\(\int \frac {1}{\sqrt {c x (a+b x)}} \, dx\) [993]

Optimal result

Integrand size = 12, antiderivative size = 40 \[ \int \frac {1}{\sqrt {c x (a+b x)}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c} x}{\sqrt {a c x+b c x^2}}\right )}{\sqrt {b} \sqrt {c}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1976, 634, 212} \[ \int \frac {1}{\sqrt {c x (a+b x)}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c} x}{\sqrt {a c x+b c x^2}}\right )}{\sqrt {b} \sqrt {c}} \]

[In]

Int[1/Sqrt[c*x*(a + b*x)],x]

[Out]

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

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 1976

Int[(u_.)*((e_.)*((a_.) + (b_.)*(x_)^(n_.))*((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> Int[u*(a*c*e + (b*c
+ a*d)*e*x^n + b*d*e*x^(2*n))^p, x] /; FreeQ[{a, b, c, d, e, n, p}, x]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.40 \[ \int \frac {1}{\sqrt {c x (a+b x)}} \, dx=-\frac {2 \sqrt {x} \sqrt {a+b x} \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right )}{\sqrt {b} \sqrt {c x (a+b x)}} \]

[In]

Integrate[1/Sqrt[c*x*(a + b*x)],x]

[Out]

(-2*Sqrt[x]*Sqrt[a + b*x]*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[a + b*x]])/(Sqrt[b]*Sqrt[c*x*(a + b*x)])

Maple [A] (verified)

Time = 1.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.70

method	result	size
pseudoelliptic	\(\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {c x \left (b x +a \right )}}{x \sqrt {b c}}\right )}{\sqrt {b c}}\)	\(28\)
default	\(\frac {\ln \left (\frac {\frac {1}{2} a c +b c x}{\sqrt {b c}}+\sqrt {b c \,x^{2}+a c x}\right )}{\sqrt {b c}}\)	\(37\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 87, normalized size of antiderivative = 2.18 \[ \int \frac {1}{\sqrt {c x (a+b x)}} \, dx=\left [\frac {\sqrt {b c} \log \left (2 \, b c x + a c + 2 \, \sqrt {b c x^{2} + a c x} \sqrt {b c}\right )}{b c}, -\frac {2 \, \sqrt {-b c} \arctan \left (\frac {\sqrt {b c x^{2} + a c x} \sqrt {-b c}}{b c x}\right )}{b c}\right ] \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 97 vs. \(2 (39) = 78\).

Time = 0.71 (sec) , antiderivative size = 97, normalized size of antiderivative = 2.42 \[ \int \frac {1}{\sqrt {c x (a+b x)}} \, dx=\begin {cases} \frac {\log {\left (a c + 2 b c x + 2 \sqrt {b c} \sqrt {a c x + b c x^{2}} \right )}}{\sqrt {b c}} & \text {for}\: b c \neq 0 \wedge \frac {a^{2} c}{b} \neq 0 \\\frac {\left (\frac {a}{2 b} + x\right ) \log {\left (\frac {a}{2 b} + x \right )}}{\sqrt {b c \left (\frac {a}{2 b} + x\right )^{2}}} & \text {for}\: b c \neq 0 \\\frac {2 \sqrt {a c x}}{a c} & \text {for}\: a c \neq 0 \\\tilde {\infty } x & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((log(a*c + 2*b*c*x + 2*sqrt(b*c)*sqrt(a*c*x + b*c*x**2))/sqrt(b*c), Ne(b*c, 0) & Ne(a**2*c/b, 0)), (
(a/(2*b) + x)*log(a/(2*b) + x)/sqrt(b*c*(a/(2*b) + x)**2), Ne(b*c, 0)), (2*sqrt(a*c*x)/(a*c), Ne(a*c, 0)), (zo
o*x, True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.90 \[ \int \frac {1}{\sqrt {c x (a+b x)}} \, dx=\frac {\log \left (2 \, b c x + a c + 2 \, \sqrt {b c x^{2} + a c x} \sqrt {b c}\right )}{\sqrt {b c}} \]

[In]

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

[Out]

log(2*b*c*x + a*c + 2*sqrt(b*c*x^2 + a*c*x)*sqrt(b*c))/sqrt(b*c)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 76 vs. \(2 (30) = 60\).

Time = 0.30 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.90 \[ \int \frac {1}{\sqrt {c x (a+b x)}} \, dx=\frac {a^{2} c \log \left ({\left | -a c - 2 \, \sqrt {b c} {\left (\sqrt {b c} x - \sqrt {b c x^{2} + a c x}\right )} \right |}\right )}{8 \, \sqrt {b c} b} + \frac {1}{4} \, \sqrt {b c x^{2} + a c x} {\left (2 \, x + \frac {a}{b}\right )} \]

[In]

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

[Out]

1/8*a^2*c*log(abs(-a*c - 2*sqrt(b*c)*(sqrt(b*c)*x - sqrt(b*c*x^2 + a*c*x))))/(sqrt(b*c)*b) + 1/4*sqrt(b*c*x^2
+ a*c*x)*(2*x + a/b)

Mupad [B] (verification not implemented)

Time = 22.70 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.82 \[ \int \frac {1}{\sqrt {c x (a+b x)}} \, dx=\frac {\ln \left (a\,c+2\,\sqrt {b\,c}\,\sqrt {c\,x\,\left (a+b\,x\right )}+2\,b\,c\,x\right )}{\sqrt {b\,c}} \]

[In]

int(1/(c*x*(a + b*x))^(1/2),x)

[Out]

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