3.675 \(\int \frac{\sqrt [3]{a+b x^3}}{x^3 (c+d x^3)} \, dx\)

Optimal. Leaf size=64 \[ -\frac{\sqrt [3]{a+b x^3} F_1\left (-\frac{2}{3};-\frac{1}{3},1;\frac{1}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{2 c x^2 \sqrt [3]{\frac{b x^3}{a}+1}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0531591, antiderivative size = 64, 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} \[ -\frac{\sqrt [3]{a+b x^3} F_1\left (-\frac{2}{3};-\frac{1}{3},1;\frac{1}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{2 c x^2 \sqrt [3]{\frac{b x^3}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [B]  time = 0.243332, size = 327, normalized size = 5.11 \[ \frac{\frac{c \left (16 a c \left (a \left (c+3 d x^3\right )+b d x^6\right ) F_1\left (\frac{1}{3};\frac{2}{3},1;\frac{4}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )-4 x^3 \left (a+b x^3\right ) \left (c+d x^3\right ) \left (3 a d F_1\left (\frac{4}{3};\frac{2}{3},2;\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )+2 b c F_1\left (\frac{4}{3};\frac{5}{3},1;\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )\right )\right )}{\left (c+d x^3\right ) \left (x^3 \left (3 a d F_1\left (\frac{4}{3};\frac{2}{3},2;\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )+2 b c F_1\left (\frac{4}{3};\frac{5}{3},1;\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )\right )-4 a c F_1\left (\frac{1}{3};\frac{2}{3},1;\frac{4}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )\right )}-b d x^6 \left (\frac{b x^3}{a}+1\right )^{2/3} F_1\left (\frac{4}{3};\frac{2}{3},1;\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{8 c^2 x^2 \left (a+b x^3\right )^{2/3}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-(b*d*x^6*(1 + (b*x^3)/a)^(2/3)*AppellF1[4/3, 2/3, 1, 7/3, -((b*x^3)/a), -((d*x^3)/c)]) + (c*(16*a*c*(b*d*x^6
+ a*(c + 3*d*x^3))*AppellF1[1/3, 2/3, 1, 4/3, -((b*x^3)/a), -((d*x^3)/c)] - 4*x^3*(a + b*x^3)*(c + d*x^3)*(3*
a*d*AppellF1[4/3, 2/3, 2, 7/3, -((b*x^3)/a), -((d*x^3)/c)] + 2*b*c*AppellF1[4/3, 5/3, 1, 7/3, -((b*x^3)/a), -(
(d*x^3)/c)])))/((c + d*x^3)*(-4*a*c*AppellF1[1/3, 2/3, 1, 4/3, -((b*x^3)/a), -((d*x^3)/c)] + x^3*(3*a*d*Appell
F1[4/3, 2/3, 2, 7/3, -((b*x^3)/a), -((d*x^3)/c)] + 2*b*c*AppellF1[4/3, 5/3, 1, 7/3, -((b*x^3)/a), -((d*x^3)/c)
]))))/(8*c^2*x^2*(a + b*x^3)^(2/3))

________________________________________________________________________________________

Maple [F]  time = 0.052, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{3} \left ( d{x}^{3}+c \right ) }\sqrt [3]{b{x}^{3}+a}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{3} + a\right )}^{\frac{1}{3}}}{{\left (d x^{3} + c\right )} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{3} + a\right )}^{\frac{1}{3}}}{{\left (d x^{3} + c\right )} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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