3.31.47 \(\int (a+b x) (c+d x)^n (e+f x)^{-n} \, dx\) [3047]

Optimal. Leaf size=134 \[ \frac {b (c+d x)^{1+n} (e+f x)^{1-n}}{2 d f}+\frac {(2 a d f-b (c f (1-n)+d e (1+n))) (c+d x)^{1+n} (e+f x)^{-n} \left (\frac {d (e+f x)}{d e-c f}\right )^n \, _2F_1\left (n,1+n;2+n;-\frac {f (c+d x)}{d e-c f}\right )}{2 d^2 f (1+n)} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {81, 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 (2 a d f-b c f (1-n)-b d e (n+1)) \, _2F_1\left (n,n+1;n+2;-\frac {f (c+d x)}{d e-c f}\right )}{2 d^2 f (n+1)}+\frac {b (c+d x)^{n+1} (e+f x)^{1-n}}{2 d f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(c + d*x)^(1 + n)*(e + f*x)^(1 - n))/(2*d*f) + ((2*a*d*f - b*c*f*(1 - n) - b*d*e*(1 + 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))])/(2*d^2*f*(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 81

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

Rubi steps

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

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Mathematica [A]
time = 0.11, size = 109, normalized size = 0.81 \begin {gather*} \frac {(c+d x)^{1+n} (e+f x)^{-n} \left (b d (e+f x)-\frac {(-2 a d f-b c f (-1+n)+b d e (1+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 )}{1+n}\right )}{2 d^2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \left (b x +a \right ) \left (d x +c \right )^{n} \left (f x +e \right )^{-n}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

integral((b*x + a)*(d*x + c)^n/(f*x + e)^n, x)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: HeuristicGCDFailed} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: HeuristicGCDFailed >> no luck

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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)^n/((f*x+e)^n),x, algorithm="giac")

[Out]

integrate((b*x + a)*(d*x + c)^n/(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}{{\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)/(e + f*x)^n,x)

[Out]

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

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