\(\int \frac {x}{\sqrt {x^3+x^4}} \, dx\) [327]

Optimal result

Integrand size = 13, antiderivative size = 28 \[ \int \frac {x}{\sqrt {x^3+x^4}} \, dx=\log (x)-\log \left (-x-2 x^2+2 \sqrt {x^3+x^4}\right ) \]

[Out]

ln(x)-ln(-x-2*x^2+2*(x^4+x^3)^(1/2))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.64, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2054, 212} \[ \int \frac {x}{\sqrt {x^3+x^4}} \, dx=2 \text {arctanh}\left (\frac {x^2}{\sqrt {x^4+x^3}}\right ) \]

[In]

Int[x/Sqrt[x^3 + x^4],x]

[Out]

2*ArcTanh[x^2/Sqrt[x^3 + x^4]]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2054

Int[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[-2/(n - j), Subst[Int[1/(1 - a*x^2
), x], x, x^(j/2)/Sqrt[a*x^j + b*x^n]], x] /; FreeQ[{a, b, j, n}, x] && EqQ[m, j/2 - 1] && NeQ[n, j]

Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x^2}{\sqrt {x^3+x^4}}\right ) \\ & = 2 \text {arctanh}\left (\frac {x^2}{\sqrt {x^3+x^4}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.46 \[ \int \frac {x}{\sqrt {x^3+x^4}} \, dx=-\frac {2 x^{3/2} \sqrt {1+x} \log \left (-\sqrt {x}+\sqrt {1+x}\right )}{\sqrt {x^3 (1+x)}} \]

[In]

Integrate[x/Sqrt[x^3 + x^4],x]

[Out]

(-2*x^(3/2)*Sqrt[1 + x]*Log[-Sqrt[x] + Sqrt[1 + x]])/Sqrt[x^3*(1 + x)]

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.25

method	result	size
meijerg	\(2 \,\operatorname {arcsinh}\left (\sqrt {x}\right )\)	\(7\)
pseudoelliptic	\(\ln \left (\frac {2 x^{2}+2 \sqrt {x^{3} \left (1+x \right )}+x}{x}\right )\)	\(24\)
trager	\(-\ln \left (\frac {-x -2 x^{2}+2 \sqrt {x^{4}+x^{3}}}{x}\right )\)	\(28\)
default	\(\frac {x \sqrt {\left (1+x \right ) x}\, \ln \left (x +\frac {1}{2}+\sqrt {x^{2}+x}\right )}{\sqrt {x^{4}+x^{3}}}\)	\(30\)

[In]

int(x/(x^4+x^3)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {x}{\sqrt {x^3+x^4}} \, dx=-\log \left (-\frac {2 \, x^{2} + x - 2 \, \sqrt {x^{4} + x^{3}}}{x}\right ) \]

[In]

integrate(x/(x^4+x^3)^(1/2),x, algorithm="fricas")

[Out]

-log(-(2*x^2 + x - 2*sqrt(x^4 + x^3))/x)

Sympy [F]

\[ \int \frac {x}{\sqrt {x^3+x^4}} \, dx=\int \frac {x}{\sqrt {x^{3} \left (x + 1\right )}}\, dx \]

[In]

integrate(x/(x**4+x**3)**(1/2),x)

[Out]

Integral(x/sqrt(x**3*(x + 1)), x)

Maxima [F]

\[ \int \frac {x}{\sqrt {x^3+x^4}} \, dx=\int { \frac {x}{\sqrt {x^{4} + x^{3}}} \,d x } \]

[In]

integrate(x/(x^4+x^3)^(1/2),x, algorithm="maxima")

[Out]

integrate(x/sqrt(x^4 + x^3), x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {x}{\sqrt {x^3+x^4}} \, dx=-\frac {\log \left ({\left | -2 \, x + 2 \, \sqrt {x^{2} + x} - 1 \right |}\right )}{\mathrm {sgn}\left (x\right )} \]

[In]

integrate(x/(x^4+x^3)^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x}{\sqrt {x^3+x^4}} \, dx=\int \frac {x}{\sqrt {x^4+x^3}} \,d x \]

[In]

int(x/(x^3 + x^4)^(1/2),x)

[Out]

int(x/(x^3 + x^4)^(1/2), x)