Integrand size = 26, antiderivative size = 146 \[ \int (e x)^m (c+d x) \left (a x^n+b x^{1+n}\right )^3 \, dx=\frac {3 a b (b c+a d) x^{3 (1+n)} (e x)^m}{3+m+3 n}+\frac {a^3 c x^{1+3 n} (e x)^m}{1+m+3 n}+\frac {a^2 (3 b c+a d) x^{2+3 n} (e x)^m}{2+m+3 n}+\frac {b^2 (b c+3 a d) x^{4+3 n} (e x)^m}{4+m+3 n}+\frac {b^3 d x^{5+3 n} (e x)^m}{5+m+3 n} \] Output:
3*a*b*(a*d+b*c)*x^(3+3*n)*(e*x)^m/(3+m+3*n)+a^3*c*x^(1+3*n)*(e*x)^m/(1+m+3 *n)+a^2*(a*d+3*b*c)*x^(2+3*n)*(e*x)^m/(2+m+3*n)+b^2*(3*a*d+b*c)*x^(4+3*n)* (e*x)^m/(4+m+3*n)+b^3*d*x^(5+3*n)*(e*x)^m/(5+m+3*n)
Time = 0.20 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.79 \[ \int (e x)^m (c+d x) \left (a x^n+b x^{1+n}\right )^3 \, dx=\frac {x^{1+3 n} (e x)^m \left (d (a+b x)^4+(-a d (1+m+3 n)+b c (5+m+3 n)) \left (\frac {a^3}{1+m+3 n}+\frac {3 a^2 b x}{2+m+3 n}+\frac {3 a b^2 x^2}{3+m+3 n}+\frac {b^3 x^3}{4+m+3 n}\right )\right )}{b (5+m+3 n)} \] Input:
Integrate[(e*x)^m*(c + d*x)*(a*x^n + b*x^(1 + n))^3,x]
Output:
(x^(1 + 3*n)*(e*x)^m*(d*(a + b*x)^4 + (-(a*d*(1 + m + 3*n)) + b*c*(5 + m + 3*n))*(a^3/(1 + m + 3*n) + (3*a^2*b*x)/(2 + m + 3*n) + (3*a*b^2*x^2)/(3 + m + 3*n) + (b^3*x^3)/(4 + m + 3*n))))/(b*(5 + m + 3*n))
Time = 0.54 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.94, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2027, 30, 85, 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.
\(\displaystyle \int (c+d x) (e x)^m \left (a x^n+b x^{n+1}\right )^3 \, dx\) |
\(\Big \downarrow \) 2027 |
\(\displaystyle \int x^{3 n} (a+b x)^3 (c+d x) (e x)^mdx\) |
\(\Big \downarrow \) 30 |
\(\displaystyle x^{-m} (e x)^m \int x^{m+3 n} (a+b x)^3 (c+d x)dx\) |
\(\Big \downarrow \) 85 |
\(\displaystyle x^{-m} (e x)^m \int \left (a^3 c x^{m+3 n}+a^2 (3 b c+a d) x^{m+3 n+1}+3 a b (b c+a d) x^{m+3 n+2}+b^2 (b c+3 a d) x^{m+3 n+3}+b^3 d x^{m+3 n+4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x^{-m} (e x)^m \left (\frac {a^3 c x^{m+3 n+1}}{m+3 n+1}+\frac {a^2 x^{m+3 n+2} (a d+3 b c)}{m+3 n+2}+\frac {b^2 x^{m+3 n+4} (3 a d+b c)}{m+3 n+4}+\frac {3 a b x^{m+3 n+3} (a d+b c)}{m+3 n+3}+\frac {b^3 d x^{m+3 n+5}}{m+3 n+5}\right )\) |
Input:
Int[(e*x)^m*(c + d*x)*(a*x^n + b*x^(1 + n))^3,x]
Output:
((e*x)^m*((a^3*c*x^(1 + m + 3*n))/(1 + m + 3*n) + (a^2*(3*b*c + a*d)*x^(2 + m + 3*n))/(2 + m + 3*n) + (3*a*b*(b*c + a*d)*x^(3 + m + 3*n))/(3 + m + 3 *n) + (b^2*(b*c + 3*a*d)*x^(4 + m + 3*n))/(4 + m + 3*n) + (b^3*d*x^(5 + m + 3*n))/(5 + m + 3*n)))/x^m
Int[(u_.)*((a_.)*(x_))^(m_.)*((b_.)*(x_)^(i_.))^(p_), x_Symbol] :> Simp[b^I ntPart[p]*((b*x^i)^FracPart[p]/(a^(i*IntPart[p])*(a*x)^(i*FracPart[p]))) Int[u*(a*x)^(m + i*p), x], x] /; FreeQ[{a, b, i, m, p}, x] && IntegerQ[i] & & !IntegerQ[p]
Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : > Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && NeQ[b*e + a* f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])
Int[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.))^(p_.), x_Symbol] :> Int[x^ (p*r)*(a + b*x^(s - r))^p*Fx, x] /; FreeQ[{a, b, r, s}, x] && IntegerQ[p] & & PosQ[s - r] && !(EqQ[p, 1] && EqQ[u, 1])
Leaf count of result is larger than twice the leaf count of optimal. \(1453\) vs. \(2(146)=292\).
Time = 7.01 (sec) , antiderivative size = 1454, normalized size of antiderivative = 9.96
method | result | size |
orering | \(\text {Expression too large to display}\) | \(1454\) |
risch | \(\text {Expression too large to display}\) | \(1478\) |
parallelrisch | \(\text {Expression too large to display}\) | \(2971\) |
Input:
int((e*x)^m*(d*x+c)*(a*x^n+b*x^(1+n))^3,x,method=_RETURNVERBOSE)
Output:
(b^3*d*m^4*x^4+12*b^3*d*m^3*n*x^4+54*b^3*d*m^2*n^2*x^4+108*b^3*d*m*n^3*x^4 +81*b^3*d*n^4*x^4+3*a*b^2*d*m^4*x^3+36*a*b^2*d*m^3*n*x^3+162*a*b^2*d*m^2*n ^2*x^3+324*a*b^2*d*m*n^3*x^3+243*a*b^2*d*n^4*x^3+b^3*c*m^4*x^3+12*b^3*c*m^ 3*n*x^3+54*b^3*c*m^2*n^2*x^3+108*b^3*c*m*n^3*x^3+81*b^3*c*n^4*x^3+10*b^3*d *m^3*x^4+90*b^3*d*m^2*n*x^4+270*b^3*d*m*n^2*x^4+270*b^3*d*n^3*x^4+3*a^2*b* d*m^4*x^2+36*a^2*b*d*m^3*n*x^2+162*a^2*b*d*m^2*n^2*x^2+324*a^2*b*d*m*n^3*x ^2+243*a^2*b*d*n^4*x^2+3*a*b^2*c*m^4*x^2+36*a*b^2*c*m^3*n*x^2+162*a*b^2*c* m^2*n^2*x^2+324*a*b^2*c*m*n^3*x^2+243*a*b^2*c*n^4*x^2+33*a*b^2*d*m^3*x^3+2 97*a*b^2*d*m^2*n*x^3+891*a*b^2*d*m*n^2*x^3+891*a*b^2*d*n^3*x^3+11*b^3*c*m^ 3*x^3+99*b^3*c*m^2*n*x^3+297*b^3*c*m*n^2*x^3+297*b^3*c*n^3*x^3+35*b^3*d*m^ 2*x^4+210*b^3*d*m*n*x^4+315*b^3*d*n^2*x^4+a^3*d*m^4*x+12*a^3*d*m^3*n*x+54* a^3*d*m^2*n^2*x+108*a^3*d*m*n^3*x+81*a^3*d*n^4*x+3*a^2*b*c*m^4*x+36*a^2*b* c*m^3*n*x+162*a^2*b*c*m^2*n^2*x+324*a^2*b*c*m*n^3*x+243*a^2*b*c*n^4*x+36*a ^2*b*d*m^3*x^2+324*a^2*b*d*m^2*n*x^2+972*a^2*b*d*m*n^2*x^2+972*a^2*b*d*n^3 *x^2+36*a*b^2*c*m^3*x^2+324*a*b^2*c*m^2*n*x^2+972*a*b^2*c*m*n^2*x^2+972*a* b^2*c*n^3*x^2+123*a*b^2*d*m^2*x^3+738*a*b^2*d*m*n*x^3+1107*a*b^2*d*n^2*x^3 +41*b^3*c*m^2*x^3+246*b^3*c*m*n*x^3+369*b^3*c*n^2*x^3+50*b^3*d*m*x^4+150*b ^3*d*n*x^4+a^3*c*m^4+12*a^3*c*m^3*n+54*a^3*c*m^2*n^2+108*a^3*c*m*n^3+81*a^ 3*c*n^4+13*a^3*d*m^3*x+117*a^3*d*m^2*n*x+351*a^3*d*m*n^2*x+351*a^3*d*n^3*x +39*a^2*b*c*m^3*x+351*a^2*b*c*m^2*n*x+1053*a^2*b*c*m*n^2*x+1053*a^2*b*c...
Leaf count of result is larger than twice the leaf count of optimal. 1171 vs. \(2 (146) = 292\).
Time = 0.14 (sec) , antiderivative size = 1171, normalized size of antiderivative = 8.02 \[ \int (e x)^m (c+d x) \left (a x^n+b x^{1+n}\right )^3 \, dx=\text {Too large to display} \] Input:
integrate((e*x)^m*(d*x+c)*(a*x^n+b*x^(1+n))^3,x, algorithm="fricas")
Output:
(a^3*c*m^4 + 81*a^3*c*n^4 + 14*a^3*c*m^3 + 71*a^3*c*m^2 + 154*a^3*c*m + (b ^3*d*m^4 + 81*b^3*d*n^4 + 10*b^3*d*m^3 + 35*b^3*d*m^2 + 50*b^3*d*m + 24*b^ 3*d + 54*(2*b^3*d*m + 5*b^3*d)*n^3 + 9*(6*b^3*d*m^2 + 30*b^3*d*m + 35*b^3* d)*n^2 + 6*(2*b^3*d*m^3 + 15*b^3*d*m^2 + 35*b^3*d*m + 25*b^3*d)*n)*x^4 + 1 20*a^3*c + 54*(2*a^3*c*m + 7*a^3*c)*n^3 + ((b^3*c + 3*a*b^2*d)*m^4 + 81*(b ^3*c + 3*a*b^2*d)*n^4 + 30*b^3*c + 90*a*b^2*d + 11*(b^3*c + 3*a*b^2*d)*m^3 + 27*(11*b^3*c + 33*a*b^2*d + 4*(b^3*c + 3*a*b^2*d)*m)*n^3 + 41*(b^3*c + 3*a*b^2*d)*m^2 + 9*(41*b^3*c + 123*a*b^2*d + 6*(b^3*c + 3*a*b^2*d)*m^2 + 3 3*(b^3*c + 3*a*b^2*d)*m)*n^2 + 61*(b^3*c + 3*a*b^2*d)*m + 3*(61*b^3*c + 18 3*a*b^2*d + 4*(b^3*c + 3*a*b^2*d)*m^3 + 33*(b^3*c + 3*a*b^2*d)*m^2 + 82*(b ^3*c + 3*a*b^2*d)*m)*n)*x^3 + 9*(6*a^3*c*m^2 + 42*a^3*c*m + 71*a^3*c)*n^2 + 3*((a*b^2*c + a^2*b*d)*m^4 + 81*(a*b^2*c + a^2*b*d)*n^4 + 40*a*b^2*c + 4 0*a^2*b*d + 12*(a*b^2*c + a^2*b*d)*m^3 + 108*(3*a*b^2*c + 3*a^2*b*d + (a*b ^2*c + a^2*b*d)*m)*n^3 + 49*(a*b^2*c + a^2*b*d)*m^2 + 9*(49*a*b^2*c + 49*a ^2*b*d + 6*(a*b^2*c + a^2*b*d)*m^2 + 36*(a*b^2*c + a^2*b*d)*m)*n^2 + 78*(a *b^2*c + a^2*b*d)*m + 6*(39*a*b^2*c + 39*a^2*b*d + 2*(a*b^2*c + a^2*b*d)*m ^3 + 18*(a*b^2*c + a^2*b*d)*m^2 + 49*(a*b^2*c + a^2*b*d)*m)*n)*x^2 + 6*(2* a^3*c*m^3 + 21*a^3*c*m^2 + 71*a^3*c*m + 77*a^3*c)*n + ((3*a^2*b*c + a^3*d) *m^4 + 81*(3*a^2*b*c + a^3*d)*n^4 + 180*a^2*b*c + 60*a^3*d + 13*(3*a^2*b*c + a^3*d)*m^3 + 27*(39*a^2*b*c + 13*a^3*d + 4*(3*a^2*b*c + a^3*d)*m)*n^...
Leaf count of result is larger than twice the leaf count of optimal. 18686 vs. \(2 (138) = 276\).
Time = 94.52 (sec) , antiderivative size = 18686, normalized size of antiderivative = 127.99 \[ \int (e x)^m (c+d x) \left (a x^n+b x^{1+n}\right )^3 \, dx=\text {Too large to display} \] Input:
integrate((e*x)**m*(d*x+c)*(a*x**n+b*x**(1+n))**3,x)
Output:
Piecewise((-a**3*c*x*x**(3*n)*(e*x)**(-3*n - 5)/4 - a**3*d*x**2*x**(3*n)*( e*x)**(-3*n - 5)/3 - a**2*b*c*x*x**(2*n)*x**(n + 1)*(e*x)**(-3*n - 5) - 3* a**2*b*d*x**2*x**(2*n)*x**(n + 1)*(e*x)**(-3*n - 5)/2 - 3*a*b**2*c*x*x**n* x**(2*n + 2)*(e*x)**(-3*n - 5)/2 - 3*a*b**2*d*x**2*x**n*x**(2*n + 2)*(e*x) **(-3*n - 5) - b**3*c*x*x**(3*n + 3)*(e*x)**(-3*n - 5) + b**3*d*x**2*x**(3 *n + 3)*(e*x)**(-3*n - 5)*log(x), Eq(m, -3*n - 5)), (-a**3*c*x*x**(3*n)*(e *x)**(-3*n - 4)/3 - a**3*d*x**2*x**(3*n)*(e*x)**(-3*n - 4)/2 - 3*a**2*b*c* x*x**(2*n)*x**(n + 1)*(e*x)**(-3*n - 4)/2 - 3*a**2*b*d*x**2*x**(2*n)*x**(n + 1)*(e*x)**(-3*n - 4) - 3*a*b**2*c*x*x**n*x**(2*n + 2)*(e*x)**(-3*n - 4) + 3*a*b**2*d*x**2*x**n*x**(2*n + 2)*(e*x)**(-3*n - 4)*log(x) + b**3*c*x*x **(3*n + 3)*(e*x)**(-3*n - 4)*log(x) + b**3*d*x**2*x**(3*n + 3)*(e*x)**(-3 *n - 4), Eq(m, -3*n - 4)), (-a**3*c*x*x**(3*n)*(e*x)**(-3*n - 3)/2 - a**3* d*x**2*x**(3*n)*(e*x)**(-3*n - 3) - 3*a**2*b*c*x*x**(2*n)*x**(n + 1)*(e*x) **(-3*n - 3) + 3*a**2*b*d*x**2*x**(2*n)*x**(n + 1)*(e*x)**(-3*n - 3)*log(x ) + 3*a*b**2*c*x*x**n*x**(2*n + 2)*(e*x)**(-3*n - 3)*log(x) + 3*a*b**2*d*x **2*x**n*x**(2*n + 2)*(e*x)**(-3*n - 3) + b**3*c*x*x**(3*n + 3)*(e*x)**(-3 *n - 3) + b**3*d*x**2*x**(3*n + 3)*(e*x)**(-3*n - 3)/2, Eq(m, -3*n - 3)), (-a**3*c*x*x**(3*n)*(e*x)**(-3*n - 2) + a**3*d*x**2*x**(3*n)*(e*x)**(-3*n - 2)*log(x) + 3*a**2*b*c*x*x**(2*n)*x**(n + 1)*(e*x)**(-3*n - 2)*log(x) + 3*a**2*b*d*x**2*x**(2*n)*x**(n + 1)*(e*x)**(-3*n - 2) + 3*a*b**2*c*x*x*...
Time = 0.06 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.69 \[ \int (e x)^m (c+d x) \left (a x^n+b x^{1+n}\right )^3 \, dx=\frac {b^{3} d e^{m} x^{5} e^{\left (m \log \left (x\right ) + 3 \, n \log \left (x\right )\right )}}{m + 3 \, n + 5} + \frac {b^{3} c e^{m} x^{4} e^{\left (m \log \left (x\right ) + 3 \, n \log \left (x\right )\right )}}{m + 3 \, n + 4} + \frac {3 \, a b^{2} d e^{m} x^{4} e^{\left (m \log \left (x\right ) + 3 \, n \log \left (x\right )\right )}}{m + 3 \, n + 4} + \frac {3 \, a b^{2} c e^{m} x^{3} e^{\left (m \log \left (x\right ) + 3 \, n \log \left (x\right )\right )}}{m + 3 \, n + 3} + \frac {3 \, a^{2} b d e^{m} x^{3} e^{\left (m \log \left (x\right ) + 3 \, n \log \left (x\right )\right )}}{m + 3 \, n + 3} + \frac {3 \, a^{2} b c e^{m} x^{2} e^{\left (m \log \left (x\right ) + 3 \, n \log \left (x\right )\right )}}{m + 3 \, n + 2} + \frac {a^{3} d e^{m} x^{2} e^{\left (m \log \left (x\right ) + 3 \, n \log \left (x\right )\right )}}{m + 3 \, n + 2} + \frac {a^{3} c e^{m} x e^{\left (m \log \left (x\right ) + 3 \, n \log \left (x\right )\right )}}{m + 3 \, n + 1} \] Input:
integrate((e*x)^m*(d*x+c)*(a*x^n+b*x^(1+n))^3,x, algorithm="maxima")
Output:
b^3*d*e^m*x^5*e^(m*log(x) + 3*n*log(x))/(m + 3*n + 5) + b^3*c*e^m*x^4*e^(m *log(x) + 3*n*log(x))/(m + 3*n + 4) + 3*a*b^2*d*e^m*x^4*e^(m*log(x) + 3*n* log(x))/(m + 3*n + 4) + 3*a*b^2*c*e^m*x^3*e^(m*log(x) + 3*n*log(x))/(m + 3 *n + 3) + 3*a^2*b*d*e^m*x^3*e^(m*log(x) + 3*n*log(x))/(m + 3*n + 3) + 3*a^ 2*b*c*e^m*x^2*e^(m*log(x) + 3*n*log(x))/(m + 3*n + 2) + a^3*d*e^m*x^2*e^(m *log(x) + 3*n*log(x))/(m + 3*n + 2) + a^3*c*e^m*x*e^(m*log(x) + 3*n*log(x) )/(m + 3*n + 1)
Leaf count of result is larger than twice the leaf count of optimal. 3372 vs. \(2 (146) = 292\).
Time = 0.36 (sec) , antiderivative size = 3372, normalized size of antiderivative = 23.10 \[ \int (e x)^m (c+d x) \left (a x^n+b x^{1+n}\right )^3 \, dx=\text {Too large to display} \] Input:
integrate((e*x)^m*(d*x+c)*(a*x^n+b*x^(1+n))^3,x, algorithm="giac")
Output:
(b^3*d*m^4*x^5*x^(3*n)*e^(m*log(e) + m*log(x)) + 12*b^3*d*m^3*n*x^5*x^(3*n )*e^(m*log(e) + m*log(x)) + 54*b^3*d*m^2*n^2*x^5*x^(3*n)*e^(m*log(e) + m*l og(x)) + 108*b^3*d*m*n^3*x^5*x^(3*n)*e^(m*log(e) + m*log(x)) + 81*b^3*d*n^ 4*x^5*x^(3*n)*e^(m*log(e) + m*log(x)) + b^3*c*m^4*x^4*x^(3*n)*e^(m*log(e) + m*log(x)) + 3*a*b^2*d*m^4*x^4*x^(3*n)*e^(m*log(e) + m*log(x)) + 12*b^3*c *m^3*n*x^4*x^(3*n)*e^(m*log(e) + m*log(x)) + 36*a*b^2*d*m^3*n*x^4*x^(3*n)* e^(m*log(e) + m*log(x)) + 54*b^3*c*m^2*n^2*x^4*x^(3*n)*e^(m*log(e) + m*log (x)) + 162*a*b^2*d*m^2*n^2*x^4*x^(3*n)*e^(m*log(e) + m*log(x)) + 108*b^3*c *m*n^3*x^4*x^(3*n)*e^(m*log(e) + m*log(x)) + 324*a*b^2*d*m*n^3*x^4*x^(3*n) *e^(m*log(e) + m*log(x)) + 81*b^3*c*n^4*x^4*x^(3*n)*e^(m*log(e) + m*log(x) ) + 243*a*b^2*d*n^4*x^4*x^(3*n)*e^(m*log(e) + m*log(x)) + 10*b^3*d*m^3*x^5 *x^(3*n)*e^(m*log(e) + m*log(x)) + 90*b^3*d*m^2*n*x^5*x^(3*n)*e^(m*log(e) + m*log(x)) + 270*b^3*d*m*n^2*x^5*x^(3*n)*e^(m*log(e) + m*log(x)) + 270*b^ 3*d*n^3*x^5*x^(3*n)*e^(m*log(e) + m*log(x)) + 3*a*b^2*c*m^4*x^3*x^(3*n)*e^ (m*log(e) + m*log(x)) + 3*a^2*b*d*m^4*x^3*x^(3*n)*e^(m*log(e) + m*log(x)) + 36*a*b^2*c*m^3*n*x^3*x^(3*n)*e^(m*log(e) + m*log(x)) + 36*a^2*b*d*m^3*n* x^3*x^(3*n)*e^(m*log(e) + m*log(x)) + 162*a*b^2*c*m^2*n^2*x^3*x^(3*n)*e^(m *log(e) + m*log(x)) + 162*a^2*b*d*m^2*n^2*x^3*x^(3*n)*e^(m*log(e) + m*log( x)) + 324*a*b^2*c*m*n^3*x^3*x^(3*n)*e^(m*log(e) + m*log(x)) + 324*a^2*b*d* m*n^3*x^3*x^(3*n)*e^(m*log(e) + m*log(x)) + 243*a*b^2*c*n^4*x^3*x^(3*n)...
Time = 9.75 (sec) , antiderivative size = 1641, normalized size of antiderivative = 11.24 \[ \int (e x)^m (c+d x) \left (a x^n+b x^{1+n}\right )^3 \, dx=\text {Too large to display} \] Input:
int((e*x)^m*(a*x^n + b*x^(n + 1))^3*(c + d*x),x)
Output:
(a^3*c*x*x^(3*n)*(e*x)^m*(154*m + 462*n + 426*m*n + 378*m*n^2 + 126*m^2*n + 108*m*n^3 + 12*m^3*n + 71*m^2 + 14*m^3 + m^4 + 639*n^2 + 378*n^3 + 81*n^ 4 + 54*m^2*n^2 + 120))/(274*m + 822*n + 1350*m*n + 2295*m*n^2 + 765*m^2*n + 1620*m*n^3 + 180*m^3*n + 405*m*n^4 + 15*m^4*n + 225*m^2 + 85*m^3 + 15*m^ 4 + m^5 + 2025*n^2 + 2295*n^3 + 1215*n^4 + 243*n^5 + 810*m^2*n^2 + 270*m^2 *n^3 + 90*m^3*n^2 + 120) + (b^3*c*x*x^(3*n + 3)*(e*x)^m*(61*m + 183*n + 24 6*m*n + 297*m*n^2 + 99*m^2*n + 108*m*n^3 + 12*m^3*n + 41*m^2 + 11*m^3 + m^ 4 + 369*n^2 + 297*n^3 + 81*n^4 + 54*m^2*n^2 + 30))/(274*m + 822*n + 1350*m *n + 2295*m*n^2 + 765*m^2*n + 1620*m*n^3 + 180*m^3*n + 405*m*n^4 + 15*m^4* n + 225*m^2 + 85*m^3 + 15*m^4 + m^5 + 2025*n^2 + 2295*n^3 + 1215*n^4 + 243 *n^5 + 810*m^2*n^2 + 270*m^2*n^3 + 90*m^3*n^2 + 120) + (a^3*d*x^(3*n)*x^2* (e*x)^m*(107*m + 321*n + 354*m*n + 351*m*n^2 + 117*m^2*n + 108*m*n^3 + 12* m^3*n + 59*m^2 + 13*m^3 + m^4 + 531*n^2 + 351*n^3 + 81*n^4 + 54*m^2*n^2 + 60))/(274*m + 822*n + 1350*m*n + 2295*m*n^2 + 765*m^2*n + 1620*m*n^3 + 180 *m^3*n + 405*m*n^4 + 15*m^4*n + 225*m^2 + 85*m^3 + 15*m^4 + m^5 + 2025*n^2 + 2295*n^3 + 1215*n^4 + 243*n^5 + 810*m^2*n^2 + 270*m^2*n^3 + 90*m^3*n^2 + 120) + (b^3*d*x^(3*n + 3)*x^2*(e*x)^m*(50*m + 150*n + 210*m*n + 270*m*n^ 2 + 90*m^2*n + 108*m*n^3 + 12*m^3*n + 35*m^2 + 10*m^3 + m^4 + 315*n^2 + 27 0*n^3 + 81*n^4 + 54*m^2*n^2 + 24))/(274*m + 822*n + 1350*m*n + 2295*m*n^2 + 765*m^2*n + 1620*m*n^3 + 180*m^3*n + 405*m*n^4 + 15*m^4*n + 225*m^2 +...
Time = 0.26 (sec) , antiderivative size = 1508, normalized size of antiderivative = 10.33 \[ \int (e x)^m (c+d x) \left (a x^n+b x^{1+n}\right )^3 \, dx =\text {Too large to display} \] Input:
int((e*x)^m*(d*x+c)*(a*x^n+b*x^(1+n))^3,x)
Output:
(x**(m + 3*n)*e**m*x*(a**3*c*m**4 + 12*a**3*c*m**3*n + 14*a**3*c*m**3 + 54 *a**3*c*m**2*n**2 + 126*a**3*c*m**2*n + 71*a**3*c*m**2 + 108*a**3*c*m*n**3 + 378*a**3*c*m*n**2 + 426*a**3*c*m*n + 154*a**3*c*m + 81*a**3*c*n**4 + 37 8*a**3*c*n**3 + 639*a**3*c*n**2 + 462*a**3*c*n + 120*a**3*c + a**3*d*m**4* x + 12*a**3*d*m**3*n*x + 13*a**3*d*m**3*x + 54*a**3*d*m**2*n**2*x + 117*a* *3*d*m**2*n*x + 59*a**3*d*m**2*x + 108*a**3*d*m*n**3*x + 351*a**3*d*m*n**2 *x + 354*a**3*d*m*n*x + 107*a**3*d*m*x + 81*a**3*d*n**4*x + 351*a**3*d*n** 3*x + 531*a**3*d*n**2*x + 321*a**3*d*n*x + 60*a**3*d*x + 3*a**2*b*c*m**4*x + 36*a**2*b*c*m**3*n*x + 39*a**2*b*c*m**3*x + 162*a**2*b*c*m**2*n**2*x + 351*a**2*b*c*m**2*n*x + 177*a**2*b*c*m**2*x + 324*a**2*b*c*m*n**3*x + 1053 *a**2*b*c*m*n**2*x + 1062*a**2*b*c*m*n*x + 321*a**2*b*c*m*x + 243*a**2*b*c *n**4*x + 1053*a**2*b*c*n**3*x + 1593*a**2*b*c*n**2*x + 963*a**2*b*c*n*x + 180*a**2*b*c*x + 3*a**2*b*d*m**4*x**2 + 36*a**2*b*d*m**3*n*x**2 + 36*a**2 *b*d*m**3*x**2 + 162*a**2*b*d*m**2*n**2*x**2 + 324*a**2*b*d*m**2*n*x**2 + 147*a**2*b*d*m**2*x**2 + 324*a**2*b*d*m*n**3*x**2 + 972*a**2*b*d*m*n**2*x* *2 + 882*a**2*b*d*m*n*x**2 + 234*a**2*b*d*m*x**2 + 243*a**2*b*d*n**4*x**2 + 972*a**2*b*d*n**3*x**2 + 1323*a**2*b*d*n**2*x**2 + 702*a**2*b*d*n*x**2 + 120*a**2*b*d*x**2 + 3*a*b**2*c*m**4*x**2 + 36*a*b**2*c*m**3*n*x**2 + 36*a *b**2*c*m**3*x**2 + 162*a*b**2*c*m**2*n**2*x**2 + 324*a*b**2*c*m**2*n*x**2 + 147*a*b**2*c*m**2*x**2 + 324*a*b**2*c*m*n**3*x**2 + 972*a*b**2*c*m*n...