Integrand size = 25, antiderivative size = 319 \[ \int \frac {x^4 (e+f x)^n}{(a+b x) (c+d x)} \, dx=\frac {e^2 (e+f x)^{1+n}}{b d f^3 (1+n)}+\frac {(b c+a d) e (e+f x)^{1+n}}{b^2 d^2 f^2 (1+n)}+\frac {\left (b^2 c^2+a b c d+a^2 d^2\right ) (e+f x)^{1+n}}{b^3 d^3 f (1+n)}-\frac {2 e (e+f x)^{2+n}}{b d f^3 (2+n)}-\frac {(b c+a d) (e+f x)^{2+n}}{b^2 d^2 f^2 (2+n)}+\frac {(e+f x)^{3+n}}{b d f^3 (3+n)}-\frac {a^4 (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {b (e+f x)}{b e-a f}\right )}{b^3 (b c-a d) (b e-a f) (1+n)}+\frac {c^4 (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {d (e+f x)}{d e-c f}\right )}{d^3 (b c-a d) (d e-c f) (1+n)} \]
e^2*(f*x+e)^(1+n)/b/d/f^3/(1+n)+(a*d+b*c)*e*(f*x+e)^(1+n)/b^2/d^2/f^2/(1+n )+(a^2*d^2+a*b*c*d+b^2*c^2)*(f*x+e)^(1+n)/b^3/d^3/f/(1+n)-2*e*(f*x+e)^(2+n )/b/d/f^3/(2+n)-(a*d+b*c)*(f*x+e)^(2+n)/b^2/d^2/f^2/(2+n)+(f*x+e)^(3+n)/b/ d/f^3/(3+n)-a^4*(f*x+e)^(1+n)*hypergeom([1, 1+n],[2+n],b*(f*x+e)/(-a*f+b*e ))/b^3/(-a*d+b*c)/(-a*f+b*e)/(1+n)+c^4*(f*x+e)^(1+n)*hypergeom([1, 1+n],[2 +n],d*(f*x+e)/(-c*f+d*e))/d^3/(-a*d+b*c)/(-c*f+d*e)/(1+n)
Time = 0.88 (sec) , antiderivative size = 285, normalized size of antiderivative = 0.89 \[ \int \frac {x^4 (e+f x)^n}{(a+b x) (c+d x)} \, dx=\frac {(e+f x)^{1+n} \left (-\frac {a^4 d^3 \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {b (e+f x)}{b e-a f}\right )}{(b c-a d) (b e-a f)}+\frac {-\left ((b c-a d) (-d e+c f) \left (a^2 d^2 f^2 \left (6+5 n+n^2\right )+a b d f (3+n) (c f (2+n)+d (e-f (1+n) x))+b^2 \left (c^2 f^2 \left (6+5 n+n^2\right )+c d f (3+n) (e-f (1+n) x)+d^2 \left (2 e^2-2 e f (1+n) x+f^2 \left (2+3 n+n^2\right ) x^2\right )\right )\right )\right )+b^3 c^4 f^3 \left (6+5 n+n^2\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {d (e+f x)}{d e-c f}\right )}{(-b c+a d) f^3 (-d e+c f) (2+n) (3+n)}\right )}{b^3 d^3 (1+n)} \]
((e + f*x)^(1 + n)*(-((a^4*d^3*Hypergeometric2F1[1, 1 + n, 2 + n, (b*(e + f*x))/(b*e - a*f)])/((b*c - a*d)*(b*e - a*f))) + (-((b*c - a*d)*(-(d*e) + c*f)*(a^2*d^2*f^2*(6 + 5*n + n^2) + a*b*d*f*(3 + n)*(c*f*(2 + n) + d*(e - f*(1 + n)*x)) + b^2*(c^2*f^2*(6 + 5*n + n^2) + c*d*f*(3 + n)*(e - f*(1 + n )*x) + d^2*(2*e^2 - 2*e*f*(1 + n)*x + f^2*(2 + 3*n + n^2)*x^2)))) + b^3*c^ 4*f^3*(6 + 5*n + n^2)*Hypergeometric2F1[1, 1 + n, 2 + n, (d*(e + f*x))/(d* e - c*f)])/((-(b*c) + a*d)*f^3*(-(d*e) + c*f)*(2 + n)*(3 + n))))/(b^3*d^3* (1 + n))
Time = 0.50 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {198, 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.
\(\displaystyle \int \frac {x^4 (e+f x)^n}{(a+b x) (c+d x)} \, dx\) |
\(\Big \downarrow \) 198 |
\(\displaystyle \int \left (\frac {a^4 (e+f x)^n}{b^3 (a+b x) (b c-a d)}+\frac {\left (a^2 d^2+a b c d+b^2 c^2\right ) (e+f x)^n}{b^3 d^3}-\frac {x (a d+b c) (e+f x)^n}{b^2 d^2}+\frac {c^4 (e+f x)^n}{d^3 (c+d x) (a d-b c)}+\frac {x^2 (e+f x)^n}{b d}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^4 (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {b (e+f x)}{b e-a f}\right )}{b^3 (n+1) (b c-a d) (b e-a f)}+\frac {\left (a^2 d^2+a b c d+b^2 c^2\right ) (e+f x)^{n+1}}{b^3 d^3 f (n+1)}+\frac {e (a d+b c) (e+f x)^{n+1}}{b^2 d^2 f^2 (n+1)}-\frac {(a d+b c) (e+f x)^{n+2}}{b^2 d^2 f^2 (n+2)}+\frac {c^4 (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {d (e+f x)}{d e-c f}\right )}{d^3 (n+1) (b c-a d) (d e-c f)}+\frac {e^2 (e+f x)^{n+1}}{b d f^3 (n+1)}-\frac {2 e (e+f x)^{n+2}}{b d f^3 (n+2)}+\frac {(e+f x)^{n+3}}{b d f^3 (n+3)}\) |
(e^2*(e + f*x)^(1 + n))/(b*d*f^3*(1 + n)) + ((b*c + a*d)*e*(e + f*x)^(1 + n))/(b^2*d^2*f^2*(1 + n)) + ((b^2*c^2 + a*b*c*d + a^2*d^2)*(e + f*x)^(1 + n))/(b^3*d^3*f*(1 + n)) - (2*e*(e + f*x)^(2 + n))/(b*d*f^3*(2 + n)) - ((b* c + a*d)*(e + f*x)^(2 + n))/(b^2*d^2*f^2*(2 + n)) + (e + f*x)^(3 + n)/(b*d *f^3*(3 + n)) - (a^4*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (b*(e + f*x))/(b*e - a*f)])/(b^3*(b*c - a*d)*(b*e - a*f)*(1 + n)) + (c^4*( e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (d*(e + f*x))/(d*e - c *f)])/(d^3*(b*c - a*d)*(d*e - c*f)*(1 + n))
3.2.12.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_))^(q_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && IntegersQ[p, q]
\[\int \frac {x^{4} \left (f x +e \right )^{n}}{\left (b x +a \right ) \left (d x +c \right )}d x\]
\[ \int \frac {x^4 (e+f x)^n}{(a+b x) (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{n} x^{4}}{{\left (b x + a\right )} {\left (d x + c\right )}} \,d x } \]
Exception generated. \[ \int \frac {x^4 (e+f x)^n}{(a+b x) (c+d x)} \, dx=\text {Exception raised: HeuristicGCDFailed} \]
\[ \int \frac {x^4 (e+f x)^n}{(a+b x) (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{n} x^{4}}{{\left (b x + a\right )} {\left (d x + c\right )}} \,d x } \]
\[ \int \frac {x^4 (e+f x)^n}{(a+b x) (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{n} x^{4}}{{\left (b x + a\right )} {\left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {x^4 (e+f x)^n}{(a+b x) (c+d x)} \, dx=\int \frac {x^4\,{\left (e+f\,x\right )}^n}{\left (a+b\,x\right )\,\left (c+d\,x\right )} \,d x \]