\(\int \frac {c+d x+e x^2+f x^3+g x^4}{(a+b x^4)^2} \, dx\) [108]

Optimal result

Integrand size = 30, antiderivative size = 269 \[ \int \frac {c+d x+e x^2+f x^3+g x^4}{\left (a+b x^4\right )^2} \, dx=\frac {x \left (b c-a g+b d x+b e x^2+b f x^3\right )}{4 a b \left (a+b x^4\right )}+\frac {d \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{4 a^{3/2} \sqrt {b}}-\frac {\left (3 b c+\sqrt {a} \sqrt {b} e+a g\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} b^{5/4}}+\frac {\left (3 b c+\sqrt {a} \sqrt {b} e+a g\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} b^{5/4}}+\frac {\left (3 b c-\sqrt {a} \sqrt {b} e+a g\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x}{\sqrt {a}+\sqrt {b} x^2}\right )}{8 \sqrt {2} a^{7/4} b^{5/4}} \] Output:

1/4*x*(b*f*x^3+b*e*x^2+b*d*x-a*g+b*c)/a/b/(b*x^4+a)+1/4*d*arctan(b^(1/2)*x 
^2/a^(1/2))/a^(3/2)/b^(1/2)+1/16*(3*b*c+a^(1/2)*b^(1/2)*e+a*g)*arctan(-1+2 
^(1/2)*b^(1/4)*x/a^(1/4))*2^(1/2)/a^(7/4)/b^(5/4)+1/16*(3*b*c+a^(1/2)*b^(1 
/2)*e+a*g)*arctan(1+2^(1/2)*b^(1/4)*x/a^(1/4))*2^(1/2)/a^(7/4)/b^(5/4)+1/1 
6*(3*b*c-a^(1/2)*b^(1/2)*e+a*g)*arctanh(2^(1/2)*a^(1/4)*b^(1/4)*x/(a^(1/2) 
+b^(1/2)*x^2))*2^(1/2)/a^(7/4)/b^(5/4)
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.19 \[ \int \frac {c+d x+e x^2+f x^3+g x^4}{\left (a+b x^4\right )^2} \, dx=\frac {-\frac {8 a^{3/4} \sqrt [4]{b} (a (f+g x)-b x (c+x (d+e x)))}{a+b x^4}-2 \left (3 \sqrt {2} b c+4 \sqrt [4]{a} b^{3/4} d+\sqrt {2} \sqrt {a} \sqrt {b} e+\sqrt {2} a g\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )+2 \left (3 \sqrt {2} b c-4 \sqrt [4]{a} b^{3/4} d+\sqrt {2} \sqrt {a} \sqrt {b} e+\sqrt {2} a g\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )+\sqrt {2} \left (-3 b c+\sqrt {a} \sqrt {b} e-a g\right ) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )+\sqrt {2} \left (3 b c-\sqrt {a} \sqrt {b} e+a g\right ) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{32 a^{7/4} b^{5/4}} \] Input:

Integrate[(c + d*x + e*x^2 + f*x^3 + g*x^4)/(a + b*x^4)^2,x]
 

Output:

((-8*a^(3/4)*b^(1/4)*(a*(f + g*x) - b*x*(c + x*(d + e*x))))/(a + b*x^4) - 
2*(3*Sqrt[2]*b*c + 4*a^(1/4)*b^(3/4)*d + Sqrt[2]*Sqrt[a]*Sqrt[b]*e + Sqrt[ 
2]*a*g)*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)] + 2*(3*Sqrt[2]*b*c - 4*a^( 
1/4)*b^(3/4)*d + Sqrt[2]*Sqrt[a]*Sqrt[b]*e + Sqrt[2]*a*g)*ArcTan[1 + (Sqrt 
[2]*b^(1/4)*x)/a^(1/4)] + Sqrt[2]*(-3*b*c + Sqrt[a]*Sqrt[b]*e - a*g)*Log[S 
qrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2] + Sqrt[2]*(3*b*c - Sqrt[ 
a]*Sqrt[b]*e + a*g)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2] 
)/(32*a^(7/4)*b^(5/4))
 

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.30, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2397, 25, 2415, 2009}

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.

\(\displaystyle \int \frac {c+d x+e x^2+f x^3+g x^4}{\left (a+b x^4\right )^2} \, dx\)

\(\Big \downarrow \) 2397

\(\displaystyle \frac {x \left (-a g+b c+b d x+b e x^2+b f x^3\right )}{4 a b \left (a+b x^4\right )}-\frac {\int -\frac {b e x^2+2 b d x+3 b c+a g}{b x^4+a}dx}{4 a b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {b e x^2+2 b d x+3 b c+a g}{b x^4+a}dx}{4 a b}+\frac {x \left (-a g+b c+b d x+b e x^2+b f x^3\right )}{4 a b \left (a+b x^4\right )}\)

\(\Big \downarrow \) 2415

\(\displaystyle \frac {\int \left (\frac {2 b d x}{b x^4+a}+\frac {b e x^2+3 b c+a g}{b x^4+a}\right )dx}{4 a b}+\frac {x \left (-a g+b c+b d x+b e x^2+b f x^3\right )}{4 a b \left (a+b x^4\right )}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (\sqrt {a} \sqrt {b} e+a g+3 b c\right )}{2 \sqrt {2} a^{3/4} \sqrt [4]{b}}+\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right ) \left (\sqrt {a} \sqrt {b} e+a g+3 b c\right )}{2 \sqrt {2} a^{3/4} \sqrt [4]{b}}-\frac {\log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right ) \left (-\sqrt {a} \sqrt {b} e+a g+3 b c\right )}{4 \sqrt {2} a^{3/4} \sqrt [4]{b}}+\frac {\log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right ) \left (-\sqrt {a} \sqrt {b} e+a g+3 b c\right )}{4 \sqrt {2} a^{3/4} \sqrt [4]{b}}+\frac {\sqrt {b} d \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{\sqrt {a}}}{4 a b}+\frac {x \left (-a g+b c+b d x+b e x^2+b f x^3\right )}{4 a b \left (a+b x^4\right )}\)

Input:

Int[(c + d*x + e*x^2 + f*x^3 + g*x^4)/(a + b*x^4)^2,x]
 

Output:

(x*(b*c - a*g + b*d*x + b*e*x^2 + b*f*x^3))/(4*a*b*(a + b*x^4)) + ((Sqrt[b 
]*d*ArcTan[(Sqrt[b]*x^2)/Sqrt[a]])/Sqrt[a] - ((3*b*c + Sqrt[a]*Sqrt[b]*e + 
 a*g)*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]*a^(3/4)*b^(1/4)) 
 + ((3*b*c + Sqrt[a]*Sqrt[b]*e + a*g)*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/ 
4)])/(2*Sqrt[2]*a^(3/4)*b^(1/4)) - ((3*b*c - Sqrt[a]*Sqrt[b]*e + a*g)*Log[ 
Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(4*Sqrt[2]*a^(3/4)*b^( 
1/4)) + ((3*b*c - Sqrt[a]*Sqrt[b]*e + a*g)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b 
^(1/4)*x + Sqrt[b]*x^2])/(4*Sqrt[2]*a^(3/4)*b^(1/4)))/(4*a*b)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2009

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

rule 2397

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = Expon[Pq, 
x]}, Module[{Q = PolynomialQuotient[b^(Floor[(q - 1)/n] + 1)*Pq, a + b*x^n, 
 x], R = PolynomialRemainder[b^(Floor[(q - 1)/n] + 1)*Pq, a + b*x^n, x]}, S 
imp[(-x)*R*((a + b*x^n)^(p + 1)/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1))), x] 
 + Simp[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1))   Int[(a + b*x^n)^(p + 1)* 
ExpandToSum[a*n*(p + 1)*Q + n*(p + 1)*R + D[x*R, x], x], x], x]] /; GeQ[q, 
n]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1]
 

rule 2415

Int[(Pq_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = Sum[x^ii*((Coeff 
[Pq, x, ii] + Coeff[Pq, x, n/2 + ii]*x^(n/2))/(a + b*x^n)), {ii, 0, n/2 - 1 
}]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n/2, 
 0] && Expon[Pq, x] < n
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.11 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.39

method	result	size
risch	\(\frac {\frac {e \,x^{3}}{4 a}+\frac {d \,x^{2}}{4 a}-\frac {\left (a g -c b \right ) x}{4 a b}-\frac {f}{4 b}}{b \,x^{4}+a}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{4}+a \right )}{\sum }\frac {\left (\textit {\_R}^{2} e +2 d \textit {\_R} +\frac {a g +3 c b}{b}\right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}}{16 b a}\)	\(105\)
default	\(\frac {\frac {e \,x^{3}}{4 a}+\frac {d \,x^{2}}{4 a}-\frac {\left (a g -c b \right ) x}{4 a b}-\frac {f}{4 b}}{b \,x^{4}+a}+\frac {\frac {\left (a g +3 c b \right ) \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}{x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{8 a}+\frac {b d \arctan \left (x^{2} \sqrt {\frac {b}{a}}\right )}{\sqrt {a b}}+\frac {e \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}{x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{8 \left (\frac {a}{b}\right )^{\frac {1}{4}}}}{4 b a}\)	\(291\)

Input:

int((g*x^4+f*x^3+e*x^2+d*x+c)/(b*x^4+a)^2,x,method=_RETURNVERBOSE)
 

Output:

(1/4*e/a*x^3+1/4*d/a*x^2-1/4*(a*g-b*c)/a/b*x-1/4*f/b)/(b*x^4+a)+1/16/b/a*s 
um((_R^2*e+2*d*_R+1/b*(a*g+3*b*c))/_R^3*ln(x-_R),_R=RootOf(_Z^4*b+a))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 24.43 (sec) , antiderivative size = 352423, normalized size of antiderivative = 1310.12 \[ \int \frac {c+d x+e x^2+f x^3+g x^4}{\left (a+b x^4\right )^2} \, dx=\text {Too large to display} \] Input:

integrate((g*x^4+f*x^3+e*x^2+d*x+c)/(b*x^4+a)^2,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {c+d x+e x^2+f x^3+g x^4}{\left (a+b x^4\right )^2} \, dx=\text {Timed out} \] Input:

integrate((g*x**4+f*x**3+e*x**2+d*x+c)/(b*x**4+a)**2,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.30 \[ \int \frac {c+d x+e x^2+f x^3+g x^4}{\left (a+b x^4\right )^2} \, dx=\frac {b e x^{3} + b d x^{2} - a f + {\left (b c - a g\right )} x}{4 \, {\left (a b^{2} x^{4} + a^{2} b\right )}} + \frac {\frac {\sqrt {2} {\left (3 \, b^{\frac {3}{2}} c - \sqrt {a} b e + a \sqrt {b} g\right )} \log \left (\sqrt {b} x^{2} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {3}{4}}} - \frac {\sqrt {2} {\left (3 \, b^{\frac {3}{2}} c - \sqrt {a} b e + a \sqrt {b} g\right )} \log \left (\sqrt {b} x^{2} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {3}{4}}} + \frac {2 \, {\left (3 \, \sqrt {2} a^{\frac {1}{4}} b^{\frac {7}{4}} c + \sqrt {2} a^{\frac {3}{4}} b^{\frac {5}{4}} e + \sqrt {2} a^{\frac {5}{4}} b^{\frac {3}{4}} g - 4 \, \sqrt {a} b^{\frac {3}{2}} d\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {b}} b^{\frac {3}{4}}} + \frac {2 \, {\left (3 \, \sqrt {2} a^{\frac {1}{4}} b^{\frac {7}{4}} c + \sqrt {2} a^{\frac {3}{4}} b^{\frac {5}{4}} e + \sqrt {2} a^{\frac {5}{4}} b^{\frac {3}{4}} g + 4 \, \sqrt {a} b^{\frac {3}{2}} d\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{a^{\frac {3}{4}} \sqrt {\sqrt {a} \sqrt {b}} b^{\frac {3}{4}}}}{32 \, a b} \] Input:

integrate((g*x^4+f*x^3+e*x^2+d*x+c)/(b*x^4+a)^2,x, algorithm="maxima")
 

Output:

1/4*(b*e*x^3 + b*d*x^2 - a*f + (b*c - a*g)*x)/(a*b^2*x^4 + a^2*b) + 1/32*( 
sqrt(2)*(3*b^(3/2)*c - sqrt(a)*b*e + a*sqrt(b)*g)*log(sqrt(b)*x^2 + sqrt(2 
)*a^(1/4)*b^(1/4)*x + sqrt(a))/(a^(3/4)*b^(3/4)) - sqrt(2)*(3*b^(3/2)*c - 
sqrt(a)*b*e + a*sqrt(b)*g)*log(sqrt(b)*x^2 - sqrt(2)*a^(1/4)*b^(1/4)*x + s 
qrt(a))/(a^(3/4)*b^(3/4)) + 2*(3*sqrt(2)*a^(1/4)*b^(7/4)*c + sqrt(2)*a^(3/ 
4)*b^(5/4)*e + sqrt(2)*a^(5/4)*b^(3/4)*g - 4*sqrt(a)*b^(3/2)*d)*arctan(1/2 
*sqrt(2)*(2*sqrt(b)*x + sqrt(2)*a^(1/4)*b^(1/4))/sqrt(sqrt(a)*sqrt(b)))/(a 
^(3/4)*sqrt(sqrt(a)*sqrt(b))*b^(3/4)) + 2*(3*sqrt(2)*a^(1/4)*b^(7/4)*c + s 
qrt(2)*a^(3/4)*b^(5/4)*e + sqrt(2)*a^(5/4)*b^(3/4)*g + 4*sqrt(a)*b^(3/2)*d 
)*arctan(1/2*sqrt(2)*(2*sqrt(b)*x - sqrt(2)*a^(1/4)*b^(1/4))/sqrt(sqrt(a)* 
sqrt(b)))/(a^(3/4)*sqrt(sqrt(a)*sqrt(b))*b^(3/4)))/(a*b)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 360, normalized size of antiderivative = 1.34 \[ \int \frac {c+d x+e x^2+f x^3+g x^4}{\left (a+b x^4\right )^2} \, dx=\frac {b e x^{3} + b d x^{2} + b c x - a g x - a f}{4 \, {\left (b x^{4} + a\right )} a b} + \frac {\sqrt {2} {\left (2 \, \sqrt {2} \sqrt {a b} b^{2} d + 3 \, \left (a b^{3}\right )^{\frac {1}{4}} b^{2} c + \left (a b^{3}\right )^{\frac {1}{4}} a b g + \left (a b^{3}\right )^{\frac {3}{4}} e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{16 \, a^{2} b^{3}} + \frac {\sqrt {2} {\left (2 \, \sqrt {2} \sqrt {a b} b^{2} d + 3 \, \left (a b^{3}\right )^{\frac {1}{4}} b^{2} c + \left (a b^{3}\right )^{\frac {1}{4}} a b g + \left (a b^{3}\right )^{\frac {3}{4}} e\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{16 \, a^{2} b^{3}} + \frac {\sqrt {2} {\left (3 \, \left (a b^{3}\right )^{\frac {1}{4}} b^{2} c + \left (a b^{3}\right )^{\frac {1}{4}} a b g - \left (a b^{3}\right )^{\frac {3}{4}} e\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{32 \, a^{2} b^{3}} - \frac {\sqrt {2} {\left (3 \, \left (a b^{3}\right )^{\frac {1}{4}} b^{2} c + \left (a b^{3}\right )^{\frac {1}{4}} a b g - \left (a b^{3}\right )^{\frac {3}{4}} e\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{32 \, a^{2} b^{3}} \] Input:

integrate((g*x^4+f*x^3+e*x^2+d*x+c)/(b*x^4+a)^2,x, algorithm="giac")
 

Output:

1/4*(b*e*x^3 + b*d*x^2 + b*c*x - a*g*x - a*f)/((b*x^4 + a)*a*b) + 1/16*sqr 
t(2)*(2*sqrt(2)*sqrt(a*b)*b^2*d + 3*(a*b^3)^(1/4)*b^2*c + (a*b^3)^(1/4)*a* 
b*g + (a*b^3)^(3/4)*e)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)*(a/b)^(1/4))/(a/b 
)^(1/4))/(a^2*b^3) + 1/16*sqrt(2)*(2*sqrt(2)*sqrt(a*b)*b^2*d + 3*(a*b^3)^( 
1/4)*b^2*c + (a*b^3)^(1/4)*a*b*g + (a*b^3)^(3/4)*e)*arctan(1/2*sqrt(2)*(2* 
x - sqrt(2)*(a/b)^(1/4))/(a/b)^(1/4))/(a^2*b^3) + 1/32*sqrt(2)*(3*(a*b^3)^ 
(1/4)*b^2*c + (a*b^3)^(1/4)*a*b*g - (a*b^3)^(3/4)*e)*log(x^2 + sqrt(2)*x*( 
a/b)^(1/4) + sqrt(a/b))/(a^2*b^3) - 1/32*sqrt(2)*(3*(a*b^3)^(1/4)*b^2*c + 
(a*b^3)^(1/4)*a*b*g - (a*b^3)^(3/4)*e)*log(x^2 - sqrt(2)*x*(a/b)^(1/4) + s 
qrt(a/b))/(a^2*b^3)
 

Mupad [B] (verification not implemented)

Time = 6.73 (sec) , antiderivative size = 1383, normalized size of antiderivative = 5.14 \[ \int \frac {c+d x+e x^2+f x^3+g x^4}{\left (a+b x^4\right )^2} \, dx =\text {Too large to display} \] Input:

int((c + d*x + e*x^2 + f*x^3 + g*x^4)/(a + b*x^4)^2,x)
 

Output:

symsum(log(- (9*b^2*c^2*e - 12*b^2*c*d^2 + a^2*e*g^2 + a*b*e^3 - 4*a*b*d^2 
*g + 6*a*b*c*e*g)/(64*a^3) - (root(65536*a^7*b^5*z^4 + 1024*a^5*b^3*e*g*z^ 
2 + 3072*a^4*b^4*c*e*z^2 + 2048*a^4*b^4*d^2*z^2 - 768*a^3*b^3*c*d*g*z - 12 
8*a^4*b^2*d*g^2*z + 128*a^3*b^3*d*e^2*z - 1152*a^2*b^4*c^2*d*z - 16*a^2*b^ 
2*d^2*e*g + 12*a^2*b^2*c*e^2*g - 48*a*b^3*c*d^2*e + 108*a*b^3*c^3*g + 12*a 
^3*b*c*g^3 + 54*a^2*b^2*c^2*g^2 + 2*a^3*b*e^2*g^2 + 18*a*b^3*c^2*e^2 + 16* 
a*b^3*d^4 + 81*b^4*c^4 + a^2*b^2*e^4 + a^4*g^4, z, k)*b*(9*b^2*c^2*x + a^2 
*g^2*x + 16*root(65536*a^7*b^5*z^4 + 1024*a^5*b^3*e*g*z^2 + 3072*a^4*b^4*c 
*e*z^2 + 2048*a^4*b^4*d^2*z^2 - 768*a^3*b^3*c*d*g*z - 128*a^4*b^2*d*g^2*z 
+ 128*a^3*b^3*d*e^2*z - 1152*a^2*b^4*c^2*d*z - 16*a^2*b^2*d^2*e*g + 12*a^2 
*b^2*c*e^2*g - 48*a*b^3*c*d^2*e + 108*a*b^3*c^3*g + 12*a^3*b*c*g^3 + 54*a^ 
2*b^2*c^2*g^2 + 2*a^3*b*e^2*g^2 + 18*a*b^3*c^2*e^2 + 16*a*b^3*d^4 + 81*b^4 
*c^4 + a^2*b^2*e^4 + a^4*g^4, z, k)*a^3*b*g - a*b*e^2*x + 48*root(65536*a^ 
7*b^5*z^4 + 1024*a^5*b^3*e*g*z^2 + 3072*a^4*b^4*c*e*z^2 + 2048*a^4*b^4*d^2 
*z^2 - 768*a^3*b^3*c*d*g*z - 128*a^4*b^2*d*g^2*z + 128*a^3*b^3*d*e^2*z - 1 
152*a^2*b^4*c^2*d*z - 16*a^2*b^2*d^2*e*g + 12*a^2*b^2*c*e^2*g - 48*a*b^3*c 
*d^2*e + 108*a*b^3*c^3*g + 12*a^3*b*c*g^3 + 54*a^2*b^2*c^2*g^2 + 2*a^3*b*e 
^2*g^2 + 18*a*b^3*c^2*e^2 + 16*a*b^3*d^4 + 81*b^4*c^4 + a^2*b^2*e^4 + a^4* 
g^4, z, k)*a^2*b^2*c + 4*a*b*d*e - 32*root(65536*a^7*b^5*z^4 + 1024*a^5*b^ 
3*e*g*z^2 + 3072*a^4*b^4*c*e*z^2 + 2048*a^4*b^4*d^2*z^2 - 768*a^3*b^3*c...
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 1082, normalized size of antiderivative = 4.02 \[ \int \frac {c+d x+e x^2+f x^3+g x^4}{\left (a+b x^4\right )^2} \, dx =\text {Too large to display} \] Input:

int((g*x^4+f*x^3+e*x^2+d*x+c)/(b*x^4+a)^2,x)
 

Output:

( - 2*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(b 
)*x)/(b**(1/4)*a**(1/4)*sqrt(2)))*a*b*e - 2*b**(1/4)*a**(3/4)*sqrt(2)*atan 
((b**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(b)*x)/(b**(1/4)*a**(1/4)*sqrt(2)))*b* 
*2*e*x**4 - 2*b**(3/4)*a**(1/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) - 
2*sqrt(b)*x)/(b**(1/4)*a**(1/4)*sqrt(2)))*a**2*g - 6*b**(3/4)*a**(1/4)*sqr 
t(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(b)*x)/(b**(1/4)*a**(1/4)*sqr 
t(2)))*a*b*c - 2*b**(3/4)*a**(1/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) 
 - 2*sqrt(b)*x)/(b**(1/4)*a**(1/4)*sqrt(2)))*a*b*g*x**4 - 6*b**(3/4)*a**(1 
/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(b)*x)/(b**(1/4)*a**(1 
/4)*sqrt(2)))*b**2*c*x**4 - 8*sqrt(b)*sqrt(a)*atan((b**(1/4)*a**(1/4)*sqrt 
(2) - 2*sqrt(b)*x)/(b**(1/4)*a**(1/4)*sqrt(2)))*a*b*d - 8*sqrt(b)*sqrt(a)* 
atan((b**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(b)*x)/(b**(1/4)*a**(1/4)*sqrt(2)) 
)*b**2*d*x**4 + 2*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2 
) + 2*sqrt(b)*x)/(b**(1/4)*a**(1/4)*sqrt(2)))*a*b*e + 2*b**(1/4)*a**(3/4)* 
sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) + 2*sqrt(b)*x)/(b**(1/4)*a**(1/4)* 
sqrt(2)))*b**2*e*x**4 + 2*b**(3/4)*a**(1/4)*sqrt(2)*atan((b**(1/4)*a**(1/4 
)*sqrt(2) + 2*sqrt(b)*x)/(b**(1/4)*a**(1/4)*sqrt(2)))*a**2*g + 6*b**(3/4)* 
a**(1/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) + 2*sqrt(b)*x)/(b**(1/4)* 
a**(1/4)*sqrt(2)))*a*b*c + 2*b**(3/4)*a**(1/4)*sqrt(2)*atan((b**(1/4)*a**( 
1/4)*sqrt(2) + 2*sqrt(b)*x)/(b**(1/4)*a**(1/4)*sqrt(2)))*a*b*g*x**4 + 6...