3.3150 \(\int (a+b x)^m (c+d x)^{-m-n} (e+f x)^{-3+n} \, dx\)

Optimal. Leaf size=237 \[ -\frac{(a+b x)^{m+1} (e+f x)^{n-1} (c+d x)^{-m-n} (a d f (m+1)-b (d e (2-n)-c f (-m-n+1))) \left (\frac{(c+d x) (b e-a f)}{(e+f x) (b c-a d)}\right )^{m+n} \, _2F_1\left (m+1,m+n;m+2;-\frac{(d e-c f) (a+b x)}{(b c-a d) (e+f x)}\right )}{(m+1) (2-n) (b e-a f)^2 (d e-c f)}-\frac{f (a+b x)^{m+1} (e+f x)^{n-2} (c+d x)^{-m-n+1}}{(2-n) (b e-a f) (d e-c f)} \]

[Out]

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

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Rubi [A]  time = 0.0894516, antiderivative size = 235, normalized size of antiderivative = 0.99, number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {96, 132} \[ -\frac{(a+b x)^{m+1} (e+f x)^{n-1} (c+d x)^{-m-n} (a d f (m+1)+b c f (-m-n+1)-b d e (2-n)) \left (\frac{(c+d x) (b e-a f)}{(e+f x) (b c-a d)}\right )^{m+n} \, _2F_1\left (m+1,m+n;m+2;-\frac{(d e-c f) (a+b x)}{(b c-a d) (e+f x)}\right )}{(m+1) (2-n) (b e-a f)^2 (d e-c f)}-\frac{f (a+b x)^{m+1} (e+f x)^{n-2} (c+d x)^{-m-n+1}}{(2-n) (b e-a f) (d e-c f)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^m*(c + d*x)^(-m - n)*(e + f*x)^(-3 + n),x]

[Out]

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

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 132

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

Rubi steps

\begin{align*} \int (a+b x)^m (c+d x)^{-m-n} (e+f x)^{-3+n} \, dx &=-\frac{f (a+b x)^{1+m} (c+d x)^{1-m-n} (e+f x)^{-2+n}}{(b e-a f) (d e-c f) (2-n)}-\frac{(a d f (1+m)-b d e (2-n)+b c f (1-m-n)) \int (a+b x)^m (c+d x)^{-m-n} (e+f x)^{-2+n} \, dx}{(b e-a f) (d e-c f) (2-n)}\\ &=-\frac{f (a+b x)^{1+m} (c+d x)^{1-m-n} (e+f x)^{-2+n}}{(b e-a f) (d e-c f) (2-n)}-\frac{(a d f (1+m)-b d e (2-n)+b c f (1-m-n)) (a+b x)^{1+m} (c+d x)^{-m-n} \left (\frac{(b e-a f) (c+d x)}{(b c-a d) (e+f x)}\right )^{m+n} (e+f x)^{-1+n} \, _2F_1\left (1+m,m+n;2+m;-\frac{(d e-c f) (a+b x)}{(b c-a d) (e+f x)}\right )}{(b e-a f)^2 (d e-c f) (1+m) (2-n)}\\ \end{align*}

Mathematica [A]  time = 0.213865, size = 186, normalized size = 0.78 \[ \frac{(a+b x)^{m+1} (e+f x)^{n-2} (c+d x)^{-m-n} \left (\frac{(e+f x) (a d f (m+1)-b c f (m+n-1)+b d e (n-2)) \left (\frac{(c+d x) (b e-a f)}{(e+f x) (b c-a d)}\right )^{m+n} \, _2F_1\left (m+1,m+n;m+2;\frac{(c f-d e) (a+b x)}{(b c-a d) (e+f x)}\right )}{(m+1) (b e-a f)}+f (c+d x)\right )}{(n-2) (b e-a f) (d e-c f)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^m*(c + d*x)^(-m - n)*(e + f*x)^(-3 + n),x]

[Out]

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

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Maple [F]  time = 0.149, size = 0, normalized size = 0. \begin{align*} \int \left ( bx+a \right ) ^{m} \left ( dx+c \right ) ^{-n-m} \left ( fx+e \right ) ^{-3+n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^m*(d*x+c)^(-n-m)*(f*x+e)^(-3+n),x)

[Out]

int((b*x+a)^m*(d*x+c)^(-n-m)*(f*x+e)^(-3+n),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m - n}{\left (f x + e\right )}^{n - 3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(-m-n)*(f*x+e)^(-3+n),x, algorithm="maxima")

[Out]

integrate((b*x + a)^m*(d*x + c)^(-m - n)*(f*x + e)^(n - 3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m - n}{\left (f x + e\right )}^{n - 3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(-m-n)*(f*x+e)^(-3+n),x, algorithm="fricas")

[Out]

integral((b*x + a)^m*(d*x + c)^(-m - n)*(f*x + e)^(n - 3), 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)**(-m-n)*(f*x+e)**(-3+n),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{-m - n}{\left (f x + e\right )}^{n - 3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(-m-n)*(f*x+e)^(-3+n),x, algorithm="giac")

[Out]

integrate((b*x + a)^m*(d*x + c)^(-m - n)*(f*x + e)^(n - 3), x)