Integrand size = 24, antiderivative size = 204 \[ \int \frac {\sqrt [3]{a+b x^3}}{x^5 \left (c+d x^3\right )} \, dx=-\frac {\sqrt [3]{a+b x^3}}{4 c x^4}-\frac {(b c-4 a d) \sqrt [3]{a+b x^3}}{4 a c^2 x}+\frac {d \sqrt [3]{b c-a d} \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} c^{7/3}}-\frac {d \sqrt [3]{b c-a d} \log \left (c+d x^3\right )}{6 c^{7/3}}+\frac {d \sqrt [3]{b c-a d} \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{7/3}} \]
-1/4*(b*x^3+a)^(1/3)/c/x^4-1/4*(-4*a*d+b*c)*(b*x^3+a)^(1/3)/a/c^2/x-1/6*d* (-a*d+b*c)^(1/3)*ln(d*x^3+c)/c^(7/3)+1/2*d*(-a*d+b*c)^(1/3)*ln((-a*d+b*c)^ (1/3)*x/c^(1/3)-(b*x^3+a)^(1/3))/c^(7/3)+1/3*d*(-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))/c^(7/3)*3^(1/2)
Result contains complex when optimal does not.
Time = 2.62 (sec) , antiderivative size = 333, normalized size of antiderivative = 1.63 \[ \int \frac {\sqrt [3]{a+b x^3}}{x^5 \left (c+d x^3\right )} \, dx=\frac {\frac {3 \sqrt [3]{c} \sqrt [3]{a+b x^3} \left (-a c-b c x^3+4 a d x^3\right )}{a x^4}-2 \sqrt {-6-6 i \sqrt {3}} d \sqrt [3]{b c-a d} \arctan \left (\frac {3 \sqrt [3]{b c-a d} x}{\sqrt {3} \sqrt [3]{b c-a d} x-\left (3 i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )+2 i \left (i+\sqrt {3}\right ) d \sqrt [3]{b c-a d} \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )+\left (1-i \sqrt {3}\right ) d \sqrt [3]{b c-a d} \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{12 c^{7/3}} \]
((3*c^(1/3)*(a + b*x^3)^(1/3)*(-(a*c) - b*c*x^3 + 4*a*d*x^3))/(a*x^4) - 2* Sqrt[-6 - (6*I)*Sqrt[3]]*d*(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) )] + (2*I)*(I + Sqrt[3])*d*(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)] + (1 - I*Sqrt[3])*d*(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)])/(12 *c^(7/3))
Time = 0.35 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.09, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {975, 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.
| \(\displaystyle \int \frac {\sqrt [3]{a+b x^3}}{x^5 \left (c+d x^3\right )} \, dx\) |
| \(\Big \downarrow \) 975 |
| \(\displaystyle \frac {\int \frac {-3 b d x^3+b c-4 a d}{x^2 \left (b x^3+a\right )^{2/3} \left (d x^3+c\right )}dx}{4 c}-\frac {\sqrt [3]{a+b x^3}}{4 c x^4}\) |
| \(\Big \downarrow \) 1053 |
| \(\displaystyle \frac {-\frac {\int \frac {4 a d (b c-a d) x}{\left (b x^3+a\right )^{2/3} \left (d x^3+c\right )}dx}{a c}-\frac {\sqrt [3]{a+b x^3} (b c-4 a d)}{a c x}}{4 c}-\frac {\sqrt [3]{a+b x^3}}{4 c x^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {4 d (b c-a d) \int \frac {x}{\left (b x^3+a\right )^{2/3} \left (d x^3+c\right )}dx}{c}-\frac {\sqrt [3]{a+b x^3} (b c-4 a d)}{a c x}}{4 c}-\frac {\sqrt [3]{a+b x^3}}{4 c x^4}\) |
| \(\Big \downarrow \) 992 |
| \(\displaystyle \frac {-\frac {4 d (b c-a d) \left (-\frac {\arctan \left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{c} (b c-a d)^{2/3}}+\frac {\log \left (c+d x^3\right )}{6 \sqrt [3]{c} (b c-a d)^{2/3}}-\frac {\log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 \sqrt [3]{c} (b c-a d)^{2/3}}\right )}{c}-\frac {\sqrt [3]{a+b x^3} (b c-4 a d)}{a c x}}{4 c}-\frac {\sqrt [3]{a+b x^3}}{4 c x^4}\) |
-1/4*(a + b*x^3)^(1/3)/(c*x^4) + (-(((b*c - 4*a*d)*(a + b*x^3)^(1/3))/(a*c *x)) - (4*d*(b*c - a*d)*(-(ArcTan[(1 + (2*(b*c - a*d)^(1/3)*x)/(c^(1/3)*(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*c^(1/3)*(b*c - a*d)^(2/3))) + Log[c + d*x^3]/(6*c^(1/3)*(b*c - a*d)^(2/3)) - Log[((b*c - a*d)^(1/3)*x)/c^(1/3) - (a + b*x^3)^(1/3)]/(2*c^(1/3)*(b*c - a*d)^(2/3))))/c)/(4*c)
3.7.69.3.1 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 = 4.81 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.20
| method | result | size |
| pseudoelliptic | \(\frac {-2 \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) a \left (a d -b c \right ) d \,x^{4}-\frac {3 \left (\left (-4 a d +b c \right ) x^{3}+a c \right ) \left (b \,x^{3}+a \right )^{\frac {1}{3}} c \left (\frac {a d -b c}{c}\right )^{\frac {2}{3}}}{2}+x^{4} \left (2 \arctan \left (\frac {\sqrt {3}\, \left (\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x -2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x}\right ) \sqrt {3}+\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )\right ) d a \left (a d -b c \right )}{6 \left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{4} c^{3} a}\) | \(245\) |
1/6/((a*d-b*c)/c)^(2/3)*(-2*ln((((a*d-b*c)/c)^(1/3)*x+(b*x^3+a)^(1/3))/x)* a*(a*d-b*c)*d*x^4-3/2*((-4*a*d+b*c)*x^3+a*c)*(b*x^3+a)^(1/3)*c*((a*d-b*c)/ c)^(2/3)+x^4*(2*arctan(1/3*3^(1/2)*(((a*d-b*c)/c)^(1/3)*x-2*(b*x^3+a)^(1/3 ))/((a*d-b*c)/c)^(1/3)/x)*3^(1/2)+ln((((a*d-b*c)/c)^(2/3)*x^2-((a*d-b*c)/c )^(1/3)*(b*x^3+a)^(1/3)*x+(b*x^3+a)^(2/3))/x^2))*d*a*(a*d-b*c))/x^4/c^3/a
Timed out. \[ \int \frac {\sqrt [3]{a+b x^3}}{x^5 \left (c+d x^3\right )} \, dx=\text {Timed out} \]
\[ \int \frac {\sqrt [3]{a+b x^3}}{x^5 \left (c+d x^3\right )} \, dx=\int \frac {\sqrt [3]{a + b x^{3}}}{x^{5} \left (c + d x^{3}\right )}\, dx \]
\[ \int \frac {\sqrt [3]{a+b x^3}}{x^5 \left (c+d x^3\right )} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{{\left (d x^{3} + c\right )} x^{5}} \,d x } \]
\[ \int \frac {\sqrt [3]{a+b x^3}}{x^5 \left (c+d x^3\right )} \, dx=\int { \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{{\left (d x^{3} + c\right )} x^{5}} \,d x } \]
Timed out. \[ \int \frac {\sqrt [3]{a+b x^3}}{x^5 \left (c+d x^3\right )} \, dx=\int \frac {{\left (b\,x^3+a\right )}^{1/3}}{x^5\,\left (d\,x^3+c\right )} \,d x \]