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

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

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1385, 510} \[ -\frac {\sqrt {\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^3}{\sqrt {b^2-4 a c}+b}+1} F_1\left (-\frac {2}{3};\frac {1}{2},\frac {1}{2};\frac {1}{3};-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{2 x^2 \sqrt {a+b x^3+c x^6}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(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, 1/3, (-2*c*x^3)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^3)/(b + Sqrt[b^2 - 4*a*c])])/(2*x^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 {1}{x^3 \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 {1}{x^3 \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 {\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 {1}{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 x^2 \sqrt {a+b x^3+c x^6}}\\ \end {align*}

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Mathematica [B]  time = 0.30, size = 342, normalized size = 2.44 \[ \frac {-2 b x^3 \sqrt {\frac {-\sqrt {b^2-4 a c}+b+2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^3}{\sqrt {b^2-4 a c}+b}} F_1\left (\frac {1}{3};\frac {1}{2},\frac {1}{2};\frac {4}{3};-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{\sqrt {b^2-4 a c}-b}\right )+c x^6 \sqrt {\frac {-\sqrt {b^2-4 a c}+b+2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^3}{\sqrt {b^2-4 a c}+b}} F_1\left (\frac {4}{3};\frac {1}{2},\frac {1}{2};\frac {7}{3};-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{\sqrt {b^2-4 a c}-b}\right )-4 \left (a+b x^3+c x^6\right )}{8 a x^2 \sqrt {a+b x^3+c x^6}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-4*(a + b*x^3 + c*x^6) - 2*b*x^3*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^3)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[(b + Sq
rt[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])] + c*x^6*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[4/3, 1/2, 1/2, 7/3, (-2*c*x^3)/(
b + Sqrt[b^2 - 4*a*c]), (2*c*x^3)/(-b + Sqrt[b^2 - 4*a*c])])/(8*a*x^2*Sqrt[a + b*x^3 + c*x^6])

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fricas [F]  time = 1.12, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{6} + b x^{3} + a}}{c x^{9} + b x^{6} + a x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^6 + b*x^3 + a)/(c*x^9 + b*x^6 + a*x^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {c x^{6} + b x^{3} + a} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {c \,x^{6}+b \,x^{3}+a}\, x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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