\(\int \frac {x}{\sqrt {b x+c x^2}} \, dx\) [45]

Optimal result

Integrand size = 15, antiderivative size = 47 \[ \int \frac {x}{\sqrt {b x+c x^2}} \, dx=\frac {\sqrt {b x+c x^2}}{c}-\frac {b \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {654, 634, 212} \[ \int \frac {x}{\sqrt {b x+c x^2}} \, dx=\frac {\sqrt {b x+c x^2}}{c}-\frac {b \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{3/2}} \]

[In]

Int[x/Sqrt[b*x + c*x^2],x]

[Out]

Sqrt[b*x + c*x^2]/c - (b*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/c^(3/2)

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 634

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 654

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

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

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.66 \[ \int \frac {x}{\sqrt {b x+c x^2}} \, dx=\frac {\sqrt {c} x (b+c x)+2 b \sqrt {x} \sqrt {b+c x} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}-\sqrt {b+c x}}\right )}{c^{3/2} \sqrt {x (b+c x)}} \]

[In]

Integrate[x/Sqrt[b*x + c*x^2],x]

[Out]

(Sqrt[c]*x*(b + c*x) + 2*b*Sqrt[x]*Sqrt[b + c*x]*ArcTanh[(Sqrt[c]*Sqrt[x])/(Sqrt[b] - Sqrt[b + c*x])])/(c^(3/2
)*Sqrt[x*(b + c*x)])

Maple [A] (verified)

Time = 2.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.83

method	result	size
pseudoelliptic	\(\frac {-\operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right ) b +\sqrt {x \left (c x +b \right )}\, \sqrt {c}}{c^{\frac {3}{2}}}\)	\(39\)
default	\(\frac {\sqrt {c \,x^{2}+b x}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {3}{2}}}\)	\(47\)
risch	\(\frac {x \left (c x +b \right )}{c \sqrt {x \left (c x +b \right )}}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {3}{2}}}\)	\(51\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 98, normalized size of antiderivative = 2.09 \[ \int \frac {x}{\sqrt {b x+c x^2}} \, dx=\left [\frac {b \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, \sqrt {c x^{2} + b x} c}{2 \, c^{2}}, \frac {b \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + \sqrt {c x^{2} + b x} c}{c^{2}}\right ] \]

[In]

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

[Out]

[1/2*(b*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*sqrt(c*x^2 + b*x)*c)/c^2, (b*sqrt(-c)*arctan(
sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + sqrt(c*x^2 + b*x)*c)/c^2]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 99 vs. \(2 (39) = 78\).

Time = 0.29 (sec) , antiderivative size = 99, normalized size of antiderivative = 2.11 \[ \int \frac {x}{\sqrt {b x+c x^2}} \, dx=\begin {cases} - \frac {b \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{2 c} + \frac {\sqrt {b x + c x^{2}}}{c} & \text {for}\: c \neq 0 \\\frac {2 \left (b x\right )^{\frac {3}{2}}}{3 b^{2}} & \text {for}\: b \neq 0 \\\tilde {\infty } x^{2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-b*Piecewise((log(b + 2*sqrt(c)*sqrt(b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c) + x)
*log(b/(2*c) + x)/sqrt(c*(b/(2*c) + x)**2), True))/(2*c) + sqrt(b*x + c*x**2)/c, Ne(c, 0)), (2*(b*x)**(3/2)/(3
*b**2), Ne(b, 0)), (zoo*x**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.96 \[ \int \frac {x}{\sqrt {b x+c x^2}} \, dx=-\frac {b \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2 \, c^{\frac {3}{2}}} + \frac {\sqrt {c x^{2} + b x}}{c} \]

[In]

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

[Out]

-1/2*b*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(3/2) + sqrt(c*x^2 + b*x)/c

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.06 \[ \int \frac {x}{\sqrt {b x+c x^2}} \, dx=\frac {b \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{2 \, c^{\frac {3}{2}}} + \frac {\sqrt {c x^{2} + b x}}{c} \]

[In]

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

[Out]

1/2*b*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/c^(3/2) + sqrt(c*x^2 + b*x)/c

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.98 \[ \int \frac {x}{\sqrt {b x+c x^2}} \, dx=\frac {\sqrt {c\,x^2+b\,x}}{c}-\frac {b\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x}\right )}{2\,c^{3/2}} \]

[In]

int(x/(b*x + c*x^2)^(1/2),x)

[Out]

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