Integrand size = 28, antiderivative size = 268 \[ \int \frac {x^7 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=-\frac {7 a x^2 \sqrt [3]{a+b x^3}}{18 b^2 d}-\frac {x^5 \sqrt [3]{a+b x^3}}{6 b d}+\frac {11 a^2 \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{9 \sqrt {3} b^{8/3} d}-\frac {\sqrt [3]{2} a^2 \arctan \left (\frac {1+\frac {2 \sqrt [3]{2} \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} b^{8/3} d}+\frac {a^2 \log \left (a d-b d x^3\right )}{3\ 2^{2/3} b^{8/3} d}+\frac {11 a^2 \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{18 b^{8/3} d}-\frac {a^2 \log \left (\sqrt [3]{2} \sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2^{2/3} b^{8/3} d} \]
-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+11/ 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)
Time = 1.47 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.22 \[ \int \frac {x^7 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=-\frac {21 a b^{2/3} x^2 \sqrt [3]{a+b x^3}+9 b^{5/3} x^5 \sqrt [3]{a+b x^3}-22 \sqrt {3} a^2 \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )+18 \sqrt [3]{2} \sqrt {3} a^2 \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2^{2/3} \sqrt [3]{a+b x^3}}\right )-22 a^2 \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )+18 \sqrt [3]{2} a^2 \log \left (-2 \sqrt [3]{b} x+2^{2/3} \sqrt [3]{a+b x^3}\right )+11 a^2 \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )-9 \sqrt [3]{2} a^2 \log \left (2 b^{2/3} x^2+2^{2/3} \sqrt [3]{b} x \sqrt [3]{a+b x^3}+\sqrt [3]{2} \left (a+b x^3\right )^{2/3}\right )}{54 b^{8/3} d} \]
-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)
Time = 0.43 (sec) , antiderivative size = 248, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {978, 27, 1052, 27, 1054, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {x^7 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx\) |
| \(\Big \downarrow \) 978 |
| \(\displaystyle \frac {\int \frac {a x^4 \left (7 b x^3+5 a\right )}{\left (a-b x^3\right ) \left (b x^3+a\right )^{2/3}}dx}{6 b d}-\frac {x^5 \sqrt [3]{a+b x^3}}{6 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a \int \frac {x^4 \left (7 b x^3+5 a\right )}{\left (a-b x^3\right ) \left (b x^3+a\right )^{2/3}}dx}{6 b d}-\frac {x^5 \sqrt [3]{a+b x^3}}{6 b d}\) |
| \(\Big \downarrow \) 1052 |
| \(\displaystyle \frac {a \left (\frac {\int \frac {2 a b x \left (11 b x^3+7 a\right )}{\left (a-b x^3\right ) \left (b x^3+a\right )^{2/3}}dx}{3 b^2}-\frac {7 x^2 \sqrt [3]{a+b x^3}}{3 b}\right )}{6 b d}-\frac {x^5 \sqrt [3]{a+b x^3}}{6 b d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a \left (\frac {2 a \int \frac {x \left (11 b x^3+7 a\right )}{\left (a-b x^3\right ) \left (b x^3+a\right )^{2/3}}dx}{3 b}-\frac {7 x^2 \sqrt [3]{a+b x^3}}{3 b}\right )}{6 b d}-\frac {x^5 \sqrt [3]{a+b x^3}}{6 b d}\) |
| \(\Big \downarrow \) 1054 |
| \(\displaystyle \frac {a \left (\frac {2 a \int \left (\frac {18 a x}{\left (a-b x^3\right ) \left (b x^3+a\right )^{2/3}}-\frac {11 x}{\left (b x^3+a\right )^{2/3}}\right )dx}{3 b}-\frac {7 x^2 \sqrt [3]{a+b x^3}}{3 b}\right )}{6 b d}-\frac {x^5 \sqrt [3]{a+b x^3}}{6 b d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a \left (\frac {2 a \left (\frac {11 \arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} b^{2/3}}-\frac {3 \sqrt [3]{2} \sqrt {3} \arctan \left (\frac {\frac {2 \sqrt [3]{2} \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{b^{2/3}}+\frac {3 \log \left (a-b x^3\right )}{2^{2/3} b^{2/3}}+\frac {11 \log \left (\sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2 b^{2/3}}-\frac {9 \log \left (\sqrt [3]{2} \sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2^{2/3} b^{2/3}}\right )}{3 b}-\frac {7 x^2 \sqrt [3]{a+b x^3}}{3 b}\right )}{6 b d}-\frac {x^5 \sqrt [3]{a+b x^3}}{6 b d}\) |
-1/6*(x^5*(a + b*x^3)^(1/3))/(b*d) + (a*((-7*x^2*(a + b*x^3)^(1/3))/(3*b) + (2*a*((11*ArcTan[(1 + (2*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt[3]])/(Sqrt[3 ]*b^(2/3)) - (3*2^(1/3)*Sqrt[3]*ArcTan[(1 + (2*2^(1/3)*b^(1/3)*x)/(a + b*x ^3)^(1/3))/Sqrt[3]])/b^(2/3) + (3*Log[a - b*x^3])/(2^(2/3)*b^(2/3)) + (11* Log[b^(1/3)*x - (a + b*x^3)^(1/3)])/(2*b^(2/3)) - (9*Log[2^(1/3)*b^(1/3)*x - (a + b*x^3)^(1/3)])/(2^(2/3)*b^(2/3))))/(3*b)))/(6*b*d)
3.6.74.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] - Simp[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]
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] - Simp[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]
Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n _)))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> 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]
Time = 7.07 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.02
| method | result | size |
| pseudoelliptic | \(\frac {-9 x^{5} \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{\frac {5}{3}}-21 a \,x^{2} \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{\frac {2}{3}}+18 \,2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (2^{\frac {2}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+b^{\frac {1}{3}} x \right )}{3 b^{\frac {1}{3}} x}\right ) a^{2}-18 \,2^{\frac {1}{3}} \ln \left (\frac {-2^{\frac {1}{3}} b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) a^{2}+9 \,2^{\frac {1}{3}} \ln \left (\frac {2^{\frac {2}{3}} b^{\frac {2}{3}} x^{2}+2^{\frac {1}{3}} b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) a^{2}-22 a^{2} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right )+22 a^{2} \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )-11 a^{2} \ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )}{54 b^{\frac {8}{3}} d}\) | \(273\) |
1/54/b^(8/3)*(-9*x^5*(b*x^3+a)^(1/3)*b^(5/3)-21*a*x^2*(b*x^3+a)^(1/3)*b^(2 /3)+18*2^(1/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2^(2/3)*(b*x^3+a)^(1/3)+b^(1/3) *x)/b^(1/3)/x)*a^2-18*2^(1/3)*ln((-2^(1/3)*b^(1/3)*x+(b*x^3+a)^(1/3))/x)*a ^2+9*2^(1/3)*ln((2^(2/3)*b^(2/3)*x^2+2^(1/3)*b^(1/3)*(b*x^3+a)^(1/3)*x+(b* x^3+a)^(2/3))/x^2)*a^2-22*a^2*3^(1/2)*arctan(1/3*3^(1/2)*(b^(1/3)*x+2*(b*x ^3+a)^(1/3))/b^(1/3)/x)+22*a^2*ln((-b^(1/3)*x+(b*x^3+a)^(1/3))/x)-11*a^2*l n((b^(2/3)*x^2+b^(1/3)*(b*x^3+a)^(1/3)*x+(b*x^3+a)^(2/3))/x^2))/d
Time = 0.27 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.35 \[ \int \frac {x^7 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=-\frac {18 \, \sqrt {3} 2^{\frac {1}{3}} a^{2} b^{2} \left (-\frac {1}{b^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} b \left (-\frac {1}{b^{2}}\right )^{\frac {2}{3}} + \sqrt {3} x}{3 \, x}\right ) - 18 \cdot 2^{\frac {1}{3}} a^{2} b^{2} \left (-\frac {1}{b^{2}}\right )^{\frac {1}{3}} \log \left (\frac {2^{\frac {1}{3}} b x \left (-\frac {1}{b^{2}}\right )^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right ) + 9 \cdot 2^{\frac {1}{3}} a^{2} b^{2} \left (-\frac {1}{b^{2}}\right )^{\frac {1}{3}} \log \left (\frac {2^{\frac {2}{3}} b^{2} x^{2} \left (-\frac {1}{b^{2}}\right )^{\frac {2}{3}} - 2^{\frac {1}{3}} {\left (b x^{3} + a\right )}^{\frac {1}{3}} b x \left (-\frac {1}{b^{2}}\right )^{\frac {1}{3}} + {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right ) + 22 \, \sqrt {3} a^{2} {\left (b^{2}\right )}^{\frac {1}{6}} b \arctan \left (\frac {{\left (\sqrt {3} {\left (b^{2}\right )}^{\frac {1}{3}} b x + 2 \, \sqrt {3} {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b^{2}\right )}^{\frac {2}{3}}\right )} {\left (b^{2}\right )}^{\frac {1}{6}}}{3 \, b^{2} x}\right ) - 22 \, a^{2} {\left (b^{2}\right )}^{\frac {2}{3}} \log \left (-\frac {{\left (b^{2}\right )}^{\frac {2}{3}} x - {\left (b x^{3} + a\right )}^{\frac {1}{3}} b}{x}\right ) + 11 \, a^{2} {\left (b^{2}\right )}^{\frac {2}{3}} \log \left (\frac {{\left (b^{2}\right )}^{\frac {1}{3}} b x^{2} + {\left (b x^{3} + a\right )}^{\frac {1}{3}} {\left (b^{2}\right )}^{\frac {2}{3}} x + {\left (b x^{3} + a\right )}^{\frac {2}{3}} b}{x^{2}}\right ) + 3 \, {\left (3 \, b^{3} x^{5} + 7 \, a b^{2} x^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{54 \, b^{4} d} \]
-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*s qrt(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)*lo g(-((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)
\[ \int \frac {x^7 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=- \frac {\int \frac {x^{7} \sqrt [3]{a + b x^{3}}}{- a + b x^{3}}\, dx}{d} \]
\[ \int \frac {x^7 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=\int { -\frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} x^{7}}{b d x^{3} - a d} \,d x } \]
\[ \int \frac {x^7 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=\int { -\frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} x^{7}}{b d x^{3} - a d} \,d x } \]
Timed out. \[ \int \frac {x^7 \sqrt [3]{a+b x^3}}{a d-b d x^3} \, dx=\int \frac {x^7\,{\left (b\,x^3+a\right )}^{1/3}}{a\,d-b\,d\,x^3} \,d x \]