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

Optimal. Leaf size=86 \[ -\frac {\sqrt {\frac {b x^3}{a}+1} \sqrt {\frac {d x^3}{c}+1} F_1\left (-\frac {1}{3};\frac {1}{2},\frac {1}{2};\frac {2}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{x \sqrt {a+b x^3} \sqrt {c+d x^3}} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {511, 510} \[ -\frac {\sqrt {\frac {b x^3}{a}+1} \sqrt {\frac {d x^3}{c}+1} F_1\left (-\frac {1}{3};\frac {1}{2},\frac {1}{2};\frac {2}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{x \sqrt {a+b x^3} \sqrt {c+d x^3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[1 + (b*x^3)/a]*Sqrt[1 + (d*x^3)/c]*AppellF1[-1/3, 1/2, 1/2, 2/3, -((b*x^3)/a), -((d*x^3)/c)])/(x*Sqrt[
a + b*x^3]*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 \sqrt {a+b x^3} \sqrt {c+d x^3}} \, dx &=\frac {\sqrt {1+\frac {b x^3}{a}} \int \frac {1}{x^2 \sqrt {1+\frac {b x^3}{a}} \sqrt {c+d x^3}} \, dx}{\sqrt {a+b x^3}}\\ &=\frac {\left (\sqrt {1+\frac {b x^3}{a}} \sqrt {1+\frac {d x^3}{c}}\right ) \int \frac {1}{x^2 \sqrt {1+\frac {b x^3}{a}} \sqrt {1+\frac {d x^3}{c}}} \, dx}{\sqrt {a+b x^3} \sqrt {c+d x^3}}\\ &=-\frac {\sqrt {1+\frac {b x^3}{a}} \sqrt {1+\frac {d x^3}{c}} F_1\left (-\frac {1}{3};\frac {1}{2},\frac {1}{2};\frac {2}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{x \sqrt {a+b x^3} \sqrt {c+d x^3}}\\ \end {align*}

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Mathematica [B]  time = 0.26, size = 189, normalized size = 2.20 \[ \frac {5 x^3 \sqrt {\frac {b x^3}{a}+1} \sqrt {\frac {d x^3}{c}+1} (a d+b c) F_1\left (\frac {2}{3};\frac {1}{2},\frac {1}{2};\frac {5}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+8 b d x^6 \sqrt {\frac {b x^3}{a}+1} \sqrt {\frac {d x^3}{c}+1} F_1\left (\frac {5}{3};\frac {1}{2},\frac {1}{2};\frac {8}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )-20 \left (a+b x^3\right ) \left (c+d x^3\right )}{20 a c x \sqrt {a+b x^3} \sqrt {c+d x^3}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.85, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x^{3} + a} \sqrt {d x^{3} + c}}{b d x^{8} + {\left (b c + a d\right )} x^{5} + a c x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*x^3 + a)*sqrt(d*x^3 + c)/(b*d*x^8 + (b*c + a*d)*x^5 + a*c*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.18, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b \,x^{3}+a}\, \sqrt {d \,x^{3}+c}\, x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^2\,\sqrt {b\,x^3+a}\,\sqrt {d\,x^3+c}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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