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

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Rubi [A]  time = 0.07, antiderivative size = 158, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.17, size = 142, normalized size = 0.90 \[ \frac {(a+b x)^{m+1} (c+d x)^{-m} \left (\frac {(b c-a d) (b (c f m+d e)-a d f (m+1)) \, _2F_1\left (2,m+1;m+2;\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )}{(m+1) (c+d x) (b e-a f)^2}-\frac {d}{e+f x}\right )}{m (b c-a d) (c f-d e)} \]

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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fricas [F]  time = 1.09, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 1}}{f^{2} x^{2} + 2 \, e f x + e^{2}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 1}}{{\left (f x + e\right )}^{2}}\,{d x} \]

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)

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maple [F]  time = 0.24, size = 0, normalized size = 0.00 \[ \int \frac {\left (b x +a \right )^{m} \left (d x +c \right )^{-m -1}}{\left (f x +e \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m - 1}}{{\left (f x + e\right )}^{2}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,x\right )}^m}{{\left (e+f\,x\right )}^2\,{\left (c+d\,x\right )}^{m+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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