Integrand size = 19, antiderivative size = 112 \[ \int \left (a+b x^n\right )^2 \left (d+e x^n\right )^2 \, dx=a^2 d^2 x+\frac {2 a d (b d+a e) x^{1+n}}{1+n}+\frac {\left (b^2 d^2+4 a b d e+a^2 e^2\right ) x^{1+2 n}}{1+2 n}+\frac {2 b e (b d+a e) x^{1+3 n}}{1+3 n}+\frac {b^2 e^2 x^{1+4 n}}{1+4 n} \]
a^2*d^2*x+2*a*d*(a*e+b*d)*x^(1+n)/(1+n)+(a^2*e^2+4*a*b*d*e+b^2*d^2)*x^(1+2 *n)/(1+2*n)+2*b*e*(a*e+b*d)*x^(1+3*n)/(1+3*n)+b^2*e^2*x^(1+4*n)/(1+4*n)
Time = 0.76 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.94 \[ \int \left (a+b x^n\right )^2 \left (d+e x^n\right )^2 \, dx=x \left (a^2 d^2+\frac {2 a d (b d+a e) x^n}{1+n}+\frac {\left (b^2 d^2+4 a b d e+a^2 e^2\right ) x^{2 n}}{1+2 n}+\frac {2 b e (b d+a e) x^{3 n}}{1+3 n}+\frac {b^2 e^2 x^{4 n}}{1+4 n}\right ) \]
x*(a^2*d^2 + (2*a*d*(b*d + a*e)*x^n)/(1 + n) + ((b^2*d^2 + 4*a*b*d*e + a^2 *e^2)*x^(2*n))/(1 + 2*n) + (2*b*e*(b*d + a*e)*x^(3*n))/(1 + 3*n) + (b^2*e^ 2*x^(4*n))/(1 + 4*n))
Time = 0.27 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {897, 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.
\(\displaystyle \int \left (a+b x^n\right )^2 \left (d+e x^n\right )^2 \, dx\) |
\(\Big \downarrow \) 897 |
\(\displaystyle \int \left (x^{2 n} \left (a^2 e^2+4 a b d e+b^2 d^2\right )+a^2 d^2+2 b e x^{3 n} (a e+b d)+2 a d x^n (a e+b d)+b^2 e^2 x^{4 n}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x^{2 n+1} \left (a^2 e^2+4 a b d e+b^2 d^2\right )}{2 n+1}+a^2 d^2 x+\frac {2 a d x^{n+1} (a e+b d)}{n+1}+\frac {2 b e x^{3 n+1} (a e+b d)}{3 n+1}+\frac {b^2 e^2 x^{4 n+1}}{4 n+1}\) |
a^2*d^2*x + (2*a*d*(b*d + a*e)*x^(1 + n))/(1 + n) + ((b^2*d^2 + 4*a*b*d*e + a^2*e^2)*x^(1 + 2*n))/(1 + 2*n) + (2*b*e*(b*d + a*e)*x^(1 + 3*n))/(1 + 3 *n) + (b^2*e^2*x^(1 + 4*n))/(1 + 4*n)
3.3.93.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol ] :> Int[ExpandIntegrand[(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b , c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]
Time = 4.00 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.97
method | result | size |
risch | \(a^{2} d^{2} x +\frac {\left (a^{2} e^{2}+4 a b d e +b^{2} d^{2}\right ) x \,x^{2 n}}{1+2 n}+\frac {b^{2} e^{2} x \,x^{4 n}}{1+4 n}+\frac {2 a d \left (a e +b d \right ) x \,x^{n}}{1+n}+\frac {2 b e \left (a e +b d \right ) x \,x^{3 n}}{1+3 n}\) | \(109\) |
norman | \(a^{2} d^{2} x +\frac {\left (a^{2} e^{2}+4 a b d e +b^{2} d^{2}\right ) x \,{\mathrm e}^{2 n \ln \left (x \right )}}{1+2 n}+\frac {b^{2} e^{2} x \,{\mathrm e}^{4 n \ln \left (x \right )}}{1+4 n}+\frac {2 a d \left (a e +b d \right ) x \,{\mathrm e}^{n \ln \left (x \right )}}{1+n}+\frac {2 b e \left (a e +b d \right ) x \,{\mathrm e}^{3 n \ln \left (x \right )}}{1+3 n}\) | \(117\) |
parallelrisch | \(\frac {28 x \,x^{3 n} b^{2} d e \,n^{2}+14 x \,x^{3 n} a b \,e^{2} n +52 x \,x^{n} a^{2} d e \,n^{2}+35 x \,a^{2} d^{2} n^{2}+10 x \,a^{2} d^{2} n +b^{2} e^{2} x \,x^{4 n}+18 x \,x^{n} a^{2} d e n +4 x \,x^{2 n} a b d e +48 x \,x^{n} a^{2} d e \,n^{3}+2 x \,x^{n} a^{2} d e +2 x \,x^{n} a b \,d^{2}+a^{2} d^{2} x +16 x \,x^{3 n} b^{2} d e \,n^{3}+14 x \,x^{3 n} b^{2} d e n +2 x \,x^{3 n} b^{2} d e +8 x \,x^{2 n} a^{2} e^{2} n +8 x \,x^{2 n} b^{2} d^{2} n +11 x \,x^{4 n} b^{2} e^{2} n^{2}+6 x \,x^{4 n} b^{2} e^{2} n +52 x \,x^{n} a b \,d^{2} n^{2}+48 x \,x^{n} a b \,d^{2} n^{3}+24 x \,a^{2} d^{2} n^{4}+50 x \,a^{2} d^{2} n^{3}+18 x \,x^{n} a b \,d^{2} n +48 x \,x^{2 n} a b d e \,n^{3}+76 x \,x^{2 n} a b d e \,n^{2}+32 x \,x^{2 n} a b d e n +28 x \,x^{3 n} a b \,e^{2} n^{2}+16 x \,x^{3 n} a b \,e^{2} n^{3}+6 x \,x^{4 n} b^{2} e^{2} n^{3}+12 x \,x^{2 n} a^{2} e^{2} n^{3}+12 x \,x^{2 n} b^{2} d^{2} n^{3}+19 x \,x^{2 n} a^{2} e^{2} n^{2}+19 x \,x^{2 n} b^{2} d^{2} n^{2}+2 x \,x^{3 n} a b \,e^{2}+x \,x^{2 n} a^{2} e^{2}+x \,x^{2 n} b^{2} d^{2}}{\left (1+2 n \right ) \left (1+4 n \right ) \left (1+n \right ) \left (1+3 n \right )}\) | \(544\) |
a^2*d^2*x+(a^2*e^2+4*a*b*d*e+b^2*d^2)/(1+2*n)*x*(x^n)^2+b^2*e^2/(1+4*n)*x* (x^n)^4+2*a*d*(a*e+b*d)/(1+n)*x*x^n+2*b*e*(a*e+b*d)/(1+3*n)*x*(x^n)^3
Leaf count of result is larger than twice the leaf count of optimal. 370 vs. \(2 (112) = 224\).
Time = 0.25 (sec) , antiderivative size = 370, normalized size of antiderivative = 3.30 \[ \int \left (a+b x^n\right )^2 \left (d+e x^n\right )^2 \, dx=\frac {{\left (6 \, b^{2} e^{2} n^{3} + 11 \, b^{2} e^{2} n^{2} + 6 \, b^{2} e^{2} n + b^{2} e^{2}\right )} x x^{4 \, n} + 2 \, {\left (b^{2} d e + a b e^{2} + 8 \, {\left (b^{2} d e + a b e^{2}\right )} n^{3} + 14 \, {\left (b^{2} d e + a b e^{2}\right )} n^{2} + 7 \, {\left (b^{2} d e + a b e^{2}\right )} n\right )} x x^{3 \, n} + {\left (b^{2} d^{2} + 4 \, a b d e + a^{2} e^{2} + 12 \, {\left (b^{2} d^{2} + 4 \, a b d e + a^{2} e^{2}\right )} n^{3} + 19 \, {\left (b^{2} d^{2} + 4 \, a b d e + a^{2} e^{2}\right )} n^{2} + 8 \, {\left (b^{2} d^{2} + 4 \, a b d e + a^{2} e^{2}\right )} n\right )} x x^{2 \, n} + 2 \, {\left (a b d^{2} + a^{2} d e + 24 \, {\left (a b d^{2} + a^{2} d e\right )} n^{3} + 26 \, {\left (a b d^{2} + a^{2} d e\right )} n^{2} + 9 \, {\left (a b d^{2} + a^{2} d e\right )} n\right )} x x^{n} + {\left (24 \, a^{2} d^{2} n^{4} + 50 \, a^{2} d^{2} n^{3} + 35 \, a^{2} d^{2} n^{2} + 10 \, a^{2} d^{2} n + a^{2} d^{2}\right )} x}{24 \, n^{4} + 50 \, n^{3} + 35 \, n^{2} + 10 \, n + 1} \]
((6*b^2*e^2*n^3 + 11*b^2*e^2*n^2 + 6*b^2*e^2*n + b^2*e^2)*x*x^(4*n) + 2*(b ^2*d*e + a*b*e^2 + 8*(b^2*d*e + a*b*e^2)*n^3 + 14*(b^2*d*e + a*b*e^2)*n^2 + 7*(b^2*d*e + a*b*e^2)*n)*x*x^(3*n) + (b^2*d^2 + 4*a*b*d*e + a^2*e^2 + 12 *(b^2*d^2 + 4*a*b*d*e + a^2*e^2)*n^3 + 19*(b^2*d^2 + 4*a*b*d*e + a^2*e^2)* n^2 + 8*(b^2*d^2 + 4*a*b*d*e + a^2*e^2)*n)*x*x^(2*n) + 2*(a*b*d^2 + a^2*d* e + 24*(a*b*d^2 + a^2*d*e)*n^3 + 26*(a*b*d^2 + a^2*d*e)*n^2 + 9*(a*b*d^2 + a^2*d*e)*n)*x*x^n + (24*a^2*d^2*n^4 + 50*a^2*d^2*n^3 + 35*a^2*d^2*n^2 + 1 0*a^2*d^2*n + a^2*d^2)*x)/(24*n^4 + 50*n^3 + 35*n^2 + 10*n + 1)
Leaf count of result is larger than twice the leaf count of optimal. 1760 vs. \(2 (104) = 208\).
Time = 2.96 (sec) , antiderivative size = 1760, normalized size of antiderivative = 15.71 \[ \int \left (a+b x^n\right )^2 \left (d+e x^n\right )^2 \, dx=\text {Too large to display} \]
Piecewise((a**2*d**2*x + 2*a**2*d*e*log(x) - a**2*e**2/x + 2*a*b*d**2*log( x) - 4*a*b*d*e/x - a*b*e**2/x**2 - b**2*d**2/x - b**2*d*e/x**2 - b**2*e**2 /(3*x**3), Eq(n, -1)), (a**2*d**2*x + 4*a**2*d*e*sqrt(x) + a**2*e**2*log(x ) + 4*a*b*d**2*sqrt(x) + 4*a*b*d*e*log(x) - 4*a*b*e**2/sqrt(x) + b**2*d**2 *log(x) - 4*b**2*d*e/sqrt(x) - b**2*e**2/x, Eq(n, -1/2)), (a**2*d**2*x + 3 *a**2*d*e*x**(2/3) + 3*a**2*e**2*x**(1/3) + 3*a*b*d**2*x**(2/3) + 12*a*b*d *e*x**(1/3) + 2*a*b*e**2*log(x) + 3*b**2*d**2*x**(1/3) + 2*b**2*d*e*log(x) - 3*b**2*e**2/x**(1/3), Eq(n, -1/3)), (a**2*d**2*x + 8*a*d*x**(3/4)*(a*e + b*d)/3 + 4*b**2*e**2*log(x**(1/4)) + 8*b*e*x**(1/4)*(a*e + b*d) - 2*sqrt (x)*(-a**2*e**2 - 4*a*b*d*e - b**2*d**2), Eq(n, -1/4)), (24*a**2*d**2*n**4 *x/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 50*a**2*d**2*n**3*x/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 35*a**2*d**2*n**2*x/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 10*a**2*d**2*n*x/(24*n**4 + 50*n**3 + 35*n**2 + 10 *n + 1) + a**2*d**2*x/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 48*a**2*d *e*n**3*x*x**n/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 52*a**2*d*e*n**2 *x*x**n/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 18*a**2*d*e*n*x*x**n/(2 4*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 2*a**2*d*e*x*x**n/(24*n**4 + 50*n **3 + 35*n**2 + 10*n + 1) + 12*a**2*e**2*n**3*x*x**(2*n)/(24*n**4 + 50*n** 3 + 35*n**2 + 10*n + 1) + 19*a**2*e**2*n**2*x*x**(2*n)/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 8*a**2*e**2*n*x*x**(2*n)/(24*n**4 + 50*n**3 + 3...
Time = 0.27 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.50 \[ \int \left (a+b x^n\right )^2 \left (d+e x^n\right )^2 \, dx=a^{2} d^{2} x + \frac {b^{2} e^{2} x^{4 \, n + 1}}{4 \, n + 1} + \frac {2 \, b^{2} d e x^{3 \, n + 1}}{3 \, n + 1} + \frac {2 \, a b e^{2} x^{3 \, n + 1}}{3 \, n + 1} + \frac {b^{2} d^{2} x^{2 \, n + 1}}{2 \, n + 1} + \frac {4 \, a b d e x^{2 \, n + 1}}{2 \, n + 1} + \frac {a^{2} e^{2} x^{2 \, n + 1}}{2 \, n + 1} + \frac {2 \, a b d^{2} x^{n + 1}}{n + 1} + \frac {2 \, a^{2} d e x^{n + 1}}{n + 1} \]
a^2*d^2*x + b^2*e^2*x^(4*n + 1)/(4*n + 1) + 2*b^2*d*e*x^(3*n + 1)/(3*n + 1 ) + 2*a*b*e^2*x^(3*n + 1)/(3*n + 1) + b^2*d^2*x^(2*n + 1)/(2*n + 1) + 4*a* b*d*e*x^(2*n + 1)/(2*n + 1) + a^2*e^2*x^(2*n + 1)/(2*n + 1) + 2*a*b*d^2*x^ (n + 1)/(n + 1) + 2*a^2*d*e*x^(n + 1)/(n + 1)
Leaf count of result is larger than twice the leaf count of optimal. 539 vs. \(2 (112) = 224\).
Time = 0.30 (sec) , antiderivative size = 539, normalized size of antiderivative = 4.81 \[ \int \left (a+b x^n\right )^2 \left (d+e x^n\right )^2 \, dx=\frac {24 \, a^{2} d^{2} n^{4} x + 6 \, b^{2} e^{2} n^{3} x x^{4 \, n} + 16 \, b^{2} d e n^{3} x x^{3 \, n} + 16 \, a b e^{2} n^{3} x x^{3 \, n} + 12 \, b^{2} d^{2} n^{3} x x^{2 \, n} + 48 \, a b d e n^{3} x x^{2 \, n} + 12 \, a^{2} e^{2} n^{3} x x^{2 \, n} + 48 \, a b d^{2} n^{3} x x^{n} + 48 \, a^{2} d e n^{3} x x^{n} + 50 \, a^{2} d^{2} n^{3} x + 11 \, b^{2} e^{2} n^{2} x x^{4 \, n} + 28 \, b^{2} d e n^{2} x x^{3 \, n} + 28 \, a b e^{2} n^{2} x x^{3 \, n} + 19 \, b^{2} d^{2} n^{2} x x^{2 \, n} + 76 \, a b d e n^{2} x x^{2 \, n} + 19 \, a^{2} e^{2} n^{2} x x^{2 \, n} + 52 \, a b d^{2} n^{2} x x^{n} + 52 \, a^{2} d e n^{2} x x^{n} + 35 \, a^{2} d^{2} n^{2} x + 6 \, b^{2} e^{2} n x x^{4 \, n} + 14 \, b^{2} d e n x x^{3 \, n} + 14 \, a b e^{2} n x x^{3 \, n} + 8 \, b^{2} d^{2} n x x^{2 \, n} + 32 \, a b d e n x x^{2 \, n} + 8 \, a^{2} e^{2} n x x^{2 \, n} + 18 \, a b d^{2} n x x^{n} + 18 \, a^{2} d e n x x^{n} + 10 \, a^{2} d^{2} n x + b^{2} e^{2} x x^{4 \, n} + 2 \, b^{2} d e x x^{3 \, n} + 2 \, a b e^{2} x x^{3 \, n} + b^{2} d^{2} x x^{2 \, n} + 4 \, a b d e x x^{2 \, n} + a^{2} e^{2} x x^{2 \, n} + 2 \, a b d^{2} x x^{n} + 2 \, a^{2} d e x x^{n} + a^{2} d^{2} x}{24 \, n^{4} + 50 \, n^{3} + 35 \, n^{2} + 10 \, n + 1} \]
(24*a^2*d^2*n^4*x + 6*b^2*e^2*n^3*x*x^(4*n) + 16*b^2*d*e*n^3*x*x^(3*n) + 1 6*a*b*e^2*n^3*x*x^(3*n) + 12*b^2*d^2*n^3*x*x^(2*n) + 48*a*b*d*e*n^3*x*x^(2 *n) + 12*a^2*e^2*n^3*x*x^(2*n) + 48*a*b*d^2*n^3*x*x^n + 48*a^2*d*e*n^3*x*x ^n + 50*a^2*d^2*n^3*x + 11*b^2*e^2*n^2*x*x^(4*n) + 28*b^2*d*e*n^2*x*x^(3*n ) + 28*a*b*e^2*n^2*x*x^(3*n) + 19*b^2*d^2*n^2*x*x^(2*n) + 76*a*b*d*e*n^2*x *x^(2*n) + 19*a^2*e^2*n^2*x*x^(2*n) + 52*a*b*d^2*n^2*x*x^n + 52*a^2*d*e*n^ 2*x*x^n + 35*a^2*d^2*n^2*x + 6*b^2*e^2*n*x*x^(4*n) + 14*b^2*d*e*n*x*x^(3*n ) + 14*a*b*e^2*n*x*x^(3*n) + 8*b^2*d^2*n*x*x^(2*n) + 32*a*b*d*e*n*x*x^(2*n ) + 8*a^2*e^2*n*x*x^(2*n) + 18*a*b*d^2*n*x*x^n + 18*a^2*d*e*n*x*x^n + 10*a ^2*d^2*n*x + b^2*e^2*x*x^(4*n) + 2*b^2*d*e*x*x^(3*n) + 2*a*b*e^2*x*x^(3*n) + b^2*d^2*x*x^(2*n) + 4*a*b*d*e*x*x^(2*n) + a^2*e^2*x*x^(2*n) + 2*a*b*d^2 *x*x^n + 2*a^2*d*e*x*x^n + a^2*d^2*x)/(24*n^4 + 50*n^3 + 35*n^2 + 10*n + 1 )
Time = 5.79 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.96 \[ \int \left (a+b x^n\right )^2 \left (d+e x^n\right )^2 \, dx=a^2\,d^2\,x+\frac {x\,x^{2\,n}\,\left (a^2\,e^2+4\,a\,b\,d\,e+b^2\,d^2\right )}{2\,n+1}+\frac {b^2\,e^2\,x\,x^{4\,n}}{4\,n+1}+\frac {2\,b\,e\,x\,x^{3\,n}\,\left (a\,e+b\,d\right )}{3\,n+1}+\frac {2\,a\,d\,x\,x^n\,\left (a\,e+b\,d\right )}{n+1} \]