3.4.94 \(\int \frac {x}{(a+b x^3) (c+d x^3)^{3/2}} \, dx\) [394]

Optimal. Leaf size=67 \[ \frac {x^2 \sqrt {1+\frac {d x^3}{c}} F_1\left (\frac {2}{3};1,\frac {3}{2};\frac {5}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 a c \sqrt {c+d x^3}} \]

[Out]

1/2*x^2*AppellF1(2/3,1,3/2,5/3,-b*x^3/a,-d*x^3/c)*(1+d*x^3/c)^(1/2)/a/c/(d*x^3+c)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {525, 524} \begin {gather*} \frac {x^2 \sqrt {\frac {d x^3}{c}+1} F_1\left (\frac {2}{3};1,\frac {3}{2};\frac {5}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 a c \sqrt {c+d x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 524

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)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], 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 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[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 {x}{\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 {x}{\left (a+b x^3\right ) \left (1+\frac {d x^3}{c}\right )^{3/2}} \, dx}{c \sqrt {c+d x^3}}\\ &=\frac {x^2 \sqrt {1+\frac {d x^3}{c}} F_1\left (\frac {2}{3};1,\frac {3}{2};\frac {5}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 a c \sqrt {c+d x^3}}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(142\) vs. \(2(67)=134\).
time = 10.08, size = 142, normalized size = 2.12 \begin {gather*} \frac {x^2 \left (-20 a d+5 (3 b c+a d) \sqrt {1+\frac {d x^3}{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 )+2 b d x^3 \sqrt {1+\frac {d x^3}{c}} F_1\left (\frac {5}{3};\frac {1}{2},1;\frac {8}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )}{30 a c (b c-a d) \sqrt {c+d x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*(-20*a*d + 5*(3*b*c + a*d)*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^3*Sqrt[1 + (d*x^3)/c]*AppellF1[5/3, 1/2, 1, 8/3, -((d*x^3)/c), -((b*x^3)/a)]))/(30*a*c*(b*c - a*d)*Sqrt
[c + d*x^3])

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 6.
time = 0.33, size = 907, normalized size = 13.54

method	result	size
default	\(\text {Expression too large to display}\)	\(907\)
elliptic	\(\text {Expression too large to display}\)	\(907\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(b*x^3+a)/(d*x^3+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/3*d*x^2/c/(a*d-b*c)/((x^3+c/d)*d)^(1/2)+2/9*I/c/(a*d-b*c)*3^(1/2)*(-c*d^2)^(1/3)*(I*(x+1/2/d*(-c*d^2)^(1/3)-
1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2)*((x-1/d*(-c*d^2)^(1/3))/(-3/2/d*(-c*d^2)^(1/3)
+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3)))^(1/2)*(-I*(x+1/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/
(-c*d^2)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*((-3/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*EllipticE(1/3*3^
(1/2)*(I*(x+1/2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2),(I*3^(1/2)/d*
(-c*d^2)^(1/3)/(-3/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3)))^(1/2))+1/d*(-c*d^2)^(1/3)*EllipticF(1/3
*3^(1/2)*(I*(x+1/2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2),(I*3^(1/2)
/d*(-c*d^2)^(1/3)/(-3/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3)))^(1/2)))+1/3*I*b/d^2*2^(1/2)*sum(1/(a
*d-b*c)^2/_alpha*(-c*d^2)^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^
(1/2)*(d*(x-1/d*(-c*d^2)^(1/3))/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1
/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-c*d^2)^(1/3)*_alpha*3^(1/2)*d-I
*3^(1/2)*(-c*d^2)^(2/3)+2*_alpha^2*d^2-(-c*d^2)^(1/3)*_alpha*d-(-c*d^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/
2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2),1/2*b/d*(2*I*(-c*d^2)^(1/3)
*3^(1/2)*_alpha^2*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-c*d^2)^(2/3)*_alpha-3*c*d)/(a*d-b*c),(I*
3^(1/2)/d*(-c*d^2)^(1/3)/(-3/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*b+a
))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\left (a + b x^{3}\right ) \left (c + d x^{3}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x}{\left (b\,x^3+a\right )\,{\left (d\,x^3+c\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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