3.1230 \(\int \frac{(A+B x) (d+e x)^{5/2}}{b x+c x^2} \, dx\)

Optimal. Leaf size=173 \[ \frac{2 (d+e x)^{3/2} (A c e-b B e+B c d)}{3 c^2}+\frac{2 \sqrt{d+e x} \left (A c e (2 c d-b e)+B (c d-b e)^2\right )}{c^3}-\frac{2 (b B-A c) (c d-b e)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b c^{7/2}}-\frac{2 A d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b}+\frac{2 B (d+e x)^{5/2}}{5 c} \]

[Out]

(2*(B*(c*d - b*e)^2 + A*c*e*(2*c*d - b*e))*Sqrt[d + e*x])/c^3 + (2*(B*c*d - b*B*e + A*c*e)*(d + e*x)^(3/2))/(3
*c^2) + (2*B*(d + e*x)^(5/2))/(5*c) - (2*A*d^(5/2)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/b - (2*(b*B - A*c)*(c*d - b
*e)^(5/2)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(b*c^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.362408, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {824, 826, 1166, 208} \[ \frac{2 (d+e x)^{3/2} (A c e-b B e+B c d)}{3 c^2}+\frac{2 \sqrt{d+e x} \left (A c e (2 c d-b e)+B (c d-b e)^2\right )}{c^3}-\frac{2 (b B-A c) (c d-b e)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b c^{7/2}}-\frac{2 A d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b}+\frac{2 B (d+e x)^{5/2}}{5 c} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(d + e*x)^(5/2))/(b*x + c*x^2),x]

[Out]

(2*(B*(c*d - b*e)^2 + A*c*e*(2*c*d - b*e))*Sqrt[d + e*x])/c^3 + (2*(B*c*d - b*B*e + A*c*e)*(d + e*x)^(3/2))/(3
*c^2) + (2*B*(d + e*x)^(5/2))/(5*c) - (2*A*d^(5/2)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/b - (2*(b*B - A*c)*(c*d - b
*e)^(5/2)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(b*c^(7/2))

Rule 824

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

Rule 826

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

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{(A+B x) (d+e x)^{5/2}}{b x+c x^2} \, dx &=\frac{2 B (d+e x)^{5/2}}{5 c}+\frac{\int \frac{(d+e x)^{3/2} (A c d+(B c d-b B e+A c e) x)}{b x+c x^2} \, dx}{c}\\ &=\frac{2 (B c d-b B e+A c e) (d+e x)^{3/2}}{3 c^2}+\frac{2 B (d+e x)^{5/2}}{5 c}+\frac{\int \frac{\sqrt{d+e x} \left (A c^2 d^2+\left (B (c d-b e)^2+A c e (2 c d-b e)\right ) x\right )}{b x+c x^2} \, dx}{c^2}\\ &=\frac{2 \left (B (c d-b e)^2+A c e (2 c d-b e)\right ) \sqrt{d+e x}}{c^3}+\frac{2 (B c d-b B e+A c e) (d+e x)^{3/2}}{3 c^2}+\frac{2 B (d+e x)^{5/2}}{5 c}+\frac{\int \frac{A c^3 d^3+\left (B (c d-b e)^3+A c e \left (3 c^2 d^2-3 b c d e+b^2 e^2\right )\right ) x}{\sqrt{d+e x} \left (b x+c x^2\right )} \, dx}{c^3}\\ &=\frac{2 \left (B (c d-b e)^2+A c e (2 c d-b e)\right ) \sqrt{d+e x}}{c^3}+\frac{2 (B c d-b B e+A c e) (d+e x)^{3/2}}{3 c^2}+\frac{2 B (d+e x)^{5/2}}{5 c}+\frac{2 \operatorname{Subst}\left (\int \frac{A c^3 d^3 e-d \left (B (c d-b e)^3+A c e \left (3 c^2 d^2-3 b c d e+b^2 e^2\right )\right )+\left (B (c d-b e)^3+A c e \left (3 c^2 d^2-3 b c d e+b^2 e^2\right )\right ) x^2}{c d^2-b d e+(-2 c d+b e) x^2+c x^4} \, dx,x,\sqrt{d+e x}\right )}{c^3}\\ &=\frac{2 \left (B (c d-b e)^2+A c e (2 c d-b e)\right ) \sqrt{d+e x}}{c^3}+\frac{2 (B c d-b B e+A c e) (d+e x)^{3/2}}{3 c^2}+\frac{2 B (d+e x)^{5/2}}{5 c}+\frac{\left (2 A c d^3\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{b e}{2}+\frac{1}{2} (-2 c d+b e)+c x^2} \, dx,x,\sqrt{d+e x}\right )}{b}+\frac{\left (2 (b B-A c) (c d-b e)^3\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{b e}{2}+\frac{1}{2} (-2 c d+b e)+c x^2} \, dx,x,\sqrt{d+e x}\right )}{b c^3}\\ &=\frac{2 \left (B (c d-b e)^2+A c e (2 c d-b e)\right ) \sqrt{d+e x}}{c^3}+\frac{2 (B c d-b B e+A c e) (d+e x)^{3/2}}{3 c^2}+\frac{2 B (d+e x)^{5/2}}{5 c}-\frac{2 A d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b}-\frac{2 (b B-A c) (c d-b e)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.235833, size = 174, normalized size = 1.01 \[ \frac{2 \left (\frac{(b B-A c) \left (5 (c d-b e) \left (\sqrt{c} \sqrt{d+e x} (-3 b e+4 c d+c e x)-3 (c d-b e)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )\right )+3 c^{5/2} (d+e x)^{5/2}\right )}{c^{7/2}}+A \sqrt{d+e x} \left (23 d^2+11 d e x+3 e^2 x^2\right )-15 A d^{5/2} \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )\right )}{15 b} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(d + e*x)^(5/2))/(b*x + c*x^2),x]

[Out]

(2*(A*Sqrt[d + e*x]*(23*d^2 + 11*d*e*x + 3*e^2*x^2) - 15*A*d^(5/2)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] + ((b*B - A*
c)*(3*c^(5/2)*(d + e*x)^(5/2) + 5*(c*d - b*e)*(Sqrt[c]*Sqrt[d + e*x]*(4*c*d - 3*b*e + c*e*x) - 3*(c*d - b*e)^(
3/2)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])))/c^(7/2)))/(15*b)

________________________________________________________________________________________

Maple [B]  time = 0.014, size = 516, normalized size = 3. \begin{align*}{\frac{2\,B}{5\,c} \left ( ex+d \right ) ^{{\frac{5}{2}}}}+{\frac{2\,Ae}{3\,c} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{2\,bBe}{3\,{c}^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}+{\frac{2\,Bd}{3\,c} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-2\,{\frac{Ab{e}^{2}\sqrt{ex+d}}{{c}^{2}}}+4\,{\frac{Ade\sqrt{ex+d}}{c}}+2\,{\frac{B{e}^{2}{b}^{2}\sqrt{ex+d}}{{c}^{3}}}-4\,{\frac{bBde\sqrt{ex+d}}{{c}^{2}}}+2\,{\frac{B{d}^{2}\sqrt{ex+d}}{c}}-2\,{\frac{A{d}^{5/2}}{b}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) }+2\,{\frac{A{b}^{2}{e}^{3}}{{c}^{2}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-6\,{\frac{Abd{e}^{2}}{c\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }+6\,{\frac{A{d}^{2}e}{\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-2\,{\frac{A{d}^{3}c}{b\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-2\,{\frac{B{e}^{3}{b}^{3}}{{c}^{3}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }+6\,{\frac{{b}^{2}Bd{e}^{2}}{{c}^{2}\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-6\,{\frac{bB{d}^{2}e}{c\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }+2\,{\frac{B{d}^{3}}{\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^(5/2)/(c*x^2+b*x),x)

[Out]

2/5*B*(e*x+d)^(5/2)/c+2/3/c*A*(e*x+d)^(3/2)*e-2/3/c^2*B*(e*x+d)^(3/2)*b*e+2/3/c*B*(e*x+d)^(3/2)*d-2/c^2*A*b*e^
2*(e*x+d)^(1/2)+4/c*A*d*e*(e*x+d)^(1/2)+2/c^3*B*e^2*b^2*(e*x+d)^(1/2)-4/c^2*B*b*d*e*(e*x+d)^(1/2)+2/c*B*d^2*(e
*x+d)^(1/2)-2*A*d^(5/2)*arctanh((e*x+d)^(1/2)/d^(1/2))/b+2/c^2*b^2/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/
((b*e-c*d)*c)^(1/2))*A*e^3-6/c*b/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*A*d*e^2+6/((b
*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*A*d^2*e-2*c/b/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^
(1/2)*c/((b*e-c*d)*c)^(1/2))*A*d^3-2/c^3*b^3/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*B
*e^3+6/c^2*b^2/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*B*d*e^2-6/c*b/((b*e-c*d)*c)^(1/
2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*B*d^2*e+2/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)
*c)^(1/2))*B*d^3

________________________________________________________________________________________

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((B*x+A)*(e*x+d)^(5/2)/(c*x^2+b*x),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 20.3992, size = 2187, normalized size = 12.64 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(5/2)/(c*x^2+b*x),x, algorithm="fricas")

[Out]

[1/15*(15*A*c^3*d^(5/2)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + 15*((B*b*c^2 - A*c^3)*d^2 - 2*(B*b^2*c
- A*b*c^2)*d*e + (B*b^3 - A*b^2*c)*e^2)*sqrt((c*d - b*e)/c)*log((c*e*x + 2*c*d - b*e - 2*sqrt(e*x + d)*c*sqrt(
(c*d - b*e)/c))/(c*x + b)) + 2*(3*B*b*c^2*e^2*x^2 + 23*B*b*c^2*d^2 - 35*(B*b^2*c - A*b*c^2)*d*e + 15*(B*b^3 -
A*b^2*c)*e^2 + (11*B*b*c^2*d*e - 5*(B*b^2*c - A*b*c^2)*e^2)*x)*sqrt(e*x + d))/(b*c^3), 1/15*(15*A*c^3*d^(5/2)*
log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) - 30*((B*b*c^2 - A*c^3)*d^2 - 2*(B*b^2*c - A*b*c^2)*d*e + (B*b^3
- A*b^2*c)*e^2)*sqrt(-(c*d - b*e)/c)*arctan(-sqrt(e*x + d)*c*sqrt(-(c*d - b*e)/c)/(c*d - b*e)) + 2*(3*B*b*c^2*
e^2*x^2 + 23*B*b*c^2*d^2 - 35*(B*b^2*c - A*b*c^2)*d*e + 15*(B*b^3 - A*b^2*c)*e^2 + (11*B*b*c^2*d*e - 5*(B*b^2*
c - A*b*c^2)*e^2)*x)*sqrt(e*x + d))/(b*c^3), 1/15*(30*A*c^3*sqrt(-d)*d^2*arctan(sqrt(e*x + d)*sqrt(-d)/d) + 15
*((B*b*c^2 - A*c^3)*d^2 - 2*(B*b^2*c - A*b*c^2)*d*e + (B*b^3 - A*b^2*c)*e^2)*sqrt((c*d - b*e)/c)*log((c*e*x +
2*c*d - b*e - 2*sqrt(e*x + d)*c*sqrt((c*d - b*e)/c))/(c*x + b)) + 2*(3*B*b*c^2*e^2*x^2 + 23*B*b*c^2*d^2 - 35*(
B*b^2*c - A*b*c^2)*d*e + 15*(B*b^3 - A*b^2*c)*e^2 + (11*B*b*c^2*d*e - 5*(B*b^2*c - A*b*c^2)*e^2)*x)*sqrt(e*x +
d))/(b*c^3), 2/15*(15*A*c^3*sqrt(-d)*d^2*arctan(sqrt(e*x + d)*sqrt(-d)/d) - 15*((B*b*c^2 - A*c^3)*d^2 - 2*(B*
b^2*c - A*b*c^2)*d*e + (B*b^3 - A*b^2*c)*e^2)*sqrt(-(c*d - b*e)/c)*arctan(-sqrt(e*x + d)*c*sqrt(-(c*d - b*e)/c
)/(c*d - b*e)) + (3*B*b*c^2*e^2*x^2 + 23*B*b*c^2*d^2 - 35*(B*b^2*c - A*b*c^2)*d*e + 15*(B*b^3 - A*b^2*c)*e^2 +
(11*B*b*c^2*d*e - 5*(B*b^2*c - A*b*c^2)*e^2)*x)*sqrt(e*x + d))/(b*c^3)]

________________________________________________________________________________________

Sympy [A]  time = 121.973, size = 199, normalized size = 1.15 \begin{align*} \frac{2 A d^{3} \operatorname{atan}{\left (\frac{\sqrt{d + e x}}{\sqrt{- d}} \right )}}{b \sqrt{- d}} + \frac{2 B \left (d + e x\right )^{\frac{5}{2}}}{5 c} + \frac{\left (d + e x\right )^{\frac{3}{2}} \left (2 A c e - 2 B b e + 2 B c d\right )}{3 c^{2}} + \frac{\sqrt{d + e x} \left (- 2 A b c e^{2} + 4 A c^{2} d e + 2 B b^{2} e^{2} - 4 B b c d e + 2 B c^{2} d^{2}\right )}{c^{3}} - \frac{2 \left (- A c + B b\right ) \left (b e - c d\right )^{3} \operatorname{atan}{\left (\frac{\sqrt{d + e x}}{\sqrt{\frac{b e - c d}{c}}} \right )}}{b c^{4} \sqrt{\frac{b e - c d}{c}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**(5/2)/(c*x**2+b*x),x)

[Out]

2*A*d**3*atan(sqrt(d + e*x)/sqrt(-d))/(b*sqrt(-d)) + 2*B*(d + e*x)**(5/2)/(5*c) + (d + e*x)**(3/2)*(2*A*c*e -
2*B*b*e + 2*B*c*d)/(3*c**2) + sqrt(d + e*x)*(-2*A*b*c*e**2 + 4*A*c**2*d*e + 2*B*b**2*e**2 - 4*B*b*c*d*e + 2*B*
c**2*d**2)/c**3 - 2*(-A*c + B*b)*(b*e - c*d)**3*atan(sqrt(d + e*x)/sqrt((b*e - c*d)/c))/(b*c**4*sqrt((b*e - c*
d)/c))

________________________________________________________________________________________

Giac [B]  time = 1.45418, size = 427, normalized size = 2.47 \begin{align*} \frac{2 \, A d^{3} \arctan \left (\frac{\sqrt{x e + d}}{\sqrt{-d}}\right )}{b \sqrt{-d}} + \frac{2 \,{\left (B b c^{3} d^{3} - A c^{4} d^{3} - 3 \, B b^{2} c^{2} d^{2} e + 3 \, A b c^{3} d^{2} e + 3 \, B b^{3} c d e^{2} - 3 \, A b^{2} c^{2} d e^{2} - B b^{4} e^{3} + A b^{3} c e^{3}\right )} \arctan \left (\frac{\sqrt{x e + d} c}{\sqrt{-c^{2} d + b c e}}\right )}{\sqrt{-c^{2} d + b c e} b c^{3}} + \frac{2 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} B c^{4} + 5 \,{\left (x e + d\right )}^{\frac{3}{2}} B c^{4} d + 15 \, \sqrt{x e + d} B c^{4} d^{2} - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} B b c^{3} e + 5 \,{\left (x e + d\right )}^{\frac{3}{2}} A c^{4} e - 30 \, \sqrt{x e + d} B b c^{3} d e + 30 \, \sqrt{x e + d} A c^{4} d e + 15 \, \sqrt{x e + d} B b^{2} c^{2} e^{2} - 15 \, \sqrt{x e + d} A b c^{3} e^{2}\right )}}{15 \, c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(5/2)/(c*x^2+b*x),x, algorithm="giac")

[Out]

2*A*d^3*arctan(sqrt(x*e + d)/sqrt(-d))/(b*sqrt(-d)) + 2*(B*b*c^3*d^3 - A*c^4*d^3 - 3*B*b^2*c^2*d^2*e + 3*A*b*c
^3*d^2*e + 3*B*b^3*c*d*e^2 - 3*A*b^2*c^2*d*e^2 - B*b^4*e^3 + A*b^3*c*e^3)*arctan(sqrt(x*e + d)*c/sqrt(-c^2*d +
b*c*e))/(sqrt(-c^2*d + b*c*e)*b*c^3) + 2/15*(3*(x*e + d)^(5/2)*B*c^4 + 5*(x*e + d)^(3/2)*B*c^4*d + 15*sqrt(x*
e + d)*B*c^4*d^2 - 5*(x*e + d)^(3/2)*B*b*c^3*e + 5*(x*e + d)^(3/2)*A*c^4*e - 30*sqrt(x*e + d)*B*b*c^3*d*e + 30
*sqrt(x*e + d)*A*c^4*d*e + 15*sqrt(x*e + d)*B*b^2*c^2*e^2 - 15*sqrt(x*e + d)*A*b*c^3*e^2)/c^5