Integrand size = 26, antiderivative size = 127 \[ \int (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )^p \, dx=\frac {(b d-a e)^2 (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^p}{b^3 (1+2 p)}+\frac {e (b d-a e) (a+b x)^2 \left (a^2+2 a b x+b^2 x^2\right )^p}{b^3 (1+p)}+\frac {e^2 (a+b x)^3 \left (a^2+2 a b x+b^2 x^2\right )^p}{b^3 (3+2 p)} \]
(-a*e+b*d)^2*(b*x+a)*(b^2*x^2+2*a*b*x+a^2)^p/b^3/(1+2*p)+e*(-a*e+b*d)*(b*x +a)^2*(b^2*x^2+2*a*b*x+a^2)^p/b^3/(p+1)+e^2*(b*x+a)^3*(b^2*x^2+2*a*b*x+a^2 )^p/b^3/(3+2*p)
Time = 0.08 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.59 \[ \int (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )^p \, dx=\frac {(a+b x) \left ((a+b x)^2\right )^p \left (\frac {(b d-a e)^2}{1+2 p}+\frac {e (b d-a e) (a+b x)}{1+p}+\frac {e^2 (a+b x)^2}{3+2 p}\right )}{b^3} \]
((a + b*x)*((a + b*x)^2)^p*((b*d - a*e)^2/(1 + 2*p) + (e*(b*d - a*e)*(a + b*x))/(1 + p) + (e^2*(a + b*x)^2)/(3 + 2*p)))/b^3
Time = 0.29 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1102, 53, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )^p \, dx\) |
| \(\Big \downarrow \) 1102 |
| \(\displaystyle \left (a b+b^2 x\right )^{-2 p} \left (a^2+2 a b x+b^2 x^2\right )^p \int \left (x b^2+a b\right )^{2 p} (d+e x)^2dx\) |
| \(\Big \downarrow \) 53 |
| \(\displaystyle \left (a b+b^2 x\right )^{-2 p} \left (a^2+2 a b x+b^2 x^2\right )^p \int \left (\frac {(b d-a e)^2 \left (x b^2+a b\right )^{2 p}}{b^2}+\frac {2 e (b d-a e) \left (x b^2+a b\right )^{2 p+1}}{b^3}+\frac {e^2 \left (x b^2+a b\right )^{2 p+2}}{b^4}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \left (a b+b^2 x\right )^{-2 p} \left (a^2+2 a b x+b^2 x^2\right )^p \left (\frac {e^2 \left (a b+b^2 x\right )^{2 p+3}}{b^6 (2 p+3)}+\frac {e (b d-a e) \left (a b+b^2 x\right )^{2 (p+1)}}{b^5 (p+1)}+\frac {(b d-a e)^2 \left (a b+b^2 x\right )^{2 p+1}}{b^4 (2 p+1)}\right )\) |
((a^2 + 2*a*b*x + b^2*x^2)^p*((e*(b*d - a*e)*(a*b + b^2*x)^(2*(1 + p)))/(b ^5*(1 + p)) + ((b*d - a*e)^2*(a*b + b^2*x)^(1 + 2*p))/(b^4*(1 + 2*p)) + (e ^2*(a*b + b^2*x)^(3 + 2*p))/(b^6*(3 + 2*p))))/(a*b + b^2*x)^(2*p)
3.18.45.3.1 Defintions of rubi rules used
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] && 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])
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*F racPart[p])) Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0]
Time = 2.68 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.38
| method | result | size |
| gosper | \(\frac {\left (b x +a \right ) \left (2 b^{2} e^{2} p^{2} x^{2}+4 b^{2} d e \,p^{2} x +3 b^{2} e^{2} p \,x^{2}-2 a b \,e^{2} p x +2 b^{2} d^{2} p^{2}+8 b^{2} d e p x +x^{2} b^{2} e^{2}-2 a b d e p -x a b \,e^{2}+5 b^{2} d^{2} p +3 b^{2} d e x +a^{2} e^{2}-3 a b d e +3 b^{2} d^{2}\right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p}}{b^{3} \left (4 p^{3}+12 p^{2}+11 p +3\right )}\) | \(175\) |
| risch | \(\frac {\left (2 b^{3} e^{2} p^{2} x^{3}+2 a \,b^{2} e^{2} p^{2} x^{2}+4 b^{3} d e \,p^{2} x^{2}+3 b^{3} e^{2} p \,x^{3}+4 a \,b^{2} d e \,p^{2} x +a \,b^{2} e^{2} p \,x^{2}+2 b^{3} d^{2} p^{2} x +8 b^{3} d e p \,x^{2}+e^{2} x^{3} b^{3}-2 a^{2} b \,e^{2} p x +2 a \,b^{2} d^{2} p^{2}+6 a \,b^{2} d e p x +5 b^{3} d^{2} p x +3 b^{3} d e \,x^{2}-2 a^{2} b d e p +5 a \,b^{2} d^{2} p +3 b^{3} d^{2} x +a^{3} e^{2}-3 a^{2} b d e +3 a \,b^{2} d^{2}\right ) \left (\left (b x +a \right )^{2}\right )^{p}}{\left (1+p \right ) \left (3+2 p \right ) \left (1+2 p \right ) b^{3}}\) | \(250\) |
| norman | \(\frac {e^{2} x^{3} {\mathrm e}^{p \ln \left (b^{2} x^{2}+2 a b x +a^{2}\right )}}{3+2 p}+\frac {a \left (2 b^{2} d^{2} p^{2}-2 a b d e p +5 b^{2} d^{2} p +a^{2} e^{2}-3 a b d e +3 b^{2} d^{2}\right ) {\mathrm e}^{p \ln \left (b^{2} x^{2}+2 a b x +a^{2}\right )}}{b^{3} \left (4 p^{3}+12 p^{2}+11 p +3\right )}+\frac {\left (a e p +2 b d p +3 b d \right ) e \,x^{2} {\mathrm e}^{p \ln \left (b^{2} x^{2}+2 a b x +a^{2}\right )}}{b \left (2 p^{2}+5 p +3\right )}-\frac {\left (-4 a b d e \,p^{2}-2 b^{2} d^{2} p^{2}+2 a^{2} e^{2} p -6 a b d e p -5 b^{2} d^{2} p -3 b^{2} d^{2}\right ) x \,{\mathrm e}^{p \ln \left (b^{2} x^{2}+2 a b x +a^{2}\right )}}{b^{2} \left (4 p^{3}+12 p^{2}+11 p +3\right )}\) | \(278\) |
| parallelrisch | \(\frac {2 x^{3} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a \,b^{3} e^{2} p^{2}+3 x^{3} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a \,b^{3} e^{2} p +2 x^{2} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a^{2} b^{2} e^{2} p^{2}+4 x^{2} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a \,b^{3} d e \,p^{2}+x^{3} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a \,b^{3} e^{2}+x^{2} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a^{2} b^{2} e^{2} p +8 x^{2} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a \,b^{3} d e p +4 x \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a^{2} b^{2} d e \,p^{2}+2 x \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a \,b^{3} d^{2} p^{2}+3 x^{2} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a \,b^{3} d e -2 x \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a^{3} b \,e^{2} p +6 x \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a^{2} b^{2} d e p +5 x \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a \,b^{3} d^{2} p +2 \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a^{2} b^{2} d^{2} p^{2}+3 x \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a \,b^{3} d^{2}-2 \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a^{3} b d e p +5 \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a^{2} b^{2} d^{2} p +\left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a^{4} e^{2}-3 \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a^{3} b d e +3 \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{p} a^{2} b^{2} d^{2}}{\left (3+2 p \right ) \left (1+p \right ) \left (1+2 p \right ) a \,b^{3}}\) | \(627\) |
(b*x+a)*(2*b^2*e^2*p^2*x^2+4*b^2*d*e*p^2*x+3*b^2*e^2*p*x^2-2*a*b*e^2*p*x+2 *b^2*d^2*p^2+8*b^2*d*e*p*x+b^2*e^2*x^2-2*a*b*d*e*p-a*b*e^2*x+5*b^2*d^2*p+3 *b^2*d*e*x+a^2*e^2-3*a*b*d*e+3*b^2*d^2)*(b^2*x^2+2*a*b*x+a^2)^p/b^3/(4*p^3 +12*p^2+11*p+3)
Time = 0.41 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.96 \[ \int (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )^p \, dx=\frac {{\left (2 \, a b^{2} d^{2} p^{2} + 3 \, a b^{2} d^{2} - 3 \, a^{2} b d e + a^{3} e^{2} + {\left (2 \, b^{3} e^{2} p^{2} + 3 \, b^{3} e^{2} p + b^{3} e^{2}\right )} x^{3} + {\left (3 \, b^{3} d e + 2 \, {\left (2 \, b^{3} d e + a b^{2} e^{2}\right )} p^{2} + {\left (8 \, b^{3} d e + a b^{2} e^{2}\right )} p\right )} x^{2} + {\left (5 \, a b^{2} d^{2} - 2 \, a^{2} b d e\right )} p + {\left (3 \, b^{3} d^{2} + 2 \, {\left (b^{3} d^{2} + 2 \, a b^{2} d e\right )} p^{2} + {\left (5 \, b^{3} d^{2} + 6 \, a b^{2} d e - 2 \, a^{2} b e^{2}\right )} p\right )} x\right )} {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p}}{4 \, b^{3} p^{3} + 12 \, b^{3} p^{2} + 11 \, b^{3} p + 3 \, b^{3}} \]
(2*a*b^2*d^2*p^2 + 3*a*b^2*d^2 - 3*a^2*b*d*e + a^3*e^2 + (2*b^3*e^2*p^2 + 3*b^3*e^2*p + b^3*e^2)*x^3 + (3*b^3*d*e + 2*(2*b^3*d*e + a*b^2*e^2)*p^2 + (8*b^3*d*e + a*b^2*e^2)*p)*x^2 + (5*a*b^2*d^2 - 2*a^2*b*d*e)*p + (3*b^3*d^ 2 + 2*(b^3*d^2 + 2*a*b^2*d*e)*p^2 + (5*b^3*d^2 + 6*a*b^2*d*e - 2*a^2*b*e^2 )*p)*x)*(b^2*x^2 + 2*a*b*x + a^2)^p/(4*b^3*p^3 + 12*b^3*p^2 + 11*b^3*p + 3 *b^3)
\[ \int (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )^p \, dx=\text {Too large to display} \]
Piecewise(((d**2*x + d*e*x**2 + e**2*x**3/3)*(a**2)**p, Eq(b, 0)), (Integr al((d + e*x)**2/((a + b*x)**2)**(3/2), x), Eq(p, -3/2)), (-2*a**2*e**2*log (a/b + x)/(a*b**3 + b**4*x) - 2*a**2*e**2/(a*b**3 + b**4*x) + 2*a*b*d*e*lo g(a/b + x)/(a*b**3 + b**4*x) + 2*a*b*d*e/(a*b**3 + b**4*x) - 2*a*b*e**2*x* log(a/b + x)/(a*b**3 + b**4*x) - b**2*d**2/(a*b**3 + b**4*x) + 2*b**2*d*e* x*log(a/b + x)/(a*b**3 + b**4*x) + b**2*e**2*x**2/(a*b**3 + b**4*x), Eq(p, -1)), (Piecewise(((e**2*x/(2*b**2) + (-3*a*e**2/(2*b) + 2*d*e)/b**2)*sqrt (a**2 + 2*a*b*x + b**2*x**2) + (a/b + x)*(-a**2*e**2/(2*b**2) - a*(-3*a*e* *2/(2*b) + 2*d*e)/b + d**2)*log(a/b + x)/sqrt(b**2*(a/b + x)**2), Ne(b**2, 0)), ((sqrt(a**2 + 2*a*b*x)*(a**2*e**2 - 4*a*b*d*e + 4*b**2*d**2)/(4*b**2 ) + (a**2 + 2*a*b*x)**(3/2)*(-a*e**2 + 2*b*d*e)/(6*a*b**2) + e**2*(a**2 + 2*a*b*x)**(5/2)/(20*a**2*b**2))/(a*b), Ne(a*b, 0)), (Piecewise((d**2*x, Eq (e, 0)), ((d + e*x)**3/(3*e), True))/sqrt(a**2), True)), Eq(p, -1/2)), (a* *3*e**2*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**3*p**3 + 12*b**3*p**2 + 11*b **3*p + 3*b**3) - 2*a**2*b*d*e*p*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**3*p **3 + 12*b**3*p**2 + 11*b**3*p + 3*b**3) - 3*a**2*b*d*e*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**3*p**3 + 12*b**3*p**2 + 11*b**3*p + 3*b**3) - 2*a**2*b *e**2*p*x*(a**2 + 2*a*b*x + b**2*x**2)**p/(4*b**3*p**3 + 12*b**3*p**2 + 11 *b**3*p + 3*b**3) + 2*a*b**2*d**2*p**2*(a**2 + 2*a*b*x + b**2*x**2)**p/(4* b**3*p**3 + 12*b**3*p**2 + 11*b**3*p + 3*b**3) + 5*a*b**2*d**2*p*(a**2 ...
Time = 0.23 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.24 \[ \int (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )^p \, dx=\frac {{\left (b x + a\right )} {\left (b x + a\right )}^{2 \, p} d^{2}}{b {\left (2 \, p + 1\right )}} + \frac {{\left (b^{2} {\left (2 \, p + 1\right )} x^{2} + 2 \, a b p x - a^{2}\right )} {\left (b x + a\right )}^{2 \, p} d e}{{\left (2 \, p^{2} + 3 \, p + 1\right )} b^{2}} + \frac {{\left ({\left (2 \, p^{2} + 3 \, p + 1\right )} b^{3} x^{3} + {\left (2 \, p^{2} + p\right )} a b^{2} x^{2} - 2 \, a^{2} b p x + a^{3}\right )} {\left (b x + a\right )}^{2 \, p} e^{2}}{{\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} b^{3}} \]
(b*x + a)*(b*x + a)^(2*p)*d^2/(b*(2*p + 1)) + (b^2*(2*p + 1)*x^2 + 2*a*b*p *x - a^2)*(b*x + a)^(2*p)*d*e/((2*p^2 + 3*p + 1)*b^2) + ((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)*e^2/ ((4*p^3 + 12*p^2 + 11*p + 3)*b^3)
Leaf count of result is larger than twice the leaf count of optimal. 608 vs. \(2 (127) = 254\).
Time = 0.29 (sec) , antiderivative size = 608, normalized size of antiderivative = 4.79 \[ \int (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )^p \, dx=\frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} e^{2} p^{2} x^{3} + 4 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} d e p^{2} x^{2} + 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b^{2} e^{2} p^{2} x^{2} + 3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} e^{2} p x^{3} + 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} d^{2} p^{2} x + 4 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b^{2} d e p^{2} x + 8 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} d e p x^{2} + {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b^{2} e^{2} p x^{2} + {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} e^{2} x^{3} + 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b^{2} d^{2} p^{2} + 5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} d^{2} p x + 6 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b^{2} d e p x - 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a^{2} b e^{2} p x + 3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} d e x^{2} + 5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b^{2} d^{2} p - 2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a^{2} b d e p + 3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} b^{3} d^{2} x + 3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a b^{2} d^{2} - 3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a^{2} b d e + {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{p} a^{3} e^{2}}{4 \, b^{3} p^{3} + 12 \, b^{3} p^{2} + 11 \, b^{3} p + 3 \, b^{3}} \]
(2*(b^2*x^2 + 2*a*b*x + a^2)^p*b^3*e^2*p^2*x^3 + 4*(b^2*x^2 + 2*a*b*x + a^ 2)^p*b^3*d*e*p^2*x^2 + 2*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^2*e^2*p^2*x^2 + 3 *(b^2*x^2 + 2*a*b*x + a^2)^p*b^3*e^2*p*x^3 + 2*(b^2*x^2 + 2*a*b*x + a^2)^p *b^3*d^2*p^2*x + 4*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^2*d*e*p^2*x + 8*(b^2*x^ 2 + 2*a*b*x + a^2)^p*b^3*d*e*p*x^2 + (b^2*x^2 + 2*a*b*x + a^2)^p*a*b^2*e^2 *p*x^2 + (b^2*x^2 + 2*a*b*x + a^2)^p*b^3*e^2*x^3 + 2*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^2*d^2*p^2 + 5*(b^2*x^2 + 2*a*b*x + a^2)^p*b^3*d^2*p*x + 6*(b^2* x^2 + 2*a*b*x + a^2)^p*a*b^2*d*e*p*x - 2*(b^2*x^2 + 2*a*b*x + a^2)^p*a^2*b *e^2*p*x + 3*(b^2*x^2 + 2*a*b*x + a^2)^p*b^3*d*e*x^2 + 5*(b^2*x^2 + 2*a*b* x + a^2)^p*a*b^2*d^2*p - 2*(b^2*x^2 + 2*a*b*x + a^2)^p*a^2*b*d*e*p + 3*(b^ 2*x^2 + 2*a*b*x + a^2)^p*b^3*d^2*x + 3*(b^2*x^2 + 2*a*b*x + a^2)^p*a*b^2*d ^2 - 3*(b^2*x^2 + 2*a*b*x + a^2)^p*a^2*b*d*e + (b^2*x^2 + 2*a*b*x + a^2)^p *a^3*e^2)/(4*b^3*p^3 + 12*b^3*p^2 + 11*b^3*p + 3*b^3)
Time = 9.90 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.97 \[ \int (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )^p \, dx={\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^p\,\left (\frac {e^2\,x^3\,\left (2\,p^2+3\,p+1\right )}{4\,p^3+12\,p^2+11\,p+3}+\frac {x\,\left (-2\,a^2\,b\,e^2\,p+4\,a\,b^2\,d\,e\,p^2+6\,a\,b^2\,d\,e\,p+2\,b^3\,d^2\,p^2+5\,b^3\,d^2\,p+3\,b^3\,d^2\right )}{b^3\,\left (4\,p^3+12\,p^2+11\,p+3\right )}+\frac {a\,\left (a^2\,e^2-2\,a\,b\,d\,e\,p-3\,a\,b\,d\,e+2\,b^2\,d^2\,p^2+5\,b^2\,d^2\,p+3\,b^2\,d^2\right )}{b^3\,\left (4\,p^3+12\,p^2+11\,p+3\right )}+\frac {e\,x^2\,\left (2\,p+1\right )\,\left (3\,b\,d+a\,e\,p+2\,b\,d\,p\right )}{b\,\left (4\,p^3+12\,p^2+11\,p+3\right )}\right ) \]
(a^2 + b^2*x^2 + 2*a*b*x)^p*((e^2*x^3*(3*p + 2*p^2 + 1))/(11*p + 12*p^2 + 4*p^3 + 3) + (x*(3*b^3*d^2 + 5*b^3*d^2*p + 2*b^3*d^2*p^2 - 2*a^2*b*e^2*p + 4*a*b^2*d*e*p^2 + 6*a*b^2*d*e*p))/(b^3*(11*p + 12*p^2 + 4*p^3 + 3)) + (a* (a^2*e^2 + 3*b^2*d^2 + 5*b^2*d^2*p + 2*b^2*d^2*p^2 - 3*a*b*d*e - 2*a*b*d*e *p))/(b^3*(11*p + 12*p^2 + 4*p^3 + 3)) + (e*x^2*(2*p + 1)*(3*b*d + a*e*p + 2*b*d*p))/(b*(11*p + 12*p^2 + 4*p^3 + 3)))