Integrand size = 17, antiderivative size = 105 \[ \int x^{-7-2 p} \left (a+b x^2\right )^p \, dx=-\frac {b^2 x^{-2 (1+p)} \left (a+b x^2\right )^{1+p}}{a^3 (1+p) (2+p) (3+p)}+\frac {b x^{-2 (2+p)} \left (a+b x^2\right )^{1+p}}{a^2 (2+p) (3+p)}-\frac {x^{-2 (3+p)} \left (a+b x^2\right )^{1+p}}{2 a (3+p)} \] Output:
-b^2*(b*x^2+a)^(p+1)/a^3/(p+1)/(2+p)/(3+p)/(x^(2*p+2))+b*(b*x^2+a)^(p+1)/a ^2/(2+p)/(3+p)/(x^(4+2*p))-1/2*(b*x^2+a)^(p+1)/a/(3+p)/(x^(6+2*p))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.03 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.59 \[ \int x^{-7-2 p} \left (a+b x^2\right )^p \, dx=-\frac {x^{-2 (3+p)} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-3-p,-p,-2-p,-\frac {b x^2}{a}\right )}{2 (3+p)} \] Input:
Integrate[x^(-7 - 2*p)*(a + b*x^2)^p,x]
Output:
-1/2*((a + b*x^2)^p*Hypergeometric2F1[-3 - p, -p, -2 - p, -((b*x^2)/a)])/( (3 + p)*x^(2*(3 + p))*(1 + (b*x^2)/a)^p)
Time = 0.22 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {245, 245, 242}
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.
\(\displaystyle \int x^{-2 p-7} \left (a+b x^2\right )^p \, dx\) |
\(\Big \downarrow \) 245 |
\(\displaystyle -\frac {2 b \int x^{-2 p-5} \left (b x^2+a\right )^pdx}{a (p+3)}-\frac {x^{-2 (p+3)} \left (a+b x^2\right )^{p+1}}{2 a (p+3)}\) |
\(\Big \downarrow \) 245 |
\(\displaystyle -\frac {2 b \left (-\frac {b \int x^{-2 p-3} \left (b x^2+a\right )^pdx}{a (p+2)}-\frac {x^{-2 (p+2)} \left (a+b x^2\right )^{p+1}}{2 a (p+2)}\right )}{a (p+3)}-\frac {x^{-2 (p+3)} \left (a+b x^2\right )^{p+1}}{2 a (p+3)}\) |
\(\Big \downarrow \) 242 |
\(\displaystyle -\frac {2 b \left (\frac {b x^{-2 (p+1)} \left (a+b x^2\right )^{p+1}}{2 a^2 (p+1) (p+2)}-\frac {x^{-2 (p+2)} \left (a+b x^2\right )^{p+1}}{2 a (p+2)}\right )}{a (p+3)}-\frac {x^{-2 (p+3)} \left (a+b x^2\right )^{p+1}}{2 a (p+3)}\) |
Input:
Int[x^(-7 - 2*p)*(a + b*x^2)^p,x]
Output:
-1/2*(a + b*x^2)^(1 + p)/(a*(3 + p)*x^(2*(3 + p))) - (2*b*((b*(a + b*x^2)^ (1 + p))/(2*a^2*(1 + p)*(2 + p)*x^(2*(1 + p))) - (a + b*x^2)^(1 + p)/(2*a* (2 + p)*x^(2*(2 + p)))))/(a*(3 + p))
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] /; FreeQ[{a, b, c, m, p}, x ] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + 2*(p + 1) + 1)/(a*(m + 1))) Int[x^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, m, p}, x] && ILtQ[Si mplify[(m + 1)/2 + p + 1], 0] && NeQ[m, -1]
Time = 0.44 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.77
method | result | size |
gosper | \(-\frac {x^{-6-2 p} \left (b \,x^{2}+a \right )^{p +1} \left (2 b^{2} x^{4}-2 a b p \,x^{2}+a^{2} p^{2}-2 a b \,x^{2}+3 a^{2} p +2 a^{2}\right )}{2 a^{3} \left (2+p \right ) \left (3+p \right ) \left (p +1\right )}\) | \(81\) |
orering | \(-\frac {\left (b \,x^{2}+a \right ) x \left (2 b^{2} x^{4}-2 a b p \,x^{2}+a^{2} p^{2}-2 a b \,x^{2}+3 a^{2} p +2 a^{2}\right ) x^{-7-2 p} \left (b \,x^{2}+a \right )^{p}}{2 \left (3+p \right ) \left (2+p \right ) \left (p +1\right ) a^{3}}\) | \(87\) |
Input:
int(x^(-7-2*p)*(b*x^2+a)^p,x,method=_RETURNVERBOSE)
Output:
-1/2*x^(-6-2*p)/a^3/(2+p)/(3+p)/(p+1)*(b*x^2+a)^(p+1)*(2*b^2*x^4-2*a*b*p*x ^2+a^2*p^2-2*a*b*x^2+3*a^2*p+2*a^2)
Time = 0.10 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.01 \[ \int x^{-7-2 p} \left (a+b x^2\right )^p \, dx=-\frac {{\left (2 \, b^{3} x^{7} - 2 \, a b^{2} p x^{5} + {\left (a^{2} b p^{2} + a^{2} b p\right )} x^{3} + {\left (a^{3} p^{2} + 3 \, a^{3} p + 2 \, a^{3}\right )} x\right )} {\left (b x^{2} + a\right )}^{p} x^{-2 \, p - 7}}{2 \, {\left (a^{3} p^{3} + 6 \, a^{3} p^{2} + 11 \, a^{3} p + 6 \, a^{3}\right )}} \] Input:
integrate(x^(-7-2*p)*(b*x^2+a)^p,x, algorithm="fricas")
Output:
-1/2*(2*b^3*x^7 - 2*a*b^2*p*x^5 + (a^2*b*p^2 + a^2*b*p)*x^3 + (a^3*p^2 + 3 *a^3*p + 2*a^3)*x)*(b*x^2 + a)^p*x^(-2*p - 7)/(a^3*p^3 + 6*a^3*p^2 + 11*a^ 3*p + 6*a^3)
Leaf count of result is larger than twice the leaf count of optimal. 452 vs. \(2 (90) = 180\).
Time = 9.04 (sec) , antiderivative size = 452, normalized size of antiderivative = 4.30 \[ \int x^{-7-2 p} \left (a+b x^2\right )^p \, dx=\frac {a^{2} a^{p} p^{2} x^{- 2 p - 6} \left (1 + \frac {b x^{2}}{a}\right )^{p + 3} \Gamma \left (- p - 3\right )}{2 a^{2} \Gamma \left (- p\right ) + 4 a b x^{2} \Gamma \left (- p\right ) + 2 b^{2} x^{4} \Gamma \left (- p\right )} + \frac {3 a^{2} a^{p} p x^{- 2 p - 6} \left (1 + \frac {b x^{2}}{a}\right )^{p + 3} \Gamma \left (- p - 3\right )}{2 a^{2} \Gamma \left (- p\right ) + 4 a b x^{2} \Gamma \left (- p\right ) + 2 b^{2} x^{4} \Gamma \left (- p\right )} + \frac {2 a^{2} a^{p} x^{- 2 p - 6} \left (1 + \frac {b x^{2}}{a}\right )^{p + 3} \Gamma \left (- p - 3\right )}{2 a^{2} \Gamma \left (- p\right ) + 4 a b x^{2} \Gamma \left (- p\right ) + 2 b^{2} x^{4} \Gamma \left (- p\right )} - \frac {2 a a^{p} b p x^{2} x^{- 2 p - 6} \left (1 + \frac {b x^{2}}{a}\right )^{p + 3} \Gamma \left (- p - 3\right )}{2 a^{2} \Gamma \left (- p\right ) + 4 a b x^{2} \Gamma \left (- p\right ) + 2 b^{2} x^{4} \Gamma \left (- p\right )} - \frac {2 a a^{p} b x^{2} x^{- 2 p - 6} \left (1 + \frac {b x^{2}}{a}\right )^{p + 3} \Gamma \left (- p - 3\right )}{2 a^{2} \Gamma \left (- p\right ) + 4 a b x^{2} \Gamma \left (- p\right ) + 2 b^{2} x^{4} \Gamma \left (- p\right )} + \frac {2 a^{p} b^{2} x^{4} x^{- 2 p - 6} \left (1 + \frac {b x^{2}}{a}\right )^{p + 3} \Gamma \left (- p - 3\right )}{2 a^{2} \Gamma \left (- p\right ) + 4 a b x^{2} \Gamma \left (- p\right ) + 2 b^{2} x^{4} \Gamma \left (- p\right )} \] Input:
integrate(x**(-7-2*p)*(b*x**2+a)**p,x)
Output:
a**2*a**p*p**2*x**(-2*p - 6)*(1 + b*x**2/a)**(p + 3)*gamma(-p - 3)/(2*a**2 *gamma(-p) + 4*a*b*x**2*gamma(-p) + 2*b**2*x**4*gamma(-p)) + 3*a**2*a**p*p *x**(-2*p - 6)*(1 + b*x**2/a)**(p + 3)*gamma(-p - 3)/(2*a**2*gamma(-p) + 4 *a*b*x**2*gamma(-p) + 2*b**2*x**4*gamma(-p)) + 2*a**2*a**p*x**(-2*p - 6)*( 1 + b*x**2/a)**(p + 3)*gamma(-p - 3)/(2*a**2*gamma(-p) + 4*a*b*x**2*gamma( -p) + 2*b**2*x**4*gamma(-p)) - 2*a*a**p*b*p*x**2*x**(-2*p - 6)*(1 + b*x**2 /a)**(p + 3)*gamma(-p - 3)/(2*a**2*gamma(-p) + 4*a*b*x**2*gamma(-p) + 2*b* *2*x**4*gamma(-p)) - 2*a*a**p*b*x**2*x**(-2*p - 6)*(1 + b*x**2/a)**(p + 3) *gamma(-p - 3)/(2*a**2*gamma(-p) + 4*a*b*x**2*gamma(-p) + 2*b**2*x**4*gamm a(-p)) + 2*a**p*b**2*x**4*x**(-2*p - 6)*(1 + b*x**2/a)**(p + 3)*gamma(-p - 3)/(2*a**2*gamma(-p) + 4*a*b*x**2*gamma(-p) + 2*b**2*x**4*gamma(-p))
Time = 0.04 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.80 \[ \int x^{-7-2 p} \left (a+b x^2\right )^p \, dx=-\frac {{\left (2 \, b^{3} x^{6} - 2 \, a b^{2} p x^{4} + {\left (p^{2} + p\right )} a^{2} b x^{2} + {\left (p^{2} + 3 \, p + 2\right )} a^{3}\right )} e^{\left (p \log \left (b x^{2} + a\right ) - 2 \, p \log \left (x\right )\right )}}{2 \, {\left (p^{3} + 6 \, p^{2} + 11 \, p + 6\right )} a^{3} x^{6}} \] Input:
integrate(x^(-7-2*p)*(b*x^2+a)^p,x, algorithm="maxima")
Output:
-1/2*(2*b^3*x^6 - 2*a*b^2*p*x^4 + (p^2 + p)*a^2*b*x^2 + (p^2 + 3*p + 2)*a^ 3)*e^(p*log(b*x^2 + a) - 2*p*log(x))/((p^3 + 6*p^2 + 11*p + 6)*a^3*x^6)
\[ \int x^{-7-2 p} \left (a+b x^2\right )^p \, dx=\int { {\left (b x^{2} + a\right )}^{p} x^{-2 \, p - 7} \,d x } \] Input:
integrate(x^(-7-2*p)*(b*x^2+a)^p,x, algorithm="giac")
Output:
integrate((b*x^2 + a)^p*x^(-2*p - 7), x)
Time = 0.58 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.47 \[ \int x^{-7-2 p} \left (a+b x^2\right )^p \, dx=-{\left (b\,x^2+a\right )}^p\,\left (\frac {x\,\left (p^2+3\,p+2\right )}{2\,x^{2\,p+7}\,\left (p^3+6\,p^2+11\,p+6\right )}+\frac {b^3\,x^7}{a^3\,x^{2\,p+7}\,\left (p^3+6\,p^2+11\,p+6\right )}-\frac {b^2\,p\,x^5}{a^2\,x^{2\,p+7}\,\left (p^3+6\,p^2+11\,p+6\right )}+\frac {b\,p\,x^3\,\left (p+1\right )}{2\,a\,x^{2\,p+7}\,\left (p^3+6\,p^2+11\,p+6\right )}\right ) \] Input:
int((a + b*x^2)^p/x^(2*p + 7),x)
Output:
-(a + b*x^2)^p*((x*(3*p + p^2 + 2))/(2*x^(2*p + 7)*(11*p + 6*p^2 + p^3 + 6 )) + (b^3*x^7)/(a^3*x^(2*p + 7)*(11*p + 6*p^2 + p^3 + 6)) - (b^2*p*x^5)/(a ^2*x^(2*p + 7)*(11*p + 6*p^2 + p^3 + 6)) + (b*p*x^3*(p + 1))/(2*a*x^(2*p + 7)*(11*p + 6*p^2 + p^3 + 6)))
Time = 0.21 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.94 \[ \int x^{-7-2 p} \left (a+b x^2\right )^p \, dx=\frac {\left (b \,x^{2}+a \right )^{p} \left (-2 b^{3} x^{6}+2 a \,b^{2} p \,x^{4}-a^{2} b \,p^{2} x^{2}-a^{2} b p \,x^{2}-a^{3} p^{2}-3 a^{3} p -2 a^{3}\right )}{2 x^{2 p} a^{3} x^{6} \left (p^{3}+6 p^{2}+11 p +6\right )} \] Input:
int(x^(-7-2*p)*(b*x^2+a)^p,x)
Output:
((a + b*x**2)**p*( - a**3*p**2 - 3*a**3*p - 2*a**3 - a**2*b*p**2*x**2 - a* *2*b*p*x**2 + 2*a*b**2*p*x**4 - 2*b**3*x**6))/(2*x**(2*p)*a**3*x**6*(p**3 + 6*p**2 + 11*p + 6))