3.535 \(\int (d+e x)^3 (a+c x^2)^{3/2} \, dx\)

Optimal. Leaf size=180 \[ \frac{3 a^2 d \left (2 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{16 c^{3/2}}+\frac{e \left (a+c x^2\right )^{5/2} \left (4 \left (8 c d^2-a e^2\right )+15 c d e x\right )}{70 c^2}+\frac{d x \left (a+c x^2\right )^{3/2} \left (2 c d^2-a e^2\right )}{8 c}+\frac{3 a d x \sqrt{a+c x^2} \left (2 c d^2-a e^2\right )}{16 c}+\frac{e \left (a+c x^2\right )^{5/2} (d+e x)^2}{7 c} \]

[Out]

(3*a*d*(2*c*d^2 - a*e^2)*x*Sqrt[a + c*x^2])/(16*c) + (d*(2*c*d^2 - a*e^2)*x*(a + c*x^2)^(3/2))/(8*c) + (e*(d +
e*x)^2*(a + c*x^2)^(5/2))/(7*c) + (e*(4*(8*c*d^2 - a*e^2) + 15*c*d*e*x)*(a + c*x^2)^(5/2))/(70*c^2) + (3*a^2*
d*(2*c*d^2 - a*e^2)*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(16*c^(3/2))

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Rubi [A]  time = 0.145263, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {743, 780, 195, 217, 206} \[ \frac{3 a^2 d \left (2 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{16 c^{3/2}}+\frac{e \left (a+c x^2\right )^{5/2} \left (4 \left (8 c d^2-a e^2\right )+15 c d e x\right )}{70 c^2}+\frac{d x \left (a+c x^2\right )^{3/2} \left (2 c d^2-a e^2\right )}{8 c}+\frac{3 a d x \sqrt{a+c x^2} \left (2 c d^2-a e^2\right )}{16 c}+\frac{e \left (a+c x^2\right )^{5/2} (d+e x)^2}{7 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3*(a + c*x^2)^(3/2),x]

[Out]

(3*a*d*(2*c*d^2 - a*e^2)*x*Sqrt[a + c*x^2])/(16*c) + (d*(2*c*d^2 - a*e^2)*x*(a + c*x^2)^(3/2))/(8*c) + (e*(d +
e*x)^2*(a + c*x^2)^(5/2))/(7*c) + (e*(4*(8*c*d^2 - a*e^2) + 15*c*d*e*x)*(a + c*x^2)^(5/2))/(70*c^2) + (3*a^2*
d*(2*c*d^2 - a*e^2)*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(16*c^(3/2))

Rule 743

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)*(a + c*x^2)^(p
+ 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - a*e^
2*(m - 1) + 2*c*d*e*(m + p)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2,
0] && If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, 0, c, d, e, m
, p, x]

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 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 (d+e x)^3 \left (a+c x^2\right )^{3/2} \, dx &=\frac{e (d+e x)^2 \left (a+c x^2\right )^{5/2}}{7 c}+\frac{\int (d+e x) \left (7 c d^2-2 a e^2+9 c d e x\right ) \left (a+c x^2\right )^{3/2} \, dx}{7 c}\\ &=\frac{e (d+e x)^2 \left (a+c x^2\right )^{5/2}}{7 c}+\frac{e \left (4 \left (8 c d^2-a e^2\right )+15 c d e x\right ) \left (a+c x^2\right )^{5/2}}{70 c^2}+\frac{\left (d \left (2 c d^2-a e^2\right )\right ) \int \left (a+c x^2\right )^{3/2} \, dx}{2 c}\\ &=\frac{d \left (2 c d^2-a e^2\right ) x \left (a+c x^2\right )^{3/2}}{8 c}+\frac{e (d+e x)^2 \left (a+c x^2\right )^{5/2}}{7 c}+\frac{e \left (4 \left (8 c d^2-a e^2\right )+15 c d e x\right ) \left (a+c x^2\right )^{5/2}}{70 c^2}+\frac{\left (3 a d \left (2 c d^2-a e^2\right )\right ) \int \sqrt{a+c x^2} \, dx}{8 c}\\ &=\frac{3 a d \left (2 c d^2-a e^2\right ) x \sqrt{a+c x^2}}{16 c}+\frac{d \left (2 c d^2-a e^2\right ) x \left (a+c x^2\right )^{3/2}}{8 c}+\frac{e (d+e x)^2 \left (a+c x^2\right )^{5/2}}{7 c}+\frac{e \left (4 \left (8 c d^2-a e^2\right )+15 c d e x\right ) \left (a+c x^2\right )^{5/2}}{70 c^2}+\frac{\left (3 a^2 d \left (2 c d^2-a e^2\right )\right ) \int \frac{1}{\sqrt{a+c x^2}} \, dx}{16 c}\\ &=\frac{3 a d \left (2 c d^2-a e^2\right ) x \sqrt{a+c x^2}}{16 c}+\frac{d \left (2 c d^2-a e^2\right ) x \left (a+c x^2\right )^{3/2}}{8 c}+\frac{e (d+e x)^2 \left (a+c x^2\right )^{5/2}}{7 c}+\frac{e \left (4 \left (8 c d^2-a e^2\right )+15 c d e x\right ) \left (a+c x^2\right )^{5/2}}{70 c^2}+\frac{\left (3 a^2 d \left (2 c d^2-a e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{a+c x^2}}\right )}{16 c}\\ &=\frac{3 a d \left (2 c d^2-a e^2\right ) x \sqrt{a+c x^2}}{16 c}+\frac{d \left (2 c d^2-a e^2\right ) x \left (a+c x^2\right )^{3/2}}{8 c}+\frac{e (d+e x)^2 \left (a+c x^2\right )^{5/2}}{7 c}+\frac{e \left (4 \left (8 c d^2-a e^2\right )+15 c d e x\right ) \left (a+c x^2\right )^{5/2}}{70 c^2}+\frac{3 a^2 d \left (2 c d^2-a e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{16 c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.121345, size = 174, normalized size = 0.97 \[ \frac{\sqrt{a+c x^2} \left (a^2 c e \left (336 d^2+105 d e x+16 e^2 x^2\right )-32 a^3 e^3+2 a c^2 x \left (336 d^2 e x+175 d^3+245 d e^2 x^2+64 e^3 x^3\right )+4 c^3 x^3 \left (84 d^2 e x+35 d^3+70 d e^2 x^2+20 e^3 x^3\right )\right )-105 a^2 \sqrt{c} d \left (a e^2-2 c d^2\right ) \log \left (\sqrt{c} \sqrt{a+c x^2}+c x\right )}{560 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + c*x^2]*(-32*a^3*e^3 + a^2*c*e*(336*d^2 + 105*d*e*x + 16*e^2*x^2) + 4*c^3*x^3*(35*d^3 + 84*d^2*e*x +
70*d*e^2*x^2 + 20*e^3*x^3) + 2*a*c^2*x*(175*d^3 + 336*d^2*e*x + 245*d*e^2*x^2 + 64*e^3*x^3)) - 105*a^2*Sqrt[c]
*d*(-2*c*d^2 + a*e^2)*Log[c*x + Sqrt[c]*Sqrt[a + c*x^2]])/(560*c^2)

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Maple [A]  time = 0.052, size = 205, normalized size = 1.1 \begin{align*}{\frac{{e}^{3}{x}^{2}}{7\,c} \left ( c{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{2\,a{e}^{3}}{35\,{c}^{2}} \left ( c{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{d{e}^{2}x}{2\,c} \left ( c{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{ad{e}^{2}x}{8\,c} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{3\,d{e}^{2}{a}^{2}x}{16\,c}\sqrt{c{x}^{2}+a}}-{\frac{3\,d{e}^{2}{a}^{3}}{16}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){c}^{-{\frac{3}{2}}}}+{\frac{3\,{d}^{2}e}{5\,c} \left ( c{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{{d}^{3}x}{4} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{3\,{d}^{3}ax}{8}\sqrt{c{x}^{2}+a}}+{\frac{3\,{d}^{3}{a}^{2}}{8}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){\frac{1}{\sqrt{c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*e^3*x^2*(c*x^2+a)^(5/2)/c-2/35*e^3*a/c^2*(c*x^2+a)^(5/2)+1/2*d*e^2*x*(c*x^2+a)^(5/2)/c-1/8*d*e^2*a/c*x*(c*
x^2+a)^(3/2)-3/16*d*e^2*a^2/c*x*(c*x^2+a)^(1/2)-3/16*d*e^2*a^3/c^(3/2)*ln(x*c^(1/2)+(c*x^2+a)^(1/2))+3/5*d^2*e
*(c*x^2+a)^(5/2)/c+1/4*d^3*x*(c*x^2+a)^(3/2)+3/8*d^3*a*x*(c*x^2+a)^(1/2)+3/8*d^3*a^2/c^(1/2)*ln(x*c^(1/2)+(c*x
^2+a)^(1/2))

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.27818, size = 902, normalized size = 5.01 \begin{align*} \left [\frac{105 \,{\left (2 \, a^{2} c d^{3} - a^{3} d e^{2}\right )} \sqrt{c} \log \left (-2 \, c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{c} x - a\right ) + 2 \,{\left (80 \, c^{3} e^{3} x^{6} + 280 \, c^{3} d e^{2} x^{5} + 336 \, a^{2} c d^{2} e - 32 \, a^{3} e^{3} + 16 \,{\left (21 \, c^{3} d^{2} e + 8 \, a c^{2} e^{3}\right )} x^{4} + 70 \,{\left (2 \, c^{3} d^{3} + 7 \, a c^{2} d e^{2}\right )} x^{3} + 16 \,{\left (42 \, a c^{2} d^{2} e + a^{2} c e^{3}\right )} x^{2} + 35 \,{\left (10 \, a c^{2} d^{3} + 3 \, a^{2} c d e^{2}\right )} x\right )} \sqrt{c x^{2} + a}}{1120 \, c^{2}}, -\frac{105 \,{\left (2 \, a^{2} c d^{3} - a^{3} d e^{2}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{-c} x}{\sqrt{c x^{2} + a}}\right ) -{\left (80 \, c^{3} e^{3} x^{6} + 280 \, c^{3} d e^{2} x^{5} + 336 \, a^{2} c d^{2} e - 32 \, a^{3} e^{3} + 16 \,{\left (21 \, c^{3} d^{2} e + 8 \, a c^{2} e^{3}\right )} x^{4} + 70 \,{\left (2 \, c^{3} d^{3} + 7 \, a c^{2} d e^{2}\right )} x^{3} + 16 \,{\left (42 \, a c^{2} d^{2} e + a^{2} c e^{3}\right )} x^{2} + 35 \,{\left (10 \, a c^{2} d^{3} + 3 \, a^{2} c d e^{2}\right )} x\right )} \sqrt{c x^{2} + a}}{560 \, c^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/1120*(105*(2*a^2*c*d^3 - a^3*d*e^2)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(80*c^3*e^3
*x^6 + 280*c^3*d*e^2*x^5 + 336*a^2*c*d^2*e - 32*a^3*e^3 + 16*(21*c^3*d^2*e + 8*a*c^2*e^3)*x^4 + 70*(2*c^3*d^3
+ 7*a*c^2*d*e^2)*x^3 + 16*(42*a*c^2*d^2*e + a^2*c*e^3)*x^2 + 35*(10*a*c^2*d^3 + 3*a^2*c*d*e^2)*x)*sqrt(c*x^2 +
a))/c^2, -1/560*(105*(2*a^2*c*d^3 - a^3*d*e^2)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (80*c^3*e^3*x^6
+ 280*c^3*d*e^2*x^5 + 336*a^2*c*d^2*e - 32*a^3*e^3 + 16*(21*c^3*d^2*e + 8*a*c^2*e^3)*x^4 + 70*(2*c^3*d^3 + 7*a
*c^2*d*e^2)*x^3 + 16*(42*a*c^2*d^2*e + a^2*c*e^3)*x^2 + 35*(10*a*c^2*d^3 + 3*a^2*c*d*e^2)*x)*sqrt(c*x^2 + a))/
c^2]

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Sympy [A]  time = 17.0949, size = 551, normalized size = 3.06 \begin{align*} \frac{3 a^{\frac{5}{2}} d e^{2} x}{16 c \sqrt{1 + \frac{c x^{2}}{a}}} + \frac{a^{\frac{3}{2}} d^{3} x \sqrt{1 + \frac{c x^{2}}{a}}}{2} + \frac{a^{\frac{3}{2}} d^{3} x}{8 \sqrt{1 + \frac{c x^{2}}{a}}} + \frac{17 a^{\frac{3}{2}} d e^{2} x^{3}}{16 \sqrt{1 + \frac{c x^{2}}{a}}} + \frac{3 \sqrt{a} c d^{3} x^{3}}{8 \sqrt{1 + \frac{c x^{2}}{a}}} + \frac{11 \sqrt{a} c d e^{2} x^{5}}{8 \sqrt{1 + \frac{c x^{2}}{a}}} - \frac{3 a^{3} d e^{2} \operatorname{asinh}{\left (\frac{\sqrt{c} x}{\sqrt{a}} \right )}}{16 c^{\frac{3}{2}}} + \frac{3 a^{2} d^{3} \operatorname{asinh}{\left (\frac{\sqrt{c} x}{\sqrt{a}} \right )}}{8 \sqrt{c}} + 3 a d^{2} e \left (\begin{cases} \frac{\sqrt{a} x^{2}}{2} & \text{for}\: c = 0 \\\frac{\left (a + c x^{2}\right )^{\frac{3}{2}}}{3 c} & \text{otherwise} \end{cases}\right ) + a e^{3} \left (\begin{cases} - \frac{2 a^{2} \sqrt{a + c x^{2}}}{15 c^{2}} + \frac{a x^{2} \sqrt{a + c x^{2}}}{15 c} + \frac{x^{4} \sqrt{a + c x^{2}}}{5} & \text{for}\: c \neq 0 \\\frac{\sqrt{a} x^{4}}{4} & \text{otherwise} \end{cases}\right ) + 3 c d^{2} e \left (\begin{cases} - \frac{2 a^{2} \sqrt{a + c x^{2}}}{15 c^{2}} + \frac{a x^{2} \sqrt{a + c x^{2}}}{15 c} + \frac{x^{4} \sqrt{a + c x^{2}}}{5} & \text{for}\: c \neq 0 \\\frac{\sqrt{a} x^{4}}{4} & \text{otherwise} \end{cases}\right ) + c e^{3} \left (\begin{cases} \frac{8 a^{3} \sqrt{a + c x^{2}}}{105 c^{3}} - \frac{4 a^{2} x^{2} \sqrt{a + c x^{2}}}{105 c^{2}} + \frac{a x^{4} \sqrt{a + c x^{2}}}{35 c} + \frac{x^{6} \sqrt{a + c x^{2}}}{7} & \text{for}\: c \neq 0 \\\frac{\sqrt{a} x^{6}}{6} & \text{otherwise} \end{cases}\right ) + \frac{c^{2} d^{3} x^{5}}{4 \sqrt{a} \sqrt{1 + \frac{c x^{2}}{a}}} + \frac{c^{2} d e^{2} x^{7}}{2 \sqrt{a} \sqrt{1 + \frac{c x^{2}}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*a**(5/2)*d*e**2*x/(16*c*sqrt(1 + c*x**2/a)) + a**(3/2)*d**3*x*sqrt(1 + c*x**2/a)/2 + a**(3/2)*d**3*x/(8*sqrt
(1 + c*x**2/a)) + 17*a**(3/2)*d*e**2*x**3/(16*sqrt(1 + c*x**2/a)) + 3*sqrt(a)*c*d**3*x**3/(8*sqrt(1 + c*x**2/a
)) + 11*sqrt(a)*c*d*e**2*x**5/(8*sqrt(1 + c*x**2/a)) - 3*a**3*d*e**2*asinh(sqrt(c)*x/sqrt(a))/(16*c**(3/2)) +
3*a**2*d**3*asinh(sqrt(c)*x/sqrt(a))/(8*sqrt(c)) + 3*a*d**2*e*Piecewise((sqrt(a)*x**2/2, Eq(c, 0)), ((a + c*x*
*2)**(3/2)/(3*c), True)) + a*e**3*Piecewise((-2*a**2*sqrt(a + c*x**2)/(15*c**2) + a*x**2*sqrt(a + c*x**2)/(15*
c) + x**4*sqrt(a + c*x**2)/5, Ne(c, 0)), (sqrt(a)*x**4/4, True)) + 3*c*d**2*e*Piecewise((-2*a**2*sqrt(a + c*x*
*2)/(15*c**2) + a*x**2*sqrt(a + c*x**2)/(15*c) + x**4*sqrt(a + c*x**2)/5, Ne(c, 0)), (sqrt(a)*x**4/4, True)) +
c*e**3*Piecewise((8*a**3*sqrt(a + c*x**2)/(105*c**3) - 4*a**2*x**2*sqrt(a + c*x**2)/(105*c**2) + a*x**4*sqrt(
a + c*x**2)/(35*c) + x**6*sqrt(a + c*x**2)/7, Ne(c, 0)), (sqrt(a)*x**6/6, True)) + c**2*d**3*x**5/(4*sqrt(a)*s
qrt(1 + c*x**2/a)) + c**2*d*e**2*x**7/(2*sqrt(a)*sqrt(1 + c*x**2/a))

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Giac [A]  time = 1.29844, size = 286, normalized size = 1.59 \begin{align*} \frac{1}{560} \, \sqrt{c x^{2} + a}{\left ({\left (2 \,{\left ({\left (4 \,{\left (5 \,{\left (2 \, c x e^{3} + 7 \, c d e^{2}\right )} x + \frac{2 \,{\left (21 \, c^{6} d^{2} e + 8 \, a c^{5} e^{3}\right )}}{c^{5}}\right )} x + \frac{35 \,{\left (2 \, c^{6} d^{3} + 7 \, a c^{5} d e^{2}\right )}}{c^{5}}\right )} x + \frac{8 \,{\left (42 \, a c^{5} d^{2} e + a^{2} c^{4} e^{3}\right )}}{c^{5}}\right )} x + \frac{35 \,{\left (10 \, a c^{5} d^{3} + 3 \, a^{2} c^{4} d e^{2}\right )}}{c^{5}}\right )} x + \frac{16 \,{\left (21 \, a^{2} c^{4} d^{2} e - 2 \, a^{3} c^{3} e^{3}\right )}}{c^{5}}\right )} - \frac{3 \,{\left (2 \, a^{2} c d^{3} - a^{3} d e^{2}\right )} \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + a} \right |}\right )}{16 \, c^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/560*sqrt(c*x^2 + a)*((2*((4*(5*(2*c*x*e^3 + 7*c*d*e^2)*x + 2*(21*c^6*d^2*e + 8*a*c^5*e^3)/c^5)*x + 35*(2*c^6
*d^3 + 7*a*c^5*d*e^2)/c^5)*x + 8*(42*a*c^5*d^2*e + a^2*c^4*e^3)/c^5)*x + 35*(10*a*c^5*d^3 + 3*a^2*c^4*d*e^2)/c
^5)*x + 16*(21*a^2*c^4*d^2*e - 2*a^3*c^3*e^3)/c^5) - 3/16*(2*a^2*c*d^3 - a^3*d*e^2)*log(abs(-sqrt(c)*x + sqrt(
c*x^2 + a)))/c^(3/2)