\(\int \frac {1}{\sqrt {a+b x^3+c x^6}} \, dx\) [231]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 16, antiderivative size = 135 \[ \int \frac {1}{\sqrt {a+b x^3+c x^6}} \, dx=\frac {x \sqrt {1+\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},\frac {1}{2},\frac {4}{3},-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {a+b x^3+c x^6}} \]

[Out]

x*AppellF1(1/3,1/2,1/2,4/3,-2*c*x^3/(b-(-4*a*c+b^2)^(1/2)),-2*c*x^3/(b+(-4*a*c+b^2)^(1/2)))*(1+2*c*x^3/(b-(-4*
a*c+b^2)^(1/2)))^(1/2)*(1+2*c*x^3/(b+(-4*a*c+b^2)^(1/2)))^(1/2)/(c*x^6+b*x^3+a)^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1362, 440} \[ \int \frac {1}{\sqrt {a+b x^3+c x^6}} \, dx=\frac {x \sqrt {\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^3}{\sqrt {b^2-4 a c}+b}+1} \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},\frac {1}{2},\frac {4}{3},-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {a+b x^3+c x^6}} \]

[In]

Int[1/Sqrt[a + b*x^3 + c*x^6],x]

[Out]

(x*Sqrt[1 + (2*c*x^3)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^3)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[1/3, 1/2,
1/2, 4/3, (-2*c*x^3)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^3)/(b + Sqrt[b^2 - 4*a*c])])/Sqrt[a + b*x^3 + c*x^6]

Rule 440

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
 -q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 1362

Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n + c*x^(2*n))
^FracPart[p]/((1 + 2*c*(x^n/(b + Rt[b^2 - 4*a*c, 2])))^FracPart[p]*(1 + 2*c*(x^n/(b - Rt[b^2 - 4*a*c, 2])))^Fr
acPart[p])), Int[(1 + 2*c*(x^n/(b + Sqrt[b^2 - 4*a*c])))^p*(1 + 2*c*(x^n/(b - Sqrt[b^2 - 4*a*c])))^p, x], x] /
; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {1+\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}}\right ) \int \frac {1}{\sqrt {1+\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}}} \, dx}{\sqrt {a+b x^3+c x^6}} \\ & = \frac {x \sqrt {1+\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}} F_1\left (\frac {1}{3};\frac {1}{2},\frac {1}{2};\frac {4}{3};-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {a+b x^3+c x^6}} \\ \end{align*}

Mathematica [A] (verified)

Time = 10.05 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.21 \[ \int \frac {1}{\sqrt {a+b x^3+c x^6}} \, dx=\frac {x \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^3}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},\frac {1}{2},\frac {4}{3},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{-b+\sqrt {b^2-4 a c}}\right )}{\sqrt {a+b x^3+c x^6}} \]

[In]

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

[Out]

(x*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^3)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^3)/(b +
 Sqrt[b^2 - 4*a*c])]*AppellF1[1/3, 1/2, 1/2, 4/3, (-2*c*x^3)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^3)/(-b + Sqrt[b^2
 - 4*a*c])])/Sqrt[a + b*x^3 + c*x^6]

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/(a + b*x^3 + c*x^6)^(1/2),x)

[Out]

int(1/(a + b*x^3 + c*x^6)^(1/2), x)