Integrand size = 22, antiderivative size = 318 \[ \int \frac {1}{x^5 \left (a+b x^3\right ) \left (c+d x^3\right )} \, dx=-\frac {1}{4 a c x^4}+\frac {b c+a d}{a^2 c^2 x}-\frac {b^{7/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{7/3} (b c-a d)}+\frac {d^{7/3} \arctan \left (\frac {\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )}{\sqrt {3} c^{7/3} (b c-a d)}-\frac {b^{7/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{7/3} (b c-a d)}+\frac {d^{7/3} \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{3 c^{7/3} (b c-a d)}+\frac {b^{7/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{7/3} (b c-a d)}-\frac {d^{7/3} \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{6 c^{7/3} (b c-a d)} \] Output:
-1/4/a/c/x^4+(a*d+b*c)/a^2/c^2/x-1/3*b^(7/3)*arctan(1/3*(a^(1/3)-2*b^(1/3) *x)*3^(1/2)/a^(1/3))*3^(1/2)/a^(7/3)/(-a*d+b*c)+1/3*d^(7/3)*arctan(1/3*(c^ (1/3)-2*d^(1/3)*x)*3^(1/2)/c^(1/3))*3^(1/2)/c^(7/3)/(-a*d+b*c)-1/3*b^(7/3) *ln(a^(1/3)+b^(1/3)*x)/a^(7/3)/(-a*d+b*c)+1/3*d^(7/3)*ln(c^(1/3)+d^(1/3)*x )/c^(7/3)/(-a*d+b*c)+1/6*b^(7/3)*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2) /a^(7/3)/(-a*d+b*c)-1/6*d^(7/3)*ln(c^(2/3)-c^(1/3)*d^(1/3)*x+d^(2/3)*x^2)/ c^(7/3)/(-a*d+b*c)
Time = 0.27 (sec) , antiderivative size = 282, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x^5 \left (a+b x^3\right ) \left (c+d x^3\right )} \, dx=\frac {\frac {3 b}{a}-\frac {3 d}{c}-\frac {12 b^2 x^3}{a^2}+\frac {12 d^2 x^3}{c^2}+\frac {4 \sqrt {3} b^{7/3} x^4 \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{a^{7/3}}-\frac {4 \sqrt {3} d^{7/3} x^4 \arctan \left (\frac {1-\frac {2 \sqrt [3]{d} x}{\sqrt [3]{c}}}{\sqrt {3}}\right )}{c^{7/3}}+\frac {4 b^{7/3} x^4 \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{a^{7/3}}-\frac {4 d^{7/3} x^4 \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{c^{7/3}}-\frac {2 b^{7/3} x^4 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{a^{7/3}}+\frac {2 d^{7/3} x^4 \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{c^{7/3}}}{12 (-b c+a d) x^4} \] Input:
Integrate[1/(x^5*(a + b*x^3)*(c + d*x^3)),x]
Output:
((3*b)/a - (3*d)/c - (12*b^2*x^3)/a^2 + (12*d^2*x^3)/c^2 + (4*Sqrt[3]*b^(7 /3)*x^4*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]])/a^(7/3) - (4*Sqrt[3]* d^(7/3)*x^4*ArcTan[(1 - (2*d^(1/3)*x)/c^(1/3))/Sqrt[3]])/c^(7/3) + (4*b^(7 /3)*x^4*Log[a^(1/3) + b^(1/3)*x])/a^(7/3) - (4*d^(7/3)*x^4*Log[c^(1/3) + d ^(1/3)*x])/c^(7/3) - (2*b^(7/3)*x^4*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2 /3)*x^2])/a^(7/3) + (2*d^(7/3)*x^4*Log[c^(2/3) - c^(1/3)*d^(1/3)*x + d^(2/ 3)*x^2])/c^(7/3))/(12*(-(b*c) + a*d)*x^4)
Time = 0.95 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.12, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {980, 27, 1053, 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.
\(\displaystyle \int \frac {1}{x^5 \left (a+b x^3\right ) \left (c+d x^3\right )} \, dx\) |
\(\Big \downarrow \) 980 |
\(\displaystyle \frac {\int -\frac {4 \left (b d x^3+b c+a d\right )}{x^2 \left (b x^3+a\right ) \left (d x^3+c\right )}dx}{4 a c}-\frac {1}{4 a c x^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {b d x^3+b c+a d}{x^2 \left (b x^3+a\right ) \left (d x^3+c\right )}dx}{a c}-\frac {1}{4 a c x^4}\) |
\(\Big \downarrow \) 1053 |
\(\displaystyle -\frac {-\frac {\int \frac {x \left (b d (b c+a d) x^3+b^2 c^2+a^2 d^2+a b c d\right )}{\left (b x^3+a\right ) \left (d x^3+c\right )}dx}{a c}-\frac {a d+b c}{a c x}}{a c}-\frac {1}{4 a c x^4}\) |
\(\Big \downarrow \) 1054 |
\(\displaystyle -\frac {-\frac {\int \left (\frac {c^2 x b^3}{(b c-a d) \left (b x^3+a\right )}+\frac {a^2 d^3 x}{(a d-b c) \left (d x^3+c\right )}\right )dx}{a c}-\frac {a d+b c}{a c x}}{a c}-\frac {1}{4 a c x^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {\frac {b^{7/3} c^2 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 \sqrt [3]{a} (b c-a d)}+\frac {a^2 d^{7/3} \arctan \left (\frac {\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt {3} \sqrt [3]{c}}\right )}{\sqrt {3} \sqrt [3]{c} (b c-a d)}-\frac {a^2 d^{7/3} \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{6 \sqrt [3]{c} (b c-a d)}+\frac {a^2 d^{7/3} \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{3 \sqrt [3]{c} (b c-a d)}-\frac {b^{7/3} c^2 \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} \sqrt [3]{a} (b c-a d)}-\frac {b^{7/3} c^2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} (b c-a d)}}{a c}-\frac {a d+b c}{a c x}}{a c}-\frac {1}{4 a c x^4}\) |
Input:
Int[1/(x^5*(a + b*x^3)*(c + d*x^3)),x]
Output:
-1/4*1/(a*c*x^4) - (-((b*c + a*d)/(a*c*x)) - (-((b^(7/3)*c^2*ArcTan[(a^(1/ 3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(1/3)*(b*c - a*d))) + (a^ 2*d^(7/3)*ArcTan[(c^(1/3) - 2*d^(1/3)*x)/(Sqrt[3]*c^(1/3))])/(Sqrt[3]*c^(1 /3)*(b*c - a*d)) - (b^(7/3)*c^2*Log[a^(1/3) + b^(1/3)*x])/(3*a^(1/3)*(b*c - a*d)) + (a^2*d^(7/3)*Log[c^(1/3) + d^(1/3)*x])/(3*c^(1/3)*(b*c - a*d)) + (b^(7/3)*c^2*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(6*a^(1/3)*( b*c - a*d)) - (a^2*d^(7/3)*Log[c^(2/3) - c^(1/3)*d^(1/3)*x + d^(2/3)*x^2]) /(6*c^(1/3)*(b*c - a*d)))/(a*c))/(a*c)
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 + 1)/(a*c*e*(m + 1))), x] - Simp[1/(a*c*e^n*(m + 1)) Int[(e*x)^(m + n)*( a + b*x^n)^p*(c + d*x^n)^q*Simp[(b*c + a*d)*(m + n + 1) + n*(b*c*p + a*d*q) + b*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -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]
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 = 1.13 (sec) , antiderivative size = 248, normalized size of antiderivative = 0.78
method | result | size |
default | \(\frac {\left (-\frac {\ln \left (x +\left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}{3 d \left (\frac {c}{d}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {c}{d}\right )^{\frac {1}{3}} x +\left (\frac {c}{d}\right )^{\frac {2}{3}}\right )}{6 d \left (\frac {c}{d}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {c}{d}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 d \left (\frac {c}{d}\right )^{\frac {1}{3}}}\right ) d^{3}}{c^{2} \left (a d -b c \right )}-\frac {1}{4 a c \,x^{4}}-\frac {-a d -b c}{a^{2} c^{2} x}-\frac {\left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right ) b^{3}}{a^{2} \left (a d -b c \right )}\) | \(248\) |
risch | \(\frac {\frac {\left (a d +b c \right ) x^{3}}{a^{2} c^{2}}-\frac {1}{4 a c}}{x^{4}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (d^{3} a^{10}-3 c \,d^{2} a^{9} b +3 c^{2} d \,a^{8} b^{2}-a^{7} b^{3} c^{3}\right ) \textit {\_Z}^{3}-b^{7}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (-4 a^{13} c^{7} d^{6}+22 a^{12} b \,c^{8} d^{5}-52 a^{11} b^{2} c^{9} d^{4}+68 a^{10} b^{3} c^{10} d^{3}-52 a^{9} b^{4} c^{11} d^{2}+22 a^{8} b^{5} c^{12} d -4 a^{7} b^{6} c^{13}\right ) \textit {\_R}^{6}+\left (-3 a^{10} d^{10}+9 a^{9} b c \,d^{9}-10 a^{8} c^{2} d^{8} b^{2}+4 a^{7} c^{3} d^{7} b^{3}+4 a^{3} c^{7} d^{3} b^{7}-10 a^{2} c^{8} d^{2} b^{8}+9 a \,c^{9} d \,b^{9}-3 c^{10} b^{10}\right ) \textit {\_R}^{3}+3 b^{7} d^{7}\right ) x +\left (a^{12} c^{5} d^{7}-3 a^{11} b \,c^{6} d^{6}+3 a^{10} b^{2} c^{7} d^{5}-a^{9} b^{3} c^{8} d^{4}-a^{8} b^{4} c^{9} d^{3}+3 a^{7} b^{5} c^{10} d^{2}-3 a^{6} b^{6} c^{11} d +a^{5} b^{7} c^{12}\right ) \textit {\_R}^{5}+\left (-a^{5} b^{4} c^{2} d^{7}-a^{2} b^{7} c^{5} d^{4}\right ) \textit {\_R}^{2}\right )\right )}{3}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (d^{3} c^{7} a^{3}-3 a^{2} b \,c^{8} d^{2}+3 a \,b^{2} c^{9} d -b^{3} c^{10}\right ) \textit {\_Z}^{3}+d^{7}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (-4 a^{13} c^{7} d^{6}+22 a^{12} b \,c^{8} d^{5}-52 a^{11} b^{2} c^{9} d^{4}+68 a^{10} b^{3} c^{10} d^{3}-52 a^{9} b^{4} c^{11} d^{2}+22 a^{8} b^{5} c^{12} d -4 a^{7} b^{6} c^{13}\right ) \textit {\_R}^{6}+\left (-3 a^{10} d^{10}+9 a^{9} b c \,d^{9}-10 a^{8} c^{2} d^{8} b^{2}+4 a^{7} c^{3} d^{7} b^{3}+4 a^{3} c^{7} d^{3} b^{7}-10 a^{2} c^{8} d^{2} b^{8}+9 a \,c^{9} d \,b^{9}-3 c^{10} b^{10}\right ) \textit {\_R}^{3}+3 b^{7} d^{7}\right ) x +\left (a^{12} c^{5} d^{7}-3 a^{11} b \,c^{6} d^{6}+3 a^{10} b^{2} c^{7} d^{5}-a^{9} b^{3} c^{8} d^{4}-a^{8} b^{4} c^{9} d^{3}+3 a^{7} b^{5} c^{10} d^{2}-3 a^{6} b^{6} c^{11} d +a^{5} b^{7} c^{12}\right ) \textit {\_R}^{5}+\left (-a^{5} b^{4} c^{2} d^{7}-a^{2} b^{7} c^{5} d^{4}\right ) \textit {\_R}^{2}\right )\right )}{3}\) | \(830\) |
Input:
int(1/x^5/(b*x^3+a)/(d*x^3+c),x,method=_RETURNVERBOSE)
Output:
(-1/3/d/(c/d)^(1/3)*ln(x+(c/d)^(1/3))+1/6/d/(c/d)^(1/3)*ln(x^2-(c/d)^(1/3) *x+(c/d)^(2/3))+1/3*3^(1/2)/d/(c/d)^(1/3)*arctan(1/3*3^(1/2)*(2/(c/d)^(1/3 )*x-1)))*d^3/c^2/(a*d-b*c)-1/4/a/c/x^4-1/a^2/c^2*(-a*d-b*c)/x-(-1/3/b/(a/b )^(1/3)*ln(x+(a/b)^(1/3))+1/6/b/(a/b)^(1/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/ 3))+1/3*3^(1/2)/b/(a/b)^(1/3)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1)))*b^3 /a^2/(a*d-b*c)
Time = 0.72 (sec) , antiderivative size = 305, normalized size of antiderivative = 0.96 \[ \int \frac {1}{x^5 \left (a+b x^3\right ) \left (c+d x^3\right )} \, dx=\frac {4 \, \sqrt {3} b^{2} c^{2} x^{4} \left (\frac {b}{a}\right )^{\frac {1}{3}} \arctan \left (\frac {2}{3} \, \sqrt {3} x \left (\frac {b}{a}\right )^{\frac {1}{3}} - \frac {1}{3} \, \sqrt {3}\right ) - 4 \, \sqrt {3} a^{2} d^{2} x^{4} \left (-\frac {d}{c}\right )^{\frac {1}{3}} \arctan \left (\frac {2}{3} \, \sqrt {3} x \left (-\frac {d}{c}\right )^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) + 2 \, b^{2} c^{2} x^{4} \left (\frac {b}{a}\right )^{\frac {1}{3}} \log \left (b x^{2} - a x \left (\frac {b}{a}\right )^{\frac {2}{3}} + a \left (\frac {b}{a}\right )^{\frac {1}{3}}\right ) + 2 \, a^{2} d^{2} x^{4} \left (-\frac {d}{c}\right )^{\frac {1}{3}} \log \left (d x^{2} - c x \left (-\frac {d}{c}\right )^{\frac {2}{3}} - c \left (-\frac {d}{c}\right )^{\frac {1}{3}}\right ) - 4 \, b^{2} c^{2} x^{4} \left (\frac {b}{a}\right )^{\frac {1}{3}} \log \left (b x + a \left (\frac {b}{a}\right )^{\frac {2}{3}}\right ) - 4 \, a^{2} d^{2} x^{4} \left (-\frac {d}{c}\right )^{\frac {1}{3}} \log \left (d x + c \left (-\frac {d}{c}\right )^{\frac {2}{3}}\right ) - 3 \, a b c^{2} + 3 \, a^{2} c d + 12 \, {\left (b^{2} c^{2} - a^{2} d^{2}\right )} x^{3}}{12 \, {\left (a^{2} b c^{3} - a^{3} c^{2} d\right )} x^{4}} \] Input:
integrate(1/x^5/(b*x^3+a)/(d*x^3+c),x, algorithm="fricas")
Output:
1/12*(4*sqrt(3)*b^2*c^2*x^4*(b/a)^(1/3)*arctan(2/3*sqrt(3)*x*(b/a)^(1/3) - 1/3*sqrt(3)) - 4*sqrt(3)*a^2*d^2*x^4*(-d/c)^(1/3)*arctan(2/3*sqrt(3)*x*(- d/c)^(1/3) + 1/3*sqrt(3)) + 2*b^2*c^2*x^4*(b/a)^(1/3)*log(b*x^2 - a*x*(b/a )^(2/3) + a*(b/a)^(1/3)) + 2*a^2*d^2*x^4*(-d/c)^(1/3)*log(d*x^2 - c*x*(-d/ c)^(2/3) - c*(-d/c)^(1/3)) - 4*b^2*c^2*x^4*(b/a)^(1/3)*log(b*x + a*(b/a)^( 2/3)) - 4*a^2*d^2*x^4*(-d/c)^(1/3)*log(d*x + c*(-d/c)^(2/3)) - 3*a*b*c^2 + 3*a^2*c*d + 12*(b^2*c^2 - a^2*d^2)*x^3)/((a^2*b*c^3 - a^3*c^2*d)*x^4)
Timed out. \[ \int \frac {1}{x^5 \left (a+b x^3\right ) \left (c+d x^3\right )} \, dx=\text {Timed out} \] Input:
integrate(1/x**5/(b*x**3+a)/(d*x**3+c),x)
Output:
Timed out
Time = 0.11 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.07 \[ \int \frac {1}{x^5 \left (a+b x^3\right ) \left (c+d x^3\right )} \, dx=\frac {\sqrt {3} b^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, {\left (a^{2} b c - a^{3} d\right )} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {\sqrt {3} d^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {c}{d}\right )^{\frac {1}{3}}}\right )}{3 \, {\left (b c^{3} - a c^{2} d\right )} \left (\frac {c}{d}\right )^{\frac {1}{3}}} + \frac {b^{2} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, {\left (a^{2} b c \left (\frac {a}{b}\right )^{\frac {1}{3}} - a^{3} d \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}} - \frac {d^{2} \log \left (x^{2} - x \left (\frac {c}{d}\right )^{\frac {1}{3}} + \left (\frac {c}{d}\right )^{\frac {2}{3}}\right )}{6 \, {\left (b c^{3} \left (\frac {c}{d}\right )^{\frac {1}{3}} - a c^{2} d \left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}} - \frac {b^{2} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, {\left (a^{2} b c \left (\frac {a}{b}\right )^{\frac {1}{3}} - a^{3} d \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}} + \frac {d^{2} \log \left (x + \left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}{3 \, {\left (b c^{3} \left (\frac {c}{d}\right )^{\frac {1}{3}} - a c^{2} d \left (\frac {c}{d}\right )^{\frac {1}{3}}\right )}} + \frac {4 \, {\left (b c + a d\right )} x^{3} - a c}{4 \, a^{2} c^{2} x^{4}} \] Input:
integrate(1/x^5/(b*x^3+a)/(d*x^3+c),x, algorithm="maxima")
Output:
1/3*sqrt(3)*b^2*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/((a^2* b*c - a^3*d)*(a/b)^(1/3)) - 1/3*sqrt(3)*d^2*arctan(1/3*sqrt(3)*(2*x - (c/d )^(1/3))/(c/d)^(1/3))/((b*c^3 - a*c^2*d)*(c/d)^(1/3)) + 1/6*b^2*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(a^2*b*c*(a/b)^(1/3) - a^3*d*(a/b)^(1/3)) - 1 /6*d^2*log(x^2 - x*(c/d)^(1/3) + (c/d)^(2/3))/(b*c^3*(c/d)^(1/3) - a*c^2*d *(c/d)^(1/3)) - 1/3*b^2*log(x + (a/b)^(1/3))/(a^2*b*c*(a/b)^(1/3) - a^3*d* (a/b)^(1/3)) + 1/3*d^2*log(x + (c/d)^(1/3))/(b*c^3*(c/d)^(1/3) - a*c^2*d*( c/d)^(1/3)) + 1/4*(4*(b*c + a*d)*x^3 - a*c)/(a^2*c^2*x^4)
Time = 0.15 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.03 \[ \int \frac {1}{x^5 \left (a+b x^3\right ) \left (c+d x^3\right )} \, dx=-\frac {b^{3} \left (-\frac {a}{b}\right )^{\frac {2}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{3 \, {\left (a^{3} b c - a^{4} d\right )}} + \frac {d^{3} \left (-\frac {c}{d}\right )^{\frac {2}{3}} \log \left ({\left | x - \left (-\frac {c}{d}\right )^{\frac {1}{3}} \right |}\right )}{3 \, {\left (b c^{4} - a c^{3} d\right )}} - \frac {\left (-a b^{2}\right )^{\frac {2}{3}} b \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{\sqrt {3} a^{3} b c - \sqrt {3} a^{4} d} + \frac {\left (-c d^{2}\right )^{\frac {2}{3}} d \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {c}{d}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {c}{d}\right )^{\frac {1}{3}}}\right )}{\sqrt {3} b c^{4} - \sqrt {3} a c^{3} d} + \frac {\left (-a b^{2}\right )^{\frac {2}{3}} b \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, {\left (a^{3} b c - a^{4} d\right )}} - \frac {\left (-c d^{2}\right )^{\frac {2}{3}} d \log \left (x^{2} + x \left (-\frac {c}{d}\right )^{\frac {1}{3}} + \left (-\frac {c}{d}\right )^{\frac {2}{3}}\right )}{6 \, {\left (b c^{4} - a c^{3} d\right )}} + \frac {4 \, b c x^{3} + 4 \, a d x^{3} - a c}{4 \, a^{2} c^{2} x^{4}} \] Input:
integrate(1/x^5/(b*x^3+a)/(d*x^3+c),x, algorithm="giac")
Output:
-1/3*b^3*(-a/b)^(2/3)*log(abs(x - (-a/b)^(1/3)))/(a^3*b*c - a^4*d) + 1/3*d ^3*(-c/d)^(2/3)*log(abs(x - (-c/d)^(1/3)))/(b*c^4 - a*c^3*d) - (-a*b^2)^(2 /3)*b*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/(sqrt(3)*a^3*b *c - sqrt(3)*a^4*d) + (-c*d^2)^(2/3)*d*arctan(1/3*sqrt(3)*(2*x + (-c/d)^(1 /3))/(-c/d)^(1/3))/(sqrt(3)*b*c^4 - sqrt(3)*a*c^3*d) + 1/6*(-a*b^2)^(2/3)* b*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/(a^3*b*c - a^4*d) - 1/6*(-c*d^2 )^(2/3)*d*log(x^2 + x*(-c/d)^(1/3) + (-c/d)^(2/3))/(b*c^4 - a*c^3*d) + 1/4 *(4*b*c*x^3 + 4*a*d*x^3 - a*c)/(a^2*c^2*x^4)
Time = 10.18 (sec) , antiderivative size = 1734, normalized size of antiderivative = 5.45 \[ \int \frac {1}{x^5 \left (a+b x^3\right ) \left (c+d x^3\right )} \, dx=\text {Too large to display} \] Input:
int(1/(x^5*(a + b*x^3)*(c + d*x^3)),x)
Output:
log(((b^7/(a^7*(a*d - b*c)^3))^(2/3)*(((27*a^14*b^3*c^14*d^3*x*(a^6*d^6 + b^6*c^6)*(a*d - b*c)^2 + 27*a^19*b^3*c^19*d^3*(a*d + b*c)*(a*d - b*c)^4*(b ^7/(a^7*(a*d - b*c)^3))^(2/3))*(b^7/(a^7*(a*d - b*c)^3))^(1/3))/3 + 9*a^13 *b^11*c^20*d^4 - 9*a^14*b^10*c^19*d^5 - 9*a^19*b^5*c^14*d^10 + 9*a^20*b^4* c^13*d^11))/9 + a^13*b^9*c^13*d^9*x)*(b^7/(27*a^10*d^3 - 27*a^7*b^3*c^3 + 81*a^8*b^2*c^2*d - 81*a^9*b*c*d^2))^(1/3) + log(((-d^7/(c^7*(a*d - b*c)^3) )^(2/3)*(((27*a^14*b^3*c^14*d^3*x*(a^6*d^6 + b^6*c^6)*(a*d - b*c)^2 + 27*a ^19*b^3*c^19*d^3*(a*d + b*c)*(a*d - b*c)^4*(-d^7/(c^7*(a*d - b*c)^3))^(2/3 ))*(-d^7/(c^7*(a*d - b*c)^3))^(1/3))/3 + 9*a^13*b^11*c^20*d^4 - 9*a^14*b^1 0*c^19*d^5 - 9*a^19*b^5*c^14*d^10 + 9*a^20*b^4*c^13*d^11))/9 + a^13*b^9*c^ 13*d^9*x)*(d^7/(27*b^3*c^10 - 27*a^3*c^7*d^3 + 81*a^2*b*c^8*d^2 - 81*a*b^2 *c^9*d))^(1/3) - (1/(4*a*c) - (x^3*(a*d + b*c))/(a^2*c^2))/x^4 + (log(((3^ (1/2)*1i - 1)^2*(b^7/(a^7*(a*d - b*c)^3))^(2/3)*(((3^(1/2)*1i - 1)*(27*a^1 4*b^3*c^14*d^3*x*(a^6*d^6 + b^6*c^6)*(a*d - b*c)^2 + (27*a^19*b^3*c^19*d^3 *(3^(1/2)*1i - 1)^2*(a*d + b*c)*(a*d - b*c)^4*(b^7/(a^7*(a*d - b*c)^3))^(2 /3))/4)*(b^7/(a^7*(a*d - b*c)^3))^(1/3))/6 + 9*a^13*b^11*c^20*d^4 - 9*a^14 *b^10*c^19*d^5 - 9*a^19*b^5*c^14*d^10 + 9*a^20*b^4*c^13*d^11))/36 + a^13*b ^9*c^13*d^9*x)*(b^7/(27*a^10*d^3 - 27*a^7*b^3*c^3 + 81*a^8*b^2*c^2*d - 81* a^9*b*c*d^2))^(1/3)*(3^(1/2)*1i - 1))/2 - (log(((3^(1/2)*1i + 1)^2*(b^7/(a ^7*(a*d - b*c)^3))^(2/3)*(((3^(1/2)*1i + 1)*(27*a^14*b^3*c^14*d^3*x*(a^...
Time = 0.24 (sec) , antiderivative size = 278, normalized size of antiderivative = 0.87 \[ \int \frac {1}{x^5 \left (a+b x^3\right ) \left (c+d x^3\right )} \, dx=\frac {4 d^{\frac {2}{3}} c^{\frac {7}{3}} \sqrt {3}\, \mathit {atan} \left (\frac {a^{\frac {1}{3}}-2 b^{\frac {1}{3}} x}{a^{\frac {1}{3}} \sqrt {3}}\right ) b^{3} x^{4}-4 b^{\frac {2}{3}} a^{\frac {7}{3}} \sqrt {3}\, \mathit {atan} \left (\frac {c^{\frac {1}{3}}-2 d^{\frac {1}{3}} x}{c^{\frac {1}{3}} \sqrt {3}}\right ) d^{3} x^{4}-3 d^{\frac {5}{3}} c^{\frac {4}{3}} b^{\frac {2}{3}} a^{\frac {7}{3}}+12 d^{\frac {8}{3}} c^{\frac {1}{3}} b^{\frac {2}{3}} a^{\frac {7}{3}} x^{3}+3 d^{\frac {2}{3}} c^{\frac {7}{3}} b^{\frac {5}{3}} a^{\frac {4}{3}}-12 d^{\frac {2}{3}} c^{\frac {7}{3}} b^{\frac {8}{3}} a^{\frac {1}{3}} x^{3}+2 b^{\frac {2}{3}} a^{\frac {7}{3}} \mathrm {log}\left (c^{\frac {2}{3}}-d^{\frac {1}{3}} c^{\frac {1}{3}} x +d^{\frac {2}{3}} x^{2}\right ) d^{3} x^{4}-4 b^{\frac {2}{3}} a^{\frac {7}{3}} \mathrm {log}\left (c^{\frac {1}{3}}+d^{\frac {1}{3}} x \right ) d^{3} x^{4}-2 d^{\frac {2}{3}} c^{\frac {7}{3}} \mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) b^{3} x^{4}+4 d^{\frac {2}{3}} c^{\frac {7}{3}} \mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) b^{3} x^{4}}{12 d^{\frac {2}{3}} c^{\frac {7}{3}} b^{\frac {2}{3}} a^{\frac {7}{3}} x^{4} \left (a d -b c \right )} \] Input:
int(1/x^5/(b*x^3+a)/(d*x^3+c),x)
Output:
(4*d**(2/3)*c**(1/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt (3)))*b**3*c**2*x**4 - 4*b**(2/3)*a**(1/3)*sqrt(3)*atan((c**(1/3) - 2*d**( 1/3)*x)/(c**(1/3)*sqrt(3)))*a**2*d**3*x**4 - 3*d**(2/3)*c**(1/3)*b**(2/3)* a**(1/3)*a**2*c*d + 12*d**(2/3)*c**(1/3)*b**(2/3)*a**(1/3)*a**2*d**2*x**3 + 3*d**(2/3)*c**(1/3)*b**(2/3)*a**(1/3)*a*b*c**2 - 12*d**(2/3)*c**(1/3)*b* *(2/3)*a**(1/3)*b**2*c**2*x**3 + 2*b**(2/3)*a**(1/3)*log(c**(2/3) - d**(1/ 3)*c**(1/3)*x + d**(2/3)*x**2)*a**2*d**3*x**4 - 4*b**(2/3)*a**(1/3)*log(c* *(1/3) + d**(1/3)*x)*a**2*d**3*x**4 - 2*d**(2/3)*c**(1/3)*log(a**(2/3) - b **(1/3)*a**(1/3)*x + b**(2/3)*x**2)*b**3*c**2*x**4 + 4*d**(2/3)*c**(1/3)*l og(a**(1/3) + b**(1/3)*x)*b**3*c**2*x**4)/(12*d**(2/3)*c**(1/3)*b**(2/3)*a **(1/3)*a**2*c**2*x**4*(a*d - b*c))