x]

[Out]

(-2*(e*f - d*g)^2*(c*f^2 + a*g^2))/(g^5*Sqrt[f + g*x]) - (4*(e*f - d*g)*(a*e*g^2 + c*f*(2*e*f - d*g))*Sqrt[f +
g*x])/g^5 + (2*(a*e^2*g^2 + c*(6*e^2*f^2 - 6*d*e*f*g + d^2*g^2))*(f + g*x)^(3/2))/(3*g^5) - (4*c*e*(2*e*f - d
*g)*(f + g*x)^(5/2))/(5*g^5) + (2*c*e^2*(f + g*x)^(7/2))/(7*g^5)

Rule 912

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> With[{q = De
nominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + g*(x^q/e))^n*((c*d^2 + a*e^2)/e^2 - 2*c*
d*(x^q/e^2) + c*(x^(2*q)/e^2))^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*
g, 0] && NeQ[c*d^2 + a*e^2, 0] && IntegersQ[n, p] && FractionQ[m]

Rule 1275

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rubi steps

\begin {align*} \int \frac {(d+e x)^2 \left (a+c x^2\right )}{(f+g x)^{3/2}} \, dx &=\frac {2 \text {Subst}\left (\int \frac {\left (\frac {-e f+d g}{g}+\frac {e x^2}{g}\right )^2 \left (\frac {c f^2+a g^2}{g^2}-\frac {2 c f x^2}{g^2}+\frac {c x^4}{g^2}\right )}{x^2} \, dx,x,\sqrt {f+g x}\right )}{g}\\ &=\frac {2 \text {Subst}\left (\int \left (\frac {2 (e f-d g) \left (-a e g^2-c f (2 e f-d g)\right )}{g^4}+\frac {(-e f+d g)^2 \left (c f^2+a g^2\right )}{g^4 x^2}+\frac {\left (a e^2 g^2+c \left (6 e^2 f^2-6 d e f g+d^2 g^2\right )\right ) x^2}{g^4}-\frac {2 c e (2 e f-d g) x^4}{g^4}+\frac {c e^2 x^6}{g^4}\right ) \, dx,x,\sqrt {f+g x}\right )}{g}\\ &=-\frac {2 (e f-d g)^2 \left (c f^2+a g^2\right )}{g^5 \sqrt {f+g x}}-\frac {4 (e f-d g) \left (a e g^2+c f (2 e f-d g)\right ) \sqrt {f+g x}}{g^5}+\frac {2 \left (a e^2 g^2+c \left (6 e^2 f^2-6 d e f g+d^2 g^2\right )\right ) (f+g x)^{3/2}}{3 g^5}-\frac {4 c e (2 e f-d g) (f+g x)^{5/2}}{5 g^5}+\frac {2 c e^2 (f+g x)^{7/2}}{7 g^5}\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 177, normalized size = 1.02 \begin {gather*} \frac {-70 a g^2 \left (3 d^2 g^2-6 d e g (2 f+g x)+e^2 \left (8 f^2+4 f g x-g^2 x^2\right )\right )+2 c \left (35 d^2 g^2 \left (-8 f^2-4 f g x+g^2 x^2\right )+42 d e g \left (16 f^3+8 f^2 g x-2 f g^2 x^2+g^3 x^3\right )-3 e^2 \left (128 f^4+64 f^3 g x-16 f^2 g^2 x^2+8 f g^3 x^3-5 g^4 x^4\right )\right )}{105 g^5 \sqrt {f+g x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)^2*(a + c*x^2))/(f + g*x)^(3/2),x]

[Out]

(-70*a*g^2*(3*d^2*g^2 - 6*d*e*g*(2*f + g*x) + e^2*(8*f^2 + 4*f*g*x - g^2*x^2)) + 2*c*(35*d^2*g^2*(-8*f^2 - 4*f
*g*x + g^2*x^2) + 42*d*e*g*(16*f^3 + 8*f^2*g*x - 2*f*g^2*x^2 + g^3*x^3) - 3*e^2*(128*f^4 + 64*f^3*g*x - 16*f^2
*g^2*x^2 + 8*f*g^3*x^3 - 5*g^4*x^4)))/(105*g^5*Sqrt[f + g*x])

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Maple [A]
time = 0.09, size = 256, normalized size = 1.48

method	result	size
risch	\(\frac {2 \left (15 c \,e^{2} x^{3} g^{3}+42 c d e \,g^{3} x^{2}-39 c \,e^{2} f \,g^{2} x^{2}+35 a \,e^{2} g^{3} x +35 c \,d^{2} g^{3} x -126 c d e f \,g^{2} x +87 c \,e^{2} f^{2} g x +210 a d e \,g^{3}-175 a \,e^{2} f \,g^{2}-175 c \,d^{2} f \,g^{2}+462 c d e \,f^{2} g -279 c \,e^{2} f^{3}\right ) \sqrt {g x +f}}{105 g^{5}}-\frac {2 \left (a \,d^{2} g^{4}-2 a d e f \,g^{3}+a \,e^{2} f^{2} g^{2}+c \,d^{2} f^{2} g^{2}-2 c d e \,f^{3} g +c \,e^{2} f^{4}\right )}{g^{5} \sqrt {g x +f}}\)	\(207\)
gosper	\(-\frac {2 \left (-15 c \,e^{2} x^{4} g^{4}-42 c d e \,g^{4} x^{3}+24 c \,e^{2} f \,g^{3} x^{3}-35 a \,e^{2} g^{4} x^{2}-35 c \,d^{2} g^{4} x^{2}+84 c d e f \,g^{3} x^{2}-48 c \,e^{2} f^{2} g^{2} x^{2}-210 a d e \,g^{4} x +140 a \,e^{2} f \,g^{3} x +140 c \,d^{2} f \,g^{3} x -336 c d e \,f^{2} g^{2} x +192 c \,e^{2} f^{3} g x +105 a \,d^{2} g^{4}-420 a d e f \,g^{3}+280 a \,e^{2} f^{2} g^{2}+280 c \,d^{2} f^{2} g^{2}-672 c d e \,f^{3} g +384 c \,e^{2} f^{4}\right )}{105 \sqrt {g x +f}\, g^{5}}\)	\(215\)
trager	\(-\frac {2 \left (-15 c \,e^{2} x^{4} g^{4}-42 c d e \,g^{4} x^{3}+24 c \,e^{2} f \,g^{3} x^{3}-35 a \,e^{2} g^{4} x^{2}-35 c \,d^{2} g^{4} x^{2}+84 c d e f \,g^{3} x^{2}-48 c \,e^{2} f^{2} g^{2} x^{2}-210 a d e \,g^{4} x +140 a \,e^{2} f \,g^{3} x +140 c \,d^{2} f \,g^{3} x -336 c d e \,f^{2} g^{2} x +192 c \,e^{2} f^{3} g x +105 a \,d^{2} g^{4}-420 a d e f \,g^{3}+280 a \,e^{2} f^{2} g^{2}+280 c \,d^{2} f^{2} g^{2}-672 c d e \,f^{3} g +384 c \,e^{2} f^{4}\right )}{105 \sqrt {g x +f}\, g^{5}}\)	\(215\)
derivativedivides	\(\frac {\frac {2 c \,e^{2} \left (g x +f \right )^{\frac {7}{2}}}{7}+\frac {4 c d e g \left (g x +f \right )^{\frac {5}{2}}}{5}-\frac {8 c \,e^{2} f \left (g x +f \right )^{\frac {5}{2}}}{5}+\frac {2 a \,e^{2} g^{2} \left (g x +f \right )^{\frac {3}{2}}}{3}+\frac {2 c \,d^{2} g^{2} \left (g x +f \right )^{\frac {3}{2}}}{3}-4 c d e f g \left (g x +f \right )^{\frac {3}{2}}+4 c \,e^{2} f^{2} \left (g x +f \right )^{\frac {3}{2}}+4 a d e \,g^{3} \sqrt {g x +f}-4 a \,e^{2} f \,g^{2} \sqrt {g x +f}-4 c \,d^{2} f \,g^{2} \sqrt {g x +f}+12 c d e \,f^{2} g \sqrt {g x +f}-8 c \,e^{2} f^{3} \sqrt {g x +f}-\frac {2 \left (a \,d^{2} g^{4}-2 a d e f \,g^{3}+a \,e^{2} f^{2} g^{2}+c \,d^{2} f^{2} g^{2}-2 c d e \,f^{3} g +c \,e^{2} f^{4}\right )}{\sqrt {g x +f}}}{g^{5}}\)	\(256\)
default	\(\frac {\frac {2 c \,e^{2} \left (g x +f \right )^{\frac {7}{2}}}{7}+\frac {4 c d e g \left (g x +f \right )^{\frac {5}{2}}}{5}-\frac {8 c \,e^{2} f \left (g x +f \right )^{\frac {5}{2}}}{5}+\frac {2 a \,e^{2} g^{2} \left (g x +f \right )^{\frac {3}{2}}}{3}+\frac {2 c \,d^{2} g^{2} \left (g x +f \right )^{\frac {3}{2}}}{3}-4 c d e f g \left (g x +f \right )^{\frac {3}{2}}+4 c \,e^{2} f^{2} \left (g x +f \right )^{\frac {3}{2}}+4 a d e \,g^{3} \sqrt {g x +f}-4 a \,e^{2} f \,g^{2} \sqrt {g x +f}-4 c \,d^{2} f \,g^{2} \sqrt {g x +f}+12 c d e \,f^{2} g \sqrt {g x +f}-8 c \,e^{2} f^{3} \sqrt {g x +f}-\frac {2 \left (a \,d^{2} g^{4}-2 a d e f \,g^{3}+a \,e^{2} f^{2} g^{2}+c \,d^{2} f^{2} g^{2}-2 c d e \,f^{3} g +c \,e^{2} f^{4}\right )}{\sqrt {g x +f}}}{g^{5}}\)	\(256\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^2*(c*x^2+a)/(g*x+f)^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/g^5*(1/7*c*e^2*(g*x+f)^(7/2)+2/5*c*d*e*g*(g*x+f)^(5/2)-4/5*c*e^2*f*(g*x+f)^(5/2)+1/3*a*e^2*g^2*(g*x+f)^(3/2)
+1/3*c*d^2*g^2*(g*x+f)^(3/2)-2*c*d*e*f*g*(g*x+f)^(3/2)+2*c*e^2*f^2*(g*x+f)^(3/2)+2*a*d*e*g^3*(g*x+f)^(1/2)-2*a
*e^2*f*g^2*(g*x+f)^(1/2)-2*c*d^2*f*g^2*(g*x+f)^(1/2)+6*c*d*e*f^2*g*(g*x+f)^(1/2)-4*c*e^2*f^3*(g*x+f)^(1/2)-(a*
d^2*g^4-2*a*d*e*f*g^3+a*e^2*f^2*g^2+c*d^2*f^2*g^2-2*c*d*e*f^3*g+c*e^2*f^4)/(g*x+f)^(1/2))

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Maxima [A]
time = 0.31, size = 203, normalized size = 1.17 \begin {gather*} \frac {2 \, {\left (\frac {15 \, {\left (g x + f\right )}^{\frac {7}{2}} c e^{2} + 42 \, {\left (c d g e - 2 \, c f e^{2}\right )} {\left (g x + f\right )}^{\frac {5}{2}} - 35 \, {\left (6 \, c d f g e - 6 \, c f^{2} e^{2} - {\left (c d^{2} + a e^{2}\right )} g^{2}\right )} {\left (g x + f\right )}^{\frac {3}{2}} + 210 \, {\left (3 \, c d f^{2} g e + a d g^{3} e - 2 \, c f^{3} e^{2} - {\left (c d^{2} + a e^{2}\right )} f g^{2}\right )} \sqrt {g x + f}}{g^{4}} - \frac {105 \, {\left (a d^{2} g^{4} - 2 \, c d f^{3} g e - 2 \, a d f g^{3} e + c f^{4} e^{2} + {\left (c d^{2} + a e^{2}\right )} f^{2} g^{2}\right )}}{\sqrt {g x + f} g^{4}}\right )}}{105 \, g} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2*(c*x^2+a)/(g*x+f)^(3/2),x, algorithm="maxima")

[Out]

2/105*((15*(g*x + f)^(7/2)*c*e^2 + 42*(c*d*g*e - 2*c*f*e^2)*(g*x + f)^(5/2) - 35*(6*c*d*f*g*e - 6*c*f^2*e^2 -
(c*d^2 + a*e^2)*g^2)*(g*x + f)^(3/2) + 210*(3*c*d*f^2*g*e + a*d*g^3*e - 2*c*f^3*e^2 - (c*d^2 + a*e^2)*f*g^2)*s
qrt(g*x + f))/g^4 - 105*(a*d^2*g^4 - 2*c*d*f^3*g*e - 2*a*d*f*g^3*e + c*f^4*e^2 + (c*d^2 + a*e^2)*f^2*g^2)/(sqr
t(g*x + f)*g^4))/g

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Fricas [A]
time = 3.97, size = 204, normalized size = 1.18 \begin {gather*} \frac {2 \, {\left (35 \, c d^{2} g^{4} x^{2} - 140 \, c d^{2} f g^{3} x - 280 \, c d^{2} f^{2} g^{2} - 105 \, a d^{2} g^{4} + {\left (15 \, c g^{4} x^{4} - 24 \, c f g^{3} x^{3} - 384 \, c f^{4} - 280 \, a f^{2} g^{2} + {\left (48 \, c f^{2} g^{2} + 35 \, a g^{4}\right )} x^{2} - 4 \, {\left (48 \, c f^{3} g + 35 \, a f g^{3}\right )} x\right )} e^{2} + 42 \, {\left (c d g^{4} x^{3} - 2 \, c d f g^{3} x^{2} + 16 \, c d f^{3} g + 10 \, a d f g^{3} + {\left (8 \, c d f^{2} g^{2} + 5 \, a d g^{4}\right )} x\right )} e\right )} \sqrt {g x + f}}{105 \, {\left (g^{6} x + f g^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2*(c*x^2+a)/(g*x+f)^(3/2),x, algorithm="fricas")

[Out]

2/105*(35*c*d^2*g^4*x^2 - 140*c*d^2*f*g^3*x - 280*c*d^2*f^2*g^2 - 105*a*d^2*g^4 + (15*c*g^4*x^4 - 24*c*f*g^3*x
^3 - 384*c*f^4 - 280*a*f^2*g^2 + (48*c*f^2*g^2 + 35*a*g^4)*x^2 - 4*(48*c*f^3*g + 35*a*f*g^3)*x)*e^2 + 42*(c*d*
g^4*x^3 - 2*c*d*f*g^3*x^2 + 16*c*d*f^3*g + 10*a*d*f*g^3 + (8*c*d*f^2*g^2 + 5*a*d*g^4)*x)*e)*sqrt(g*x + f)/(g^6
*x + f*g^5)

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Sympy [A]
time = 15.17, size = 204, normalized size = 1.18 \begin {gather*} \frac {2 c e^{2} \left (f + g x\right )^{\frac {7}{2}}}{7 g^{5}} + \frac {\left (f + g x\right )^{\frac {5}{2}} \cdot \left (4 c d e g - 8 c e^{2} f\right )}{5 g^{5}} + \frac {\left (f + g x\right )^{\frac {3}{2}} \cdot \left (2 a e^{2} g^{2} + 2 c d^{2} g^{2} - 12 c d e f g + 12 c e^{2} f^{2}\right )}{3 g^{5}} + \frac {\sqrt {f + g x} \left (4 a d e g^{3} - 4 a e^{2} f g^{2} - 4 c d^{2} f g^{2} + 12 c d e f^{2} g - 8 c e^{2} f^{3}\right )}{g^{5}} - \frac {2 \left (a g^{2} + c f^{2}\right ) \left (d g - e f\right )^{2}}{g^{5} \sqrt {f + g x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**2*(c*x**2+a)/(g*x+f)**(3/2),x)

[Out]

2*c*e**2*(f + g*x)**(7/2)/(7*g**5) + (f + g*x)**(5/2)*(4*c*d*e*g - 8*c*e**2*f)/(5*g**5) + (f + g*x)**(3/2)*(2*
a*e**2*g**2 + 2*c*d**2*g**2 - 12*c*d*e*f*g + 12*c*e**2*f**2)/(3*g**5) + sqrt(f + g*x)*(4*a*d*e*g**3 - 4*a*e**2
*f*g**2 - 4*c*d**2*f*g**2 + 12*c*d*e*f**2*g - 8*c*e**2*f**3)/g**5 - 2*(a*g**2 + c*f**2)*(d*g - e*f)**2/(g**5*s
qrt(f + g*x))

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Giac [A]
time = 1.70, size = 275, normalized size = 1.59 \begin {gather*} -\frac {2 \, {\left (c d^{2} f^{2} g^{2} + a d^{2} g^{4} - 2 \, c d f^{3} g e - 2 \, a d f g^{3} e + c f^{4} e^{2} + a f^{2} g^{2} e^{2}\right )}}{\sqrt {g x + f} g^{5}} + \frac {2 \, {\left (35 \, {\left (g x + f\right )}^{\frac {3}{2}} c d^{2} g^{32} - 210 \, \sqrt {g x + f} c d^{2} f g^{32} + 42 \, {\left (g x + f\right )}^{\frac {5}{2}} c d g^{31} e - 210 \, {\left (g x + f\right )}^{\frac {3}{2}} c d f g^{31} e + 630 \, \sqrt {g x + f} c d f^{2} g^{31} e + 210 \, \sqrt {g x + f} a d g^{33} e + 15 \, {\left (g x + f\right )}^{\frac {7}{2}} c g^{30} e^{2} - 84 \, {\left (g x + f\right )}^{\frac {5}{2}} c f g^{30} e^{2} + 210 \, {\left (g x + f\right )}^{\frac {3}{2}} c f^{2} g^{30} e^{2} - 420 \, \sqrt {g x + f} c f^{3} g^{30} e^{2} + 35 \, {\left (g x + f\right )}^{\frac {3}{2}} a g^{32} e^{2} - 210 \, \sqrt {g x + f} a f g^{32} e^{2}\right )}}{105 \, g^{35}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2*(c*x^2+a)/(g*x+f)^(3/2),x, algorithm="giac")

[Out]

-2*(c*d^2*f^2*g^2 + a*d^2*g^4 - 2*c*d*f^3*g*e - 2*a*d*f*g^3*e + c*f^4*e^2 + a*f^2*g^2*e^2)/(sqrt(g*x + f)*g^5)
+ 2/105*(35*(g*x + f)^(3/2)*c*d^2*g^32 - 210*sqrt(g*x + f)*c*d^2*f*g^32 + 42*(g*x + f)^(5/2)*c*d*g^31*e - 210
*(g*x + f)^(3/2)*c*d*f*g^31*e + 630*sqrt(g*x + f)*c*d*f^2*g^31*e + 210*sqrt(g*x + f)*a*d*g^33*e + 15*(g*x + f)
^(7/2)*c*g^30*e^2 - 84*(g*x + f)^(5/2)*c*f*g^30*e^2 + 210*(g*x + f)^(3/2)*c*f^2*g^30*e^2 - 420*sqrt(g*x + f)*c
*f^3*g^30*e^2 + 35*(g*x + f)^(3/2)*a*g^32*e^2 - 210*sqrt(g*x + f)*a*f*g^32*e^2)/g^35

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Mupad [B]
time = 2.66, size = 199, normalized size = 1.15 \begin {gather*} \frac {{\left (f+g\,x\right )}^{3/2}\,\left (2\,c\,d^2\,g^2-12\,c\,d\,e\,f\,g+12\,c\,e^2\,f^2+2\,a\,e^2\,g^2\right )}{3\,g^5}-\frac {2\,c\,d^2\,f^2\,g^2+2\,a\,d^2\,g^4-4\,c\,d\,e\,f^3\,g-4\,a\,d\,e\,f\,g^3+2\,c\,e^2\,f^4+2\,a\,e^2\,f^2\,g^2}{g^5\,\sqrt {f+g\,x}}+\frac {4\,\sqrt {f+g\,x}\,\left (d\,g-e\,f\right )\,\left (2\,c\,e\,f^2-c\,d\,f\,g+a\,e\,g^2\right )}{g^5}+\frac {2\,c\,e^2\,{\left (f+g\,x\right )}^{7/2}}{7\,g^5}+\frac {4\,c\,e\,{\left (f+g\,x\right )}^{5/2}\,\left (d\,g-2\,e\,f\right )}{5\,g^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + c*x^2)*(d + e*x)^2)/(f + g*x)^(3/2),x)

[Out]

((f + g*x)^(3/2)*(2*a*e^2*g^2 + 2*c*d^2*g^2 + 12*c*e^2*f^2 - 12*c*d*e*f*g))/(3*g^5) - (2*a*d^2*g^4 + 2*c*e^2*f
^4 + 2*a*e^2*f^2*g^2 + 2*c*d^2*f^2*g^2 - 4*a*d*e*f*g^3 - 4*c*d*e*f^3*g)/(g^5*(f + g*x)^(1/2)) + (4*(f + g*x)^(
1/2)*(d*g - e*f)*(a*e*g^2 + 2*c*e*f^2 - c*d*f*g))/g^5 + (2*c*e^2*(f + g*x)^(7/2))/(7*g^5) + (4*c*e*(f + g*x)^(
5/2)*(d*g - 2*e*f))/(5*g^5)

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