3.230 \(\int \frac {x}{\sqrt {a+b x^3+c x^6}} \, dx\)

Optimal. Leaf size=140 \[ \frac {x^2 \sqrt {\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^3}{\sqrt {b^2-4 a c}+b}+1} F_1\left (\frac {2}{3};\frac {1}{2},\frac {1}{2};\frac {5}{3};-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{2 \sqrt {a+b x^3+c x^6}} \]

[Out]

1/2*x^2*AppellF1(2/3,1/2,1/2,5/3,-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)

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Rubi [A]  time = 0.09, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1385, 510} \[ \frac {x^2 \sqrt {\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^3}{\sqrt {b^2-4 a c}+b}+1} F_1\left (\frac {2}{3};\frac {1}{2},\frac {1}{2};\frac {5}{3};-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{2 \sqrt {a+b x^3+c x^6}} \]

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[a + b*x^3 + c*x^6],x]

[Out]

(x^2*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[2/3, 1/2
, 1/2, 5/3, (-2*c*x^3)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^3)/(b + Sqrt[b^2 - 4*a*c])])/(2*Sqrt[a + b*x^3 + c*x^6
])

Rule 510

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

Rule 1385

Int[((d_.)*(x_))^(m_.)*((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]))^FracPart[p]), Int[(d*x)^m*(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, d, m, n, p}, x] && EqQ[n2, 2*n]

Rubi steps

\begin {align*} \int \frac {x}{\sqrt {a+b x^3+c x^6}} \, dx &=\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 {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}}}} \, dx}{\sqrt {a+b x^3+c x^6}}\\ &=\frac {x^2 \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 {2}{3};\frac {1}{2},\frac {1}{2};\frac {5}{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 )}{2 \sqrt {a+b x^3+c x^6}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 168, normalized size = 1.20 \[ \frac {x^2 \sqrt {\frac {-\sqrt {b^2-4 a c}+b+2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^3}{\sqrt {b^2-4 a c}+b}} F_1\left (\frac {2}{3};\frac {1}{2},\frac {1}{2};\frac {5}{3};-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{\sqrt {b^2-4 a c}-b}\right )}{2 \sqrt {a+b x^3+c x^6}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x^2*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[2/3, 1/2, 1/2, 5/3, (-2*c*x^3)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^3)/(-b + Sqrt[b
^2 - 4*a*c])])/(2*Sqrt[a + b*x^3 + c*x^6])

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fricas [F]  time = 1.31, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x}{\sqrt {c x^{6} + b x^{3} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\sqrt {c x^{6} + b x^{3} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[ \int \frac {x}{\sqrt {c \,x^{6}+b \,x^{3}+a}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\sqrt {c x^{6} + b x^{3} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x}{\sqrt {c\,x^6+b\,x^3+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*x^3 + c*x^6)^(1/2),x)

[Out]

int(x/(a + b*x^3 + c*x^6)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\sqrt {a + b x^{3} + c x^{6}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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