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

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

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 62, normalized size of antiderivative = 1.00, 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} \[ -\frac {\sqrt {\frac {d x^3}{c}+1} F_1\left (-\frac {1}{3};2,\frac {1}{2};\frac {2}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{a^2 x \sqrt {c+d x^3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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])

Rubi steps

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

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Mathematica [B]  time = 0.25, size = 226, normalized size = 3.65 \[ \frac {-5 x^3 \left (a+b x^3\right ) \sqrt {\frac {d x^3}{c}+1} \left (3 a^2 d^2-15 a b c d+8 b^2 c^2\right ) F_1\left (\frac {2}{3};\frac {1}{2},1;\frac {5}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+20 a \left (c+d x^3\right ) \left (3 a^2 d-3 a b \left (c-d x^3\right )-4 b^2 c x^3\right )+2 b d x^6 \left (a+b x^3\right ) \sqrt {\frac {d x^3}{c}+1} (4 b c-3 a d) F_1\left (\frac {5}{3};\frac {1}{2},1;\frac {8}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )}{60 a^3 c x \left (a+b x^3\right ) \sqrt {c+d x^3} (b c-a d)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(20*a*(c + d*x^3)*(3*a^2*d - 4*b^2*c*x^3 - 3*a*b*(c - d*x^3)) - 5*(8*b^2*c^2 - 15*a*b*c*d + 3*a^2*d^2)*x^3*(a
+ b*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*(4*b*c - 3*a*d)*x^
6*(a + b*x^3)*Sqrt[1 + (d*x^3)/c]*AppellF1[5/3, 1/2, 1, 8/3, -((d*x^3)/c), -((b*x^3)/a)])/(60*a^3*c*(b*c - a*d
)*x*(a + b*x^3)*Sqrt[c + d*x^3])

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.27, size = 1818, normalized size = 29.32 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*I/a^2*b/d^2*2^(1/2)*sum(1/_alpha/(a*d-b*c)*(-c*d^2)^(1/3)*(1/2*I*(2*x+(-I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^
(1/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)*((x-(-c*d^2)^(1/3)/d)/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2)^(1/3))*d)^(1/2)*
(-1/2*I*(2*x+(I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)/(d*x^3+c)^(1/2)*(2*_alpha^2*
d^2+I*(-c*d^2)^(1/3)*3^(1/2)*_alpha*d-(-c*d^2)^(1/3)*_alpha*d-I*3^(1/2)*(-c*d^2)^(2/3)-(-c*d^2)^(2/3))*Ellipti
cPi(1/3*3^(1/2)*(I*(x+1/2*(-c*d^2)^(1/3)/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2),1/2
*(2*I*(-c*d^2)^(1/3)*3^(1/2)*_alpha^2*d+I*3^(1/2)*c*d-3*c*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha-3*(-c*d^2)^(2/3)*_
alpha)/(a*d-b*c)*b/d,(I*3^(1/2)*(-c*d^2)^(1/3)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)/d)^(1/2)
),_alpha=RootOf(_Z^3*b+a))-1/a*b*(-1/3/(a*d-b*c)*(d*x^3+c)^(1/2)/(b*x^3+a)/a*b*x^2-1/9*I/(a*d-b*c)/a*3^(1/2)*(
-c*d^2)^(1/3)*(I*(x+1/2*(-c*d^2)^(1/3)/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2)*((x-(
-c*d^2)^(1/3)/d)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d))^(1/2)*(-I*(x+1/2*(-c*d^2)^(1/3)/d+1/2
*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2)/(d*x^3+c)^(1/2)*((-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(
1/2)*(-c*d^2)^(1/3)/d)*EllipticE(1/3*3^(1/2)*(I*(x+1/2*(-c*d^2)^(1/3)/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2
)/(-c*d^2)^(1/3)*d)^(1/2),(I*3^(1/2)*(-c*d^2)^(1/3)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)/d)^
(1/2))+(-c*d^2)^(1/3)/d*EllipticF(1/3*3^(1/2)*(I*(x+1/2*(-c*d^2)^(1/3)/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/
2)/(-c*d^2)^(1/3)*d)^(1/2),(I*3^(1/2)*(-c*d^2)^(1/3)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)/d)
^(1/2)))+1/18*I/a/d^2*2^(1/2)*sum((-5*a*d+2*b*c)/(a*d-b*c)^2/_alpha*(-c*d^2)^(1/3)*(1/2*I*(2*x+(-I*3^(1/2)*(-c
*d^2)^(1/3)+(-c*d^2)^(1/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)*((x-(-c*d^2)^(1/3)/d)/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*
d^2)^(1/3))*d)^(1/2)*(-1/2*I*(2*x+(I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)/(d*x^3+
c)^(1/2)*(2*_alpha^2*d^2+I*(-c*d^2)^(1/3)*3^(1/2)*_alpha*d-(-c*d^2)^(1/3)*_alpha*d-I*3^(1/2)*(-c*d^2)^(2/3)-(-
c*d^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2*(-c*d^2)^(1/3)/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^
2)^(1/3)*d)^(1/2),1/2*(2*I*(-c*d^2)^(1/3)*3^(1/2)*_alpha^2*d+I*3^(1/2)*c*d-3*c*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alp
ha-3*(-c*d^2)^(2/3)*_alpha)/(a*d-b*c)*b/d,(I*3^(1/2)*(-c*d^2)^(1/3)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d
^2)^(1/3)/d)/d)^(1/2)),_alpha=RootOf(_Z^3*b+a)))+1/a^2*(-(d*x^3+c)^(1/2)/c/x-1/3*I/c*3^(1/2)*(-c*d^2)^(1/3)*(I
*(x+1/2*(-c*d^2)^(1/3)/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2)*((x-(-c*d^2)^(1/3)/d)
/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d))^(1/2)*(-I*(x+1/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d
^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2)/(d*x^3+c)^(1/2)*((-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1
/3)/d)*EllipticE(1/3*3^(1/2)*(I*(x+1/2*(-c*d^2)^(1/3)/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)
*d)^(1/2),(I*3^(1/2)*(-c*d^2)^(1/3)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)/d)^(1/2))+(-c*d^2)^
(1/3)/d*EllipticF(1/3*3^(1/2)*(I*(x+1/2*(-c*d^2)^(1/3)/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3
)*d)^(1/2),(I*3^(1/2)*(-c*d^2)^(1/3)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)/d)^(1/2))))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{x^2\,{\left (b\,x^3+a\right )}^2\,\sqrt {d\,x^3+c}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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