3.3048 \(\int (a+b x) (c+d x)^{-1+n} (e+f x)^{-n} \, dx\)

Optimal. Leaf size=152 \[ -\frac{(c+d x)^{n+1} (e+f x)^{-n} \left (\frac{d (e+f x)}{d e-c f}\right )^n (a d f-b (c f (1-n)+d e n)) \, _2F_1\left (n,n+1;n+2;-\frac{f (c+d x)}{d e-c f}\right )}{d^2 n (n+1) (d e-c f)}-\frac{(b c-a d) (c+d x)^n (e+f x)^{1-n}}{d n (d e-c f)} \]

[Out]

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

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Rubi [A]  time = 0.0740748, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {79, 70, 69} \[ -\frac{(c+d x)^{n+1} (e+f x)^{-n} \left (\frac{d (e+f x)}{d e-c f}\right )^n (a d f-b c f (1-n)-b d e n) \, _2F_1\left (n,n+1;n+2;-\frac{f (c+d x)}{d e-c f}\right )}{d^2 n (n+1) (d e-c f)}-\frac{(b c-a d) (c+d x)^n (e+f x)^{1-n}}{d n (d e-c f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 79

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

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), 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*Simp[(b*c)/(b*c - a*d) + (b*d*x)/(b*c -
a*d), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

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

Mathematica [A]  time = 0.0950399, size = 124, normalized size = 0.82 \[ \frac{(c+d x)^n (e+f x)^{-n} \left (\frac{(c+d x) \left (\frac{d (e+f x)}{d e-c f}\right )^n (-a d f+b c (f-f n)+b d e n) \, _2F_1\left (n,n+1;n+2;\frac{f (c+d x)}{c f-d e}\right )}{n+1}+d (e+f x) (a d-b c)\right )}{d^2 n (d e-c f)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.052, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) \left ( dx+c \right ) ^{-1+n}}{ \left ( fx+e \right ) ^{n}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}{\left (d x + c\right )}^{n - 1}}{{\left (f x + e\right )}^{n}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(d*x+c)^(-1+n)/((f*x+e)^n),x, algorithm="maxima")

[Out]

integrate((b*x + a)*(d*x + c)^(n - 1)/(f*x + e)^n, 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 )}{\left (d x + c\right )}^{n - 1}}{{\left (f x + e\right )}^{n}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(d*x+c)^(-1+n)/((f*x+e)^n),x, algorithm="fricas")

[Out]

integral((b*x + a)*(d*x + c)^(n - 1)/(f*x + e)^n, 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)*(d*x+c)**(-1+n)/((f*x+e)**n),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 )}{\left (d x + c\right )}^{n - 1}}{{\left (f x + e\right )}^{n}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(d*x+c)^(-1+n)/((f*x+e)^n),x, algorithm="giac")

[Out]

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