\(\int \frac {x}{\sqrt {a x^2+b x^3}} \, dx\) [255]

Optimal result

Integrand size = 17, antiderivative size = 23 \[ \int \frac {x}{\sqrt {a x^2+b x^3}} \, dx=\frac {2 \sqrt {a x^2+b x^3}}{b x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {1602} \[ \int \frac {x}{\sqrt {a x^2+b x^3}} \, dx=\frac {2 \sqrt {a x^2+b x^3}}{b x} \]

[In]

Int[x/Sqrt[a*x^2 + b*x^3],x]

[Out]

(2*Sqrt[a*x^2 + b*x^3])/(b*x)

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}& = \frac {2 \sqrt {a x^2+b x^3}}{b x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {x}{\sqrt {a x^2+b x^3}} \, dx=\frac {2 \sqrt {x^2 (a+b x)}}{b x} \]

[In]

Integrate[x/Sqrt[a*x^2 + b*x^3],x]

[Out]

(2*Sqrt[x^2*(a + b*x)])/(b*x)

Maple [A] (verified)

Time = 2.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91

method	result	size
pseudoelliptic	\(-\frac {2 \sqrt {b x +a}\, \left (-b x +2 a \right )}{3 b^{2}}\)	\(21\)
trager	\(\frac {2 \sqrt {b \,x^{3}+a \,x^{2}}}{b x}\)	\(22\)
risch	\(\frac {2 x \left (b x +a \right )}{\sqrt {x^{2} \left (b x +a \right )}\, b}\)	\(23\)
gosper	\(\frac {2 x \left (b x +a \right )}{b \sqrt {b \,x^{3}+a \,x^{2}}}\)	\(25\)
default	\(\frac {2 x \left (b x +a \right )}{b \sqrt {b \,x^{3}+a \,x^{2}}}\)	\(25\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {x}{\sqrt {a x^2+b x^3}} \, dx=\frac {2 \, \sqrt {b x^{3} + a x^{2}}}{b x} \]

[In]

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

[Out]

2*sqrt(b*x^3 + a*x^2)/(b*x)

Sympy [F]

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

[In]

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

[Out]

Integral(x/sqrt(x**2*(a + b*x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.52 \[ \int \frac {x}{\sqrt {a x^2+b x^3}} \, dx=\frac {2 \, \sqrt {b x + a}}{b} \]

[In]

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

[Out]

2*sqrt(b*x + a)/b

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {x}{\sqrt {a x^2+b x^3}} \, dx=-\frac {2 \, \sqrt {a} \mathrm {sgn}\left (x\right )}{b} + \frac {2 \, \sqrt {b x + a}}{b \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

-2*sqrt(a)*sgn(x)/b + 2*sqrt(b*x + a)/(b*sgn(x))

Mupad [B] (verification not implemented)

Time = 8.88 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {x}{\sqrt {a x^2+b x^3}} \, dx=\frac {2\,\left |x\right |\,\sqrt {a+b\,x}}{b\,x} \]

[In]

int(x/(a*x^2 + b*x^3)^(1/2),x)

[Out]

(2*abs(x)*(a + b*x)^(1/2))/(b*x)