\(\int \frac {1}{x^2 \sqrt {a x+b x^4}} \, dx\) [94]

Optimal result

Integrand size = 17, antiderivative size = 23 \[ \int \frac {1}{x^2 \sqrt {a x+b x^4}} \, dx=-\frac {2 \sqrt {a x+b x^4}}{3 a x^2} \]

[Out]

-2/3*(b*x^4+a*x)^(1/2)/a/x^2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 23, 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 = {2039} \[ \int \frac {1}{x^2 \sqrt {a x+b x^4}} \, dx=-\frac {2 \sqrt {a x+b x^4}}{3 a x^2} \]

[In]

Int[1/(x^2*Sqrt[a*x + b*x^4]),x]

[Out]

(-2*Sqrt[a*x + b*x^4])/(3*a*x^2)

Rule 2039

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \sqrt {a x+b x^4}}{3 a x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^2 \sqrt {a x+b x^4}} \, dx=-\frac {2 \sqrt {x \left (a+b x^3\right )}}{3 a x^2} \]

[In]

Integrate[1/(x^2*Sqrt[a*x + b*x^4]),x]

[Out]

(-2*Sqrt[x*(a + b*x^3)])/(3*a*x^2)

Maple [A] (verified)

Time = 2.41 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87

method	result	size
default	\(-\frac {2 \sqrt {b \,x^{4}+a x}}{3 a \,x^{2}}\)	\(20\)
trager	\(-\frac {2 \sqrt {b \,x^{4}+a x}}{3 a \,x^{2}}\)	\(20\)
elliptic	\(-\frac {2 \sqrt {b \,x^{4}+a x}}{3 a \,x^{2}}\)	\(20\)
pseudoelliptic	\(-\frac {2 \sqrt {x \left (b \,x^{3}+a \right )}}{3 a \,x^{2}}\)	\(20\)
gosper	\(-\frac {2 \left (b \,x^{3}+a \right )}{3 x a \sqrt {b \,x^{4}+a x}}\)	\(27\)
risch	\(-\frac {2 \left (b \,x^{3}+a \right )}{3 a x \sqrt {x \left (b \,x^{3}+a \right )}}\)	\(27\)

[In]

int(1/x^2/(b*x^4+a*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-2/3*(b*x^4+a*x)^(1/2)/a/x^2

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^2 \sqrt {a x+b x^4}} \, dx=-\frac {2 \, \sqrt {b x^{4} + a x}}{3 \, a x^{2}} \]

[In]

integrate(1/x^2/(b*x^4+a*x)^(1/2),x, algorithm="fricas")

[Out]

-2/3*sqrt(b*x^4 + a*x)/(a*x^2)

Sympy [F]

\[ \int \frac {1}{x^2 \sqrt {a x+b x^4}} \, dx=\int \frac {1}{x^{2} \sqrt {x \left (a + b x^{3}\right )}}\, dx \]

[In]

integrate(1/x**2/(b*x**4+a*x)**(1/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {1}{x^2 \sqrt {a x+b x^4}} \, dx=-\frac {2 \, {\left (b x^{4} + a x\right )}}{3 \, \sqrt {b x^{3} + a} a x^{\frac {5}{2}}} \]

[In]

integrate(1/x^2/(b*x^4+a*x)^(1/2),x, algorithm="maxima")

[Out]

-2/3*(b*x^4 + a*x)/(sqrt(b*x^3 + a)*a*x^(5/2))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.61 \[ \int \frac {1}{x^2 \sqrt {a x+b x^4}} \, dx=-\frac {2 \, \sqrt {b + \frac {a}{x^{3}}}}{3 \, a} \]

[In]

integrate(1/x^2/(b*x^4+a*x)^(1/2),x, algorithm="giac")

[Out]

-2/3*sqrt(b + a/x^3)/a

Mupad [B] (verification not implemented)

Time = 9.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^2 \sqrt {a x+b x^4}} \, dx=-\frac {2\,\sqrt {b\,x^4+a\,x}}{3\,a\,x^2} \]

[In]

int(1/(x^2*(a*x + b*x^4)^(1/2)),x)

[Out]

-(2*(a*x + b*x^4)^(1/2))/(3*a*x^2)