\(\int \frac {1}{x \sqrt {b x+c x^2}} \, dx\) [47]

Optimal result

Integrand size = 17, antiderivative size = 21 \[ \int \frac {1}{x \sqrt {b x+c x^2}} \, dx=-\frac {2 \sqrt {b x+c x^2}}{b x} \]

[Out]

-2*(c*x^2+b*x)^(1/2)/b/x

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, 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 = {664} \[ \int \frac {1}{x \sqrt {b x+c x^2}} \, dx=-\frac {2 \sqrt {b x+c x^2}}{b x} \]

[In]

Int[1/(x*Sqrt[b*x + c*x^2]),x]

[Out]

(-2*Sqrt[b*x + c*x^2])/(b*x)

Rule 664

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^m*((a +
b*x + c*x^2)^(p + 1)/((p + 1)*(2*c*d - b*e))), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] &&
EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \sqrt {b x+c x^2}}{b x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \sqrt {b x+c x^2}} \, dx=-\frac {2 (b+c x)}{b \sqrt {x (b+c x)}} \]

[In]

Integrate[1/(x*Sqrt[b*x + c*x^2]),x]

[Out]

(-2*(b + c*x))/(b*Sqrt[x*(b + c*x)])

Maple [A] (verified)

Time = 1.96 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86

method	result	size
pseudoelliptic	\(-\frac {2 \sqrt {x \left (c x +b \right )}}{b x}\)	\(18\)
default	\(-\frac {2 \sqrt {c \,x^{2}+b x}}{b x}\)	\(20\)
trager	\(-\frac {2 \sqrt {c \,x^{2}+b x}}{b x}\)	\(20\)
risch	\(-\frac {2 \left (c x +b \right )}{b \sqrt {x \left (c x +b \right )}}\)	\(20\)
gosper	\(-\frac {2 \left (c x +b \right )}{b \sqrt {c \,x^{2}+b x}}\)	\(22\)

[In]

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

[Out]

-2*(x*(c*x+b))^(1/2)/b/x

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x \sqrt {b x+c x^2}} \, dx=-\frac {2 \, \sqrt {c x^{2} + b x}}{b x} \]

[In]

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

[Out]

-2*sqrt(c*x^2 + b*x)/(b*x)

Sympy [F]

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

[In]

integrate(1/x/(c*x**2+b*x)**(1/2),x)

[Out]

Integral(1/(x*sqrt(x*(b + c*x))), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x \sqrt {b x+c x^2}} \, dx=-\frac {2 \, \sqrt {c x^{2} + b x}}{b x} \]

[In]

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

[Out]

-2*sqrt(c*x^2 + b*x)/(b*x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {1}{x \sqrt {b x+c x^2}} \, dx=\frac {2}{\sqrt {c} x - \sqrt {c x^{2} + b x}} \]

[In]

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

[Out]

2/(sqrt(c)*x - sqrt(c*x^2 + b*x))

Mupad [B] (verification not implemented)

Time = 8.98 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x \sqrt {b x+c x^2}} \, dx=-\frac {2\,\sqrt {c\,x^2+b\,x}}{b\,x} \]

[In]

int(1/(x*(b*x + c*x^2)^(1/2)),x)

[Out]

-(2*(b*x + c*x^2)^(1/2))/(b*x)