3.6.8 \(\int x^{-1+2 n} (a^2+2 a b x^n+b^2 x^{2 n})^{5/2} \, dx\) [508]

Optimal. Leaf size=112 \[ -\frac {a \left (a+b x^n\right )^6 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{6 n \left (a b^2+b^3 x^n\right )}+\frac {\left (a+b x^n\right )^7 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{7 n \left (a b^2+b^3 x^n\right )} \]

[Out]

-1/6*a*(a+b*x^n)^6*(a^2+2*a*b*x^n+b^2*x^(2*n))^(1/2)/n/(a*b^2+b^3*x^n)+1/7*(a+b*x^n)^7*(a^2+2*a*b*x^n+b^2*x^(2
*n))^(1/2)/n/(a*b^2+b^3*x^n)

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Rubi [A]
time = 0.03, antiderivative size = 112, 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 {\left (a+b x^n\right )^7 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{7 n \left (a b^2+b^3 x^n\right )}-\frac {a \left (a+b x^n\right )^6 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{6 n \left (a b^2+b^3 x^n\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(-1 + 2*n)*(a^2 + 2*a*b*x^n + b^2*x^(2*n))^(5/2),x]

[Out]

-1/6*(a*(a + b*x^n)^6*Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)])/(n*(a*b^2 + b^3*x^n)) + ((a + b*x^n)^7*Sqrt[a^2 + 2
*a*b*x^n + b^2*x^(2*n)])/(7*n*(a*b^2 + b^3*x^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 x^{-1+2 n} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{5/2} \, dx &=\frac {\sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \int x^{-1+2 n} \left (a b+b^2 x^n\right )^5 \, dx}{b^4 \left (a b+b^2 x^n\right )}\\ &=\frac {\sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \text {Subst}\left (\int x \left (a b+b^2 x\right )^5 \, dx,x,x^n\right )}{b^4 n \left (a b+b^2 x^n\right )}\\ &=\frac {\sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \text {Subst}\left (\int \left (-\frac {a \left (a b+b^2 x\right )^5}{b}+\frac {\left (a b+b^2 x\right )^6}{b^2}\right ) \, dx,x,x^n\right )}{b^4 n \left (a b+b^2 x^n\right )}\\ &=-\frac {a \left (a+b x^n\right )^6 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{6 n \left (a b^2+b^3 x^n\right )}+\frac {\left (a+b x^n\right )^7 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{7 n \left (a b^2+b^3 x^n\right )}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 96, normalized size = 0.86 \begin {gather*} \frac {x^{2 n} \left (\left (a+b x^n\right )^2\right )^{5/2} \left (21 a^5+70 a^4 b x^n+105 a^3 b^2 x^{2 n}+84 a^2 b^3 x^{3 n}+35 a b^4 x^{4 n}+6 b^5 x^{5 n}\right )}{42 n \left (a+b x^n\right )^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 + 2*n)*(a^2 + 2*a*b*x^n + b^2*x^(2*n))^(5/2),x]

[Out]

(x^(2*n)*((a + b*x^n)^2)^(5/2)*(21*a^5 + 70*a^4*b*x^n + 105*a^3*b^2*x^(2*n) + 84*a^2*b^3*x^(3*n) + 35*a*b^4*x^
(4*n) + 6*b^5*x^(5*n)))/(42*n*(a + b*x^n)^5)

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Maple [A]
time = 0.04, size = 208, normalized size = 1.86

method	result	size
risch	\(\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, b^{5} x^{7 n}}{7 \left (a +b \,x^{n}\right ) n}+\frac {5 \sqrt {\left (a +b \,x^{n}\right )^{2}}\, b^{4} a \,x^{6 n}}{6 \left (a +b \,x^{n}\right ) n}+\frac {2 \sqrt {\left (a +b \,x^{n}\right )^{2}}\, a^{2} b^{3} x^{5 n}}{\left (a +b \,x^{n}\right ) n}+\frac {5 \sqrt {\left (a +b \,x^{n}\right )^{2}}\, b^{2} a^{3} x^{4 n}}{2 \left (a +b \,x^{n}\right ) n}+\frac {5 \sqrt {\left (a +b \,x^{n}\right )^{2}}\, b \,a^{4} x^{3 n}}{3 \left (a +b \,x^{n}\right ) n}+\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, a^{5} x^{2 n}}{2 \left (a +b \,x^{n}\right ) n}\)	\(208\)

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))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/7*((a+b*x^n)^2)^(1/2)/(a+b*x^n)*b^5/n*(x^n)^7+5/6*((a+b*x^n)^2)^(1/2)/(a+b*x^n)*b^4*a/n*(x^n)^6+2*((a+b*x^n)
^2)^(1/2)/(a+b*x^n)*a^2*b^3/n*(x^n)^5+5/2*((a+b*x^n)^2)^(1/2)/(a+b*x^n)*b^2*a^3/n*(x^n)^4+5/3*((a+b*x^n)^2)^(1
/2)/(a+b*x^n)*b*a^4/n*(x^n)^3+1/2*((a+b*x^n)^2)^(1/2)/(a+b*x^n)*a^5/n*(x^n)^2

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Maxima [A]
time = 0.32, size = 74, normalized size = 0.66 \begin {gather*} \frac {6 \, b^{5} x^{7 \, n} + 35 \, a b^{4} x^{6 \, n} + 84 \, a^{2} b^{3} x^{5 \, n} + 105 \, a^{3} b^{2} x^{4 \, n} + 70 \, a^{4} b x^{3 \, n} + 21 \, a^{5} x^{2 \, n}}{42 \, n} \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))^(5/2),x, algorithm="maxima")

[Out]

1/42*(6*b^5*x^(7*n) + 35*a*b^4*x^(6*n) + 84*a^2*b^3*x^(5*n) + 105*a^3*b^2*x^(4*n) + 70*a^4*b*x^(3*n) + 21*a^5*
x^(2*n))/n

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Fricas [A]
time = 0.35, size = 74, normalized size = 0.66 \begin {gather*} \frac {6 \, b^{5} x^{7 \, n} + 35 \, a b^{4} x^{6 \, n} + 84 \, a^{2} b^{3} x^{5 \, n} + 105 \, a^{3} b^{2} x^{4 \, n} + 70 \, a^{4} b x^{3 \, n} + 21 \, a^{5} x^{2 \, n}}{42 \, n} \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))^(5/2),x, algorithm="fricas")

[Out]

1/42*(6*b^5*x^(7*n) + 35*a*b^4*x^(6*n) + 84*a^2*b^3*x^(5*n) + 105*a^3*b^2*x^(4*n) + 70*a^4*b*x^(3*n) + 21*a^5*
x^(2*n))/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))**(5/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 8569 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))^(5/2),x, algorithm="giac")

[Out]

integrate((b^2*x^(2*n) + 2*a*b*x^n + a^2)^(5/2)*x^(2*n - 1), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^{2\,n-1}\,{\left (a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n\right )}^{5/2} \,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)^(5/2),x)

[Out]

int(x^(2*n - 1)*(a^2 + b^2*x^(2*n) + 2*a*b*x^n)^(5/2), x)

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