3.6.96 \(\int \frac {(a+b x^3)^{2/3}}{x^3 (a d-b d x^3)} \, dx\) [596]

Optimal. Leaf size=157 \[ -\frac {\left (a+b x^3\right )^{2/3}}{2 a d x^2}+\frac {2^{2/3} b^{2/3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{2} \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} a d}+\frac {b^{2/3} \log \left (a d-b d x^3\right )}{3 \sqrt [3]{2} a d}-\frac {b^{2/3} \log \left (\sqrt [3]{2} \sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{\sqrt [3]{2} a d} \]

[Out]

-1/2*(b*x^3+a)^(2/3)/a/d/x^2+1/6*b^(2/3)*ln(-b*d*x^3+a*d)*2^(2/3)/a/d-1/2*b^(2/3)*ln(2^(1/3)*b^(1/3)*x-(b*x^3+
a)^(1/3))*2^(2/3)/a/d+1/3*2^(2/3)*b^(2/3)*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 = 0.05, antiderivative size = 157, 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, 384} \begin {gather*} \frac {2^{2/3} b^{2/3} \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 {b^{2/3} \log \left (a d-b d x^3\right )}{3 \sqrt [3]{2} a d}-\frac {b^{2/3} \log \left (\sqrt [3]{2} \sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{\sqrt [3]{2} a d}-\frac {\left (a+b x^3\right )^{2/3}}{2 a d x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(a + b*x^3)^(2/3)/(a*d*x^2) + (2^(2/3)*b^(2/3)*ArcTan[(1 + (2*2^(1/3)*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[
3]])/(Sqrt[3]*a*d) + (b^(2/3)*Log[a*d - b*d*x^3])/(3*2^(1/3)*a*d) - (b^(2/3)*Log[2^(1/3)*b^(1/3)*x - (a + b*x^
3)^(1/3)])/(2^(1/3)*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 384

Int[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/c, 3]}, Simp[
ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*c*q), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*c*q
), x] + Simp[Log[c + d*x^3]/(6*c*q), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

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]

Rubi steps

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

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Mathematica [A]
time = 0.32, size = 195, normalized size = 1.24 \begin {gather*} \frac {-3 \left (a+b x^3\right )^{2/3}+2\ 2^{2/3} \sqrt {3} b^{2/3} x^2 \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\ 2^{2/3} b^{2/3} x^2 \log \left (-2 \sqrt [3]{b} x+2^{2/3} \sqrt [3]{a+b x^3}\right )+2^{2/3} b^{2/3} x^2 \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^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 434 vs. \(2 (125) = 250\).
time = 82.54, size = 434, normalized size = 2.76 \begin {gather*} -\frac {2 \cdot 4^{\frac {1}{3}} \sqrt {3} \left (-b^{2}\right )^{\frac {1}{3}} x^{2} \arctan \left (\frac {3 \cdot 4^{\frac {2}{3}} \sqrt {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^{2}\right )^{\frac {2}{3}} + 6 \cdot 4^{\frac {1}{3}} \sqrt {3} {\left (19 \, b^{3} x^{8} + 16 \, a b^{2} x^{5} + a^{2} b x^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {1}{3}} - \sqrt {3} {\left (71 \, b^{4} x^{9} + 111 \, a b^{3} x^{6} + 33 \, a^{2} b^{2} x^{3} + a^{3} b\right )}}{3 \, {\left (109 \, b^{4} x^{9} + 105 \, a b^{3} x^{6} + 3 \, a^{2} b^{2} x^{3} - a^{3} b\right )}}\right ) - 2 \cdot 4^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {1}{3}} x^{2} \log \left (\frac {3 \cdot 4^{\frac {2}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {2}{3}} x^{2} - 6 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b x + 4^{\frac {1}{3}} {\left (b x^{3} - a\right )} \left (-b^{2}\right )^{\frac {1}{3}}}{b x^{3} - a}\right ) + 4^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {1}{3}} x^{2} \log \left (-\frac {6 \cdot 4^{\frac {1}{3}} {\left (5 \, b^{2} x^{4} + a b x\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}} \left (-b^{2}\right )^{\frac {1}{3}} - 4^{\frac {2}{3}} {\left (19 \, b^{2} x^{6} + 16 \, a b x^{3} + a^{2}\right )} \left (-b^{2}\right )^{\frac {2}{3}} - 24 \, {\left (2 \, b^{3} x^{5} + a b^{2} x^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{b^{2} x^{6} - 2 \, a b x^{3} + a^{2}}\right ) + 9 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{18 \, a d x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/18*(2*4^(1/3)*sqrt(3)*(-b^2)^(1/3)*x^2*arctan(1/3*(3*4^(2/3)*sqrt(3)*(5*b^2*x^7 - 4*a*b*x^4 - a^2*x)*(b*x^3
+ a)^(2/3)*(-b^2)^(2/3) + 6*4^(1/3)*sqrt(3)*(19*b^3*x^8 + 16*a*b^2*x^5 + a^2*b*x^2)*(b*x^3 + a)^(1/3)*(-b^2)^
(1/3) - sqrt(3)*(71*b^4*x^9 + 111*a*b^3*x^6 + 33*a^2*b^2*x^3 + a^3*b))/(109*b^4*x^9 + 105*a*b^3*x^6 + 3*a^2*b^
2*x^3 - a^3*b)) - 2*4^(1/3)*(-b^2)^(1/3)*x^2*log((3*4^(2/3)*(b*x^3 + a)^(1/3)*(-b^2)^(2/3)*x^2 - 6*(b*x^3 + a)
^(2/3)*b*x + 4^(1/3)*(b*x^3 - a)*(-b^2)^(1/3))/(b*x^3 - a)) + 4^(1/3)*(-b^2)^(1/3)*x^2*log(-(6*4^(1/3)*(5*b^2*
x^4 + a*b*x)*(b*x^3 + a)^(2/3)*(-b^2)^(1/3) - 4^(2/3)*(19*b^2*x^6 + 16*a*b*x^3 + a^2)*(-b^2)^(2/3) - 24*(2*b^3
*x^5 + a*b^2*x^2)*(b*x^3 + a)^(1/3))/(b^2*x^6 - 2*a*b*x^3 + a^2)) + 9*(b*x^3 + a)^(2/3))/(a*d*x^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {\left (a + b x^{3}\right )^{\frac {2}{3}}}{- a x^{3} + b x^{6}}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral((a + b*x**3)**(2/3)/(-a*x**3 + b*x**6), 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)^(2/3)/x^3/(-b*d*x^3+a*d),x, algorithm="giac")

[Out]

integrate(-(b*x^3 + a)^(2/3)/((b*d*x^3 - a*d)*x^3), 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 )}^{2/3}}{x^3\,\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)^(2/3)/(x^3*(a*d - b*d*x^3)),x)

[Out]

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

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