3.7.66 \(\int \frac {x^4 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx\) [666]

Optimal. Leaf size=276 \[ \frac {x^2 \sqrt [3]{a+b x^3}}{3 d}+\frac {(3 b c-a d) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} b^{2/3} d^2}-\frac {c^{2/3} \sqrt [3]{b c-a d} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} d^2}+\frac {c^{2/3} \sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 d^2}+\frac {(3 b c-a d) \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{6 b^{2/3} d^2}-\frac {c^{2/3} \sqrt [3]{b c-a d} \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 d^2} \]

[Out]

1/3*x^2*(b*x^3+a)^(1/3)/d+1/6*c^(2/3)*(-a*d+b*c)^(1/3)*ln(d*x^3+c)/d^2+1/6*(-a*d+3*b*c)*ln(b^(1/3)*x-(b*x^3+a)
^(1/3))/b^(2/3)/d^2-1/2*c^(2/3)*(-a*d+b*c)^(1/3)*ln((-a*d+b*c)^(1/3)*x/c^(1/3)-(b*x^3+a)^(1/3))/d^2+1/9*(-a*d+
3*b*c)*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))/b^(2/3)/d^2*3^(1/2)-1/3*c^(2/3)*(-a*d+b*c)^(1/3)*ar
ctan(1/3*(1+2*(-a*d+b*c)^(1/3)*x/c^(1/3)/(b*x^3+a)^(1/3))*3^(1/2))/d^2*3^(1/2)

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Rubi [A]
time = 0.16, antiderivative size = 276, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {489, 598, 337, 503} \begin {gather*} \frac {\text {ArcTan}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right ) (3 b c-a d)}{3 \sqrt {3} b^{2/3} d^2}-\frac {c^{2/3} \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} d^2}+\frac {(3 b c-a d) \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{6 b^{2/3} d^2}+\frac {c^{2/3} \sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 d^2}-\frac {c^{2/3} \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 d^2}+\frac {x^2 \sqrt [3]{a+b x^3}}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*x^3)^(1/3))/(c + d*x^3),x]

[Out]

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

Rule 337

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

Rule 489

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

Rule 598

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)*(x_)^(n_)), x_Sy
mbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^n)^p*((e + f*x^n)/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e,
f, g, m, p}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {x^4 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx &=\frac {\sqrt [3]{a+b x^3} \int \frac {x^4 \sqrt [3]{1+\frac {b x^3}{a}}}{c+d x^3} \, dx}{\sqrt [3]{1+\frac {b x^3}{a}}}\\ &=\frac {x^5 \sqrt [3]{a+b x^3} F_1\left (\frac {5}{3};-\frac {1}{3},1;\frac {8}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{5 c \sqrt [3]{1+\frac {b x^3}{a}}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 2.76, size = 467, normalized size = 1.69 \begin {gather*} \frac {12 d x^2 \sqrt [3]{a+b x^3}+\frac {4 \sqrt {3} (3 b c-a d) \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )}{b^{2/3}}+6 \sqrt {-6-6 i \sqrt {3}} c^{2/3} \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 )+\frac {4 (3 b c-a d) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{b^{2/3}}+6 \left (1-i \sqrt {3}\right ) c^{2/3} \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 )+\frac {2 (-3 b c+a d) \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{b^{2/3}}+3 i \left (i+\sqrt {3}\right ) c^{2/3} \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 )}{36 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(a + b*x^3)^(1/3))/(c + d*x^3),x]

[Out]

(12*d*x^2*(a + b*x^3)^(1/3) + (4*Sqrt[3]*(3*b*c - a*d)*ArcTan[(Sqrt[3]*b^(1/3)*x)/(b^(1/3)*x + 2*(a + b*x^3)^(
1/3))])/b^(2/3) + 6*Sqrt[-6 - (6*I)*Sqrt[3]]*c^(2/3)*(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))] + (4*(3*b*c - a*d)*Log[-(b^(1/3)*x) + (a +
b*x^3)^(1/3)])/b^(2/3) + 6*(1 - I*Sqrt[3])*c^(2/3)*(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)] + (2*(-3*b*c + a*d)*Log[b^(2/3)*x^2 + b^(1/3)*x*(a + b*x^3)^(1/3) + (a + b*x^3)^
(2/3)])/b^(2/3) + (3*I)*(I + Sqrt[3])*c^(2/3)*(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)])/(36*d^2)

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {x^{4} \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{d \,x^{3}+c}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(b*x^3+a)^(1/3)/(d*x^3+c),x)

[Out]

int(x^4*(b*x^3+a)^(1/3)/(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(x^4*(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="maxima")

[Out]

integrate((b*x^3 + a)^(1/3)*x^4/(d*x^3 + c), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 452 vs. \(2 (222) = 444\).
time = 4.30, size = 452, normalized size = 1.64 \begin {gather*} \frac {6 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{2} d x^{2} + 6 \, \sqrt {3} {\left (-b c^{3} + a c^{2} d\right )}^{\frac {1}{3}} b^{2} \arctan \left (-\frac {\sqrt {3} {\left (b c^{2} - a c d\right )} x + 2 \, \sqrt {3} {\left (-b c^{3} + a c^{2} d\right )}^{\frac {2}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{3 \, {\left (b c^{2} - a c d\right )} x}\right ) + 6 \, {\left (-b c^{3} + a c^{2} d\right )}^{\frac {1}{3}} b^{2} \log \left (\frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} c + {\left (-b c^{3} + a c^{2} d\right )}^{\frac {1}{3}} x}{x}\right ) - 3 \, {\left (-b c^{3} + a c^{2} d\right )}^{\frac {1}{3}} b^{2} \log \left (\frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}} c^{2} - {\left (-b c^{3} + a c^{2} d\right )}^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} c x + {\left (-b c^{3} + a c^{2} d\right )}^{\frac {2}{3}} x^{2}}{x^{2}}\right ) - 2 \, \sqrt {3} {\left (3 \, b^{2} c - a b d\right )} \sqrt {-\left (-b^{2}\right )^{\frac {1}{3}}} \arctan \left (-\frac {{\left (\sqrt {3} \left (-b^{2}\right )^{\frac {1}{3}} b x - 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {2}{3}}\right )} \sqrt {-\left (-b^{2}\right )^{\frac {1}{3}}}}{3 \, b^{2} x}\right ) + 2 \, \left (-b^{2}\right )^{\frac {2}{3}} {\left (3 \, b c - a d\right )} \log \left (-\frac {\left (-b^{2}\right )^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{x}\right ) - \left (-b^{2}\right )^{\frac {2}{3}} {\left (3 \, b c - a d\right )} \log \left (-\frac {\left (-b^{2}\right )^{\frac {1}{3}} b x^{2} - {\left (b x^{3} + a\right )}^{\frac {1}{3}} \left (-b^{2}\right )^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {2}{3}} b}{x^{2}}\right )}{18 \, b^{2} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

1/18*(6*(b*x^3 + a)^(1/3)*b^2*d*x^2 + 6*sqrt(3)*(-b*c^3 + a*c^2*d)^(1/3)*b^2*arctan(-1/3*(sqrt(3)*(b*c^2 - a*c
*d)*x + 2*sqrt(3)*(-b*c^3 + a*c^2*d)^(2/3)*(b*x^3 + a)^(1/3))/((b*c^2 - a*c*d)*x)) + 6*(-b*c^3 + a*c^2*d)^(1/3
)*b^2*log(((b*x^3 + a)^(1/3)*c + (-b*c^3 + a*c^2*d)^(1/3)*x)/x) - 3*(-b*c^3 + a*c^2*d)^(1/3)*b^2*log(((b*x^3 +
a)^(2/3)*c^2 - (-b*c^3 + a*c^2*d)^(1/3)*(b*x^3 + a)^(1/3)*c*x + (-b*c^3 + a*c^2*d)^(2/3)*x^2)/x^2) - 2*sqrt(3
)*(3*b^2*c - a*b*d)*sqrt(-(-b^2)^(1/3))*arctan(-1/3*(sqrt(3)*(-b^2)^(1/3)*b*x - 2*sqrt(3)*(b*x^3 + a)^(1/3)*(-
b^2)^(2/3))*sqrt(-(-b^2)^(1/3))/(b^2*x)) + 2*(-b^2)^(2/3)*(3*b*c - a*d)*log(-((-b^2)^(2/3)*x - (b*x^3 + a)^(1/
3)*b)/x) - (-b^2)^(2/3)*(3*b*c - a*d)*log(-((-b^2)^(1/3)*b*x^2 - (b*x^3 + a)^(1/3)*(-b^2)^(2/3)*x - (b*x^3 + a
)^(2/3)*b)/x^2))/(b^2*d^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4} \sqrt [3]{a + b x^{3}}}{c + d x^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(b*x**3+a)**(1/3)/(d*x**3+c),x)

[Out]

Integral(x**4*(a + b*x**3)**(1/3)/(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(x^4*(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(1/3)*x^4/(d*x^3 + c), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^4\,{\left (b\,x^3+a\right )}^{1/3}}{d\,x^3+c} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(a + b*x^3)^(1/3))/(c + d*x^3),x)

[Out]

int((x^4*(a + b*x^3)^(1/3))/(c + d*x^3), x)

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