3.3085 \(\int \frac{(a+b x)^m (c+d x)^{-2-m}}{(e+f x)^2} \, dx\)

Optimal. Leaf size=233 \[ -\frac{f (a+b x)^m (c+d x)^{-m} (a d f (m+2)-b (c f m+2 d e)) \, _2F_1\left (1,-m;1-m;\frac{(b e-a f) (c+d x)}{(d e-c f) (a+b x)}\right )}{m (b e-a f) (d e-c f)^3}-\frac{d (a+b x)^{m+1} (c+d x)^{-m-1} (a d f (m+2)-b (c f (m+1)+d e))}{(m+1) (b c-a d) (b e-a f) (d e-c f)^2}-\frac{f (a+b x)^{m+1} (c+d x)^{-m-1}}{(e+f x) (b e-a f) (d e-c f)} \]

[Out]

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

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Rubi [A]  time = 0.203663, antiderivative size = 243, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {129, 151, 12, 131} \[ \frac{f (a+b x)^{m+1} (c+d x)^{-m-1} (a d f (m+2)-b (c f m+2 d e)) \, _2F_1\left (1,m+1;m+2;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{(m+1) (b e-a f)^2 (d e-c f)^2}+\frac{d (a+b x)^{m+1} (c+d x)^{-m-1}}{(m+1) (e+f x) (b c-a d) (d e-c f)}+\frac{f (a+b x)^{m+1} (c+d x)^{-m} (-a d f (m+2)+b c f (m+1)+b d e)}{(m+1) (e+f x) (b c-a d) (b e-a f) (d e-c f)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 129

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && ILtQ[m + n
+ p + 2, 0] && NeQ[m, -1] && (SumSimplerQ[m, 1] || ( !(NeQ[n, -1] && SumSimplerQ[n, 1]) &&  !(NeQ[p, -1] && S
umSimplerQ[p, 1])))

Rule 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 131

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((b*c -
a*d)^n*(a + b*x)^(m + 1)*Hypergeometric2F1[m + 1, -n, m + 2, -(((d*e - c*f)*(a + b*x))/((b*c - a*d)*(e + f*x))
)])/((m + 1)*(b*e - a*f)^(n + 1)*(e + f*x)^(m + 1)), x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p
+ 2, 0] && ILtQ[n, 0]

Rubi steps

\begin{align*} \int \frac{(a+b x)^m (c+d x)^{-2-m}}{(e+f x)^2} \, dx &=\frac{d (a+b x)^{1+m} (c+d x)^{-1-m}}{(b c-a d) (d e-c f) (1+m) (e+f x)}+\frac{\int \frac{(a+b x)^m (c+d x)^{-1-m} (-f (b c (1+m)-a d (2+m))+b d f x)}{(e+f x)^2} \, dx}{(b c-a d) (d e-c f) (1+m)}\\ &=\frac{d (a+b x)^{1+m} (c+d x)^{-1-m}}{(b c-a d) (d e-c f) (1+m) (e+f x)}+\frac{f (b d e+b c f (1+m)-a d f (2+m)) (a+b x)^{1+m} (c+d x)^{-m}}{(b c-a d) (b e-a f) (d e-c f)^2 (1+m) (e+f x)}+\frac{\int \frac{(b c-a d) f (1+m) (a d f (2+m)-b (2 d e+c f m)) (a+b x)^m (c+d x)^{-1-m}}{e+f x} \, dx}{(b c-a d) (b e-a f) (d e-c f)^2 (1+m)}\\ &=\frac{d (a+b x)^{1+m} (c+d x)^{-1-m}}{(b c-a d) (d e-c f) (1+m) (e+f x)}+\frac{f (b d e+b c f (1+m)-a d f (2+m)) (a+b x)^{1+m} (c+d x)^{-m}}{(b c-a d) (b e-a f) (d e-c f)^2 (1+m) (e+f x)}+\frac{(f (a d f (2+m)-b (2 d e+c f m))) \int \frac{(a+b x)^m (c+d x)^{-1-m}}{e+f x} \, dx}{(b e-a f) (d e-c f)^2}\\ &=\frac{d (a+b x)^{1+m} (c+d x)^{-1-m}}{(b c-a d) (d e-c f) (1+m) (e+f x)}+\frac{f (b d e+b c f (1+m)-a d f (2+m)) (a+b x)^{1+m} (c+d x)^{-m}}{(b c-a d) (b e-a f) (d e-c f)^2 (1+m) (e+f x)}+\frac{f (a d f (2+m)-b (2 d e+c f m)) (a+b x)^{1+m} (c+d x)^{-1-m} \, _2F_1\left (1,1+m;2+m;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{(b e-a f)^2 (d e-c f)^2 (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.250391, size = 193, normalized size = 0.83 \[ \frac{(a+b x)^{m+1} (c+d x)^{-m-1} \left (f (e+f x) (b c-a d) (a d f (m+2)-b (c f m+2 d e)) \, _2F_1\left (1,m+1;m+2;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )+f (c+d x) (b e-a f) (-a d f (m+2)+b c f (m+1)+b d e)+d (b e-a f)^2 (d e-c f)\right )}{(m+1) (e+f x) (b c-a d) (b e-a f)^2 (d e-c f)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.07, 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)

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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 - 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)

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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 - 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)

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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)**(-2-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 - 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)