Integrand size = 31, antiderivative size = 380 \[ \int \frac {(e x)^m \left (a+b x^n\right )^4 \left (A+B x^n\right )}{c+d x^n} \, dx=\frac {b \left (4 a^3 B d^3-b^3 c^2 (B c-A d)+4 a b^2 c d (B c-A d)-6 a^2 b d^2 (B c-A d)\right ) x^{1+n} (e x)^m}{d^4 (1+m+n)}+\frac {b^2 \left (6 a^2 B d^2+b^2 c (B c-A d)-4 a b d (B c-A d)\right ) x^{1+2 n} (e x)^m}{d^3 (1+m+2 n)}-\frac {b^3 (b B c-A b d-4 a B d) x^{1+3 n} (e x)^m}{d^2 (1+m+3 n)}+\frac {b^4 B x^{1+4 n} (e x)^m}{d (1+m+4 n)}+\frac {\left (a^4 B d^4+b^4 c^3 (B c-A d)-4 a b^3 c^2 d (B c-A d)+6 a^2 b^2 c d^2 (B c-A d)-4 a^3 b d^3 (B c-A d)\right ) (e x)^{1+m}}{d^5 e (1+m)}-\frac {(b c-a d)^4 (B c-A d) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{c d^5 e (1+m)} \]
b*(4*a^3*B*d^3-b^3*c^2*(-A*d+B*c)+4*a*b^2*c*d*(-A*d+B*c)-6*a^2*b*d^2*(-A*d +B*c))*x^(1+n)*(e*x)^m/d^4/(1+m+n)+b^2*(6*a^2*B*d^2+b^2*c*(-A*d+B*c)-4*a*b *d*(-A*d+B*c))*x^(1+2*n)*(e*x)^m/d^3/(1+m+2*n)-b^3*(-A*b*d-4*B*a*d+B*b*c)* x^(1+3*n)*(e*x)^m/d^2/(1+m+3*n)+b^4*B*x^(1+4*n)*(e*x)^m/d/(1+m+4*n)+(a^4*B *d^4+b^4*c^3*(-A*d+B*c)-4*a*b^3*c^2*d*(-A*d+B*c)+6*a^2*b^2*c*d^2*(-A*d+B*c )-4*a^3*b*d^3*(-A*d+B*c))*(e*x)^(1+m)/d^5/e/(1+m)-(-a*d+b*c)^4*(-A*d+B*c)* (e*x)^(1+m)*hypergeom([1, (1+m)/n],[(1+m+n)/n],-d*x^n/c)/c/d^5/e/(1+m)
Time = 1.31 (sec) , antiderivative size = 332, normalized size of antiderivative = 0.87 \[ \int \frac {(e x)^m \left (a+b x^n\right )^4 \left (A+B x^n\right )}{c+d x^n} \, dx=\frac {x (e x)^m \left (\frac {a^4 B d^4+b^4 c^3 (B c-A d)+6 a^2 b^2 c d^2 (B c-A d)+4 a b^3 c^2 d (-B c+A d)+4 a^3 b d^3 (-B c+A d)}{1+m}+\frac {b d \left (4 a^3 B d^3+4 a b^2 c d (B c-A d)+b^3 c^2 (-B c+A d)+6 a^2 b d^2 (-B c+A d)\right ) x^n}{1+m+n}+\frac {b^2 d^2 \left (6 a^2 B d^2+b^2 c (B c-A d)+4 a b d (-B c+A d)\right ) x^{2 n}}{1+m+2 n}+\frac {b^3 d^3 (-b B c+A b d+4 a B d) x^{3 n}}{1+m+3 n}+\frac {b^4 B d^4 x^{4 n}}{1+m+4 n}-\frac {(b c-a d)^4 (B c-A d) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{n},\frac {1+m+n}{n},-\frac {d x^n}{c}\right )}{c (1+m)}\right )}{d^5} \]
(x*(e*x)^m*((a^4*B*d^4 + b^4*c^3*(B*c - A*d) + 6*a^2*b^2*c*d^2*(B*c - A*d) + 4*a*b^3*c^2*d*(-(B*c) + A*d) + 4*a^3*b*d^3*(-(B*c) + A*d))/(1 + m) + (b *d*(4*a^3*B*d^3 + 4*a*b^2*c*d*(B*c - A*d) + b^3*c^2*(-(B*c) + A*d) + 6*a^2 *b*d^2*(-(B*c) + A*d))*x^n)/(1 + m + n) + (b^2*d^2*(6*a^2*B*d^2 + b^2*c*(B *c - A*d) + 4*a*b*d*(-(B*c) + A*d))*x^(2*n))/(1 + m + 2*n) + (b^3*d^3*(-(b *B*c) + A*b*d + 4*a*B*d)*x^(3*n))/(1 + m + 3*n) + (b^4*B*d^4*x^(4*n))/(1 + m + 4*n) - ((b*c - a*d)^4*(B*c - A*d)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, -((d*x^n)/c)])/(c*(1 + m))))/d^5
Time = 0.79 (sec) , antiderivative size = 380, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {1040, 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.
| \(\displaystyle \int \frac {(e x)^m \left (a+b x^n\right )^4 \left (A+B x^n\right )}{c+d x^n} \, dx\) |
| \(\Big \downarrow \) 1040 |
| \(\displaystyle \int \left (\frac {b^2 x^{2 n} (e x)^m \left (6 a^2 B d^2-4 a b d (B c-A d)+b^2 c (B c-A d)\right )}{d^3}+\frac {b x^n (e x)^m \left (4 a^3 B d^3-6 a^2 b d^2 (B c-A d)+4 a b^2 c d (B c-A d)+b^3 \left (-c^2\right ) (B c-A d)\right )}{d^4}+\frac {(e x)^m \left (a^4 B d^4-4 a^3 b d^3 (B c-A d)+6 a^2 b^2 c d^2 (B c-A d)-4 a b^3 c^2 d (B c-A d)+b^4 c^3 (B c-A d)\right )}{d^5}+\frac {b^3 x^{3 n} (e x)^m (4 a B d+A b d-b B c)}{d^2}+\frac {(e x)^m (a d-b c)^4 (A d-B c)}{d^5 \left (c+d x^n\right )}+\frac {b^4 B x^{4 n} (e x)^m}{d}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b^2 x^{2 n+1} (e x)^m \left (6 a^2 B d^2-4 a b d (B c-A d)+b^2 c (B c-A d)\right )}{d^3 (m+2 n+1)}+\frac {b x^{n+1} (e x)^m \left (4 a^3 B d^3-6 a^2 b d^2 (B c-A d)+4 a b^2 c d (B c-A d)+b^3 \left (-c^2\right ) (B c-A d)\right )}{d^4 (m+n+1)}+\frac {(e x)^{m+1} \left (a^4 B d^4-4 a^3 b d^3 (B c-A d)+6 a^2 b^2 c d^2 (B c-A d)-4 a b^3 c^2 d (B c-A d)+b^4 c^3 (B c-A d)\right )}{d^5 e (m+1)}-\frac {b^3 x^{3 n+1} (e x)^m (-4 a B d-A b d+b B c)}{d^2 (m+3 n+1)}-\frac {(e x)^{m+1} (b c-a d)^4 (B c-A d) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{n},\frac {m+n+1}{n},-\frac {d x^n}{c}\right )}{c d^5 e (m+1)}+\frac {b^4 B x^{4 n+1} (e x)^m}{d (m+4 n+1)}\) |
(b*(4*a^3*B*d^3 - b^3*c^2*(B*c - A*d) + 4*a*b^2*c*d*(B*c - A*d) - 6*a^2*b* d^2*(B*c - A*d))*x^(1 + n)*(e*x)^m)/(d^4*(1 + m + n)) + (b^2*(6*a^2*B*d^2 + b^2*c*(B*c - A*d) - 4*a*b*d*(B*c - A*d))*x^(1 + 2*n)*(e*x)^m)/(d^3*(1 + m + 2*n)) - (b^3*(b*B*c - A*b*d - 4*a*B*d)*x^(1 + 3*n)*(e*x)^m)/(d^2*(1 + m + 3*n)) + (b^4*B*x^(1 + 4*n)*(e*x)^m)/(d*(1 + m + 4*n)) + ((a^4*B*d^4 + b^4*c^3*(B*c - A*d) - 4*a*b^3*c^2*d*(B*c - A*d) + 6*a^2*b^2*c*d^2*(B*c - A *d) - 4*a^3*b*d^3*(B*c - A*d))*(e*x)^(1 + m))/(d^5*e*(1 + m)) - ((b*c - a* d)^4*(B*c - A*d)*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n) /n, -((d*x^n)/c)])/(c*d^5*e*(1 + m))
3.1.21.3.1 Defintions of rubi rules used
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _))^(q_.)*((e_) + (f_.)*(x_)^(n_))^(r_.), x_Symbol] :> Int[ExpandIntegrand[ (g*x)^m*(a + b*x^n)^p*(c + d*x^n)^q*(e + f*x^n)^r, x], x] /; FreeQ[{a, b, c , d, e, f, g, m, n}, x] && IGtQ[p, -2] && IGtQ[q, 0] && IGtQ[r, 0]
\[\int \frac {\left (e x \right )^{m} \left (a +b \,x^{n}\right )^{4} \left (A +B \,x^{n}\right )}{c +d \,x^{n}}d x\]
\[ \int \frac {(e x)^m \left (a+b x^n\right )^4 \left (A+B x^n\right )}{c+d x^n} \, dx=\int { \frac {{\left (B x^{n} + A\right )} {\left (b x^{n} + a\right )}^{4} \left (e x\right )^{m}}{d x^{n} + c} \,d x } \]
integral((B*b^4*x^(5*n) + A*a^4 + (4*B*a*b^3 + A*b^4)*x^(4*n) + 2*(3*B*a^2 *b^2 + 2*A*a*b^3)*x^(3*n) + 2*(2*B*a^3*b + 3*A*a^2*b^2)*x^(2*n) + (B*a^4 + 4*A*a^3*b)*x^n)*(e*x)^m/(d*x^n + c), x)
Result contains complex when optimal does not.
Time = 28.35 (sec) , antiderivative size = 2463, normalized size of antiderivative = 6.48 \[ \int \frac {(e x)^m \left (a+b x^n\right )^4 \left (A+B x^n\right )}{c+d x^n} \, dx=\text {Too large to display} \]
A*a**4*c**(m/n + 1/n)*c**(-m/n - 1 - 1/n)*e**m*m*x**(m + 1)*lerchphi(d*x** n*exp_polar(I*pi)/c, 1, m/n + 1/n)*gamma(m/n + 1/n)/(n**2*gamma(m/n + 1 + 1/n)) + A*a**4*c**(m/n + 1/n)*c**(-m/n - 1 - 1/n)*e**m*x**(m + 1)*lerchphi (d*x**n*exp_polar(I*pi)/c, 1, m/n + 1/n)*gamma(m/n + 1/n)/(n**2*gamma(m/n + 1 + 1/n)) + 4*A*a**3*b*c**(-m/n - 2 - 1/n)*c**(m/n + 1 + 1/n)*e**m*m*x** (m + n + 1)*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 1 + 1/n)*gamma(m/n + 1 + 1/n)/(n**2*gamma(m/n + 2 + 1/n)) + 4*A*a**3*b*c**(-m/n - 2 - 1/n)*c **(m/n + 1 + 1/n)*e**m*x**(m + n + 1)*lerchphi(d*x**n*exp_polar(I*pi)/c, 1 , m/n + 1 + 1/n)*gamma(m/n + 1 + 1/n)/(n*gamma(m/n + 2 + 1/n)) + 4*A*a**3* b*c**(-m/n - 2 - 1/n)*c**(m/n + 1 + 1/n)*e**m*x**(m + n + 1)*lerchphi(d*x* *n*exp_polar(I*pi)/c, 1, m/n + 1 + 1/n)*gamma(m/n + 1 + 1/n)/(n**2*gamma(m /n + 2 + 1/n)) + 6*A*a**2*b**2*c**(-m/n - 3 - 1/n)*c**(m/n + 2 + 1/n)*e**m *m*x**(m + 2*n + 1)*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 2 + 1/n)*g amma(m/n + 2 + 1/n)/(n**2*gamma(m/n + 3 + 1/n)) + 12*A*a**2*b**2*c**(-m/n - 3 - 1/n)*c**(m/n + 2 + 1/n)*e**m*x**(m + 2*n + 1)*lerchphi(d*x**n*exp_po lar(I*pi)/c, 1, m/n + 2 + 1/n)*gamma(m/n + 2 + 1/n)/(n*gamma(m/n + 3 + 1/n )) + 6*A*a**2*b**2*c**(-m/n - 3 - 1/n)*c**(m/n + 2 + 1/n)*e**m*x**(m + 2*n + 1)*lerchphi(d*x**n*exp_polar(I*pi)/c, 1, m/n + 2 + 1/n)*gamma(m/n + 2 + 1/n)/(n**2*gamma(m/n + 3 + 1/n)) + 4*A*a*b**3*c**(-m/n - 4 - 1/n)*c**(m/n + 3 + 1/n)*e**m*m*x**(m + 3*n + 1)*lerchphi(d*x**n*exp_polar(I*pi)/c, ...
\[ \int \frac {(e x)^m \left (a+b x^n\right )^4 \left (A+B x^n\right )}{c+d x^n} \, dx=\int { \frac {{\left (B x^{n} + A\right )} {\left (b x^{n} + a\right )}^{4} \left (e x\right )^{m}}{d x^{n} + c} \,d x } \]
((b^4*c^4*d*e^m - 4*a*b^3*c^3*d^2*e^m + 6*a^2*b^2*c^2*d^3*e^m - 4*a^3*b*c* d^4*e^m + a^4*d^5*e^m)*A - (b^4*c^5*e^m - 4*a*b^3*c^4*d*e^m + 6*a^2*b^2*c^ 3*d^2*e^m - 4*a^3*b*c^2*d^3*e^m + a^4*c*d^4*e^m)*B)*integrate(x^m/(d^6*x^n + c*d^5), x) + ((m^4 + 2*m^3*(3*n + 2) + (11*n^2 + 18*n + 6)*m^2 + 6*n^3 + 2*(3*n^3 + 11*n^2 + 9*n + 2)*m + 11*n^2 + 6*n + 1)*B*b^4*d^4*e^m*x*e^(m* log(x) + 4*n*log(x)) - (((m^4 + 2*m^3*(5*n + 2) + 24*n^4 + (35*n^2 + 30*n + 6)*m^2 + 50*n^3 + 2*(25*n^3 + 35*n^2 + 15*n + 2)*m + 35*n^2 + 10*n + 1)* b^4*c^3*d*e^m - 4*(m^4 + 2*m^3*(5*n + 2) + 24*n^4 + (35*n^2 + 30*n + 6)*m^ 2 + 50*n^3 + 2*(25*n^3 + 35*n^2 + 15*n + 2)*m + 35*n^2 + 10*n + 1)*a*b^3*c ^2*d^2*e^m + 6*(m^4 + 2*m^3*(5*n + 2) + 24*n^4 + (35*n^2 + 30*n + 6)*m^2 + 50*n^3 + 2*(25*n^3 + 35*n^2 + 15*n + 2)*m + 35*n^2 + 10*n + 1)*a^2*b^2*c* d^3*e^m - 4*(m^4 + 2*m^3*(5*n + 2) + 24*n^4 + (35*n^2 + 30*n + 6)*m^2 + 50 *n^3 + 2*(25*n^3 + 35*n^2 + 15*n + 2)*m + 35*n^2 + 10*n + 1)*a^3*b*d^4*e^m )*A - ((m^4 + 2*m^3*(5*n + 2) + 24*n^4 + (35*n^2 + 30*n + 6)*m^2 + 50*n^3 + 2*(25*n^3 + 35*n^2 + 15*n + 2)*m + 35*n^2 + 10*n + 1)*b^4*c^4*e^m - 4*(m ^4 + 2*m^3*(5*n + 2) + 24*n^4 + (35*n^2 + 30*n + 6)*m^2 + 50*n^3 + 2*(25*n ^3 + 35*n^2 + 15*n + 2)*m + 35*n^2 + 10*n + 1)*a*b^3*c^3*d*e^m + 6*(m^4 + 2*m^3*(5*n + 2) + 24*n^4 + (35*n^2 + 30*n + 6)*m^2 + 50*n^3 + 2*(25*n^3 + 35*n^2 + 15*n + 2)*m + 35*n^2 + 10*n + 1)*a^2*b^2*c^2*d^2*e^m - 4*(m^4 + 2 *m^3*(5*n + 2) + 24*n^4 + (35*n^2 + 30*n + 6)*m^2 + 50*n^3 + 2*(25*n^3 ...
\[ \int \frac {(e x)^m \left (a+b x^n\right )^4 \left (A+B x^n\right )}{c+d x^n} \, dx=\int { \frac {{\left (B x^{n} + A\right )} {\left (b x^{n} + a\right )}^{4} \left (e x\right )^{m}}{d x^{n} + c} \,d x } \]
Timed out. \[ \int \frac {(e x)^m \left (a+b x^n\right )^4 \left (A+B x^n\right )}{c+d x^n} \, dx=\int \frac {{\left (e\,x\right )}^m\,\left (A+B\,x^n\right )\,{\left (a+b\,x^n\right )}^4}{c+d\,x^n} \,d x \]