\(\int \frac {x+3 x^2}{\sqrt {x^2+2 x^3}} \, dx\) [606]

Optimal result

Integrand size = 21, antiderivative size = 13 \[ \int \frac {x+3 x^2}{\sqrt {x^2+2 x^3}} \, dx=\sqrt {x^2+2 x^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {1602} \[ \int \frac {x+3 x^2}{\sqrt {x^2+2 x^3}} \, dx=\sqrt {2 x^3+x^2} \]

[In]

Int[(x + 3*x^2)/Sqrt[x^2 + 2*x^3],x]

[Out]

Sqrt[x^2 + 2*x^3]

Rule 1602

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*x^(p - q +
1)*(Qq^(m + 1)/((p + m*q + 1)*Coeff[Qq, x, q])), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \sqrt {x^2+2 x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {x+3 x^2}{\sqrt {x^2+2 x^3}} \, dx=\sqrt {x^2 (1+2 x)} \]

[In]

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

[Out]

Sqrt[x^2*(1 + 2*x)]

Maple [A] (verified)

Time = 1.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92

method	result	size
trager	\(\sqrt {2 x^{3}+x^{2}}\)	\(12\)
pseudoelliptic	\(\sqrt {x^{2} \left (1+2 x \right )}\)	\(12\)
gosper	\(\frac {x^{2} \left (1+2 x \right )}{\sqrt {2 x^{3}+x^{2}}}\)	\(21\)
default	\(\frac {x^{2} \left (1+2 x \right )}{\sqrt {2 x^{3}+x^{2}}}\)	\(21\)
risch	\(\frac {x^{2} \left (1+2 x \right )}{\sqrt {x^{2} \left (1+2 x \right )}}\)	\(21\)
meijerg	\(\frac {\sqrt {\pi }-\frac {\sqrt {\pi }\, \left (-8 x +8\right ) \sqrt {1+2 x}}{8}}{\sqrt {\pi }}+\frac {-2 \sqrt {\pi }+2 \sqrt {\pi }\, \sqrt {1+2 x}}{2 \sqrt {\pi }}\)	\(53\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {x+3 x^2}{\sqrt {x^2+2 x^3}} \, dx=\sqrt {2 \, x^{3} + x^{2}} \]

[In]

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

[Out]

sqrt(2*x^3 + x^2)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int \frac {x+3 x^2}{\sqrt {x^2+2 x^3}} \, dx=\sqrt {2 x^{3} + x^{2}} \]

[In]

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

[Out]

sqrt(2*x**3 + x**2)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {x+3 x^2}{\sqrt {x^2+2 x^3}} \, dx=\sqrt {2 \, x^{3} + x^{2}} \]

[In]

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

[Out]

sqrt(2*x^3 + x^2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 23 vs. \(2 (11) = 22\).

Time = 0.37 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.77 \[ \int \frac {x+3 x^2}{\sqrt {x^2+2 x^3}} \, dx=\frac {{\left (2 \, x + 1\right )}^{\frac {3}{2}} - \sqrt {2 \, x + 1}}{2 \, \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

1/2*((2*x + 1)^(3/2) - sqrt(2*x + 1))/sgn(x)

Mupad [B] (verification not implemented)

Time = 17.10 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int \frac {x+3 x^2}{\sqrt {x^2+2 x^3}} \, dx=\left |x\right |\,\sqrt {2\,x+1} \]

[In]

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

[Out]

abs(x)*(2*x + 1)^(1/2)