Integrand size = 15, antiderivative size = 110 \[ \int x^{-4+p} (a+b x)^{-p} \, dx=-\frac {x^{-3+p} (a+b x)^{1-p}}{a (3-p)}+\frac {2 b x^{-2+p} (a+b x)^{1-p}}{a^2 (2-p) (3-p)}-\frac {2 b^2 x^{-1+p} (a+b x)^{1-p}}{a^3 (1-p) (2-p) (3-p)} \] Output:
-x^(-3+p)*(b*x+a)^(1-p)/a/(3-p)+2*b*x^(-2+p)*(b*x+a)^(1-p)/a^2/(2-p)/(3-p) -2*b^2*x^(-1+p)*(b*x+a)^(1-p)/a^3/(1-p)/(2-p)/(3-p)
Time = 0.04 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.58 \[ \int x^{-4+p} (a+b x)^{-p} \, dx=\frac {x^{-3+p} (a+b x)^{1-p} \left (a^2 \left (2-3 p+p^2\right )+2 a b (-1+p) x+2 b^2 x^2\right )}{a^3 (-3+p) (-2+p) (-1+p)} \] Input:
Integrate[x^(-4 + p)/(a + b*x)^p,x]
Output:
(x^(-3 + p)*(a + b*x)^(1 - p)*(a^2*(2 - 3*p + p^2) + 2*a*b*(-1 + p)*x + 2* b^2*x^2))/(a^3*(-3 + p)*(-2 + p)*(-1 + p))
Time = 0.20 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {55, 55, 48}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^{p-4} (a+b x)^{-p} \, dx\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {2 b \int x^{p-3} (a+b x)^{-p}dx}{a (3-p)}-\frac {x^{p-3} (a+b x)^{1-p}}{a (3-p)}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle -\frac {2 b \left (-\frac {b \int x^{p-2} (a+b x)^{-p}dx}{a (2-p)}-\frac {x^{p-2} (a+b x)^{1-p}}{a (2-p)}\right )}{a (3-p)}-\frac {x^{p-3} (a+b x)^{1-p}}{a (3-p)}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle -\frac {2 b \left (\frac {b x^{p-1} (a+b x)^{1-p}}{a^2 (1-p) (2-p)}-\frac {x^{p-2} (a+b x)^{1-p}}{a (2-p)}\right )}{a (3-p)}-\frac {x^{p-3} (a+b x)^{1-p}}{a (3-p)}\) |
Input:
Int[x^(-4 + p)/(a + b*x)^p,x]
Output:
-((x^(-3 + p)*(a + b*x)^(1 - p))/(a*(3 - p))) - (2*b*(-((x^(-2 + p)*(a + b *x)^(1 - p))/(a*(2 - p))) + (b*x^(-1 + p)*(a + b*x)^(1 - p))/(a^2*(1 - p)* (2 - p))))/(a*(3 - p))
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp [(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(S implify[m + n + 2]/((b*c - a*d)*(m + 1))) Int[(a + b*x)^Simplify[m + 1]*( c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 2], 0] && NeQ[m, -1] && !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] || !SumSimp lerQ[n, 1])
Time = 0.18 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.70
method | result | size |
gosper | \(\frac {x^{-3+p} \left (b x +a \right ) \left (b x +a \right )^{-p} \left (a^{2} p^{2}+2 a b p x +2 b^{2} x^{2}-3 a^{2} p -2 a b x +2 a^{2}\right )}{a^{3} \left (-3+p \right ) \left (-2+p \right ) \left (-1+p \right )}\) | \(77\) |
orering | \(\frac {\left (b x +a \right ) x \left (a^{2} p^{2}+2 a b p x +2 b^{2} x^{2}-3 a^{2} p -2 a b x +2 a^{2}\right ) x^{-4+p} \left (b x +a \right )^{-p}}{\left (-3+p \right ) \left (-2+p \right ) \left (-1+p \right ) a^{3}}\) | \(78\) |
Input:
int(x^(-4+p)/((b*x+a)^p),x,method=_RETURNVERBOSE)
Output:
x^(-3+p)/a^3/(-3+p)/(-2+p)/(-1+p)*(b*x+a)/((b*x+a)^p)*(a^2*p^2+2*a*b*p*x+2 *b^2*x^2-3*a^2*p-2*a*b*x+2*a^2)
Time = 0.09 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.95 \[ \int x^{-4+p} (a+b x)^{-p} \, dx=\frac {{\left (2 \, a b^{2} p x^{3} + 2 \, b^{3} x^{4} + {\left (a^{2} b p^{2} - a^{2} b p\right )} x^{2} + {\left (a^{3} p^{2} - 3 \, a^{3} p + 2 \, a^{3}\right )} x\right )} x^{p - 4}}{{\left (a^{3} p^{3} - 6 \, a^{3} p^{2} + 11 \, a^{3} p - 6 \, a^{3}\right )} {\left (b x + a\right )}^{p}} \] Input:
integrate(x^(-4+p)/((b*x+a)^p),x, algorithm="fricas")
Output:
(2*a*b^2*p*x^3 + 2*b^3*x^4 + (a^2*b*p^2 - a^2*b*p)*x^2 + (a^3*p^2 - 3*a^3* p + 2*a^3)*x)*x^(p - 4)/((a^3*p^3 - 6*a^3*p^2 + 11*a^3*p - 6*a^3)*(b*x + a )^p)
Leaf count of result is larger than twice the leaf count of optimal. 388 vs. \(2 (76) = 152\).
Time = 10.74 (sec) , antiderivative size = 388, normalized size of antiderivative = 3.53 \[ \int x^{-4+p} (a+b x)^{-p} \, dx=\frac {a^{2} p^{2} x^{p - 3} \left (1 + \frac {b x}{a}\right )^{3 - p} \Gamma \left (p - 3\right )}{a^{2} a^{p} \Gamma \left (p\right ) + 2 a a^{p} b x \Gamma \left (p\right ) + a^{p} b^{2} x^{2} \Gamma \left (p\right )} - \frac {3 a^{2} p x^{p - 3} \left (1 + \frac {b x}{a}\right )^{3 - p} \Gamma \left (p - 3\right )}{a^{2} a^{p} \Gamma \left (p\right ) + 2 a a^{p} b x \Gamma \left (p\right ) + a^{p} b^{2} x^{2} \Gamma \left (p\right )} + \frac {2 a^{2} x^{p - 3} \left (1 + \frac {b x}{a}\right )^{3 - p} \Gamma \left (p - 3\right )}{a^{2} a^{p} \Gamma \left (p\right ) + 2 a a^{p} b x \Gamma \left (p\right ) + a^{p} b^{2} x^{2} \Gamma \left (p\right )} + \frac {2 a b p x x^{p - 3} \left (1 + \frac {b x}{a}\right )^{3 - p} \Gamma \left (p - 3\right )}{a^{2} a^{p} \Gamma \left (p\right ) + 2 a a^{p} b x \Gamma \left (p\right ) + a^{p} b^{2} x^{2} \Gamma \left (p\right )} - \frac {2 a b x x^{p - 3} \left (1 + \frac {b x}{a}\right )^{3 - p} \Gamma \left (p - 3\right )}{a^{2} a^{p} \Gamma \left (p\right ) + 2 a a^{p} b x \Gamma \left (p\right ) + a^{p} b^{2} x^{2} \Gamma \left (p\right )} + \frac {2 b^{2} x^{2} x^{p - 3} \left (1 + \frac {b x}{a}\right )^{3 - p} \Gamma \left (p - 3\right )}{a^{2} a^{p} \Gamma \left (p\right ) + 2 a a^{p} b x \Gamma \left (p\right ) + a^{p} b^{2} x^{2} \Gamma \left (p\right )} \] Input:
integrate(x**(-4+p)/((b*x+a)**p),x)
Output:
a**2*p**2*x**(p - 3)*(1 + b*x/a)**(3 - p)*gamma(p - 3)/(a**2*a**p*gamma(p) + 2*a*a**p*b*x*gamma(p) + a**p*b**2*x**2*gamma(p)) - 3*a**2*p*x**(p - 3)* (1 + b*x/a)**(3 - p)*gamma(p - 3)/(a**2*a**p*gamma(p) + 2*a*a**p*b*x*gamma (p) + a**p*b**2*x**2*gamma(p)) + 2*a**2*x**(p - 3)*(1 + b*x/a)**(3 - p)*ga mma(p - 3)/(a**2*a**p*gamma(p) + 2*a*a**p*b*x*gamma(p) + a**p*b**2*x**2*ga mma(p)) + 2*a*b*p*x*x**(p - 3)*(1 + b*x/a)**(3 - p)*gamma(p - 3)/(a**2*a** p*gamma(p) + 2*a*a**p*b*x*gamma(p) + a**p*b**2*x**2*gamma(p)) - 2*a*b*x*x* *(p - 3)*(1 + b*x/a)**(3 - p)*gamma(p - 3)/(a**2*a**p*gamma(p) + 2*a*a**p* b*x*gamma(p) + a**p*b**2*x**2*gamma(p)) + 2*b**2*x**2*x**(p - 3)*(1 + b*x/ a)**(3 - p)*gamma(p - 3)/(a**2*a**p*gamma(p) + 2*a*a**p*b*x*gamma(p) + a** p*b**2*x**2*gamma(p))
\[ \int x^{-4+p} (a+b x)^{-p} \, dx=\int { \frac {x^{p - 4}}{{\left (b x + a\right )}^{p}} \,d x } \] Input:
integrate(x^(-4+p)/((b*x+a)^p),x, algorithm="maxima")
Output:
integrate(x^(p - 4)/(b*x + a)^p, x)
\[ \int x^{-4+p} (a+b x)^{-p} \, dx=\int { \frac {x^{p - 4}}{{\left (b x + a\right )}^{p}} \,d x } \] Input:
integrate(x^(-4+p)/((b*x+a)^p),x, algorithm="giac")
Output:
integrate(x^(p - 4)/(b*x + a)^p, x)
Time = 0.33 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.24 \[ \int x^{-4+p} (a+b x)^{-p} \, dx=\frac {\frac {x\,x^{p-4}\,\left (p^2-3\,p+2\right )}{p^3-6\,p^2+11\,p-6}+\frac {2\,b^3\,x^{p-4}\,x^4}{a^3\,\left (p^3-6\,p^2+11\,p-6\right )}+\frac {2\,b^2\,p\,x^{p-4}\,x^3}{a^2\,\left (p^3-6\,p^2+11\,p-6\right )}+\frac {b\,p\,x^{p-4}\,x^2\,\left (p-1\right )}{a\,\left (p^3-6\,p^2+11\,p-6\right )}}{{\left (a+b\,x\right )}^p} \] Input:
int(x^(p - 4)/(a + b*x)^p,x)
Output:
((x*x^(p - 4)*(p^2 - 3*p + 2))/(11*p - 6*p^2 + p^3 - 6) + (2*b^3*x^(p - 4) *x^4)/(a^3*(11*p - 6*p^2 + p^3 - 6)) + (2*b^2*p*x^(p - 4)*x^3)/(a^2*(11*p - 6*p^2 + p^3 - 6)) + (b*p*x^(p - 4)*x^2*(p - 1))/(a*(11*p - 6*p^2 + p^3 - 6)))/(a + b*x)^p
\[ \int x^{-4+p} (a+b x)^{-p} \, dx=\int \frac {x^{p}}{\left (b x +a \right )^{p} x^{4}}d x \] Input:
int(x^(-4+p)/((b*x+a)^p),x)
Output:
int(x**p/((a + b*x)**p*x**4),x)