\(\int \frac {a^2 e+6 a b e x^2+5 b^2 e x^4}{c+d x^2 (a+b x^2)^4} \, dx\) [259]

Optimal result

Integrand size = 42, antiderivative size = 34 \[ \int \frac {a^2 e+6 a b e x^2+5 b^2 e x^4}{c+d x^2 \left (a+b x^2\right )^4} \, dx=\frac {e \arctan \left (\frac {\sqrt {d} x \left (a+b x^2\right )^2}{\sqrt {c}}\right )}{\sqrt {c} \sqrt {d}} \] Output:

e*arctan(d^(1/2)*x*(b*x^2+a)^2/c^(1/2))/c^(1/2)/d^(1/2)
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {a^2 e+6 a b e x^2+5 b^2 e x^4}{c+d x^2 \left (a+b x^2\right )^4} \, dx=\frac {e \arctan \left (\frac {\sqrt {d} x \left (a+b x^2\right )^2}{\sqrt {c}}\right )}{\sqrt {c} \sqrt {d}} \] Input:

Integrate[(a^2*e + 6*a*b*e*x^2 + 5*b^2*e*x^4)/(c + d*x^2*(a + b*x^2)^4),x]
 

Output:

(e*ArcTan[(Sqrt[d]*x*(a + b*x^2)^2)/Sqrt[c]])/(Sqrt[c]*Sqrt[d])
 

Rubi [F]

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.

\(\displaystyle \int \frac {a^2 e+6 a b e x^2+5 b^2 e x^4}{d x^2 \left (a+b x^2\right )^4+c} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {5 b^2 e x^4}{a^4 d x^2+4 a^3 b d x^4+6 a^2 b^2 d x^6+4 a b^3 d x^8+b^4 d x^{10}+c}+\frac {6 a b e x^2}{a^4 d x^2+4 a^3 b d x^4+6 a^2 b^2 d x^6+4 a b^3 d x^8+b^4 d x^{10}+c}+\frac {a^2 e}{a^4 d x^2+4 a^3 b d x^4+6 a^2 b^2 d x^6+4 a b^3 d x^8+b^4 d x^{10}+c}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle a^2 e \int \frac {1}{d x^2 \left (b x^2+a\right )^4+c}dx+5 b^2 e \int \frac {x^4}{d x^2 \left (b x^2+a\right )^4+c}dx+6 a b e \int \frac {x^2}{d x^2 \left (b x^2+a\right )^4+c}dx\)

Input:

Int[(a^2*e + 6*a*b*e*x^2 + 5*b^2*e*x^4)/(c + d*x^2*(a + b*x^2)^4),x]
 

Output:

$Aborted
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(108\) vs. \(2(26)=52\).

Time = 0.07 (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((5*b^2*e*x^4+6*a*b*e*x^2+a^2*e)/(c+d*x^2*(b*x^2+a)^4),x,method=_RETURN 
VERBOSE)
 

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)
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 193, normalized size of antiderivative = 5.68 \[ \int \frac {a^2 e+6 a b e x^2+5 b^2 e x^4}{c+d x^2 \left (a+b x^2\right )^4} \, dx=\left [-\frac {\sqrt {-c d} e \log \left (\frac {b^{4} d x^{10} + 4 \, a b^{3} d x^{8} + 6 \, a^{2} b^{2} d x^{6} + 4 \, a^{3} b d x^{4} + a^{4} d x^{2} - 2 \, {\left (b^{2} x^{5} + 2 \, a b x^{3} + a^{2} x\right )} \sqrt {-c d} - c}{b^{4} d x^{10} + 4 \, a b^{3} d x^{8} + 6 \, a^{2} b^{2} d x^{6} + 4 \, a^{3} b d x^{4} + a^{4} d x^{2} + c}\right )}{2 \, c d}, \frac {\sqrt {c d} e \arctan \left (\frac {{\left (b^{2} x^{5} + 2 \, a b x^{3} + a^{2} x\right )} \sqrt {c d}}{c}\right )}{c d}\right ] \] Input:

integrate((5*b^2*e*x^4+6*a*b*e*x^2+a^2*e)/(c+d*x^2*(b*x^2+a)^4),x, algorit 
hm="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)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 99 vs. \(2 (32) = 64\).

Time = 0.63 (sec) , antiderivative size = 99, normalized size of antiderivative = 2.91 \[ \int \frac {a^2 e+6 a b e x^2+5 b^2 e x^4}{c+d x^2 \left (a+b x^2\right )^4} \, dx=e \left (- \frac {\sqrt {- \frac {1}{c d}} \log {\left (\frac {a^{2} x}{b^{2}} + \frac {2 a x^{3}}{b} + x^{5} - \frac {c \sqrt {- \frac {1}{c d}}}{b^{2}} \right )}}{2} + \frac {\sqrt {- \frac {1}{c d}} \log {\left (\frac {a^{2} x}{b^{2}} + \frac {2 a x^{3}}{b} + x^{5} + \frac {c \sqrt {- \frac {1}{c d}}}{b^{2}} \right )}}{2}\right ) \] Input:

integrate((5*b**2*e*x**4+6*a*b*e*x**2+a**2*e)/(c+d*x**2*(b*x**2+a)**4),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)
 

Maxima [F]

\[ \int \frac {a^2 e+6 a b e x^2+5 b^2 e x^4}{c+d x^2 \left (a+b x^2\right )^4} \, dx=\int { \frac {5 \, b^{2} e x^{4} + 6 \, a b e x^{2} + a^{2} e}{{\left (b x^{2} + a\right )}^{4} d x^{2} + c} \,d x } \] Input:

integrate((5*b^2*e*x^4+6*a*b*e*x^2+a^2*e)/(c+d*x^2*(b*x^2+a)^4),x, algorit 
hm="maxima")
 

Output:

integrate((5*b^2*e*x^4 + 6*a*b*e*x^2 + a^2*e)/((b*x^2 + a)^4*d*x^2 + c), x 
)
 

Giac [F(-1)]

Timed out. \[ \int \frac {a^2 e+6 a b e x^2+5 b^2 e x^4}{c+d x^2 \left (a+b x^2\right )^4} \, dx=\text {Timed out} \] Input:

integrate((5*b^2*e*x^4+6*a*b*e*x^2+a^2*e)/(c+d*x^2*(b*x^2+a)^4),x, algorit 
hm="giac")
 

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.24 \[ \int \frac {a^2 e+6 a b e x^2+5 b^2 e x^4}{c+d x^2 \left (a+b x^2\right )^4} \, dx=\frac {e\,\mathrm {atan}\left (\frac {a^2\,\sqrt {d}\,x+b^2\,\sqrt {d}\,x^5+2\,a\,b\,\sqrt {d}\,x^3}{\sqrt {c}}\right )}{\sqrt {c}\,\sqrt {d}} \] Input:

int((a^2*e + 5*b^2*e*x^4 + 6*a*b*e*x^2)/(c + d*x^2*(a + b*x^2)^4),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))
 

Reduce [F]

\[ \int \frac {a^2 e+6 a b e x^2+5 b^2 e x^4}{c+d x^2 \left (a+b x^2\right )^4} \, dx=e \left (5 \left (\int \frac {x^{4}}{b^{4} d \,x^{10}+4 a \,b^{3} d \,x^{8}+6 a^{2} b^{2} d \,x^{6}+4 a^{3} b d \,x^{4}+a^{4} d \,x^{2}+c}d x \right ) b^{2}+6 \left (\int \frac {x^{2}}{b^{4} d \,x^{10}+4 a \,b^{3} d \,x^{8}+6 a^{2} b^{2} d \,x^{6}+4 a^{3} b d \,x^{4}+a^{4} d \,x^{2}+c}d x \right ) a b +\left (\int \frac {1}{b^{4} d \,x^{10}+4 a \,b^{3} d \,x^{8}+6 a^{2} b^{2} d \,x^{6}+4 a^{3} b d \,x^{4}+a^{4} d \,x^{2}+c}d x \right ) a^{2}\right ) \] Input:

int((5*b^2*e*x^4+6*a*b*e*x^2+a^2*e)/(c+d*x^2*(b*x^2+a)^4),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)