3.712 \(\int \frac {(a+b x^3)^{4/3}}{x^3 (c+d x^3)} \, dx\)

Optimal. Leaf size=65 \[ -\frac {a \sqrt [3]{a+b x^3} F_1\left (-\frac {2}{3};-\frac {4}{3},1;\frac {1}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 c x^2 \sqrt [3]{\frac {b x^3}{a}+1}} \]

[Out]

-1/2*a*(b*x^3+a)^(1/3)*AppellF1(-2/3,-4/3,1,1/3,-b*x^3/a,-d*x^3/c)/c/x^2/(1+b*x^3/a)^(1/3)

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Rubi [A]  time = 0.06, antiderivative size = 65, 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} \[ -\frac {a \sqrt [3]{a+b x^3} F_1\left (-\frac {2}{3};-\frac {4}{3},1;\frac {1}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 c x^2 \sqrt [3]{\frac {b x^3}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(4/3)/(x^3*(c + d*x^3)),x]

[Out]

-(a*(a + b*x^3)^(1/3)*AppellF1[-2/3, -4/3, 1, 1/3, -((b*x^3)/a), -((d*x^3)/c)])/(2*c*x^2*(1 + (b*x^3)/a)^(1/3)
)

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 )^{4/3}}{x^3 \left (c+d x^3\right )} \, dx &=\frac {\left (a \sqrt [3]{a+b x^3}\right ) \int \frac {\left (1+\frac {b x^3}{a}\right )^{4/3}}{x^3 \left (c+d x^3\right )} \, dx}{\sqrt [3]{1+\frac {b x^3}{a}}}\\ &=-\frac {a \sqrt [3]{a+b x^3} F_1\left (-\frac {2}{3};-\frac {4}{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*}

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Mathematica [B]  time = 0.38, size = 341, normalized size = 5.25 \[ -\frac {b x^6 \left (\frac {b x^3}{a}+1\right )^{2/3} (a d-2 b c) F_1\left (\frac {4}{3};\frac {2}{3},1;\frac {7}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+\frac {4 a c \left (x^3 \left (a+b x^3\right ) \left (c+d x^3\right ) \left (3 a d F_1\left (\frac {4}{3};\frac {2}{3},2;\frac {7}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+2 b c F_1\left (\frac {4}{3};\frac {5}{3},1;\frac {7}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )-4 a c \left (a c+3 a d x^3-2 b c x^3+b d x^6\right ) F_1\left (\frac {1}{3};\frac {2}{3},1;\frac {4}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )}{\left (c+d x^3\right ) \left (x^3 \left (3 a d F_1\left (\frac {4}{3};\frac {2}{3},2;\frac {7}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )+2 b c F_1\left (\frac {4}{3};\frac {5}{3},1;\frac {7}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )-4 a c F_1\left (\frac {1}{3};\frac {2}{3},1;\frac {4}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )\right )}}{8 c^2 x^2 \left (a+b x^3\right )^{2/3}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*x^3)^(4/3)/(x^3*(c + d*x^3)),x]

[Out]

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

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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)^(4/3)/x^3/(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 \[ \int \frac {{\left (b x^{3} + a\right )}^{\frac {4}{3}}}{{\left (d x^{3} + c\right )} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(4/3)/x^3/(d*x^3+c),x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(4/3)/((d*x^3 + c)*x^3), x)

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maple [F]  time = 0.64, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{3}+a \right )^{\frac {4}{3}}}{\left (d \,x^{3}+c \right ) x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(4/3)/x^3/(d*x^3+c),x)

[Out]

int((b*x^3+a)^(4/3)/x^3/(d*x^3+c),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{3} + a\right )}^{\frac {4}{3}}}{{\left (d x^{3} + c\right )} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(4/3)/x^3/(d*x^3+c),x, algorithm="maxima")

[Out]

integrate((b*x^3 + a)^(4/3)/((d*x^3 + c)*x^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (b\,x^3+a\right )}^{4/3}}{x^3\,\left (d\,x^3+c\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^(4/3)/(x^3*(c + d*x^3)),x)

[Out]

int((a + b*x^3)^(4/3)/(x^3*(c + d*x^3)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x^{3}\right )^{\frac {4}{3}}}{x^{3} \left (c + d x^{3}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**(4/3)/x**3/(d*x**3+c),x)

[Out]

Integral((a + b*x**3)**(4/3)/(x**3*(c + d*x**3)), x)

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