3.2158 \(\int (a+b x) (d+e x)^3 (a^2+2 a b x+b^2 x^2)^p \, dx\)

Optimal. Leaf size=183 \[ \frac{3 e^2 (a+b x)^4 (b d-a e) \left (a^2+2 a b x+b^2 x^2\right )^p}{2 b^4 (p+2)}+\frac{3 e (a+b x)^3 (b d-a e)^2 \left (a^2+2 a b x+b^2 x^2\right )^p}{b^4 (2 p+3)}+\frac{(a+b x)^2 (b d-a e)^3 \left (a^2+2 a b x+b^2 x^2\right )^p}{2 b^4 (p+1)}+\frac{e^3 (a+b x)^5 \left (a^2+2 a b x+b^2 x^2\right )^p}{b^4 (2 p+5)} \]

[Out]

((b*d - a*e)^3*(a + b*x)^2*(a^2 + 2*a*b*x + b^2*x^2)^p)/(2*b^4*(1 + p)) + (3*e*(b*d - a*e)^2*(a + b*x)^3*(a^2
+ 2*a*b*x + b^2*x^2)^p)/(b^4*(3 + 2*p)) + (3*e^2*(b*d - a*e)*(a + b*x)^4*(a^2 + 2*a*b*x + b^2*x^2)^p)/(2*b^4*(
2 + p)) + (e^3*(a + b*x)^5*(a^2 + 2*a*b*x + b^2*x^2)^p)/(b^4*(5 + 2*p))

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Rubi [A]  time = 0.119249, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {770, 21, 43} \[ \frac{3 e^2 (a+b x)^4 (b d-a e) \left (a^2+2 a b x+b^2 x^2\right )^p}{2 b^4 (p+2)}+\frac{3 e (a+b x)^3 (b d-a e)^2 \left (a^2+2 a b x+b^2 x^2\right )^p}{b^4 (2 p+3)}+\frac{(a+b x)^2 (b d-a e)^3 \left (a^2+2 a b x+b^2 x^2\right )^p}{2 b^4 (p+1)}+\frac{e^3 (a+b x)^5 \left (a^2+2 a b x+b^2 x^2\right )^p}{b^4 (2 p+5)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)*(d + e*x)^3*(a^2 + 2*a*b*x + b^2*x^2)^p,x]

[Out]

((b*d - a*e)^3*(a + b*x)^2*(a^2 + 2*a*b*x + b^2*x^2)^p)/(2*b^4*(1 + p)) + (3*e*(b*d - a*e)^2*(a + b*x)^3*(a^2
+ 2*a*b*x + b^2*x^2)^p)/(b^4*(3 + 2*p)) + (3*e^2*(b*d - a*e)*(a + b*x)^4*(a^2 + 2*a*b*x + b^2*x^2)^p)/(2*b^4*(
2 + p)) + (e^3*(a + b*x)^5*(a^2 + 2*a*b*x + b^2*x^2)^p)/(b^4*(5 + 2*p))

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int (a+b x) (d+e x)^3 \left (a^2+2 a b x+b^2 x^2\right )^p \, dx &=\left (\left (a b+b^2 x\right )^{-2 p} \left (a^2+2 a b x+b^2 x^2\right )^p\right ) \int (a+b x) \left (a b+b^2 x\right )^{2 p} (d+e x)^3 \, dx\\ &=\frac{\left (\left (a b+b^2 x\right )^{-2 p} \left (a^2+2 a b x+b^2 x^2\right )^p\right ) \int \left (a b+b^2 x\right )^{1+2 p} (d+e x)^3 \, dx}{b}\\ &=\frac{\left (\left (a b+b^2 x\right )^{-2 p} \left (a^2+2 a b x+b^2 x^2\right )^p\right ) \int \left (\frac{(b d-a e)^3 \left (a b+b^2 x\right )^{1+2 p}}{b^3}+\frac{3 e (b d-a e)^2 \left (a b+b^2 x\right )^{2+2 p}}{b^4}+\frac{3 e^2 (b d-a e) \left (a b+b^2 x\right )^{3+2 p}}{b^5}+\frac{e^3 \left (a b+b^2 x\right )^{4+2 p}}{b^6}\right ) \, dx}{b}\\ &=\frac{(b d-a e)^3 (a+b x)^2 \left (a^2+2 a b x+b^2 x^2\right )^p}{2 b^4 (1+p)}+\frac{3 e (b d-a e)^2 (a+b x)^3 \left (a^2+2 a b x+b^2 x^2\right )^p}{b^4 (3+2 p)}+\frac{3 e^2 (b d-a e) (a+b x)^4 \left (a^2+2 a b x+b^2 x^2\right )^p}{2 b^4 (2+p)}+\frac{e^3 (a+b x)^5 \left (a^2+2 a b x+b^2 x^2\right )^p}{b^4 (5+2 p)}\\ \end{align*}

Mathematica [A]  time = 0.0890187, size = 104, normalized size = 0.57 \[ \frac{\left ((a+b x)^2\right )^{p+1} \left (\frac{3 e^2 (a+b x)^2 (b d-a e)}{p+2}+\frac{6 e (a+b x) (b d-a e)^2}{2 p+3}+\frac{(b d-a e)^3}{p+1}+\frac{2 e^3 (a+b x)^3}{2 p+5}\right )}{2 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)*(d + e*x)^3*(a^2 + 2*a*b*x + b^2*x^2)^p,x]

[Out]

(((a + b*x)^2)^(1 + p)*((b*d - a*e)^3/(1 + p) + (6*e*(b*d - a*e)^2*(a + b*x))/(3 + 2*p) + (3*e^2*(b*d - a*e)*(
a + b*x)^2)/(2 + p) + (2*e^3*(a + b*x)^3)/(5 + 2*p)))/(2*b^4)

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Maple [B]  time = 0.008, size = 407, normalized size = 2.2 \begin{align*} -{\frac{ \left ({b}^{2}{x}^{2}+2\,abx+{a}^{2} \right ) ^{p} \left ( -4\,{b}^{3}{e}^{3}{p}^{3}{x}^{3}-12\,{b}^{3}d{e}^{2}{p}^{3}{x}^{2}-18\,{b}^{3}{e}^{3}{p}^{2}{x}^{3}+6\,a{b}^{2}{e}^{3}{p}^{2}{x}^{2}-12\,{b}^{3}{d}^{2}e{p}^{3}x-60\,{b}^{3}d{e}^{2}{p}^{2}{x}^{2}-26\,{b}^{3}{e}^{3}p{x}^{3}+12\,a{b}^{2}d{e}^{2}{p}^{2}x+15\,a{b}^{2}{e}^{3}p{x}^{2}-4\,{b}^{3}{d}^{3}{p}^{3}-66\,{b}^{3}{d}^{2}e{p}^{2}x-93\,{b}^{3}d{e}^{2}p{x}^{2}-12\,{x}^{3}{b}^{3}{e}^{3}-6\,{a}^{2}b{e}^{3}px+6\,a{b}^{2}{d}^{2}e{p}^{2}+42\,a{b}^{2}d{e}^{2}px+9\,{x}^{2}a{b}^{2}{e}^{3}-24\,{b}^{3}{d}^{3}{p}^{2}-114\,{b}^{3}{d}^{2}epx-45\,{x}^{2}{b}^{3}d{e}^{2}-6\,{a}^{2}bd{e}^{2}p-6\,x{a}^{2}b{e}^{3}+27\,a{b}^{2}{d}^{2}ep+30\,xa{b}^{2}d{e}^{2}-47\,{b}^{3}{d}^{3}p-60\,x{b}^{3}{d}^{2}e+3\,{e}^{3}{a}^{3}-15\,d{e}^{2}{a}^{2}b+30\,a{d}^{2}e{b}^{2}-30\,{d}^{3}{b}^{3} \right ) \left ( bx+a \right ) ^{2}}{2\,{b}^{4} \left ( 4\,{p}^{4}+28\,{p}^{3}+71\,{p}^{2}+77\,p+30 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*(e*x+d)^3*(b^2*x^2+2*a*b*x+a^2)^p,x)

[Out]

-1/2*(b^2*x^2+2*a*b*x+a^2)^p*(-4*b^3*e^3*p^3*x^3-12*b^3*d*e^2*p^3*x^2-18*b^3*e^3*p^2*x^3+6*a*b^2*e^3*p^2*x^2-1
2*b^3*d^2*e*p^3*x-60*b^3*d*e^2*p^2*x^2-26*b^3*e^3*p*x^3+12*a*b^2*d*e^2*p^2*x+15*a*b^2*e^3*p*x^2-4*b^3*d^3*p^3-
66*b^3*d^2*e*p^2*x-93*b^3*d*e^2*p*x^2-12*b^3*e^3*x^3-6*a^2*b*e^3*p*x+6*a*b^2*d^2*e*p^2+42*a*b^2*d*e^2*p*x+9*a*
b^2*e^3*x^2-24*b^3*d^3*p^2-114*b^3*d^2*e*p*x-45*b^3*d*e^2*x^2-6*a^2*b*d*e^2*p-6*a^2*b*e^3*x+27*a*b^2*d^2*e*p+3
0*a*b^2*d*e^2*x-47*b^3*d^3*p-60*b^3*d^2*e*x+3*a^3*e^3-15*a^2*b*d*e^2+30*a*b^2*d^2*e-30*b^3*d^3)*(b*x+a)^2/b^4/
(4*p^4+28*p^3+71*p^2+77*p+30)

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Maxima [B]  time = 1.28394, size = 917, normalized size = 5.01 \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)^3*(b^2*x^2+2*a*b*x+a^2)^p,x, algorithm="maxima")

[Out]

(b*x + a)*(b*x + a)^(2*p)*a*d^3/(b*(2*p + 1)) + 1/2*(b^2*(2*p + 1)*x^2 + 2*a*b*p*x - a^2)*(b*x + a)^(2*p)*d^3/
((2*p^2 + 3*p + 1)*b) + 3/2*(b^2*(2*p + 1)*x^2 + 2*a*b*p*x - a^2)*(b*x + a)^(2*p)*a*d^2*e/((2*p^2 + 3*p + 1)*b
^2) + 3*((2*p^2 + 3*p + 1)*b^3*x^3 + (2*p^2 + p)*a*b^2*x^2 - 2*a^2*b*p*x + a^3)*(b*x + a)^(2*p)*d^2*e/((4*p^3
+ 12*p^2 + 11*p + 3)*b^2) + 3*((2*p^2 + 3*p + 1)*b^3*x^3 + (2*p^2 + p)*a*b^2*x^2 - 2*a^2*b*p*x + a^3)*(b*x + a
)^(2*p)*a*d*e^2/((4*p^3 + 12*p^2 + 11*p + 3)*b^3) + 3/2*((4*p^3 + 12*p^2 + 11*p + 3)*b^4*x^4 + 2*(2*p^3 + 3*p^
2 + p)*a*b^3*x^3 - 3*(2*p^2 + p)*a^2*b^2*x^2 + 6*a^3*b*p*x - 3*a^4)*(b*x + a)^(2*p)*d*e^2/((4*p^4 + 20*p^3 + 3
5*p^2 + 25*p + 6)*b^3) + 1/2*((4*p^3 + 12*p^2 + 11*p + 3)*b^4*x^4 + 2*(2*p^3 + 3*p^2 + p)*a*b^3*x^3 - 3*(2*p^2
+ p)*a^2*b^2*x^2 + 6*a^3*b*p*x - 3*a^4)*(b*x + a)^(2*p)*a*e^3/((4*p^4 + 20*p^3 + 35*p^2 + 25*p + 6)*b^4) + ((
4*p^4 + 20*p^3 + 35*p^2 + 25*p + 6)*b^5*x^5 + (4*p^4 + 12*p^3 + 11*p^2 + 3*p)*a*b^4*x^4 - 4*(2*p^3 + 3*p^2 + p
)*a^2*b^3*x^3 + 6*(2*p^2 + p)*a^3*b^2*x^2 - 12*a^4*b*p*x + 6*a^5)*(b*x + a)^(2*p)*e^3/((8*p^5 + 60*p^4 + 170*p
^3 + 225*p^2 + 137*p + 30)*b^4)

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Fricas [B]  time = 1.15083, size = 1476, normalized size = 8.07 \begin{align*} \frac{{\left (4 \, a^{2} b^{3} d^{3} p^{3} + 30 \, a^{2} b^{3} d^{3} - 30 \, a^{3} b^{2} d^{2} e + 15 \, a^{4} b d e^{2} - 3 \, a^{5} e^{3} + 2 \,{\left (2 \, b^{5} e^{3} p^{3} + 9 \, b^{5} e^{3} p^{2} + 13 \, b^{5} e^{3} p + 6 \, b^{5} e^{3}\right )} x^{5} +{\left (45 \, b^{5} d e^{2} + 15 \, a b^{4} e^{3} + 4 \,{\left (3 \, b^{5} d e^{2} + 2 \, a b^{4} e^{3}\right )} p^{3} + 30 \,{\left (2 \, b^{5} d e^{2} + a b^{4} e^{3}\right )} p^{2} +{\left (93 \, b^{5} d e^{2} + 37 \, a b^{4} e^{3}\right )} p\right )} x^{4} + 2 \,{\left (30 \, b^{5} d^{2} e + 30 \, a b^{4} d e^{2} + 2 \,{\left (3 \, b^{5} d^{2} e + 6 \, a b^{4} d e^{2} + a^{2} b^{3} e^{3}\right )} p^{3} + 3 \,{\left (11 \, b^{5} d^{2} e + 18 \, a b^{4} d e^{2} + a^{2} b^{3} e^{3}\right )} p^{2} +{\left (57 \, b^{5} d^{2} e + 72 \, a b^{4} d e^{2} + a^{2} b^{3} e^{3}\right )} p\right )} x^{3} + 6 \,{\left (4 \, a^{2} b^{3} d^{3} - a^{3} b^{2} d^{2} e\right )} p^{2} +{\left (30 \, b^{5} d^{3} + 90 \, a b^{4} d^{2} e + 4 \,{\left (b^{5} d^{3} + 6 \, a b^{4} d^{2} e + 3 \, a^{2} b^{3} d e^{2}\right )} p^{3} + 6 \,{\left (4 \, b^{5} d^{3} + 21 \, a b^{4} d^{2} e + 6 \, a^{2} b^{3} d e^{2} - a^{3} b^{2} e^{3}\right )} p^{2} +{\left (47 \, b^{5} d^{3} + 201 \, a b^{4} d^{2} e + 15 \, a^{2} b^{3} d e^{2} - 3 \, a^{3} b^{2} e^{3}\right )} p\right )} x^{2} +{\left (47 \, a^{2} b^{3} d^{3} - 27 \, a^{3} b^{2} d^{2} e + 6 \, a^{4} b d e^{2}\right )} p + 2 \,{\left (30 \, a b^{4} d^{3} + 2 \,{\left (2 \, a b^{4} d^{3} + 3 \, a^{2} b^{3} d^{2} e\right )} p^{3} + 3 \,{\left (8 \, a b^{4} d^{3} + 9 \, a^{2} b^{3} d^{2} e - 2 \, a^{3} b^{2} d e^{2}\right )} p^{2} +{\left (47 \, a b^{4} d^{3} + 30 \, a^{2} b^{3} d^{2} e - 15 \, a^{3} b^{2} d e^{2} + 3 \, a^{4} b e^{3}\right )} p\right )} x\right )}{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p}}{2 \,{\left (4 \, b^{4} p^{4} + 28 \, b^{4} p^{3} + 71 \, b^{4} p^{2} + 77 \, b^{4} p + 30 \, b^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*a^2*b^3*d^3*p^3 + 30*a^2*b^3*d^3 - 30*a^3*b^2*d^2*e + 15*a^4*b*d*e^2 - 3*a^5*e^3 + 2*(2*b^5*e^3*p^3 + 9
*b^5*e^3*p^2 + 13*b^5*e^3*p + 6*b^5*e^3)*x^5 + (45*b^5*d*e^2 + 15*a*b^4*e^3 + 4*(3*b^5*d*e^2 + 2*a*b^4*e^3)*p^
3 + 30*(2*b^5*d*e^2 + a*b^4*e^3)*p^2 + (93*b^5*d*e^2 + 37*a*b^4*e^3)*p)*x^4 + 2*(30*b^5*d^2*e + 30*a*b^4*d*e^2
+ 2*(3*b^5*d^2*e + 6*a*b^4*d*e^2 + a^2*b^3*e^3)*p^3 + 3*(11*b^5*d^2*e + 18*a*b^4*d*e^2 + a^2*b^3*e^3)*p^2 + (
57*b^5*d^2*e + 72*a*b^4*d*e^2 + a^2*b^3*e^3)*p)*x^3 + 6*(4*a^2*b^3*d^3 - a^3*b^2*d^2*e)*p^2 + (30*b^5*d^3 + 90
*a*b^4*d^2*e + 4*(b^5*d^3 + 6*a*b^4*d^2*e + 3*a^2*b^3*d*e^2)*p^3 + 6*(4*b^5*d^3 + 21*a*b^4*d^2*e + 6*a^2*b^3*d
*e^2 - a^3*b^2*e^3)*p^2 + (47*b^5*d^3 + 201*a*b^4*d^2*e + 15*a^2*b^3*d*e^2 - 3*a^3*b^2*e^3)*p)*x^2 + (47*a^2*b
^3*d^3 - 27*a^3*b^2*d^2*e + 6*a^4*b*d*e^2)*p + 2*(30*a*b^4*d^3 + 2*(2*a*b^4*d^3 + 3*a^2*b^3*d^2*e)*p^3 + 3*(8*
a*b^4*d^3 + 9*a^2*b^3*d^2*e - 2*a^3*b^2*d*e^2)*p^2 + (47*a*b^4*d^3 + 30*a^2*b^3*d^2*e - 15*a^3*b^2*d*e^2 + 3*a
^4*b*e^3)*p)*x)*(b^2*x^2 + 2*a*b*x + a^2)^p/(4*b^4*p^4 + 28*b^4*p^3 + 71*b^4*p^2 + 77*b^4*p + 30*b^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)**3*(b**2*x**2+2*a*b*x+a**2)**p,x)

[Out]

Exception raised: TypeError

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Giac [B]  time = 1.23705, size = 2437, normalized size = 13.32 \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)^3*(b^2*x^2+2*a*b*x+a^2)^p,x, algorithm="giac")

[Out]

1/2*(4*(b^2*x^2 + 2*a*b*x + a^2)^p*b^5*p^3*x^5*e^3 + 12*(b^2*x^2 + 2*a*b*x + a^2)^p*b^5*d*p^3*x^4*e^2 + 12*(b^
2*x^2 + 2*a*b*x + a^2)^p*b^5*d^2*p^3*x^3*e + 4*(b^2*x^2 + 2*a*b*x + a^2)^p*b^5*d^3*p^3*x^2 + 8*(b^2*x^2 + 2*a*
b*x + a^2)^p*a*b^4*p^3*x^4*e^3 + 18*(b^2*x^2 + 2*a*b*x + a^2)^p*b^5*p^2*x^5*e^3 + 24*(b^2*x^2 + 2*a*b*x + a^2)
^p*a*b^4*d*p^3*x^3*e^2 + 60*(b^2*x^2 + 2*a*b*x + a^2)^p*b^5*d*p^2*x^4*e^2 + 24*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b
^4*d^2*p^3*x^2*e + 66*(b^2*x^2 + 2*a*b*x + a^2)^p*b^5*d^2*p^2*x^3*e + 8*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^4*d^3*
p^3*x + 24*(b^2*x^2 + 2*a*b*x + a^2)^p*b^5*d^3*p^2*x^2 + 4*(b^2*x^2 + 2*a*b*x + a^2)^p*a^2*b^3*p^3*x^3*e^3 + 3
0*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^4*p^2*x^4*e^3 + 26*(b^2*x^2 + 2*a*b*x + a^2)^p*b^5*p*x^5*e^3 + 12*(b^2*x^2 +
2*a*b*x + a^2)^p*a^2*b^3*d*p^3*x^2*e^2 + 108*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^4*d*p^2*x^3*e^2 + 93*(b^2*x^2 +
2*a*b*x + a^2)^p*b^5*d*p*x^4*e^2 + 12*(b^2*x^2 + 2*a*b*x + a^2)^p*a^2*b^3*d^2*p^3*x*e + 126*(b^2*x^2 + 2*a*b*x
+ a^2)^p*a*b^4*d^2*p^2*x^2*e + 114*(b^2*x^2 + 2*a*b*x + a^2)^p*b^5*d^2*p*x^3*e + 4*(b^2*x^2 + 2*a*b*x + a^2)^
p*a^2*b^3*d^3*p^3 + 48*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^4*d^3*p^2*x + 47*(b^2*x^2 + 2*a*b*x + a^2)^p*b^5*d^3*p*
x^2 + 6*(b^2*x^2 + 2*a*b*x + a^2)^p*a^2*b^3*p^2*x^3*e^3 + 37*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^4*p*x^4*e^3 + 12*
(b^2*x^2 + 2*a*b*x + a^2)^p*b^5*x^5*e^3 + 36*(b^2*x^2 + 2*a*b*x + a^2)^p*a^2*b^3*d*p^2*x^2*e^2 + 144*(b^2*x^2
+ 2*a*b*x + a^2)^p*a*b^4*d*p*x^3*e^2 + 45*(b^2*x^2 + 2*a*b*x + a^2)^p*b^5*d*x^4*e^2 + 54*(b^2*x^2 + 2*a*b*x +
a^2)^p*a^2*b^3*d^2*p^2*x*e + 201*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^4*d^2*p*x^2*e + 60*(b^2*x^2 + 2*a*b*x + a^2)^
p*b^5*d^2*x^3*e + 24*(b^2*x^2 + 2*a*b*x + a^2)^p*a^2*b^3*d^3*p^2 + 94*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^4*d^3*p*
x + 30*(b^2*x^2 + 2*a*b*x + a^2)^p*b^5*d^3*x^2 - 6*(b^2*x^2 + 2*a*b*x + a^2)^p*a^3*b^2*p^2*x^2*e^3 + 2*(b^2*x^
2 + 2*a*b*x + a^2)^p*a^2*b^3*p*x^3*e^3 + 15*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^4*x^4*e^3 - 12*(b^2*x^2 + 2*a*b*x
+ a^2)^p*a^3*b^2*d*p^2*x*e^2 + 15*(b^2*x^2 + 2*a*b*x + a^2)^p*a^2*b^3*d*p*x^2*e^2 + 60*(b^2*x^2 + 2*a*b*x + a^
2)^p*a*b^4*d*x^3*e^2 - 6*(b^2*x^2 + 2*a*b*x + a^2)^p*a^3*b^2*d^2*p^2*e + 60*(b^2*x^2 + 2*a*b*x + a^2)^p*a^2*b^
3*d^2*p*x*e + 90*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^4*d^2*x^2*e + 47*(b^2*x^2 + 2*a*b*x + a^2)^p*a^2*b^3*d^3*p +
60*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^4*d^3*x - 3*(b^2*x^2 + 2*a*b*x + a^2)^p*a^3*b^2*p*x^2*e^3 - 30*(b^2*x^2 + 2
*a*b*x + a^2)^p*a^3*b^2*d*p*x*e^2 - 27*(b^2*x^2 + 2*a*b*x + a^2)^p*a^3*b^2*d^2*p*e + 30*(b^2*x^2 + 2*a*b*x + a
^2)^p*a^2*b^3*d^3 + 6*(b^2*x^2 + 2*a*b*x + a^2)^p*a^4*b*p*x*e^3 + 6*(b^2*x^2 + 2*a*b*x + a^2)^p*a^4*b*d*p*e^2
- 30*(b^2*x^2 + 2*a*b*x + a^2)^p*a^3*b^2*d^2*e + 15*(b^2*x^2 + 2*a*b*x + a^2)^p*a^4*b*d*e^2 - 3*(b^2*x^2 + 2*a
*b*x + a^2)^p*a^5*e^3)/(4*b^4*p^4 + 28*b^4*p^3 + 71*b^4*p^2 + 77*b^4*p + 30*b^4)