Integrand size = 26, antiderivative size = 650 \[ \int (a+b x)^m (c+d x)^{-5-m} (e+f x)^4 \, dx=\frac {(d e-c f)^4 (a+b x)^{1+m} (c+d x)^{-4-m}}{d^4 (b c-a d) (4+m)}+\frac {4 f (d e-c f)^3 (a+b x)^{1+m} (c+d x)^{-3-m}}{d^4 (b c-a d) (3+m)}+\frac {3 b (d e-c f)^4 (a+b x)^{1+m} (c+d x)^{-3-m}}{d^4 (b c-a d)^2 (3+m) (4+m)}+\frac {6 f^2 (d e-c f)^2 (a+b x)^{1+m} (c+d x)^{-2-m}}{d^4 (b c-a d) (2+m)}+\frac {8 b f (d e-c f)^3 (a+b x)^{1+m} (c+d x)^{-2-m}}{d^4 (b c-a d)^2 (2+m) (3+m)}+\frac {6 b^2 (d e-c f)^4 (a+b x)^{1+m} (c+d x)^{-2-m}}{d^4 (b c-a d)^3 (2+m) (3+m) (4+m)}+\frac {4 f^3 (d e-c f) (a+b x)^{1+m} (c+d x)^{-1-m}}{d^4 (b c-a d) (1+m)}+\frac {6 b f^2 (d e-c f)^2 (a+b x)^{1+m} (c+d x)^{-1-m}}{d^4 (b c-a d)^2 (1+m) (2+m)}+\frac {8 b^2 f (d e-c f)^3 (a+b x)^{1+m} (c+d x)^{-1-m}}{d^4 (b c-a d)^3 (1+m) (2+m) (3+m)}+\frac {6 b^3 (d e-c f)^4 (a+b x)^{1+m} (c+d x)^{-1-m}}{d^4 (b c-a d)^4 (1+m) (2+m) (3+m) (4+m)}-\frac {f^4 (a+b x)^m \left (-\frac {d (a+b x)}{b c-a d}\right )^{-m} (c+d x)^{-m} \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,\frac {b (c+d x)}{b c-a d}\right )}{d^5 m} \]
(-c*f+d*e)^4*(b*x+a)^(1+m)*(d*x+c)^(-4-m)/d^4/(-a*d+b*c)/(4+m)+4*f*(-c*f+d *e)^3*(b*x+a)^(1+m)*(d*x+c)^(-3-m)/d^4/(-a*d+b*c)/(3+m)+3*b*(-c*f+d*e)^4*( b*x+a)^(1+m)*(d*x+c)^(-3-m)/d^4/(-a*d+b*c)^2/(3+m)/(4+m)+6*f^2*(-c*f+d*e)^ 2*(b*x+a)^(1+m)*(d*x+c)^(-2-m)/d^4/(-a*d+b*c)/(2+m)+8*b*f*(-c*f+d*e)^3*(b* x+a)^(1+m)*(d*x+c)^(-2-m)/d^4/(-a*d+b*c)^2/(2+m)/(3+m)+6*b^2*(-c*f+d*e)^4* (b*x+a)^(1+m)*(d*x+c)^(-2-m)/d^4/(-a*d+b*c)^3/(2+m)/(3+m)/(4+m)+4*f^3*(-c* f+d*e)*(b*x+a)^(1+m)*(d*x+c)^(-1-m)/d^4/(-a*d+b*c)/(1+m)+6*b*f^2*(-c*f+d*e )^2*(b*x+a)^(1+m)*(d*x+c)^(-1-m)/d^4/(-a*d+b*c)^2/(1+m)/(2+m)+8*b^2*f*(-c* f+d*e)^3*(b*x+a)^(1+m)*(d*x+c)^(-1-m)/d^4/(-a*d+b*c)^3/(1+m)/(2+m)/(3+m)+6 *b^3*(-c*f+d*e)^4*(b*x+a)^(1+m)*(d*x+c)^(-1-m)/d^4/(-a*d+b*c)^4/(1+m)/(2+m )/(3+m)/(4+m)-f^4*(b*x+a)^m*hypergeom([-m, -m],[1-m],b*(d*x+c)/(-a*d+b*c)) /d^5/m/((-d*(b*x+a)/(-a*d+b*c))^m)/((d*x+c)^m)
Time = 15.28 (sec) , antiderivative size = 523, normalized size of antiderivative = 0.80 \[ \int (a+b x)^m (c+d x)^{-5-m} (e+f x)^4 \, dx=(a+b x)^m (c+d x)^{-m} \left (-\frac {24 f^3 (a+b x)^4 (e+f x)}{(b c-a d)^4 (1+m) (2+m) (3+m) (4+m) (c+d x)}+\frac {12 f^2 (a+b x)^3 (-a d (1+m)+b c (4+m)+3 b d x) (e+f x)^2}{(b c-a d)^4 (1+m) (2+m) (3+m) (4+m) (c+d x)^2}+\frac {(e+f x)^4 \left (\frac {6 b^4}{(b c-a d)^4 (1+m) (2+m) (3+m) (4+m)}-\frac {1}{(4+m) (c+d x)^4}+\frac {b m}{(b c-a d) (3+m) (4+m) (c+d x)^3}+\frac {3 b^2 m}{(b c-a d)^2 (2+m) \left (12+7 m+m^2\right ) (c+d x)^2}+\frac {6 b^3 m}{(b c-a d)^3 (1+m) \left (24+26 m+9 m^2+m^3\right ) (c+d x)}\right )}{d}-\frac {4 f (a+b x)^2 (e+f x)^3 \left (a^2 d^2 \left (2+3 m+m^2\right )-2 a b d (1+m) (c (4+m)+2 d x)+b^2 \left (c^2 \left (12+7 m+m^2\right )+4 c d (4+m) x+6 d^2 x^2\right )\right )}{(b c-a d)^4 (1+m) (2+m) (3+m) (4+m) (c+d x)^3}-\frac {24 f^4 \left (\frac {d (a+b x)}{-b c+a d}\right )^{-m} \operatorname {Hypergeometric2F1}\left (-4-m,-m,1-m,\frac {b (c+d x)}{b c-a d}\right )}{d^5 m (1+m) (2+m) (3+m) (4+m)}\right ) \]
((a + b*x)^m*((-24*f^3*(a + b*x)^4*(e + f*x))/((b*c - a*d)^4*(1 + m)*(2 + m)*(3 + m)*(4 + m)*(c + d*x)) + (12*f^2*(a + b*x)^3*(-(a*d*(1 + m)) + b*c* (4 + m) + 3*b*d*x)*(e + f*x)^2)/((b*c - a*d)^4*(1 + m)*(2 + m)*(3 + m)*(4 + m)*(c + d*x)^2) + ((e + f*x)^4*((6*b^4)/((b*c - a*d)^4*(1 + m)*(2 + m)*( 3 + m)*(4 + m)) - 1/((4 + m)*(c + d*x)^4) + (b*m)/((b*c - a*d)*(3 + m)*(4 + m)*(c + d*x)^3) + (3*b^2*m)/((b*c - a*d)^2*(2 + m)*(12 + 7*m + m^2)*(c + d*x)^2) + (6*b^3*m)/((b*c - a*d)^3*(1 + m)*(24 + 26*m + 9*m^2 + m^3)*(c + d*x))))/d - (4*f*(a + b*x)^2*(e + f*x)^3*(a^2*d^2*(2 + 3*m + m^2) - 2*a*b *d*(1 + m)*(c*(4 + m) + 2*d*x) + b^2*(c^2*(12 + 7*m + m^2) + 4*c*d*(4 + m) *x + 6*d^2*x^2)))/((b*c - a*d)^4*(1 + m)*(2 + m)*(3 + m)*(4 + m)*(c + d*x) ^3) - (24*f^4*Hypergeometric2F1[-4 - m, -m, 1 - m, (b*(c + d*x))/(b*c - a* d)])/(d^5*m*(1 + m)*(2 + m)*(3 + m)*(4 + m)*((d*(a + b*x))/(-(b*c) + a*d)) ^m)))/(c + d*x)^m
Time = 0.82 (sec) , antiderivative size = 650, 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.
| \(\displaystyle \int (e+f x)^4 (a+b x)^m (c+d x)^{-m-5} \, dx\) |
| \(\Big \downarrow \) 137 |
| \(\displaystyle \int \left (\frac {4 f^3 (a+b x)^m (d e-c f) (c+d x)^{-m-2}}{d^4}+\frac {6 f^2 (a+b x)^m (d e-c f)^2 (c+d x)^{-m-3}}{d^4}+\frac {(a+b x)^m (d e-c f)^4 (c+d x)^{-m-5}}{d^4}+\frac {4 f (a+b x)^m (d e-c f)^3 (c+d x)^{-m-4}}{d^4}+\frac {f^4 (a+b x)^m (c+d x)^{-m-1}}{d^4}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {6 b^3 (a+b x)^{m+1} (d e-c f)^4 (c+d x)^{-m-1}}{d^4 (m+1) (m+2) (m+3) (m+4) (b c-a d)^4}+\frac {6 b^2 (a+b x)^{m+1} (d e-c f)^4 (c+d x)^{-m-2}}{d^4 (m+2) (m+3) (m+4) (b c-a d)^3}+\frac {8 b^2 f (a+b x)^{m+1} (d e-c f)^3 (c+d x)^{-m-1}}{d^4 (m+1) (m+2) (m+3) (b c-a d)^3}-\frac {f^4 (a+b x)^m (c+d x)^{-m} \left (-\frac {d (a+b x)}{b c-a d}\right )^{-m} \operatorname {Hypergeometric2F1}\left (-m,-m,1-m,\frac {b (c+d x)}{b c-a d}\right )}{d^5 m}+\frac {4 f^3 (a+b x)^{m+1} (d e-c f) (c+d x)^{-m-1}}{d^4 (m+1) (b c-a d)}+\frac {6 f^2 (a+b x)^{m+1} (d e-c f)^2 (c+d x)^{-m-2}}{d^4 (m+2) (b c-a d)}+\frac {6 b f^2 (a+b x)^{m+1} (d e-c f)^2 (c+d x)^{-m-1}}{d^4 (m+1) (m+2) (b c-a d)^2}+\frac {(a+b x)^{m+1} (d e-c f)^4 (c+d x)^{-m-4}}{d^4 (m+4) (b c-a d)}+\frac {4 f (a+b x)^{m+1} (d e-c f)^3 (c+d x)^{-m-3}}{d^4 (m+3) (b c-a d)}+\frac {3 b (a+b x)^{m+1} (d e-c f)^4 (c+d x)^{-m-3}}{d^4 (m+3) (m+4) (b c-a d)^2}+\frac {8 b f (a+b x)^{m+1} (d e-c f)^3 (c+d x)^{-m-2}}{d^4 (m+2) (m+3) (b c-a d)^2}\) |
((d*e - c*f)^4*(a + b*x)^(1 + m)*(c + d*x)^(-4 - m))/(d^4*(b*c - a*d)*(4 + m)) + (4*f*(d*e - c*f)^3*(a + b*x)^(1 + m)*(c + d*x)^(-3 - m))/(d^4*(b*c - a*d)*(3 + m)) + (3*b*(d*e - c*f)^4*(a + b*x)^(1 + m)*(c + d*x)^(-3 - m)) /(d^4*(b*c - a*d)^2*(3 + m)*(4 + m)) + (6*f^2*(d*e - c*f)^2*(a + b*x)^(1 + m)*(c + d*x)^(-2 - m))/(d^4*(b*c - a*d)*(2 + m)) + (8*b*f*(d*e - c*f)^3*( a + b*x)^(1 + m)*(c + d*x)^(-2 - m))/(d^4*(b*c - a*d)^2*(2 + m)*(3 + m)) + (6*b^2*(d*e - c*f)^4*(a + b*x)^(1 + m)*(c + d*x)^(-2 - m))/(d^4*(b*c - a* d)^3*(2 + m)*(3 + m)*(4 + m)) + (4*f^3*(d*e - c*f)*(a + b*x)^(1 + m)*(c + d*x)^(-1 - m))/(d^4*(b*c - a*d)*(1 + m)) + (6*b*f^2*(d*e - c*f)^2*(a + b*x )^(1 + m)*(c + d*x)^(-1 - m))/(d^4*(b*c - a*d)^2*(1 + m)*(2 + m)) + (8*b^2 *f*(d*e - c*f)^3*(a + b*x)^(1 + m)*(c + d*x)^(-1 - m))/(d^4*(b*c - a*d)^3* (1 + m)*(2 + m)*(3 + m)) + (6*b^3*(d*e - c*f)^4*(a + b*x)^(1 + m)*(c + d*x )^(-1 - m))/(d^4*(b*c - a*d)^4*(1 + m)*(2 + m)*(3 + m)*(4 + m)) - (f^4*(a + b*x)^m*Hypergeometric2F1[-m, -m, 1 - m, (b*(c + d*x))/(b*c - a*d)])/(d^5 *m*(-((d*(a + b*x))/(b*c - a*d)))^m*(c + d*x)^m)
3.32.5.3.1 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]))
\[\int \left (b x +a \right )^{m} \left (d x +c \right )^{-5-m} \left (f x +e \right )^{4}d x\]
\[ \int (a+b x)^m (c+d x)^{-5-m} (e+f x)^4 \, dx=\int { {\left (f x + e\right )}^{4} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 5} \,d x } \]
integral((f^4*x^4 + 4*e*f^3*x^3 + 6*e^2*f^2*x^2 + 4*e^3*f*x + e^4)*(b*x + a)^m*(d*x + c)^(-m - 5), x)
Exception generated. \[ \int (a+b x)^m (c+d x)^{-5-m} (e+f x)^4 \, dx=\text {Exception raised: HeuristicGCDFailed} \]
\[ \int (a+b x)^m (c+d x)^{-5-m} (e+f x)^4 \, dx=\int { {\left (f x + e\right )}^{4} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 5} \,d x } \]
\[ \int (a+b x)^m (c+d x)^{-5-m} (e+f x)^4 \, dx=\int { {\left (f x + e\right )}^{4} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 5} \,d x } \]
Timed out. \[ \int (a+b x)^m (c+d x)^{-5-m} (e+f x)^4 \, dx=\int \frac {{\left (e+f\,x\right )}^4\,{\left (a+b\,x\right )}^m}{{\left (c+d\,x\right )}^{m+5}} \,d x \]