3.704 \(\int \frac{(a+b x^3)^{4/3}}{x^2 (c+d x^3)} \, dx\)

Optimal. Leaf size=254 \[ -\frac{b^{4/3} \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2 d}-\frac{b^{4/3} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{\sqrt{3} d}-\frac{(b c-a d)^{4/3} \log \left (c+d x^3\right )}{6 c^{4/3} d}+\frac{(b c-a d)^{4/3} \log \left (\frac{x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{4/3} d}+\frac{(b c-a d)^{4/3} \tan ^{-1}\left (\frac{\frac{2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{\sqrt{3} c^{4/3} d}-\frac{a \sqrt [3]{a+b x^3}}{c x} \]

[Out]

-((a*(a + b*x^3)^(1/3))/(c*x)) - (b^(4/3)*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(Sqrt[3]*d) +
((b*c - a*d)^(4/3)*ArcTan[(1 + (2*(b*c - a*d)^(1/3)*x)/(c^(1/3)*(a + b*x^3)^(1/3)))/Sqrt[3]])/(Sqrt[3]*c^(4/3
)*d) - ((b*c - a*d)^(4/3)*Log[c + d*x^3])/(6*c^(4/3)*d) - (b^(4/3)*Log[b^(1/3)*x - (a + b*x^3)^(1/3)])/(2*d) +
((b*c - a*d)^(4/3)*Log[((b*c - a*d)^(1/3)*x)/c^(1/3) - (a + b*x^3)^(1/3)])/(2*c^(4/3)*d)

________________________________________________________________________________________

Rubi [C]  time = 0.0590058, antiderivative size = 63, normalized size of antiderivative = 0.25, 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} \[ -\frac{a \sqrt [3]{a+b x^3} F_1\left (-\frac{1}{3};-\frac{4}{3},1;\frac{2}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{c x \sqrt [3]{\frac{b x^3}{a}+1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((a*(a + b*x^3)^(1/3)*AppellF1[-1/3, -4/3, 1, 2/3, -((b*x^3)/a), -((d*x^3)/c)])/(c*x*(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{\left (a+b x^3\right )^{4/3}}{x^2 \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^2 \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{1}{3};-\frac{4}{3},1;\frac{2}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{c x \sqrt [3]{1+\frac{b x^3}{a}}}\\ \end{align*}

Mathematica [C]  time = 0.323071, size = 161, normalized size = 0.63 \[ \frac{2 b^2 c x^6 \left (\frac{b x^3}{a}+1\right )^{2/3} F_1\left (\frac{5}{3};\frac{2}{3},1;\frac{8}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )-\frac{5 a x^3 \left (\frac{b x^3}{a}+1\right )^{2/3} (a d-2 b c) \, _2F_1\left (\frac{2}{3},\frac{2}{3};\frac{5}{3};\frac{(a d-b c) x^3}{a \left (d x^3+c\right )}\right )}{\left (\frac{d x^3}{c}+1\right )^{2/3}}-10 a c \left (a+b x^3\right )}{10 c^2 x \left (a+b x^3\right )^{2/3}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-10*a*c*(a + b*x^3) + 2*b^2*c*x^6*(1 + (b*x^3)/a)^(2/3)*AppellF1[5/3, 2/3, 1, 8/3, -((b*x^3)/a), -((d*x^3)/c)
] - (5*a*(-2*b*c + a*d)*x^3*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1[2/3, 2/3, 5/3, ((-(b*c) + a*d)*x^3)/(a*(c
+ d*x^3))])/(1 + (d*x^3)/c)^(2/3))/(10*c^2*x*(a + b*x^3)^(2/3))

________________________________________________________________________________________

Maple [F]  time = 0.055, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{2} \left ( d{x}^{3}+c \right ) } \left ( b{x}^{3}+a \right ) ^{{\frac{4}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(4/3)/x^2/(d*x^3+c),x)

[Out]

int((b*x^3+a)^(4/3)/x^2/(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{4}{3}}}{{\left (d x^{3} + c\right )} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(4/3)/x^2/(d*x^3+c),x, algorithm="maxima")

[Out]

integrate((b*x^3 + a)^(4/3)/((d*x^3 + c)*x^2), 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)^(4/3)/x^2/(d*x^3+c),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x^{3}\right )^{\frac{4}{3}}}{x^{2} \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)**(4/3)/x**2/(d*x**3+c),x)

[Out]

Integral((a + b*x**3)**(4/3)/(x**2*(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{4}{3}}}{{\left (d x^{3} + c\right )} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(4/3)/x^2/(d*x^3+c),x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(4/3)/((d*x^3 + c)*x^2), x)