3.6.77 \(\int \frac {\sqrt [3]{a+b x^3}}{x^2 (a d-b d x^3)} \, dx\) [577]

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

[Out]

-(b*x^3+a)^(1/3)/a/d/x+1/6*b^(1/3)*ln(-b*d*x^3+a*d)*2^(1/3)/a/d-1/2*b^(1/3)*ln(2^(1/3)*b^(1/3)*x-(b*x^3+a)^(1/
3))*2^(1/3)/a/d-1/3*2^(1/3)*b^(1/3)*arctan(1/3*(1+2*2^(1/3)*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))/a/d*3^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {486, 12, 503} \begin {gather*} -\frac {\sqrt [3]{2} \sqrt [3]{b} \text {ArcTan}\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}-\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 486

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(a*e*(m + 1))), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*
x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*b*(m + 1) + n*(b*c*(p + 1) + a*d*q) + d*(b*(m + 1) + b*n*(p + q + 1))*x^n, x
], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[0, q, 1] && LtQ[m, -1] &&
IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 503

Int[(x_)/(((a_) + (b_.)*(x_)^3)^(2/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/c, 3]}, Si
mp[-ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*c*q^2), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/
(2*c*q^2), x] + Simp[Log[c + d*x^3]/(6*c*q^2), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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 [A]
time = 0.31, size = 190, normalized size = 1.22 \begin {gather*} -\frac {6 \sqrt [3]{a+b x^3}+2 \sqrt [3]{2} \sqrt {3} \sqrt [3]{b} x \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2^{2/3} \sqrt [3]{a+b x^3}}\right )+2 \sqrt [3]{2} \sqrt [3]{b} x \log \left (-2 \sqrt [3]{b} x+2^{2/3} \sqrt [3]{a+b x^3}\right )-\sqrt [3]{2} \sqrt [3]{b} x \log \left (2 b^{2/3} x^2+2^{2/3} \sqrt [3]{b} x \sqrt [3]{a+b x^3}+\sqrt [3]{2} \left (a+b x^3\right )^{2/3}\right )}{6 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]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x^{2} \left (-b d \,x^{3}+a d \right )}\, dx\]

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*} \text {Failed to integrate} \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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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 395 vs. \(2 (125) = 250\).
time = 186.28, size = 395, normalized size = 2.53 \begin {gather*} -\frac {2 \, \sqrt {3} 2^{\frac {1}{3}} \left (-b\right )^{\frac {1}{3}} x \arctan \left (\frac {6 \, \sqrt {3} 2^{\frac {2}{3}} {\left (19 \, b^{2} x^{8} + 16 \, a b x^{5} + a^{2} x^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {2}{3}} + 6 \, \sqrt {3} 2^{\frac {1}{3}} {\left (5 \, b^{2} x^{7} - 4 \, a b x^{4} - a^{2} x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}} \left (-b\right )^{\frac {1}{3}} + \sqrt {3} {\left (71 \, b^{3} x^{9} + 111 \, a b^{2} x^{6} + 33 \, a^{2} b x^{3} + a^{3}\right )}}{3 \, {\left (109 \, b^{3} x^{9} + 105 \, a b^{2} x^{6} + 3 \, a^{2} b x^{3} - a^{3}\right )}}\right ) - 2 \cdot 2^{\frac {1}{3}} \left (-b\right )^{\frac {1}{3}} x \log \left (-\frac {6 \cdot 2^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b\right )^{\frac {1}{3}} b x^{2} + 6 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b x + 2^{\frac {2}{3}} {\left (b x^{3} - a\right )} \left (-b\right )^{\frac {2}{3}}}{b x^{3} - a}\right ) + 2^{\frac {1}{3}} \left (-b\right )^{\frac {1}{3}} x \log \left (\frac {3 \cdot 2^{\frac {2}{3}} {\left (5 \, b x^{4} + a x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}} \left (-b\right )^{\frac {2}{3}} - 2^{\frac {1}{3}} {\left (19 \, b^{2} x^{6} + 16 \, a b x^{3} + a^{2}\right )} \left (-b\right )^{\frac {1}{3}} + 12 \, {\left (2 \, b^{2} x^{5} + a b x^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{b^{2} x^{6} - 2 \, a b x^{3} + a^{2}}\right ) + 18 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{18 \, a 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="fricas")

[Out]

-1/18*(2*sqrt(3)*2^(1/3)*(-b)^(1/3)*x*arctan(1/3*(6*sqrt(3)*2^(2/3)*(19*b^2*x^8 + 16*a*b*x^5 + a^2*x^2)*(b*x^3
+ a)^(1/3)*(-b)^(2/3) + 6*sqrt(3)*2^(1/3)*(5*b^2*x^7 - 4*a*b*x^4 - a^2*x)*(b*x^3 + a)^(2/3)*(-b)^(1/3) + sqrt
(3)*(71*b^3*x^9 + 111*a*b^2*x^6 + 33*a^2*b*x^3 + a^3))/(109*b^3*x^9 + 105*a*b^2*x^6 + 3*a^2*b*x^3 - a^3)) - 2*
2^(1/3)*(-b)^(1/3)*x*log(-(6*2^(1/3)*(b*x^3 + a)^(1/3)*(-b)^(1/3)*b*x^2 + 6*(b*x^3 + a)^(2/3)*b*x + 2^(2/3)*(b
*x^3 - a)*(-b)^(2/3))/(b*x^3 - a)) + 2^(1/3)*(-b)^(1/3)*x*log((3*2^(2/3)*(5*b*x^4 + a*x)*(b*x^3 + a)^(2/3)*(-b
)^(2/3) - 2^(1/3)*(19*b^2*x^6 + 16*a*b*x^3 + a^2)*(-b)^(1/3) + 12*(2*b^2*x^5 + a*b*x^2)*(b*x^3 + a)^(1/3))/(b^
2*x^6 - 2*a*b*x^3 + a^2)) + 18*(b*x^3 + a)^(1/3))/(a*d*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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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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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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