Integrand size = 28, antiderivative size = 237 \[ \int \frac {\sqrt [3]{a+b x^3}}{x^{11} \left (a d-b d x^3\right )} \, dx=-\frac {\sqrt [3]{a+b x^3}}{10 a d x^{10}}-\frac {11 b \sqrt [3]{a+b x^3}}{70 a^2 d x^7}-\frac {37 b^2 \sqrt [3]{a+b x^3}}{140 a^3 d x^4}-\frac {169 b^3 \sqrt [3]{a+b x^3}}{140 a^4 d x}-\frac {\sqrt [3]{2} b^{10/3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{2} \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} a^4 d}+\frac {b^{10/3} \log \left (a d-b d x^3\right )}{3\ 2^{2/3} a^4 d}-\frac {b^{10/3} \log \left (\sqrt [3]{2} \sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2^{2/3} a^4 d} \]
-1/10*(b*x^3+a)^(1/3)/a/d/x^10-11/70*b*(b*x^3+a)^(1/3)/a^2/d/x^7-37/140*b^ 2*(b*x^3+a)^(1/3)/a^3/d/x^4-169/140*b^3*(b*x^3+a)^(1/3)/a^4/d/x+1/6*b^(10/ 3)*ln(-b*d*x^3+a*d)*2^(1/3)/a^4/d-1/2*b^(10/3)*ln(2^(1/3)*b^(1/3)*x-(b*x^3 +a)^(1/3))*2^(1/3)/a^4/d-1/3*2^(1/3)*b^(10/3)*arctan(1/3*(1+2*2^(1/3)*b^(1 /3)*x/(b*x^3+a)^(1/3))*3^(1/2))/a^4/d*3^(1/2)
Time = 0.76 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.92 \[ \int \frac {\sqrt [3]{a+b x^3}}{x^{11} \left (a d-b d x^3\right )} \, dx=-\frac {\frac {3 \sqrt [3]{a+b x^3} \left (14 a^3+22 a^2 b x^3+37 a b^2 x^6+169 b^3 x^9\right )}{x^{10}}+140 \sqrt [3]{2} \sqrt {3} b^{10/3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2^{2/3} \sqrt [3]{a+b x^3}}\right )+140 \sqrt [3]{2} b^{10/3} \log \left (-2 \sqrt [3]{b} x+2^{2/3} \sqrt [3]{a+b x^3}\right )-70 \sqrt [3]{2} b^{10/3} \log \left (2 b^{2/3} x^2+2^{2/3} \sqrt [3]{b} x \sqrt [3]{a+b x^3}+\sqrt [3]{2} \left (a+b x^3\right )^{2/3}\right )}{420 a^4 d} \]
-1/420*((3*(a + b*x^3)^(1/3)*(14*a^3 + 22*a^2*b*x^3 + 37*a*b^2*x^6 + 169*b ^3*x^9))/x^10 + 140*2^(1/3)*Sqrt[3]*b^(10/3)*ArcTan[(Sqrt[3]*b^(1/3)*x)/(b ^(1/3)*x + 2^(2/3)*(a + b*x^3)^(1/3))] + 140*2^(1/3)*b^(10/3)*Log[-2*b^(1/ 3)*x + 2^(2/3)*(a + b*x^3)^(1/3)] - 70*2^(1/3)*b^(10/3)*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)])/(a^4*d )
Time = 0.47 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {975, 27, 1053, 27, 1053, 25, 27, 1053, 27, 992}
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.
\(\displaystyle \int \frac {\sqrt [3]{a+b x^3}}{x^{11} \left (a d-b d x^3\right )} \, dx\) |
\(\Big \downarrow \) 975 |
\(\displaystyle \frac {\int \frac {b \left (9 b x^3+11 a\right )}{x^8 \left (a-b x^3\right ) \left (b x^3+a\right )^{2/3}}dx}{10 a d}-\frac {\sqrt [3]{a+b x^3}}{10 a d x^{10}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b \int \frac {9 b x^3+11 a}{x^8 \left (a-b x^3\right ) \left (b x^3+a\right )^{2/3}}dx}{10 a d}-\frac {\sqrt [3]{a+b x^3}}{10 a d x^{10}}\) |
\(\Big \downarrow \) 1053 |
\(\displaystyle \frac {b \left (-\frac {\int -\frac {2 a b \left (33 b x^3+37 a\right )}{x^5 \left (a-b x^3\right ) \left (b x^3+a\right )^{2/3}}dx}{7 a^2}-\frac {11 \sqrt [3]{a+b x^3}}{7 a x^7}\right )}{10 a d}-\frac {\sqrt [3]{a+b x^3}}{10 a d x^{10}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b \left (\frac {2 b \int \frac {33 b x^3+37 a}{x^5 \left (a-b x^3\right ) \left (b x^3+a\right )^{2/3}}dx}{7 a}-\frac {11 \sqrt [3]{a+b x^3}}{7 a x^7}\right )}{10 a d}-\frac {\sqrt [3]{a+b x^3}}{10 a d x^{10}}\) |
\(\Big \downarrow \) 1053 |
\(\displaystyle \frac {b \left (\frac {2 b \left (-\frac {\int -\frac {a b \left (111 b x^3+169 a\right )}{x^2 \left (a-b x^3\right ) \left (b x^3+a\right )^{2/3}}dx}{4 a^2}-\frac {37 \sqrt [3]{a+b x^3}}{4 a x^4}\right )}{7 a}-\frac {11 \sqrt [3]{a+b x^3}}{7 a x^7}\right )}{10 a d}-\frac {\sqrt [3]{a+b x^3}}{10 a d x^{10}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {b \left (\frac {2 b \left (\frac {\int \frac {a b \left (111 b x^3+169 a\right )}{x^2 \left (a-b x^3\right ) \left (b x^3+a\right )^{2/3}}dx}{4 a^2}-\frac {37 \sqrt [3]{a+b x^3}}{4 a x^4}\right )}{7 a}-\frac {11 \sqrt [3]{a+b x^3}}{7 a x^7}\right )}{10 a d}-\frac {\sqrt [3]{a+b x^3}}{10 a d x^{10}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b \left (\frac {2 b \left (\frac {b \int \frac {111 b x^3+169 a}{x^2 \left (a-b x^3\right ) \left (b x^3+a\right )^{2/3}}dx}{4 a}-\frac {37 \sqrt [3]{a+b x^3}}{4 a x^4}\right )}{7 a}-\frac {11 \sqrt [3]{a+b x^3}}{7 a x^7}\right )}{10 a d}-\frac {\sqrt [3]{a+b x^3}}{10 a d x^{10}}\) |
\(\Big \downarrow \) 1053 |
\(\displaystyle \frac {b \left (\frac {2 b \left (\frac {b \left (-\frac {\int -\frac {280 a^2 b x}{\left (a-b x^3\right ) \left (b x^3+a\right )^{2/3}}dx}{a^2}-\frac {169 \sqrt [3]{a+b x^3}}{a x}\right )}{4 a}-\frac {37 \sqrt [3]{a+b x^3}}{4 a x^4}\right )}{7 a}-\frac {11 \sqrt [3]{a+b x^3}}{7 a x^7}\right )}{10 a d}-\frac {\sqrt [3]{a+b x^3}}{10 a d x^{10}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b \left (\frac {2 b \left (\frac {b \left (280 b \int \frac {x}{\left (a-b x^3\right ) \left (b x^3+a\right )^{2/3}}dx-\frac {169 \sqrt [3]{a+b x^3}}{a x}\right )}{4 a}-\frac {37 \sqrt [3]{a+b x^3}}{4 a x^4}\right )}{7 a}-\frac {11 \sqrt [3]{a+b x^3}}{7 a x^7}\right )}{10 a d}-\frac {\sqrt [3]{a+b x^3}}{10 a d x^{10}}\) |
\(\Big \downarrow \) 992 |
\(\displaystyle \frac {b \left (\frac {2 b \left (\frac {b \left (280 b \left (-\frac {\arctan \left (\frac {\frac {2 \sqrt [3]{2} \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3} a b^{2/3}}+\frac {\log \left (a-b x^3\right )}{6\ 2^{2/3} a b^{2/3}}-\frac {\log \left (\sqrt [3]{2} \sqrt [3]{b} x-\sqrt [3]{a+b x^3}\right )}{2\ 2^{2/3} a b^{2/3}}\right )-\frac {169 \sqrt [3]{a+b x^3}}{a x}\right )}{4 a}-\frac {37 \sqrt [3]{a+b x^3}}{4 a x^4}\right )}{7 a}-\frac {11 \sqrt [3]{a+b x^3}}{7 a x^7}\right )}{10 a d}-\frac {\sqrt [3]{a+b x^3}}{10 a d x^{10}}\) |
-1/10*(a + b*x^3)^(1/3)/(a*d*x^10) + (b*((-11*(a + b*x^3)^(1/3))/(7*a*x^7) + (2*b*((-37*(a + b*x^3)^(1/3))/(4*a*x^4) + (b*((-169*(a + b*x^3)^(1/3))/ (a*x) + 280*b*(-(ArcTan[(1 + (2*2^(1/3)*b^(1/3)*x)/(a + b*x^3)^(1/3))/Sqrt [3]]/(2^(2/3)*Sqrt[3]*a*b^(2/3))) + Log[a - b*x^3]/(6*2^(2/3)*a*b^(2/3)) - Log[2^(1/3)*b^(1/3)*x - (a + b*x^3)^(1/3)]/(2*2^(2/3)*a*b^(2/3)))))/(4*a) ))/(7*a)))/(10*a*d)
3.6.80.3.1 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*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/ (a*e*(m + 1))), x] - Simp[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] && IntBinomi alQ[a, b, c, d, e, m, n, p, q, x]
Int[(x_)/(((a_) + (b_.)*(x_)^3)^(2/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/c, 3]}, Simp[-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]
Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_ ))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b *x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*c*g*(m + 1))), x] + Simp[1/(a*c*g^n*( m + 1)) Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2 ) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && IGtQ[n, 0] && LtQ[m, -1]
Time = 4.93 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.72
method | result | size |
pseudoelliptic | \(\frac {\left (-507 b^{3} x^{9}-111 a \,b^{2} x^{6}-66 a^{2} b \,x^{3}-42 a^{3}\right ) \left (b \,x^{3}+a \right )^{\frac {1}{3}}+70 \,2^{\frac {1}{3}} b^{\frac {10}{3}} x^{10} \left (2 \arctan \left (\frac {\sqrt {3}\, \left (2^{\frac {2}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}}+b^{\frac {1}{3}} x \right )}{3 b^{\frac {1}{3}} x}\right ) \sqrt {3}+\ln \left (\frac {2^{\frac {2}{3}} b^{\frac {2}{3}} x^{2}+2^{\frac {1}{3}} b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )-2 \ln \left (\frac {-2^{\frac {1}{3}} b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )\right )}{420 x^{10} a^{4} d}\) | \(171\) |
1/420*((-507*b^3*x^9-111*a*b^2*x^6-66*a^2*b*x^3-42*a^3)*(b*x^3+a)^(1/3)+70 *2^(1/3)*b^(10/3)*x^10*(2*arctan(1/3*3^(1/2)*(2^(2/3)*(b*x^3+a)^(1/3)+b^(1 /3)*x)/b^(1/3)/x)*3^(1/2)+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)-2*ln((-2^(1/3)*b^(1/3)*x+(b*x^3+a)^(1/3))/ x)))/x^10/a^4/d
Timed out. \[ \int \frac {\sqrt [3]{a+b x^3}}{x^{11} \left (a d-b d x^3\right )} \, dx=\text {Timed out} \]
\[ \int \frac {\sqrt [3]{a+b x^3}}{x^{11} \left (a d-b d x^3\right )} \, dx=- \frac {\int \frac {\sqrt [3]{a + b x^{3}}}{- a x^{11} + b x^{14}}\, dx}{d} \]
\[ \int \frac {\sqrt [3]{a+b x^3}}{x^{11} \left (a d-b d x^3\right )} \, dx=\int { -\frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{{\left (b d x^{3} - a d\right )} x^{11}} \,d x } \]
\[ \int \frac {\sqrt [3]{a+b x^3}}{x^{11} \left (a d-b d x^3\right )} \, dx=\int { -\frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{{\left (b d x^{3} - a d\right )} x^{11}} \,d x } \]
Timed out. \[ \int \frac {\sqrt [3]{a+b x^3}}{x^{11} \left (a d-b d x^3\right )} \, dx=\int \frac {{\left (b\,x^3+a\right )}^{1/3}}{x^{11}\,\left (a\,d-b\,d\,x^3\right )} \,d x \]