3.6.74 \(\int \frac {x^7 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx\) [574]

Optimal. Leaf size=268 \[ -\frac {7 a x^2 \sqrt [3]{a+b x^3}}{18 b^2 d}-\frac {x^5 \sqrt [3]{a+b x^3}}{6 b d}+\frac {11 a^2 \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{9 \sqrt {3} b^{8/3} d}-\frac {\sqrt [3]{2} a^2 \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{2} \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} b^{8/3} d}+\frac {a^2 \log \left (a d-b d x^3\right )}{3\ 2^{2/3} b^{8/3} d}+\frac {11 a^2 \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{18 b^{8/3} d}-\frac {a^2 \log \left (\sqrt [3]{2} \sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2^{2/3} b^{8/3} d} \]

[Out]

-7/18*a*x^2*(b*x^3+a)^(1/3)/b^2/d-1/6*x^5*(b*x^3+a)^(1/3)/b/d+1/6*a^2*ln(-b*d*x^3+a*d)*2^(1/3)/b^(8/3)/d+11/18
*a^2*ln(b^(1/3)*x-(b*x^3+a)^(1/3))/b^(8/3)/d-1/2*a^2*ln(2^(1/3)*b^(1/3)*x-(b*x^3+a)^(1/3))*2^(1/3)/b^(8/3)/d+1
1/27*a^2*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))/b^(8/3)/d*3^(1/2)-1/3*2^(1/3)*a^2*arctan(1/3*(1+2
*2^(1/3)*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))/b^(8/3)/d*3^(1/2)

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Rubi [A]
time = 0.19, antiderivative size = 268, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {489, 596, 598, 337, 503} \begin {gather*} \frac {11 a^2 \text {ArcTan}\left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{9 \sqrt {3} b^{8/3} d}-\frac {\sqrt [3]{2} a^2 \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} b^{8/3} d}+\frac {a^2 \log \left (a d-b d x^3\right )}{3\ 2^{2/3} b^{8/3} d}+\frac {11 a^2 \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{18 b^{8/3} d}-\frac {a^2 \log \left (\sqrt [3]{2} \sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2^{2/3} b^{8/3} d}-\frac {7 a x^2 \sqrt [3]{a+b x^3}}{18 b^2 d}-\frac {x^5 \sqrt [3]{a+b x^3}}{6 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-7*a*x^2*(a + b*x^3)^(1/3))/(18*b^2*d) - (x^5*(a + b*x^3)^(1/3))/(6*b*d) + (11*a^2*ArcTan[(1 + (2*b^(1/3)*x)/
(a + b*x^3)^(1/3))/Sqrt[3]])/(9*Sqrt[3]*b^(8/3)*d) - (2^(1/3)*a^2*ArcTan[(1 + (2*2^(1/3)*b^(1/3)*x)/(a + b*x^3
)^(1/3))/Sqrt[3]])/(Sqrt[3]*b^(8/3)*d) + (a^2*Log[a*d - b*d*x^3])/(3*2^(2/3)*b^(8/3)*d) + (11*a^2*Log[b^(1/3)*
x - (a + b*x^3)^(1/3)])/(18*b^(8/3)*d) - (a^2*Log[2^(1/3)*b^(1/3)*x - (a + b*x^3)^(1/3)])/(2^(2/3)*b^(8/3)*d)

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}}{a d-b d x^3} \, dx &=\frac {\sqrt [3]{a+b x^3} \int \frac {x^7 \sqrt [3]{1+\frac {b x^3}{a}}}{a d-b 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 {b x^3}{a}\right )}{8 a d \sqrt [3]{1+\frac {b x^3}{a}}}\\ \end {align*}

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Mathematica [A]
time = 0.87, size = 327, normalized size = 1.22 \begin {gather*} -\frac {21 a b^{2/3} x^2 \sqrt [3]{a+b x^3}+9 b^{5/3} x^5 \sqrt [3]{a+b x^3}-22 \sqrt {3} a^2 \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )+18 \sqrt [3]{2} \sqrt {3} a^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 )-22 a^2 \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )+18 \sqrt [3]{2} a^2 \log \left (-2 \sqrt [3]{b} x+2^{2/3} \sqrt [3]{a+b x^3}\right )+11 a^2 \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 )-9 \sqrt [3]{2} a^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 )}{54 b^{8/3} d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/54*(21*a*b^(2/3)*x^2*(a + b*x^3)^(1/3) + 9*b^(5/3)*x^5*(a + b*x^3)^(1/3) - 22*Sqrt[3]*a^2*ArcTan[(Sqrt[3]*b
^(1/3)*x)/(b^(1/3)*x + 2*(a + b*x^3)^(1/3))] + 18*2^(1/3)*Sqrt[3]*a^2*ArcTan[(Sqrt[3]*b^(1/3)*x)/(b^(1/3)*x +
2^(2/3)*(a + b*x^3)^(1/3))] - 22*a^2*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)] + 18*2^(1/3)*a^2*Log[-2*b^(1/3)*x +
2^(2/3)*(a + b*x^3)^(1/3)] + 11*a^2*Log[b^(2/3)*x^2 + b^(1/3)*x*(a + b*x^3)^(1/3) + (a + b*x^3)^(2/3)] - 9*2^
(1/3)*a^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)])/(b^(8/3)*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [A]
time = 2.88, size = 362, normalized size = 1.35 \begin {gather*} -\frac {18 \, \sqrt {3} 2^{\frac {1}{3}} a^{2} b^{2} \left (-\frac {1}{b^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} b \left (-\frac {1}{b^{2}}\right )^{\frac {2}{3}} + \sqrt {3} x}{3 \, x}\right ) - 18 \cdot 2^{\frac {1}{3}} a^{2} b^{2} \left (-\frac {1}{b^{2}}\right )^{\frac {1}{3}} \log \left (\frac {2^{\frac {1}{3}} b x \left (-\frac {1}{b^{2}}\right )^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) + 9 \cdot 2^{\frac {1}{3}} a^{2} b^{2} \left (-\frac {1}{b^{2}}\right )^{\frac {1}{3}} \log \left (\frac {2^{\frac {2}{3}} b^{2} x^{2} \left (-\frac {1}{b^{2}}\right )^{\frac {2}{3}} - 2^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} b x \left (-\frac {1}{b^{2}}\right )^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) + 22 \, \sqrt {3} a^{2} {\left (b^{2}\right )}^{\frac {1}{6}} b \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 ) - 22 \, a^{2} {\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 ) + 11 \, a^{2} {\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} x^{5} + 7 \, a b^{2} x^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{54 \, b^{4} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/54*(18*sqrt(3)*2^(1/3)*a^2*b^2*(-1/b^2)^(1/3)*arctan(1/3*(sqrt(3)*2^(2/3)*(b*x^3 + a)^(1/3)*b*(-1/b^2)^(2/3
) + sqrt(3)*x)/x) - 18*2^(1/3)*a^2*b^2*(-1/b^2)^(1/3)*log((2^(1/3)*b*x*(-1/b^2)^(1/3) + (b*x^3 + a)^(1/3))/x)
+ 9*2^(1/3)*a^2*b^2*(-1/b^2)^(1/3)*log((2^(2/3)*b^2*x^2*(-1/b^2)^(2/3) - 2^(1/3)*(b*x^3 + a)^(1/3)*b*x*(-1/b^2
)^(1/3) + (b*x^3 + a)^(2/3))/x^2) + 22*sqrt(3)*a^2*(b^2)^(1/6)*b*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)) - 22*a^2*(b^2)^(2/3)*log(-((b^2)^(2/3)*x - (b*x^3 + a)^
(1/3)*b)/x) + 11*a^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*x^5 + 7*a*b^2*x^2)*(b*x^3 + a)^(1/3))/(b^4*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {x^{7} \sqrt [3]{a + b x^{3}}}{- a + b x^{3}}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate(-(b*x^3 + a)^(1/3)*x^7/(b*d*x^3 - a*d), 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}}{a\,d-b\,d\,x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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