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

Optimal. Leaf size=64 \[ \frac {x^2 \left (a+b x^3\right )^{2/3} F_1\left (\frac {2}{3};-\frac {2}{3},1;\frac {5}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 c \left (\frac {b x^3}{a}+1\right )^{2/3}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 64, 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 = {511, 510} \[ \frac {x^2 \left (a+b x^3\right )^{2/3} F_1\left (\frac {2}{3};-\frac {2}{3},1;\frac {5}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 c \left (\frac {b x^3}{a}+1\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*(a + b*x^3)^(2/3)*AppellF1[2/3, -2/3, 1, 5/3, -((b*x^3)/a), -((d*x^3)/c)])/(2*c*(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 {x \left (a+b x^3\right )^{2/3}}{c+d x^3} \, dx &=\frac {\left (a+b x^3\right )^{2/3} \int \frac {x \left (1+\frac {b x^3}{a}\right )^{2/3}}{c+d x^3} \, dx}{\left (1+\frac {b x^3}{a}\right )^{2/3}}\\ &=\frac {x^2 \left (a+b x^3\right )^{2/3} F_1\left (\frac {2}{3};-\frac {2}{3},1;\frac {5}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 c \left (1+\frac {b x^3}{a}\right )^{2/3}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 65, normalized size = 1.02 \[ \frac {x^2 \left (a+b x^3\right )^{2/3} F_1\left (\frac {2}{3};-\frac {2}{3},1;\frac {5}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 c \left (\frac {a+b x^3}{a}\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*(a + b*x^3)^(2/3)*AppellF1[2/3, -2/3, 1, 5/3, -((b*x^3)/a), -((d*x^3)/c)])/(2*c*((a + b*x^3)/a)^(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(x*(b*x^3+a)^(2/3)/(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 \[ \int \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}} x}{d x^{3} + c}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.57, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} x}{d \,x^{3}+c}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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