Integrand size = 26, antiderivative size = 133 \[ \int (a+b x)^m (c+d x)^{-3-m} (e+f x)^p \, dx=\frac {b^2 (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m (e+f x)^p \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} \operatorname {AppellF1}\left (1+m,3+m,-p,2+m,-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{(b c-a d)^3 (1+m)} \]
b^2*(b*x+a)^(1+m)*(b*(d*x+c)/(-a*d+b*c))^m*(f*x+e)^p*AppellF1(1+m,3+m,-p,2 +m,-d*(b*x+a)/(-a*d+b*c),-f*(b*x+a)/(-a*f+b*e))/(-a*d+b*c)^3/(1+m)/((d*x+c )^m)/((b*(f*x+e)/(-a*f+b*e))^p)
Time = 0.31 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.98 \[ \int (a+b x)^m (c+d x)^{-3-m} (e+f x)^p \, dx=\frac {b^2 (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m (e+f x)^p \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} \operatorname {AppellF1}\left (1+m,3+m,-p,2+m,\frac {d (a+b x)}{-b c+a d},\frac {f (a+b x)}{-b e+a f}\right )}{(b c-a d)^3 (1+m)} \]
(b^2*(a + b*x)^(1 + m)*((b*(c + d*x))/(b*c - a*d))^m*(e + f*x)^p*AppellF1[ 1 + m, 3 + m, -p, 2 + m, (d*(a + b*x))/(-(b*c) + a*d), (f*(a + b*x))/(-(b* e) + a*f)])/((b*c - a*d)^3*(1 + m)*(c + d*x)^m*((b*(e + f*x))/(b*e - a*f)) ^p)
Time = 0.27 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {157, 156, 155}
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.
\(\displaystyle \int (a+b x)^m (c+d x)^{-m-3} (e+f x)^p \, dx\) |
\(\Big \downarrow \) 157 |
\(\displaystyle \frac {b^3 (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \int (a+b x)^m \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{-m-3} (e+f x)^pdx}{(b c-a d)^3}\) |
\(\Big \downarrow \) 156 |
\(\displaystyle \frac {b^3 (c+d x)^{-m} (e+f x)^p \left (\frac {b (c+d x)}{b c-a d}\right )^m \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} \int (a+b x)^m \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{-m-3} \left (\frac {b e}{b e-a f}+\frac {b f x}{b e-a f}\right )^pdx}{(b c-a d)^3}\) |
\(\Big \downarrow \) 155 |
\(\displaystyle \frac {b^2 (a+b x)^{m+1} (c+d x)^{-m} (e+f x)^p \left (\frac {b (c+d x)}{b c-a d}\right )^m \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} \operatorname {AppellF1}\left (m+1,m+3,-p,m+2,-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{(m+1) (b c-a d)^3}\) |
(b^2*(a + b*x)^(1 + m)*((b*(c + d*x))/(b*c - a*d))^m*(e + f*x)^p*AppellF1[ 1 + m, 3 + m, -p, 2 + m, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b* e - a*f))])/((b*c - a*d)^3*(1 + m)*(c + d*x)^m*((b*(e + f*x))/(b*e - a*f)) ^p)
3.31.87.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* Simplify[b/(b*e - a*f)]^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/ (b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] && GtQ[Sim plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] && !(GtQ[Simpl ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d *x, a + b*x]) && !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c - e*d)], 0] && SimplerQ[e + f*x, a + b*x])
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[(e + f*x)^FracPart[p]/(Simplify[b/(b*e - a*f)]^IntPart[p ]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p]) Int[(a + b*x)^m*(c + d*x)^n*Si mp[b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)), x]^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] & & GtQ[Simplify[b/(b*c - a*d)], 0] && !GtQ[Simplify[b/(b*e - a*f)], 0]
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[(c + d*x)^FracPart[n]/(Simplify[b/(b*c - a*d)]^IntPart[n ]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n]) Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d)), x]^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] & & !GtQ[Simplify[b/(b*c - a*d)], 0] && !SimplerQ[c + d*x, a + b*x] && !Si mplerQ[e + f*x, a + b*x]
\[\int \left (b x +a \right )^{m} \left (d x +c \right )^{-3-m} \left (f x +e \right )^{p}d x\]
\[ \int (a+b x)^m (c+d x)^{-3-m} (e+f x)^p \, dx=\int { {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 3} {\left (f x + e\right )}^{p} \,d x } \]
Timed out. \[ \int (a+b x)^m (c+d x)^{-3-m} (e+f x)^p \, dx=\text {Timed out} \]
\[ \int (a+b x)^m (c+d x)^{-3-m} (e+f x)^p \, dx=\int { {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 3} {\left (f x + e\right )}^{p} \,d x } \]
\[ \int (a+b x)^m (c+d x)^{-3-m} (e+f x)^p \, dx=\int { {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 3} {\left (f x + e\right )}^{p} \,d x } \]
Timed out. \[ \int (a+b x)^m (c+d x)^{-3-m} (e+f x)^p \, dx=\int \frac {{\left (e+f\,x\right )}^p\,{\left (a+b\,x\right )}^m}{{\left (c+d\,x\right )}^{m+3}} \,d x \]