\(\int x^2 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \, dx\) [516]

Optimal result

Integrand size = 28, antiderivative size = 93 \[ \int x^2 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \, dx=\frac {a x^3 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{3 \left (a+b x^n\right )}+\frac {b^2 x^{3+n} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(3+n) \left (a b+b^2 x^n\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {1369, 14} \[ \int x^2 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \, dx=\frac {b^2 x^{n+3} \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{(n+3) \left (a b+b^2 x^n\right )}+\frac {a x^3 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{3 \left (a+b x^n\right )} \]

[In]

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

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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*} \text {integral}& = \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 ) \, dx}{a b+b^2 x^n} \\ & = \frac {\sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \int \left (a b x^2+b^2 x^{2+n}\right ) \, dx}{a b+b^2 x^n} \\ & = \frac {a x^3 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}}{3 \left (a+b x^n\right )}+\frac {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 )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.49 \[ \int x^2 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \, dx=\frac {x^3 \sqrt {\left (a+b x^n\right )^2} \left (a (3+n)+3 b x^n\right )}{3 (3+n) \left (a+b x^n\right )} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.66

method	result	size
risch	\(\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, a \,x^{3}}{3 a +3 b \,x^{n}}+\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, b \,x^{3} x^{n}}{\left (a +b \,x^{n}\right ) \left (3+n \right )}\)	\(61\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.30 \[ \int x^2 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \, dx=\frac {3 \, b x^{3} x^{n} + {\left (a n + 3 \, a\right )} x^{3}}{3 \, {\left (n + 3\right )}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int x^2 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \, dx=\int x^{2} \sqrt {\left (a + b x^{n}\right )^{2}}\, dx \]

[In]

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

[Out]

Integral(x**2*sqrt((a + b*x**n)**2), x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.27 \[ \int x^2 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \, dx=\frac {3 \, b x^{3} x^{n} + a {\left (n + 3\right )} x^{3}}{3 \, {\left (n + 3\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.57 \[ \int x^2 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \, dx=\frac {3 \, b x^{3} x^{n} \mathrm {sgn}\left (b x^{n} + a\right ) + a n x^{3} \mathrm {sgn}\left (b x^{n} + a\right ) + 3 \, a x^{3} \mathrm {sgn}\left (b x^{n} + a\right )}{3 \, {\left (n + 3\right )}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int x^2 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}} \, dx=\int x^2\,\sqrt {a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n} \,d x \]

[In]

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

[Out]

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