Integrand size = 25, antiderivative size = 375 \[ \int \frac {x^2 \left (d+e x^4\right )}{a+b x^4+c x^8} \, dx=\frac {\left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{-b-\sqrt {b^2-4 a c}}}+\frac {\left (e+\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{-b+\sqrt {b^2-4 a c}}}-\frac {\left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{-b-\sqrt {b^2-4 a c}}}-\frac {\left (e+\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{-b+\sqrt {b^2-4 a c}}} \] Output:
1/4*(e-(-b*e+2*c*d)/(-4*a*c+b^2)^(1/2))*arctan(2^(1/4)*c^(1/4)*x/(-b-(-4*a *c+b^2)^(1/2))^(1/4))*2^(1/4)/c^(3/4)/(-b-(-4*a*c+b^2)^(1/2))^(1/4)+1/4*(e +(-b*e+2*c*d)/(-4*a*c+b^2)^(1/2))*arctan(2^(1/4)*c^(1/4)*x/(-b+(-4*a*c+b^2 )^(1/2))^(1/4))*2^(1/4)/c^(3/4)/(-b+(-4*a*c+b^2)^(1/2))^(1/4)-1/4*(e-(-b*e +2*c*d)/(-4*a*c+b^2)^(1/2))*arctanh(2^(1/4)*c^(1/4)*x/(-b-(-4*a*c+b^2)^(1/ 2))^(1/4))*2^(1/4)/c^(3/4)/(-b-(-4*a*c+b^2)^(1/2))^(1/4)-1/4*(e+(-b*e+2*c* d)/(-4*a*c+b^2)^(1/2))*arctanh(2^(1/4)*c^(1/4)*x/(-b+(-4*a*c+b^2)^(1/2))^( 1/4))*2^(1/4)/c^(3/4)/(-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 = 0.05 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.16 \[ \int \frac {x^2 \left (d+e x^4\right )}{a+b x^4+c x^8} \, dx=\frac {1}{4} \text {RootSum}\left [a+b \text {$\#$1}^4+c \text {$\#$1}^8\&,\frac {d \log (x-\text {$\#$1})+e \log (x-\text {$\#$1}) \text {$\#$1}^4}{b \text {$\#$1}+2 c \text {$\#$1}^5}\&\right ] \] Input:
Integrate[(x^2*(d + e*x^4))/(a + b*x^4 + c*x^8),x]
Output:
RootSum[a + b*#1^4 + c*#1^8 & , (d*Log[x - #1] + e*Log[x - #1]*#1^4)/(b*#1 + 2*c*#1^5) & ]/4
Time = 0.52 (sec) , antiderivative size = 330, normalized size of antiderivative = 0.88, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {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.
\(\displaystyle \int \frac {x^2 \left (d+e x^4\right )}{a+b x^4+c x^8} \, dx\) |
\(\Big \downarrow \) 1834 |
\(\displaystyle \frac {1}{2} \left (\frac {2 c d-b e}{\sqrt {b^2-4 a c}}+e\right ) \int \frac {2 x^2}{2 c x^4+b-\sqrt {b^2-4 a c}}dx+\frac {1}{2} \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \int \frac {2 x^2}{2 c x^4+b+\sqrt {b^2-4 a c}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \left (\frac {2 c d-b e}{\sqrt {b^2-4 a c}}+e\right ) \int \frac {x^2}{2 c x^4+b-\sqrt {b^2-4 a c}}dx+\left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \int \frac {x^2}{2 c x^4+b+\sqrt {b^2-4 a c}}dx\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \left (\frac {\int \frac {1}{\sqrt {2} \sqrt {c} x^2+\sqrt {-b-\sqrt {b^2-4 a c}}}dx}{2 \sqrt {2} \sqrt {c}}-\frac {\int \frac {1}{\sqrt {-b-\sqrt {b^2-4 a c}}-\sqrt {2} \sqrt {c} x^2}dx}{2 \sqrt {2} \sqrt {c}}\right )+\left (\frac {2 c d-b e}{\sqrt {b^2-4 a c}}+e\right ) \left (\frac {\int \frac {1}{\sqrt {2} \sqrt {c} x^2+\sqrt {\sqrt {b^2-4 a c}-b}}dx}{2 \sqrt {2} \sqrt {c}}-\frac {\int \frac {1}{\sqrt {\sqrt {b^2-4 a c}-b}-\sqrt {2} \sqrt {c} x^2}dx}{2 \sqrt {2} \sqrt {c}}\right )\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \left (\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{-\sqrt {b^2-4 a c}-b}}-\frac {\int \frac {1}{\sqrt {-b-\sqrt {b^2-4 a c}}-\sqrt {2} \sqrt {c} x^2}dx}{2 \sqrt {2} \sqrt {c}}\right )+\left (\frac {2 c d-b e}{\sqrt {b^2-4 a c}}+e\right ) \left (\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{\sqrt {b^2-4 a c}-b}}-\frac {\int \frac {1}{\sqrt {\sqrt {b^2-4 a c}-b}-\sqrt {2} \sqrt {c} x^2}dx}{2 \sqrt {2} \sqrt {c}}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \left (\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{-\sqrt {b^2-4 a c}-b}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )+\left (\frac {2 c d-b e}{\sqrt {b^2-4 a c}}+e\right ) \left (\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{\sqrt {b^2-4 a c}-b}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} x}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )\) |
Input:
Int[(x^2*(d + e*x^4))/(a + b*x^4 + c*x^8),x]
Output:
(e - (2*c*d - b*e)/Sqrt[b^2 - 4*a*c])*(ArcTan[(2^(1/4)*c^(1/4)*x)/(-b - Sq rt[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))) + (e + (2*c*d - b*e)/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)))
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[(((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 = 0.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.14
method | result | size |
default | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8} c +\textit {\_Z}^{4} b +a \right )}{\sum }\frac {\left (\textit {\_R}^{6} e +\textit {\_R}^{2} d \right ) \ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{7} c +\textit {\_R}^{3} b}\right )}{4}\) | \(51\) |
risch | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8} c +\textit {\_Z}^{4} b +a \right )}{\sum }\frac {\left (\textit {\_R}^{6} e +\textit {\_R}^{2} d \right ) \ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{7} c +\textit {\_R}^{3} b}\right )}{4}\) | \(51\) |
Input:
int(x^2*(e*x^4+d)/(c*x^8+b*x^4+a),x,method=_RETURNVERBOSE)
Output:
1/4*sum((_R^6*e+_R^2*d)/(2*_R^7*c+_R^3*b)*ln(x-_R),_R=RootOf(_Z^8*c+_Z^4*b +a))
Leaf count of result is larger than twice the leaf count of optimal. 15561 vs. \(2 (295) = 590\).
Time = 42.82 (sec) , antiderivative size = 15561, normalized size of antiderivative = 41.50 \[ \int \frac {x^2 \left (d+e x^4\right )}{a+b x^4+c x^8} \, dx=\text {Too large to display} \] Input:
integrate(x^2*(e*x^4+d)/(c*x^8+b*x^4+a),x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \frac {x^2 \left (d+e x^4\right )}{a+b x^4+c x^8} \, dx=\text {Timed out} \] Input:
integrate(x**2*(e*x**4+d)/(c*x**8+b*x**4+a),x)
Output:
Timed out
\[ \int \frac {x^2 \left (d+e x^4\right )}{a+b x^4+c x^8} \, dx=\int { \frac {{\left (e x^{4} + d\right )} x^{2}}{c x^{8} + b x^{4} + a} \,d x } \] Input:
integrate(x^2*(e*x^4+d)/(c*x^8+b*x^4+a),x, algorithm="maxima")
Output:
integrate((e*x^4 + d)*x^2/(c*x^8 + b*x^4 + a), x)
Timed out. \[ \int \frac {x^2 \left (d+e x^4\right )}{a+b x^4+c x^8} \, dx=\text {Timed out} \] Input:
integrate(x^2*(e*x^4+d)/(c*x^8+b*x^4+a),x, algorithm="giac")
Output:
Timed out
Time = 28.43 (sec) , antiderivative size = 29445, normalized size of antiderivative = 78.52 \[ \int \frac {x^2 \left (d+e x^4\right )}{a+b x^4+c x^8} \, dx=\text {Too large to display} \] Input:
int((x^2*(d + e*x^4))/(a + b*x^4 + c*x^8),x)
Output:
2*atan(((x*(4*a^3*b^3*c*e^6 - 12*a^4*b*c^2*e^6 + 16*a^2*c^5*d^5*e + 16*a^4 *c^3*d*e^5 + 32*a^3*c^4*d^3*e^3 + 4*a*b*c^5*d^6 + 16*a^2*b^2*c^3*d^3*e^3 + 12*a^2*b^3*c^2*d^2*e^4 - 16*a*b^2*c^4*d^5*e + 4*a*b^5*c*d^2*e^4 - 8*a^2*b ^4*c*d*e^5 + 24*a*b^3*c^3*d^4*e^2 - 16*a*b^4*c^2*d^3*e^3 - 36*a^2*b*c^4*d^ 4*e^2 - 52*a^3*b*c^3*d^2*e^4 + 16*a^3*b^2*c^2*d*e^5) + (-(a*b^7*e^4 + b^5* c^3*d^4 + c^3*d^4*(-(4*a*c - b^2)^5)^(1/2) - 8*a*b^3*c^4*d^4 + 16*a^2*b*c^ 5*d^4 - a*b^2*e^4*(-(4*a*c - b^2)^5)^(1/2) - 11*a^2*b^5*c*e^4 - 48*a^4*b*c ^3*e^4 + a^2*c*e^4*(-(4*a*c - b^2)^5)^(1/2) - 128*a^3*c^5*d^3*e + 128*a^4* c^4*d*e^3 + 40*a^3*b^3*c^2*e^4 - 4*a*b^6*c*d*e^3 - 48*a^2*b^3*c^3*d^2*e^2 - 8*a*b^4*c^3*d^3*e + 6*a*b^5*c^2*d^2*e^2 + 64*a^2*b^2*c^4*d^3*e + 40*a^2* b^4*c^2*d*e^3 + 96*a^3*b*c^4*d^2*e^2 - 128*a^3*b^2*c^3*d*e^3 - 6*a*c^2*d^2 *e^2*(-(4*a*c - b^2)^5)^(1/2) + 4*a*b*c*d*e^3*(-(4*a*c - b^2)^5)^(1/2))/(5 12*(256*a^5*c^7 + a*b^8*c^3 - 16*a^2*b^6*c^4 + 96*a^3*b^4*c^5 - 256*a^4*b^ 2*c^6)))^(3/4)*(x*(-(a*b^7*e^4 + b^5*c^3*d^4 + c^3*d^4*(-(4*a*c - b^2)^5)^ (1/2) - 8*a*b^3*c^4*d^4 + 16*a^2*b*c^5*d^4 - a*b^2*e^4*(-(4*a*c - b^2)^5)^ (1/2) - 11*a^2*b^5*c*e^4 - 48*a^4*b*c^3*e^4 + a^2*c*e^4*(-(4*a*c - b^2)^5) ^(1/2) - 128*a^3*c^5*d^3*e + 128*a^4*c^4*d*e^3 + 40*a^3*b^3*c^2*e^4 - 4*a* b^6*c*d*e^3 - 48*a^2*b^3*c^3*d^2*e^2 - 8*a*b^4*c^3*d^3*e + 6*a*b^5*c^2*d^2 *e^2 + 64*a^2*b^2*c^4*d^3*e + 40*a^2*b^4*c^2*d*e^3 + 96*a^3*b*c^4*d^2*e^2 - 128*a^3*b^2*c^3*d*e^3 - 6*a*c^2*d^2*e^2*(-(4*a*c - b^2)^5)^(1/2) + 4*...
\[ \int \frac {x^2 \left (d+e x^4\right )}{a+b x^4+c x^8} \, dx=\int \frac {x^{2} \left (e \,x^{4}+d \right )}{c \,x^{8}+b \,x^{4}+a}d x \] Input:
int(x^2*(e*x^4+d)/(c*x^8+b*x^4+a),x)
Output:
int(x^2*(e*x^4+d)/(c*x^8+b*x^4+a),x)