3.3164 \(\int \frac{(a+b x)^m (c+d x)^n}{(e+f x)^2} \, dx\)

Optimal. Leaf size=101 \[ \frac{b (a+b x)^{m+1} (c+d x)^n \left (\frac{b (c+d x)}{b c-a d}\right )^{-n} F_1\left (m+1;-n,2;m+2;-\frac{d (a+b x)}{b c-a d},-\frac{f (a+b x)}{b e-a f}\right )}{(m+1) (b e-a f)^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0452203, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {137, 136} \[ \frac{b (a+b x)^{m+1} (c+d x)^n \left (\frac{b (c+d x)}{b c-a d}\right )^{-n} F_1\left (m+1;-n,2;m+2;-\frac{d (a+b x)}{b c-a d},-\frac{f (a+b x)}{b e-a f}\right )}{(m+1) (b e-a f)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 137

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}, x] &&  !IntegerQ[m] &&
!IntegerQ[n] && IntegerQ[p] &&  !GtQ[b/(b*c - a*d), 0] &&  !SimplerQ[c + d*x, a + b*x]

Rule 136

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

Rubi steps

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

Mathematica [A]  time = 0.0936833, size = 99, normalized size = 0.98 \[ \frac{b (a+b x)^{m+1} (c+d x)^n \left (\frac{b (c+d x)}{b c-a d}\right )^{-n} F_1\left (m+1;-n,2;m+2;\frac{d (a+b x)}{a d-b c},\frac{f (a+b x)}{a f-b e}\right )}{(m+1) (b e-a f)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.069, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) ^{m} \left ( dx+c \right ) ^{n}}{ \left ( fx+e \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{n}}{{\left (f x + e\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{n}}{f^{2} x^{2} + 2 \, e f x + e^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{n}}{{\left (f x + e\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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