3.578 \(\int \frac{\sqrt [3]{a+b x^3}}{x^5 (a d-b d x^3)} \, dx\)

Optimal. Leaf size=183 \[ \frac{b^{4/3} \log \left (a d-b d x^3\right )}{3\ 2^{2/3} a^2 d}-\frac{b^{4/3} \log \left (\sqrt [3]{2} \sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2^{2/3} a^2 d}-\frac{\sqrt [3]{2} b^{4/3} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{2} \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt{3}}\right )}{\sqrt{3} a^2 d}-\frac{5 b \sqrt [3]{a+b x^3}}{4 a^2 d x}-\frac{\sqrt [3]{a+b x^3}}{4 a d x^4} \]

[Out]

-(a + b*x^3)^(1/3)/(4*a*d*x^4) - (5*b*(a + b*x^3)^(1/3))/(4*a^2*d*x) - (2^(1/3)*b^(4/3)*ArcTan[(1 + (2*2^(1/3)
*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(Sqrt[3]*a^2*d) + (b^(4/3)*Log[a*d - b*d*x^3])/(3*2^(2/3)*a^2*d) - (b
^(4/3)*Log[2^(1/3)*b^(1/3)*x - (a + b*x^3)^(1/3)])/(2^(2/3)*a^2*d)

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Rubi [C]  time = 0.422017, antiderivative size = 117, normalized size of antiderivative = 0.64, number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {511, 510} \[ -\frac{a^2-b x^3 \left (a+3 b x^3\right ) \, _2F_1\left (\frac{2}{3},1;\frac{5}{3};\frac{2 b x^3}{b x^3+a}\right )+3 b x^3 \left (a-b x^3\right ) \, _2F_1\left (\frac{2}{3},2;\frac{5}{3};\frac{2 b x^3}{b x^3+a}\right )+4 a b x^3+3 b^2 x^6}{4 a^2 d x^4 \left (a+b x^3\right )^{2/3}} \]

Warning: Unable to verify antiderivative.

[In]

Int[(a + b*x^3)^(1/3)/(x^5*(a*d - b*d*x^3)),x]

[Out]

-(a^2 + 4*a*b*x^3 + 3*b^2*x^6 - b*x^3*(a + 3*b*x^3)*Hypergeometric2F1[2/3, 1, 5/3, (2*b*x^3)/(a + b*x^3)] + 3*
b*x^3*(a - b*x^3)*Hypergeometric2F1[2/3, 2, 5/3, (2*b*x^3)/(a + b*x^3)])/(4*a^2*d*x^4*(a + b*x^3)^(2/3))

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^5 \left (a d-b d x^3\right )} \, dx &=\frac{\sqrt [3]{a+b x^3} \int \frac{\sqrt [3]{1+\frac{b x^3}{a}}}{x^5 \left (a d-b d x^3\right )} \, dx}{\sqrt [3]{1+\frac{b x^3}{a}}}\\ &=-\frac{a^2+4 a b x^3+3 b^2 x^6-b x^3 \left (a+3 b x^3\right ) \, _2F_1\left (\frac{2}{3},1;\frac{5}{3};\frac{2 b x^3}{a+b x^3}\right )+3 b x^3 \left (a-b x^3\right ) \, _2F_1\left (\frac{2}{3},2;\frac{5}{3};\frac{2 b x^3}{a+b x^3}\right )}{4 a^2 d x^4 \left (a+b x^3\right )^{2/3}}\\ \end{align*}

Mathematica [C]  time = 4.85741, size = 112, normalized size = 0.61 \[ -\frac{\left (a+b x^3\right )^{4/3} \left (\left (a^2-4 a b x^3+3 b^2 x^6\right ) \, _2F_1\left (1,1;\frac{2}{3};-\frac{2 b x^3}{a-b x^3}\right )+9 b x^3 \left (a+b x^3\right ) \, _2F_1\left (2,2;\frac{5}{3};-\frac{2 b x^3}{a-b x^3}\right )\right )}{4 a^2 d x^4 \left (a-b x^3\right )^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*x^3)^(1/3)/(x^5*(a*d - b*d*x^3)),x]

[Out]

-((a + b*x^3)^(4/3)*((a^2 - 4*a*b*x^3 + 3*b^2*x^6)*Hypergeometric2F1[1, 1, 2/3, (-2*b*x^3)/(a - b*x^3)] + 9*b*
x^3*(a + b*x^3)*Hypergeometric2F1[2, 2, 5/3, (-2*b*x^3)/(a - b*x^3)]))/(4*a^2*d*x^4*(a - b*x^3)^2)

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Maple [F]  time = 0.046, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{5} \left ( -bd{x}^{3}+ad \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^5/(-b*d*x^3+a*d),x)

[Out]

int((b*x^3+a)^(1/3)/x^5/(-b*d*x^3+a*d),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{{\left (b x^{3} + a\right )}^{\frac{1}{3}}}{{\left (b d x^{3} - a d\right )} x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(1/3)/x^5/(-b*d*x^3+a*d),x, algorithm="maxima")

[Out]

-integrate((b*x^3 + a)^(1/3)/((b*d*x^3 - a*d)*x^5), x)

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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^5/(-b*d*x^3+a*d),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int \frac{\sqrt [3]{a + b x^{3}}}{- a x^{5} + b x^{8}}\, dx}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**(1/3)/x**5/(-b*d*x**3+a*d),x)

[Out]

-Integral((a + b*x**3)**(1/3)/(-a*x**5 + b*x**8), x)/d

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{{\left (b x^{3} + a\right )}^{\frac{1}{3}}}{{\left (b d x^{3} - a d\right )} x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(1/3)/x^5/(-b*d*x^3+a*d),x, algorithm="giac")

[Out]

integrate(-(b*x^3 + a)^(1/3)/((b*d*x^3 - a*d)*x^5), x)