3.974 \(\int \frac{(a+b x)^n (c+d x)^{-n}}{x} \, dx\)

Optimal. Leaf size=108 \[ \frac{(a+b x)^n (c+d x)^{-n} \left (\frac{b (c+d x)}{b c-a d}\right )^n \, _2F_1\left (n,n;n+1;-\frac{d (a+b x)}{b c-a d}\right )}{n}-\frac{(a+b x)^n (c+d x)^{-n} \, _2F_1\left (1,n;n+1;\frac{c (a+b x)}{a (c+d x)}\right )}{n} \]

[Out]

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

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Rubi [A]  time = 0.0492434, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {105, 70, 69, 131} \[ \frac{(a+b x)^n (c+d x)^{-n} \left (\frac{b (c+d x)}{b c-a d}\right )^n \, _2F_1\left (n,n;n+1;-\frac{d (a+b x)}{b c-a d}\right )}{n}-\frac{(a+b x)^n (c+d x)^{-n} \, _2F_1\left (1,n;n+1;\frac{c (a+b x)}{a (c+d x)}\right )}{n} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^n/(x*(c + d*x)^n),x]

[Out]

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

Rule 105

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Dist[b/f, Int[(a
+ b*x)^(m - 1)*(c + d*x)^n, x], x] - Dist[(b*e - a*f)/f, Int[((a + b*x)^(m - 1)*(c + d*x)^n)/(e + f*x), x], x]
/; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[Simplify[m + n + 1], 0] && (GtQ[m, 0] || ( !RationalQ[m] && (Su
mSimplerQ[m, -1] ||  !SumSimplerQ[n, -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]))

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

Mathematica [A]  time = 0.0295555, size = 89, normalized size = 0.82 \[ \frac{(a+b x)^n (c+d x)^{-n} \left (\left (\frac{b (c+d x)}{b c-a d}\right )^n \, _2F_1\left (n,n;n+1;\frac{d (a+b x)}{a d-b c}\right )-\, _2F_1\left (1,n;n+1;\frac{c (a+b x)}{a (c+d x)}\right )\right )}{n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)^n*(-Hypergeometric2F1[1, n, 1 + n, (c*(a + b*x))/(a*(c + d*x))] + ((b*(c + d*x))/(b*c - a*d))^n*Hyp
ergeometric2F1[n, n, 1 + n, (d*(a + b*x))/(-(b*c) + a*d)]))/(n*(c + d*x)^n)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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