3.694 \(\int \frac {(a+b x^3)^{2/3}}{x^2 (c+d x^3)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.06, 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} \[ -\frac {\left (a+b x^3\right )^{2/3} F_1\left (-\frac {1}{3};-\frac {2}{3},1;\frac {2}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{c x \left (\frac {b x^3}{a}+1\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [B]  time = 0.14, size = 138, normalized size = 2.23 \[ \frac {-5 x^3 \sqrt [3]{\frac {b x^3}{a}+1} (a d-2 b c) F_1\left (\frac {2}{3};\frac {1}{3},1;\frac {5}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+2 b d x^6 \sqrt [3]{\frac {b x^3}{a}+1} F_1\left (\frac {5}{3};\frac {1}{3},1;\frac {8}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )-10 c \left (a+b x^3\right )}{10 c^2 x \sqrt [3]{a+b x^3}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-10*c*(a + b*x^3) - 5*(-2*b*c + a*d)*x^3*(1 + (b*x^3)/a)^(1/3)*AppellF1[2/3, 1/3, 1, 5/3, -((b*x^3)/a), -((d*
x^3)/c)] + 2*b*d*x^6*(1 + (b*x^3)/a)^(1/3)*AppellF1[5/3, 1/3, 1, 8/3, -((b*x^3)/a), -((d*x^3)/c)])/(10*c^2*x*(
a + b*x^3)^(1/3))

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.69, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{\left (d \,x^{3}+c \right ) x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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