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

Optimal. Leaf size=130 \[ \frac{2 (p+1) x \left (a+b x^{-\frac{1}{2 (p+1)}}\right ) \left (a^2+2 a b x^{-\frac{1}{2 (p+1)}}+b^2 x^{-\frac{1}{p+1}}\right )^p}{a (2 p+1)}-\frac{x \left (a+b x^{-\frac{1}{2 (p+1)}}\right )^2 \left (a^2+2 a b x^{-\frac{1}{2 (p+1)}}+b^2 x^{-\frac{1}{p+1}}\right )^p}{a^2 (2 p+1)} \]

[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))

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Rubi [A]  time = 0.0552074, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.088, Rules used = {1343, 192, 191} \[ \frac{2 (p+1) x \left (a+b x^{-\frac{1}{2 (p+1)}}\right ) \left (a^2+2 a b x^{-\frac{1}{2 (p+1)}}+b^2 x^{-\frac{1}{p+1}}\right )^p}{a (2 p+1)}-\frac{x \left (a+b x^{-\frac{1}{2 (p+1)}}\right )^2 \left (a^2+2 a b x^{-\frac{1}{2 (p+1)}}+b^2 x^{-\frac{1}{p+1}}\right )^p}{a^2 (2 p+1)} \]

Antiderivative was successfully verified.

[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 1343

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]

Rule 192

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 191

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]

Rubi steps

\begin{align*} \int \left (a^2+b^2 x^{-\frac{1}{1+p}}+2 a b x^{-\frac{1}{2 (1+p)}}\right )^p \, dx &=\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]  time = 0.0647322, size = 80, normalized size = 0.62 \[ \frac{x^{\frac{p}{p+1}} \left (a x^{\frac{1}{2 p+2}}+b\right ) \left (x^{-\frac{1}{p+1}} \left (a x^{\frac{1}{2 p+2}}+b\right )^2\right )^p \left (a (2 p+1) x^{\frac{1}{2 p+2}}-b\right )}{a^2 (2 p+1)} \]

Antiderivative was successfully verified.

[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))

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Maple [F]  time = 0.241, size = 0, normalized size = 0. \begin{align*} \int \left ({a}^{2}+{\frac{{b}^{2}}{{x}^{ \left ( 1+p \right ) ^{-1}}}}+2\,{ab \left ({x}^{1/2\, \left ( 1+p \right ) ^{-1}} \right ) ^{-1}} \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.69303, size = 231, normalized size = 1.78 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)