\(\int (a+b x)^m (c+d x)^{2-m} (e+f x)^p \, dx\) [3122]

Optimal result

Integrand size = 26, antiderivative size = 133 \[ \int (a+b x)^m (c+d x)^{2-m} (e+f x)^p \, dx=\frac {(b c-a d)^2 (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m (e+f x)^p \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} \operatorname {AppellF1}\left (1+m,-2+m,-p,2+m,-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{b^3 (1+m)} \]

[Out]

(-a*d+b*c)^2*(b*x+a)^(1+m)*(b*(d*x+c)/(-a*d+b*c))^m*(f*x+e)^p*AppellF1(1+m,-2+m,-p,2+m,-d*(b*x+a)/(-a*d+b*c),-
f*(b*x+a)/(-a*f+b*e))/b^3/(1+m)/((d*x+c)^m)/((b*(f*x+e)/(-a*f+b*e))^p)

Rubi [A] (verified)

Time = 0.07 (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 = {145, 144, 143} \[ \int (a+b x)^m (c+d x)^{2-m} (e+f x)^p \, dx=\frac {(b c-a d)^2 (a+b x)^{m+1} (c+d x)^{-m} (e+f x)^p \left (\frac {b (c+d x)}{b c-a d}\right )^m \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} \operatorname {AppellF1}\left (m+1,m-2,-p,m+2,-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{b^3 (m+1)} \]

[In]

Int[(a + b*x)^m*(c + d*x)^(2 - m)*(e + f*x)^p,x]

[Out]

((b*c - a*d)^2*(a + b*x)^(1 + m)*((b*(c + d*x))/(b*c - a*d))^m*(e + f*x)^p*AppellF1[1 + m, -2 + m, -p, 2 + m,
-((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b^3*(1 + m)*(c + d*x)^m*((b*(e + f*x))/(b*e - a*
f))^p)

Rule 143

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n*(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] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rule 144

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((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*
(b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !GtQ[b/(b*e - a*f), 0]

Rule 145

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c + d*x)^
FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n]), Int[(a + b*x)^m*(b*(c/(b*c -
a*d)) + b*d*(x/(b*c - a*d)))^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[b/(b*c - a*d), 0] &&  !SimplerQ[c + d*x, a + b*x] &&  !SimplerQ[e +
f*x, a + b*x]

Rubi steps \begin{align*} \text {integral}& = \frac {\left ((b c-a d)^2 (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m\right ) \int (a+b x)^m \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{2-m} (e+f x)^p \, dx}{b^2} \\ & = \frac {\left ((b c-a d)^2 (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}\right ) \int (a+b x)^m \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{2-m} \left (\frac {b e}{b e-a f}+\frac {b f x}{b e-a f}\right )^p \, dx}{b^2} \\ & = \frac {(b c-a d)^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} F_1\left (1+m;-2+m,-p;2+m;-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{b^3 (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.98 \[ \int (a+b x)^m (c+d x)^{2-m} (e+f x)^p \, dx=\frac {(b c-a d)^2 (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m (e+f x)^p \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} \operatorname {AppellF1}\left (1+m,-2+m,-p,2+m,\frac {d (a+b x)}{-b c+a d},\frac {f (a+b x)}{-b e+a f}\right )}{b^3 (1+m)} \]

[In]

Integrate[(a + b*x)^m*(c + d*x)^(2 - m)*(e + f*x)^p,x]

[Out]

((b*c - a*d)^2*(a + b*x)^(1 + m)*((b*(c + d*x))/(b*c - a*d))^m*(e + f*x)^p*AppellF1[1 + m, -2 + m, -p, 2 + m,
(d*(a + b*x))/(-(b*c) + a*d), (f*(a + b*x))/(-(b*e) + a*f)])/(b^3*(1 + m)*(c + d*x)^m*((b*(e + f*x))/(b*e - a*
f))^p)

Maple [F]

\[\int \left (b x +a \right )^{m} \left (d x +c \right )^{2-m} \left (f x +e \right )^{p}d x\]

[In]

int((b*x+a)^m*(d*x+c)^(2-m)*(f*x+e)^p,x)

[Out]

int((b*x+a)^m*(d*x+c)^(2-m)*(f*x+e)^p,x)

Fricas [F]

\[ \int (a+b x)^m (c+d x)^{2-m} (e+f x)^p \, dx=\int { {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2} {\left (f x + e\right )}^{p} \,d x } \]

[In]

integrate((b*x+a)^m*(d*x+c)^(2-m)*(f*x+e)^p,x, algorithm="fricas")

[Out]

integral((b*x + a)^m*(d*x + c)^(-m + 2)*(f*x + e)^p, x)

Sympy [F(-1)]

Timed out. \[ \int (a+b x)^m (c+d x)^{2-m} (e+f x)^p \, dx=\text {Timed out} \]

[In]

integrate((b*x+a)**m*(d*x+c)**(2-m)*(f*x+e)**p,x)

[Out]

Timed out

Maxima [F]

\[ \int (a+b x)^m (c+d x)^{2-m} (e+f x)^p \, dx=\int { {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2} {\left (f x + e\right )}^{p} \,d x } \]

[In]

integrate((b*x+a)^m*(d*x+c)^(2-m)*(f*x+e)^p,x, algorithm="maxima")

[Out]

integrate((b*x + a)^m*(d*x + c)^(-m + 2)*(f*x + e)^p, x)

Giac [F]

\[ \int (a+b x)^m (c+d x)^{2-m} (e+f x)^p \, dx=\int { {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2} {\left (f x + e\right )}^{p} \,d x } \]

[In]

integrate((b*x+a)^m*(d*x+c)^(2-m)*(f*x+e)^p,x, algorithm="giac")

[Out]

integrate((b*x + a)^m*(d*x + c)^(-m + 2)*(f*x + e)^p, x)

Mupad [F(-1)]

Timed out. \[ \int (a+b x)^m (c+d x)^{2-m} (e+f x)^p \, dx=\int {\left (e+f\,x\right )}^p\,{\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^{2-m} \,d x \]

[In]

int((e + f*x)^p*(a + b*x)^m*(c + d*x)^(2 - m),x)

[Out]

int((e + f*x)^p*(a + b*x)^m*(c + d*x)^(2 - m), x)