3.7.65 \(\int \frac {x^7 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx\) [665]

Optimal. Leaf size=336 \[ -\frac {(6 b c-a d) x^2 \sqrt [3]{a+b x^3}}{18 b d^2}+\frac {x^5 \sqrt [3]{a+b x^3}}{6 d}-\frac {\left (9 b^2 c^2-3 a b c d-a^2 d^2\right ) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{9 \sqrt {3} b^{5/3} d^3}+\frac {c^{5/3} \sqrt [3]{b c-a d} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} d^3}-\frac {c^{5/3} \sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 d^3}-\frac {\left (9 b^2 c^2-3 a b c d-a^2 d^2\right ) \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{18 b^{5/3} d^3}+\frac {c^{5/3} \sqrt [3]{b c-a d} \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 d^3} \]

[Out]

-1/18*(-a*d+6*b*c)*x^2*(b*x^3+a)^(1/3)/b/d^2+1/6*x^5*(b*x^3+a)^(1/3)/d-1/6*c^(5/3)*(-a*d+b*c)^(1/3)*ln(d*x^3+c
)/d^3-1/18*(-a^2*d^2-3*a*b*c*d+9*b^2*c^2)*ln(b^(1/3)*x-(b*x^3+a)^(1/3))/b^(5/3)/d^3+1/2*c^(5/3)*(-a*d+b*c)^(1/
3)*ln((-a*d+b*c)^(1/3)*x/c^(1/3)-(b*x^3+a)^(1/3))/d^3-1/27*(-a^2*d^2-3*a*b*c*d+9*b^2*c^2)*arctan(1/3*(1+2*b^(1
/3)*x/(b*x^3+a)^(1/3))*3^(1/2))/b^(5/3)/d^3*3^(1/2)+1/3*c^(5/3)*(-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))/d^3*3^(1/2)

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Rubi [A]
time = 0.26, antiderivative size = 336, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {489, 596, 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 ) \left (-a^2 d^2-3 a b c d+9 b^2 c^2\right )}{9 \sqrt {3} b^{5/3} d^3}-\frac {\left (-a^2 d^2-3 a b c d+9 b^2 c^2\right ) \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{18 b^{5/3} d^3}+\frac {c^{5/3} \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^3}-\frac {c^{5/3} \sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 d^3}+\frac {c^{5/3} \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^3}-\frac {x^2 \sqrt [3]{a+b x^3} (6 b c-a d)}{18 b d^2}+\frac {x^5 \sqrt [3]{a+b x^3}}{6 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^7*(a + b*x^3)^(1/3))/(c + d*x^3),x]

[Out]

-1/18*((6*b*c - a*d)*x^2*(a + b*x^3)^(1/3))/(b*d^2) + (x^5*(a + b*x^3)^(1/3))/(6*d) - ((9*b^2*c^2 - 3*a*b*c*d
- a^2*d^2)*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(9*Sqrt[3]*b^(5/3)*d^3) + (c^(5/3)*(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^3) - (c^(5/3)*(
b*c - a*d)^(1/3)*Log[c + d*x^3])/(6*d^3) - ((9*b^2*c^2 - 3*a*b*c*d - a^2*d^2)*Log[b^(1/3)*x - (a + b*x^3)^(1/3
)])/(18*b^(5/3)*d^3) + (c^(5/3)*(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
^3)

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 596

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

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^7 \sqrt [3]{a+b x^3}}{c+d x^3} \, dx &=\frac {\sqrt [3]{a+b x^3} \int \frac {x^7 \sqrt [3]{1+\frac {b x^3}{a}}}{c+d x^3} \, dx}{\sqrt [3]{1+\frac {b x^3}{a}}}\\ &=\frac {x^8 \sqrt [3]{a+b x^3} F_1\left (\frac {8}{3};-\frac {1}{3},1;\frac {11}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{8 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 = 4.11, size = 527, normalized size = 1.57 \begin {gather*} \frac {\frac {6 d x^2 \sqrt [3]{a+b x^3} \left (-6 b c+a d+3 b d x^3\right )}{b}-\frac {4 \sqrt {3} \left (9 b^2 c^2-3 a b c d-a^2 d^2\right ) \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )}{b^{5/3}}-18 \sqrt {-6-6 i \sqrt {3}} c^{5/3} \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 \left (-9 b^2 c^2+3 a b c d+a^2 d^2\right ) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{b^{5/3}}+18 i \left (i+\sqrt {3}\right ) c^{5/3} \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 \left (9 b^2 c^2-3 a b c d-a^2 d^2\right ) \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^{5/3}}+9 \left (1-i \sqrt {3}\right ) c^{5/3} \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 )}{108 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^7*(a + b*x^3)^(1/3))/(c + d*x^3),x]

[Out]

((6*d*x^2*(a + b*x^3)^(1/3)*(-6*b*c + a*d + 3*b*d*x^3))/b - (4*Sqrt[3]*(9*b^2*c^2 - 3*a*b*c*d - a^2*d^2)*ArcTa
n[(Sqrt[3]*b^(1/3)*x)/(b^(1/3)*x + 2*(a + b*x^3)^(1/3))])/b^(5/3) - 18*Sqrt[-6 - (6*I)*Sqrt[3]]*c^(5/3)*(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*(-9*b^2*c^2 + 3*a*b*c*d + a^2*d^2)*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)])/b^(5/3) + (18*I)*(I + S
qrt[3])*c^(5/3)*(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*
(9*b^2*c^2 - 3*a*b*c*d - a^2*d^2)*Log[b^(2/3)*x^2 + b^(1/3)*x*(a + b*x^3)^(1/3) + (a + b*x^3)^(2/3)])/b^(5/3)
+ 9*(1 - I*Sqrt[3])*c^(5/3)*(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)])/(108*d^3)

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {x^{7} \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{d \,x^{3}+c}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(b*x^3+a)^(1/3)/(d*x^3+c),x)

[Out]

int(x^7*(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^7*(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="maxima")

[Out]

integrate((b*x^3 + a)^(1/3)*x^7/(d*x^3 + c), x)

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Fricas [A]
time = 4.98, size = 494, normalized size = 1.47 \begin {gather*} \frac {18 \, \sqrt {3} {\left (b c^{3} - a c^{2} d\right )}^{\frac {1}{3}} b^{3} c \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 ) + 18 \, {\left (b c^{3} - a c^{2} d\right )}^{\frac {1}{3}} b^{3} c \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 ) - 9 \, {\left (b c^{3} - a c^{2} d\right )}^{\frac {1}{3}} b^{3} c \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 (9 \, b^{3} c^{2} - 3 \, a b^{2} c d - a^{2} b d^{2}\right )} {\left (b^{2}\right )}^{\frac {1}{6}} \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 )} {\left (b^{2}\right )}^{\frac {1}{6}}}{3 \, b^{2} x}\right ) - 2 \, {\left (9 \, b^{2} c^{2} - 3 \, a b c d - a^{2} d^{2}\right )} {\left (b^{2}\right )}^{\frac {2}{3}} \log \left (-\frac {{\left (b^{2}\right )}^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{x}\right ) + {\left (9 \, b^{2} c^{2} - 3 \, a b c d - a^{2} d^{2}\right )} {\left (b^{2}\right )}^{\frac {2}{3}} \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 ) + 3 \, {\left (3 \, b^{3} d^{2} x^{5} - {\left (6 \, b^{3} c d - a b^{2} d^{2}\right )} x^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{54 \, b^{3} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="fricas")

[Out]

1/54*(18*sqrt(3)*(b*c^3 - a*c^2*d)^(1/3)*b^3*c*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)) + 18*(b*c^3 - a*c^2*d)^(1/3)*b^3*c*log(((b*x^3 + a)^(1/3)*
c - (b*c^3 - a*c^2*d)^(1/3)*x)/x) - 9*(b*c^3 - a*c^2*d)^(1/3)*b^3*c*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)*(9*b^3*c^2 - 3*a*b^2*c*d - a^
2*b*d^2)*(b^2)^(1/6)*arctan(1/3*(sqrt(3)*(b^2)^(1/3)*b*x + 2*sqrt(3)*(b*x^3 + a)^(1/3)*(b^2)^(2/3))*(b^2)^(1/6
)/(b^2*x)) - 2*(9*b^2*c^2 - 3*a*b*c*d - a^2*d^2)*(b^2)^(2/3)*log(-((b^2)^(2/3)*x - (b*x^3 + a)^(1/3)*b)/x) + (
9*b^2*c^2 - 3*a*b*c*d - a^2*d^2)*(b^2)^(2/3)*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) + 3*(3*b^3*d^2*x^5 - (6*b^3*c*d - a*b^2*d^2)*x^2)*(b*x^3 + a)^(1/3))/(b^3*d^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{7} \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**7*(b*x**3+a)**(1/3)/(d*x**3+c),x)

[Out]

Integral(x**7*(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^7*(b*x^3+a)^(1/3)/(d*x^3+c),x, algorithm="giac")

[Out]

integrate((b*x^3 + a)^(1/3)*x^7/(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^7\,{\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^7*(a + b*x^3)^(1/3))/(c + d*x^3),x)

[Out]

int((x^7*(a + b*x^3)^(1/3))/(c + d*x^3), x)

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