Integrand size = 22, antiderivative size = 129 \[ \int (d x)^m \left (a+b x^n+c x^{2 n}\right )^2 \, dx=\frac {a^2 (d x)^{1+m}}{d (1+m)}+\frac {2 a b x^n (d x)^{1+m}}{d (1+m+n)}+\frac {\left (b^2+2 a c\right ) x^{2 n} (d x)^{1+m}}{d (1+m+2 n)}+\frac {2 b c x^{3 n} (d x)^{1+m}}{d (1+m+3 n)}+\frac {c^2 x^{4 n} (d x)^{1+m}}{d (1+m+4 n)} \] Output:
a^2*(d*x)^(1+m)/d/(1+m)+2*a*b*x^n*(d*x)^(1+m)/d/(1+m+n)+(2*a*c+b^2)*x^(2*n )*(d*x)^(1+m)/d/(1+m+2*n)+2*b*c*x^(3*n)*(d*x)^(1+m)/d/(1+m+3*n)+c^2*x^(4*n )*(d*x)^(1+m)/d/(1+m+4*n)
Time = 0.24 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.67 \[ \int (d x)^m \left (a+b x^n+c x^{2 n}\right )^2 \, dx=x (d x)^m \left (\frac {a^2}{1+m}+\frac {2 a b x^n}{1+m+n}+\frac {\left (b^2+2 a c\right ) x^{2 n}}{1+m+2 n}+\frac {2 b c x^{3 n}}{1+m+3 n}+\frac {c^2 x^{4 n}}{1+m+4 n}\right ) \] Input:
Integrate[(d*x)^m*(a + b*x^n + c*x^(2*n))^2,x]
Output:
x*(d*x)^m*(a^2/(1 + m) + (2*a*b*x^n)/(1 + m + n) + ((b^2 + 2*a*c)*x^(2*n)) /(1 + m + 2*n) + (2*b*c*x^(3*n))/(1 + m + 3*n) + (c^2*x^(4*n))/(1 + m + 4* n))
Time = 0.27 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.91, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1691, 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.
| \(\displaystyle \int (d x)^m \left (a+b x^n+c x^{2 n}\right )^2 \, dx\) |
| \(\Big \downarrow \) 1691 |
| \(\displaystyle \int \left (a^2 (d x)^m+b^2 x^{2 n} \left (\frac {2 a c}{b^2}+1\right ) (d x)^m+2 a b x^n (d x)^m+2 b c x^{3 n} (d x)^m+c^2 x^{4 n} (d x)^m\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^2 (d x)^{m+1}}{d (m+1)}+\frac {x^{2 n+1} \left (2 a c+b^2\right ) (d x)^m}{m+2 n+1}+\frac {2 a b x^{n+1} (d x)^m}{m+n+1}+\frac {2 b c x^{3 n+1} (d x)^m}{m+3 n+1}+\frac {c^2 x^{4 n+1} (d x)^m}{m+4 n+1}\) |
Input:
Int[(d*x)^m*(a + b*x^n + c*x^(2*n))^2,x]
Output:
(2*a*b*x^(1 + n)*(d*x)^m)/(1 + m + n) + ((b^2 + 2*a*c)*x^(1 + 2*n)*(d*x)^m )/(1 + m + 2*n) + (2*b*c*x^(1 + 3*n)*(d*x)^m)/(1 + m + 3*n) + (c^2*x^(1 + 4*n)*(d*x)^m)/(1 + m + 4*n) + (a^2*(d*x)^(1 + m))/(d*(1 + m))
Int[((d_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d*x)^m*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[n2, 2*n] && IGtQ[p, 0] && !IntegerQ [Simplify[(m + 1)/n]]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.53 (sec) , antiderivative size = 1032, normalized size of antiderivative = 8.00
| method | result | size |
| risch | \(\text {Expression too large to display}\) | \(1032\) |
| parallelrisch | \(\text {Expression too large to display}\) | \(1566\) |
| orering | \(\text {Expression too large to display}\) | \(2292\) |
Input:
int((d*x)^m*(a+b*x^n+c*x^(2*n))^2,x,method=_RETURNVERBOSE)
Output:
x*(a^2+35*a^2*n^2+38*b^2*m*n^2*(x^n)^2+19*b^2*m^2*n^2*(x^n)^2+12*b^2*m*n^3 *(x^n)^2+8*b*c*m^3*(x^n)^3+16*b*c*n^3*(x^n)^3+18*c^2*m*n*(x^n)^4+2*a*b*m^4 *x^n+8*a*c*m^3*(x^n)^2+24*a*c*n^3*(x^n)^2+24*b^2*m^2*n*(x^n)^2+2*a*c*m^4*( x^n)^2+8*b^2*m^3*n*(x^n)^2+6*c^2*m^3*n*(x^n)^4+11*c^2*m^2*n^2*(x^n)^4+6*c^ 2*m*n^3*(x^n)^4+2*b*c*m^4*(x^n)^3+18*c^2*m^2*n*(x^n)^4+22*c^2*m*n^2*(x^n)^ 4+16*a*c*(x^n)^2*n+24*b^2*m*n*(x^n)^2+8*m*b*c*(x^n)^3+14*b*c*(x^n)^3*n+12* a*b*m^2*x^n+52*a*b*n^2*x^n+8*a*c*(x^n)^2*m+12*b*c*m^2*(x^n)^3+28*b*c*n^2*( x^n)^3+8*a*b*m^3*x^n+48*a*b*n^3*x^n+12*a*c*m^2*(x^n)^2+38*a*c*n^2*(x^n)^2+ 8*x^n*a*b*m+18*x^n*n*a*b+54*a*b*m^2*n*x^n+104*a*b*m*n^2*x^n+48*a*c*m*n*(x^ n)^2+54*a*b*m*n*x^n+16*a*c*m^3*n*(x^n)^2+38*a*c*m^2*n^2*(x^n)^2+24*a*c*m*n ^3*(x^n)^2+42*b*c*m^2*n*(x^n)^3+76*a*c*m*n^2*(x^n)^2+42*b*c*m*n*(x^n)^3+14 *b*c*m^3*n*(x^n)^3+28*b*c*m^2*n^2*(x^n)^3+16*b*c*m*n^3*(x^n)^3+2*(x^n)^2*a *c+50*a^2*n^3+6*a^2*m^2+56*b*c*m*n^2*(x^n)^3+18*a*b*m^3*n*x^n+52*a*b*m^2*n ^2*x^n+48*a*b*m*n^3*x^n+48*a*c*m^2*n*(x^n)^2+4*b^2*m^3*(x^n)^2+10*a^2*n+4* a^2*m+(x^n)^2*b^2+24*a^2*n^4+6*b^2*m^2*(x^n)^2+4*b^2*(x^n)^2*m+a^2*m^4+4*a ^2*m^3+6*c^2*m^2*(x^n)^4+(x^n)^4*c^2+11*c^2*n^2*(x^n)^4+12*b^2*n^3*(x^n)^2 +6*c^2*(x^n)^4*n+19*b^2*n^2*(x^n)^2+b^2*m^4*(x^n)^2+8*b^2*(x^n)^2*n+30*a^2 *m^2*n+70*a^2*m*n^2+30*a^2*m*n+35*a^2*m^2*n^2+50*a^2*m*n^3+4*m*c^2*(x^n)^4 +4*c^2*m^3*(x^n)^4+c^2*m^4*(x^n)^4+6*c^2*n^3*(x^n)^4+2*x^n*a*b+10*a^2*m^3* n+2*(x^n)^3*b*c)/(1+m)/(1+m+n)/(1+m+2*n)/(1+m+3*n)/(1+m+4*n)*x^m*d^m*ex...
Leaf count of result is larger than twice the leaf count of optimal. 706 vs. \(2 (129) = 258\).
Time = 0.21 (sec) , antiderivative size = 706, normalized size of antiderivative = 5.47 \[ \int (d x)^m \left (a+b x^n+c x^{2 n}\right )^2 \, dx =\text {Too large to display} \] Input:
integrate((d*x)^m*(a+b*x^n+c*x^(2*n))^2,x, algorithm="fricas")
Output:
((c^2*m^4 + 4*c^2*m^3 + 6*c^2*m^2 + 6*(c^2*m + c^2)*n^3 + 4*c^2*m + 11*(c^ 2*m^2 + 2*c^2*m + c^2)*n^2 + c^2 + 6*(c^2*m^3 + 3*c^2*m^2 + 3*c^2*m + c^2) *n)*x*x^(4*n)*e^(m*log(d) + m*log(x)) + 2*(b*c*m^4 + 4*b*c*m^3 + 6*b*c*m^2 + 8*(b*c*m + b*c)*n^3 + 4*b*c*m + 14*(b*c*m^2 + 2*b*c*m + b*c)*n^2 + b*c + 7*(b*c*m^3 + 3*b*c*m^2 + 3*b*c*m + b*c)*n)*x*x^(3*n)*e^(m*log(d) + m*log (x)) + ((b^2 + 2*a*c)*m^4 + 4*(b^2 + 2*a*c)*m^3 + 12*(b^2 + 2*a*c + (b^2 + 2*a*c)*m)*n^3 + 6*(b^2 + 2*a*c)*m^2 + 19*((b^2 + 2*a*c)*m^2 + b^2 + 2*a*c + 2*(b^2 + 2*a*c)*m)*n^2 + b^2 + 2*a*c + 4*(b^2 + 2*a*c)*m + 8*((b^2 + 2* a*c)*m^3 + 3*(b^2 + 2*a*c)*m^2 + b^2 + 2*a*c + 3*(b^2 + 2*a*c)*m)*n)*x*x^( 2*n)*e^(m*log(d) + m*log(x)) + 2*(a*b*m^4 + 4*a*b*m^3 + 6*a*b*m^2 + 24*(a* b*m + a*b)*n^3 + 4*a*b*m + 26*(a*b*m^2 + 2*a*b*m + a*b)*n^2 + a*b + 9*(a*b *m^3 + 3*a*b*m^2 + 3*a*b*m + a*b)*n)*x*x^n*e^(m*log(d) + m*log(x)) + (a^2* m^4 + 24*a^2*n^4 + 4*a^2*m^3 + 6*a^2*m^2 + 50*(a^2*m + a^2)*n^3 + 4*a^2*m + 35*(a^2*m^2 + 2*a^2*m + a^2)*n^2 + a^2 + 10*(a^2*m^3 + 3*a^2*m^2 + 3*a^2 *m + a^2)*n)*x*e^(m*log(d) + m*log(x)))/(m^5 + 24*(m + 1)*n^4 + 5*m^4 + 50 *(m^2 + 2*m + 1)*n^3 + 10*m^3 + 35*(m^3 + 3*m^2 + 3*m + 1)*n^2 + 10*m^2 + 10*(m^4 + 4*m^3 + 6*m^2 + 4*m + 1)*n + 5*m + 1)
Leaf count of result is larger than twice the leaf count of optimal. 12323 vs. \(2 (114) = 228\).
Time = 16.31 (sec) , antiderivative size = 12323, normalized size of antiderivative = 95.53 \[ \int (d x)^m \left (a+b x^n+c x^{2 n}\right )^2 \, dx=\text {Too large to display} \] Input:
integrate((d*x)**m*(a+b*x**n+c*x**(2*n))**2,x)
Output:
Piecewise(((a + b + c)**2*log(x)/d, Eq(m, -1) & Eq(n, 0)), ((a**2*log(x) + 2*a*b*x**n/n + a*c*x**(2*n)/n + b**2*x**(2*n)/(2*n) + 2*b*c*x**(3*n)/(3*n ) + c**2*x**(4*n)/(4*n))/d, Eq(m, -1)), (a**2*Piecewise((0**(-4*n - 1)*x, Eq(d, 0)), (Piecewise((-1/(4*n*(d*x)**(4*n)), Ne(n, 0)), (log(d*x), True)) /d, True)) + 2*a*b*Piecewise((-x*x**n*(d*x)**(-4*n - 1)/(3*n), Ne(n, 0)), (x*x**n*(d*x)**(-4*n - 1)*log(x), True)) + 2*a*c*Piecewise((-x*x**(2*n)*(d *x)**(-4*n - 1)/(2*n), Ne(n, 0)), (x*x**(2*n)*(d*x)**(-4*n - 1)*log(x), Tr ue)) + b**2*Piecewise((-x*x**(2*n)*(d*x)**(-4*n - 1)/(2*n), Ne(n, 0)), (x* x**(2*n)*(d*x)**(-4*n - 1)*log(x), True)) + 2*b*c*Piecewise((-x*x**(3*n)*( d*x)**(-4*n - 1)/n, Ne(n, 0)), (x*x**(3*n)*(d*x)**(-4*n - 1)*log(x), True) ) + c**2*x*x**(4*n)*(d*x)**(-4*n - 1)*log(x), Eq(m, -4*n - 1)), (a**2*Piec ewise((0**(-3*n - 1)*x, Eq(d, 0)), (Piecewise((-1/(3*n*(d*x)**(3*n)), Ne(n , 0)), (log(d*x), True))/d, True)) + 2*a*b*Piecewise((-x*x**n*(d*x)**(-3*n - 1)/(2*n), Ne(n, 0)), (x*x**n*(d*x)**(-3*n - 1)*log(x), True)) + 2*a*c*P iecewise((-x*x**(2*n)*(d*x)**(-3*n - 1)/n, Ne(n, 0)), (x*x**(2*n)*(d*x)**( -3*n - 1)*log(x), True)) + b**2*Piecewise((-x*x**(2*n)*(d*x)**(-3*n - 1)/n , Ne(n, 0)), (x*x**(2*n)*(d*x)**(-3*n - 1)*log(x), True)) + 2*b*c*x*x**(3* n)*(d*x)**(-3*n - 1)*log(x) + c**2*Piecewise((x*x**(4*n)*(d*x)**(-3*n - 1) /n, Ne(n, 0)), (x*x**(4*n)*(d*x)**(-3*n - 1)*log(x), True)), Eq(m, -3*n - 1)), (a**2*Piecewise((0**(-2*n - 1)*x, Eq(d, 0)), (Piecewise((-1/(2*n*(...
Time = 0.05 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.18 \[ \int (d x)^m \left (a+b x^n+c x^{2 n}\right )^2 \, dx=\frac {c^{2} d^{m} x e^{\left (m \log \left (x\right ) + 4 \, n \log \left (x\right )\right )}}{m + 4 \, n + 1} + \frac {2 \, b c d^{m} x e^{\left (m \log \left (x\right ) + 3 \, n \log \left (x\right )\right )}}{m + 3 \, n + 1} + \frac {b^{2} d^{m} x e^{\left (m \log \left (x\right ) + 2 \, n \log \left (x\right )\right )}}{m + 2 \, n + 1} + \frac {2 \, a c d^{m} x e^{\left (m \log \left (x\right ) + 2 \, n \log \left (x\right )\right )}}{m + 2 \, n + 1} + \frac {2 \, a b d^{m} x e^{\left (m \log \left (x\right ) + n \log \left (x\right )\right )}}{m + n + 1} + \frac {\left (d x\right )^{m + 1} a^{2}}{d {\left (m + 1\right )}} \] Input:
integrate((d*x)^m*(a+b*x^n+c*x^(2*n))^2,x, algorithm="maxima")
Output:
c^2*d^m*x*e^(m*log(x) + 4*n*log(x))/(m + 4*n + 1) + 2*b*c*d^m*x*e^(m*log(x ) + 3*n*log(x))/(m + 3*n + 1) + b^2*d^m*x*e^(m*log(x) + 2*n*log(x))/(m + 2 *n + 1) + 2*a*c*d^m*x*e^(m*log(x) + 2*n*log(x))/(m + 2*n + 1) + 2*a*b*d^m* x*e^(m*log(x) + n*log(x))/(m + n + 1) + (d*x)^(m + 1)*a^2/(d*(m + 1))
Leaf count of result is larger than twice the leaf count of optimal. 5454 vs. \(2 (129) = 258\).
Time = 0.17 (sec) , antiderivative size = 5454, normalized size of antiderivative = 42.28 \[ \int (d x)^m \left (a+b x^n+c x^{2 n}\right )^2 \, dx=\text {Too large to display} \] Input:
integrate((d*x)^m*(a+b*x^n+c*x^(2*n))^2,x, algorithm="giac")
Output:
(c^2*m^4*x*x^(4*n)*e^(m*log(d) + m*log(x)) + 6*c^2*m^3*n*x*x^(4*n)*e^(m*lo g(d) + m*log(x)) + 11*c^2*m^2*n^2*x*x^(4*n)*e^(m*log(d) + m*log(x)) + 6*c^ 2*m*n^3*x*x^(4*n)*e^(m*log(d) + m*log(x)) + 2*b*c*m^4*x*x^(3*n)*e^(m*log(d ) + m*log(x)) + c^2*m^4*x*x^(3*n)*e^(m*log(d) + m*log(x)) + 14*b*c*m^3*n*x *x^(3*n)*e^(m*log(d) + m*log(x)) + 6*c^2*m^3*n*x*x^(3*n)*e^(m*log(d) + m*l og(x)) + 28*b*c*m^2*n^2*x*x^(3*n)*e^(m*log(d) + m*log(x)) + 11*c^2*m^2*n^2 *x*x^(3*n)*e^(m*log(d) + m*log(x)) + 16*b*c*m*n^3*x*x^(3*n)*e^(m*log(d) + m*log(x)) + 6*c^2*m*n^3*x*x^(3*n)*e^(m*log(d) + m*log(x)) + b^2*m^4*x*x^(2 *n)*e^(m*log(d) + m*log(x)) + 2*a*c*m^4*x*x^(2*n)*e^(m*log(d) + m*log(x)) + 2*b*c*m^4*x*x^(2*n)*e^(m*log(d) + m*log(x)) + c^2*m^4*x*x^(2*n)*e^(m*log (d) + m*log(x)) + 8*b^2*m^3*n*x*x^(2*n)*e^(m*log(d) + m*log(x)) + 16*a*c*m ^3*n*x*x^(2*n)*e^(m*log(d) + m*log(x)) + 14*b*c*m^3*n*x*x^(2*n)*e^(m*log(d ) + m*log(x)) + 6*c^2*m^3*n*x*x^(2*n)*e^(m*log(d) + m*log(x)) + 19*b^2*m^2 *n^2*x*x^(2*n)*e^(m*log(d) + m*log(x)) + 38*a*c*m^2*n^2*x*x^(2*n)*e^(m*log (d) + m*log(x)) + 28*b*c*m^2*n^2*x*x^(2*n)*e^(m*log(d) + m*log(x)) + 11*c^ 2*m^2*n^2*x*x^(2*n)*e^(m*log(d) + m*log(x)) + 12*b^2*m*n^3*x*x^(2*n)*e^(m* log(d) + m*log(x)) + 24*a*c*m*n^3*x*x^(2*n)*e^(m*log(d) + m*log(x)) + 16*b *c*m*n^3*x*x^(2*n)*e^(m*log(d) + m*log(x)) + 6*c^2*m*n^3*x*x^(2*n)*e^(m*lo g(d) + m*log(x)) + 2*a*b*m^4*x*x^n*e^(m*log(d) + m*log(x)) + b^2*m^4*x*x^n *e^(m*log(d) + m*log(x)) + 2*a*c*m^4*x*x^n*e^(m*log(d) + m*log(x)) + 2*...
Time = 10.79 (sec) , antiderivative size = 543, normalized size of antiderivative = 4.21 \[ \int (d x)^m \left (a+b x^n+c x^{2 n}\right )^2 \, dx=\frac {a^2\,x\,{\left (d\,x\right )}^m}{m+1}+\frac {x\,x^{2\,n}\,{\left (d\,x\right )}^m\,\left (b^2+2\,a\,c\right )\,\left (m^3+8\,m^2\,n+3\,m^2+19\,m\,n^2+16\,m\,n+3\,m+12\,n^3+19\,n^2+8\,n+1\right )}{m^4+10\,m^3\,n+4\,m^3+35\,m^2\,n^2+30\,m^2\,n+6\,m^2+50\,m\,n^3+70\,m\,n^2+30\,m\,n+4\,m+24\,n^4+50\,n^3+35\,n^2+10\,n+1}+\frac {c^2\,x\,x^{4\,n}\,{\left (d\,x\right )}^m\,\left (m^3+6\,m^2\,n+3\,m^2+11\,m\,n^2+12\,m\,n+3\,m+6\,n^3+11\,n^2+6\,n+1\right )}{m^4+10\,m^3\,n+4\,m^3+35\,m^2\,n^2+30\,m^2\,n+6\,m^2+50\,m\,n^3+70\,m\,n^2+30\,m\,n+4\,m+24\,n^4+50\,n^3+35\,n^2+10\,n+1}+\frac {2\,a\,b\,x\,x^n\,{\left (d\,x\right )}^m\,\left (m^3+9\,m^2\,n+3\,m^2+26\,m\,n^2+18\,m\,n+3\,m+24\,n^3+26\,n^2+9\,n+1\right )}{m^4+10\,m^3\,n+4\,m^3+35\,m^2\,n^2+30\,m^2\,n+6\,m^2+50\,m\,n^3+70\,m\,n^2+30\,m\,n+4\,m+24\,n^4+50\,n^3+35\,n^2+10\,n+1}+\frac {2\,b\,c\,x\,x^{3\,n}\,{\left (d\,x\right )}^m\,\left (m^3+7\,m^2\,n+3\,m^2+14\,m\,n^2+14\,m\,n+3\,m+8\,n^3+14\,n^2+7\,n+1\right )}{m^4+10\,m^3\,n+4\,m^3+35\,m^2\,n^2+30\,m^2\,n+6\,m^2+50\,m\,n^3+70\,m\,n^2+30\,m\,n+4\,m+24\,n^4+50\,n^3+35\,n^2+10\,n+1} \] Input:
int((d*x)^m*(a + b*x^n + c*x^(2*n))^2,x)
Output:
(a^2*x*(d*x)^m)/(m + 1) + (x*x^(2*n)*(d*x)^m*(2*a*c + b^2)*(3*m + 8*n + 16 *m*n + 19*m*n^2 + 8*m^2*n + 3*m^2 + m^3 + 19*n^2 + 12*n^3 + 1))/(4*m + 10* n + 30*m*n + 70*m*n^2 + 30*m^2*n + 50*m*n^3 + 10*m^3*n + 6*m^2 + 4*m^3 + m ^4 + 35*n^2 + 50*n^3 + 24*n^4 + 35*m^2*n^2 + 1) + (c^2*x*x^(4*n)*(d*x)^m*( 3*m + 6*n + 12*m*n + 11*m*n^2 + 6*m^2*n + 3*m^2 + m^3 + 11*n^2 + 6*n^3 + 1 ))/(4*m + 10*n + 30*m*n + 70*m*n^2 + 30*m^2*n + 50*m*n^3 + 10*m^3*n + 6*m^ 2 + 4*m^3 + m^4 + 35*n^2 + 50*n^3 + 24*n^4 + 35*m^2*n^2 + 1) + (2*a*b*x*x^ n*(d*x)^m*(3*m + 9*n + 18*m*n + 26*m*n^2 + 9*m^2*n + 3*m^2 + m^3 + 26*n^2 + 24*n^3 + 1))/(4*m + 10*n + 30*m*n + 70*m*n^2 + 30*m^2*n + 50*m*n^3 + 10* m^3*n + 6*m^2 + 4*m^3 + m^4 + 35*n^2 + 50*n^3 + 24*n^4 + 35*m^2*n^2 + 1) + (2*b*c*x*x^(3*n)*(d*x)^m*(3*m + 7*n + 14*m*n + 14*m*n^2 + 7*m^2*n + 3*m^2 + m^3 + 14*n^2 + 8*n^3 + 1))/(4*m + 10*n + 30*m*n + 70*m*n^2 + 30*m^2*n + 50*m*n^3 + 10*m^3*n + 6*m^2 + 4*m^3 + m^4 + 35*n^2 + 50*n^3 + 24*n^4 + 35 *m^2*n^2 + 1)
Time = 0.20 (sec) , antiderivative size = 1063, normalized size of antiderivative = 8.24 \[ \int (d x)^m \left (a+b x^n+c x^{2 n}\right )^2 \, dx =\text {Too large to display} \] Input:
int((d*x)^m*(a+b*x^n+c*x^(2*n))^2,x)
Output:
(x**m*d**m*x*(x**(4*n)*c**2*m**4 + 6*x**(4*n)*c**2*m**3*n + 4*x**(4*n)*c** 2*m**3 + 11*x**(4*n)*c**2*m**2*n**2 + 18*x**(4*n)*c**2*m**2*n + 6*x**(4*n) *c**2*m**2 + 6*x**(4*n)*c**2*m*n**3 + 22*x**(4*n)*c**2*m*n**2 + 18*x**(4*n )*c**2*m*n + 4*x**(4*n)*c**2*m + 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 + 2*x**(3*n)*b*c*m**4 + 14*x**(3*n )*b*c*m**3*n + 8*x**(3*n)*b*c*m**3 + 28*x**(3*n)*b*c*m**2*n**2 + 42*x**(3* n)*b*c*m**2*n + 12*x**(3*n)*b*c*m**2 + 16*x**(3*n)*b*c*m*n**3 + 56*x**(3*n )*b*c*m*n**2 + 42*x**(3*n)*b*c*m*n + 8*x**(3*n)*b*c*m + 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 + 2*x**(2*n )*a*c*m**4 + 16*x**(2*n)*a*c*m**3*n + 8*x**(2*n)*a*c*m**3 + 38*x**(2*n)*a* c*m**2*n**2 + 48*x**(2*n)*a*c*m**2*n + 12*x**(2*n)*a*c*m**2 + 24*x**(2*n)* a*c*m*n**3 + 76*x**(2*n)*a*c*m*n**2 + 48*x**(2*n)*a*c*m*n + 8*x**(2*n)*a*c *m + 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 + x**(2*n)*b**2*m**4 + 8*x**(2*n)*b**2*m**3*n + 4*x**(2*n)*b** 2*m**3 + 19*x**(2*n)*b**2*m**2*n**2 + 24*x**(2*n)*b**2*m**2*n + 6*x**(2*n) *b**2*m**2 + 12*x**(2*n)*b**2*m*n**3 + 38*x**(2*n)*b**2*m*n**2 + 24*x**(2* n)*b**2*m*n + 4*x**(2*n)*b**2*m + 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 + 2*x**n*a*b*m**4 + 18*x**n*a*b* m**3*n + 8*x**n*a*b*m**3 + 52*x**n*a*b*m**2*n**2 + 54*x**n*a*b*m**2*n + 12 *x**n*a*b*m**2 + 48*x**n*a*b*m*n**3 + 104*x**n*a*b*m*n**2 + 54*x**n*a*b...