Integrand size = 22, antiderivative size = 96 \[ \int \frac {x^3}{\left (d+e x^2\right ) \left (a+c x^4\right )} \, dx=\frac {\sqrt {a} e \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{2 \sqrt {c} \left (c d^2+a e^2\right )}-\frac {d \log \left (d+e x^2\right )}{2 \left (c d^2+a e^2\right )}+\frac {d \log \left (a+c x^4\right )}{4 \left (c d^2+a e^2\right )} \]
-1/2*d*ln(e*x^2+d)/(a*e^2+c*d^2)+1/4*d*ln(c*x^4+a)/(a*e^2+c*d^2)+1/2*e*arc tan(x^2*c^(1/2)/a^(1/2))*a^(1/2)/(a*e^2+c*d^2)/c^(1/2)
Time = 0.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.69 \[ \int \frac {x^3}{\left (d+e x^2\right ) \left (a+c x^4\right )} \, dx=\frac {\frac {2 \sqrt {a} e \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{\sqrt {c}}-2 d \log \left (d+e x^2\right )+d \log \left (a+c x^4\right )}{4 c d^2+4 a e^2} \]
((2*Sqrt[a]*e*ArcTan[(Sqrt[c]*x^2)/Sqrt[a]])/Sqrt[c] - 2*d*Log[d + e*x^2] + d*Log[a + c*x^4])/(4*c*d^2 + 4*a*e^2)
Time = 0.22 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.88, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1579, 587, 16, 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.
| \(\displaystyle \int \frac {x^3}{\left (a+c x^4\right ) \left (d+e x^2\right )} \, dx\) |
| \(\Big \downarrow \) 1579 |
| \(\displaystyle \frac {1}{2} \int \frac {x^2}{\left (e x^2+d\right ) \left (c x^4+a\right )}dx^2\) |
| \(\Big \downarrow \) 587 |
| \(\displaystyle \frac {1}{2} \left (\frac {\int \frac {c d x^2+a e}{c x^4+a}dx^2}{a e^2+c d^2}-\frac {d e \int \frac {1}{e x^2+d}dx^2}{a e^2+c d^2}\right )\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {1}{2} \left (\frac {\int \frac {c d x^2+a e}{c x^4+a}dx^2}{a e^2+c d^2}-\frac {d \log \left (d+e x^2\right )}{a e^2+c d^2}\right )\) |
| \(\Big \downarrow \) 452 |
| \(\displaystyle \frac {1}{2} \left (\frac {c d \int \frac {x^2}{c x^4+a}dx^2+a e \int \frac {1}{c x^4+a}dx^2}{a e^2+c d^2}-\frac {d \log \left (d+e x^2\right )}{a e^2+c d^2}\right )\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {1}{2} \left (\frac {c d \int \frac {x^2}{c x^4+a}dx^2+\frac {\sqrt {a} e \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{\sqrt {c}}}{a e^2+c d^2}-\frac {d \log \left (d+e x^2\right )}{a e^2+c d^2}\right )\) |
| \(\Big \downarrow \) 240 |
| \(\displaystyle \frac {1}{2} \left (\frac {\frac {\sqrt {a} e \arctan \left (\frac {\sqrt {c} x^2}{\sqrt {a}}\right )}{\sqrt {c}}+\frac {1}{2} d \log \left (a+c x^4\right )}{a e^2+c d^2}-\frac {d \log \left (d+e x^2\right )}{a e^2+c d^2}\right )\) |
(-((d*Log[d + e*x^2])/(c*d^2 + a*e^2)) + ((Sqrt[a]*e*ArcTan[(Sqrt[c]*x^2)/ Sqrt[a]])/Sqrt[c] + (d*Log[a + c*x^4])/2)/(c*d^2 + a*e^2))/2
3.3.32.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
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[(x_.)/(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)), x_Symbol] :> Simp[(- c)*(d/(b*c^2 + a*d^2)) Int[1/(c + d*x), x], x] + Simp[1/(b*c^2 + a*d^2) Int[(a*d + b*c*x)/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c ^2 + a*d^2, 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 = 0.39 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.75
| method | result | size |
| default | \(\frac {\frac {d \ln \left (c \,x^{4}+a \right )}{2}+\frac {a e \arctan \left (\frac {c \,x^{2}}{\sqrt {a c}}\right )}{\sqrt {a c}}}{2 a \,e^{2}+2 c \,d^{2}}-\frac {d \ln \left (e \,x^{2}+d \right )}{2 \left (a \,e^{2}+c \,d^{2}\right )}\) | \(72\) |
| risch | \(\frac {\ln \left (\left (\sqrt {-a c}\, a \,e^{3}-7 \sqrt {-a c}\, c \,d^{2} e +5 a c d \,e^{2}-3 c^{2} d^{3}\right ) x^{2}+5 \sqrt {-a c}\, a d \,e^{2}-3 \sqrt {-a c}\, c \,d^{3}-e^{3} a^{2}+7 a c \,d^{2} e \right ) e \sqrt {-a c}}{4 c \left (a \,e^{2}+c \,d^{2}\right )}+\frac {\ln \left (\left (\sqrt {-a c}\, a \,e^{3}-7 \sqrt {-a c}\, c \,d^{2} e +5 a c d \,e^{2}-3 c^{2} d^{3}\right ) x^{2}+5 \sqrt {-a c}\, a d \,e^{2}-3 \sqrt {-a c}\, c \,d^{3}-e^{3} a^{2}+7 a c \,d^{2} e \right ) d}{4 a \,e^{2}+4 c \,d^{2}}-\frac {\ln \left (\left (-\sqrt {-a c}\, a \,e^{3}+7 \sqrt {-a c}\, c \,d^{2} e +5 a c d \,e^{2}-3 c^{2} d^{3}\right ) x^{2}-5 \sqrt {-a c}\, a d \,e^{2}+3 \sqrt {-a c}\, c \,d^{3}-e^{3} a^{2}+7 a c \,d^{2} e \right ) e \sqrt {-a c}}{4 c \left (a \,e^{2}+c \,d^{2}\right )}+\frac {\ln \left (\left (-\sqrt {-a c}\, a \,e^{3}+7 \sqrt {-a c}\, c \,d^{2} e +5 a c d \,e^{2}-3 c^{2} d^{3}\right ) x^{2}-5 \sqrt {-a c}\, a d \,e^{2}+3 \sqrt {-a c}\, c \,d^{3}-e^{3} a^{2}+7 a c \,d^{2} e \right ) d}{4 a \,e^{2}+4 c \,d^{2}}-\frac {d \ln \left (e \,x^{2}+d \right )}{2 \left (a \,e^{2}+c \,d^{2}\right )}\) | \(462\) |
1/2/(a*e^2+c*d^2)*(1/2*d*ln(c*x^4+a)+a*e/(a*c)^(1/2)*arctan(c*x^2/(a*c)^(1 /2)))-1/2*d*ln(e*x^2+d)/(a*e^2+c*d^2)
Time = 0.37 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.51 \[ \int \frac {x^3}{\left (d+e x^2\right ) \left (a+c x^4\right )} \, dx=\left [\frac {e \sqrt {-\frac {a}{c}} \log \left (\frac {c x^{4} + 2 \, c x^{2} \sqrt {-\frac {a}{c}} - a}{c x^{4} + a}\right ) + d \log \left (c x^{4} + a\right ) - 2 \, d \log \left (e x^{2} + d\right )}{4 \, {\left (c d^{2} + a e^{2}\right )}}, \frac {2 \, e \sqrt {\frac {a}{c}} \arctan \left (\frac {c x^{2} \sqrt {\frac {a}{c}}}{a}\right ) + d \log \left (c x^{4} + a\right ) - 2 \, d \log \left (e x^{2} + d\right )}{4 \, {\left (c d^{2} + a e^{2}\right )}}\right ] \]
[1/4*(e*sqrt(-a/c)*log((c*x^4 + 2*c*x^2*sqrt(-a/c) - a)/(c*x^4 + a)) + d*l og(c*x^4 + a) - 2*d*log(e*x^2 + d))/(c*d^2 + a*e^2), 1/4*(2*e*sqrt(a/c)*ar ctan(c*x^2*sqrt(a/c)/a) + d*log(c*x^4 + a) - 2*d*log(e*x^2 + d))/(c*d^2 + a*e^2)]
Timed out. \[ \int \frac {x^3}{\left (d+e x^2\right ) \left (a+c x^4\right )} \, dx=\text {Timed out} \]
Time = 0.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.85 \[ \int \frac {x^3}{\left (d+e x^2\right ) \left (a+c x^4\right )} \, dx=\frac {a e \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{2 \, {\left (c d^{2} + a e^{2}\right )} \sqrt {a c}} + \frac {d \log \left (c x^{4} + a\right )}{4 \, {\left (c d^{2} + a e^{2}\right )}} - \frac {d \log \left (e x^{2} + d\right )}{2 \, {\left (c d^{2} + a e^{2}\right )}} \]
1/2*a*e*arctan(c*x^2/sqrt(a*c))/((c*d^2 + a*e^2)*sqrt(a*c)) + 1/4*d*log(c* x^4 + a)/(c*d^2 + a*e^2) - 1/2*d*log(e*x^2 + d)/(c*d^2 + a*e^2)
Time = 0.27 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.89 \[ \int \frac {x^3}{\left (d+e x^2\right ) \left (a+c x^4\right )} \, dx=-\frac {d e \log \left ({\left | e x^{2} + d \right |}\right )}{2 \, {\left (c d^{2} e + a e^{3}\right )}} + \frac {a e \arctan \left (\frac {c x^{2}}{\sqrt {a c}}\right )}{2 \, {\left (c d^{2} + a e^{2}\right )} \sqrt {a c}} + \frac {d \log \left (c x^{4} + a\right )}{4 \, {\left (c d^{2} + a e^{2}\right )}} \]
-1/2*d*e*log(abs(e*x^2 + d))/(c*d^2*e + a*e^3) + 1/2*a*e*arctan(c*x^2/sqrt (a*c))/((c*d^2 + a*e^2)*sqrt(a*c)) + 1/4*d*log(c*x^4 + a)/(c*d^2 + a*e^2)
Time = 8.74 (sec) , antiderivative size = 944, normalized size of antiderivative = 9.83 \[ \int \frac {x^3}{\left (d+e x^2\right ) \left (a+c x^4\right )} \, dx=\frac {c\,d\,\ln \left (a^4\,e^6-9\,a\,c^3\,d^6-39\,a^3\,c\,d^2\,e^4+a^3\,e^6\,x^2\,\sqrt {-a\,c}-9\,c^3\,d^6\,x^2\,\sqrt {-a\,c}+79\,a^2\,c^2\,d^4\,e^2-42\,c\,d^5\,e\,{\left (-a\,c\right )}^{3/2}+76\,a\,d^3\,e^3\,{\left (-a\,c\right )}^{3/2}+10\,a^3\,d\,e^5\,\sqrt {-a\,c}+76\,a^2\,c^2\,d^3\,e^3\,x^2-42\,a\,c^3\,d^5\,e\,x^2-10\,a^3\,c\,d\,e^5\,x^2+39\,a\,d^2\,e^4\,x^2\,{\left (-a\,c\right )}^{3/2}-79\,c\,d^4\,e^2\,x^2\,{\left (-a\,c\right )}^{3/2}\right )}{4\,c^2\,d^2+4\,a\,c\,e^2}-\frac {d\,\ln \left (e\,x^2+d\right )}{2\,\left (c\,d^2+a\,e^2\right )}+\frac {c\,d\,\ln \left (9\,a\,c^3\,d^6-a^4\,e^6+39\,a^3\,c\,d^2\,e^4+a^3\,e^6\,x^2\,\sqrt {-a\,c}-9\,c^3\,d^6\,x^2\,\sqrt {-a\,c}-79\,a^2\,c^2\,d^4\,e^2+10\,a^3\,d\,e^5\,\sqrt {-a\,c}+42\,a\,c^2\,d^5\,e\,\sqrt {-a\,c}-76\,a^2\,c^2\,d^3\,e^3\,x^2+42\,a\,c^3\,d^5\,e\,x^2+10\,a^3\,c\,d\,e^5\,x^2-76\,a^2\,c\,d^3\,e^3\,\sqrt {-a\,c}+79\,a\,c^2\,d^4\,e^2\,x^2\,\sqrt {-a\,c}-39\,a^2\,c\,d^2\,e^4\,x^2\,\sqrt {-a\,c}\right )}{4\,c^2\,d^2+4\,a\,c\,e^2}-\frac {e\,\ln \left (a^4\,e^6-9\,a\,c^3\,d^6-39\,a^3\,c\,d^2\,e^4+a^3\,e^6\,x^2\,\sqrt {-a\,c}-9\,c^3\,d^6\,x^2\,\sqrt {-a\,c}+79\,a^2\,c^2\,d^4\,e^2-42\,c\,d^5\,e\,{\left (-a\,c\right )}^{3/2}+76\,a\,d^3\,e^3\,{\left (-a\,c\right )}^{3/2}+10\,a^3\,d\,e^5\,\sqrt {-a\,c}+76\,a^2\,c^2\,d^3\,e^3\,x^2-42\,a\,c^3\,d^5\,e\,x^2-10\,a^3\,c\,d\,e^5\,x^2+39\,a\,d^2\,e^4\,x^2\,{\left (-a\,c\right )}^{3/2}-79\,c\,d^4\,e^2\,x^2\,{\left (-a\,c\right )}^{3/2}\right )\,\sqrt {-a\,c}}{4\,c^2\,d^2+4\,a\,c\,e^2}+\frac {e\,\ln \left (9\,a\,c^3\,d^6-a^4\,e^6+39\,a^3\,c\,d^2\,e^4+a^3\,e^6\,x^2\,\sqrt {-a\,c}-9\,c^3\,d^6\,x^2\,\sqrt {-a\,c}-79\,a^2\,c^2\,d^4\,e^2+10\,a^3\,d\,e^5\,\sqrt {-a\,c}+42\,a\,c^2\,d^5\,e\,\sqrt {-a\,c}-76\,a^2\,c^2\,d^3\,e^3\,x^2+42\,a\,c^3\,d^5\,e\,x^2+10\,a^3\,c\,d\,e^5\,x^2-76\,a^2\,c\,d^3\,e^3\,\sqrt {-a\,c}+79\,a\,c^2\,d^4\,e^2\,x^2\,\sqrt {-a\,c}-39\,a^2\,c\,d^2\,e^4\,x^2\,\sqrt {-a\,c}\right )\,\sqrt {-a\,c}}{4\,c^2\,d^2+4\,a\,c\,e^2} \]
(c*d*log(a^4*e^6 - 9*a*c^3*d^6 - 39*a^3*c*d^2*e^4 + a^3*e^6*x^2*(-a*c)^(1/ 2) - 9*c^3*d^6*x^2*(-a*c)^(1/2) + 79*a^2*c^2*d^4*e^2 - 42*c*d^5*e*(-a*c)^( 3/2) + 76*a*d^3*e^3*(-a*c)^(3/2) + 10*a^3*d*e^5*(-a*c)^(1/2) + 76*a^2*c^2* d^3*e^3*x^2 - 42*a*c^3*d^5*e*x^2 - 10*a^3*c*d*e^5*x^2 + 39*a*d^2*e^4*x^2*( -a*c)^(3/2) - 79*c*d^4*e^2*x^2*(-a*c)^(3/2)))/(4*c^2*d^2 + 4*a*c*e^2) - (d *log(d + e*x^2))/(2*(a*e^2 + c*d^2)) + (c*d*log(9*a*c^3*d^6 - a^4*e^6 + 39 *a^3*c*d^2*e^4 + a^3*e^6*x^2*(-a*c)^(1/2) - 9*c^3*d^6*x^2*(-a*c)^(1/2) - 7 9*a^2*c^2*d^4*e^2 + 10*a^3*d*e^5*(-a*c)^(1/2) + 42*a*c^2*d^5*e*(-a*c)^(1/2 ) - 76*a^2*c^2*d^3*e^3*x^2 + 42*a*c^3*d^5*e*x^2 + 10*a^3*c*d*e^5*x^2 - 76* a^2*c*d^3*e^3*(-a*c)^(1/2) + 79*a*c^2*d^4*e^2*x^2*(-a*c)^(1/2) - 39*a^2*c* d^2*e^4*x^2*(-a*c)^(1/2)))/(4*c^2*d^2 + 4*a*c*e^2) - (e*log(a^4*e^6 - 9*a* c^3*d^6 - 39*a^3*c*d^2*e^4 + a^3*e^6*x^2*(-a*c)^(1/2) - 9*c^3*d^6*x^2*(-a* c)^(1/2) + 79*a^2*c^2*d^4*e^2 - 42*c*d^5*e*(-a*c)^(3/2) + 76*a*d^3*e^3*(-a *c)^(3/2) + 10*a^3*d*e^5*(-a*c)^(1/2) + 76*a^2*c^2*d^3*e^3*x^2 - 42*a*c^3* d^5*e*x^2 - 10*a^3*c*d*e^5*x^2 + 39*a*d^2*e^4*x^2*(-a*c)^(3/2) - 79*c*d^4* e^2*x^2*(-a*c)^(3/2))*(-a*c)^(1/2))/(4*c^2*d^2 + 4*a*c*e^2) + (e*log(9*a*c ^3*d^6 - a^4*e^6 + 39*a^3*c*d^2*e^4 + a^3*e^6*x^2*(-a*c)^(1/2) - 9*c^3*d^6 *x^2*(-a*c)^(1/2) - 79*a^2*c^2*d^4*e^2 + 10*a^3*d*e^5*(-a*c)^(1/2) + 42*a* c^2*d^5*e*(-a*c)^(1/2) - 76*a^2*c^2*d^3*e^3*x^2 + 42*a*c^3*d^5*e*x^2 + 10* a^3*c*d*e^5*x^2 - 76*a^2*c*d^3*e^3*(-a*c)^(1/2) + 79*a*c^2*d^4*e^2*x^2*...