\(\int \frac {(3+2 x^2) \sqrt {x+2 x^3}}{(1+2 x^2)^2} \, dx\) [213]

Optimal result

Integrand size = 28, antiderivative size = 23 \[ \int \frac {\left (3+2 x^2\right ) \sqrt {x+2 x^3}}{\left (1+2 x^2\right )^2} \, dx=\frac {2 x \sqrt {x+2 x^3}}{1+2 x^2} \]

[Out]

2*x*(2*x^3+x)^(1/2)/(2*x^2+1)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 23, 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 = {2081, 460} \[ \int \frac {\left (3+2 x^2\right ) \sqrt {x+2 x^3}}{\left (1+2 x^2\right )^2} \, dx=\frac {2 x \sqrt {2 x^3+x}}{2 x^2+1} \]

[In]

Int[((3 + 2*x^2)*Sqrt[x + 2*x^3])/(1 + 2*x^2)^2,x]

[Out]

(2*x*Sqrt[x + 2*x^3])/(1 + 2*x^2)

Rule 460

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[
a*d*(m + 1) - b*c*(m + n*(p + 1) + 1), 0] && NeQ[m, -1]

Rule 2081

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {x+2 x^3} \int \frac {\sqrt {x} \left (3+2 x^2\right )}{\left (1+2 x^2\right )^{3/2}} \, dx}{\sqrt {x} \sqrt {1+2 x^2}} \\ & = \frac {2 x \sqrt {x+2 x^3}}{1+2 x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 10.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {\left (3+2 x^2\right ) \sqrt {x+2 x^3}}{\left (1+2 x^2\right )^2} \, dx=\frac {2 x \sqrt {x+2 x^3}}{1+2 x^2} \]

[In]

Integrate[((3 + 2*x^2)*Sqrt[x + 2*x^3])/(1 + 2*x^2)^2,x]

[Out]

(2*x*Sqrt[x + 2*x^3])/(1 + 2*x^2)

Maple [A] (verified)

Time = 1.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74

method	result	size
default	\(\frac {x^{2} \sqrt {2}}{\sqrt {x \left (x^{2}+\frac {1}{2}\right )}}\)	\(17\)
elliptic	\(\frac {x^{2} \sqrt {2}}{\sqrt {x \left (x^{2}+\frac {1}{2}\right )}}\)	\(17\)
gosper	\(\frac {2 x \sqrt {2 x^{3}+x}}{2 x^{2}+1}\)	\(22\)
trager	\(\frac {2 x \sqrt {2 x^{3}+x}}{2 x^{2}+1}\)	\(22\)
meijerg	\(2 x^{\frac {3}{2}} \operatorname {hypergeom}\left (\left [\frac {3}{4}, \frac {3}{2}\right ], \left [\frac {7}{4}\right ], -2 x^{2}\right )+\frac {4 x^{\frac {7}{2}} \operatorname {hypergeom}\left (\left [\frac {3}{2}, \frac {7}{4}\right ], \left [\frac {11}{4}\right ], -2 x^{2}\right )}{7}\)	\(34\)

[In]

int((2*x^2+3)*(2*x^3+x)^(1/2)/(2*x^2+1)^2,x,method=_RETURNVERBOSE)

[Out]

x^2*2^(1/2)/(x*(x^2+1/2))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {\left (3+2 x^2\right ) \sqrt {x+2 x^3}}{\left (1+2 x^2\right )^2} \, dx=\frac {2 \, \sqrt {2 \, x^{3} + x} x}{2 \, x^{2} + 1} \]

[In]

integrate((2*x^2+3)*(2*x^3+x)^(1/2)/(2*x^2+1)^2,x, algorithm="fricas")

[Out]

2*sqrt(2*x^3 + x)*x/(2*x^2 + 1)

Sympy [F]

\[ \int \frac {\left (3+2 x^2\right ) \sqrt {x+2 x^3}}{\left (1+2 x^2\right )^2} \, dx=\int \frac {\sqrt {x \left (2 x^{2} + 1\right )} \left (2 x^{2} + 3\right )}{\left (2 x^{2} + 1\right )^{2}}\, dx \]

[In]

integrate((2*x**2+3)*(2*x**3+x)**(1/2)/(2*x**2+1)**2,x)

[Out]

Integral(sqrt(x*(2*x**2 + 1))*(2*x**2 + 3)/(2*x**2 + 1)**2, x)

Maxima [F]

\[ \int \frac {\left (3+2 x^2\right ) \sqrt {x+2 x^3}}{\left (1+2 x^2\right )^2} \, dx=\int { \frac {\sqrt {2 \, x^{3} + x} {\left (2 \, x^{2} + 3\right )}}{{\left (2 \, x^{2} + 1\right )}^{2}} \,d x } \]

[In]

integrate((2*x^2+3)*(2*x^3+x)^(1/2)/(2*x^2+1)^2,x, algorithm="maxima")

[Out]

integrate(sqrt(2*x^3 + x)*(2*x^2 + 3)/(2*x^2 + 1)^2, x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.57 \[ \int \frac {\left (3+2 x^2\right ) \sqrt {x+2 x^3}}{\left (1+2 x^2\right )^2} \, dx=\frac {2}{\sqrt {\frac {2}{x} + \frac {1}{x^{3}}}} \]

[In]

integrate((2*x^2+3)*(2*x^3+x)^(1/2)/(2*x^2+1)^2,x, algorithm="giac")

[Out]

2/sqrt(2/x + 1/x^3)

Mupad [B] (verification not implemented)

Time = 5.38 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.61 \[ \int \frac {\left (3+2 x^2\right ) \sqrt {x+2 x^3}}{\left (1+2 x^2\right )^2} \, dx=\frac {2\,x^2}{\sqrt {2\,x^3+x}} \]

[In]

int(((x + 2*x^3)^(1/2)*(2*x^2 + 3))/(2*x^2 + 1)^2,x)

[Out]

(2*x^2)/(x + 2*x^3)^(1/2)