Integrand size = 18, antiderivative size = 365 \[ \int \frac {1}{x^4 \left (a+b x^4+c x^8\right )} \, dx=-\frac {1}{3 a x^3}+\frac {c^{3/4} \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{2 \sqrt [4]{2} a \left (-b-\sqrt {b^2-4 a c}\right )^{3/4}}+\frac {c^{3/4} \left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{2 \sqrt [4]{2} a \left (-b+\sqrt {b^2-4 a c}\right )^{3/4}}+\frac {c^{3/4} \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{2 \sqrt [4]{2} a \left (-b-\sqrt {b^2-4 a c}\right )^{3/4}}+\frac {c^{3/4} \left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{2 \sqrt [4]{2} a \left (-b+\sqrt {b^2-4 a c}\right )^{3/4}} \]
-1/3/a/x^3+1/4*c^(3/4)*arctan(2^(1/4)*c^(1/4)*x/(-b-(-4*a*c+b^2)^(1/2))^(1 /4))*(1-b/(-4*a*c+b^2)^(1/2))*2^(3/4)/a/(-b-(-4*a*c+b^2)^(1/2))^(3/4)+1/4* c^(3/4)*arctanh(2^(1/4)*c^(1/4)*x/(-b-(-4*a*c+b^2)^(1/2))^(1/4))*(1-b/(-4* a*c+b^2)^(1/2))*2^(3/4)/a/(-b-(-4*a*c+b^2)^(1/2))^(3/4)+1/4*c^(3/4)*arctan (2^(1/4)*c^(1/4)*x/(-b+(-4*a*c+b^2)^(1/2))^(1/4))*(1+b/(-4*a*c+b^2)^(1/2)) *2^(3/4)/a/(-b+(-4*a*c+b^2)^(1/2))^(3/4)+1/4*c^(3/4)*arctanh(2^(1/4)*c^(1/ 4)*x/(-b+(-4*a*c+b^2)^(1/2))^(1/4))*(1+b/(-4*a*c+b^2)^(1/2))*2^(3/4)/a/(-b +(-4*a*c+b^2)^(1/2))^(3/4)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.21 \[ \int \frac {1}{x^4 \left (a+b x^4+c x^8\right )} \, dx=-\frac {1}{3 a x^3}-\frac {\text {RootSum}\left [a+b \text {$\#$1}^4+c \text {$\#$1}^8\&,\frac {b \log (x-\text {$\#$1})+c \log (x-\text {$\#$1}) \text {$\#$1}^4}{b \text {$\#$1}^3+2 c \text {$\#$1}^7}\&\right ]}{4 a} \]
-1/3*1/(a*x^3) - RootSum[a + b*#1^4 + c*#1^8 & , (b*Log[x - #1] + c*Log[x - #1]*#1^4)/(b*#1^3 + 2*c*#1^7) & ]/(4*a)
Time = 0.47 (sec) , antiderivative size = 330, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1704, 27, 1752, 756, 218, 221}
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.
| \(\displaystyle \int \frac {1}{x^4 \left (a+b x^4+c x^8\right )} \, dx\) |
| \(\Big \downarrow \) 1704 |
| \(\displaystyle \frac {\int -\frac {3 \left (c x^4+b\right )}{c x^8+b x^4+a}dx}{3 a}-\frac {1}{3 a x^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {c x^4+b}{c x^8+b x^4+a}dx}{a}-\frac {1}{3 a x^3}\) |
| \(\Big \downarrow \) 1752 |
| \(\displaystyle -\frac {\frac {1}{2} c \left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \int \frac {1}{c x^4+\frac {1}{2} \left (b-\sqrt {b^2-4 a c}\right )}dx+\frac {1}{2} c \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \int \frac {1}{c x^4+\frac {1}{2} \left (b+\sqrt {b^2-4 a c}\right )}dx}{a}-\frac {1}{3 a x^3}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle -\frac {\frac {1}{2} c \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \left (-\frac {\int \frac {1}{\sqrt {-b-\sqrt {b^2-4 a c}}-\sqrt {2} \sqrt {c} x^2}dx}{\sqrt {-\sqrt {b^2-4 a c}-b}}-\frac {\int \frac {1}{\sqrt {2} \sqrt {c} x^2+\sqrt {-b-\sqrt {b^2-4 a c}}}dx}{\sqrt {-\sqrt {b^2-4 a c}-b}}\right )+\frac {1}{2} c \left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \left (-\frac {\int \frac {1}{\sqrt {\sqrt {b^2-4 a c}-b}-\sqrt {2} \sqrt {c} x^2}dx}{\sqrt {\sqrt {b^2-4 a c}-b}}-\frac {\int \frac {1}{\sqrt {2} \sqrt {c} x^2+\sqrt {\sqrt {b^2-4 a c}-b}}dx}{\sqrt {\sqrt {b^2-4 a c}-b}}\right )}{a}-\frac {1}{3 a x^3}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle -\frac {\frac {1}{2} c \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \left (-\frac {\int \frac {1}{\sqrt {-b-\sqrt {b^2-4 a c}}-\sqrt {2} \sqrt {c} x^2}dx}{\sqrt {-\sqrt {b^2-4 a c}-b}}-\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{\sqrt [4]{2} \sqrt [4]{c} \left (-\sqrt {b^2-4 a c}-b\right )^{3/4}}\right )+\frac {1}{2} c \left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \left (-\frac {\int \frac {1}{\sqrt {\sqrt {b^2-4 a c}-b}-\sqrt {2} \sqrt {c} x^2}dx}{\sqrt {\sqrt {b^2-4 a c}-b}}-\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{\sqrt [4]{2} \sqrt [4]{c} \left (\sqrt {b^2-4 a c}-b\right )^{3/4}}\right )}{a}-\frac {1}{3 a x^3}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {\frac {1}{2} c \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \left (-\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{\sqrt [4]{2} \sqrt [4]{c} \left (-\sqrt {b^2-4 a c}-b\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{\sqrt [4]{2} \sqrt [4]{c} \left (-\sqrt {b^2-4 a c}-b\right )^{3/4}}\right )+\frac {1}{2} c \left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \left (-\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{\sqrt [4]{2} \sqrt [4]{c} \left (\sqrt {b^2-4 a c}-b\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{\sqrt [4]{2} \sqrt [4]{c} \left (\sqrt {b^2-4 a c}-b\right )^{3/4}}\right )}{a}-\frac {1}{3 a x^3}\) |
-1/3*1/(a*x^3) - ((c*(1 - b/Sqrt[b^2 - 4*a*c])*(-(ArcTan[(2^(1/4)*c^(1/4)* x)/(-b - Sqrt[b^2 - 4*a*c])^(1/4)]/(2^(1/4)*c^(1/4)*(-b - Sqrt[b^2 - 4*a*c ])^(3/4))) - ArcTanh[(2^(1/4)*c^(1/4)*x)/(-b - Sqrt[b^2 - 4*a*c])^(1/4)]/( 2^(1/4)*c^(1/4)*(-b - Sqrt[b^2 - 4*a*c])^(3/4))))/2 + (c*(1 + b/Sqrt[b^2 - 4*a*c])*(-(ArcTan[(2^(1/4)*c^(1/4)*x)/(-b + Sqrt[b^2 - 4*a*c])^(1/4)]/(2^ (1/4)*c^(1/4)*(-b + Sqrt[b^2 - 4*a*c])^(3/4))) - ArcTanh[(2^(1/4)*c^(1/4)* x)/(-b + Sqrt[b^2 - 4*a*c])^(1/4)]/(2^(1/4)*c^(1/4)*(-b + Sqrt[b^2 - 4*a*c ])^(3/4))))/2)/a
3.4.26.3.1 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[((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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[((d_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_ Symbol] :> Simp[(d*x)^(m + 1)*((a + b*x^n + c*x^(2*n))^(p + 1)/(a*d*(m + 1) )), x] - Simp[1/(a*d^n*(m + 1)) Int[(d*x)^(m + n)*(b*(m + n*(p + 1) + 1) + c*(m + 2*n*(p + 1) + 1)*x^n)*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{ a, b, c, d, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && LtQ[m, -1] && IntegerQ[p]
Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x _Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) Int[1/(b/2 - q/2 + c*x^n), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q)) I nt[1/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2 , 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && (PosQ[b^2 - 4*a*c] || !IGtQ[n/2, 0])
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.08 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.17
| method | result | size |
| default | \(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{8}+\textit {\_Z}^{4} b +a \right )}{\sum }\frac {\left (-\textit {\_R}^{4} c -b \right ) \ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{7} c +\textit {\_R}^{3} b}}{4 a}-\frac {1}{3 a \,x^{3}}\) | \(62\) |
| risch | \(-\frac {1}{3 a \,x^{3}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (256 c^{4} a^{11}-256 a^{10} b^{2} c^{3}+96 a^{9} b^{4} c^{2}-16 a^{8} b^{6} c +a^{7} b^{8}\right ) \textit {\_Z}^{8}+\left (-112 b \,c^{5} a^{5}+280 b^{3} c^{4} a^{4}-231 c^{3} b^{5} a^{3}+86 b^{7} c^{2} a^{2}-15 b^{9} c a +b^{11}\right ) \textit {\_Z}^{4}+c^{7}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (-1152 c^{4} a^{11}+1184 a^{10} b^{2} c^{3}-456 a^{9} b^{4} c^{2}+78 a^{8} b^{6} c -5 a^{7} b^{8}\right ) \textit {\_R}^{8}+\left (468 b \,c^{5} a^{5}-1145 b^{3} c^{4} a^{4}+933 c^{3} b^{5} a^{3}-345 b^{7} c^{2} a^{2}+60 b^{9} c a -4 b^{11}\right ) \textit {\_R}^{4}-4 c^{7}\right ) x +\left (16 a^{7} c^{5}-104 a^{6} b^{2} c^{4}+129 a^{5} b^{4} c^{3}-62 a^{4} b^{6} c^{2}+13 a^{3} b^{8} c -a^{2} b^{10}\right ) \textit {\_R}^{5}\right )\right )}{4}\) | \(314\) |
Leaf count of result is larger than twice the leaf count of optimal. 5030 vs. \(2 (281) = 562\).
Time = 0.62 (sec) , antiderivative size = 5030, normalized size of antiderivative = 13.78 \[ \int \frac {1}{x^4 \left (a+b x^4+c x^8\right )} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {1}{x^4 \left (a+b x^4+c x^8\right )} \, dx=\text {Timed out} \]
\[ \int \frac {1}{x^4 \left (a+b x^4+c x^8\right )} \, dx=\int { \frac {1}{{\left (c x^{8} + b x^{4} + a\right )} x^{4}} \,d x } \]
\[ \int \frac {1}{x^4 \left (a+b x^4+c x^8\right )} \, dx=\int { \frac {1}{{\left (c x^{8} + b x^{4} + a\right )} x^{4}} \,d x } \]
Time = 10.98 (sec) , antiderivative size = 16497, normalized size of antiderivative = 45.20 \[ \int \frac {1}{x^4 \left (a+b x^4+c x^8\right )} \, dx=\text {Too large to display} \]
2*atan(-(((-(b^11 + b^6*(-(4*a*c - b^2)^5)^(1/2) - 112*a^5*b*c^5 + 86*a^2* b^7*c^2 - 231*a^3*b^5*c^3 + 280*a^4*b^3*c^4 - a^3*c^3*(-(4*a*c - b^2)^5)^( 1/2) - 15*a*b^9*c + 6*a^2*b^2*c^2*(-(4*a*c - b^2)^5)^(1/2) - 5*a*b^4*c*(-( 4*a*c - b^2)^5)^(1/2))/(512*(a^7*b^8 + 256*a^11*c^4 - 16*a^8*b^6*c + 96*a^ 9*b^4*c^2 - 256*a^10*b^2*c^3)))^(1/4)*((x*(81920*a^15*b*c^8 + 1024*a^11*b^ 9*c^4 - 13312*a^12*b^7*c^5 + 62464*a^13*b^5*c^6 - 122880*a^14*b^3*c^7) - ( -(b^11 + b^6*(-(4*a*c - b^2)^5)^(1/2) - 112*a^5*b*c^5 + 86*a^2*b^7*c^2 - 2 31*a^3*b^5*c^3 + 280*a^4*b^3*c^4 - a^3*c^3*(-(4*a*c - b^2)^5)^(1/2) - 15*a *b^9*c + 6*a^2*b^2*c^2*(-(4*a*c - b^2)^5)^(1/2) - 5*a*b^4*c*(-(4*a*c - b^2 )^5)^(1/2))/(512*(a^7*b^8 + 256*a^11*c^4 - 16*a^8*b^6*c + 96*a^9*b^4*c^2 - 256*a^10*b^2*c^3)))^(1/4)*(262144*a^17*c^8 + 4096*a^13*b^8*c^4 - 53248*a^ 14*b^6*c^5 + 245760*a^15*b^4*c^6 - 458752*a^16*b^2*c^7)*1i)*(-(b^11 + b^6* (-(4*a*c - b^2)^5)^(1/2) - 112*a^5*b*c^5 + 86*a^2*b^7*c^2 - 231*a^3*b^5*c^ 3 + 280*a^4*b^3*c^4 - a^3*c^3*(-(4*a*c - b^2)^5)^(1/2) - 15*a*b^9*c + 6*a^ 2*b^2*c^2*(-(4*a*c - b^2)^5)^(1/2) - 5*a*b^4*c*(-(4*a*c - b^2)^5)^(1/2))/( 512*(a^7*b^8 + 256*a^11*c^4 - 16*a^8*b^6*c + 96*a^9*b^4*c^2 - 256*a^10*b^2 *c^3)))^(3/4)*1i - 128*a^11*b*c^9 - 16*a^9*b^5*c^7 + 96*a^10*b^3*c^8)*1i + x*(8*a^10*c^10 - 4*a^9*b^2*c^9))*(-(b^11 + b^6*(-(4*a*c - b^2)^5)^(1/2) - 112*a^5*b*c^5 + 86*a^2*b^7*c^2 - 231*a^3*b^5*c^3 + 280*a^4*b^3*c^4 - a^3* c^3*(-(4*a*c - b^2)^5)^(1/2) - 15*a*b^9*c + 6*a^2*b^2*c^2*(-(4*a*c - b^...