Integrand size = 20, antiderivative size = 96 \[ \int \frac {x}{\left (d+e x^2\right ) \left (a+c x^4\right )} \, dx=\frac {\sqrt {c} d \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{2 \sqrt {a} \left (c d^2+a e^2\right )}+\frac {e \log \left (d+e x^2\right )}{2 \left (c d^2+a e^2\right )}-\frac {e \log \left (a+c x^4\right )}{4 \left (c d^2+a e^2\right )} \]
1/2*e*ln(e*x^2+d)/(a*e^2+c*d^2)-1/4*e*ln(c*x^4+a)/(a*e^2+c*d^2)+1/2*d*arct an(x^2*c^(1/2)/a^(1/2))*c^(1/2)/(a*e^2+c*d^2)/a^(1/2)
Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.70 \[ \int \frac {x}{\left (d+e x^2\right ) \left (a+c x^4\right )} \, dx=\frac {\frac {2 \sqrt {c} d \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{\sqrt {a}}+2 e \log \left (d+e x^2\right )-e \log \left (a+c x^4\right )}{4 c d^2+4 a e^2} \]
((2*Sqrt[c]*d*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/Sqrt[a] + 2*e*Log[d + e*x^2] - e*Log[a + c*x^4])/(4*c*d^2 + 4*a*e^2)
Time = 0.22 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.91, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1577, 479, 452, 218, 240}
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.
\(\displaystyle \int \frac {x}{\left (a+c x^4\right ) \left (d+e x^2\right )} \, dx\) |
\(\Big \downarrow \) 1577 |
\(\displaystyle \frac {1}{2} \int \frac {1}{\left (e x^2+d\right ) \left (c x^4+a\right )}dx^2\) |
\(\Big \downarrow \) 479 |
\(\displaystyle \frac {1}{2} \left (\frac {c \int \frac {d-e x^2}{c x^4+a}dx^2}{a e^2+c d^2}+\frac {e \log \left (d+e x^2\right )}{a e^2+c d^2}\right )\) |
\(\Big \downarrow \) 452 |
\(\displaystyle \frac {1}{2} \left (\frac {c \left (d \int \frac {1}{c x^4+a}dx^2-e \int \frac {x^2}{c x^4+a}dx^2\right )}{a e^2+c d^2}+\frac {e \log \left (d+e x^2\right )}{a e^2+c d^2}\right )\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {1}{2} \left (\frac {c \left (\frac {d \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {c}}-e \int \frac {x^2}{c x^4+a}dx^2\right )}{a e^2+c d^2}+\frac {e \log \left (d+e x^2\right )}{a e^2+c d^2}\right )\) |
\(\Big \downarrow \) 240 |
\(\displaystyle \frac {1}{2} \left (\frac {c \left (\frac {d \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {c}}-\frac {e \log \left (a+c x^4\right )}{2 c}\right )}{a e^2+c d^2}+\frac {e \log \left (d+e x^2\right )}{a e^2+c d^2}\right )\) |
((e*Log[d + e*x^2])/(c*d^2 + a*e^2) + (c*((d*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]] )/(Sqrt[a]*Sqrt[c]) - (e*Log[a + c*x^4])/(2*c)))/(c*d^2 + a*e^2))/2
3.3.33.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x ^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
Int[((c_) + (d_.)*(x_))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[c Int[1/ (a + b*x^2), x], x] + Simp[d Int[x/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c^2 + a*d^2, 0]
Int[1/(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)), x_Symbol] :> Simp[d*(Log [RemoveContent[c + d*x, x]]/(b*c^2 + a*d^2)), x] + Simp[b/(b*c^2 + a*d^2) Int[(c - d*x)/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d}, x]
Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[1/2 Subst[Int[(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; Free Q[{a, c, d, e, p, q}, x]
Time = 0.38 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.78
method | result | size |
default | \(\frac {c \left (-\frac {e \ln \left (c \,x^{4}+a \right )}{2 c}+\frac {d \arctan \left (\frac {c \,x^{2}}{\sqrt {a c}}\right )}{\sqrt {a c}}\right )}{2 a \,e^{2}+2 c \,d^{2}}+\frac {e \ln \left (e \,x^{2}+d \right )}{2 a \,e^{2}+2 c \,d^{2}}\) | \(75\) |
risch | \(\frac {\ln \left (\left (-3 a^{2} c \,e^{3}+5 a \,c^{2} d^{2} e +7 \sqrt {-a c}\, a c d \,e^{2}-\sqrt {-a c}\, c^{2} d^{3}\right ) x^{2}-7 a^{2} c d \,e^{2}+a \,c^{2} d^{3}-3 \sqrt {-a c}\, a^{2} e^{3}+5 \sqrt {-a c}\, a c \,d^{2} e \right ) d \sqrt {-a c}}{4 a \left (a \,e^{2}+c \,d^{2}\right )}-\frac {\ln \left (\left (-3 a^{2} c \,e^{3}+5 a \,c^{2} d^{2} e +7 \sqrt {-a c}\, a c d \,e^{2}-\sqrt {-a c}\, c^{2} d^{3}\right ) x^{2}-7 a^{2} c d \,e^{2}+a \,c^{2} d^{3}-3 \sqrt {-a c}\, a^{2} e^{3}+5 \sqrt {-a c}\, a c \,d^{2} e \right ) e}{4 \left (a \,e^{2}+c \,d^{2}\right )}-\frac {\ln \left (\left (-3 a^{2} c \,e^{3}+5 a \,c^{2} d^{2} e -7 \sqrt {-a c}\, a c d \,e^{2}+\sqrt {-a c}\, c^{2} d^{3}\right ) x^{2}-7 a^{2} c d \,e^{2}+a \,c^{2} d^{3}+3 \sqrt {-a c}\, a^{2} e^{3}-5 \sqrt {-a c}\, a c \,d^{2} e \right ) d \sqrt {-a c}}{4 a \left (a \,e^{2}+c \,d^{2}\right )}-\frac {\ln \left (\left (-3 a^{2} c \,e^{3}+5 a \,c^{2} d^{2} e -7 \sqrt {-a c}\, a c d \,e^{2}+\sqrt {-a c}\, c^{2} d^{3}\right ) x^{2}-7 a^{2} c d \,e^{2}+a \,c^{2} d^{3}+3 \sqrt {-a c}\, a^{2} e^{3}-5 \sqrt {-a c}\, a c \,d^{2} e \right ) e}{4 \left (a \,e^{2}+c \,d^{2}\right )}+\frac {e \ln \left (e \,x^{2}+d \right )}{2 a \,e^{2}+2 c \,d^{2}}\) | \(506\) |
1/2*c/(a*e^2+c*d^2)*(-1/2*e/c*ln(c*x^4+a)+d/(a*c)^(1/2)*arctan(c*x^2/(a*c) ^(1/2)))+1/2*e*ln(e*x^2+d)/(a*e^2+c*d^2)
Time = 0.43 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.52 \[ \int \frac {x}{\left (d+e x^2\right ) \left (a+c x^4\right )} \, dx=\left [\frac {d \sqrt {-\frac {c}{a}} \log \left (\frac {c x^{4} + 2 \, a x^{2} \sqrt {-\frac {c}{a}} - a}{c x^{4} + a}\right ) - e \log \left (c x^{4} + a\right ) + 2 \, e \log \left (e x^{2} + d\right )}{4 \, {\left (c d^{2} + a e^{2}\right )}}, -\frac {2 \, d \sqrt {\frac {c}{a}} \arctan \left (\frac {a \sqrt {\frac {c}{a}}}{c x^{2}}\right ) + e \log \left (c x^{4} + a\right ) - 2 \, e \log \left (e x^{2} + d\right )}{4 \, {\left (c d^{2} + a e^{2}\right )}}\right ] \]
[1/4*(d*sqrt(-c/a)*log((c*x^4 + 2*a*x^2*sqrt(-c/a) - a)/(c*x^4 + a)) - e*l og(c*x^4 + a) + 2*e*log(e*x^2 + d))/(c*d^2 + a*e^2), -1/4*(2*d*sqrt(c/a)*a rctan(a*sqrt(c/a)/(c*x^2)) + e*log(c*x^4 + a) - 2*e*log(e*x^2 + d))/(c*d^2 + a*e^2)]
Timed out. \[ \int \frac {x}{\left (d+e x^2\right ) \left (a+c x^4\right )} \, dx=\text {Timed out} \]
Time = 0.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.85 \[ \int \frac {x}{\left (d+e x^2\right ) \left (a+c x^4\right )} \, dx=\frac {c d \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{2 \, {\left (c d^{2} + a e^{2}\right )} \sqrt {a c}} - \frac {e \log \left (c x^{4} + a\right )}{4 \, {\left (c d^{2} + a e^{2}\right )}} + \frac {e \log \left (e x^{2} + d\right )}{2 \, {\left (c d^{2} + a e^{2}\right )}} \]
1/2*c*d*arctan(c*x^2/sqrt(a*c))/((c*d^2 + a*e^2)*sqrt(a*c)) - 1/4*e*log(c* x^4 + a)/(c*d^2 + a*e^2) + 1/2*e*log(e*x^2 + d)/(c*d^2 + a*e^2)
Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.90 \[ \int \frac {x}{\left (d+e x^2\right ) \left (a+c x^4\right )} \, dx=\frac {e^{2} \log \left ({\left | e x^{2} + d \right |}\right )}{2 \, {\left (c d^{2} e + a e^{3}\right )}} + \frac {c d \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{2 \, {\left (c d^{2} + a e^{2}\right )} \sqrt {a c}} - \frac {e \log \left (c x^{4} + a\right )}{4 \, {\left (c d^{2} + a e^{2}\right )}} \]
1/2*e^2*log(abs(e*x^2 + d))/(c*d^2*e + a*e^3) + 1/2*c*d*arctan(c*x^2/sqrt( a*c))/((c*d^2 + a*e^2)*sqrt(a*c)) - 1/4*e*log(c*x^4 + a)/(c*d^2 + a*e^2)
Time = 8.31 (sec) , antiderivative size = 328, normalized size of antiderivative = 3.42 \[ \int \frac {x}{\left (d+e x^2\right ) \left (a+c x^4\right )} \, dx=\frac {e\,\ln \left (e\,x^2+d\right )}{2\,c\,d^2+2\,a\,e^2}-\frac {\ln \left (a\,c^5\,d^6\,x^2-c^3\,d^6\,{\left (-a\,c\right )}^{3/2}-9\,a^3\,e^6\,{\left (-a\,c\right )}^{3/2}+9\,a^4\,c^2\,e^6\,x^2+19\,a\,d^2\,e^4\,{\left (-a\,c\right )}^{5/2}+11\,c\,d^4\,e^2\,{\left (-a\,c\right )}^{5/2}+11\,a^2\,c^4\,d^4\,e^2\,x^2+19\,a^3\,c^3\,d^2\,e^4\,x^2\right )\,\left (a\,e-d\,\sqrt {-a\,c}\right )}{4\,\left (a^2\,e^2+c\,a\,d^2\right )}-\frac {\ln \left (9\,a^3\,e^6\,{\left (-a\,c\right )}^{3/2}+c^3\,d^6\,{\left (-a\,c\right )}^{3/2}+a\,c^5\,d^6\,x^2+9\,a^4\,c^2\,e^6\,x^2-19\,a\,d^2\,e^4\,{\left (-a\,c\right )}^{5/2}-11\,c\,d^4\,e^2\,{\left (-a\,c\right )}^{5/2}+11\,a^2\,c^4\,d^4\,e^2\,x^2+19\,a^3\,c^3\,d^2\,e^4\,x^2\right )\,\left (a\,e+d\,\sqrt {-a\,c}\right )}{4\,\left (a^2\,e^2+c\,a\,d^2\right )} \]
(e*log(d + e*x^2))/(2*a*e^2 + 2*c*d^2) - (log(a*c^5*d^6*x^2 - c^3*d^6*(-a* c)^(3/2) - 9*a^3*e^6*(-a*c)^(3/2) + 9*a^4*c^2*e^6*x^2 + 19*a*d^2*e^4*(-a*c )^(5/2) + 11*c*d^4*e^2*(-a*c)^(5/2) + 11*a^2*c^4*d^4*e^2*x^2 + 19*a^3*c^3* d^2*e^4*x^2)*(a*e - d*(-a*c)^(1/2)))/(4*(a^2*e^2 + a*c*d^2)) - (log(9*a^3* e^6*(-a*c)^(3/2) + c^3*d^6*(-a*c)^(3/2) + a*c^5*d^6*x^2 + 9*a^4*c^2*e^6*x^ 2 - 19*a*d^2*e^4*(-a*c)^(5/2) - 11*c*d^4*e^2*(-a*c)^(5/2) + 11*a^2*c^4*d^4 *e^2*x^2 + 19*a^3*c^3*d^2*e^4*x^2)*(a*e + d*(-a*c)^(1/2)))/(4*(a^2*e^2 + a *c*d^2))