3.523 \(\int x^2 (a^2+2 a b x^n+b^2 x^{2 n})^{3/2} \, dx\)

Optimal. Leaf size=212 \[ \frac {3 a^2 b^2 x^{n+3} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(n+3) \left (a b+b^2 x^n\right )}+\frac {b^4 x^{3 (n+1)} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{3 (n+1) \left (a b+b^2 x^n\right )}+\frac {3 a b^3 x^{2 n+3} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(2 n+3) \left (a b+b^2 x^n\right )}+\frac {a^3 x^3 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{3 \left (a+b x^n\right )} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {1355, 270} \[ \frac {b^4 x^{3 (n+1)} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{3 (n+1) \left (a b+b^2 x^n\right )}+\frac {3 a^2 b^2 x^{n+3} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(n+3) \left (a b+b^2 x^n\right )}+\frac {3 a b^3 x^{2 n+3} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(2 n+3) \left (a b+b^2 x^n\right )}+\frac {a^3 x^3 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{3 \left (a+b x^n\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*x^3*Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)])/(3*(a + b*x^n)) + (b^4*x^(3*(1 + n))*Sqrt[a^2 + 2*a*b*x^n + b^2*
x^(2*n)])/(3*(1 + n)*(a*b + b^2*x^n)) + (3*a^2*b^2*x^(3 + n)*Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)])/((3 + n)*(a*
b + b^2*x^n)) + (3*a*b^3*x^(3 + 2*n)*Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)])/((3 + 2*n)*(a*b + b^2*x^n))

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 1355

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

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Mathematica [A]  time = 0.08, size = 123, normalized size = 0.58 \[ \frac {x^3 \sqrt {\left (a+b x^n\right )^2} \left (a^3 \left (2 n^3+11 n^2+18 n+9\right )+9 a^2 b \left (2 n^2+5 n+3\right ) x^n+9 a b^2 \left (n^2+4 n+3\right ) x^{2 n}+b^3 \left (2 n^2+9 n+9\right ) x^{3 n}\right )}{3 (n+1) (n+3) (2 n+3) \left (a+b x^n\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*Sqrt[(a + b*x^n)^2]*(a^3*(9 + 18*n + 11*n^2 + 2*n^3) + 9*a^2*b*(3 + 5*n + 2*n^2)*x^n + 9*a*b^2*(3 + 4*n +
n^2)*x^(2*n) + b^3*(9 + 9*n + 2*n^2)*x^(3*n)))/(3*(1 + n)*(3 + n)*(3 + 2*n)*(a + b*x^n))

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fricas [A]  time = 0.74, size = 144, normalized size = 0.68 \[ \frac {{\left (2 \, b^{3} n^{2} + 9 \, b^{3} n + 9 \, b^{3}\right )} x^{3} x^{3 \, n} + 9 \, {\left (a b^{2} n^{2} + 4 \, a b^{2} n + 3 \, a b^{2}\right )} x^{3} x^{2 \, n} + 9 \, {\left (2 \, a^{2} b n^{2} + 5 \, a^{2} b n + 3 \, a^{2} b\right )} x^{3} x^{n} + {\left (2 \, a^{3} n^{3} + 11 \, a^{3} n^{2} + 18 \, a^{3} n + 9 \, a^{3}\right )} x^{3}}{3 \, {\left (2 \, n^{3} + 11 \, n^{2} + 18 \, n + 9\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((2*b^3*n^2 + 9*b^3*n + 9*b^3)*x^3*x^(3*n) + 9*(a*b^2*n^2 + 4*a*b^2*n + 3*a*b^2)*x^3*x^(2*n) + 9*(2*a^2*b*
n^2 + 5*a^2*b*n + 3*a^2*b)*x^3*x^n + (2*a^3*n^3 + 11*a^3*n^2 + 18*a^3*n + 9*a^3)*x^3)/(2*n^3 + 11*n^2 + 18*n +
9)

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giac [A]  time = 0.43, size = 292, normalized size = 1.38 \[ \frac {2 \, b^{3} n^{2} x^{3} x^{3 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 9 \, a b^{2} n^{2} x^{3} x^{2 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 18 \, a^{2} b n^{2} x^{3} x^{n} \mathrm {sgn}\left (b x^{n} + a\right ) + 2 \, a^{3} n^{3} x^{3} \mathrm {sgn}\left (b x^{n} + a\right ) + 9 \, b^{3} n x^{3} x^{3 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 36 \, a b^{2} n x^{3} x^{2 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 45 \, a^{2} b n x^{3} x^{n} \mathrm {sgn}\left (b x^{n} + a\right ) + 11 \, a^{3} n^{2} x^{3} \mathrm {sgn}\left (b x^{n} + a\right ) + 9 \, b^{3} x^{3} x^{3 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 27 \, a b^{2} x^{3} x^{2 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 27 \, a^{2} b x^{3} x^{n} \mathrm {sgn}\left (b x^{n} + a\right ) + 18 \, a^{3} n x^{3} \mathrm {sgn}\left (b x^{n} + a\right ) + 9 \, a^{3} x^{3} \mathrm {sgn}\left (b x^{n} + a\right )}{3 \, {\left (2 \, n^{3} + 11 \, n^{2} + 18 \, n + 9\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(2*b^3*n^2*x^3*x^(3*n)*sgn(b*x^n + a) + 9*a*b^2*n^2*x^3*x^(2*n)*sgn(b*x^n + a) + 18*a^2*b*n^2*x^3*x^n*sgn(
b*x^n + a) + 2*a^3*n^3*x^3*sgn(b*x^n + a) + 9*b^3*n*x^3*x^(3*n)*sgn(b*x^n + a) + 36*a*b^2*n*x^3*x^(2*n)*sgn(b*
x^n + a) + 45*a^2*b*n*x^3*x^n*sgn(b*x^n + a) + 11*a^3*n^2*x^3*sgn(b*x^n + a) + 9*b^3*x^3*x^(3*n)*sgn(b*x^n + a
) + 27*a*b^2*x^3*x^(2*n)*sgn(b*x^n + a) + 27*a^2*b*x^3*x^n*sgn(b*x^n + a) + 18*a^3*n*x^3*sgn(b*x^n + a) + 9*a^
3*x^3*sgn(b*x^n + a))/(2*n^3 + 11*n^2 + 18*n + 9)

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maple [A]  time = 0.02, size = 146, normalized size = 0.69 \[ \frac {3 \sqrt {\left (b \,x^{n}+a \right )^{2}}\, a^{2} b \,x^{3} x^{n}}{\left (b \,x^{n}+a \right ) \left (n +3\right )}+\frac {3 \sqrt {\left (b \,x^{n}+a \right )^{2}}\, a \,b^{2} x^{3} x^{2 n}}{\left (b \,x^{n}+a \right ) \left (2 n +3\right )}+\frac {\sqrt {\left (b \,x^{n}+a \right )^{2}}\, b^{3} x^{3} x^{3 n}}{3 \left (b \,x^{n}+a \right ) \left (n +1\right )}+\frac {\sqrt {\left (b \,x^{n}+a \right )^{2}}\, a^{3} x^{3}}{3 b \,x^{n}+3 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.93, size = 108, normalized size = 0.51 \[ \frac {{\left (2 \, n^{2} + 9 \, n + 9\right )} b^{3} x^{3} x^{3 \, n} + 9 \, {\left (n^{2} + 4 \, n + 3\right )} a b^{2} x^{3} x^{2 \, n} + 9 \, {\left (2 \, n^{2} + 5 \, n + 3\right )} a^{2} b x^{3} x^{n} + {\left (2 \, n^{3} + 11 \, n^{2} + 18 \, n + 9\right )} a^{3} x^{3}}{3 \, {\left (2 \, n^{3} + 11 \, n^{2} + 18 \, n + 9\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((2*n^2 + 9*n + 9)*b^3*x^3*x^(3*n) + 9*(n^2 + 4*n + 3)*a*b^2*x^3*x^(2*n) + 9*(2*n^2 + 5*n + 3)*a^2*b*x^3*x
^n + (2*n^3 + 11*n^2 + 18*n + 9)*a^3*x^3)/(2*n^3 + 11*n^2 + 18*n + 9)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^2\,{\left (a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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