### 3.108 $$\int \frac{(g+h x)^3 (d+e x+f x^2)}{(a+c x^2)^{3/2}} \, dx$$

Optimal. Leaf size=229 $-\frac{h \sqrt{a+c x^2} \left (4 \left (4 a^2 f h^2-a c \left (3 h (d h+3 e g)+7 f g^2\right )+3 c^2 d g^2\right )+c h x (-9 a e h-11 a f g+6 c d g)\right )}{6 a c^3}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right ) \left (3 a h^2 (e h+3 f g)-2 c g \left (3 h (d h+e g)+f g^2\right )\right )}{2 c^{5/2}}-\frac{h \sqrt{a+c x^2} (g+h x)^2 (3 c d-4 a f)}{3 a c^2}-\frac{(g+h x)^3 (a e-x (c d-a f))}{a c \sqrt{a+c x^2}}$

[Out]

-(((a*e - (c*d - a*f)*x)*(g + h*x)^3)/(a*c*Sqrt[a + c*x^2])) - ((3*c*d - 4*a*f)*h*(g + h*x)^2*Sqrt[a + c*x^2])
/(3*a*c^2) - (h*(4*(3*c^2*d*g^2 + 4*a^2*f*h^2 - a*c*(7*f*g^2 + 3*h*(3*e*g + d*h))) + c*h*(6*c*d*g - 11*a*f*g -
9*a*e*h)*x)*Sqrt[a + c*x^2])/(6*a*c^3) - ((3*a*h^2*(3*f*g + e*h) - 2*c*g*(f*g^2 + 3*h*(e*g + d*h)))*ArcTanh[(
Sqrt[c]*x)/Sqrt[a + c*x^2]])/(2*c^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 0.324773, antiderivative size = 228, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 29, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.172, Rules used = {1645, 833, 780, 217, 206} $-\frac{h \sqrt{a+c x^2} \left (4 \left (4 a^2 f h^2-a c \left (3 h (d h+3 e g)+7 f g^2\right )+3 c^2 d g^2\right )+c h x (-9 a e h-11 a f g+6 c d g)\right )}{6 a c^3}+\frac{\tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right ) \left (-3 a h^2 (e h+3 f g)+6 c g h (d h+e g)+2 c f g^3\right )}{2 c^{5/2}}-\frac{h \sqrt{a+c x^2} (g+h x)^2 (3 c d-4 a f)}{3 a c^2}-\frac{(g+h x)^3 (a e-x (c d-a f))}{a c \sqrt{a+c x^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(((a*e - (c*d - a*f)*x)*(g + h*x)^3)/(a*c*Sqrt[a + c*x^2])) - ((3*c*d - 4*a*f)*h*(g + h*x)^2*Sqrt[a + c*x^2])
/(3*a*c^2) - (h*(4*(3*c^2*d*g^2 + 4*a^2*f*h^2 - a*c*(7*f*g^2 + 3*h*(3*e*g + d*h))) + c*h*(6*c*d*g - 11*a*f*g -
9*a*e*h)*x)*Sqrt[a + c*x^2])/(6*a*c^3) + ((2*c*f*g^3 + 6*c*g*h*(e*g + d*h) - 3*a*h^2*(3*f*g + e*h))*ArcTanh[(
Sqrt[c]*x)/Sqrt[a + c*x^2]])/(2*c^(5/2))

Rule 1645

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a + c*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + c
*x^2, x], x, 1]}, Simp[((d + e*x)^m*(a + c*x^2)^(p + 1)*(a*g - c*f*x))/(2*a*c*(p + 1)), x] + Dist[1/(2*a*c*(p
+ 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*c*(p + 1)*(d + e*x)*Q - a*e*g*m + c*d*f*(2*p
+ 3) + c*e*f*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && GtQ[m, 0] &&  !(IGtQ[m, 0] && RationalQ[a, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

Rule 833

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(g*(d + e*x)
^m*(a + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*
Simp[c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p
}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p]) &&  !(IGtQ[m, 0] && EqQ[f, 0])

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{(g+h x)^3 \left (d+e x+f x^2\right )}{\left (a+c x^2\right )^{3/2}} \, dx &=-\frac{(a e-(c d-a f) x) (g+h x)^3}{a c \sqrt{a+c x^2}}-\frac{\int \frac{(g+h x)^2 (-a (f g+3 e h)+(3 c d-4 a f) h x)}{\sqrt{a+c x^2}} \, dx}{a c}\\ &=-\frac{(a e-(c d-a f) x) (g+h x)^3}{a c \sqrt{a+c x^2}}-\frac{(3 c d-4 a f) h (g+h x)^2 \sqrt{a+c x^2}}{3 a c^2}-\frac{\int \frac{(g+h x) \left (-a \left (2 (3 c d-4 a f) h^2+3 c g (f g+3 e h)\right )+c h (6 c d g-11 a f g-9 a e h) x\right )}{\sqrt{a+c x^2}} \, dx}{3 a c^2}\\ &=-\frac{(a e-(c d-a f) x) (g+h x)^3}{a c \sqrt{a+c x^2}}-\frac{(3 c d-4 a f) h (g+h x)^2 \sqrt{a+c x^2}}{3 a c^2}-\frac{h \left (4 \left (3 c^2 d g^2+4 a^2 f h^2-a c \left (7 f g^2+3 h (3 e g+d h)\right )\right )+c h (6 c d g-11 a f g-9 a e h) x\right ) \sqrt{a+c x^2}}{6 a c^3}+\frac{\left (2 c f g^3+6 c g h (e g+d h)-3 a h^2 (3 f g+e h)\right ) \int \frac{1}{\sqrt{a+c x^2}} \, dx}{2 c^2}\\ &=-\frac{(a e-(c d-a f) x) (g+h x)^3}{a c \sqrt{a+c x^2}}-\frac{(3 c d-4 a f) h (g+h x)^2 \sqrt{a+c x^2}}{3 a c^2}-\frac{h \left (4 \left (3 c^2 d g^2+4 a^2 f h^2-a c \left (7 f g^2+3 h (3 e g+d h)\right )\right )+c h (6 c d g-11 a f g-9 a e h) x\right ) \sqrt{a+c x^2}}{6 a c^3}+\frac{\left (2 c f g^3+6 c g h (e g+d h)-3 a h^2 (3 f g+e h)\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{a+c x^2}}\right )}{2 c^2}\\ &=-\frac{(a e-(c d-a f) x) (g+h x)^3}{a c \sqrt{a+c x^2}}-\frac{(3 c d-4 a f) h (g+h x)^2 \sqrt{a+c x^2}}{3 a c^2}-\frac{h \left (4 \left (3 c^2 d g^2+4 a^2 f h^2-a c \left (7 f g^2+3 h (3 e g+d h)\right )\right )+c h (6 c d g-11 a f g-9 a e h) x\right ) \sqrt{a+c x^2}}{6 a c^3}+\frac{\left (2 c f g^3+6 c g h (e g+d h)-3 a h^2 (3 f g+e h)\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{2 c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.503031, size = 246, normalized size = 1.07 $\frac{\frac{a^2 c h \left (3 h (4 d h+3 e (4 g+h x))+f \left (36 g^2+27 g h x-8 h^2 x^2\right )\right )-16 a^3 f h^3+a c^2 \left (6 d h \left (-3 g^2-3 g h x+h^2 x^2\right )-3 e \left (6 g^2 h x+2 g^3-6 g h^2 x^2-h^3 x^3\right )+f x \left (18 g^2 h x-6 g^3+9 g h^2 x^2+2 h^3 x^3\right )\right )+6 c^3 d g^3 x}{a \sqrt{a+c x^2}}+3 \sqrt{c} \log \left (\sqrt{c} \sqrt{a+c x^2}+c x\right ) \left (-3 a h^2 (e h+3 f g)+6 c g h (d h+e g)+2 c f g^3\right )}{6 c^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((-16*a^3*f*h^3 + 6*c^3*d*g^3*x + a*c^2*(6*d*h*(-3*g^2 - 3*g*h*x + h^2*x^2) - 3*e*(2*g^3 + 6*g^2*h*x - 6*g*h^2
*x^2 - h^3*x^3) + f*x*(-6*g^3 + 18*g^2*h*x + 9*g*h^2*x^2 + 2*h^3*x^3)) + a^2*c*h*(f*(36*g^2 + 27*g*h*x - 8*h^2
*x^2) + 3*h*(4*d*h + 3*e*(4*g + h*x))))/(a*Sqrt[a + c*x^2]) + 3*Sqrt[c]*(2*c*f*g^3 + 6*c*g*h*(e*g + d*h) - 3*a
*h^2*(3*f*g + e*h))*Log[c*x + Sqrt[c]*Sqrt[a + c*x^2]])/(6*c^3)

________________________________________________________________________________________

Maple [B]  time = 0.06, size = 516, normalized size = 2.3 \begin{align*}{\frac{{h}^{3}f{x}^{4}}{3\,c}{\frac{1}{\sqrt{c{x}^{2}+a}}}}-{\frac{4\,a{h}^{3}f{x}^{2}}{3\,{c}^{2}}{\frac{1}{\sqrt{c{x}^{2}+a}}}}-{\frac{8\,{a}^{2}f{h}^{3}}{3\,{c}^{3}}{\frac{1}{\sqrt{c{x}^{2}+a}}}}+{\frac{{x}^{3}{h}^{3}e}{2\,c}{\frac{1}{\sqrt{c{x}^{2}+a}}}}+{\frac{3\,{x}^{3}g{h}^{2}f}{2\,c}{\frac{1}{\sqrt{c{x}^{2}+a}}}}+{\frac{3\,ax{h}^{3}e}{2\,{c}^{2}}{\frac{1}{\sqrt{c{x}^{2}+a}}}}+{\frac{9\,axg{h}^{2}f}{2\,{c}^{2}}{\frac{1}{\sqrt{c{x}^{2}+a}}}}-{\frac{3\,ae{h}^{3}}{2}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){c}^{-{\frac{5}{2}}}}-{\frac{9\,ag{h}^{2}f}{2}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){c}^{-{\frac{5}{2}}}}+{\frac{{x}^{2}{h}^{3}d}{c}{\frac{1}{\sqrt{c{x}^{2}+a}}}}+3\,{\frac{{x}^{2}g{h}^{2}e}{c\sqrt{c{x}^{2}+a}}}+3\,{\frac{{g}^{2}{x}^{2}hf}{c\sqrt{c{x}^{2}+a}}}+2\,{\frac{a{h}^{3}d}{{c}^{2}\sqrt{c{x}^{2}+a}}}+6\,{\frac{ag{h}^{2}e}{{c}^{2}\sqrt{c{x}^{2}+a}}}+6\,{\frac{a{g}^{2}hf}{{c}^{2}\sqrt{c{x}^{2}+a}}}-3\,{\frac{gx{h}^{2}d}{c\sqrt{c{x}^{2}+a}}}-3\,{\frac{x{g}^{2}he}{c\sqrt{c{x}^{2}+a}}}-{\frac{x{g}^{3}f}{c}{\frac{1}{\sqrt{c{x}^{2}+a}}}}+3\,{\frac{\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ) g{h}^{2}d}{{c}^{3/2}}}+3\,{\frac{\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){g}^{2}he}{{c}^{3/2}}}+{{g}^{3}f\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){c}^{-{\frac{3}{2}}}}-3\,{\frac{{g}^{2}hd}{c\sqrt{c{x}^{2}+a}}}-{\frac{{g}^{3}e}{c}{\frac{1}{\sqrt{c{x}^{2}+a}}}}+{\frac{{g}^{3}dx}{a}{\frac{1}{\sqrt{c{x}^{2}+a}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*h^3*f*x^4/c/(c*x^2+a)^(1/2)-4/3*h^3*f*a/c^2*x^2/(c*x^2+a)^(1/2)-8/3*h^3*f*a^2/c^3/(c*x^2+a)^(1/2)+1/2*x^3/
c/(c*x^2+a)^(1/2)*h^3*e+3/2*x^3/c/(c*x^2+a)^(1/2)*g*h^2*f+3/2*a/c^2*x/(c*x^2+a)^(1/2)*h^3*e+9/2*a/c^2*x/(c*x^2
+a)^(1/2)*g*h^2*f-3/2*a/c^(5/2)*ln(x*c^(1/2)+(c*x^2+a)^(1/2))*h^3*e-9/2*a/c^(5/2)*ln(x*c^(1/2)+(c*x^2+a)^(1/2)
)*g*h^2*f+x^2/c/(c*x^2+a)^(1/2)*h^3*d+3*x^2/c/(c*x^2+a)^(1/2)*g*h^2*e+3*x^2/c/(c*x^2+a)^(1/2)*g^2*h*f+2*a/c^2/
(c*x^2+a)^(1/2)*h^3*d+6*a/c^2/(c*x^2+a)^(1/2)*g*h^2*e+6*a/c^2/(c*x^2+a)^(1/2)*g^2*h*f-3*x/c/(c*x^2+a)^(1/2)*g*
h^2*d-3*x/c/(c*x^2+a)^(1/2)*g^2*h*e-x/c/(c*x^2+a)^(1/2)*g^3*f+3/c^(3/2)*ln(x*c^(1/2)+(c*x^2+a)^(1/2))*g*h^2*d+
3/c^(3/2)*ln(x*c^(1/2)+(c*x^2+a)^(1/2))*g^2*h*e+1/c^(3/2)*ln(x*c^(1/2)+(c*x^2+a)^(1/2))*g^3*f-3/c/(c*x^2+a)^(1
/2)*g^2*h*d-1/c/(c*x^2+a)^(1/2)*g^3*e+g^3*d*x/a/(c*x^2+a)^(1/2)

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.20619, size = 1635, normalized size = 7.14 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/12*(3*(2*a^2*c*f*g^3 + 6*a^2*c*e*g^2*h - 3*a^3*e*h^3 + 3*(2*a^2*c*d - 3*a^3*f)*g*h^2 + (2*a*c^2*f*g^3 + 6*
a*c^2*e*g^2*h - 3*a^2*c*e*h^3 + 3*(2*a*c^2*d - 3*a^2*c*f)*g*h^2)*x^2)*sqrt(c)*log(-2*c*x^2 + 2*sqrt(c*x^2 + a)
*sqrt(c)*x - a) - 2*(2*a*c^2*f*h^3*x^4 - 6*a*c^2*e*g^3 + 36*a^2*c*e*g*h^2 - 18*(a*c^2*d - 2*a^2*c*f)*g^2*h + 4
*(3*a^2*c*d - 4*a^3*f)*h^3 + 3*(3*a*c^2*f*g*h^2 + a*c^2*e*h^3)*x^3 + 2*(9*a*c^2*f*g^2*h + 9*a*c^2*e*g*h^2 + (3
*a*c^2*d - 4*a^2*c*f)*h^3)*x^2 - 3*(6*a*c^2*e*g^2*h - 3*a^2*c*e*h^3 - 2*(c^3*d - a*c^2*f)*g^3 + 3*(2*a*c^2*d -
3*a^2*c*f)*g*h^2)*x)*sqrt(c*x^2 + a))/(a*c^4*x^2 + a^2*c^3), -1/6*(3*(2*a^2*c*f*g^3 + 6*a^2*c*e*g^2*h - 3*a^3
*e*h^3 + 3*(2*a^2*c*d - 3*a^3*f)*g*h^2 + (2*a*c^2*f*g^3 + 6*a*c^2*e*g^2*h - 3*a^2*c*e*h^3 + 3*(2*a*c^2*d - 3*a
^2*c*f)*g*h^2)*x^2)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (2*a*c^2*f*h^3*x^4 - 6*a*c^2*e*g^3 + 36*a^2*
c*e*g*h^2 - 18*(a*c^2*d - 2*a^2*c*f)*g^2*h + 4*(3*a^2*c*d - 4*a^3*f)*h^3 + 3*(3*a*c^2*f*g*h^2 + a*c^2*e*h^3)*x
^3 + 2*(9*a*c^2*f*g^2*h + 9*a*c^2*e*g*h^2 + (3*a*c^2*d - 4*a^2*c*f)*h^3)*x^2 - 3*(6*a*c^2*e*g^2*h - 3*a^2*c*e*
h^3 - 2*(c^3*d - a*c^2*f)*g^3 + 3*(2*a*c^2*d - 3*a^2*c*f)*g*h^2)*x)*sqrt(c*x^2 + a))/(a*c^4*x^2 + a^2*c^3)]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (g + h x\right )^{3} \left (d + e x + f x^{2}\right )}{\left (a + c x^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.20149, size = 458, normalized size = 2. \begin{align*} \frac{{\left ({\left ({\left (\frac{2 \, f h^{3} x}{c} + \frac{3 \,{\left (3 \, a c^{4} f g h^{2} + a c^{4} h^{3} e\right )}}{a c^{5}}\right )} x + \frac{2 \,{\left (9 \, a c^{4} f g^{2} h + 3 \, a c^{4} d h^{3} - 4 \, a^{2} c^{3} f h^{3} + 9 \, a c^{4} g h^{2} e\right )}}{a c^{5}}\right )} x + \frac{3 \,{\left (2 \, c^{5} d g^{3} - 2 \, a c^{4} f g^{3} - 6 \, a c^{4} d g h^{2} + 9 \, a^{2} c^{3} f g h^{2} - 6 \, a c^{4} g^{2} h e + 3 \, a^{2} c^{3} h^{3} e\right )}}{a c^{5}}\right )} x - \frac{2 \,{\left (9 \, a c^{4} d g^{2} h - 18 \, a^{2} c^{3} f g^{2} h - 6 \, a^{2} c^{3} d h^{3} + 8 \, a^{3} c^{2} f h^{3} + 3 \, a c^{4} g^{3} e - 18 \, a^{2} c^{3} g h^{2} e\right )}}{a c^{5}}}{6 \, \sqrt{c x^{2} + a}} - \frac{{\left (2 \, c f g^{3} + 6 \, c d g h^{2} - 9 \, a f g h^{2} + 6 \, c g^{2} h e - 3 \, a h^{3} e\right )} \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + a} \right |}\right )}{2 \, c^{\frac{5}{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*((((2*f*h^3*x/c + 3*(3*a*c^4*f*g*h^2 + a*c^4*h^3*e)/(a*c^5))*x + 2*(9*a*c^4*f*g^2*h + 3*a*c^4*d*h^3 - 4*a^
2*c^3*f*h^3 + 9*a*c^4*g*h^2*e)/(a*c^5))*x + 3*(2*c^5*d*g^3 - 2*a*c^4*f*g^3 - 6*a*c^4*d*g*h^2 + 9*a^2*c^3*f*g*h
^2 - 6*a*c^4*g^2*h*e + 3*a^2*c^3*h^3*e)/(a*c^5))*x - 2*(9*a*c^4*d*g^2*h - 18*a^2*c^3*f*g^2*h - 6*a^2*c^3*d*h^3
+ 8*a^3*c^2*f*h^3 + 3*a*c^4*g^3*e - 18*a^2*c^3*g*h^2*e)/(a*c^5))/sqrt(c*x^2 + a) - 1/2*(2*c*f*g^3 + 6*c*d*g*h
^2 - 9*a*f*g*h^2 + 6*c*g^2*h*e - 3*a*h^3*e)*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/c^(5/2)