\(\int \frac {(a-b x)^{-n} (a+b x)^{1+n}}{x^4} \, dx\) [1009]

Optimal result

Integrand size = 23, antiderivative size = 101 \[ \int \frac {(a-b x)^{-n} (a+b x)^{1+n}}{x^4} \, dx=-\frac {(a-b x)^{1-n} (a+b x)^{2+n}}{3 a^2 x^3}-\frac {4 b^3 (1+2 n) (a-b x)^{1-n} (a+b x)^{-1+n} \operatorname {Hypergeometric2F1}\left (3,1-n,2-n,\frac {a-b x}{a+b x}\right )}{3 a^2 (1-n)} \]

[Out]

-1/3*(-b*x+a)^(1-n)*(b*x+a)^(2+n)/a^2/x^3-4/3*b^3*(1+2*n)*(-b*x+a)^(1-n)*(b*x+a)^(-1+n)*hypergeom([3, 1-n],[2-
n],(-b*x+a)/(b*x+a))/a^2/(1-n)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {98, 133} \[ \int \frac {(a-b x)^{-n} (a+b x)^{1+n}}{x^4} \, dx=-\frac {4 b^3 (2 n+1) (a+b x)^{n-1} (a-b x)^{1-n} \operatorname {Hypergeometric2F1}\left (3,1-n,2-n,\frac {a-b x}{a+b x}\right )}{3 a^2 (1-n)}-\frac {(a+b x)^{n+2} (a-b x)^{1-n}}{3 a^2 x^3} \]

[In]

Int[(a + b*x)^(1 + n)/(x^4*(a - b*x)^n),x]

[Out]

-1/3*((a - b*x)^(1 - n)*(a + b*x)^(2 + n))/(a^2*x^3) - (4*b^3*(1 + 2*n)*(a - b*x)^(1 - n)*(a + b*x)^(-1 + n)*H
ypergeometric2F1[3, 1 - n, 2 - n, (a - b*x)/(a + b*x)])/(3*a^2*(1 - n))

Rule 98

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 133

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)/((m + 1)*(b*e - a*f)^(n + 1)*(e + f*x)^(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2,
(-(d*e - c*f))*((a + b*x)/((b*c - a*d)*(e + f*x)))], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p
+ 2, 0] && ILtQ[n, 0] && (SumSimplerQ[m, 1] ||  !SumSimplerQ[p, 1]) &&  !ILtQ[m, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {(a-b x)^{1-n} (a+b x)^{2+n}}{3 a^2 x^3}+\frac {(b (1+2 n)) \int \frac {(a-b x)^{-n} (a+b x)^{1+n}}{x^3} \, dx}{3 a} \\ & = -\frac {(a-b x)^{1-n} (a+b x)^{2+n}}{3 a^2 x^3}-\frac {4 b^3 (1+2 n) (a-b x)^{1-n} (a+b x)^{-1+n} \, _2F_1\left (3,1-n;2-n;\frac {a-b x}{a+b x}\right )}{3 a^2 (1-n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.87 \[ \int \frac {(a-b x)^{-n} (a+b x)^{1+n}}{x^4} \, dx=\frac {(a-b x)^{1-n} (a+b x)^{-1+n} \left (-\left ((-1+n) (a+b x)^3\right )+4 b^3 (1+2 n) x^3 \operatorname {Hypergeometric2F1}\left (3,1-n,2-n,\frac {a-b x}{a+b x}\right )\right )}{3 a^2 (-1+n) x^3} \]

[In]

Integrate[(a + b*x)^(1 + n)/(x^4*(a - b*x)^n),x]

[Out]

((a - b*x)^(1 - n)*(a + b*x)^(-1 + n)*(-((-1 + n)*(a + b*x)^3) + 4*b^3*(1 + 2*n)*x^3*Hypergeometric2F1[3, 1 -
n, 2 - n, (a - b*x)/(a + b*x)]))/(3*a^2*(-1 + n)*x^3)

Maple [F]

\[\int \frac {\left (b x +a \right )^{1+n} \left (-b x +a \right )^{-n}}{x^{4}}d x\]

[In]

int((b*x+a)^(1+n)/x^4/((-b*x+a)^n),x)

[Out]

int((b*x+a)^(1+n)/x^4/((-b*x+a)^n),x)

Fricas [F]

\[ \int \frac {(a-b x)^{-n} (a+b x)^{1+n}}{x^4} \, dx=\int { \frac {{\left (b x + a\right )}^{n + 1}}{{\left (-b x + a\right )}^{n} x^{4}} \,d x } \]

[In]

integrate((b*x+a)^(1+n)/x^4/((-b*x+a)^n),x, algorithm="fricas")

[Out]

integral((b*x + a)^(n + 1)/((-b*x + a)^n*x^4), x)

Sympy [F]

\[ \int \frac {(a-b x)^{-n} (a+b x)^{1+n}}{x^4} \, dx=\int \frac {\left (a - b x\right )^{- n} \left (a + b x\right )^{n + 1}}{x^{4}}\, dx \]

[In]

integrate((b*x+a)**(1+n)/x**4/((-b*x+a)**n),x)

[Out]

Integral((a + b*x)**(n + 1)/(x**4*(a - b*x)**n), x)

Maxima [F]

\[ \int \frac {(a-b x)^{-n} (a+b x)^{1+n}}{x^4} \, dx=\int { \frac {{\left (b x + a\right )}^{n + 1}}{{\left (-b x + a\right )}^{n} x^{4}} \,d x } \]

[In]

integrate((b*x+a)^(1+n)/x^4/((-b*x+a)^n),x, algorithm="maxima")

[Out]

integrate((b*x + a)^(n + 1)/((-b*x + a)^n*x^4), x)

Giac [F]

\[ \int \frac {(a-b x)^{-n} (a+b x)^{1+n}}{x^4} \, dx=\int { \frac {{\left (b x + a\right )}^{n + 1}}{{\left (-b x + a\right )}^{n} x^{4}} \,d x } \]

[In]

integrate((b*x+a)^(1+n)/x^4/((-b*x+a)^n),x, algorithm="giac")

[Out]

integrate((b*x + a)^(n + 1)/((-b*x + a)^n*x^4), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a-b x)^{-n} (a+b x)^{1+n}}{x^4} \, dx=\int \frac {{\left (a+b\,x\right )}^{n+1}}{x^4\,{\left (a-b\,x\right )}^n} \,d x \]

[In]

int((a + b*x)^(n + 1)/(x^4*(a - b*x)^n),x)

[Out]

int((a + b*x)^(n + 1)/(x^4*(a - b*x)^n), x)