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*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])

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Rubi [A]  time = 0.0672494, antiderivative size = 138, normalized size of antiderivative = 1., 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 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]

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])

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*}

Mathematica [B]  time = 0.468416, size = 359, normalized size = 2.6 \[ \frac{x \left (b c x^3 \sqrt{\frac{-\sqrt{b^2-4 a c}+b+2 c x^3}{b-\sqrt{b^2-4 a c}}} \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^3}{\sqrt{b^2-4 a c}+b}} F_1\left (\frac{4}{3};\frac{1}{2},\frac{1}{2};\frac{7}{3};-\frac{2 c x^3}{b+\sqrt{b^2-4 a c}},\frac{2 c x^3}{\sqrt{b^2-4 a c}-b}\right )-2 \left (b^2-8 a c\right ) \sqrt{\frac{-\sqrt{b^2-4 a c}+b+2 c x^3}{b-\sqrt{b^2-4 a c}}} \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^3}{\sqrt{b^2-4 a c}+b}} F_1\left (\frac{1}{3};\frac{1}{2},\frac{1}{2};\frac{4}{3};-\frac{2 c x^3}{b+\sqrt{b^2-4 a c}},\frac{2 c x^3}{\sqrt{b^2-4 a c}-b}\right )-4 \left (-2 a c+b^2+b c x^3\right )\right )}{6 a \left (4 a c-b^2\right ) \sqrt{a+b x^3+c x^6}} \]

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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Maple [F]  time = 0.013, size = 0, normalized size = 0. \begin{align*} \int \left ( c{x}^{6}+b{x}^{3}+a \right ) ^{-{\frac{3}{2}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{6} + b x^{3} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b x^{3} + c x^{6}\right )^{\frac{3}{2}}}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{6} + b x^{3} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

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)