\(\int \frac {1}{x^4 \sqrt {a+b x^3+c x^6}} \, dx\) [225]

Optimal result

Integrand size = 20, antiderivative size = 72 \[ \int \frac {1}{x^4 \sqrt {a+b x^3+c x^6}} \, dx=-\frac {\sqrt {a+b x^3+c x^6}}{3 a x^3}+\frac {b \text {arctanh}\left (\frac {2 a+b x^3}{2 \sqrt {a} \sqrt {a+b x^3+c x^6}}\right )}{6 a^{3/2}} \]

[Out]

1/6*b*arctanh(1/2*(b*x^3+2*a)/a^(1/2)/(c*x^6+b*x^3+a)^(1/2))/a^(3/2)-1/3*(c*x^6+b*x^3+a)^(1/2)/a/x^3

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1371, 744, 738, 212} \[ \int \frac {1}{x^4 \sqrt {a+b x^3+c x^6}} \, dx=\frac {b \text {arctanh}\left (\frac {2 a+b x^3}{2 \sqrt {a} \sqrt {a+b x^3+c x^6}}\right )}{6 a^{3/2}}-\frac {\sqrt {a+b x^3+c x^6}}{3 a x^3} \]

[In]

Int[1/(x^4*Sqrt[a + b*x^3 + c*x^6]),x]

[Out]

-1/3*Sqrt[a + b*x^3 + c*x^6]/(a*x^3) + (b*ArcTanh[(2*a + b*x^3)/(2*Sqrt[a]*Sqrt[a + b*x^3 + c*x^6])])/(6*a^(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 738

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 744

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m + 1)*
((a + b*x + c*x^2)^(p + 1)/((m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Dist[(2*c*d - b*e)/(2*(c*d^2 - b*d*e + a*e
^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c,
0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && EqQ[m + 2*p + 3, 0]

Rule 1371

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[
b^2 - 4*a*c, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x+c x^2}} \, dx,x,x^3\right ) \\ & = -\frac {\sqrt {a+b x^3+c x^6}}{3 a x^3}-\frac {b \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx,x,x^3\right )}{6 a} \\ & = -\frac {\sqrt {a+b x^3+c x^6}}{3 a x^3}+\frac {b \text {Subst}\left (\int \frac {1}{4 a-x^2} \, dx,x,\frac {2 a+b x^3}{\sqrt {a+b x^3+c x^6}}\right )}{3 a} \\ & = -\frac {\sqrt {a+b x^3+c x^6}}{3 a x^3}+\frac {b \tanh ^{-1}\left (\frac {2 a+b x^3}{2 \sqrt {a} \sqrt {a+b x^3+c x^6}}\right )}{6 a^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^4 \sqrt {a+b x^3+c x^6}} \, dx=-\frac {\sqrt {a+b x^3+c x^6}}{3 a x^3}-\frac {b \text {arctanh}\left (\frac {\sqrt {c} x^3-\sqrt {a+b x^3+c x^6}}{\sqrt {a}}\right )}{3 a^{3/2}} \]

[In]

Integrate[1/(x^4*Sqrt[a + b*x^3 + c*x^6]),x]

[Out]

-1/3*Sqrt[a + b*x^3 + c*x^6]/(a*x^3) - (b*ArcTanh[(Sqrt[c]*x^3 - Sqrt[a + b*x^3 + c*x^6])/Sqrt[a]])/(3*a^(3/2)
)

Maple [F]

\[\int \frac {1}{x^{4} \sqrt {c \,x^{6}+b \,x^{3}+a}}d x\]

[In]

int(1/x^4/(c*x^6+b*x^3+a)^(1/2),x)

[Out]

int(1/x^4/(c*x^6+b*x^3+a)^(1/2),x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 179, normalized size of antiderivative = 2.49 \[ \int \frac {1}{x^4 \sqrt {a+b x^3+c x^6}} \, dx=\left [\frac {\sqrt {a} b x^{3} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{6} + 8 \, a b x^{3} + 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left (b x^{3} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{6}}\right ) - 4 \, \sqrt {c x^{6} + b x^{3} + a} a}{12 \, a^{2} x^{3}}, -\frac {\sqrt {-a} b x^{3} \arctan \left (\frac {\sqrt {c x^{6} + b x^{3} + a} {\left (b x^{3} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{6} + a b x^{3} + a^{2}\right )}}\right ) + 2 \, \sqrt {c x^{6} + b x^{3} + a} a}{6 \, a^{2} x^{3}}\right ] \]

[In]

integrate(1/x^4/(c*x^6+b*x^3+a)^(1/2),x, algorithm="fricas")

[Out]

[1/12*(sqrt(a)*b*x^3*log(-((b^2 + 4*a*c)*x^6 + 8*a*b*x^3 + 4*sqrt(c*x^6 + b*x^3 + a)*(b*x^3 + 2*a)*sqrt(a) + 8
*a^2)/x^6) - 4*sqrt(c*x^6 + b*x^3 + a)*a)/(a^2*x^3), -1/6*(sqrt(-a)*b*x^3*arctan(1/2*sqrt(c*x^6 + b*x^3 + a)*(
b*x^3 + 2*a)*sqrt(-a)/(a*c*x^6 + a*b*x^3 + a^2)) + 2*sqrt(c*x^6 + b*x^3 + a)*a)/(a^2*x^3)]

Sympy [F]

\[ \int \frac {1}{x^4 \sqrt {a+b x^3+c x^6}} \, dx=\int \frac {1}{x^{4} \sqrt {a + b x^{3} + c x^{6}}}\, dx \]

[In]

integrate(1/x**4/(c*x**6+b*x**3+a)**(1/2),x)

[Out]

Integral(1/(x**4*sqrt(a + b*x**3 + c*x**6)), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{x^4 \sqrt {a+b x^3+c x^6}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/x^4/(c*x^6+b*x^3+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta

Giac [F]

\[ \int \frac {1}{x^4 \sqrt {a+b x^3+c x^6}} \, dx=\int { \frac {1}{\sqrt {c x^{6} + b x^{3} + a} x^{4}} \,d x } \]

[In]

integrate(1/x^4/(c*x^6+b*x^3+a)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(c*x^6 + b*x^3 + a)*x^4), x)

Mupad [B] (verification not implemented)

Time = 8.51 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x^4 \sqrt {a+b x^3+c x^6}} \, dx=\frac {b\,\mathrm {atanh}\left (\frac {\frac {b\,x^3}{2}+a}{\sqrt {a}\,\sqrt {c\,x^6+b\,x^3+a}}\right )}{6\,a^{3/2}}-\frac {\sqrt {c\,x^6+b\,x^3+a}}{3\,a\,x^3} \]

[In]

int(1/(x^4*(a + b*x^3 + c*x^6)^(1/2)),x)

[Out]

(b*atanh((a + (b*x^3)/2)/(a^(1/2)*(a + b*x^3 + c*x^6)^(1/2))))/(6*a^(3/2)) - (a + b*x^3 + c*x^6)^(1/2)/(3*a*x^
3)