3.245 \(\int \frac {1}{(a+b x^3+c x^6)^{3/2}} \, dx\)

Optimal. Leaf size=138 \[ \frac {x \sqrt {\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^3}{\sqrt {b^2-4 a c}+b}+1} F_1\left (\frac {1}{3};\frac {3}{2},\frac {3}{2};\frac {4}{3};-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{a \sqrt {a+b x^3+c x^6}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1348, 429} \[ \frac {x \sqrt {\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^3}{\sqrt {b^2-4 a c}+b}+1} F_1\left (\frac {1}{3};\frac {3}{2},\frac {3}{2};\frac {4}{3};-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{a \sqrt {a+b x^3+c x^6}} \]

Antiderivative was successfully verified.

[In]

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

Rule 429

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 1348

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]))^F
racPart[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*} \int \frac {1}{\left (a+b x^3+c x^6\right )^{3/2}} \, 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}{\left (1+\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}\right )^{3/2} \left (1+\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )^{3/2}} \, dx}{a \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 {3}{2},\frac {3}{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 )}{a \sqrt {a+b x^3+c x^6}}\\ \end {align*}

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Mathematica [B]  time = 0.47, size = 359, normalized size = 2.60 \[ \frac {x \left (b c x^3 \sqrt {\frac {-\sqrt {b^2-4 a c}+b+2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^3}{\sqrt {b^2-4 a c}+b}} F_1\left (\frac {4}{3};\frac {1}{2},\frac {1}{2};\frac {7}{3};-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{\sqrt {b^2-4 a c}-b}\right )-2 \left (b^2-8 a c\right ) \sqrt {\frac {-\sqrt {b^2-4 a c}+b+2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^3}{\sqrt {b^2-4 a c}+b}} F_1\left (\frac {1}{3};\frac {1}{2},\frac {1}{2};\frac {4}{3};-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{\sqrt {b^2-4 a c}-b}\right )-4 \left (-2 a c+b^2+b c x^3\right )\right )}{6 a \left (4 a c-b^2\right ) \sqrt {a+b x^3+c x^6}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*x^3 + c*x^6)^(-3/2),x]

[Out]

(x*(-4*(b^2 - 2*a*c + b*c*x^3) - 2*(b^2 - 8*a*c)*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])] + b*c*x^3*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])]))/(6*a*(-b^2 + 4*a*c)*Sqrt[a + b*
x^3 + c*x^6])

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fricas [F]  time = 0.87, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{6} + b x^{3} + a}}{c^{2} x^{12} + 2 \, b c x^{9} + {\left (b^{2} + 2 \, a c\right )} x^{6} + 2 \, a b x^{3} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^6 + b*x^3 + a)/(c^2*x^12 + 2*b*c*x^9 + (b^2 + 2*a*c)*x^6 + 2*a*b*x^3 + a^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (c x^{6} + b x^{3} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (c \,x^{6}+b \,x^{3}+a \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (c x^{6} + b x^{3} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (c\,x^6+b\,x^3+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*x^3 + c*x^6)^(3/2),x)

[Out]

int(1/(a + b*x^3 + c*x^6)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b x^{3} + c x^{6}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*x**3 + c*x**6)**(-3/2), x)

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