\(\int \frac {1}{\sqrt {a x^2+b x^3}} \, dx\) [256]

Optimal result

Integrand size = 15, antiderivative size = 30 \[ \int \frac {1}{\sqrt {a x^2+b x^3}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{\sqrt {a}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2033, 212} \[ \int \frac {1}{\sqrt {a x^2+b x^3}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{\sqrt {a}} \]

[In]

Int[1/Sqrt[a*x^2 + b*x^3],x]

[Out]

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

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 2033

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.53 \[ \int \frac {1}{\sqrt {a x^2+b x^3}} \, dx=-\frac {2 x \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {x^2 (a+b x)}} \]

[In]

Integrate[1/Sqrt[a*x^2 + b*x^3],x]

[Out]

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

Maple [A] (verified)

Time = 1.83 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.43

method	result	size
pseudoelliptic	\(\frac {2 \sqrt {b x +a}}{b}\)	\(13\)
default	\(-\frac {2 x \sqrt {b x +a}\, \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {b \,x^{3}+a \,x^{2}}\, \sqrt {a}}\)	\(39\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 74, normalized size of antiderivative = 2.47 \[ \int \frac {1}{\sqrt {a x^2+b x^3}} \, dx=\left [\frac {\log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right )}{\sqrt {a}}, \frac {2 \, \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{a x}\right )}{a}\right ] \]

[In]

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

[Out]

[log((b*x^2 + 2*a*x - 2*sqrt(b*x^3 + a*x^2)*sqrt(a))/x^2)/sqrt(a), 2*sqrt(-a)*arctan(sqrt(b*x^3 + a*x^2)*sqrt(
-a)/(a*x))/a]

Sympy [F]

\[ \int \frac {1}{\sqrt {a x^2+b x^3}} \, dx=\int \frac {1}{\sqrt {a x^{2} + b x^{3}}}\, dx \]

[In]

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

[Out]

Integral(1/sqrt(a*x**2 + b*x**3), x)

Maxima [F]

\[ \int \frac {1}{\sqrt {a x^2+b x^3}} \, dx=\int { \frac {1}{\sqrt {b x^{3} + a x^{2}}} \,d x } \]

[In]

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

[Out]

integrate(1/sqrt(b*x^3 + a*x^2), x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.50 \[ \int \frac {1}{\sqrt {a x^2+b x^3}} \, dx=-\frac {2 \, \arctan \left (\frac {\sqrt {a}}{\sqrt {-a}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-a}} + \frac {2 \, \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {a x^2+b x^3}} \, dx=\int \frac {1}{\sqrt {b\,x^3+a\,x^2}} \,d x \]

[In]

int(1/(a*x^2 + b*x^3)^(1/2),x)

[Out]

int(1/(a*x^2 + b*x^3)^(1/2), x)