3.2707 \(\int (a+b x^n)^{-2-\frac {1}{n}} \, dx\)

Optimal. Leaf size=50 \[ \frac {n x \left (a+b x^n\right )^{-1/n}}{a^2 (n+1)}+\frac {x \left (a+b x^n\right )^{-\frac {1}{n}-1}}{a (n+1)} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {192, 191} \[ \frac {n x \left (a+b x^n\right )^{-1/n}}{a^2 (n+1)}+\frac {x \left (a+b x^n\right )^{-\frac {1}{n}-1}}{a (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^n)^(-2 - n^(-1)),x]

[Out]

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

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]

Rubi steps

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

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Mathematica [C]  time = 0.05, size = 55, normalized size = 1.10 \[ \frac {x \left (a+b x^n\right )^{-1/n} \left (\frac {b x^n}{a}+1\right )^{\frac {1}{n}} \, _2F_1\left (2+\frac {1}{n},\frac {1}{n};1+\frac {1}{n};-\frac {b x^n}{a}\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(1 + (b*x^n)/a)^n^(-1)*Hypergeometric2F1[2 + n^(-1), n^(-1), 1 + n^(-1), -((b*x^n)/a)])/(a^2*(a + b*x^n)^n^
(-1))

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fricas [A]  time = 0.78, size = 68, normalized size = 1.36 \[ \frac {b^{2} n x x^{2 \, n} + {\left (2 \, a b n + a b\right )} x x^{n} + {\left (a^{2} n + a^{2}\right )} x}{{\left (a^{2} n + a^{2}\right )} {\left (b x^{n} + a\right )}^{\frac {2 \, n + 1}{n}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^2*n*x*x^(2*n) + (2*a*b*n + a*b)*x*x^n + (a^2*n + a^2)*x)/((a^2*n + a^2)*(b*x^n + a)^((2*n + 1)/n))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{n} + a\right )}^{-\frac {1}{n} - 2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^n + a)^(-1/n - 2), x)

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maple [F]  time = 0.22, size = 0, normalized size = 0.00 \[ \int \left (b \,x^{n}+a \right )^{-\frac {1}{n}-2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{n} + a\right )}^{-\frac {1}{n} - 2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^n + a)^(-1/n - 2), x)

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mupad [B]  time = 1.44, size = 64, normalized size = 1.28 \[ -\frac {x^{1-2\,n}\,{\left (\frac {a}{b\,x^n}+1\right )}^{1/n}\,{{}}_2{\mathrm {F}}_1\left (2,\frac {1}{n}+2;\ 3;\ -\frac {a}{b\,x^n}\right )}{2\,b^2\,n\,{\left (a+b\,x^n\right )}^{1/n}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*x^n)^(1/n + 2),x)

[Out]

-(x^(1 - 2*n)*(a/(b*x^n) + 1)^(1/n)*hypergeom([2, 1/n + 2], 3, -a/(b*x^n)))/(2*b^2*n*(a + b*x^n)^(1/n))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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