3.3128 \(\int \frac{(a+b x)^m (c+d x)^{2-m}}{(e+f x)^2} \, dx\)

Optimal. Leaf size=316 \[ \frac{d^2 (a+b x)^{m+1} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m (b (2 d e-c f (2-m))-a d f m) \, _2F_1\left (m,m+1;m+2;-\frac{d (a+b x)}{b c-a d}\right )}{b f^3 m (m+1) (b c-a d)}-\frac{2 d^2 (a+b x)^{m+1} (d e-c f) (c+d x)^{-m}}{f^3 m (b c-a d)}+\frac{(a+b x)^m (d e-c f) (c+d x)^{-m} (a d f (2-m)-b (2 d e-c f m)) \, _2F_1\left (1,-m;1-m;\frac{(b e-a f) (c+d x)}{(d e-c f) (a+b x)}\right )}{f^3 m (b e-a f)}+\frac{(a+b x)^{m+1} (d e-c f)^2 (c+d x)^{-m}}{f^2 (e+f x) (b e-a f)} \]

[Out]

(-2*d^2*(d*e - c*f)*(a + b*x)^(1 + m))/((b*c - a*d)*f^3*m*(c + d*x)^m) + ((d*e - c*f)^2*(a + b*x)^(1 + m))/(f^
2*(b*e - a*f)*(c + d*x)^m*(e + f*x)) + ((d*e - c*f)*(a*d*f*(2 - m) - b*(2*d*e - c*f*m))*(a + b*x)^m*Hypergeome
tric2F1[1, -m, 1 - m, ((b*e - a*f)*(c + d*x))/((d*e - c*f)*(a + b*x))])/(f^3*(b*e - a*f)*m*(c + d*x)^m) + (d^2
*(b*(2*d*e - c*f*(2 - m)) - a*d*f*m)*(a + b*x)^(1 + m)*((b*(c + d*x))/(b*c - a*d))^m*Hypergeometric2F1[m, 1 +
m, 2 + m, -((d*(a + b*x))/(b*c - a*d))])/(b*(b*c - a*d)*f^3*m*(1 + m)*(c + d*x)^m)

________________________________________________________________________________________

Rubi [C]  time = 0.0556324, antiderivative size = 113, normalized size of antiderivative = 0.36, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {137, 136} \[ \frac{(b c-a d)^2 (a+b x)^{m+1} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m F_1\left (m+1;m-2,2;m+2;-\frac{d (a+b x)}{b c-a d},-\frac{f (a+b x)}{b e-a f}\right )}{b (m+1) (b e-a f)^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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

Mathematica [C]  time = 0.190933, size = 103, normalized size = 0.33 \[ \frac{b (a+b x)^{m+1} (c+d x)^{2-m} \left (\frac{b (c+d x)}{b c-a d}\right )^{m-2} F_1\left (m+1;m-2,2;m+2;\frac{d (a+b x)}{a d-b c},\frac{f (a+b x)}{a f-b e}\right )}{(m+1) (b e-a f)^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

int((b*x+a)^m*(d*x+c)^(2-m)/(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 )}^{-m + 2}}{{\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)^(2-m)/(f*x+e)^2,x, algorithm="maxima")

[Out]

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

[Out]

integral((b*x + a)^m*(d*x + c)^(-m + 2)/(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)**(2-m)/(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 )}^{-m + 2}}{{\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)^(2-m)/(f*x+e)^2,x, algorithm="giac")

[Out]

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