### 3.285 $$\int (d+e x)^3 \sqrt{b x+c x^2} \, dx$$

Optimal. Leaf size=210 $\frac{e \left (b x+c x^2\right )^{3/2} \left (35 b^2 e^2+42 c e x (2 c d-b e)-150 b c d e+192 c^2 d^2\right )}{240 c^3}+\frac{(b+2 c x) \sqrt{b x+c x^2} (2 c d-b e) \left (7 b^2 e^2-16 b c d e+16 c^2 d^2\right )}{128 c^4}-\frac{b^2 (2 c d-b e) \left (7 b^2 e^2-16 b c d e+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{128 c^{9/2}}+\frac{e \left (b x+c x^2\right )^{3/2} (d+e x)^2}{5 c}$

[Out]

((2*c*d - b*e)*(16*c^2*d^2 - 16*b*c*d*e + 7*b^2*e^2)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(128*c^4) + (e*(d + e*x)^2
*(b*x + c*x^2)^(3/2))/(5*c) + (e*(192*c^2*d^2 - 150*b*c*d*e + 35*b^2*e^2 + 42*c*e*(2*c*d - b*e)*x)*(b*x + c*x^
2)^(3/2))/(240*c^3) - (b^2*(2*c*d - b*e)*(16*c^2*d^2 - 16*b*c*d*e + 7*b^2*e^2)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x +
c*x^2]])/(128*c^(9/2))

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Rubi [A]  time = 0.269013, antiderivative size = 210, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.238, Rules used = {742, 779, 612, 620, 206} $\frac{e \left (b x+c x^2\right )^{3/2} \left (35 b^2 e^2+42 c e x (2 c d-b e)-150 b c d e+192 c^2 d^2\right )}{240 c^3}+\frac{(b+2 c x) \sqrt{b x+c x^2} (2 c d-b e) \left (7 b^2 e^2-16 b c d e+16 c^2 d^2\right )}{128 c^4}-\frac{b^2 (2 c d-b e) \left (7 b^2 e^2-16 b c d e+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{128 c^{9/2}}+\frac{e \left (b x+c x^2\right )^{3/2} (d+e x)^2}{5 c}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^3*Sqrt[b*x + c*x^2],x]

[Out]

((2*c*d - b*e)*(16*c^2*d^2 - 16*b*c*d*e + 7*b^2*e^2)*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(128*c^4) + (e*(d + e*x)^2
*(b*x + c*x^2)^(3/2))/(5*c) + (e*(192*c^2*d^2 - 150*b*c*d*e + 35*b^2*e^2 + 42*c*e*(2*c*d - b*e)*x)*(b*x + c*x^
2)^(3/2))/(240*c^3) - (b^2*(2*c*d - b*e)*(16*c^2*d^2 - 16*b*c*d*e + 7*b^2*e^2)*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x +
c*x^2]])/(128*c^(9/2))

Rule 742

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + 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) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

Rule 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rule 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

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 \sqrt{b x+c x^2} \, dx &=\frac{e (d+e x)^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac{\int (d+e x) \left (\frac{1}{2} d (10 c d-3 b e)+\frac{7}{2} e (2 c d-b e) x\right ) \sqrt{b x+c x^2} \, dx}{5 c}\\ &=\frac{e (d+e x)^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac{e \left (192 c^2 d^2-150 b c d e+35 b^2 e^2+42 c e (2 c d-b e) x\right ) \left (b x+c x^2\right )^{3/2}}{240 c^3}+\frac{\left ((2 c d-b e) \left (16 c^2 d^2-16 b c d e+7 b^2 e^2\right )\right ) \int \sqrt{b x+c x^2} \, dx}{32 c^3}\\ &=\frac{(2 c d-b e) \left (16 c^2 d^2-16 b c d e+7 b^2 e^2\right ) (b+2 c x) \sqrt{b x+c x^2}}{128 c^4}+\frac{e (d+e x)^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac{e \left (192 c^2 d^2-150 b c d e+35 b^2 e^2+42 c e (2 c d-b e) x\right ) \left (b x+c x^2\right )^{3/2}}{240 c^3}-\frac{\left (b^2 (2 c d-b e) \left (16 c^2 d^2-16 b c d e+7 b^2 e^2\right )\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{256 c^4}\\ &=\frac{(2 c d-b e) \left (16 c^2 d^2-16 b c d e+7 b^2 e^2\right ) (b+2 c x) \sqrt{b x+c x^2}}{128 c^4}+\frac{e (d+e x)^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac{e \left (192 c^2 d^2-150 b c d e+35 b^2 e^2+42 c e (2 c d-b e) x\right ) \left (b x+c x^2\right )^{3/2}}{240 c^3}-\frac{\left (b^2 (2 c d-b e) \left (16 c^2 d^2-16 b c d e+7 b^2 e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{128 c^4}\\ &=\frac{(2 c d-b e) \left (16 c^2 d^2-16 b c d e+7 b^2 e^2\right ) (b+2 c x) \sqrt{b x+c x^2}}{128 c^4}+\frac{e (d+e x)^2 \left (b x+c x^2\right )^{3/2}}{5 c}+\frac{e \left (192 c^2 d^2-150 b c d e+35 b^2 e^2+42 c e (2 c d-b e) x\right ) \left (b x+c x^2\right )^{3/2}}{240 c^3}-\frac{b^2 (2 c d-b e) \left (16 c^2 d^2-16 b c d e+7 b^2 e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{128 c^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.536631, size = 230, normalized size = 1.1 $\frac{\sqrt{x (b+c x)} \left (\sqrt{c} \left (-4 b^2 c^2 e \left (180 d^2+75 d e x+14 e^2 x^2\right )+10 b^3 c e^2 (45 d+7 e x)-105 b^4 e^3+48 b c^3 \left (10 d^2 e x+10 d^3+5 d e^2 x^2+e^3 x^3\right )+96 c^4 x \left (20 d^2 e x+10 d^3+15 d e^2 x^2+4 e^3 x^3\right )\right )+\frac{15 b^{3/2} \left (-30 b^2 c d e^2+7 b^3 e^3+48 b c^2 d^2 e-32 c^3 d^3\right ) \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{x} \sqrt{\frac{c x}{b}+1}}\right )}{1920 c^{9/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^3*Sqrt[b*x + c*x^2],x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(-105*b^4*e^3 + 10*b^3*c*e^2*(45*d + 7*e*x) - 4*b^2*c^2*e*(180*d^2 + 75*d*e*x + 14
*e^2*x^2) + 48*b*c^3*(10*d^3 + 10*d^2*e*x + 5*d*e^2*x^2 + e^3*x^3) + 96*c^4*x*(10*d^3 + 20*d^2*e*x + 15*d*e^2*
x^2 + 4*e^3*x^3)) + (15*b^(3/2)*(-32*c^3*d^3 + 48*b*c^2*d^2*e - 30*b^2*c*d*e^2 + 7*b^3*e^3)*ArcSinh[(Sqrt[c]*S
qrt[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(1920*c^(9/2))

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Maple [B]  time = 0.053, size = 444, normalized size = 2.1 \begin{align*}{\frac{{e}^{3}{x}^{2}}{5\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{7\,b{e}^{3}x}{40\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{7\,{e}^{3}{b}^{2}}{48\,{c}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{7\,{b}^{3}{e}^{3}x}{64\,{c}^{3}}\sqrt{c{x}^{2}+bx}}-{\frac{7\,{e}^{3}{b}^{4}}{128\,{c}^{4}}\sqrt{c{x}^{2}+bx}}+{\frac{7\,{e}^{3}{b}^{5}}{256}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{9}{2}}}}+{\frac{3\,d{e}^{2}x}{4\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{5\,bd{e}^{2}}{8\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{15\,{b}^{2}d{e}^{2}x}{32\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{15\,{b}^{3}d{e}^{2}}{64\,{c}^{3}}\sqrt{c{x}^{2}+bx}}-{\frac{15\,d{e}^{2}{b}^{4}}{128}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{7}{2}}}}+{\frac{{d}^{2}e}{c} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{3\,b{d}^{2}ex}{4\,c}\sqrt{c{x}^{2}+bx}}-{\frac{3\,{d}^{2}e{b}^{2}}{8\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{3\,{d}^{2}e{b}^{3}}{16}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{5}{2}}}}+{\frac{{d}^{3}x}{2}\sqrt{c{x}^{2}+bx}}+{\frac{{d}^{3}b}{4\,c}\sqrt{c{x}^{2}+bx}}-{\frac{{d}^{3}{b}^{2}}{8}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*e^3*x^2*(c*x^2+b*x)^(3/2)/c-7/40*e^3*b/c^2*x*(c*x^2+b*x)^(3/2)+7/48*e^3*b^2/c^3*(c*x^2+b*x)^(3/2)-7/64*e^3
*b^3/c^3*x*(c*x^2+b*x)^(1/2)-7/128*e^3*b^4/c^4*(c*x^2+b*x)^(1/2)+7/256*e^3*b^5/c^(9/2)*ln((1/2*b+c*x)/c^(1/2)+
(c*x^2+b*x)^(1/2))+3/4*d*e^2*x*(c*x^2+b*x)^(3/2)/c-5/8*d*e^2*b/c^2*(c*x^2+b*x)^(3/2)+15/32*d*e^2*b^2/c^2*x*(c*
x^2+b*x)^(1/2)+15/64*d*e^2*b^3/c^3*(c*x^2+b*x)^(1/2)-15/128*d*e^2*b^4/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*
x)^(1/2))+d^2*e*(c*x^2+b*x)^(3/2)/c-3/4*d^2*e*b/c*x*(c*x^2+b*x)^(1/2)-3/8*d^2*e*b^2/c^2*(c*x^2+b*x)^(1/2)+3/16
*d^2*e*b^3/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))+1/2*d^3*x*(c*x^2+b*x)^(1/2)+1/4*d^3/c*(c*x^2+b*x)
^(1/2)*b-1/8*d^3*b^2/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(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+b*x)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.76282, size = 1112, normalized size = 5.3 \begin{align*} \left [-\frac{15 \,{\left (32 \, b^{2} c^{3} d^{3} - 48 \, b^{3} c^{2} d^{2} e + 30 \, b^{4} c d e^{2} - 7 \, b^{5} e^{3}\right )} \sqrt{c} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) - 2 \,{\left (384 \, c^{5} e^{3} x^{4} + 480 \, b c^{4} d^{3} - 720 \, b^{2} c^{3} d^{2} e + 450 \, b^{3} c^{2} d e^{2} - 105 \, b^{4} c e^{3} + 48 \,{\left (30 \, c^{5} d e^{2} + b c^{4} e^{3}\right )} x^{3} + 8 \,{\left (240 \, c^{5} d^{2} e + 30 \, b c^{4} d e^{2} - 7 \, b^{2} c^{3} e^{3}\right )} x^{2} + 10 \,{\left (96 \, c^{5} d^{3} + 48 \, b c^{4} d^{2} e - 30 \, b^{2} c^{3} d e^{2} + 7 \, b^{3} c^{2} e^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{3840 \, c^{5}}, \frac{15 \,{\left (32 \, b^{2} c^{3} d^{3} - 48 \, b^{3} c^{2} d^{2} e + 30 \, b^{4} c d e^{2} - 7 \, b^{5} e^{3}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) +{\left (384 \, c^{5} e^{3} x^{4} + 480 \, b c^{4} d^{3} - 720 \, b^{2} c^{3} d^{2} e + 450 \, b^{3} c^{2} d e^{2} - 105 \, b^{4} c e^{3} + 48 \,{\left (30 \, c^{5} d e^{2} + b c^{4} e^{3}\right )} x^{3} + 8 \,{\left (240 \, c^{5} d^{2} e + 30 \, b c^{4} d e^{2} - 7 \, b^{2} c^{3} e^{3}\right )} x^{2} + 10 \,{\left (96 \, c^{5} d^{3} + 48 \, b c^{4} d^{2} e - 30 \, b^{2} c^{3} d e^{2} + 7 \, b^{3} c^{2} e^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{1920 \, c^{5}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/3840*(15*(32*b^2*c^3*d^3 - 48*b^3*c^2*d^2*e + 30*b^4*c*d*e^2 - 7*b^5*e^3)*sqrt(c)*log(2*c*x + b + 2*sqrt(c
*x^2 + b*x)*sqrt(c)) - 2*(384*c^5*e^3*x^4 + 480*b*c^4*d^3 - 720*b^2*c^3*d^2*e + 450*b^3*c^2*d*e^2 - 105*b^4*c*
e^3 + 48*(30*c^5*d*e^2 + b*c^4*e^3)*x^3 + 8*(240*c^5*d^2*e + 30*b*c^4*d*e^2 - 7*b^2*c^3*e^3)*x^2 + 10*(96*c^5*
d^3 + 48*b*c^4*d^2*e - 30*b^2*c^3*d*e^2 + 7*b^3*c^2*e^3)*x)*sqrt(c*x^2 + b*x))/c^5, 1/1920*(15*(32*b^2*c^3*d^3
- 48*b^3*c^2*d^2*e + 30*b^4*c*d*e^2 - 7*b^5*e^3)*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (384*c^5
*e^3*x^4 + 480*b*c^4*d^3 - 720*b^2*c^3*d^2*e + 450*b^3*c^2*d*e^2 - 105*b^4*c*e^3 + 48*(30*c^5*d*e^2 + b*c^4*e^
3)*x^3 + 8*(240*c^5*d^2*e + 30*b*c^4*d*e^2 - 7*b^2*c^3*e^3)*x^2 + 10*(96*c^5*d^3 + 48*b*c^4*d^2*e - 30*b^2*c^3
*d*e^2 + 7*b^3*c^2*e^3)*x)*sqrt(c*x^2 + b*x))/c^5]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{x \left (b + c x\right )} \left (d + e x\right )^{3}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(x*(b + c*x))*(d + e*x)**3, x)

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Giac [A]  time = 1.31892, size = 338, normalized size = 1.61 \begin{align*} \frac{1}{1920} \, \sqrt{c x^{2} + b x}{\left (2 \,{\left (4 \,{\left (6 \,{\left (8 \, x e^{3} + \frac{30 \, c^{4} d e^{2} + b c^{3} e^{3}}{c^{4}}\right )} x + \frac{240 \, c^{4} d^{2} e + 30 \, b c^{3} d e^{2} - 7 \, b^{2} c^{2} e^{3}}{c^{4}}\right )} x + \frac{5 \,{\left (96 \, c^{4} d^{3} + 48 \, b c^{3} d^{2} e - 30 \, b^{2} c^{2} d e^{2} + 7 \, b^{3} c e^{3}\right )}}{c^{4}}\right )} x + \frac{15 \,{\left (32 \, b c^{3} d^{3} - 48 \, b^{2} c^{2} d^{2} e + 30 \, b^{3} c d e^{2} - 7 \, b^{4} e^{3}\right )}}{c^{4}}\right )} + \frac{{\left (32 \, b^{2} c^{3} d^{3} - 48 \, b^{3} c^{2} d^{2} e + 30 \, b^{4} c d e^{2} - 7 \, b^{5} e^{3}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{256 \, c^{\frac{9}{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1920*sqrt(c*x^2 + b*x)*(2*(4*(6*(8*x*e^3 + (30*c^4*d*e^2 + b*c^3*e^3)/c^4)*x + (240*c^4*d^2*e + 30*b*c^3*d*e
^2 - 7*b^2*c^2*e^3)/c^4)*x + 5*(96*c^4*d^3 + 48*b*c^3*d^2*e - 30*b^2*c^2*d*e^2 + 7*b^3*c*e^3)/c^4)*x + 15*(32*
b*c^3*d^3 - 48*b^2*c^2*d^2*e + 30*b^3*c*d*e^2 - 7*b^4*e^3)/c^4) + 1/256*(32*b^2*c^3*d^3 - 48*b^3*c^2*d^2*e + 3
0*b^4*c*d*e^2 - 7*b^5*e^3)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(9/2)