3.386 \(\int \frac{1}{x^2 (a+b x^3) \sqrt{c+d x^3}} \, dx\)

Optimal. Leaf size=62 \[ -\frac{\sqrt{\frac{d x^3}{c}+1} F_1\left (-\frac{1}{3};1,\frac{1}{2};\frac{2}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{a x \sqrt{c+d x^3}} \]

[Out]

-((Sqrt[1 + (d*x^3)/c]*AppellF1[-1/3, 1, 1/2, 2/3, -((b*x^3)/a), -((d*x^3)/c)])/(a*x*Sqrt[c + d*x^3]))

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Rubi [A]  time = 0.0534893, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {511, 510} \[ -\frac{\sqrt{\frac{d x^3}{c}+1} F_1\left (-\frac{1}{3};1,\frac{1}{2};\frac{2}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{a x \sqrt{c+d x^3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[1 + (d*x^3)/c]*AppellF1[-1/3, 1, 1/2, 2/3, -((b*x^3)/a), -((d*x^3)/c)])/(a*x*Sqrt[c + d*x^3]))

Rule 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{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])

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\right ) \sqrt{c+d x^3}} \, dx &=\frac{\sqrt{1+\frac{d x^3}{c}} \int \frac{1}{x^2 \left (a+b x^3\right ) \sqrt{1+\frac{d x^3}{c}}} \, dx}{\sqrt{c+d x^3}}\\ &=-\frac{\sqrt{1+\frac{d x^3}{c}} F_1\left (-\frac{1}{3};1,\frac{1}{2};\frac{2}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{a x \sqrt{c+d x^3}}\\ \end{align*}

Mathematica [B]  time = 0.10002, size = 141, normalized size = 2.27 \[ \frac{2 b d x^6 \sqrt{\frac{d x^3}{c}+1} F_1\left (\frac{5}{3};\frac{1}{2},1;\frac{8}{3};-\frac{d x^3}{c},-\frac{b x^3}{a}\right )+5 x^3 \sqrt{\frac{d x^3}{c}+1} (a d-2 b c) F_1\left (\frac{2}{3};\frac{1}{2},1;\frac{5}{3};-\frac{d x^3}{c},-\frac{b x^3}{a}\right )-20 a \left (c+d x^3\right )}{20 a^2 c x \sqrt{c+d x^3}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-20*a*(c + d*x^3) + 5*(-2*b*c + a*d)*x^3*Sqrt[1 + (d*x^3)/c]*AppellF1[2/3, 1/2, 1, 5/3, -((d*x^3)/c), -((b*x^
3)/a)] + 2*b*d*x^6*Sqrt[1 + (d*x^3)/c]*AppellF1[5/3, 1/2, 1, 8/3, -((d*x^3)/c), -((b*x^3)/a)])/(20*a^2*c*x*Sqr
t[c + d*x^3])

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Maple [C]  time = 0.009, size = 890, normalized size = 14.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a*(-(d*x^3+c)^(1/2)/c/x-1/3*I/c*3^(1/2)*(-d^2*c)^(1/3)*(I*(x+1/2/d*(-d^2*c)^(1/3)-1/2*I*3^(1/2)/d*(-d^2*c)^(
1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2)*((x-1/d*(-d^2*c)^(1/3))/(-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^
(1/3)))^(1/2)*(-I*(x+1/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2)/(d*x
^3+c)^(1/2)*((-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*EllipticE(1/3*3^(1/2)*(I*(x+1/2/d*(-d^2*c)
^(1/3)-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2),(I*3^(1/2)/d*(-d^2*c)^(1/3)/(-3/2/d*(-d
^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^(1/2))+1/d*(-d^2*c)^(1/3)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/d*(-d^2
*c)^(1/3)-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2),(I*3^(1/2)/d*(-d^2*c)^(1/3)/(-3/2/d*
(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^(1/2))))+1/3*I*b/a/d^2*2^(1/2)*sum(1/_alpha/(a*d-b*c)*(-d^2*c)
^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)*(d*(x-1/d*(-d^2*c)^
(1/3))/(-3*(-d^2*c)^(1/3)+I*3^(1/2)*(-d^2*c)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*
c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-d^2*c)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-d^2*c)^(2/3)+2
*_alpha^2*d^2-(-d^2*c)^(1/3)*_alpha*d-(-d^2*c)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-d^2*c)^(1/3)-1/2*I*
3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2),1/2*b/d*(2*I*(-d^2*c)^(1/3)*3^(1/2)*_alpha^2*d-I*(-d
^2*c)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-d^2*c)^(2/3)*_alpha-3*c*d)/(a*d-b*c),(I*3^(1/2)/d*(-d^2*c)^(1/3)/
(-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*b+a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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}\right ) \sqrt{c + d x^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x**2*(a + b*x**3)*sqrt(c + d*x**3)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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