3.6.39 \(\int \frac {1}{(a^2+2 a b x^n+b^2 x^{2 n})^{3/2}} \, dx\) [539]

Optimal. Leaf size=57 \[ \frac {x \left (a+b x^n\right )^3 \, _2F_1\left (3,\frac {1}{n};1+\frac {1}{n};-\frac {b x^n}{a}\right )}{a^3 \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1357, 251} \begin {gather*} \frac {x \left (a+b x^n\right )^3 \, _2F_1\left (3,\frac {1}{n};1+\frac {1}{n};-\frac {b x^n}{a}\right )}{a^3 \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a + b*x^n)^3*Hypergeometric2F1[3, n^(-1), 1 + n^(-1), -((b*x^n)/a)])/(a^3*(a^2 + 2*a*b*x^n + b^2*x^(2*n))^
(3/2))

Rule 251

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

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*} \int \frac {1}{\left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}} \, dx &=\frac {\left (2 a b+2 b^2 x^n\right )^3 \int \frac {1}{\left (2 a b+2 b^2 x^n\right )^3} \, dx}{\left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}}\\ &=\frac {x \left (a+b x^n\right )^3 \, _2F_1\left (3,\frac {1}{n};1+\frac {1}{n};-\frac {b x^n}{a}\right )}{a^3 \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 46, normalized size = 0.81 \begin {gather*} \frac {x \left (a+b x^n\right )^3 \, _2F_1\left (3,\frac {1}{n};1+\frac {1}{n};-\frac {b x^n}{a}\right )}{a^3 \left (\left (a+b x^n\right )^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a + b*x^n)^3*Hypergeometric2F1[3, n^(-1), 1 + n^(-1), -((b*x^n)/a)])/(a^3*((a + b*x^n)^2)^(3/2))

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Maple [F]
time = 0.00, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (a^{2}+2 a b \,x^{n}+b^{2} x^{2 n}\right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a^2+2*a*b*x^n+b^2*x^(2*n))^(3/2),x)

[Out]

int(1/(a^2+2*a*b*x^n+b^2*x^(2*n))^(3/2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a^2+2*a*b*x^n+b^2*x^(2*n))^(3/2),x, algorithm="maxima")

[Out]

(2*n^2 - 3*n + 1)*integrate(1/2/(a^2*b*n^2*x^n + a^3*n^2), x) + 1/2*(b*(2*n - 1)*x*x^n + a*(3*n - 1)*x)/(a^2*b
^2*n^2*x^(2*n) + 2*a^3*b*n^2*x^n + a^4*n^2)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a^2+2*a*b*x^n+b^2*x^(2*n))^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(b^2*x^(2*n) + 2*a*b*x^n + a^2)/(b^4*x^(4*n) + 4*a^2*b^2*x^(2*n) + 4*a^3*b*x^n + a^4 + 2*(2*a*b^3
*x^n + a^2*b^2)*x^(2*n)), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a^{2} + 2 a b x^{n} + b^{2} x^{2 n}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a**2+2*a*b*x**n+b**2*x**(2*n))**(3/2),x)

[Out]

Integral((a**2 + 2*a*b*x**n + b**2*x**(2*n))**(-3/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a^2+2*a*b*x^n+b^2*x^(2*n))^(3/2),x, algorithm="giac")

[Out]

integrate((b^2*x^(2*n) + 2*a*b*x^n + a^2)^(-3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{{\left (a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a^2 + b^2*x^(2*n) + 2*a*b*x^n)^(3/2),x)

[Out]

int(1/(a^2 + b^2*x^(2*n) + 2*a*b*x^n)^(3/2), x)

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