Integrand size = 24, antiderivative size = 229 \[ \int \frac {1}{x^3 \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\frac {b}{a (b c-a d) x^2 \sqrt [3]{a+b x^3}}-\frac {(3 b c-a d) \left (a+b x^3\right )^{2/3}}{2 a^2 c (b c-a d) x^2}+\frac {d^2 \arctan \left (\frac {1+\frac {2 \sqrt [3]{b c-a d} x}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} c^{5/3} (b c-a d)^{4/3}}+\frac {d^2 \log \left (c+d x^3\right )}{6 c^{5/3} (b c-a d)^{4/3}}-\frac {d^2 \log \left (\frac {\sqrt [3]{b c-a d} x}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{5/3} (b c-a d)^{4/3}} \]
b/a/(-a*d+b*c)/x^2/(b*x^3+a)^(1/3)-1/2*(-a*d+3*b*c)*(b*x^3+a)^(2/3)/a^2/c/ (-a*d+b*c)/x^2+1/6*d^2*ln(d*x^3+c)/c^(5/3)/(-a*d+b*c)^(4/3)-1/2*d^2*ln((-a *d+b*c)^(1/3)*x/c^(1/3)-(b*x^3+a)^(1/3))/c^(5/3)/(-a*d+b*c)^(4/3)+1/3*d^2* 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^(5/ 3)/(-a*d+b*c)^(4/3)*3^(1/2)
Result contains complex when optimal does not.
Time = 3.74 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.56 \[ \int \frac {1}{x^3 \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\frac {\frac {6 c^{2/3} \left (-a^2 d+3 b^2 c x^3+a b \left (c-d x^3\right )\right )}{a^2 (-b c+a d) x^2 \sqrt [3]{a+b x^3}}-\frac {2 \sqrt {-6+6 i \sqrt {3}} d^2 \arctan \left (\frac {3 \sqrt [3]{b c-a d} x}{\sqrt {3} \sqrt [3]{b c-a d} x-\left (3 i+\sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}}\right )}{(b c-a d)^{4/3}}+\frac {2 \left (1+i \sqrt {3}\right ) d^2 \log \left (2 \sqrt [3]{b c-a d} x+\left (1+i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{a+b x^3}\right )}{(b c-a d)^{4/3}}-\frac {i \left (-i+\sqrt {3}\right ) d^2 \log \left (2 (b c-a d)^{2/3} x^2+\left (-1-i \sqrt {3}\right ) \sqrt [3]{c} \sqrt [3]{b c-a d} x \sqrt [3]{a+b x^3}+i \left (i+\sqrt {3}\right ) c^{2/3} \left (a+b x^3\right )^{2/3}\right )}{(b c-a d)^{4/3}}}{12 c^{5/3}} \]
((6*c^(2/3)*(-(a^2*d) + 3*b^2*c*x^3 + a*b*(c - d*x^3)))/(a^2*(-(b*c) + a*d )*x^2*(a + b*x^3)^(1/3)) - (2*Sqrt[-6 + (6*I)*Sqrt[3]]*d^2*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))])/(b*c - a*d)^(4/3) + (2*(1 + I*Sqrt[3])*d^2*Log[2*(b*c - a*d)^(1/3)*x + (1 + I*Sqrt[3])*c^(1/3)*(a + b*x^3)^(1/3)])/(b*c - a*d)^(4/ 3) - (I*(-I + Sqrt[3])*d^2*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)])/(b*c - a*d)^(4/3))/(12*c^(5/3))
Time = 0.37 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {972, 25, 1053, 27, 901}
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.
| \(\displaystyle \int \frac {1}{x^3 \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx\) |
| \(\Big \downarrow \) 972 |
| \(\displaystyle \frac {b}{a x^2 \sqrt [3]{a+b x^3} (b c-a d)}-\frac {\int -\frac {3 b d x^3+3 b c-a d}{x^3 \sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{a (b c-a d)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {3 b d x^3+3 b c-a d}{x^3 \sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{a (b c-a d)}+\frac {b}{a x^2 \sqrt [3]{a+b x^3} (b c-a d)}\) |
| \(\Big \downarrow \) 1053 |
| \(\displaystyle \frac {-\frac {\int -\frac {2 a^2 d^2}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{2 a c}-\frac {\left (a+b x^3\right )^{2/3} (3 b c-a d)}{2 a c x^2}}{a (b c-a d)}+\frac {b}{a x^2 \sqrt [3]{a+b x^3} (b c-a d)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {a d^2 \int \frac {1}{\sqrt [3]{b x^3+a} \left (d x^3+c\right )}dx}{c}-\frac {\left (a+b x^3\right )^{2/3} (3 b c-a d)}{2 a c x^2}}{a (b c-a d)}+\frac {b}{a x^2 \sqrt [3]{a+b x^3} (b c-a d)}\) |
| \(\Big \downarrow \) 901 |
| \(\displaystyle \frac {\frac {a d^2 \left (\frac {\arctan \left (\frac {\frac {2 x \sqrt [3]{b c-a d}}{\sqrt [3]{c} \sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} c^{2/3} \sqrt [3]{b c-a d}}+\frac {\log \left (c+d x^3\right )}{6 c^{2/3} \sqrt [3]{b c-a d}}-\frac {\log \left (\frac {x \sqrt [3]{b c-a d}}{\sqrt [3]{c}}-\sqrt [3]{a+b x^3}\right )}{2 c^{2/3} \sqrt [3]{b c-a d}}\right )}{c}-\frac {\left (a+b x^3\right )^{2/3} (3 b c-a d)}{2 a c x^2}}{a (b c-a d)}+\frac {b}{a x^2 \sqrt [3]{a+b x^3} (b c-a d)}\) |
b/(a*(b*c - a*d)*x^2*(a + b*x^3)^(1/3)) + (-1/2*((3*b*c - a*d)*(a + b*x^3) ^(2/3))/(a*c*x^2) + (a*d^2*(ArcTan[(1 + (2*(b*c - a*d)^(1/3)*x)/(c^(1/3)*( a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*c^(2/3)*(b*c - a*d)^(1/3)) + Log[c + d*x^3]/(6*c^(2/3)*(b*c - a*d)^(1/3)) - Log[((b*c - a*d)^(1/3)*x)/c^(1/3) - (a + b*x^3)^(1/3)]/(2*c^(2/3)*(b*c - a*d)^(1/3))))/c)/(a*(b*c - a*d))
3.8.58.3.1 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[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> Wit h[{q = Rt[(b*c - a*d)/c, 3]}, Simp[ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/S qrt[3]]/(Sqrt[3]*c*q), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*c*q), x] + Simp[Log[c + d*x^3]/(6*c*q), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[(-b)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x ^n)^(q + 1)/(a*e*n*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*n*(b*c - a*d)*(p + 1)) Int[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*b*(m + 1) + n*( b*c - a*d)*(p + 1) + d*b*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{ a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] & & 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[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 = 4.76 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.12
| method | result | size |
| pseudoelliptic | \(\frac {-3 \left (\left (a b d -3 b^{2} c \right ) x^{3}+a^{2} d -a b c \right ) c \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}}+a^{2} d^{2} x^{2} \left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (-2 \arctan \left (\frac {\sqrt {3}\, \left (\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x -2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x}\right ) \sqrt {3}+\ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {2}{3}} x^{2}-\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )-2 \ln \left (\frac {\left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )\right )}{6 \left (\frac {a d -b c}{c}\right )^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} c^{2} x^{2} \left (a d -b c \right ) a^{2}}\) | \(257\) |
1/6*(-3*((a*b*d-3*b^2*c)*x^3+a^2*d-a*b*c)*c*((a*d-b*c)/c)^(1/3)+a^2*d^2*x^ 2*(b*x^3+a)^(1/3)*(-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)-2*ln((((a*d-b*c)/c)^ (1/3)*x+(b*x^3+a)^(1/3))/x)))/((a*d-b*c)/c)^(1/3)/(b*x^3+a)^(1/3)/c^2/x^2/ (a*d-b*c)/a^2
Timed out. \[ \int \frac {1}{x^3 \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\text {Timed out} \]
\[ \int \frac {1}{x^3 \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\int \frac {1}{x^{3} \left (a + b x^{3}\right )^{\frac {4}{3}} \left (c + d x^{3}\right )}\, dx \]
\[ \int \frac {1}{x^3 \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {4}{3}} {\left (d x^{3} + c\right )} x^{3}} \,d x } \]
\[ \int \frac {1}{x^3 \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {4}{3}} {\left (d x^{3} + c\right )} x^{3}} \,d x } \]
Timed out. \[ \int \frac {1}{x^3 \left (a+b x^3\right )^{4/3} \left (c+d x^3\right )} \, dx=\int \frac {1}{x^3\,{\left (b\,x^3+a\right )}^{4/3}\,\left (d\,x^3+c\right )} \,d x \]