Integrand size = 34, antiderivative size = 130 \[ \int \left (a^2+b^2 x^{-\frac {1}{1+p}}+2 a b x^{-\frac {1}{2 (1+p)}}\right )^p \, dx=\frac {2 (1+p) x \left (a+b x^{-\frac {1}{2 (1+p)}}\right ) \left (a^2+b^2 x^{-\frac {1}{1+p}}+2 a b x^{-\frac {1}{2 (1+p)}}\right )^p}{a (1+2 p)}-\frac {x \left (a+b x^{-\frac {1}{2 (1+p)}}\right )^2 \left (a^2+b^2 x^{-\frac {1}{1+p}}+2 a b x^{-\frac {1}{2 (1+p)}}\right )^p}{a^2 (1+2 p)} \]
2*(p+1)*x*(a+b/(x^(1/2/(p+1))))*(a^2+b^2/(x^(1/(p+1)))+2*a*b/(x^(1/2/(p+1) )))^p/a/(1+2*p)-x*(a+b/(x^(1/2/(p+1))))^2*(a^2+b^2/(x^(1/(p+1)))+2*a*b/(x^ (1/2/(p+1))))^p/a^2/(1+2*p)
Time = 0.14 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.62 \[ \int \left (a^2+b^2 x^{-\frac {1}{1+p}}+2 a b x^{-\frac {1}{2 (1+p)}}\right )^p \, dx=\frac {x^{\frac {p}{1+p}} \left (b+a x^{\frac {1}{2+2 p}}\right ) \left (x^{-\frac {1}{1+p}} \left (b+a x^{\frac {1}{2+2 p}}\right )^2\right )^p \left (-b+a (1+2 p) x^{\frac {1}{2+2 p}}\right )}{a^2 (1+2 p)} \]
(x^(p/(1 + p))*(b + a*x^(2 + 2*p)^(-1))*((b + a*x^(2 + 2*p)^(-1))^2/x^(1 + p)^(-1))^p*(-b + a*(1 + 2*p)*x^(2 + 2*p)^(-1)))/(a^2*(1 + 2*p))
Time = 0.26 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {1385, 777, 746}
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.
\(\displaystyle \int \left (a^2+2 a b x^{-\frac {1}{2 (p+1)}}+b^2 x^{-\frac {1}{p+1}}\right )^p \, dx\) |
\(\Big \downarrow \) 1385 |
\(\displaystyle \left (\frac {b x^{-\frac {1}{2 (p+1)}}}{a}+1\right )^{-2 p} \left (a^2+2 a b x^{-\frac {1}{2 (p+1)}}+b^2 x^{-\frac {1}{p+1}}\right )^p \int \left (\frac {b x^{-\frac {1}{2 (p+1)}}}{a}+1\right )^{2 p}dx\) |
\(\Big \downarrow \) 777 |
\(\displaystyle \left (\frac {b x^{-\frac {1}{2 (p+1)}}}{a}+1\right )^{-2 p} \left (a^2+2 a b x^{-\frac {1}{2 (p+1)}}+b^2 x^{-\frac {1}{p+1}}\right )^p \left (\frac {2 (p+1) x \left (\frac {b x^{-\frac {1}{2 (p+1)}}}{a}+1\right )^{2 p+1}}{2 p+1}-\frac {\int \left (\frac {b x^{-\frac {1}{2 (p+1)}}}{a}+1\right )^{2 p+1}dx}{2 p+1}\right )\) |
\(\Big \downarrow \) 746 |
\(\displaystyle \left (\frac {b x^{-\frac {1}{2 (p+1)}}}{a}+1\right )^{-2 p} \left (a^2+2 a b x^{-\frac {1}{2 (p+1)}}+b^2 x^{-\frac {1}{p+1}}\right )^p \left (\frac {2 (p+1) x \left (\frac {b x^{-\frac {1}{2 (p+1)}}}{a}+1\right )^{2 p+1}}{2 p+1}-\frac {x \left (\frac {b x^{-\frac {1}{2 (p+1)}}}{a}+1\right )^{2 (p+1)}}{2 p+1}\right )\) |
((a^2 + b^2/x^(1 + p)^(-1) + (2*a*b)/x^(1/(2*(1 + p))))^p*(-((x*(1 + b/(a* x^(1/(2*(1 + p)))))^(2*(1 + p)))/(1 + 2*p)) + (2*(1 + p)*x*(1 + b/(a*x^(1/ (2*(1 + p)))))^(1 + 2*p))/(1 + 2*p)))/(1 + b/(a*x^(1/(2*(1 + p)))))^(2*p)
3.6.45.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1) /a), x] /; FreeQ[{a, b, n, p}, x] && EqQ[1/n + p + 1, 0]
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Simp[(n*(p + 1) + 1)/(a*n*(p + 1)) Int[(a + b*x^ n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1], 0] && NeQ[p, -1]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> S imp[a^IntPart[p]*((a + b*x^n + c*x^(2*n))^FracPart[p]/(1 + 2*c*(x^n/b))^(2* FracPart[p])) Int[u*(1 + 2*c*(x^n/b))^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[2*p] && NeQ[u, x^(n - 1)] && NeQ[u, x^(2*n - 1)]
\[\int \left (a^{2}+b^{2} x^{-\frac {1}{1+p}}+2 a b \,x^{-\frac {1}{2 \left (1+p \right )}}\right )^{p}d x\]
Time = 0.28 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.79 \[ \int \left (a^2+b^2 x^{-\frac {1}{1+p}}+2 a b x^{-\frac {1}{2 (1+p)}}\right )^p \, dx=\frac {{\left (2 \, a b p x x^{\frac {1}{2 \, {\left (p + 1\right )}}} - b^{2} x + {\left (2 \, a^{2} p + a^{2}\right )} x x^{\left (\frac {1}{p + 1}\right )}\right )} \left (\frac {2 \, a b x^{\frac {1}{2 \, {\left (p + 1\right )}}} + a^{2} x^{\left (\frac {1}{p + 1}\right )} + b^{2}}{x^{\left (\frac {1}{p + 1}\right )}}\right )^{p}}{{\left (2 \, a^{2} p + a^{2}\right )} x^{\left (\frac {1}{p + 1}\right )}} \]
(2*a*b*p*x*x^(1/2/(p + 1)) - b^2*x + (2*a^2*p + a^2)*x*x^(1/(p + 1)))*((2* a*b*x^(1/2/(p + 1)) + a^2*x^(1/(p + 1)) + b^2)/x^(1/(p + 1)))^p/((2*a^2*p + a^2)*x^(1/(p + 1)))
Timed out. \[ \int \left (a^2+b^2 x^{-\frac {1}{1+p}}+2 a b x^{-\frac {1}{2 (1+p)}}\right )^p \, dx=\text {Timed out} \]
\[ \int \left (a^2+b^2 x^{-\frac {1}{1+p}}+2 a b x^{-\frac {1}{2 (1+p)}}\right )^p \, dx=\int { {\left (a^{2} + \frac {2 \, a b}{x^{\frac {1}{2 \, {\left (p + 1\right )}}}} + \frac {b^{2}}{x^{\left (\frac {1}{p + 1}\right )}}\right )}^{p} \,d x } \]
\[ \int \left (a^2+b^2 x^{-\frac {1}{1+p}}+2 a b x^{-\frac {1}{2 (1+p)}}\right )^p \, dx=\int { {\left (a^{2} + \frac {2 \, a b}{x^{\frac {1}{2 \, {\left (p + 1\right )}}}} + \frac {b^{2}}{x^{\left (\frac {1}{p + 1}\right )}}\right )}^{p} \,d x } \]
Timed out. \[ \int \left (a^2+b^2 x^{-\frac {1}{1+p}}+2 a b x^{-\frac {1}{2 (1+p)}}\right )^p \, dx=\int {\left (\frac {b^2}{x^{\frac {1}{p+1}}}+a^2+\frac {2\,a\,b}{x^{\frac {1}{2\,\left (p+1\right )}}}\right )}^p \,d x \]