3.7.68 \(\int \frac {\sqrt [3]{a+b x^3}}{x^2 (c+d x^3)} \, dx\) [668]

Optimal. Leaf size=168 \[ -\frac {\sqrt [3]{a+b x^3}}{c x}-\frac {\sqrt [3]{b c-a d} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} c^{4/3}}+\frac {\sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 c^{4/3}}-\frac {\sqrt [3]{b c-a d} \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{4/3}} \]

[Out]

-(b*x^3+a)^(1/3)/c/x+1/6*(-a*d+b*c)^(1/3)*ln(d*x^3+c)/c^(4/3)-1/2*(-a*d+b*c)^(1/3)*ln((-a*d+b*c)^(1/3)*x/c^(1/
3)-(b*x^3+a)^(1/3))/c^(4/3)-1/3*(-a*d+b*c)^(1/3)*arctan(1/3*(1+2*(-a*d+b*c)^(1/3)*x/c^(1/3)/(b*x^3+a)^(1/3))*3
^(1/2))/c^(4/3)*3^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {486, 12, 503} \begin {gather*} -\frac {\sqrt [3]{b c-a d} \text {ArcTan}\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}}+\frac {\sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 c^{4/3}}-\frac {\sqrt [3]{b c-a d} \log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{4/3}}-\frac {\sqrt [3]{a+b x^3}}{c x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [C] Result contains complex when optimal does not.
time = 1.46, size = 309, normalized size = 1.84 \begin {gather*} \frac {-\frac {12 \sqrt [3]{c} \sqrt [3]{a+b x^3}}{x}+2 \sqrt {-6-6 i \sqrt {3}} \sqrt [3]{b c-a d} \tan ^{-1}\left (\frac {3 \sqrt [3]{b c-a d} x}{\sqrt {3} \sqrt [3]{b c-a d} x-\left (3 i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )+2 \left (1-i \sqrt {3}\right ) \sqrt [3]{b c-a d} \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )+i \left (i+\sqrt {3}\right ) \sqrt [3]{b c-a d} \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{12 c^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-12*c^(1/3)*(a + b*x^3)^(1/3))/x + 2*Sqrt[-6 - (6*I)*Sqrt[3]]*(b*c - a*d)^(1/3)*ArcTan[(3*(b*c - a*d)^(1/3)*
x)/(Sqrt[3]*(b*c - a*d)^(1/3)*x - (3*I + Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/3))] + 2*(1 - I*Sqrt[3])*(b*c - a*d)^
(1/3)*Log[2*(b*c - a*d)^(1/3)*x + (1 + I*Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/3)] + I*(I + Sqrt[3])*(b*c - a*d)^(1/
3)*Log[2*(b*c - a*d)^(2/3)*x^2 + (-1 - I*Sqrt[3])*c^(1/3)*(b*c - a*d)^(1/3)*x*(a + b*x^3)^(1/3) + I*(I + Sqrt[
3])*c^(2/3)*(a + b*x^3)^(2/3)])/(12*c^(4/3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [F(-1)] Timed out
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/(d*x^3+c),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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/(d*x^3+c),x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(1/3)/((d*x^3 + c)*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 (d\,x^3+c\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*(c + d*x^3)),x)

[Out]

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

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