\(\int \frac {x^2 (-f+(\frac {A}{(B+C) F H}+\frac {D}{F H}+\frac {G}{H}) x^4)}{(-c+a x^4) (d+b x^4)} \, dx\) [108]

Optimal result

Integrand size = 59, antiderivative size = 503 \[ \int \frac {x^2 \left (-f+\left (\frac {A}{(B+C) F H}+\frac {D}{F H}+\frac {G}{H}\right ) x^4\right )}{\left (-c+a x^4\right ) \left (d+b x^4\right )} \, dx=\frac {\left (-a f+c \left (\frac {A}{(B+C) F H}+\frac {D}{F H}+\frac {G}{H}\right )\right ) \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{c}}\right )}{2 a^{3/4} \sqrt [4]{c} (b c+a d)}-\frac {\left (b f+d \left (\frac {A}{(B+C) F H}+\frac {D}{F H}+\frac {G}{H}\right )\right ) \arctan \left (\frac {\sqrt [4]{d}-\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{d}}\right )}{2 \sqrt {2} b^{3/4} \sqrt [4]{d} (b c+a d)}+\frac {\left (b f+d \left (\frac {A}{(B+C) F H}+\frac {D}{F H}+\frac {G}{H}\right )\right ) \arctan \left (\frac {\sqrt [4]{d}+\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{d}}\right )}{2 \sqrt {2} b^{3/4} \sqrt [4]{d} (b c+a d)}+\frac {\left (a f-c \left (\frac {A}{(B+C) F H}+\frac {D}{F H}+\frac {G}{H}\right )\right ) \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{c}}\right )}{2 a^{3/4} \sqrt [4]{c} (b c+a d)}+\frac {\left (b f+d \left (\frac {A}{(B+C) F H}+\frac {D}{F H}+\frac {G}{H}\right )\right ) \log \left (\sqrt {d}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} x+\sqrt {b} x^2\right )}{4 \sqrt {2} b^{3/4} \sqrt [4]{d} (b c+a d)}-\frac {\left (b f+d \left (\frac {A}{(B+C) F H}+\frac {D}{F H}+\frac {G}{H}\right )\right ) \log \left (\sqrt {d}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} x+\sqrt {b} x^2\right )}{4 \sqrt {2} b^{3/4} \sqrt [4]{d} (b c+a d)} \] Output:

1/2*(-a*f+c*(A/(B+C)/F/H+D/F/H+G/H))*arctan(a^(1/4)*x/c^(1/4))/a^(3/4)/c^( 
1/4)/(a*d+b*c)-1/4*(b*f+d*(A/(B+C)/F/H+D/F/H+G/H))*arctan((d^(1/4)-2^(1/2) 
*b^(1/4)*x)/d^(1/4))*2^(1/2)/b^(3/4)/d^(1/4)/(a*d+b*c)+1/4*(b*f+d*(A/(B+C) 
/F/H+D/F/H+G/H))*arctan((d^(1/4)+2^(1/2)*b^(1/4)*x)/d^(1/4))*2^(1/2)/b^(3/ 
4)/d^(1/4)/(a*d+b*c)+1/2*(a*f-c*(A/(B+C)/F/H+D/F/H+G/H))*arctanh(a^(1/4)*x 
/c^(1/4))/a^(3/4)/c^(1/4)/(a*d+b*c)+1/8*(b*f+d*(A/(B+C)/F/H+D/F/H+G/H))*ln 
(d^(1/2)-2^(1/2)*b^(1/4)*d^(1/4)*x+b^(1/2)*x^2)*2^(1/2)/b^(3/4)/d^(1/4)/(a 
*d+b*c)-1/8*(b*f+d*(A/(B+C)/F/H+D/F/H+G/H))*ln(d^(1/2)+2^(1/2)*b^(1/4)*d^( 
1/4)*x+b^(1/2)*x^2)*2^(1/2)/b^(3/4)/d^(1/4)/(a*d+b*c)
 

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 428, normalized size of antiderivative = 0.85 \[ \int \frac {x^2 \left (-f+\left (\frac {A}{(B+C) F H}+\frac {D}{F H}+\frac {G}{H}\right ) x^4\right )}{\left (-c+a x^4\right ) \left (d+b x^4\right )} \, dx=\frac {\frac {4 (A c+(B+C) (c (D+F G)-a f F H)) \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{c}}\right )}{a^{3/4} \sqrt [4]{c}}-\frac {2 \sqrt {2} (A d+(B+C) (d (D+F G)+b f F H)) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{d}}\right )}{b^{3/4} \sqrt [4]{d}}+\frac {2 \sqrt {2} (A d+(B+C) (d (D+F G)+b f F H)) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{d}}\right )}{b^{3/4} \sqrt [4]{d}}+\frac {2 (A c+(B+C) (c (D+F G)-a f F H)) \log \left (\sqrt [4]{c}-\sqrt [4]{a} x\right )}{a^{3/4} \sqrt [4]{c}}-\frac {2 (A c+(B+C) (c (D+F G)-a f F H)) \log \left (\sqrt [4]{c}+\sqrt [4]{a} x\right )}{a^{3/4} \sqrt [4]{c}}+\frac {\sqrt {2} (A d+(B+C) (d (D+F G)+b f F H)) \log \left (\sqrt {d}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} x+\sqrt {b} x^2\right )}{b^{3/4} \sqrt [4]{d}}-\frac {\sqrt {2} (A d+(B+C) (d (D+F G)+b f F H)) \log \left (\sqrt {d}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} x+\sqrt {b} x^2\right )}{b^{3/4} \sqrt [4]{d}}}{8 (B+C) (b c+a d) F H} \] Input:

Integrate[(x^2*(-f + (A/((B + C)*F*H) + D/(F*H) + G/H)*x^4))/((-c + a*x^4) 
*(d + b*x^4)),x]
 

Output:

((4*(A*c + (B + C)*(c*(D + F*G) - a*f*F*H))*ArcTan[(a^(1/4)*x)/c^(1/4)])/( 
a^(3/4)*c^(1/4)) - (2*Sqrt[2]*(A*d + (B + C)*(d*(D + F*G) + b*f*F*H))*ArcT 
an[1 - (Sqrt[2]*b^(1/4)*x)/d^(1/4)])/(b^(3/4)*d^(1/4)) + (2*Sqrt[2]*(A*d + 
 (B + C)*(d*(D + F*G) + b*f*F*H))*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/d^(1/4)]) 
/(b^(3/4)*d^(1/4)) + (2*(A*c + (B + C)*(c*(D + F*G) - a*f*F*H))*Log[c^(1/4 
) - a^(1/4)*x])/(a^(3/4)*c^(1/4)) - (2*(A*c + (B + C)*(c*(D + F*G) - a*f*F 
*H))*Log[c^(1/4) + a^(1/4)*x])/(a^(3/4)*c^(1/4)) + (Sqrt[2]*(A*d + (B + C) 
*(d*(D + F*G) + b*f*F*H))*Log[Sqrt[d] - Sqrt[2]*b^(1/4)*d^(1/4)*x + Sqrt[b 
]*x^2])/(b^(3/4)*d^(1/4)) - (Sqrt[2]*(A*d + (B + C)*(d*(D + F*G) + b*f*F*H 
))*Log[Sqrt[d] + Sqrt[2]*b^(1/4)*d^(1/4)*x + Sqrt[b]*x^2])/(b^(3/4)*d^(1/4 
)))/(8*(B + C)*(b*c + a*d)*F*H)
 

Rubi [A] (verified)

Time = 1.38 (sec) , antiderivative size = 487, normalized size of antiderivative = 0.97, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {1054, 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.

\(\displaystyle \int \frac {x^2 \left (x^4 \left (\frac {A}{F H (B+C)}+\frac {D}{F H}+\frac {G}{H}\right )-f\right )}{\left (a x^4-c\right ) \left (b x^4+d\right )} \, dx\)

\(\Big \downarrow \) 1054

\(\displaystyle \int \left (\frac {x^2 (-(B+C) (c (D+F G)-a f F H)-A c)}{F H (B+C) \left (c-a x^4\right ) (a d+b c)}+\frac {x^2 (A d+(B+C) (b f F H+d (D+F G)))}{F H (B+C) \left (b x^4+d\right ) (a d+b c)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{c}}\right ) ((B+C) (c (D+F G)-a f F H)+A c)}{2 a^{3/4} \sqrt [4]{c} F H (B+C) (a d+b c)}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{c}}\right ) ((B+C) (c (D+F G)-a f F H)+A c)}{2 a^{3/4} \sqrt [4]{c} F H (B+C) (a d+b c)}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{d}}\right ) (A d+(B+C) (b f F H+d (D+F G)))}{2 \sqrt {2} b^{3/4} \sqrt [4]{d} F H (B+C) (a d+b c)}+\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{d}}+1\right ) (A d+(B+C) (b f F H+d (D+F G)))}{2 \sqrt {2} b^{3/4} \sqrt [4]{d} F H (B+C) (a d+b c)}+\frac {\log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} x+\sqrt {b} x^2+\sqrt {d}\right ) (A d+(B+C) (b f F H+d (D+F G)))}{4 \sqrt {2} b^{3/4} \sqrt [4]{d} F H (B+C) (a d+b c)}-\frac {\log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{d} x+\sqrt {b} x^2+\sqrt {d}\right ) (A d+(B+C) (b f F H+d (D+F G)))}{4 \sqrt {2} b^{3/4} \sqrt [4]{d} F H (B+C) (a d+b c)}\)

Input:

Int[(x^2*(-f + (A/((B + C)*F*H) + D/(F*H) + G/H)*x^4))/((-c + a*x^4)*(d + 
b*x^4)),x]
 

Output:

((A*c + (B + C)*(c*(D + F*G) - a*f*F*H))*ArcTan[(a^(1/4)*x)/c^(1/4)])/(2*a 
^(3/4)*c^(1/4)*(B + C)*(b*c + a*d)*F*H) - ((A*d + (B + C)*(d*(D + F*G) + b 
*f*F*H))*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/d^(1/4)])/(2*Sqrt[2]*b^(3/4)*(B + 
C)*d^(1/4)*(b*c + a*d)*F*H) + ((A*d + (B + C)*(d*(D + F*G) + b*f*F*H))*Arc 
Tan[1 + (Sqrt[2]*b^(1/4)*x)/d^(1/4)])/(2*Sqrt[2]*b^(3/4)*(B + C)*d^(1/4)*( 
b*c + a*d)*F*H) - ((A*c + (B + C)*(c*(D + F*G) - a*f*F*H))*ArcTanh[(a^(1/4 
)*x)/c^(1/4)])/(2*a^(3/4)*c^(1/4)*(B + C)*(b*c + a*d)*F*H) + ((A*d + (B + 
C)*(d*(D + F*G) + b*f*F*H))*Log[Sqrt[d] - Sqrt[2]*b^(1/4)*d^(1/4)*x + Sqrt 
[b]*x^2])/(4*Sqrt[2]*b^(3/4)*(B + C)*d^(1/4)*(b*c + a*d)*F*H) - ((A*d + (B 
 + C)*(d*(D + F*G) + b*f*F*H))*Log[Sqrt[d] + Sqrt[2]*b^(1/4)*d^(1/4)*x + S 
qrt[b]*x^2])/(4*Sqrt[2]*b^(3/4)*(B + C)*d^(1/4)*(b*c + a*d)*F*H)
 

Defintions of rubi rules used

rule 1054

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n 
_)))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a 
+ b*x^n)^p*((e + f*x^n)/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, g, 
 m, p}, x] && IGtQ[n, 0]
 

rule 2009

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

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 254, normalized size of antiderivative = 0.50

method	result	size
default	\(\frac {\frac {\left (-B F H a f -C F H a f +B F G c +C F G c +B D c +C D c +A c \right ) \left (2 \arctan \left (\frac {x}{\left (\frac {c}{a}\right )^{\frac {1}{4}}}\right )-\ln \left (\frac {x +\left (\frac {c}{a}\right )^{\frac {1}{4}}}{x -\left (\frac {c}{a}\right )^{\frac {1}{4}}}\right )\right )}{4 \left (a d +b c \right ) a \left (\frac {c}{a}\right )^{\frac {1}{4}}}+\frac {\left (B F H b f +C F H b f +B F G d +C F G d +B D d +C D d +A d \right ) \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {d}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {d}{b}}}{x^{2}+\left (\frac {d}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {d}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {d}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {d}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{8 \left (a d +b c \right ) b \left (\frac {d}{b}\right )^{\frac {1}{4}}}}{\left (B +C \right ) F H}\)	\(254\)

Input:

int(x^2*(-f+(A/(B+C)/F/H+D/F/H+G/H)*x^4)/(a*x^4-c)/(b*x^4+d),x,method=_RET 
URNVERBOSE)
 

Output:

1/(B+C)/F/H*(1/4*(-B*F*H*a*f-C*F*H*a*f+B*F*G*c+C*F*G*c+B*D*c+C*D*c+A*c)/(a 
*d+b*c)/a/(c/a)^(1/4)*(2*arctan(x/(c/a)^(1/4))-ln((x+(c/a)^(1/4))/(x-(c/a) 
^(1/4))))+1/8*(B*F*H*b*f+C*F*H*b*f+B*F*G*d+C*F*G*d+B*D*d+C*D*d+A*d)/(a*d+b 
*c)/b/(d/b)^(1/4)*2^(1/2)*(ln((x^2-(d/b)^(1/4)*x*2^(1/2)+(d/b)^(1/2))/(x^2 
+(d/b)^(1/4)*x*2^(1/2)+(d/b)^(1/2)))+2*arctan(2^(1/2)/(d/b)^(1/4)*x+1)+2*a 
rctan(2^(1/2)/(d/b)^(1/4)*x-1)))
 

Fricas [F(-1)]

Timed out. \[ \int \frac {x^2 \left (-f+\left (\frac {A}{(B+C) F H}+\frac {D}{F H}+\frac {G}{H}\right ) x^4\right )}{\left (-c+a x^4\right ) \left (d+b x^4\right )} \, dx=\text {Timed out} \] Input:

integrate(x^2*(-f+(A/(B+C)/F/H+D/F/H+G/H)*x^4)/(a*x^4-c)/(b*x^4+d),x, algo 
rithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^2 \left (-f+\left (\frac {A}{(B+C) F H}+\frac {D}{F H}+\frac {G}{H}\right ) x^4\right )}{\left (-c+a x^4\right ) \left (d+b x^4\right )} \, dx=\text {Timed out} \] Input:

integrate(x**2*(-f+(A/(B+C)/F/H+D/F/H+G/H)*x**4)/(a*x**4-c)/(b*x**4+d),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 339, normalized size of antiderivative = 0.67 \[ \int \frac {x^2 \left (-f+\left (\frac {A}{(B+C) F H}+\frac {D}{F H}+\frac {G}{H}\right ) x^4\right )}{\left (-c+a x^4\right ) \left (d+b x^4\right )} \, dx=\frac {{\left ({\left (B + C\right )} F H b f + {\left ({\left (B + C\right )} F G + {\left (B + C\right )} D + A\right )} d\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x + \sqrt {2} b^{\frac {1}{4}} d^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {d}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {d}}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x - \sqrt {2} b^{\frac {1}{4}} d^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {d}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {d}}} - \frac {\sqrt {2} \log \left (\sqrt {b} x^{2} + \sqrt {2} b^{\frac {1}{4}} d^{\frac {1}{4}} x + \sqrt {d}\right )}{b^{\frac {3}{4}} d^{\frac {1}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {b} x^{2} - \sqrt {2} b^{\frac {1}{4}} d^{\frac {1}{4}} x + \sqrt {d}\right )}{b^{\frac {3}{4}} d^{\frac {1}{4}}}\right )}}{8 \, {\left ({\left (B + C\right )} F H b c + {\left (B + C\right )} F H a d\right )}} - \frac {{\left ({\left (B + C\right )} F H a f - {\left ({\left (B + C\right )} F G + {\left (B + C\right )} D + A\right )} c\right )} {\left (\frac {2 \, \arctan \left (\frac {\sqrt {a} x}{\sqrt {\sqrt {a} \sqrt {c}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {c}}} + \frac {\log \left (\frac {\sqrt {a} x - \sqrt {\sqrt {a} \sqrt {c}}}{\sqrt {a} x + \sqrt {\sqrt {a} \sqrt {c}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {c}}}\right )}}{4 \, {\left ({\left (B + C\right )} F H b c + {\left (B + C\right )} F H a d\right )}} \] Input:

integrate(x^2*(-f+(A/(B+C)/F/H+D/F/H+G/H)*x^4)/(a*x^4-c)/(b*x^4+d),x, algo 
rithm="maxima")
 

Output:

1/8*((B + C)*F*H*b*f + ((B + C)*F*G + (B + C)*D + A)*d)*(2*sqrt(2)*arctan( 
1/2*sqrt(2)*(2*sqrt(b)*x + sqrt(2)*b^(1/4)*d^(1/4))/sqrt(sqrt(b)*sqrt(d))) 
/(sqrt(b)*sqrt(sqrt(b)*sqrt(d))) + 2*sqrt(2)*arctan(1/2*sqrt(2)*(2*sqrt(b) 
*x - sqrt(2)*b^(1/4)*d^(1/4))/sqrt(sqrt(b)*sqrt(d)))/(sqrt(b)*sqrt(sqrt(b) 
*sqrt(d))) - sqrt(2)*log(sqrt(b)*x^2 + sqrt(2)*b^(1/4)*d^(1/4)*x + sqrt(d) 
)/(b^(3/4)*d^(1/4)) + sqrt(2)*log(sqrt(b)*x^2 - sqrt(2)*b^(1/4)*d^(1/4)*x 
+ sqrt(d))/(b^(3/4)*d^(1/4)))/((B + C)*F*H*b*c + (B + C)*F*H*a*d) - 1/4*(( 
B + C)*F*H*a*f - ((B + C)*F*G + (B + C)*D + A)*c)*(2*arctan(sqrt(a)*x/sqrt 
(sqrt(a)*sqrt(c)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(c))) + log((sqrt(a)*x - sqrt 
(sqrt(a)*sqrt(c)))/(sqrt(a)*x + sqrt(sqrt(a)*sqrt(c))))/(sqrt(a)*sqrt(sqrt 
(a)*sqrt(c))))/((B + C)*F*H*b*c + (B + C)*F*H*a*d)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1229 vs. \(2 (417) = 834\).

Time = 0.18 (sec) , antiderivative size = 1229, normalized size of antiderivative = 2.44 \[ \int \frac {x^2 \left (-f+\left (\frac {A}{(B+C) F H}+\frac {D}{F H}+\frac {G}{H}\right ) x^4\right )}{\left (-c+a x^4\right ) \left (d+b x^4\right )} \, dx=\text {Too large to display} \] Input:

integrate(x^2*(-f+(A/(B+C)/F/H+D/F/H+G/H)*x^4)/(a*x^4-c)/(b*x^4+d),x, algo 
rithm="giac")
 

Output:

-1/2*(B*F*H*a*f + C*F*H*a*f - B*F*G*c - C*F*G*c - B*D*c - C*D*c - A*c)*arc 
tan(1/2*sqrt(2)*(2*x + sqrt(2)*(-c/a)^(1/4))/(-c/a)^(1/4))/(sqrt(2)*(-a^3* 
c)^(1/4)*B*F*H*b*c + sqrt(2)*(-a^3*c)^(1/4)*C*F*H*b*c + sqrt(2)*(-a^3*c)^( 
1/4)*B*F*H*a*d + sqrt(2)*(-a^3*c)^(1/4)*C*F*H*a*d) - 1/2*(B*F*H*a*f + C*F* 
H*a*f - B*F*G*c - C*F*G*c - B*D*c - C*D*c - A*c)*arctan(1/2*sqrt(2)*(2*x - 
 sqrt(2)*(-c/a)^(1/4))/(-c/a)^(1/4))/(sqrt(2)*(-a^3*c)^(1/4)*B*F*H*b*c + s 
qrt(2)*(-a^3*c)^(1/4)*C*F*H*b*c + sqrt(2)*(-a^3*c)^(1/4)*B*F*H*a*d + sqrt( 
2)*(-a^3*c)^(1/4)*C*F*H*a*d) + 1/2*((b^3*d)^(3/4)*B*F*H*b*f + (b^3*d)^(3/4 
)*C*F*H*b*f + (b^3*d)^(3/4)*B*F*G*d + (b^3*d)^(3/4)*C*F*G*d + (b^3*d)^(3/4 
)*B*D*d + (b^3*d)^(3/4)*C*D*d + (b^3*d)^(3/4)*A*d)*arctan(1/2*sqrt(2)*(2*x 
 + sqrt(2)*(d/b)^(1/4))/(d/b)^(1/4))/(sqrt(2)*B*F*H*b^4*c*d + sqrt(2)*C*F* 
H*b^4*c*d + sqrt(2)*B*F*H*a*b^3*d^2 + sqrt(2)*C*F*H*a*b^3*d^2) + 1/2*((b^3 
*d)^(3/4)*B*F*H*b*f + (b^3*d)^(3/4)*C*F*H*b*f + (b^3*d)^(3/4)*B*F*G*d + (b 
^3*d)^(3/4)*C*F*G*d + (b^3*d)^(3/4)*B*D*d + (b^3*d)^(3/4)*C*D*d + (b^3*d)^ 
(3/4)*A*d)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*(d/b)^(1/4))/(d/b)^(1/4))/(sq 
rt(2)*B*F*H*b^4*c*d + sqrt(2)*C*F*H*b^4*c*d + sqrt(2)*B*F*H*a*b^3*d^2 + sq 
rt(2)*C*F*H*a*b^3*d^2) + 1/4*(B*F*H*a*f + C*F*H*a*f - B*F*G*c - C*F*G*c - 
B*D*c - C*D*c - A*c)*log(x^2 + sqrt(2)*x*(-c/a)^(1/4) + sqrt(-c/a))/(sqrt( 
2)*(-a^3*c)^(1/4)*B*F*H*b*c + sqrt(2)*(-a^3*c)^(1/4)*C*F*H*b*c + sqrt(2)*( 
-a^3*c)^(1/4)*B*F*H*a*d + sqrt(2)*(-a^3*c)^(1/4)*C*F*H*a*d) - 1/4*(B*F*...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 \left (-f+\left (\frac {A}{(B+C) F H}+\frac {D}{F H}+\frac {G}{H}\right ) x^4\right )}{\left (-c+a x^4\right ) \left (d+b x^4\right )} \, dx=\int \frac {x^2\,\left (f-x^4\,\left (\frac {G}{H}+\frac {D}{F\,H}+\frac {A}{F\,H\,\left (B+C\right )}\right )\right )}{\left (c-a\,x^4\right )\,\left (b\,x^4+d\right )} \,d x \] Input:

int((x^2*(f - x^4*(G/H + D/(F*H) + A/(F*H*(B + C)))))/((c - a*x^4)*(d + b* 
x^4)),x)
 

Output:

int((x^2*(f - x^4*(G/H + D/(F*H) + A/(F*H*(B + C)))))/((c - a*x^4)*(d + b* 
x^4)), x)
 

Reduce [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 1623, normalized size of antiderivative = 3.23 \[ \int \frac {x^2 \left (-f+\left (\frac {A}{(B+C) F H}+\frac {D}{F H}+\frac {G}{H}\right ) x^4\right )}{\left (-c+a x^4\right ) \left (d+b x^4\right )} \, dx =\text {Too large to display} \] Input:

int(x^2*(-f+(A/(B+C)/F/H+D/F/H+G/H)*x^4)/(a*x^4-c)/(b*x^4+d),x)
 

Output:

( - 2*d**(3/4)*b**(1/4)*sqrt(2)*atan((d**(1/4)*b**(1/4)*sqrt(2) - 2*sqrt(b 
)*x)/(d**(1/4)*b**(1/4)*sqrt(2)))*a**2*c*d - 2*d**(3/4)*b**(1/4)*sqrt(2)*a 
tan((d**(1/4)*b**(1/4)*sqrt(2) - 2*sqrt(b)*x)/(d**(1/4)*b**(1/4)*sqrt(2))) 
*a*b**2*c*f**2*h - 2*d**(3/4)*b**(1/4)*sqrt(2)*atan((d**(1/4)*b**(1/4)*sqr 
t(2) - 2*sqrt(b)*x)/(d**(1/4)*b**(1/4)*sqrt(2)))*a*b*c**2*f**2*h - 2*d**(3 
/4)*b**(1/4)*sqrt(2)*atan((d**(1/4)*b**(1/4)*sqrt(2) - 2*sqrt(b)*x)/(d**(1 
/4)*b**(1/4)*sqrt(2)))*a*b*c*d**2 - 2*d**(3/4)*b**(1/4)*sqrt(2)*atan((d**( 
1/4)*b**(1/4)*sqrt(2) - 2*sqrt(b)*x)/(d**(1/4)*b**(1/4)*sqrt(2)))*a*b*c*d* 
f*g - 2*d**(3/4)*b**(1/4)*sqrt(2)*atan((d**(1/4)*b**(1/4)*sqrt(2) - 2*sqrt 
(b)*x)/(d**(1/4)*b**(1/4)*sqrt(2)))*a*c**2*d**2 - 2*d**(3/4)*b**(1/4)*sqrt 
(2)*atan((d**(1/4)*b**(1/4)*sqrt(2) - 2*sqrt(b)*x)/(d**(1/4)*b**(1/4)*sqrt 
(2)))*a*c**2*d*f*g + 2*d**(3/4)*b**(1/4)*sqrt(2)*atan((d**(1/4)*b**(1/4)*s 
qrt(2) + 2*sqrt(b)*x)/(d**(1/4)*b**(1/4)*sqrt(2)))*a**2*c*d + 2*d**(3/4)*b 
**(1/4)*sqrt(2)*atan((d**(1/4)*b**(1/4)*sqrt(2) + 2*sqrt(b)*x)/(d**(1/4)*b 
**(1/4)*sqrt(2)))*a*b**2*c*f**2*h + 2*d**(3/4)*b**(1/4)*sqrt(2)*atan((d**( 
1/4)*b**(1/4)*sqrt(2) + 2*sqrt(b)*x)/(d**(1/4)*b**(1/4)*sqrt(2)))*a*b*c**2 
*f**2*h + 2*d**(3/4)*b**(1/4)*sqrt(2)*atan((d**(1/4)*b**(1/4)*sqrt(2) + 2* 
sqrt(b)*x)/(d**(1/4)*b**(1/4)*sqrt(2)))*a*b*c*d**2 + 2*d**(3/4)*b**(1/4)*s 
qrt(2)*atan((d**(1/4)*b**(1/4)*sqrt(2) + 2*sqrt(b)*x)/(d**(1/4)*b**(1/4)*s 
qrt(2)))*a*b*c*d*f*g + 2*d**(3/4)*b**(1/4)*sqrt(2)*atan((d**(1/4)*b**(1...