3.3059 \(\int (a+b x)^m (c+d x)^{-m} (e+f x)^p \, dx\)

Optimal. Leaf size=121 \[ \frac{(a+b x)^{m+1} (c+d x)^{-m} (e+f x)^p \left (\frac{b (c+d x)}{b c-a d}\right )^m \left (\frac{b (e+f x)}{b e-a f}\right )^{-p} F_1\left (m+1;m,-p;m+2;-\frac{d (a+b x)}{b c-a d},-\frac{f (a+b x)}{b e-a f}\right )}{b (m+1)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.093475, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {140, 139, 138} \[ \frac{(a+b x)^{m+1} (c+d x)^{-m} (e+f x)^p \left (\frac{b (c+d x)}{b c-a d}\right )^m \left (\frac{b (e+f x)}{b e-a f}\right )^{-p} F_1\left (m+1;m,-p;m+2;-\frac{d (a+b x)}{b c-a d},-\frac{f (a+b x)}{b e-a f}\right )}{b (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 140

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]

Rule 139

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 138

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*(m + 1
)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p), 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])

Rubi steps

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

Mathematica [A]  time = 0.198775, size = 119, normalized size = 0.98 \[ \frac{(a+b x)^{m+1} (c+d x)^{-m} (e+f x)^p \left (\frac{b (c+d x)}{b c-a d}\right )^m \left (\frac{b (e+f x)}{b e-a f}\right )^{-p} F_1\left (m+1;m,-p;m+2;\frac{d (a+b x)}{a d-b c},\frac{f (a+b x)}{a f-b e}\right )}{b (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.145, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{p} \left ( bx+a \right ) ^{m}}{ \left ( dx+c \right ) ^{m}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{m}{\left (f x + e\right )}^{p}}{{\left (d x + c\right )}^{m}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x + a\right )}^{m}{\left (f x + e\right )}^{p}}{{\left (d x + c\right )}^{m}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{m}{\left (f x + e\right )}^{p}}{{\left (d x + c\right )}^{m}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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