\(\int \frac {x^2}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx\) [530]

Optimal result

Integrand size = 28, antiderivative size = 64 \[ \int \frac {x^2}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx=\frac {x^3 \left (a+b x^n\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {3}{n},\frac {3+n}{n},-\frac {b x^n}{a}\right )}{3 a \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \]

[Out]

1/3*x^3*(a+b*x^n)*hypergeom([1, 3/n],[(3+n)/n],-b*x^n/a)/a/(a^2+2*a*b*x^n+b^2*x^(2*n))^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {1369, 371} \[ \int \frac {x^2}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx=\frac {x^3 \left (a+b x^n\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {3}{n},\frac {n+3}{n},-\frac {b x^n}{a}\right )}{3 a \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \]

[In]

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

[Out]

(x^3*(a + b*x^n)*Hypergeometric2F1[1, 3/n, (3 + n)/n, -((b*x^n)/a)])/(3*a*Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)])

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 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*} \text {integral}& = \frac {\left (a b+b^2 x^n\right ) \int \frac {x^2}{a b+b^2 x^n} \, dx}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \\ & = \frac {x^3 \left (a+b x^n\right ) \, _2F_1\left (1,\frac {3}{n};\frac {3+n}{n};-\frac {b x^n}{a}\right )}{3 a \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.83 \[ \int \frac {x^2}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx=\frac {x^3 \left (a+b x^n\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {3}{n},1+\frac {3}{n},-\frac {b x^n}{a}\right )}{3 a \sqrt {\left (a+b x^n\right )^2}} \]

[In]

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

[Out]

(x^3*(a + b*x^n)*Hypergeometric2F1[1, 3/n, 1 + 3/n, -((b*x^n)/a)])/(3*a*Sqrt[(a + b*x^n)^2])

Maple [F]

\[\int \frac {x^{2}}{\sqrt {a^{2}+2 a b \,x^{n}+b^{2} x^{2 n}}}d x\]

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(x^2/sqrt(b^2*x^(2*n) + 2*a*b*x^n + a^2), x)

Sympy [F]

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

[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 [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[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)