\(\int \frac {b+a x^2}{x \sqrt {-b x+a x^3}} \, dx\) [160]

Optimal result

Integrand size = 25, antiderivative size = 19 \[ \int \frac {b+a x^2}{x \sqrt {-b x+a x^3}} \, dx=\frac {2 \sqrt {-b x+a x^3}}{x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2061} \[ \int \frac {b+a x^2}{x \sqrt {-b x+a x^3}} \, dx=\frac {2 \sqrt {a x^3-b x}}{x} \]

[In]

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

[Out]

(2*Sqrt[-(b*x) + a*x^3])/x

Rule 2061

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Sim
p[c*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(a*(m + j*p + 1))), x] /; FreeQ[{a, b, c, d, e,
j, m, n, p}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] && NeQ[b*c - a*d, 0] && EqQ[a*d*(m + j*p + 1) - b*c*(m + n
+ p*(j + n) + 1), 0] && (GtQ[e, 0] || IntegersQ[j]) && NeQ[m + j*p + 1, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {2 \sqrt {-b x+a x^3}}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {b+a x^2}{x \sqrt {-b x+a x^3}} \, dx=\frac {2 \sqrt {-b x+a x^3}}{x} \]

[In]

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

[Out]

(2*Sqrt[-(b*x) + a*x^3])/x

Maple [A] (verified)

Time = 0.85 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95

method	result	size
trager	\(\frac {2 \sqrt {a \,x^{3}-b x}}{x}\)	\(18\)
gosper	\(\frac {2 a \,x^{2}-2 b}{\sqrt {a \,x^{3}-b x}}\)	\(24\)
default	\(\frac {2 a \,x^{2}-2 b}{\sqrt {x \left (a \,x^{2}-b \right )}}\)	\(25\)
risch	\(\frac {2 a \,x^{2}-2 b}{\sqrt {x \left (a \,x^{2}-b \right )}}\)	\(25\)
elliptic	\(\frac {2 a \,x^{2}-2 b}{\sqrt {x \left (a \,x^{2}-b \right )}}\)	\(25\)
pseudoelliptic	\(\frac {2 a \,x^{2}-2 b}{\sqrt {x \left (a \,x^{2}-b \right )}}\)	\(25\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {b+a x^2}{x \sqrt {-b x+a x^3}} \, dx=\frac {2 \, \sqrt {a x^{3} - b x}}{x} \]

[In]

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

[Out]

2*sqrt(a*x^3 - b*x)/x

Sympy [F]

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

[In]

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

[Out]

Integral((a*x**2 + b)/(x*sqrt(x*(a*x**2 - b))), x)

Maxima [F]

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

[In]

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

[Out]

integrate((a*x^2 + b)/(sqrt(a*x^3 - b*x)*x), x)

Giac [F]

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

[In]

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

[Out]

integrate((a*x^2 + b)/(sqrt(a*x^3 - b*x)*x), x)

Mupad [B] (verification not implemented)

Time = 5.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {b+a x^2}{x \sqrt {-b x+a x^3}} \, dx=\frac {2\,\sqrt {a\,x^3-b\,x}}{x} \]

[In]

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

[Out]

(2*(a*x^3 - b*x)^(1/2))/x