3.4.54 \(\int \frac {\sqrt [3]{a+b x^3}}{x^2 (a d-b d x^3)} \, dx\)

Optimal. Leaf size=156 \[ -\frac {\sqrt [3]{a+b x^3}}{a d x}+\frac {\sqrt [3]{b} \log \left (a d-b d x^3\right )}{3\ 2^{2/3} a d}-\frac {\sqrt [3]{b} \log \left (\sqrt [3]{2} \sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2^{2/3} a d}-\frac {\sqrt [3]{2} \sqrt [3]{b} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{2} \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} a d} \]

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Rubi [C]  time = 0.07, antiderivative size = 77, normalized size of antiderivative = 0.49, 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} \begin {gather*} -\frac {\sqrt [3]{a+b x^3} \sqrt [3]{1-\frac {b x^3}{a}} \, _2F_1\left (-\frac {1}{3},-\frac {1}{3};\frac {2}{3};-\frac {2 b x^3}{a-b x^3}\right )}{a d x \sqrt [3]{\frac {b x^3}{a}+1}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(((a + b*x^3)^(1/3)*(1 - (b*x^3)/a)^(1/3)*Hypergeometric2F1[-1/3, -1/3, 2/3, (-2*b*x^3)/(a - b*x^3)])/(a*d*x*
(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 {\sqrt [3]{a+b x^3}}{x^2 \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^2 \left (a d-b d x^3\right )} \, dx}{\sqrt [3]{1+\frac {b x^3}{a}}}\\ &=-\frac {\sqrt [3]{a+b x^3} \sqrt [3]{1-\frac {b x^3}{a}} \, _2F_1\left (-\frac {1}{3},-\frac {1}{3};\frac {2}{3};-\frac {2 b x^3}{a-b x^3}\right )}{a d x \sqrt [3]{1+\frac {b x^3}{a}}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 45, normalized size = 0.29 \begin {gather*} -\frac {\sqrt [3]{a+b x^3} \, _2F_1\left (-\frac {1}{3},1;\frac {2}{3};\frac {2 b x^3}{b x^3+a}\right )}{a d x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.43, size = 205, normalized size = 1.31 \begin {gather*} \frac {\sqrt [3]{b} \log \left (2^{2/3} \sqrt [3]{b} x \sqrt [3]{a+b x^3}+\sqrt [3]{2} \left (a+b x^3\right )^{2/3}+2 b^{2/3} x^2\right )}{3\ 2^{2/3} a d}-\frac {\sqrt [3]{a+b x^3}}{a d x}-\frac {\sqrt [3]{2} \sqrt [3]{b} \log \left (2^{2/3} \sqrt [3]{a+b x^3}-2 \sqrt [3]{b} x\right )}{3 a d}-\frac {\sqrt [3]{2} \sqrt [3]{b} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} x}{2^{2/3} \sqrt [3]{a+b x^3}+\sqrt [3]{b} x}\right )}{\sqrt {3} a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(1/3)/x^2/(-b*d*x^3+a*d),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{{\left (b d x^{3} - a d\right )} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(1/3)/x^2/(-b*d*x^3+a*d),x, algorithm="giac")

[Out]

integrate(-(b*x^3 + a)^(1/3)/((b*d*x^3 - a*d)*x^2), x)

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maple [F]  time = 0.84, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{\left (-b d \,x^{3}+a d \right ) x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(1/3)/x^2/(-b*d*x^3+a*d),x)

[Out]

int((b*x^3+a)^(1/3)/x^2/(-b*d*x^3+a*d),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{{\left (b d x^{3} - a d\right )} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(1/3)/x^2/(-b*d*x^3+a*d),x, algorithm="maxima")

[Out]

-integrate((b*x^3 + a)^(1/3)/((b*d*x^3 - a*d)*x^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,x^3+a\right )}^{1/3}}{x^2\,\left (a\,d-b\,d\,x^3\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^(1/3)/(x^2*(a*d - b*d*x^3)),x)

[Out]

int((a + b*x^3)^(1/3)/(x^2*(a*d - b*d*x^3)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {\sqrt [3]{a + b x^{3}}}{- a x^{2} + b x^{5}}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**(1/3)/x**2/(-b*d*x**3+a*d),x)

[Out]

-Integral((a + b*x**3)**(1/3)/(-a*x**2 + b*x**5), x)/d

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