3.3076 \(\int \frac{(a+b x)^m (c+d x)^{-1-m}}{(e+f x)^2} \, dx\)

Optimal. Leaf size=158 \[ \frac{(a+b x)^{m+1} (c+d x)^{-m-1} (a d f (m+1)-b (c f m+d e)) \, _2F_1\left (2,m+1;m+2;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{m (m+1) (b e-a f)^2 (d e-c f)}+\frac{d (a+b x)^{m+1} (c+d x)^{-m}}{m (e+f x) (b c-a d) (d e-c f)} \]

[Out]

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

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Rubi [A]  time = 0.0718896, antiderivative size = 158, normalized size of antiderivative = 1., 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 = {96, 131} \[ \frac{(a+b x)^{m+1} (c+d x)^{-m-1} (a d f (m+1)-b (c f m+d e)) \, _2F_1\left (2,m+1;m+2;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{m (m+1) (b e-a f)^2 (d e-c f)}+\frac{d (a+b x)^{m+1} (c+d x)^{-m}}{m (e+f x) (b c-a d) (d e-c f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 96

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[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 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, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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

Mathematica [A]  time = 0.115825, size = 142, normalized size = 0.9 \[ \frac{(a+b x)^{m+1} (c+d x)^{-m} \left (\frac{(b c-a d) (b (c f m+d e)-a d f (m+1)) \, _2F_1\left (2,m+1;m+2;\frac{(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{(m+1) (c+d x) (b e-a f)^2}-\frac{d}{e+f x}\right )}{m (b c-a d) (c f-d e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.072, 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)