Integrand size = 71, antiderivative size = 34 \[ \int \frac {\left (a+b x^2\right ) \left (a e+5 b e x^2\right )}{c+a^4 d x^2+4 a^3 b d x^4+6 a^2 b^2 d x^6+4 a b^3 d x^8+b^4 d x^{10}} \, dx=\frac {e \arctan \left (\frac {\sqrt {d} x \left (a+b x^2\right )^2}{\sqrt {c}}\right )}{\sqrt {c} \sqrt {d}} \] Output:
e*arctan(d^(1/2)*x*(b*x^2+a)^2/c^(1/2))/c^(1/2)/d^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.08 (sec) , antiderivative size = 99, normalized size of antiderivative = 2.91 \[ \int \frac {\left (a+b x^2\right ) \left (a e+5 b e x^2\right )}{c+a^4 d x^2+4 a^3 b d x^4+6 a^2 b^2 d x^6+4 a b^3 d x^8+b^4 d x^{10}} \, dx=\frac {e \text {RootSum}\left [c+a^4 d \text {$\#$1}^2+4 a^3 b d \text {$\#$1}^4+6 a^2 b^2 d \text {$\#$1}^6+4 a b^3 d \text {$\#$1}^8+b^4 d \text {$\#$1}^{10}\&,\frac {\log (x-\text {$\#$1})}{a^2 \text {$\#$1}+2 a b \text {$\#$1}^3+b^2 \text {$\#$1}^5}\&\right ]}{2 d} \] Input:
Integrate[((a + b*x^2)*(a*e + 5*b*e*x^2))/(c + a^4*d*x^2 + 4*a^3*b*d*x^4 + 6*a^2*b^2*d*x^6 + 4*a*b^3*d*x^8 + b^4*d*x^10),x]
Output:
(e*RootSum[c + a^4*d*#1^2 + 4*a^3*b*d*#1^4 + 6*a^2*b^2*d*#1^6 + 4*a*b^3*d* #1^8 + b^4*d*#1^10 & , Log[x - #1]/(a^2*#1 + 2*a*b*#1^3 + b^2*#1^5) & ])/( 2*d)
Time = 1.14 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {7239, 27, 7260, 218}
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.
\(\displaystyle \int \frac {\left (a+b x^2\right ) \left (a e+5 b e x^2\right )}{a^4 d x^2+4 a^3 b d x^4+6 a^2 b^2 d x^6+4 a b^3 d x^8+b^4 d x^{10}+c} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e \left (a+b x^2\right ) \left (a+5 b x^2\right )}{d x^2 \left (a+b x^2\right )^4+c}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle e \int \frac {\left (b x^2+a\right ) \left (5 b x^2+a\right )}{d x^2 \left (b x^2+a\right )^4+c}dx\) |
\(\Big \downarrow \) 7260 |
\(\displaystyle e \int \frac {1}{d x^2 \left (b x^2+a\right )^4+c}d\left (x \left (b x^2+a\right )^2\right )\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {e \arctan \left (\frac {\sqrt {d} x \left (a+b x^2\right )^2}{\sqrt {c}}\right )}{\sqrt {c} \sqrt {d}}\) |
Input:
Int[((a + b*x^2)*(a*e + 5*b*e*x^2))/(c + a^4*d*x^2 + 4*a^3*b*d*x^4 + 6*a^2 *b^2*d*x^6 + 4*a*b^3*d*x^8 + b^4*d*x^10),x]
Output:
(e*ArcTan[(Sqrt[d]*x*(a + b*x^2)^2)/Sqrt[c]])/(Sqrt[c]*Sqrt[d])
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[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Int[(u_)*(v_)^(r_.)*((a_) + (b_.)*(v_)^(p_.)*(w_)^(q_.))^(m_.), x_Symbol] : > With[{c = Simplify[u/(p*w*D[v, x] + q*v*D[w, x])]}, Simp[c*(p/(r + 1)) Subst[Int[(a + b*x^(p/(r + 1)))^m, x], x, v^(r + 1)*w], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p, q, r}, x] && EqQ[p, q*(r + 1)] && NeQ[r, -1] && Integ erQ[p/(r + 1)]
Leaf count of result is larger than twice the leaf count of optimal. \(108\) vs. \(2(26)=52\).
Time = 0.08 (sec) , antiderivative size = 109, normalized size of antiderivative = 3.21
method | result | size |
risch | \(-\frac {e \ln \left (\left (-c d \right )^{\frac {3}{2}} b^{2} x^{5}+2 \left (-c d \right )^{\frac {3}{2}} a b \,x^{3}+\left (-c d \right )^{\frac {3}{2}} a^{2} x +c^{2} d \right )}{2 \sqrt {-c d}}+\frac {e \ln \left (\left (-c d \right )^{\frac {3}{2}} b^{2} x^{5}+2 \left (-c d \right )^{\frac {3}{2}} a b \,x^{3}+\left (-c d \right )^{\frac {3}{2}} a^{2} x -c^{2} d \right )}{2 \sqrt {-c d}}\) | \(109\) |
default | \(\frac {e \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b^{4} d \,\textit {\_Z}^{10}+4 d a \,b^{3} \textit {\_Z}^{8}+6 a^{2} d \,b^{2} \textit {\_Z}^{6}+4 a^{3} d b \,\textit {\_Z}^{4}+a^{4} d \,\textit {\_Z}^{2}+c \right )}{\sum }\frac {\left (5 \textit {\_R}^{4} b^{2}+6 \textit {\_R}^{2} a b +a^{2}\right ) \ln \left (x -\textit {\_R} \right )}{5 b^{4} \textit {\_R}^{9}+16 a \,b^{3} \textit {\_R}^{7}+18 a^{2} b^{2} \textit {\_R}^{5}+8 a^{3} b \,\textit {\_R}^{3}+a^{4} \textit {\_R}}\right )}{2 d}\) | \(132\) |
Input:
int((b*x^2+a)*(5*b*e*x^2+a*e)/(b^4*d*x^10+4*a*b^3*d*x^8+6*a^2*b^2*d*x^6+4* a^3*b*d*x^4+a^4*d*x^2+c),x,method=_RETURNVERBOSE)
Output:
-1/2/(-c*d)^(1/2)*e*ln((-c*d)^(3/2)*b^2*x^5+2*(-c*d)^(3/2)*a*b*x^3+(-c*d)^ (3/2)*a^2*x+c^2*d)+1/2/(-c*d)^(1/2)*e*ln((-c*d)^(3/2)*b^2*x^5+2*(-c*d)^(3/ 2)*a*b*x^3+(-c*d)^(3/2)*a^2*x-c^2*d)
Time = 0.14 (sec) , antiderivative size = 193, normalized size of antiderivative = 5.68 \[ \int \frac {\left (a+b x^2\right ) \left (a e+5 b e x^2\right )}{c+a^4 d x^2+4 a^3 b d x^4+6 a^2 b^2 d x^6+4 a b^3 d x^8+b^4 d x^{10}} \, dx=\left [-\frac {\sqrt {-c d} e \log \left (\frac {b^{4} d x^{10} + 4 \, a b^{3} d x^{8} + 6 \, a^{2} b^{2} d x^{6} + 4 \, a^{3} b d x^{4} + a^{4} d x^{2} - 2 \, {\left (b^{2} x^{5} + 2 \, a b x^{3} + a^{2} x\right )} \sqrt {-c d} - c}{b^{4} d x^{10} + 4 \, a b^{3} d x^{8} + 6 \, a^{2} b^{2} d x^{6} + 4 \, a^{3} b d x^{4} + a^{4} d x^{2} + c}\right )}{2 \, c d}, \frac {\sqrt {c d} e \arctan \left (\frac {{\left (b^{2} x^{5} + 2 \, a b x^{3} + a^{2} x\right )} \sqrt {c d}}{c}\right )}{c d}\right ] \] Input:
integrate((b*x^2+a)*(5*b*e*x^2+a*e)/(b^4*d*x^10+4*a*b^3*d*x^8+6*a^2*b^2*d* x^6+4*a^3*b*d*x^4+a^4*d*x^2+c),x, algorithm="fricas")
Output:
[-1/2*sqrt(-c*d)*e*log((b^4*d*x^10 + 4*a*b^3*d*x^8 + 6*a^2*b^2*d*x^6 + 4*a ^3*b*d*x^4 + a^4*d*x^2 - 2*(b^2*x^5 + 2*a*b*x^3 + a^2*x)*sqrt(-c*d) - c)/( b^4*d*x^10 + 4*a*b^3*d*x^8 + 6*a^2*b^2*d*x^6 + 4*a^3*b*d*x^4 + a^4*d*x^2 + c))/(c*d), sqrt(c*d)*e*arctan((b^2*x^5 + 2*a*b*x^3 + a^2*x)*sqrt(c*d)/c)/ (c*d)]
Leaf count of result is larger than twice the leaf count of optimal. 99 vs. \(2 (32) = 64\).
Time = 0.65 (sec) , antiderivative size = 99, normalized size of antiderivative = 2.91 \[ \int \frac {\left (a+b x^2\right ) \left (a e+5 b e x^2\right )}{c+a^4 d x^2+4 a^3 b d x^4+6 a^2 b^2 d x^6+4 a b^3 d x^8+b^4 d x^{10}} \, dx=e \left (- \frac {\sqrt {- \frac {1}{c d}} \log {\left (\frac {a^{2} x}{b^{2}} + \frac {2 a x^{3}}{b} + x^{5} - \frac {c \sqrt {- \frac {1}{c d}}}{b^{2}} \right )}}{2} + \frac {\sqrt {- \frac {1}{c d}} \log {\left (\frac {a^{2} x}{b^{2}} + \frac {2 a x^{3}}{b} + x^{5} + \frac {c \sqrt {- \frac {1}{c d}}}{b^{2}} \right )}}{2}\right ) \] Input:
integrate((b*x**2+a)*(5*b*e*x**2+a*e)/(b**4*d*x**10+4*a*b**3*d*x**8+6*a**2 *b**2*d*x**6+4*a**3*b*d*x**4+a**4*d*x**2+c),x)
Output:
e*(-sqrt(-1/(c*d))*log(a**2*x/b**2 + 2*a*x**3/b + x**5 - c*sqrt(-1/(c*d))/ b**2)/2 + sqrt(-1/(c*d))*log(a**2*x/b**2 + 2*a*x**3/b + x**5 + c*sqrt(-1/( c*d))/b**2)/2)
\[ \int \frac {\left (a+b x^2\right ) \left (a e+5 b e x^2\right )}{c+a^4 d x^2+4 a^3 b d x^4+6 a^2 b^2 d x^6+4 a b^3 d x^8+b^4 d x^{10}} \, dx=\int { \frac {{\left (5 \, b e x^{2} + a e\right )} {\left (b x^{2} + a\right )}}{b^{4} d x^{10} + 4 \, a b^{3} d x^{8} + 6 \, a^{2} b^{2} d x^{6} + 4 \, a^{3} b d x^{4} + a^{4} d x^{2} + c} \,d x } \] Input:
integrate((b*x^2+a)*(5*b*e*x^2+a*e)/(b^4*d*x^10+4*a*b^3*d*x^8+6*a^2*b^2*d* x^6+4*a^3*b*d*x^4+a^4*d*x^2+c),x, algorithm="maxima")
Output:
integrate((5*b*e*x^2 + a*e)*(b*x^2 + a)/(b^4*d*x^10 + 4*a*b^3*d*x^8 + 6*a^ 2*b^2*d*x^6 + 4*a^3*b*d*x^4 + a^4*d*x^2 + c), x)
Timed out. \[ \int \frac {\left (a+b x^2\right ) \left (a e+5 b e x^2\right )}{c+a^4 d x^2+4 a^3 b d x^4+6 a^2 b^2 d x^6+4 a b^3 d x^8+b^4 d x^{10}} \, dx=\text {Timed out} \] Input:
integrate((b*x^2+a)*(5*b*e*x^2+a*e)/(b^4*d*x^10+4*a*b^3*d*x^8+6*a^2*b^2*d* x^6+4*a^3*b*d*x^4+a^4*d*x^2+c),x, algorithm="giac")
Output:
Timed out
Time = 23.15 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.24 \[ \int \frac {\left (a+b x^2\right ) \left (a e+5 b e x^2\right )}{c+a^4 d x^2+4 a^3 b d x^4+6 a^2 b^2 d x^6+4 a b^3 d x^8+b^4 d x^{10}} \, dx=\frac {e\,\mathrm {atan}\left (\frac {a^2\,\sqrt {d}\,x+b^2\,\sqrt {d}\,x^5+2\,a\,b\,\sqrt {d}\,x^3}{\sqrt {c}}\right )}{\sqrt {c}\,\sqrt {d}} \] Input:
int(((a + b*x^2)*(a*e + 5*b*e*x^2))/(c + a^4*d*x^2 + b^4*d*x^10 + 6*a^2*b^ 2*d*x^6 + 4*a^3*b*d*x^4 + 4*a*b^3*d*x^8),x)
Output:
(e*atan((a^2*d^(1/2)*x + b^2*d^(1/2)*x^5 + 2*a*b*d^(1/2)*x^3)/c^(1/2)))/(c ^(1/2)*d^(1/2))
\[ \int \frac {\left (a+b x^2\right ) \left (a e+5 b e x^2\right )}{c+a^4 d x^2+4 a^3 b d x^4+6 a^2 b^2 d x^6+4 a b^3 d x^8+b^4 d x^{10}} \, dx=e \left (5 \left (\int \frac {x^{4}}{b^{4} d \,x^{10}+4 a \,b^{3} d \,x^{8}+6 a^{2} b^{2} d \,x^{6}+4 a^{3} b d \,x^{4}+a^{4} d \,x^{2}+c}d x \right ) b^{2}+6 \left (\int \frac {x^{2}}{b^{4} d \,x^{10}+4 a \,b^{3} d \,x^{8}+6 a^{2} b^{2} d \,x^{6}+4 a^{3} b d \,x^{4}+a^{4} d \,x^{2}+c}d x \right ) a b +\left (\int \frac {1}{b^{4} d \,x^{10}+4 a \,b^{3} d \,x^{8}+6 a^{2} b^{2} d \,x^{6}+4 a^{3} b d \,x^{4}+a^{4} d \,x^{2}+c}d x \right ) a^{2}\right ) \] Input:
int((b*x^2+a)*(5*b*e*x^2+a*e)/(b^4*d*x^10+4*a*b^3*d*x^8+6*a^2*b^2*d*x^6+4* a^3*b*d*x^4+a^4*d*x^2+c),x)
Output:
e*(5*int(x**4/(a**4*d*x**2 + 4*a**3*b*d*x**4 + 6*a**2*b**2*d*x**6 + 4*a*b* *3*d*x**8 + b**4*d*x**10 + c),x)*b**2 + 6*int(x**2/(a**4*d*x**2 + 4*a**3*b *d*x**4 + 6*a**2*b**2*d*x**6 + 4*a*b**3*d*x**8 + b**4*d*x**10 + c),x)*a*b + int(1/(a**4*d*x**2 + 4*a**3*b*d*x**4 + 6*a**2*b**2*d*x**6 + 4*a*b**3*d*x **8 + b**4*d*x**10 + c),x)*a**2)