3.3117 \(\int \frac{(a+b x)^m (c+d x)^{1-m}}{(e+f x)^2} \, dx\)

Optimal. Leaf size=190 \[ \frac{(a+b x)^m (c+d x)^{-m} (a d f (1-m)-b (d e-c f m)) \, _2F_1\left (1,m;m+1;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{f^2 m (b e-a f)}-\frac{(a+b x)^m (c+d x)^{1-m}}{f (e+f x)}+\frac{d (a+b x)^m (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,m;m+1;-\frac{d (a+b x)}{b c-a d}\right )}{f^2 m} \]

[Out]

-(((a + b*x)^m*(c + d*x)^(1 - m))/(f*(e + f*x))) + ((a*d*f*(1 - m) - b*(d*e - c*f*m))*(a + b*x)^m*Hypergeometr
ic2F1[1, m, 1 + m, ((d*e - c*f)*(a + b*x))/((b*e - a*f)*(c + d*x))])/(f^2*(b*e - a*f)*m*(c + d*x)^m) + (d*(a +
b*x)^m*((b*(c + d*x))/(b*c - a*d))^m*Hypergeometric2F1[m, m, 1 + m, -((d*(a + b*x))/(b*c - a*d))])/(f^2*m*(c
+ d*x)^m)

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Rubi [C]  time = 0.0512165, antiderivative size = 108, normalized size of antiderivative = 0.57, 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} \[ \frac{(b c-a d) (a+b x)^{m+1} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m F_1\left (m+1;m-1,2;m+2;-\frac{d (a+b x)}{b c-a d},-\frac{f (a+b x)}{b e-a f}\right )}{(m+1) (b e-a f)^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((b*c - a*d)*(a + b*x)^(1 + m)*((b*(c + d*x))/(b*c - a*d))^m*AppellF1[1 + m, -1 + 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)*(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)^{1-m}}{(e+f x)^2} \, dx &=\frac{\left ((b c-a d) (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 )^{1-m}}{(e+f x)^2} \, dx}{b}\\ &=\frac{(b c-a d) (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;-1+m,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 [C]  time = 0.238086, size = 203, normalized size = 1.07 \[ -\frac{(a+b x)^{m+1} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \left (\frac{b (e+f x)}{b e-a f}\right )^{-m} \left (d (e+f x) \left (\frac{b (e+f x)}{b e-a f}\right )^m F_1\left (m+1;m,1;m+2;\frac{d (a+b x)}{a d-b c},\frac{f (a+b x)}{a f-b e}\right )+(c f-d e) \, _2F_1\left (m,m+1;m+2;\frac{(c f-d e) (a+b x)}{(b c-a d) (e+f x)}\right )\right )}{f (m+1) (e+f x) (a f-b e)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(((a + b*x)^(1 + m)*((b*(c + d*x))/(b*c - a*d))^m*(d*(e + f*x)*((b*(e + f*x))/(b*e - a*f))^m*AppellF1[1 + m,
m, 1, 2 + m, (d*(a + b*x))/(-(b*c) + a*d), (f*(a + b*x))/(-(b*e) + a*f)] + (-(d*e) + c*f)*Hypergeometric2F1[m,
1 + m, 2 + m, ((-(d*e) + c*f)*(a + b*x))/((b*c - a*d)*(e + f*x))]))/(f*(-(b*e) + a*f)*(1 + m)*(c + d*x)^m*(e
+ f*x)*((b*(e + f*x))/(b*e - a*f))^m))

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Maple [F]  time = 0.068, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) ^{m} \left ( dx+c \right ) ^{1-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)^(1-m)/(f*x+e)^2,x)

[Out]

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

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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 + 1}}{{\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)^(1-m)/(f*x+e)^2,x, algorithm="maxima")

[Out]

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

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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 + 1}}{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)^(1-m)/(f*x+e)^2,x, algorithm="fricas")

[Out]

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

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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)**(1-m)/(f*x+e)**2,x)

[Out]

Timed out

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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 + 1}}{{\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)^(1-m)/(f*x+e)^2,x, algorithm="giac")

[Out]

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