3.6.14 \(\int \frac {x^{-1+2 n}}{(a^2+2 a b x^n+b^2 x^{2 n})^{7/2}} \, dx\) [514]

Optimal. Leaf size=88 \[ \frac {a}{6 b^2 n \left (a+b x^n\right )^5 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}-\frac {1}{5 b^2 n \left (a+b x^n\right )^4 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {1369, 272, 45} \begin {gather*} \frac {a}{6 b^2 n \left (a+b x^n\right )^5 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}-\frac {1}{5 b^2 n \left (a+b x^n\right )^4 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(-1 + 2*n)/(a^2 + 2*a*b*x^n + b^2*x^(2*n))^(7/2),x]

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1369

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^
(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^n)^(2*p), x], x] /; Fr
eeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps

\begin {align*} \int \frac {x^{-1+2 n}}{\left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{7/2}} \, dx &=\frac {\left (b^6 \left (a b+b^2 x^n\right )\right ) \int \frac {x^{-1+2 n}}{\left (a b+b^2 x^n\right )^7} \, dx}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}\\ &=\frac {\left (b^6 \left (a b+b^2 x^n\right )\right ) \text {Subst}\left (\int \frac {x}{\left (a b+b^2 x\right )^7} \, dx,x,x^n\right )}{n \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}\\ &=\frac {\left (b^6 \left (a b+b^2 x^n\right )\right ) \text {Subst}\left (\int \left (-\frac {a}{b^8 (a+b x)^7}+\frac {1}{b^8 (a+b x)^6}\right ) \, dx,x,x^n\right )}{n \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}\\ &=\frac {a}{6 b^2 n \left (a+b x^n\right )^5 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}-\frac {1}{5 b^2 n \left (a+b x^n\right )^4 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 40, normalized size = 0.45 \begin {gather*} \frac {\left (-a-6 b x^n\right ) \left (a+b x^n\right )}{30 b^2 n \left (\left (a+b x^n\right )^2\right )^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 + 2*n)/(a^2 + 2*a*b*x^n + b^2*x^(2*n))^(7/2),x]

[Out]

((-a - 6*b*x^n)*(a + b*x^n))/(30*b^2*n*((a + b*x^n)^2)^(7/2))

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Maple [A]
time = 0.03, size = 37, normalized size = 0.42

method	result	size
risch	\(-\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, \left (6 b \,x^{n}+a \right )}{30 \left (a +b \,x^{n}\right )^{7} b^{2} n}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(-1+2*n)/(a^2+2*a*b*x^n+b^2*x^(2*n))^(7/2),x,method=_RETURNVERBOSE)

[Out]

-1/30*((a+b*x^n)^2)^(1/2)/(a+b*x^n)^7*(6*b*x^n+a)/b^2/n

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Maxima [A]
time = 0.40, size = 97, normalized size = 1.10 \begin {gather*} -\frac {6 \, b x^{n} + a}{30 \, {\left (b^{8} n x^{6 \, n} + 6 \, a b^{7} n x^{5 \, n} + 15 \, a^{2} b^{6} n x^{4 \, n} + 20 \, a^{3} b^{5} n x^{3 \, n} + 15 \, a^{4} b^{4} n x^{2 \, n} + 6 \, a^{5} b^{3} n x^{n} + a^{6} b^{2} n\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+2*n)/(a^2+2*a*b*x^n+b^2*x^(2*n))^(7/2),x, algorithm="maxima")

[Out]

-1/30*(6*b*x^n + a)/(b^8*n*x^(6*n) + 6*a*b^7*n*x^(5*n) + 15*a^2*b^6*n*x^(4*n) + 20*a^3*b^5*n*x^(3*n) + 15*a^4*
b^4*n*x^(2*n) + 6*a^5*b^3*n*x^n + a^6*b^2*n)

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Fricas [A]
time = 0.37, size = 97, normalized size = 1.10 \begin {gather*} -\frac {6 \, b x^{n} + a}{30 \, {\left (b^{8} n x^{6 \, n} + 6 \, a b^{7} n x^{5 \, n} + 15 \, a^{2} b^{6} n x^{4 \, n} + 20 \, a^{3} b^{5} n x^{3 \, n} + 15 \, a^{4} b^{4} n x^{2 \, n} + 6 \, a^{5} b^{3} n x^{n} + a^{6} b^{2} n\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+2*n)/(a^2+2*a*b*x^n+b^2*x^(2*n))^(7/2),x, algorithm="fricas")

[Out]

-1/30*(6*b*x^n + a)/(b^8*n*x^(6*n) + 6*a*b^7*n*x^(5*n) + 15*a^2*b^6*n*x^(4*n) + 20*a^3*b^5*n*x^(3*n) + 15*a^4*
b^4*n*x^(2*n) + 6*a^5*b^3*n*x^n + a^6*b^2*n)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-1+2*n)/(a**2+2*a*b*x**n+b**2*x**(2*n))**(7/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 8570 deep

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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(x^(-1+2*n)/(a^2+2*a*b*x^n+b^2*x^(2*n))^(7/2),x, algorithm="giac")

[Out]

integrate(x^(2*n - 1)/(b^2*x^(2*n) + 2*a*b*x^n + a^2)^(7/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{2\,n-1}}{{\left (a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(2*n - 1)/(a^2 + b^2*x^(2*n) + 2*a*b*x^n)^(7/2),x)

[Out]

int(x^(2*n - 1)/(a^2 + b^2*x^(2*n) + 2*a*b*x^n)^(7/2), x)

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