3.6.24 \(\int x (a^2+2 a b x^n+b^2 x^{2 n})^{3/2} \, dx\) [524]

Optimal. Leaf size=211 \[ \frac {a^3 x^2 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{2 \left (a+b x^n\right )}+\frac {3 a b^3 x^{2 (1+n)} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{2 (1+n) \left (a b+b^2 x^n\right )}+\frac {3 a^2 b^2 x^{2+n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(2+n) \left (a b+b^2 x^n\right )}+\frac {b^4 x^{2+3 n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(2+3 n) \left (a b+b^2 x^n\right )} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1369, 276} \begin {gather*} \frac {3 a^2 b^2 x^{n+2} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(n+2) \left (a b+b^2 x^n\right )}+\frac {b^4 x^{3 n+2} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(3 n+2) \left (a b+b^2 x^n\right )}+\frac {3 a b^3 x^{2 (n+1)} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{2 (n+1) \left (a b+b^2 x^n\right )}+\frac {a^3 x^2 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{2 \left (a+b x^n\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 276

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

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Mathematica [A]
time = 0.06, size = 124, normalized size = 0.59 \begin {gather*} \frac {x^2 \sqrt {\left (a+b x^n\right )^2} \left (a^3 \left (4+12 n+11 n^2+3 n^3\right )+6 a^2 b \left (2+5 n+3 n^2\right ) x^n+3 a b^2 \left (4+8 n+3 n^2\right ) x^{2 n}+2 b^3 \left (2+3 n+n^2\right ) x^{3 n}\right )}{2 (1+n) (2+n) (2+3 n) \left (a+b x^n\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*Sqrt[(a + b*x^n)^2]*(a^3*(4 + 12*n + 11*n^2 + 3*n^3) + 6*a^2*b*(2 + 5*n + 3*n^2)*x^n + 3*a*b^2*(4 + 8*n +
3*n^2)*x^(2*n) + 2*b^3*(2 + 3*n + n^2)*x^(3*n)))/(2*(1 + n)*(2 + n)*(2 + 3*n)*(a + b*x^n))

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Maple [A]
time = 0.02, size = 145, normalized size = 0.69

method	result	size
risch	\(\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, a^{3} x^{2}}{2 a +2 b \,x^{n}}+\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, b^{3} x^{2} x^{3 n}}{\left (a +b \,x^{n}\right ) \left (2+3 n \right )}+\frac {3 \sqrt {\left (a +b \,x^{n}\right )^{2}}\, a \,b^{2} x^{2} x^{2 n}}{2 \left (a +b \,x^{n}\right ) \left (1+n \right )}+\frac {3 \sqrt {\left (a +b \,x^{n}\right )^{2}}\, a^{2} b \,x^{2} x^{n}}{\left (a +b \,x^{n}\right ) \left (2+n \right )}\)	\(145\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 109, normalized size = 0.52 \begin {gather*} \frac {2 \, {\left (n^{2} + 3 \, n + 2\right )} b^{3} x^{2} x^{3 \, n} + 3 \, {\left (3 \, n^{2} + 8 \, n + 4\right )} a b^{2} x^{2} x^{2 \, n} + 6 \, {\left (3 \, n^{2} + 5 \, n + 2\right )} a^{2} b x^{2} x^{n} + {\left (3 \, n^{3} + 11 \, n^{2} + 12 \, n + 4\right )} a^{3} x^{2}}{2 \, {\left (3 \, n^{3} + 11 \, n^{2} + 12 \, n + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(n^2 + 3*n + 2)*b^3*x^2*x^(3*n) + 3*(3*n^2 + 8*n + 4)*a*b^2*x^2*x^(2*n) + 6*(3*n^2 + 5*n + 2)*a^2*b*x^2
*x^n + (3*n^3 + 11*n^2 + 12*n + 4)*a^3*x^2)/(3*n^3 + 11*n^2 + 12*n + 4)

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Fricas [A]
time = 0.36, size = 145, normalized size = 0.69 \begin {gather*} \frac {2 \, {\left (b^{3} n^{2} + 3 \, b^{3} n + 2 \, b^{3}\right )} x^{2} x^{3 \, n} + 3 \, {\left (3 \, a b^{2} n^{2} + 8 \, a b^{2} n + 4 \, a b^{2}\right )} x^{2} x^{2 \, n} + 6 \, {\left (3 \, a^{2} b n^{2} + 5 \, a^{2} b n + 2 \, a^{2} b\right )} x^{2} x^{n} + {\left (3 \, a^{3} n^{3} + 11 \, a^{3} n^{2} + 12 \, a^{3} n + 4 \, a^{3}\right )} x^{2}}{2 \, {\left (3 \, n^{3} + 11 \, n^{2} + 12 \, n + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(b^3*n^2 + 3*b^3*n + 2*b^3)*x^2*x^(3*n) + 3*(3*a*b^2*n^2 + 8*a*b^2*n + 4*a*b^2)*x^2*x^(2*n) + 6*(3*a^2*
b*n^2 + 5*a^2*b*n + 2*a^2*b)*x^2*x^n + (3*a^3*n^3 + 11*a^3*n^2 + 12*a^3*n + 4*a^3)*x^2)/(3*n^3 + 11*n^2 + 12*n
+ 4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 2.97, size = 292, normalized size = 1.38 \begin {gather*} \frac {2 \, b^{3} n^{2} x^{2} x^{3 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 9 \, a b^{2} n^{2} x^{2} x^{2 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 18 \, a^{2} b n^{2} x^{2} x^{n} \mathrm {sgn}\left (b x^{n} + a\right ) + 3 \, a^{3} n^{3} x^{2} \mathrm {sgn}\left (b x^{n} + a\right ) + 6 \, b^{3} n x^{2} x^{3 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 24 \, a b^{2} n x^{2} x^{2 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 30 \, a^{2} b n x^{2} x^{n} \mathrm {sgn}\left (b x^{n} + a\right ) + 11 \, a^{3} n^{2} x^{2} \mathrm {sgn}\left (b x^{n} + a\right ) + 4 \, b^{3} x^{2} x^{3 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 12 \, a b^{2} x^{2} x^{2 \, n} \mathrm {sgn}\left (b x^{n} + a\right ) + 12 \, a^{2} b x^{2} x^{n} \mathrm {sgn}\left (b x^{n} + a\right ) + 12 \, a^{3} n x^{2} \mathrm {sgn}\left (b x^{n} + a\right ) + 4 \, a^{3} x^{2} \mathrm {sgn}\left (b x^{n} + a\right )}{2 \, {\left (3 \, n^{3} + 11 \, n^{2} + 12 \, n + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b^3*n^2*x^2*x^(3*n)*sgn(b*x^n + a) + 9*a*b^2*n^2*x^2*x^(2*n)*sgn(b*x^n + a) + 18*a^2*b*n^2*x^2*x^n*sgn(
b*x^n + a) + 3*a^3*n^3*x^2*sgn(b*x^n + a) + 6*b^3*n*x^2*x^(3*n)*sgn(b*x^n + a) + 24*a*b^2*n*x^2*x^(2*n)*sgn(b*
x^n + a) + 30*a^2*b*n*x^2*x^n*sgn(b*x^n + a) + 11*a^3*n^2*x^2*sgn(b*x^n + a) + 4*b^3*x^2*x^(3*n)*sgn(b*x^n + a
) + 12*a*b^2*x^2*x^(2*n)*sgn(b*x^n + a) + 12*a^2*b*x^2*x^n*sgn(b*x^n + a) + 12*a^3*n*x^2*sgn(b*x^n + a) + 4*a^
3*x^2*sgn(b*x^n + a))/(3*n^3 + 11*n^2 + 12*n + 4)

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

[Out]

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

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