3.396 \(\int \frac{1}{x^2 (a+b x^3) (c+d x^3)^{3/2}} \, dx\)
Optimal. Leaf size=65 \[ -\frac{\sqrt{\frac{d x^3}{c}+1} F_1\left (-\frac{1}{3};1,\frac{3}{2};\frac{2}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{a c x \sqrt{c+d x^3}} \]
[Out]
-((Sqrt[1 + (d*x^3)/c]*AppellF1[-1/3, 1, 3/2, 2/3, -((b*x^3)/a), -((d*x^3)/c)])/(a*c*x*Sqrt[c + d*x^3]))
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Rubi [A] time = 0.0589631, antiderivative size = 65, 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} \[ -\frac{\sqrt{\frac{d x^3}{c}+1} F_1\left (-\frac{1}{3};1,\frac{3}{2};\frac{2}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{a c x \sqrt{c+d x^3}} \]
Antiderivative was successfully verified.
[In]
Int[1/(x^2*(a + b*x^3)*(c + d*x^3)^(3/2)),x]
[Out]
-((Sqrt[1 + (d*x^3)/c]*AppellF1[-1/3, 1, 3/2, 2/3, -((b*x^3)/a), -((d*x^3)/c)])/(a*c*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 ) \left (c+d x^3\right )^{3/2}} \, dx &=\frac{\sqrt{1+\frac{d x^3}{c}} \int \frac{1}{x^2 \left (a+b x^3\right ) \left (1+\frac{d x^3}{c}\right )^{3/2}} \, dx}{c \sqrt{c+d x^3}}\\ &=-\frac{\sqrt{1+\frac{d x^3}{c}} F_1\left (-\frac{1}{3};1,\frac{3}{2};\frac{2}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{a c x \sqrt{c+d x^3}}\\ \end{align*}
Mathematica [B] time = 0.196562, size = 193, normalized size = 2.97 \[ \frac{-5 x^3 \sqrt{\frac{d x^3}{c}+1} \left (5 a^2 d^2-3 a b c d+6 b^2 c^2\right ) F_1\left (\frac{2}{3};\frac{1}{2},1;\frac{5}{3};-\frac{d x^3}{c},-\frac{b x^3}{a}\right )+2 b d x^6 \sqrt{\frac{d x^3}{c}+1} (3 b c-5 a d) F_1\left (\frac{5}{3};\frac{1}{2},1;\frac{8}{3};-\frac{d x^3}{c},-\frac{b x^3}{a}\right )+20 a \left (a d \left (3 c+5 d x^3\right )-3 b c \left (c+d x^3\right )\right )}{60 a^2 c^2 x \sqrt{c+d x^3} (b c-a d)} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[1/(x^2*(a + b*x^3)*(c + d*x^3)^(3/2)),x]
[Out]
(20*a*(-3*b*c*(c + d*x^3) + a*d*(3*c + 5*d*x^3)) - 5*(6*b^2*c^2 - 3*a*b*c*d + 5*a^2*d^2)*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*(3*b*c - 5*a*d)*x^6*Sqrt[1 + (d*x^3)/c]*Appe
llF1[5/3, 1/2, 1, 8/3, -((d*x^3)/c), -((b*x^3)/a)])/(60*a^2*c^2*(b*c - a*d)*x*Sqrt[c + d*x^3])
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Maple [C] time = 0.008, size = 1392, normalized size = 21.4 \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)^(3/2),x)
[Out]
1/a*(-(d*x^3+c)^(1/2)/c^2/x-2/3*d/c^2*x^2/((x^3+1/d*c)*d)^(1/2)-5/9*I/c^2*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))*El
lipticE(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))))-b/a*(2/3*d/c*
x^2/(a*d-b*c)/((x^3+1/d*c)*d)^(1/2)+2/9*I/c/(a*d-b*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/d^2*b*2^(1/2)*sum(1/(a*d-b*c
)^2/_alpha*(-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 )}{\left (d x^{3} + c\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/(b*x^3+a)/(d*x^3+c)^(3/2),x, algorithm="maxima")
[Out]
integrate(1/((b*x^3 + a)*(d*x^3 + c)^(3/2)*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)^(3/2),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 ) \left (c + d x^{3}\right )^{\frac{3}{2}}}\, 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)**(3/2),x)
[Out]
Integral(1/(x**2*(a + b*x**3)*(c + d*x**3)**(3/2)), 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 )}{\left (d x^{3} + c\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/(b*x^3+a)/(d*x^3+c)^(3/2),x, algorithm="giac")
[Out]
integrate(1/((b*x^3 + a)*(d*x^3 + c)^(3/2)*x^2), x)