3.2730 \(\int (a+b x^n)^{-\frac{1+4 n}{n}} \, dx\)

Optimal. Leaf size=147 \[ \frac{3 n x \left (a+b x^n\right )^{-\frac{1}{n}-2}}{a^2 \left (6 n^2+5 n+1\right )}+\frac{6 n^3 x \left (a+b x^n\right )^{-1/n}}{a^4 (n+1) (2 n+1) (3 n+1)}+\frac{6 n^2 x \left (a+b x^n\right )^{-\frac{n+1}{n}}}{a^3 (n+1) (2 n+1) (3 n+1)}+\frac{x \left (a+b x^n\right )^{-\frac{1}{n}-3}}{a (3 n+1)} \]

[Out]

(x*(a + b*x^n)^(-3 - n^(-1)))/(a*(1 + 3*n)) + (3*n*x*(a + b*x^n)^(-2 - n^(-1)))/(a^2*(1 + 5*n + 6*n^2)) + (6*n
^3*x)/(a^4*(1 + n)*(1 + 2*n)*(1 + 3*n)*(a + b*x^n)^n^(-1)) + (6*n^2*x)/(a^3*(1 + n)*(1 + 2*n)*(1 + 3*n)*(a + b
*x^n)^((1 + n)/n))

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Rubi [A]  time = 0.10286, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {192, 191} \[ \frac{3 n x \left (a+b x^n\right )^{-\frac{1}{n}-2}}{a^2 \left (6 n^2+5 n+1\right )}+\frac{6 n^3 x \left (a+b x^n\right )^{-1/n}}{a^4 (n+1) (2 n+1) (3 n+1)}+\frac{6 n^2 x \left (a+b x^n\right )^{-\frac{n+1}{n}}}{a^3 (n+1) (2 n+1) (3 n+1)}+\frac{x \left (a+b x^n\right )^{-\frac{1}{n}-3}}{a (3 n+1)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^n)^(-((1 + 4*n)/n)),x]

[Out]

(x*(a + b*x^n)^(-3 - n^(-1)))/(a*(1 + 3*n)) + (3*n*x*(a + b*x^n)^(-2 - n^(-1)))/(a^2*(1 + 5*n + 6*n^2)) + (6*n
^3*x)/(a^4*(1 + n)*(1 + 2*n)*(1 + 3*n)*(a + b*x^n)^n^(-1)) + (6*n^2*x)/(a^3*(1 + n)*(1 + 2*n)*(1 + 3*n)*(a + b
*x^n)^((1 + n)/n))

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+b x^n\right )^{-\frac{1+4 n}{n}} \, dx &=\frac{x \left (a+b x^n\right )^{-3-\frac{1}{n}}}{a (1+3 n)}+\frac{(3 n) \int \left (a+b x^n\right )^{1-\frac{1+4 n}{n}} \, dx}{a (1+3 n)}\\ &=\frac{x \left (a+b x^n\right )^{-3-\frac{1}{n}}}{a (1+3 n)}+\frac{3 n x \left (a+b x^n\right )^{-2-\frac{1}{n}}}{a^2 \left (1+5 n+6 n^2\right )}+\frac{\left (6 n^2\right ) \int \left (a+b x^n\right )^{2-\frac{1+4 n}{n}} \, dx}{a^2 \left (1+5 n+6 n^2\right )}\\ &=\frac{x \left (a+b x^n\right )^{-3-\frac{1}{n}}}{a (1+3 n)}+\frac{3 n x \left (a+b x^n\right )^{-2-\frac{1}{n}}}{a^2 \left (1+5 n+6 n^2\right )}+\frac{6 n^2 x \left (a+b x^n\right )^{-\frac{1+n}{n}}}{a^3 (1+n) \left (1+5 n+6 n^2\right )}+\frac{\left (6 n^3\right ) \int \left (a+b x^n\right )^{3-\frac{1+4 n}{n}} \, dx}{a^3 (1+n) \left (1+5 n+6 n^2\right )}\\ &=\frac{x \left (a+b x^n\right )^{-3-\frac{1}{n}}}{a (1+3 n)}+\frac{3 n x \left (a+b x^n\right )^{-2-\frac{1}{n}}}{a^2 \left (1+5 n+6 n^2\right )}+\frac{6 n^3 x \left (a+b x^n\right )^{-1/n}}{a^4 (1+n) \left (1+5 n+6 n^2\right )}+\frac{6 n^2 x \left (a+b x^n\right )^{-\frac{1+n}{n}}}{a^3 (1+n) \left (1+5 n+6 n^2\right )}\\ \end{align*}

Mathematica [C]  time = 0.0341706, size = 55, normalized size = 0.37 \[ \frac{x \left (a+b x^n\right )^{-1/n} \left (\frac{b x^n}{a}+1\right )^{\frac{1}{n}} \, _2F_1\left (4+\frac{1}{n},\frac{1}{n};1+\frac{1}{n};-\frac{b x^n}{a}\right )}{a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^n)^(-((1 + 4*n)/n)),x]

[Out]

(x*(1 + (b*x^n)/a)^n^(-1)*Hypergeometric2F1[4 + n^(-1), n^(-1), 1 + n^(-1), -((b*x^n)/a)])/(a^4*(a + b*x^n)^n^
(-1))

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Maple [F]  time = 0.117, size = 0, normalized size = 0. \begin{align*} \int \left ( \left ( a+b{x}^{n} \right ) ^{{\frac{1+4\,n}{n}}} \right ) ^{-1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a+b*x^n)^((1+4*n)/n)),x)

[Out]

int(1/((a+b*x^n)^((1+4*n)/n)),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{n} + a\right )}^{\frac{4 \, n + 1}{n}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a+b*x^n)^((1+4*n)/n)),x, algorithm="maxima")

[Out]

integrate(1/((b*x^n + a)^((4*n + 1)/n)), x)

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Fricas [A]  time = 1.38964, size = 397, normalized size = 2.7 \begin{align*} \frac{6 \, b^{4} n^{3} x x^{4 \, n} + 6 \,{\left (4 \, a b^{3} n^{3} + a b^{3} n^{2}\right )} x x^{3 \, n} + 3 \,{\left (12 \, a^{2} b^{2} n^{3} + 7 \, a^{2} b^{2} n^{2} + a^{2} b^{2} n\right )} x x^{2 \, n} +{\left (24 \, a^{3} b n^{3} + 26 \, a^{3} b n^{2} + 9 \, a^{3} b n + a^{3} b\right )} x x^{n} +{\left (6 \, a^{4} n^{3} + 11 \, a^{4} n^{2} + 6 \, a^{4} n + a^{4}\right )} x}{{\left (6 \, a^{4} n^{3} + 11 \, a^{4} n^{2} + 6 \, a^{4} n + a^{4}\right )}{\left (b x^{n} + a\right )}^{\frac{4 \, n + 1}{n}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a+b*x^n)^((1+4*n)/n)),x, algorithm="fricas")

[Out]

(6*b^4*n^3*x*x^(4*n) + 6*(4*a*b^3*n^3 + a*b^3*n^2)*x*x^(3*n) + 3*(12*a^2*b^2*n^3 + 7*a^2*b^2*n^2 + a^2*b^2*n)*
x*x^(2*n) + (24*a^3*b*n^3 + 26*a^3*b*n^2 + 9*a^3*b*n + a^3*b)*x*x^n + (6*a^4*n^3 + 11*a^4*n^2 + 6*a^4*n + a^4)
*x)/((6*a^4*n^3 + 11*a^4*n^2 + 6*a^4*n + a^4)*(b*x^n + a)^((4*n + 1)/n))

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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(1/((a+b*x**n)**((1+4*n)/n)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{n} + a\right )}^{\frac{4 \, n + 1}{n}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a+b*x^n)^((1+4*n)/n)),x, algorithm="giac")

[Out]

integrate(1/((b*x^n + a)^((4*n + 1)/n)), x)