Integrand size = 22, antiderivative size = 265 \[ \int \frac {1}{x^5 \left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx=-\frac {1}{4 a^2 d x^4}+\frac {e}{2 a^2 d^2 x^2}-\frac {c^2 \left (d-e x^2\right )}{4 a^2 \left (c d^2+a e^2\right ) \left (a+c x^4\right )}+\frac {c^{3/2} e \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{4 a^{5/2} \left (c d^2+a e^2\right )}+\frac {c^{3/2} e \left (c d^2+2 a e^2\right ) \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{2 a^{5/2} \left (c d^2+a e^2\right )^2}-\frac {\left (2 c d^2-a e^2\right ) \log (x)}{a^3 d^3}-\frac {e^6 \log \left (d+e x^2\right )}{2 d^3 \left (c d^2+a e^2\right )^2}+\frac {c^2 d \left (2 c d^2+3 a e^2\right ) \log \left (a+c x^4\right )}{4 a^3 \left (c d^2+a e^2\right )^2} \]
-1/4/a^2/d/x^4+1/2*e/a^2/d^2/x^2-1/4*c^2*(-e*x^2+d)/a^2/(a*e^2+c*d^2)/(c*x ^4+a)+1/4*c^(3/2)*e*arctan(x^2*c^(1/2)/a^(1/2))/a^(5/2)/(a*e^2+c*d^2)+1/2* c^(3/2)*e*(2*a*e^2+c*d^2)*arctan(x^2*c^(1/2)/a^(1/2))/a^(5/2)/(a*e^2+c*d^2 )^2-(-a*e^2+2*c*d^2)*ln(x)/a^3/d^3-1/2*e^6*ln(e*x^2+d)/d^3/(a*e^2+c*d^2)^2 +1/4*c^2*d*(3*a*e^2+2*c*d^2)*ln(c*x^4+a)/a^3/(a*e^2+c*d^2)^2
Time = 0.30 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.05 \[ \int \frac {1}{x^5 \left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx=\frac {1}{4} \left (-\frac {1}{a^2 d x^4}+\frac {2 e}{a^2 d^2 x^2}+\frac {c^2 \left (-d+e x^2\right )}{a^2 \left (c d^2+a e^2\right ) \left (a+c x^4\right )}-\frac {c^{3/2} e \left (3 c d^2+5 a e^2\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{a^{5/2} \left (c d^2+a e^2\right )^2}-\frac {c^{3/2} e \left (3 c d^2+5 a e^2\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} x}{\sqrt [4]{a}}\right )}{a^{5/2} \left (c d^2+a e^2\right )^2}+\frac {4 \left (-2 c d^2+a e^2\right ) \log (x)}{a^3 d^3}-\frac {2 e^6 \log \left (d+e x^2\right )}{d^3 \left (c d^2+a e^2\right )^2}+\frac {c^2 \left (2 c d^3+3 a d e^2\right ) \log \left (a+c x^4\right )}{a^3 \left (c d^2+a e^2\right )^2}\right ) \]
(-(1/(a^2*d*x^4)) + (2*e)/(a^2*d^2*x^2) + (c^2*(-d + e*x^2))/(a^2*(c*d^2 + a*e^2)*(a + c*x^4)) - (c^(3/2)*e*(3*c*d^2 + 5*a*e^2)*ArcTan[1 - (Sqrt[2]* c^(1/4)*x)/a^(1/4)])/(a^(5/2)*(c*d^2 + a*e^2)^2) - (c^(3/2)*e*(3*c*d^2 + 5 *a*e^2)*ArcTan[1 + (Sqrt[2]*c^(1/4)*x)/a^(1/4)])/(a^(5/2)*(c*d^2 + a*e^2)^ 2) + (4*(-2*c*d^2 + a*e^2)*Log[x])/(a^3*d^3) - (2*e^6*Log[d + e*x^2])/(d^3 *(c*d^2 + a*e^2)^2) + (c^2*(2*c*d^3 + 3*a*d*e^2)*Log[a + c*x^4])/(a^3*(c*d ^2 + a*e^2)^2))/4
Time = 0.49 (sec) , antiderivative size = 263, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1579, 615, 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.
| \(\displaystyle \int \frac {1}{x^5 \left (a+c x^4\right )^2 \left (d+e x^2\right )} \, dx\) |
| \(\Big \downarrow \) 1579 |
| \(\displaystyle \frac {1}{2} \int \frac {1}{x^6 \left (e x^2+d\right ) \left (c x^4+a\right )^2}dx^2\) |
| \(\Big \downarrow \) 615 |
| \(\displaystyle \frac {1}{2} \int \left (-\frac {e^7}{d^3 \left (c d^2+a e^2\right )^2 \left (e x^2+d\right )}-\frac {e}{a^2 d^2 x^4}+\frac {c^2 \left (c d \left (2 c d^2+3 a e^2\right ) x^2+a e \left (c d^2+2 a e^2\right )\right )}{a^3 \left (c d^2+a e^2\right )^2 \left (c x^4+a\right )}+\frac {a e^2-2 c d^2}{a^3 d^3 x^2}+\frac {c^2 \left (c d x^2+a e\right )}{a^2 \left (c d^2+a e^2\right ) \left (c x^4+a\right )^2}+\frac {1}{a^2 d x^6}\right )dx^2\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \left (\frac {c^{3/2} e \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right ) \left (2 a e^2+c d^2\right )}{a^{5/2} \left (a e^2+c d^2\right )^2}+\frac {c^{3/2} e \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{2 a^{5/2} \left (a e^2+c d^2\right )}+\frac {c^2 d \left (3 a e^2+2 c d^2\right ) \log \left (a+c x^4\right )}{2 a^3 \left (a e^2+c d^2\right )^2}-\frac {\log \left (x^2\right ) \left (2 c d^2-a e^2\right )}{a^3 d^3}-\frac {c^2 \left (d-e x^2\right )}{2 a^2 \left (a+c x^4\right ) \left (a e^2+c d^2\right )}+\frac {e}{a^2 d^2 x^2}-\frac {1}{2 a^2 d x^4}-\frac {e^6 \log \left (d+e x^2\right )}{d^3 \left (a e^2+c d^2\right )^2}\right )\) |
(-1/2*1/(a^2*d*x^4) + e/(a^2*d^2*x^2) - (c^2*(d - e*x^2))/(2*a^2*(c*d^2 + a*e^2)*(a + c*x^4)) + (c^(3/2)*e*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/(2*a^(5/2) *(c*d^2 + a*e^2)) + (c^(3/2)*e*(c*d^2 + 2*a*e^2)*ArcTan[(Sqrt[c]*x^2)/Sqrt [a]])/(a^(5/2)*(c*d^2 + a*e^2)^2) - ((2*c*d^2 - a*e^2)*Log[x^2])/(a^3*d^3) - (e^6*Log[d + e*x^2])/(d^3*(c*d^2 + a*e^2)^2) + (c^2*d*(2*c*d^2 + 3*a*e^ 2)*Log[a + c*x^4])/(2*a^3*(c*d^2 + a*e^2)^2))/2
3.3.51.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 0]
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(m + 1)/2]
Time = 0.48 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.79
| method | result | size |
| default | \(-\frac {1}{4 a^{2} d \,x^{4}}+\frac {\left (a \,e^{2}-2 c \,d^{2}\right ) \ln \left (x \right )}{a^{3} d^{3}}+\frac {e}{2 a^{2} d^{2} x^{2}}+\frac {c^{2} \left (\frac {\left (\frac {1}{2} e^{3} a^{2}+\frac {1}{2} a c \,d^{2} e \right ) x^{2}-\frac {a d \left (a \,e^{2}+c \,d^{2}\right )}{2}}{c \,x^{4}+a}+\frac {\left (6 a c d \,e^{2}+4 c^{2} d^{3}\right ) \ln \left (c \,x^{4}+a \right )}{4 c}+\frac {\left (5 e^{3} a^{2}+3 a c \,d^{2} e \right ) \arctan \left (\frac {c \,x^{2}}{\sqrt {a c}}\right )}{2 \sqrt {a c}}\right )}{2 \left (a \,e^{2}+c \,d^{2}\right )^{2} a^{3}}-\frac {e^{6} \ln \left (e \,x^{2}+d \right )}{2 d^{3} \left (a \,e^{2}+c \,d^{2}\right )^{2}}\) | \(209\) |
| risch | \(\frac {\frac {e c \left (2 a \,e^{2}+3 c \,d^{2}\right ) x^{6}}{4 \left (a \,e^{2}+c \,d^{2}\right ) a^{2} d^{2}}-\frac {c \left (a \,e^{2}+2 c \,d^{2}\right ) x^{4}}{4 d \,a^{2} \left (a \,e^{2}+c \,d^{2}\right )}+\frac {e \,x^{2}}{2 d^{2} a}-\frac {1}{4 d a}}{x^{4} \left (c \,x^{4}+a \right )}+\frac {\ln \left (x \right ) e^{2}}{a^{2} d^{3}}-\frac {2 \ln \left (x \right ) c}{a^{3} d}-\frac {e^{6} \ln \left (e \,x^{2}+d \right )}{2 d^{3} \left (a^{2} e^{4}+2 a c \,d^{2} e^{2}+c^{2} d^{4}\right )}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (a^{8} e^{4}+2 a^{7} c \,d^{2} e^{2}+a^{6} d^{4} c^{2}\right ) \textit {\_Z}^{2}+\left (-12 a^{4} c^{2} d \,e^{2}-8 a^{3} d^{3} c^{3}\right ) \textit {\_Z} +25 a \,c^{3} e^{2}+16 d^{2} c^{4}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (-6 a^{11} d^{4} e^{8}-19 a^{10} c \,d^{6} e^{6}-25 a^{9} c^{2} d^{8} e^{4}-17 a^{8} c^{3} d^{10} e^{2}-5 a^{7} c^{4} d^{12}\right ) \textit {\_R}^{3}+\left (-36 a^{8} c \,d^{3} e^{8}+12 a^{7} c^{2} d^{5} e^{6}+140 a^{6} c^{3} d^{7} e^{4}+112 a^{5} c^{4} d^{9} e^{2}+20 a^{4} c^{5} d^{11}\right ) \textit {\_R}^{2}+\left (-32 a^{6} c \,e^{10}+96 a^{5} c^{2} d^{2} e^{8}+154 a^{4} c^{3} d^{4} e^{6}-40 a^{3} c^{4} d^{6} e^{4}-132 a^{2} c^{5} d^{8} e^{2}\right ) \textit {\_R} -128 a^{2} c^{3} d \,e^{8}-544 a \,c^{4} d^{3} e^{6}-512 c^{5} d^{5} e^{4}\right ) x^{2}+\left (-2 a^{11} d^{5} e^{7}-2 a^{10} c \,d^{7} e^{5}+2 a^{9} c^{2} d^{9} e^{3}+2 a^{8} c^{3} d^{11} e \right ) \textit {\_R}^{3}+\left (16 a^{9} d^{2} e^{9}-8 a^{8} c \,d^{4} e^{7}-29 a^{7} c^{2} d^{6} e^{5}+22 a^{6} c^{3} d^{8} e^{3}+27 a^{5} c^{4} d^{10} e \right ) \textit {\_R}^{2}+\left (-64 a^{6} c d \,e^{9}-44 a^{5} c^{2} d^{3} e^{7}+206 a^{4} c^{3} d^{5} e^{5}+104 a^{3} c^{4} d^{7} e^{3}-80 a^{2} c^{5} d^{9} e \right ) \textit {\_R} +160 a^{3} c^{2} e^{9}+80 a^{2} c^{3} d^{2} e^{7}-544 a \,c^{4} d^{4} e^{5}-512 c^{5} d^{6} e^{3}\right )\right )}{8}\) | \(748\) |
-1/4/a^2/d/x^4+(a*e^2-2*c*d^2)/a^3/d^3*ln(x)+1/2*e/a^2/d^2/x^2+1/2*c^2/(a* e^2+c*d^2)^2/a^3*(((1/2*e^3*a^2+1/2*a*c*d^2*e)*x^2-1/2*a*d*(a*e^2+c*d^2))/ (c*x^4+a)+1/4*(6*a*c*d*e^2+4*c^2*d^3)/c*ln(c*x^4+a)+1/2*(5*a^2*e^3+3*a*c*d ^2*e)/(a*c)^(1/2)*arctan(c*x^2/(a*c)^(1/2)))-1/2*e^6*ln(e*x^2+d)/d^3/(a*e^ 2+c*d^2)^2
Timed out. \[ \int \frac {1}{x^5 \left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{x^5 \left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx=\text {Timed out} \]
Time = 0.29 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.25 \[ \int \frac {1}{x^5 \left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx=-\frac {e^{6} \log \left (e x^{2} + d\right )}{2 \, {\left (c^{2} d^{7} + 2 \, a c d^{5} e^{2} + a^{2} d^{3} e^{4}\right )}} + \frac {{\left (2 \, c^{3} d^{3} + 3 \, a c^{2} d e^{2}\right )} \log \left (c x^{4} + a\right )}{4 \, {\left (a^{3} c^{2} d^{4} + 2 \, a^{4} c d^{2} e^{2} + a^{5} e^{4}\right )}} + \frac {{\left (3 \, c^{3} d^{2} e + 5 \, a c^{2} e^{3}\right )} \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{4 \, {\left (a^{2} c^{2} d^{4} + 2 \, a^{3} c d^{2} e^{2} + a^{4} e^{4}\right )} \sqrt {a c}} + \frac {{\left (3 \, c^{2} d^{2} e + 2 \, a c e^{3}\right )} x^{6} - a c d^{3} - a^{2} d e^{2} - {\left (2 \, c^{2} d^{3} + a c d e^{2}\right )} x^{4} + 2 \, {\left (a c d^{2} e + a^{2} e^{3}\right )} x^{2}}{4 \, {\left ({\left (a^{2} c^{2} d^{4} + a^{3} c d^{2} e^{2}\right )} x^{8} + {\left (a^{3} c d^{4} + a^{4} d^{2} e^{2}\right )} x^{4}\right )}} - \frac {{\left (2 \, c d^{2} - a e^{2}\right )} \log \left (x^{2}\right )}{2 \, a^{3} d^{3}} \]
-1/2*e^6*log(e*x^2 + d)/(c^2*d^7 + 2*a*c*d^5*e^2 + a^2*d^3*e^4) + 1/4*(2*c ^3*d^3 + 3*a*c^2*d*e^2)*log(c*x^4 + a)/(a^3*c^2*d^4 + 2*a^4*c*d^2*e^2 + a^ 5*e^4) + 1/4*(3*c^3*d^2*e + 5*a*c^2*e^3)*arctan(c*x^2/sqrt(a*c))/((a^2*c^2 *d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4)*sqrt(a*c)) + 1/4*((3*c^2*d^2*e + 2*a*c*e ^3)*x^6 - a*c*d^3 - a^2*d*e^2 - (2*c^2*d^3 + a*c*d*e^2)*x^4 + 2*(a*c*d^2*e + a^2*e^3)*x^2)/((a^2*c^2*d^4 + a^3*c*d^2*e^2)*x^8 + (a^3*c*d^4 + a^4*d^2 *e^2)*x^4) - 1/2*(2*c*d^2 - a*e^2)*log(x^2)/(a^3*d^3)
Time = 0.30 (sec) , antiderivative size = 361, normalized size of antiderivative = 1.36 \[ \int \frac {1}{x^5 \left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx=-\frac {e^{7} \log \left ({\left | e x^{2} + d \right |}\right )}{2 \, {\left (c^{2} d^{7} e + 2 \, a c d^{5} e^{3} + a^{2} d^{3} e^{5}\right )}} + \frac {{\left (2 \, c^{3} d^{3} + 3 \, a c^{2} d e^{2}\right )} \log \left (c x^{4} + a\right )}{4 \, {\left (a^{3} c^{2} d^{4} + 2 \, a^{4} c d^{2} e^{2} + a^{5} e^{4}\right )}} + \frac {{\left (3 \, c^{3} d^{2} e + 5 \, a c^{2} e^{3}\right )} \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{4 \, {\left (a^{2} c^{2} d^{4} + 2 \, a^{3} c d^{2} e^{2} + a^{4} e^{4}\right )} \sqrt {a c}} - \frac {2 \, c^{4} d^{3} x^{4} + 3 \, a c^{3} d e^{2} x^{4} - a c^{3} d^{2} e x^{2} - a^{2} c^{2} e^{3} x^{2} + 3 \, a c^{3} d^{3} + 4 \, a^{2} c^{2} d e^{2}}{4 \, {\left (a^{3} c^{2} d^{4} + 2 \, a^{4} c d^{2} e^{2} + a^{5} e^{4}\right )} {\left (c x^{4} + a\right )}} - \frac {{\left (2 \, c d^{2} - a e^{2}\right )} \log \left (x^{2}\right )}{2 \, a^{3} d^{3}} + \frac {6 \, c d^{2} x^{4} - 3 \, a e^{2} x^{4} + 2 \, a d e x^{2} - a d^{2}}{4 \, a^{3} d^{3} x^{4}} \]
-1/2*e^7*log(abs(e*x^2 + d))/(c^2*d^7*e + 2*a*c*d^5*e^3 + a^2*d^3*e^5) + 1 /4*(2*c^3*d^3 + 3*a*c^2*d*e^2)*log(c*x^4 + a)/(a^3*c^2*d^4 + 2*a^4*c*d^2*e ^2 + a^5*e^4) + 1/4*(3*c^3*d^2*e + 5*a*c^2*e^3)*arctan(c*x^2/sqrt(a*c))/(( a^2*c^2*d^4 + 2*a^3*c*d^2*e^2 + a^4*e^4)*sqrt(a*c)) - 1/4*(2*c^4*d^3*x^4 + 3*a*c^3*d*e^2*x^4 - a*c^3*d^2*e*x^2 - a^2*c^2*e^3*x^2 + 3*a*c^3*d^3 + 4*a ^2*c^2*d*e^2)/((a^3*c^2*d^4 + 2*a^4*c*d^2*e^2 + a^5*e^4)*(c*x^4 + a)) - 1/ 2*(2*c*d^2 - a*e^2)*log(x^2)/(a^3*d^3) + 1/4*(6*c*d^2*x^4 - 3*a*e^2*x^4 + 2*a*d*e*x^2 - a*d^2)/(a^3*d^3*x^4)
Time = 9.49 (sec) , antiderivative size = 1545, normalized size of antiderivative = 5.83 \[ \int \frac {1}{x^5 \left (d+e x^2\right ) \left (a+c x^4\right )^2} \, dx=\text {Too large to display} \]
(log(6400*a^13*c^18*d^28*x^2 + 1024*a^27*c^4*e^28*x^2 - 6400*a^3*c^13*d^28 *(-a^7*c^3)^(3/2) + 1024*a^24*c^2*e^28*(-a^7*c^3)^(1/2) - 10688*a^6*d^8*e^ 20*(-a^7*c^3)^(5/2) - 2048*a^16*d^2*e^26*(-a^7*c^3)^(3/2) + 536959*c^6*d^2 0*e^8*(-a^7*c^3)^(5/2) + 54944*a^14*c^17*d^26*e^2*x^2 + 200881*a^15*c^16*d ^24*e^4*x^2 + 413414*a^16*c^15*d^22*e^6*x^2 + 536959*a^17*c^14*d^20*e^8*x^ 2 + 465092*a^18*c^13*d^18*e^10*x^2 + 256991*a^19*c^12*d^16*e^12*x^2 + 5282 2*a^20*c^11*d^14*e^14*x^2 - 37423*a^21*c^10*d^12*e^16*x^2 - 27472*a^22*c^9 *d^10*e^18*x^2 - 10688*a^23*c^8*d^8*e^20*x^2 - 10288*a^24*c^7*d^6*e^22*x^2 - 3584*a^25*c^6*d^4*e^24*x^2 + 2048*a^26*c^5*d^2*e^26*x^2 + 465092*a*c^5* d^18*e^10*(-a^7*c^3)^(5/2) - 27472*a^5*c*d^10*e^18*(-a^7*c^3)^(5/2) + 3584 *a^15*c*d^4*e^24*(-a^7*c^3)^(3/2) + 256991*a^2*c^4*d^16*e^12*(-a^7*c^3)^(5 /2) + 52822*a^3*c^3*d^14*e^14*(-a^7*c^3)^(5/2) - 37423*a^4*c^2*d^12*e^16*( -a^7*c^3)^(5/2) - 54944*a^4*c^12*d^26*e^2*(-a^7*c^3)^(3/2) - 200881*a^5*c^ 11*d^24*e^4*(-a^7*c^3)^(3/2) - 413414*a^6*c^10*d^22*e^6*(-a^7*c^3)^(3/2) + 10288*a^14*c^2*d^6*e^22*(-a^7*c^3)^(3/2))*(4*a^3*c^3*d^3 + 5*a*e^3*(-a^7* c^3)^(1/2) + 6*a^4*c^2*d*e^2 + 3*c*d^2*e*(-a^7*c^3)^(1/2)))/(8*(a^8*e^4 + a^6*c^2*d^4 + 2*a^7*c*d^2*e^2)) - (e^6*log(d + e*x^2))/(2*(c^2*d^7 + a^2*d ^3*e^4 + 2*a*c*d^5*e^2)) - (1/(4*a*d) - (e*x^2)/(2*a*d^2) + (x^4*(2*c^2*d^ 2 + a*c*e^2))/(4*a^2*d*(a*e^2 + c*d^2)) - (c*e*x^6*(2*a*e^2 + 3*c*d^2))/(4 *a^2*d^2*(a*e^2 + c*d^2)))/(a*x^4 + c*x^8) + (log(6400*a^13*c^18*d^28*x...