Integrand size = 16, antiderivative size = 79 \[ \int \left (a+b x^n+c x^{2 n}\right )^2 \, dx=a^2 x+\frac {2 a b x^{1+n}}{1+n}+\frac {\left (b^2+2 a c\right ) x^{1+2 n}}{1+2 n}+\frac {2 b c x^{1+3 n}}{1+3 n}+\frac {c^2 x^{1+4 n}}{1+4 n} \] Output:
a^2*x+2*a*b*x^(1+n)/(1+n)+(2*a*c+b^2)*x^(1+2*n)/(1+2*n)+2*b*c*x^(1+3*n)/(1 +3*n)+c^2*x^(1+4*n)/(1+4*n)
Time = 0.15 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.90 \[ \int \left (a+b x^n+c x^{2 n}\right )^2 \, dx=x \left (a^2+\frac {2 a b x^n}{1+n}+\frac {\left (b^2+2 a c\right ) x^{2 n}}{1+2 n}+\frac {2 b c x^{3 n}}{1+3 n}+\frac {c^2 x^{4 n}}{1+4 n}\right ) \] Input:
Integrate[(a + b*x^n + c*x^(2*n))^2,x]
Output:
x*(a^2 + (2*a*b*x^n)/(1 + n) + ((b^2 + 2*a*c)*x^(2*n))/(1 + 2*n) + (2*b*c* x^(3*n))/(1 + 3*n) + (c^2*x^(4*n))/(1 + 4*n))
Time = 0.23 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1682, 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.
\(\displaystyle \int \left (a+b x^n+c x^{2 n}\right )^2 \, dx\) |
\(\Big \downarrow \) 1682 |
\(\displaystyle \int \left (a^2+b^2 x^{2 n} \left (\frac {2 a c}{b^2}+1\right )+2 a b x^n+2 b c x^{3 n}+c^2 x^{4 n}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle a^2 x+\frac {x^{2 n+1} \left (2 a c+b^2\right )}{2 n+1}+\frac {2 a b x^{n+1}}{n+1}+\frac {2 b c x^{3 n+1}}{3 n+1}+\frac {c^2 x^{4 n+1}}{4 n+1}\) |
Input:
Int[(a + b*x^n + c*x^(2*n))^2,x]
Output:
a^2*x + (2*a*b*x^(1 + n))/(1 + n) + ((b^2 + 2*a*c)*x^(1 + 2*n))/(1 + 2*n) + (2*b*c*x^(1 + 3*n))/(1 + 3*n) + (c^2*x^(1 + 4*n))/(1 + 4*n)
Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[Exp andIntegrand[(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0]
Time = 0.19 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.96
method | result | size |
risch | \(x \,a^{2}+\frac {c^{2} x \,x^{4 n}}{1+4 n}+\frac {\left (2 a c +b^{2}\right ) x \,x^{2 n}}{1+2 n}+\frac {2 b a x \,x^{n}}{1+n}+\frac {2 b c x \,x^{3 n}}{1+3 n}\) | \(76\) |
norman | \(x \,a^{2}+\frac {c^{2} x \,{\mathrm e}^{4 n \ln \left (x \right )}}{1+4 n}+\frac {\left (2 a c +b^{2}\right ) x \,{\mathrm e}^{2 n \ln \left (x \right )}}{1+2 n}+\frac {2 b a x \,{\mathrm e}^{n \ln \left (x \right )}}{1+n}+\frac {2 b c x \,{\mathrm e}^{3 n \ln \left (x \right )}}{1+3 n}\) | \(84\) |
parallelrisch | \(\frac {12 x \,x^{2 n} b^{2} n^{3}+6 x \,x^{4 n} c^{2} n^{3}+19 x \,x^{2 n} b^{2} n^{2}+11 x \,x^{4 n} c^{2} n^{2}+8 x \,x^{2 n} b^{2} n +6 x \,x^{4 n} c^{2} n +16 x \,x^{n} x^{2 n} b c \,n^{3}+28 x \,x^{n} x^{2 n} b c \,n^{2}+14 x \,x^{n} x^{2 n} b c n +2 x \,x^{n} a b +2 x \,x^{2 n} a c +48 x \,x^{n} a b \,n^{3}+x \,a^{2}+24 x \,x^{2 n} a c \,n^{3}+52 x \,x^{n} a b \,n^{2}+38 x \,x^{2 n} a c \,n^{2}+2 x \,x^{n} x^{2 n} b c +18 x \,x^{n} a b n +16 x \,x^{2 n} a c n +c^{2} x \,x^{4 n}+x \,x^{2 n} b^{2}+24 x \,a^{2} n^{4}+50 x \,a^{2} n^{3}+35 x \,a^{2} n^{2}+10 x \,a^{2} n}{\left (1+4 n \right ) \left (1+2 n \right ) \left (1+n \right ) \left (1+3 n \right )}\) | \(321\) |
orering | \(x \left (a +b \,x^{n}+c \,x^{2 n}\right )^{2}-\frac {20 x^{2} n \left (5 n^{2}+1\right ) \left (a +b \,x^{n}+c \,x^{2 n}\right ) \left (\frac {b \,x^{n} n}{x}+\frac {2 c \,x^{2 n} n}{x}\right )}{\left (12 n^{3}+19 n^{2}+8 n +1\right ) \left (1+2 n \right )}+\frac {5 x^{3} \left (7 n^{2}-4 n +1\right ) \left (2 \left (\frac {b \,x^{n} n}{x}+\frac {2 c \,x^{2 n} n}{x}\right )^{2}+2 \left (a +b \,x^{n}+c \,x^{2 n}\right ) \left (\frac {b \,x^{n} n^{2}}{x^{2}}-\frac {b \,x^{n} n}{x^{2}}+\frac {4 c \,x^{2 n} n^{2}}{x^{2}}-\frac {2 c \,x^{2 n} n}{x^{2}}\right )\right )}{\left (12 n^{3}+19 n^{2}+8 n +1\right ) \left (1+2 n \right )}-\frac {5 x^{4} \left (-1+2 n \right ) \left (6 \left (\frac {b \,x^{n} n}{x}+\frac {2 c \,x^{2 n} n}{x}\right ) \left (\frac {b \,x^{n} n^{2}}{x^{2}}-\frac {b \,x^{n} n}{x^{2}}+\frac {4 c \,x^{2 n} n^{2}}{x^{2}}-\frac {2 c \,x^{2 n} n}{x^{2}}\right )+2 \left (a +b \,x^{n}+c \,x^{2 n}\right ) \left (\frac {b \,x^{n} n^{3}}{x^{3}}-\frac {3 b \,x^{n} n^{2}}{x^{3}}+\frac {2 b \,x^{n} n}{x^{3}}+\frac {8 c \,x^{2 n} n^{3}}{x^{3}}-\frac {12 c \,x^{2 n} n^{2}}{x^{3}}+\frac {4 c \,x^{2 n} n}{x^{3}}\right )\right )}{24 n^{4}+50 n^{3}+35 n^{2}+10 n +1}+\frac {x^{5} \left (6 \left (\frac {b \,x^{n} n^{2}}{x^{2}}-\frac {b \,x^{n} n}{x^{2}}+\frac {4 c \,x^{2 n} n^{2}}{x^{2}}-\frac {2 c \,x^{2 n} n}{x^{2}}\right )^{2}+8 \left (\frac {b \,x^{n} n}{x}+\frac {2 c \,x^{2 n} n}{x}\right ) \left (\frac {b \,x^{n} n^{3}}{x^{3}}-\frac {3 b \,x^{n} n^{2}}{x^{3}}+\frac {2 b \,x^{n} n}{x^{3}}+\frac {8 c \,x^{2 n} n^{3}}{x^{3}}-\frac {12 c \,x^{2 n} n^{2}}{x^{3}}+\frac {4 c \,x^{2 n} n}{x^{3}}\right )+2 \left (a +b \,x^{n}+c \,x^{2 n}\right ) \left (\frac {b \,x^{n} n^{4}}{x^{4}}-\frac {6 b \,x^{n} n^{3}}{x^{4}}+\frac {11 b \,x^{n} n^{2}}{x^{4}}-\frac {6 b \,x^{n} n}{x^{4}}+\frac {16 c \,x^{2 n} n^{4}}{x^{4}}-\frac {48 c \,x^{2 n} n^{3}}{x^{4}}+\frac {44 c \,x^{2 n} n^{2}}{x^{4}}-\frac {12 c \,x^{2 n} n}{x^{4}}\right )\right )}{24 n^{4}+50 n^{3}+35 n^{2}+10 n +1}\) | \(711\) |
Input:
int((a+b*x^n+c*x^(2*n))^2,x,method=_RETURNVERBOSE)
Output:
x*a^2+c^2/(1+4*n)*x*(x^n)^4+(2*a*c+b^2)/(1+2*n)*x*(x^n)^2+2*b*a/(1+n)*x*x^ n+2*b*c/(1+3*n)*x*(x^n)^3
Leaf count of result is larger than twice the leaf count of optimal. 205 vs. \(2 (79) = 158\).
Time = 0.07 (sec) , antiderivative size = 205, normalized size of antiderivative = 2.59 \[ \int \left (a+b x^n+c x^{2 n}\right )^2 \, dx=\frac {{\left (6 \, c^{2} n^{3} + 11 \, c^{2} n^{2} + 6 \, c^{2} n + c^{2}\right )} x x^{4 \, n} + 2 \, {\left (8 \, b c n^{3} + 14 \, b c n^{2} + 7 \, b c n + b c\right )} x x^{3 \, n} + {\left (12 \, {\left (b^{2} + 2 \, a c\right )} n^{3} + 19 \, {\left (b^{2} + 2 \, a c\right )} n^{2} + b^{2} + 2 \, a c + 8 \, {\left (b^{2} + 2 \, a c\right )} n\right )} x x^{2 \, n} + 2 \, {\left (24 \, a b n^{3} + 26 \, a b n^{2} + 9 \, a b n + a b\right )} x x^{n} + {\left (24 \, a^{2} n^{4} + 50 \, a^{2} n^{3} + 35 \, a^{2} n^{2} + 10 \, a^{2} n + a^{2}\right )} x}{24 \, n^{4} + 50 \, n^{3} + 35 \, n^{2} + 10 \, n + 1} \] Input:
integrate((a+b*x^n+c*x^(2*n))^2,x, algorithm="fricas")
Output:
((6*c^2*n^3 + 11*c^2*n^2 + 6*c^2*n + c^2)*x*x^(4*n) + 2*(8*b*c*n^3 + 14*b* c*n^2 + 7*b*c*n + b*c)*x*x^(3*n) + (12*(b^2 + 2*a*c)*n^3 + 19*(b^2 + 2*a*c )*n^2 + b^2 + 2*a*c + 8*(b^2 + 2*a*c)*n)*x*x^(2*n) + 2*(24*a*b*n^3 + 26*a* b*n^2 + 9*a*b*n + a*b)*x*x^n + (24*a^2*n^4 + 50*a^2*n^3 + 35*a^2*n^2 + 10* a^2*n + a^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. 1027 vs. \(2 (70) = 140\).
Time = 2.28 (sec) , antiderivative size = 1027, normalized size of antiderivative = 13.00 \[ \int \left (a+b x^n+c x^{2 n}\right )^2 \, dx =\text {Too large to display} \] Input:
integrate((a+b*x**n+c*x**(2*n))**2,x)
Output:
Piecewise((a**2*x + 2*a*b*log(x) - 2*a*c/x - b**2/x - b*c/x**2 - c**2/(3*x **3), Eq(n, -1)), (a**2*x + 4*a*b*sqrt(x) + 2*a*c*log(x) + b**2*log(x) - 4 *b*c/sqrt(x) - c**2/x, Eq(n, -1/2)), (a**2*x + 3*a*b*x**(2/3) + 6*a*c*x**( 1/3) + 3*b**2*x**(1/3) + 2*b*c*log(x) - 3*c**2/x**(1/3), Eq(n, -1/3)), (a* *2*x + 8*a*b*x**(3/4)/3 + 8*b*c*x**(1/4) + 4*c**2*log(x**(1/4)) - 2*sqrt(x )*(-2*a*c - b**2), Eq(n, -1/4)), (24*a**2*n**4*x/(24*n**4 + 50*n**3 + 35*n **2 + 10*n + 1) + 50*a**2*n**3*x/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 35*a**2*n**2*x/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 10*a**2*n*x/(2 4*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + a**2*x/(24*n**4 + 50*n**3 + 35*n* *2 + 10*n + 1) + 48*a*b*n**3*x*x**n/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 52*a*b*n**2*x*x**n/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 18*a*b* n*x*x**n/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 2*a*b*x*x**n/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 24*a*c*n**3*x*x**(2*n)/(24*n**4 + 50*n** 3 + 35*n**2 + 10*n + 1) + 38*a*c*n**2*x*x**(2*n)/(24*n**4 + 50*n**3 + 35*n **2 + 10*n + 1) + 16*a*c*n*x*x**(2*n)/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 2*a*c*x*x**(2*n)/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 12*b**2 *n**3*x*x**(2*n)/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 19*b**2*n**2*x *x**(2*n)/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + 8*b**2*n*x*x**(2*n)/( 24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1) + b**2*x*x**(2*n)/(24*n**4 + 50*n* *3 + 35*n**2 + 10*n + 1) + 16*b*c*n**3*x*x**(3*n)/(24*n**4 + 50*n**3 + ...
Time = 0.03 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.16 \[ \int \left (a+b x^n+c x^{2 n}\right )^2 \, dx=a^{2} x + \frac {c^{2} x^{4 \, n + 1}}{4 \, n + 1} + \frac {2 \, b c x^{3 \, n + 1}}{3 \, n + 1} + \frac {b^{2} x^{2 \, n + 1}}{2 \, n + 1} + 2 \, a {\left (\frac {c x^{2 \, n + 1}}{2 \, n + 1} + \frac {b x^{n + 1}}{n + 1}\right )} \] Input:
integrate((a+b*x^n+c*x^(2*n))^2,x, algorithm="maxima")
Output:
a^2*x + c^2*x^(4*n + 1)/(4*n + 1) + 2*b*c*x^(3*n + 1)/(3*n + 1) + b^2*x^(2 *n + 1)/(2*n + 1) + 2*a*(c*x^(2*n + 1)/(2*n + 1) + b*x^(n + 1)/(n + 1))
Leaf count of result is larger than twice the leaf count of optimal. 296 vs. \(2 (79) = 158\).
Time = 0.12 (sec) , antiderivative size = 296, normalized size of antiderivative = 3.75 \[ \int \left (a+b x^n+c x^{2 n}\right )^2 \, dx=\frac {24 \, a^{2} n^{4} x + 6 \, c^{2} n^{3} x x^{4 \, n} + 16 \, b c n^{3} x x^{3 \, n} + 12 \, b^{2} n^{3} x x^{2 \, n} + 24 \, a c n^{3} x x^{2 \, n} + 48 \, a b n^{3} x x^{n} + 50 \, a^{2} n^{3} x + 11 \, c^{2} n^{2} x x^{4 \, n} + 28 \, b c n^{2} x x^{3 \, n} + 19 \, b^{2} n^{2} x x^{2 \, n} + 38 \, a c n^{2} x x^{2 \, n} + 52 \, a b n^{2} x x^{n} + 35 \, a^{2} n^{2} x + 6 \, c^{2} n x x^{4 \, n} + 14 \, b c n x x^{3 \, n} + 8 \, b^{2} n x x^{2 \, n} + 16 \, a c n x x^{2 \, n} + 18 \, a b n x x^{n} + 10 \, a^{2} n x + c^{2} x x^{4 \, n} + 2 \, b c x x^{3 \, n} + b^{2} x x^{2 \, n} + 2 \, a c x x^{2 \, n} + 2 \, a b x x^{n} + a^{2} x}{24 \, n^{4} + 50 \, n^{3} + 35 \, n^{2} + 10 \, n + 1} \] Input:
integrate((a+b*x^n+c*x^(2*n))^2,x, algorithm="giac")
Output:
(24*a^2*n^4*x + 6*c^2*n^3*x*x^(4*n) + 16*b*c*n^3*x*x^(3*n) + 12*b^2*n^3*x* x^(2*n) + 24*a*c*n^3*x*x^(2*n) + 48*a*b*n^3*x*x^n + 50*a^2*n^3*x + 11*c^2* n^2*x*x^(4*n) + 28*b*c*n^2*x*x^(3*n) + 19*b^2*n^2*x*x^(2*n) + 38*a*c*n^2*x *x^(2*n) + 52*a*b*n^2*x*x^n + 35*a^2*n^2*x + 6*c^2*n*x*x^(4*n) + 14*b*c*n* x*x^(3*n) + 8*b^2*n*x*x^(2*n) + 16*a*c*n*x*x^(2*n) + 18*a*b*n*x*x^n + 10*a ^2*n*x + c^2*x*x^(4*n) + 2*b*c*x*x^(3*n) + b^2*x*x^(2*n) + 2*a*c*x*x^(2*n) + 2*a*b*x*x^n + a^2*x)/(24*n^4 + 50*n^3 + 35*n^2 + 10*n + 1)
Time = 20.56 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.95 \[ \int \left (a+b x^n+c x^{2 n}\right )^2 \, dx=a^2\,x+\frac {x\,x^{2\,n}\,\left (b^2+2\,a\,c\right )}{2\,n+1}+\frac {c^2\,x\,x^{4\,n}}{4\,n+1}+\frac {2\,a\,b\,x\,x^n}{n+1}+\frac {2\,b\,c\,x\,x^{3\,n}}{3\,n+1} \] Input:
int((a + b*x^n + c*x^(2*n))^2,x)
Output:
a^2*x + (x*x^(2*n)*(2*a*c + b^2))/(2*n + 1) + (c^2*x*x^(4*n))/(4*n + 1) + (2*a*b*x*x^n)/(n + 1) + (2*b*c*x*x^(3*n))/(3*n + 1)
Time = 0.16 (sec) , antiderivative size = 271, normalized size of antiderivative = 3.43 \[ \int \left (a+b x^n+c x^{2 n}\right )^2 \, dx=\frac {x \left (16 x^{3 n} b c \,n^{3}+28 x^{3 n} b c \,n^{2}+14 x^{3 n} b c n +24 x^{2 n} a c \,n^{3}+38 x^{2 n} a c \,n^{2}+16 x^{2 n} a c n +48 x^{n} a b \,n^{3}+52 x^{n} a b \,n^{2}+18 x^{n} a b n +x^{2 n} b^{2}+24 a^{2} n^{4}+50 a^{2} n^{3}+35 a^{2} n^{2}+10 a^{2} n +a^{2}+6 x^{4 n} c^{2} n^{3}+11 x^{4 n} c^{2} n^{2}+6 x^{4 n} c^{2} n +2 x^{3 n} b c +2 x^{2 n} a c +12 x^{2 n} b^{2} n^{3}+19 x^{2 n} b^{2} n^{2}+8 x^{2 n} b^{2} n +2 x^{n} a b +x^{4 n} c^{2}\right )}{24 n^{4}+50 n^{3}+35 n^{2}+10 n +1} \] Input:
int((a+b*x^n+c*x^(2*n))^2,x)
Output:
(x*(6*x**(4*n)*c**2*n**3 + 11*x**(4*n)*c**2*n**2 + 6*x**(4*n)*c**2*n + x** (4*n)*c**2 + 16*x**(3*n)*b*c*n**3 + 28*x**(3*n)*b*c*n**2 + 14*x**(3*n)*b*c *n + 2*x**(3*n)*b*c + 24*x**(2*n)*a*c*n**3 + 38*x**(2*n)*a*c*n**2 + 16*x** (2*n)*a*c*n + 2*x**(2*n)*a*c + 12*x**(2*n)*b**2*n**3 + 19*x**(2*n)*b**2*n* *2 + 8*x**(2*n)*b**2*n + x**(2*n)*b**2 + 48*x**n*a*b*n**3 + 52*x**n*a*b*n* *2 + 18*x**n*a*b*n + 2*x**n*a*b + 24*a**2*n**4 + 50*a**2*n**3 + 35*a**2*n* *2 + 10*a**2*n + a**2))/(24*n**4 + 50*n**3 + 35*n**2 + 10*n + 1)