Integrand size = 26, antiderivative size = 406 \[ \int (a+b x)^m (c+d x)^{-4-m} (e+f x)^3 \, dx=\frac {(d e-c f)^3 (a+b x)^{1+m} (c+d x)^{-3-m}}{d^3 (b c-a d) (3+m)}+\frac {3 f (d e-c f)^2 (a+b x)^{1+m} (c+d x)^{-2-m}}{d^3 (b c-a d) (2+m)}+\frac {2 b (d e-c f)^3 (a+b x)^{1+m} (c+d x)^{-2-m}}{d^3 (b c-a d)^2 (2+m) (3+m)}+\frac {3 f^2 (d e-c f) (a+b x)^{1+m} (c+d x)^{-1-m}}{d^3 (b c-a d) (1+m)}+\frac {3 b f (d e-c f)^2 (a+b x)^{1+m} (c+d x)^{-1-m}}{d^3 (b c-a d)^2 (1+m) (2+m)}+\frac {2 b^2 (d e-c f)^3 (a+b x)^{1+m} (c+d x)^{-1-m}}{d^3 (b c-a d)^3 (1+m) (2+m) (3+m)}-\frac {f^3 (a+b x)^m \left (-\frac {d (a+b x)}{b c-a d}\right )^{-m} (c+d x)^{-m} \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,\frac {b (c+d x)}{b c-a d}\right )}{d^4 m} \]
(-c*f+d*e)^3*(b*x+a)^(1+m)*(d*x+c)^(-3-m)/d^3/(-a*d+b*c)/(3+m)+3*f*(-c*f+d *e)^2*(b*x+a)^(1+m)*(d*x+c)^(-2-m)/d^3/(-a*d+b*c)/(2+m)+2*b*(-c*f+d*e)^3*( b*x+a)^(1+m)*(d*x+c)^(-2-m)/d^3/(-a*d+b*c)^2/(2+m)/(3+m)+3*f^2*(-c*f+d*e)* (b*x+a)^(1+m)*(d*x+c)^(-1-m)/d^3/(-a*d+b*c)/(1+m)+3*b*f*(-c*f+d*e)^2*(b*x+ a)^(1+m)*(d*x+c)^(-1-m)/d^3/(-a*d+b*c)^2/(1+m)/(2+m)+2*b^2*(-c*f+d*e)^3*(b *x+a)^(1+m)*(d*x+c)^(-1-m)/d^3/(-a*d+b*c)^3/(1+m)/(2+m)/(3+m)-f^3*(b*x+a)^ m*hypergeom([-m, -m],[1-m],b*(d*x+c)/(-a*d+b*c))/d^4/m/((-d*(b*x+a)/(-a*d+ b*c))^m)/((d*x+c)^m)
Time = 12.60 (sec) , antiderivative size = 279, normalized size of antiderivative = 0.69 \[ \int (a+b x)^m (c+d x)^{-4-m} (e+f x)^3 \, dx=\frac {(a+b x)^m (c+d x)^{-m} \left (\frac {6 f^2 (a+b x)^3 (e+f x)}{(b c-a d)^3 (c+d x)}-\frac {3 f (a+b x)^2 (-a d (1+m)+b c (3+m)+2 b d x) (e+f x)^2}{(b c-a d)^3 (c+d x)^2}+\frac {(a+b x) (e+f x)^3 \left (a^2 d^2 \left (2+3 m+m^2\right )-2 a b d (1+m) (c (3+m)+d x)+b^2 \left (c^2 \left (6+5 m+m^2\right )+2 c d (3+m) x+2 d^2 x^2\right )\right )}{(b c-a d)^3 (c+d x)^3}-\frac {6 f^3 \left (\frac {d (a+b x)}{-b c+a d}\right )^{-m} \operatorname {Hypergeometric2F1}\left (-3-m,-m,1-m,\frac {b (c+d x)}{b c-a d}\right )}{d^4 m}\right )}{(1+m) (2+m) (3+m)} \]
((a + b*x)^m*((6*f^2*(a + b*x)^3*(e + f*x))/((b*c - a*d)^3*(c + d*x)) - (3 *f*(a + b*x)^2*(-(a*d*(1 + m)) + b*c*(3 + m) + 2*b*d*x)*(e + f*x)^2)/((b*c - a*d)^3*(c + d*x)^2) + ((a + b*x)*(e + f*x)^3*(a^2*d^2*(2 + 3*m + m^2) - 2*a*b*d*(1 + m)*(c*(3 + m) + d*x) + b^2*(c^2*(6 + 5*m + m^2) + 2*c*d*(3 + m)*x + 2*d^2*x^2)))/((b*c - a*d)^3*(c + d*x)^3) - (6*f^3*Hypergeometric2F 1[-3 - m, -m, 1 - m, (b*(c + d*x))/(b*c - a*d)])/(d^4*m*((d*(a + b*x))/(-( b*c) + a*d))^m)))/((1 + m)*(2 + m)*(3 + m)*(c + d*x)^m)
Time = 0.54 (sec) , antiderivative size = 406, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {137, 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.
\(\displaystyle \int (e+f x)^3 (a+b x)^m (c+d x)^{-m-4} \, dx\) |
\(\Big \downarrow \) 137 |
\(\displaystyle \int \left (\frac {3 f^2 (a+b x)^m (d e-c f) (c+d x)^{-m-2}}{d^3}+\frac {(a+b x)^m (d e-c f)^3 (c+d x)^{-m-4}}{d^3}+\frac {3 f (a+b x)^m (d e-c f)^2 (c+d x)^{-m-3}}{d^3}+\frac {f^3 (a+b x)^m (c+d x)^{-m-1}}{d^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 b^2 (a+b x)^{m+1} (d e-c f)^3 (c+d x)^{-m-1}}{d^3 (m+1) (m+2) (m+3) (b c-a d)^3}-\frac {f^3 (a+b x)^m (c+d x)^{-m} \left (-\frac {d (a+b x)}{b c-a d}\right )^{-m} \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,\frac {b (c+d x)}{b c-a d}\right )}{d^4 m}+\frac {3 f^2 (a+b x)^{m+1} (d e-c f) (c+d x)^{-m-1}}{d^3 (m+1) (b c-a d)}+\frac {(a+b x)^{m+1} (d e-c f)^3 (c+d x)^{-m-3}}{d^3 (m+3) (b c-a d)}+\frac {3 f (a+b x)^{m+1} (d e-c f)^2 (c+d x)^{-m-2}}{d^3 (m+2) (b c-a d)}+\frac {2 b (a+b x)^{m+1} (d e-c f)^3 (c+d x)^{-m-2}}{d^3 (m+2) (m+3) (b c-a d)^2}+\frac {3 b f (a+b x)^{m+1} (d e-c f)^2 (c+d x)^{-m-1}}{d^3 (m+1) (m+2) (b c-a d)^2}\) |
((d*e - c*f)^3*(a + b*x)^(1 + m)*(c + d*x)^(-3 - m))/(d^3*(b*c - a*d)*(3 + m)) + (3*f*(d*e - c*f)^2*(a + b*x)^(1 + m)*(c + d*x)^(-2 - m))/(d^3*(b*c - a*d)*(2 + m)) + (2*b*(d*e - c*f)^3*(a + b*x)^(1 + m)*(c + d*x)^(-2 - m)) /(d^3*(b*c - a*d)^2*(2 + m)*(3 + m)) + (3*f^2*(d*e - c*f)*(a + b*x)^(1 + m )*(c + d*x)^(-1 - m))/(d^3*(b*c - a*d)*(1 + m)) + (3*b*f*(d*e - c*f)^2*(a + b*x)^(1 + m)*(c + d*x)^(-1 - m))/(d^3*(b*c - a*d)^2*(1 + m)*(2 + m)) + ( 2*b^2*(d*e - c*f)^3*(a + b*x)^(1 + m)*(c + d*x)^(-1 - m))/(d^3*(b*c - a*d) ^3*(1 + m)*(2 + m)*(3 + m)) - (f^3*(a + b*x)^m*Hypergeometric2F1[-m, -m, 1 - m, (b*(c + d*x))/(b*c - a*d)])/(d^4*m*(-((d*(a + b*x))/(b*c - a*d)))^m* (c + d*x)^m)
3.31.97.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (IGtQ[m, 0] || (ILtQ[m, 0] && ILtQ[n, 0]))
\[\int \left (b x +a \right )^{m} \left (d x +c \right )^{-4-m} \left (f x +e \right )^{3}d x\]
\[ \int (a+b x)^m (c+d x)^{-4-m} (e+f x)^3 \, dx=\int { {\left (f x + e\right )}^{3} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 4} \,d x } \]
Exception generated. \[ \int (a+b x)^m (c+d x)^{-4-m} (e+f x)^3 \, dx=\text {Exception raised: HeuristicGCDFailed} \]
\[ \int (a+b x)^m (c+d x)^{-4-m} (e+f x)^3 \, dx=\int { {\left (f x + e\right )}^{3} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 4} \,d x } \]
\[ \int (a+b x)^m (c+d x)^{-4-m} (e+f x)^3 \, dx=\int { {\left (f x + e\right )}^{3} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 4} \,d x } \]
Timed out. \[ \int (a+b x)^m (c+d x)^{-4-m} (e+f x)^3 \, dx=\int \frac {{\left (e+f\,x\right )}^3\,{\left (a+b\,x\right )}^m}{{\left (c+d\,x\right )}^{m+4}} \,d x \]