\(\int (a^2+b^2 x^{-\frac {1}{1+p}}+2 a b x^{-\frac {1}{2 (1+p)}})^p \, dx\) [545]

Optimal result

Integrand size = 34, antiderivative size = 130 \[ \int \left (a^2+b^2 x^{-\frac {1}{1+p}}+2 a b x^{-\frac {1}{2 (1+p)}}\right )^p \, dx=\frac {2 (1+p) x \left (a+b x^{-\frac {1}{2 (1+p)}}\right ) \left (a^2+b^2 x^{-\frac {1}{1+p}}+2 a b x^{-\frac {1}{2 (1+p)}}\right )^p}{a (1+2 p)}-\frac {x \left (a+b x^{-\frac {1}{2 (1+p)}}\right )^2 \left (a^2+b^2 x^{-\frac {1}{1+p}}+2 a b x^{-\frac {1}{2 (1+p)}}\right )^p}{a^2 (1+2 p)} \]

[Out]

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)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {1357, 198, 197} \[ \int \left (a^2+b^2 x^{-\frac {1}{1+p}}+2 a b x^{-\frac {1}{2 (1+p)}}\right )^p \, dx=\frac {2 (p+1) x \left (a+b x^{-\frac {1}{2 (p+1)}}\right ) \left (a^2+2 a b x^{-\frac {1}{2 (p+1)}}+b^2 x^{-\frac {1}{p+1}}\right )^p}{a (2 p+1)}-\frac {x \left (a+b x^{-\frac {1}{2 (p+1)}}\right )^2 \left (a^2+2 a b x^{-\frac {1}{2 (p+1)}}+b^2 x^{-\frac {1}{p+1}}\right )^p}{a^2 (2 p+1)} \]

[In]

Int[(a^2 + b^2/x^(1 + p)^(-1) + (2*a*b)/x^(1/(2*(1 + p))))^p,x]

[Out]

(2*(1 + p)*x*(a + b/x^(1/(2*(1 + p))))*(a^2 + b^2/x^(1 + p)^(-1) + (2*a*b)/x^(1/(2*(1 + p))))^p)/(a*(1 + 2*p))
- (x*(a + b/x^(1/(2*(1 + p))))^2*(a^2 + b^2/x^(1 + p)^(-1) + (2*a*b)/x^(1/(2*(1 + p))))^p)/(a^2*(1 + 2*p))

Rule 197

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]

Rule 198

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(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]

Rule 1357

Int[((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^(2*n))^p/(b + 2*c*x
^n)^(2*p), Int[(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0]

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.62 \[ \int \left (a^2+b^2 x^{-\frac {1}{1+p}}+2 a b x^{-\frac {1}{2 (1+p)}}\right )^p \, dx=\frac {x^{\frac {p}{1+p}} \left (b+a x^{\frac {1}{2+2 p}}\right ) \left (x^{-\frac {1}{1+p}} \left (b+a x^{\frac {1}{2+2 p}}\right )^2\right )^p \left (-b+a (1+2 p) x^{\frac {1}{2+2 p}}\right )}{a^2 (1+2 p)} \]

[In]

Integrate[(a^2 + b^2/x^(1 + p)^(-1) + (2*a*b)/x^(1/(2*(1 + p))))^p,x]

[Out]

(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))

Maple [F]

\[\int \left (a^{2}+b^{2} x^{-\frac {1}{1+p}}+2 a b \,x^{-\frac {1}{2 \left (1+p \right )}}\right )^{p}d x\]

[In]

int((a^2+b^2/(x^(1/(1+p)))+2*a*b/(x^(1/2/(1+p))))^p,x)

[Out]

int((a^2+b^2/(x^(1/(1+p)))+2*a*b/(x^(1/2/(1+p))))^p,x)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.79 \[ \int \left (a^2+b^2 x^{-\frac {1}{1+p}}+2 a b x^{-\frac {1}{2 (1+p)}}\right )^p \, dx=\frac {{\left (2 \, a b p x x^{\frac {1}{2 \, {\left (p + 1\right )}}} - b^{2} x + {\left (2 \, a^{2} p + a^{2}\right )} x x^{\left (\frac {1}{p + 1}\right )}\right )} \left (\frac {2 \, a b x^{\frac {1}{2 \, {\left (p + 1\right )}}} + a^{2} x^{\left (\frac {1}{p + 1}\right )} + b^{2}}{x^{\left (\frac {1}{p + 1}\right )}}\right )^{p}}{{\left (2 \, a^{2} p + a^{2}\right )} x^{\left (\frac {1}{p + 1}\right )}} \]

[In]

integrate((a^2+b^2/(x^(1/(1+p)))+2*a*b/(x^(1/2/(1+p))))^p,x, algorithm="fricas")

[Out]

(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)))

Sympy [F(-1)]

Timed out. \[ \int \left (a^2+b^2 x^{-\frac {1}{1+p}}+2 a b x^{-\frac {1}{2 (1+p)}}\right )^p \, dx=\text {Timed out} \]

[In]

integrate((a**2+b**2/(x**(1/(1+p)))+2*a*b/(x**(1/2/(1+p))))**p,x)

[Out]

Timed out

Maxima [F]

\[ \int \left (a^2+b^2 x^{-\frac {1}{1+p}}+2 a b x^{-\frac {1}{2 (1+p)}}\right )^p \, dx=\int { {\left (a^{2} + \frac {2 \, a b}{x^{\frac {1}{2 \, {\left (p + 1\right )}}}} + \frac {b^{2}}{x^{\left (\frac {1}{p + 1}\right )}}\right )}^{p} \,d x } \]

[In]

integrate((a^2+b^2/(x^(1/(1+p)))+2*a*b/(x^(1/2/(1+p))))^p,x, algorithm="maxima")

[Out]

integrate((a^2 + 2*a*b/x^(1/2/(p + 1)) + b^2/x^(1/(p + 1)))^p, x)

Giac [F]

\[ \int \left (a^2+b^2 x^{-\frac {1}{1+p}}+2 a b x^{-\frac {1}{2 (1+p)}}\right )^p \, dx=\int { {\left (a^{2} + \frac {2 \, a b}{x^{\frac {1}{2 \, {\left (p + 1\right )}}}} + \frac {b^{2}}{x^{\left (\frac {1}{p + 1}\right )}}\right )}^{p} \,d x } \]

[In]

integrate((a^2+b^2/(x^(1/(1+p)))+2*a*b/(x^(1/2/(1+p))))^p,x, algorithm="giac")

[Out]

integrate((a^2 + 2*a*b/x^(1/2/(p + 1)) + b^2/x^(1/(p + 1)))^p, x)

Mupad [F(-1)]

Timed out. \[ \int \left (a^2+b^2 x^{-\frac {1}{1+p}}+2 a b x^{-\frac {1}{2 (1+p)}}\right )^p \, dx=\int {\left (\frac {b^2}{x^{\frac {1}{p+1}}}+a^2+\frac {2\,a\,b}{x^{\frac {1}{2\,\left (p+1\right )}}}\right )}^p \,d x \]

[In]

int((b^2/x^(1/(p + 1)) + a^2 + (2*a*b)/x^(1/(2*(p + 1))))^p,x)

[Out]

int((b^2/x^(1/(p + 1)) + a^2 + (2*a*b)/x^(1/(2*(p + 1))))^p, x)