Integrand size = 18, antiderivative size = 363 \[ \int \frac {1}{x^2 \left (a+b x^4+c x^8\right )} \, dx=-\frac {1}{a x}-\frac {\sqrt [4]{c} \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{2\ 2^{3/4} a \sqrt [4]{-b-\sqrt {b^2-4 a c}}}-\frac {\sqrt [4]{c} \left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{2\ 2^{3/4} a \sqrt [4]{-b+\sqrt {b^2-4 a c}}}+\frac {\sqrt [4]{c} \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{2\ 2^{3/4} a \sqrt [4]{-b-\sqrt {b^2-4 a c}}}+\frac {\sqrt [4]{c} \left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{2\ 2^{3/4} a \sqrt [4]{-b+\sqrt {b^2-4 a c}}} \]
-1/a/x-1/4*c^(1/4)*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^(1/4)/a/(-b-(-4*a*c+b^2)^(1/2))^(1/4)+1/4*c^(1 /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^(1/4)/a/(-b-(-4*a*c+b^2)^(1/2))^(1/4)-1/4*c^(1/4)*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^( 1/4)/a/(-b+(-4*a*c+b^2)^(1/2))^(1/4)+1/4*c^(1/4)*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^(1/4)/a/(-b+(-4 *a*c+b^2)^(1/2))^(1/4)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.02 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.20 \[ \int \frac {1}{x^2 \left (a+b x^4+c x^8\right )} \, dx=-\frac {1}{a x}-\frac {\text {RootSum}\left [a+b \text {$\#$1}^4+c \text {$\#$1}^8\&,\frac {b \log (x-\text {$\#$1})+c \log (x-\text {$\#$1}) \text {$\#$1}^4}{b \text {$\#$1}+2 c \text {$\#$1}^5}\&\right ]}{4 a} \]
-(1/(a*x)) - RootSum[a + b*#1^4 + c*#1^8 & , (b*Log[x - #1] + c*Log[x - #1 ]*#1^4)/(b*#1 + 2*c*#1^5) & ]/(4*a)
Time = 0.49 (sec) , antiderivative size = 330, normalized size of antiderivative = 0.91, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {1704, 25, 1834, 27, 827, 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.
\(\displaystyle \int \frac {1}{x^2 \left (a+b x^4+c x^8\right )} \, dx\) |
\(\Big \downarrow \) 1704 |
\(\displaystyle \frac {\int -\frac {x^2 \left (c x^4+b\right )}{c x^8+b x^4+a}dx}{a}-\frac {1}{a x}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {x^2 \left (c x^4+b\right )}{c x^8+b x^4+a}dx}{a}-\frac {1}{a x}\) |
\(\Big \downarrow \) 1834 |
\(\displaystyle -\frac {\frac {1}{2} c \left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \int \frac {2 x^2}{2 c x^4+b-\sqrt {b^2-4 a c}}dx+\frac {1}{2} c \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \int \frac {2 x^2}{2 c x^4+b+\sqrt {b^2-4 a c}}dx}{a}-\frac {1}{a x}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {c \left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \int \frac {x^2}{2 c x^4+b-\sqrt {b^2-4 a c}}dx+c \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \int \frac {x^2}{2 c x^4+b+\sqrt {b^2-4 a c}}dx}{a}-\frac {1}{a x}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle -\frac {c \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \left (\frac {\int \frac {1}{\sqrt {2} \sqrt {c} x^2+\sqrt {-b-\sqrt {b^2-4 a c}}}dx}{2 \sqrt {2} \sqrt {c}}-\frac {\int \frac {1}{\sqrt {-b-\sqrt {b^2-4 a c}}-\sqrt {2} \sqrt {c} x^2}dx}{2 \sqrt {2} \sqrt {c}}\right )+c \left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \left (\frac {\int \frac {1}{\sqrt {2} \sqrt {c} x^2+\sqrt {\sqrt {b^2-4 a c}-b}}dx}{2 \sqrt {2} \sqrt {c}}-\frac {\int \frac {1}{\sqrt {\sqrt {b^2-4 a c}-b}-\sqrt {2} \sqrt {c} x^2}dx}{2 \sqrt {2} \sqrt {c}}\right )}{a}-\frac {1}{a x}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle -\frac {c \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \left (\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{-\sqrt {b^2-4 a c}-b}}-\frac {\int \frac {1}{\sqrt {-b-\sqrt {b^2-4 a c}}-\sqrt {2} \sqrt {c} x^2}dx}{2 \sqrt {2} \sqrt {c}}\right )+c \left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \left (\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{\sqrt {b^2-4 a c}-b}}-\frac {\int \frac {1}{\sqrt {\sqrt {b^2-4 a c}-b}-\sqrt {2} \sqrt {c} x^2}dx}{2 \sqrt {2} \sqrt {c}}\right )}{a}-\frac {1}{a x}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {c \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \left (\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{-\sqrt {b^2-4 a c}-b}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )+c \left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \left (\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{\sqrt {b^2-4 a c}-b}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{a}-\frac {1}{a x}\) |
-(1/(a*x)) - (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*2^(3/4)*c^(3/4)*(-b - Sqrt[b^2 - 4*a*c])^(1 /4)) - ArcTanh[(2^(1/4)*c^(1/4)*x)/(-b - Sqrt[b^2 - 4*a*c])^(1/4)]/(2*2^(3 /4)*c^(3/4)*(-b - Sqrt[b^2 - 4*a*c])^(1/4))) + 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*2^(3/4)*c^ (3/4)*(-b + Sqrt[b^2 - 4*a*c])^(1/4)) - ArcTanh[(2^(1/4)*c^(1/4)*x)/(-b + Sqrt[b^2 - 4*a*c])^(1/4)]/(2*2^(3/4)*c^(3/4)*(-b + Sqrt[b^2 - 4*a*c])^(1/4 ))))/a
3.4.25.3.1 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[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) 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[(((f_.)*(x_))^(m_.)*((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[(f*x)^m/(b/2 - q/2 + c*x^n), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q)) Int[(f*x)^m/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ [{a, b, c, d, e, f, m}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n , 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.09 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.17
method | result | size |
default | \(-\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{8}+\textit {\_Z}^{4} b +a \right )}{\sum }\frac {\left (\textit {\_R}^{6} c +b \,\textit {\_R}^{2}\right ) \ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{7} c +\textit {\_R}^{3} b}}{4 a}-\frac {1}{a x}\) | \(63\) |
risch | \(-\frac {1}{a x}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (256 a^{9} c^{4}-256 b^{2} c^{3} a^{8}+96 b^{4} c^{2} a^{7}-16 b^{6} c \,a^{6}+b^{8} a^{5}\right ) \textit {\_Z}^{8}+\left (80 a^{4} b \,c^{4}-120 a^{3} b^{3} c^{3}+61 a^{2} b^{5} c^{2}-13 c \,b^{7} a +b^{9}\right ) \textit {\_Z}^{4}+c^{5}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (1152 a^{9} c^{4}-1184 b^{2} c^{3} a^{8}+456 b^{4} c^{2} a^{7}-78 b^{6} c \,a^{6}+5 b^{8} a^{5}\right ) \textit {\_R}^{8}+\left (332 a^{4} b \,c^{4}-487 a^{3} b^{3} c^{3}+245 a^{2} b^{5} c^{2}-52 c \,b^{7} a +4 b^{9}\right ) \textit {\_R}^{4}+4 c^{5}\right ) x +\left (64 a^{8} c^{4}-112 a^{7} b^{2} c^{3}+60 a^{6} b^{4} c^{2}-13 a^{5} b^{6} c +a^{4} b^{8}\right ) \textit {\_R}^{7}+\left (4 a^{3} b \,c^{4}-a^{2} b^{3} c^{3}\right ) \textit {\_R}^{3}\right )\right )}{4}\) | \(305\) |
Leaf count of result is larger than twice the leaf count of optimal. 5758 vs. \(2 (281) = 562\).
Time = 0.53 (sec) , antiderivative size = 5758, normalized size of antiderivative = 15.86 \[ \int \frac {1}{x^2 \left (a+b x^4+c x^8\right )} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {1}{x^2 \left (a+b x^4+c x^8\right )} \, dx=\text {Timed out} \]
\[ \int \frac {1}{x^2 \left (a+b x^4+c x^8\right )} \, dx=\int { \frac {1}{{\left (c x^{8} + b x^{4} + a\right )} x^{2}} \,d x } \]
\[ \int \frac {1}{x^2 \left (a+b x^4+c x^8\right )} \, dx=\int { \frac {1}{{\left (c x^{8} + b x^{4} + a\right )} x^{2}} \,d x } \]
Time = 9.01 (sec) , antiderivative size = 10509, normalized size of antiderivative = 28.95 \[ \int \frac {1}{x^2 \left (a+b x^4+c x^8\right )} \, dx=\text {Too large to display} \]
2*atan((((-(b^9 + b^4*(-(4*a*c - b^2)^5)^(1/2) + 80*a^4*b*c^4 + 61*a^2*b^5 *c^2 - 120*a^3*b^3*c^3 + a^2*c^2*(-(4*a*c - b^2)^5)^(1/2) - 13*a*b^7*c - 3 *a*b^2*c*(-(4*a*c - b^2)^5)^(1/2))/(512*(a^5*b^8 + 256*a^9*c^4 - 16*a^6*b^ 6*c + 96*a^7*b^4*c^2 - 256*a^8*b^2*c^3)))^(3/4)*(4096*a^15*c^8 - x*(-(b^9 + b^4*(-(4*a*c - b^2)^5)^(1/2) + 80*a^4*b*c^4 + 61*a^2*b^5*c^2 - 120*a^3*b ^3*c^3 + a^2*c^2*(-(4*a*c - b^2)^5)^(1/2) - 13*a*b^7*c - 3*a*b^2*c*(-(4*a* c - b^2)^5)^(1/2))/(512*(a^5*b^8 + 256*a^9*c^4 - 16*a^6*b^6*c + 96*a^7*b^4 *c^2 - 256*a^8*b^2*c^3)))^(1/4)*(32768*a^16*c^8 + 1024*a^12*b^8*c^4 - 1228 8*a^13*b^6*c^5 + 51200*a^14*b^4*c^6 - 81920*a^15*b^2*c^7)*1i + 256*a^11*b^ 8*c^4 - 2816*a^12*b^6*c^5 + 10496*a^13*b^4*c^6 - 14336*a^14*b^2*c^7)*1i + 4*a^11*b*c^8*x)*(-(b^9 + b^4*(-(4*a*c - b^2)^5)^(1/2) + 80*a^4*b*c^4 + 61* a^2*b^5*c^2 - 120*a^3*b^3*c^3 + a^2*c^2*(-(4*a*c - b^2)^5)^(1/2) - 13*a*b^ 7*c - 3*a*b^2*c*(-(4*a*c - b^2)^5)^(1/2))/(512*(a^5*b^8 + 256*a^9*c^4 - 16 *a^6*b^6*c + 96*a^7*b^4*c^2 - 256*a^8*b^2*c^3)))^(1/4) - ((-(b^9 + b^4*(-( 4*a*c - b^2)^5)^(1/2) + 80*a^4*b*c^4 + 61*a^2*b^5*c^2 - 120*a^3*b^3*c^3 + a^2*c^2*(-(4*a*c - b^2)^5)^(1/2) - 13*a*b^7*c - 3*a*b^2*c*(-(4*a*c - b^2)^ 5)^(1/2))/(512*(a^5*b^8 + 256*a^9*c^4 - 16*a^6*b^6*c + 96*a^7*b^4*c^2 - 25 6*a^8*b^2*c^3)))^(3/4)*(4096*a^15*c^8 + x*(-(b^9 + b^4*(-(4*a*c - b^2)^5)^ (1/2) + 80*a^4*b*c^4 + 61*a^2*b^5*c^2 - 120*a^3*b^3*c^3 + a^2*c^2*(-(4*a*c - b^2)^5)^(1/2) - 13*a*b^7*c - 3*a*b^2*c*(-(4*a*c - b^2)^5)^(1/2))/(51...