3.31.48 \(\int (a+b x) (c+d x)^{-1+n} (e+f x)^{-n} \, dx\) [3048]

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

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 152, normalized size of antiderivative = 1.00, 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 = {80, 72, 71} \begin {gather*} -\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)} \end {gather*}

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 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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]))

Rule 72

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 80

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]

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*}

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Mathematica [A]
time = 0.13, size = 124, normalized size = 0.82 \begin {gather*} \frac {(c+d x)^n (e+f x)^{-n} \left (d (-b c+a d) (e+f x)+\frac {(-a d f+b d e n+b c (f-f n)) (c+d x) \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 )}{1+n}\right )}{d^2 (d e-c f) n} \end {gather*}

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 = 0.02, size = 0, normalized size = 0.00 \[\int \left (b x +a \right ) \left (d x +c \right )^{-1+n} \left (f x +e \right )^{-n}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (a+b\,x\right )\,{\left (c+d\,x\right )}^{n-1}}{{\left (e+f\,x\right )}^n} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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