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

Optimal. Leaf size=141 \[ -\frac{\sqrt{\frac{2 c x^3}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^3}{\sqrt{b^2-4 a c}+b}+1} F_1\left (-\frac{1}{3};\frac{3}{2},\frac{3}{2};\frac{2}{3};-\frac{2 c x^3}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^3}{b+\sqrt{b^2-4 a c}}\right )}{a x \sqrt{a+b x^3+c x^6}} \]

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

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Rubi [A]  time = 0.127745, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {1385, 510} \[ -\frac{\sqrt{\frac{2 c x^3}{b-\sqrt{b^2-4 a c}}+1} \sqrt{\frac{2 c x^3}{\sqrt{b^2-4 a c}+b}+1} F_1\left (-\frac{1}{3};\frac{3}{2},\frac{3}{2};\frac{2}{3};-\frac{2 c x^3}{b-\sqrt{b^2-4 a c}},-\frac{2 c x^3}{b+\sqrt{b^2-4 a c}}\right )}{a x \sqrt{a+b x^3+c x^6}} \]

Antiderivative was successfully verified.

[In]

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

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]

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])

Rubi steps

\begin{align*} \int \frac{1}{x^2 \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}{x^2 \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{\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{2}{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 x \sqrt{a+b x^3+c x^6}}\\ \end{align*}

Mathematica [B]  time = 0.761432, size = 407, normalized size = 2.89 \[ -\frac{5 b x^3 \left (12 a c-5 b^2\right ) \sqrt{\frac{-\sqrt{b^2-4 a c}+b+2 c x^3}{b-\sqrt{b^2-4 a c}}} \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^3}{\sqrt{b^2-4 a c}+b}} F_1\left (\frac{2}{3};\frac{1}{2},\frac{1}{2};\frac{5}{3};-\frac{2 c x^3}{b+\sqrt{b^2-4 a c}},\frac{2 c x^3}{\sqrt{b^2-4 a c}-b}\right )-4 \left (60 a^2 c+2 c x^6 \left (5 b^2-16 a c\right ) \sqrt{\frac{-\sqrt{b^2-4 a c}+b+2 c x^3}{b-\sqrt{b^2-4 a c}}} \sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x^3}{\sqrt{b^2-4 a c}+b}} F_1\left (\frac{5}{3};\frac{1}{2},\frac{1}{2};\frac{8}{3};-\frac{2 c x^3}{b+\sqrt{b^2-4 a c}},\frac{2 c x^3}{\sqrt{b^2-4 a c}-b}\right )+5 a \left (-3 b^2+18 b c x^3+16 c^2 x^6\right )-25 b^2 x^3 \left (b+c x^3\right )\right )}{60 a^2 x \left (b^2-4 a c\right ) \sqrt{a+b x^3+c x^6}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(x^2*(a + b*x^3 + c*x^6)^(3/2)),x]

[Out]

-(5*b*(-5*b^2 + 12*a*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[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])] - 4*(60*a^2*c - 25*b^2*x^3*(b + c*x^3) + 5*a*(-3*b^2 + 18*b*c*x^3 + 16*c^
2*x^6) + 2*c*(5*b^2 - 16*a*c)*x^6*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[5/3, 1/2, 1/2, 8/3, (-2*c*x^3)/(b + Sqrt[b^2 - 4*
a*c]), (2*c*x^3)/(-b + Sqrt[b^2 - 4*a*c])]))/(60*a^2*(b^2 - 4*a*c)*x*Sqrt[a + b*x^3 + c*x^6])

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Maple [F]  time = 0.044, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{2}} \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/x^2/(c*x^6+b*x^3+a)^(3/2),x)

[Out]

int(1/x^2/(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}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((c*x^6 + b*x^3 + a)^(3/2)*x^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^{14} + 2 \, b c x^{11} +{\left (b^{2} + 2 \, a c\right )} x^{8} + 2 \, a b x^{5} + a^{2} x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(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^14 + 2*b*c*x^11 + (b^2 + 2*a*c)*x^8 + 2*a*b*x^5 + a^2*x^2), x)

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

[Out]

Integral(1/(x**2*(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}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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