\(\int \frac {A+B x^4}{x^6 (a+b x^4) (c+d x^4)} \, dx\) [18]

Optimal result

Integrand size = 29, antiderivative size = 413 \[ \int \frac {A+B x^4}{x^6 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=-\frac {A}{5 a c x^5}+\frac {A b c-a B c+a A d}{a^2 c^2 x}-\frac {b^{5/4} (A b-a B) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{9/4} (b c-a d)}+\frac {b^{5/4} (A b-a B) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{9/4} (b c-a d)}-\frac {d^{5/4} (B c-A d) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{2 \sqrt {2} c^{9/4} (b c-a d)}+\frac {d^{5/4} (B c-A d) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{2 \sqrt {2} c^{9/4} (b c-a d)}-\frac {b^{5/4} (A b-a B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x}{\sqrt {a}+\sqrt {b} x^2}\right )}{2 \sqrt {2} a^{9/4} (b c-a d)}-\frac {d^{5/4} (B c-A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x}{\sqrt {c}+\sqrt {d} x^2}\right )}{2 \sqrt {2} c^{9/4} (b c-a d)} \] Output:

-1/5*A/a/c/x^5+(A*a*d+A*b*c-B*a*c)/a^2/c^2/x+1/4*b^(5/4)*(A*b-B*a)*arctan( 
-1+2^(1/2)*b^(1/4)*x/a^(1/4))*2^(1/2)/a^(9/4)/(-a*d+b*c)+1/4*b^(5/4)*(A*b- 
B*a)*arctan(1+2^(1/2)*b^(1/4)*x/a^(1/4))*2^(1/2)/a^(9/4)/(-a*d+b*c)+1/4*d^ 
(5/4)*(-A*d+B*c)*arctan(-1+2^(1/2)*d^(1/4)*x/c^(1/4))*2^(1/2)/c^(9/4)/(-a* 
d+b*c)+1/4*d^(5/4)*(-A*d+B*c)*arctan(1+2^(1/2)*d^(1/4)*x/c^(1/4))*2^(1/2)/ 
c^(9/4)/(-a*d+b*c)-1/4*b^(5/4)*(A*b-B*a)*arctanh(2^(1/2)*a^(1/4)*b^(1/4)*x 
/(a^(1/2)+b^(1/2)*x^2))*2^(1/2)/a^(9/4)/(-a*d+b*c)-1/4*d^(5/4)*(-A*d+B*c)* 
arctanh(2^(1/2)*c^(1/4)*d^(1/4)*x/(c^(1/2)+d^(1/2)*x^2))*2^(1/2)/c^(9/4)/( 
-a*d+b*c)
 

Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 526, normalized size of antiderivative = 1.27 \[ \int \frac {A+B x^4}{x^6 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {8 a^{5/4} A c^{5/4} (-b c+a d)+40 \sqrt [4]{a} \sqrt [4]{c} (b c-a d) (A b c-a B c+a A d) x^4-10 \sqrt {2} b^{5/4} (A b-a B) c^{9/4} x^5 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )+10 \sqrt {2} b^{5/4} (A b-a B) c^{9/4} x^5 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )+10 \sqrt {2} a^{9/4} d^{5/4} (-B c+A d) x^5 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )+10 \sqrt {2} a^{9/4} d^{5/4} (B c-A d) x^5 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )+5 \sqrt {2} b^{5/4} (A b-a B) c^{9/4} x^5 \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )-5 \sqrt {2} b^{5/4} (A b-a B) c^{9/4} x^5 \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )+5 \sqrt {2} a^{9/4} d^{5/4} (B c-A d) x^5 \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {d} x^2\right )+5 \sqrt {2} a^{9/4} d^{5/4} (-B c+A d) x^5 \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {d} x^2\right )}{40 a^{9/4} c^{9/4} (b c-a d) x^5} \] Input:

Integrate[(A + B*x^4)/(x^6*(a + b*x^4)*(c + d*x^4)),x]
 

Output:

(8*a^(5/4)*A*c^(5/4)*(-(b*c) + a*d) + 40*a^(1/4)*c^(1/4)*(b*c - a*d)*(A*b* 
c - a*B*c + a*A*d)*x^4 - 10*Sqrt[2]*b^(5/4)*(A*b - a*B)*c^(9/4)*x^5*ArcTan 
[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)] + 10*Sqrt[2]*b^(5/4)*(A*b - a*B)*c^(9/4) 
*x^5*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)] + 10*Sqrt[2]*a^(9/4)*d^(5/4)* 
(-(B*c) + A*d)*x^5*ArcTan[1 - (Sqrt[2]*d^(1/4)*x)/c^(1/4)] + 10*Sqrt[2]*a^ 
(9/4)*d^(5/4)*(B*c - A*d)*x^5*ArcTan[1 + (Sqrt[2]*d^(1/4)*x)/c^(1/4)] + 5* 
Sqrt[2]*b^(5/4)*(A*b - a*B)*c^(9/4)*x^5*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1 
/4)*x + Sqrt[b]*x^2] - 5*Sqrt[2]*b^(5/4)*(A*b - a*B)*c^(9/4)*x^5*Log[Sqrt[ 
a] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2] + 5*Sqrt[2]*a^(9/4)*d^(5/4)* 
(B*c - A*d)*x^5*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*x + Sqrt[d]*x^2] + 5 
*Sqrt[2]*a^(9/4)*d^(5/4)*(-(B*c) + A*d)*x^5*Log[Sqrt[c] + Sqrt[2]*c^(1/4)* 
d^(1/4)*x + Sqrt[d]*x^2])/(40*a^(9/4)*c^(9/4)*(b*c - a*d)*x^5)
 

Rubi [A] (verified)

Time = 1.39 (sec) , antiderivative size = 593, normalized size of antiderivative = 1.44, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {1053, 27, 1053, 25, 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.

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

\(\Big \downarrow \) 1053

\(\displaystyle -\frac {\int \frac {5 \left (A b d x^4+A b c-a B c+a A d\right )}{x^2 \left (b x^4+a\right ) \left (d x^4+c\right )}dx}{5 a c}-\frac {A}{5 a c x^5}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {A b d x^4+A b c-a B c+a A d}{x^2 \left (b x^4+a\right ) \left (d x^4+c\right )}dx}{a c}-\frac {A}{5 a c x^5}\)

\(\Big \downarrow \) 1053

\(\displaystyle -\frac {-\frac {\int -\frac {x^2 \left (-b d (A b c-a B c+a A d) x^4+a B c (b c+a d)-A \left (b^2 c^2+a b d c+a^2 d^2\right )\right )}{\left (b x^4+a\right ) \left (d x^4+c\right )}dx}{a c}-\frac {a A d-a B c+A b c}{a c x}}{a c}-\frac {A}{5 a c x^5}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\frac {\int \frac {x^2 \left (-b d (A b c-a B c+a A d) x^4+a B c (b c+a d)-A \left (b^2 c^2+a b d c+a^2 d^2\right )\right )}{\left (b x^4+a\right ) \left (d x^4+c\right )}dx}{a c}-\frac {a A d-a B c+A b c}{a c x}}{a c}-\frac {A}{5 a c x^5}\)

\(\Big \downarrow \) 1054

\(\displaystyle -\frac {\frac {\int \left (-\frac {b^2 (A b-a B) c^2 x^2}{(b c-a d) \left (b x^4+a\right )}-\frac {a^2 d^2 (A d-B c) x^2}{(a d-b c) \left (d x^4+c\right )}\right )dx}{a c}-\frac {a A d-a B c+A b c}{a c x}}{a c}-\frac {A}{5 a c x^5}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\frac {\frac {a^2 d^{5/4} (B c-A d) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{2 \sqrt {2} \sqrt [4]{c} (b c-a d)}-\frac {a^2 d^{5/4} (B c-A d) \arctan \left (\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}+1\right )}{2 \sqrt {2} \sqrt [4]{c} (b c-a d)}-\frac {a^2 d^{5/4} (B c-A d) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {c}+\sqrt {d} x^2\right )}{4 \sqrt {2} \sqrt [4]{c} (b c-a d)}+\frac {a^2 d^{5/4} (B c-A d) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {c}+\sqrt {d} x^2\right )}{4 \sqrt {2} \sqrt [4]{c} (b c-a d)}+\frac {b^{5/4} c^2 (A b-a B) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} \sqrt [4]{a} (b c-a d)}-\frac {b^{5/4} c^2 (A b-a B) \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{2 \sqrt {2} \sqrt [4]{a} (b c-a d)}-\frac {b^{5/4} c^2 (A b-a B) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{4 \sqrt {2} \sqrt [4]{a} (b c-a d)}+\frac {b^{5/4} c^2 (A b-a B) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{4 \sqrt {2} \sqrt [4]{a} (b c-a d)}}{a c}-\frac {a A d-a B c+A b c}{a c x}}{a c}-\frac {A}{5 a c x^5}\)

Input:

Int[(A + B*x^4)/(x^6*(a + b*x^4)*(c + d*x^4)),x]
 

Output:

-1/5*A/(a*c*x^5) - (-((A*b*c - a*B*c + a*A*d)/(a*c*x)) + ((b^(5/4)*(A*b - 
a*B)*c^2*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]*a^(1/4)*(b*c 
- a*d)) - (b^(5/4)*(A*b - a*B)*c^2*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)] 
)/(2*Sqrt[2]*a^(1/4)*(b*c - a*d)) + (a^2*d^(5/4)*(B*c - A*d)*ArcTan[1 - (S 
qrt[2]*d^(1/4)*x)/c^(1/4)])/(2*Sqrt[2]*c^(1/4)*(b*c - a*d)) - (a^2*d^(5/4) 
*(B*c - A*d)*ArcTan[1 + (Sqrt[2]*d^(1/4)*x)/c^(1/4)])/(2*Sqrt[2]*c^(1/4)*( 
b*c - a*d)) - (b^(5/4)*(A*b - a*B)*c^2*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/ 
4)*x + Sqrt[b]*x^2])/(4*Sqrt[2]*a^(1/4)*(b*c - a*d)) + (b^(5/4)*(A*b - a*B 
)*c^2*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(4*Sqrt[2]*a 
^(1/4)*(b*c - a*d)) - (a^2*d^(5/4)*(B*c - A*d)*Log[Sqrt[c] - Sqrt[2]*c^(1/ 
4)*d^(1/4)*x + Sqrt[d]*x^2])/(4*Sqrt[2]*c^(1/4)*(b*c - a*d)) + (a^2*d^(5/4 
)*(B*c - A*d)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*x + Sqrt[d]*x^2])/(4*S 
qrt[2]*c^(1/4)*(b*c - a*d)))/(a*c))/(a*c)
 

Defintions of rubi rules used

rule 25

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 1053

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_ 
))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b 
*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*c*g*(m + 1))), x] + Simp[1/(a*c*g^n*( 
m + 1))   Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) 
- e*(b*c + a*d)*(m + n + 1) - e*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2 
) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && IGtQ[n, 
0] && LtQ[m, -1]
 

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 = 0.25 (sec) , antiderivative size = 280, normalized size of antiderivative = 0.68

method	result	size
default	\(-\frac {b \left (A b -B a \right ) \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}{x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{8 a^{2} \left (a d -c b \right ) \left (\frac {a}{b}\right )^{\frac {1}{4}}}-\frac {A}{5 a c \,x^{5}}-\frac {-A a d -A b c +a B c}{a^{2} c^{2} x}+\frac {d \left (A d -B c \right ) \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {c}{d}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {c}{d}}}{x^{2}+\left (\frac {c}{d}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {c}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )\right )}{8 c^{2} \left (a d -c b \right ) \left (\frac {c}{d}\right )^{\frac {1}{4}}}\)	\(280\)
risch	\(\text {Expression too large to display}\)	\(4772\)

Input:

int((B*x^4+A)/x^6/(b*x^4+a)/(d*x^4+c),x,method=_RETURNVERBOSE)
 

Output:

-1/8*b*(A*b-B*a)/a^2/(a*d-b*c)/(a/b)^(1/4)*2^(1/2)*(ln((x^2-(a/b)^(1/4)*x* 
2^(1/2)+(a/b)^(1/2))/(x^2+(a/b)^(1/4)*x*2^(1/2)+(a/b)^(1/2)))+2*arctan(2^( 
1/2)/(a/b)^(1/4)*x+1)+2*arctan(2^(1/2)/(a/b)^(1/4)*x-1))-1/5*A/a/c/x^5-1/a 
^2/c^2*(-A*a*d-A*b*c+B*a*c)/x+1/8*d*(A*d-B*c)/c^2/(a*d-b*c)/(c/d)^(1/4)*2^ 
(1/2)*(ln((x^2-(c/d)^(1/4)*x*2^(1/2)+(c/d)^(1/2))/(x^2+(c/d)^(1/4)*x*2^(1/ 
2)+(c/d)^(1/2)))+2*arctan(2^(1/2)/(c/d)^(1/4)*x+1)+2*arctan(2^(1/2)/(c/d)^ 
(1/4)*x-1))
 

Fricas [F(-1)]

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

integrate((B*x^4+A)/x^6/(b*x^4+a)/(d*x^4+c),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

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

integrate((B*x**4+A)/x**6/(b*x**4+a)/(d*x**4+c),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 435, normalized size of antiderivative = 1.05 \[ \int \frac {A+B x^4}{x^6 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=-\frac {{\left (B a b^{2} - A b^{3}\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {b} x^{2} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {b} x^{2} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{8 \, {\left (a^{2} b c - a^{3} d\right )}} + \frac {{\left (B c d^{2} - A d^{3}\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {d} x + \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {\sqrt {c} \sqrt {d}} \sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {d} x - \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {\sqrt {c} \sqrt {d}} \sqrt {d}} - \frac {\sqrt {2} \log \left (\sqrt {d} x^{2} + \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {d} x^{2} - \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}}\right )}}{8 \, {\left (b c^{3} - a c^{2} d\right )}} + \frac {5 \, {\left (A a d - {\left (B a - A b\right )} c\right )} x^{4} - A a c}{5 \, a^{2} c^{2} x^{5}} \] Input:

integrate((B*x^4+A)/x^6/(b*x^4+a)/(d*x^4+c),x, algorithm="maxima")
 

Output:

-1/8*(B*a*b^2 - A*b^3)*(2*sqrt(2)*arctan(1/2*sqrt(2)*(2*sqrt(b)*x + sqrt(2 
)*a^(1/4)*b^(1/4))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(sqrt(a)*sqrt(b))*sqrt(b)) 
+ 2*sqrt(2)*arctan(1/2*sqrt(2)*(2*sqrt(b)*x - sqrt(2)*a^(1/4)*b^(1/4))/sqr 
t(sqrt(a)*sqrt(b)))/(sqrt(sqrt(a)*sqrt(b))*sqrt(b)) - sqrt(2)*log(sqrt(b)* 
x^2 + sqrt(2)*a^(1/4)*b^(1/4)*x + sqrt(a))/(a^(1/4)*b^(3/4)) + sqrt(2)*log 
(sqrt(b)*x^2 - sqrt(2)*a^(1/4)*b^(1/4)*x + sqrt(a))/(a^(1/4)*b^(3/4)))/(a^ 
2*b*c - a^3*d) + 1/8*(B*c*d^2 - A*d^3)*(2*sqrt(2)*arctan(1/2*sqrt(2)*(2*sq 
rt(d)*x + sqrt(2)*c^(1/4)*d^(1/4))/sqrt(sqrt(c)*sqrt(d)))/(sqrt(sqrt(c)*sq 
rt(d))*sqrt(d)) + 2*sqrt(2)*arctan(1/2*sqrt(2)*(2*sqrt(d)*x - sqrt(2)*c^(1 
/4)*d^(1/4))/sqrt(sqrt(c)*sqrt(d)))/(sqrt(sqrt(c)*sqrt(d))*sqrt(d)) - sqrt 
(2)*log(sqrt(d)*x^2 + sqrt(2)*c^(1/4)*d^(1/4)*x + sqrt(c))/(c^(1/4)*d^(3/4 
)) + sqrt(2)*log(sqrt(d)*x^2 - sqrt(2)*c^(1/4)*d^(1/4)*x + sqrt(c))/(c^(1/ 
4)*d^(3/4)))/(b*c^3 - a*c^2*d) + 1/5*(5*(A*a*d - (B*a - A*b)*c)*x^4 - A*a* 
c)/(a^2*c^2*x^5)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 637, normalized size of antiderivative = 1.54 \[ \int \frac {A+B x^4}{x^6 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx =\text {Too large to display} \] Input:

integrate((B*x^4+A)/x^6/(b*x^4+a)/(d*x^4+c),x, algorithm="giac")
 

Output:

-1/2*((a*b^3)^(3/4)*B*a - (a*b^3)^(3/4)*A*b)*arctan(1/2*sqrt(2)*(2*x + sqr 
t(2)*(a/b)^(1/4))/(a/b)^(1/4))/(sqrt(2)*a^3*b^2*c - sqrt(2)*a^4*b*d) - 1/2 
*((a*b^3)^(3/4)*B*a - (a*b^3)^(3/4)*A*b)*arctan(1/2*sqrt(2)*(2*x - sqrt(2) 
*(a/b)^(1/4))/(a/b)^(1/4))/(sqrt(2)*a^3*b^2*c - sqrt(2)*a^4*b*d) + 1/2*((c 
*d^3)^(3/4)*B*c - (c*d^3)^(3/4)*A*d)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)*(c/ 
d)^(1/4))/(c/d)^(1/4))/(sqrt(2)*b*c^4*d - sqrt(2)*a*c^3*d^2) + 1/2*((c*d^3 
)^(3/4)*B*c - (c*d^3)^(3/4)*A*d)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*(c/d)^( 
1/4))/(c/d)^(1/4))/(sqrt(2)*b*c^4*d - sqrt(2)*a*c^3*d^2) + 1/4*((a*b^3)^(3 
/4)*B*a - (a*b^3)^(3/4)*A*b)*log(x^2 + sqrt(2)*x*(a/b)^(1/4) + sqrt(a/b))/ 
(sqrt(2)*a^3*b^2*c - sqrt(2)*a^4*b*d) - 1/4*((a*b^3)^(3/4)*B*a - (a*b^3)^( 
3/4)*A*b)*log(x^2 - sqrt(2)*x*(a/b)^(1/4) + sqrt(a/b))/(sqrt(2)*a^3*b^2*c 
- sqrt(2)*a^4*b*d) - 1/4*((c*d^3)^(3/4)*B*c - (c*d^3)^(3/4)*A*d)*log(x^2 + 
 sqrt(2)*x*(c/d)^(1/4) + sqrt(c/d))/(sqrt(2)*b*c^4*d - sqrt(2)*a*c^3*d^2) 
+ 1/4*((c*d^3)^(3/4)*B*c - (c*d^3)^(3/4)*A*d)*log(x^2 - sqrt(2)*x*(c/d)^(1 
/4) + sqrt(c/d))/(sqrt(2)*b*c^4*d - sqrt(2)*a*c^3*d^2) - 1/5*(5*B*a*c*x^4 
- 5*A*b*c*x^4 - 5*A*a*d*x^4 + A*a*c)/(a^2*c^2*x^5)
 

Mupad [B] (verification not implemented)

Time = 12.99 (sec) , antiderivative size = 34477, normalized size of antiderivative = 83.48 \[ \int \frac {A+B x^4}{x^6 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\text {Too large to display} \] Input:

int((A + B*x^4)/(x^6*(a + b*x^4)*(c + d*x^4)),x)
 

Output:

- (A/(5*a*c) - (x^4*(A*a*d + A*b*c - B*a*c))/(a^2*c^2))/x^5 - 2*atan((1024 
*A^2*a^11*b^10*c^13*x*(-(A^4*b^9 + B^4*a^4*b^5 + 6*A^2*B^2*a^2*b^7 - 4*A^3 
*B*a*b^8 - 4*A*B^3*a^3*b^6)/(256*a^13*d^4 + 256*a^9*b^4*c^4 - 1024*a^10*b^ 
3*c^3*d + 1536*a^11*b^2*c^2*d^2 - 1024*a^12*b*c*d^3))^(5/4) + 4*A^6*a^11*b 
^6*d^9*x*(-(A^4*b^9 + B^4*a^4*b^5 + 6*A^2*B^2*a^2*b^7 - 4*A^3*B*a*b^8 - 4* 
A*B^3*a^3*b^6)/(256*a^13*d^4 + 256*a^9*b^4*c^4 - 1024*a^10*b^3*c^3*d + 153 
6*a^11*b^2*c^2*d^2 - 1024*a^12*b*c*d^3))^(1/4) + 1024*B^2*a^13*b^8*c^13*x* 
(-(A^4*b^9 + B^4*a^4*b^5 + 6*A^2*B^2*a^2*b^7 - 4*A^3*B*a*b^8 - 4*A*B^3*a^3 
*b^6)/(256*a^13*d^4 + 256*a^9*b^4*c^4 - 1024*a^10*b^3*c^3*d + 1536*a^11*b^ 
2*c^2*d^2 - 1024*a^12*b*c*d^3))^(5/4) + 1024*A^2*a^21*c^3*d^10*x*(-(A^4*b^ 
9 + B^4*a^4*b^5 + 6*A^2*B^2*a^2*b^7 - 4*A^3*B*a*b^8 - 4*A*B^3*a^3*b^6)/(25 
6*a^13*d^4 + 256*a^9*b^4*c^4 - 1024*a^10*b^3*c^3*d + 1536*a^11*b^2*c^2*d^2 
 - 1024*a^12*b*c*d^3))^(5/4) + 1024*B^2*a^21*c^5*d^8*x*(-(A^4*b^9 + B^4*a^ 
4*b^5 + 6*A^2*B^2*a^2*b^7 - 4*A^3*B*a*b^8 - 4*A*B^3*a^3*b^6)/(256*a^13*d^4 
 + 256*a^9*b^4*c^4 - 1024*a^10*b^3*c^3*d + 1536*a^11*b^2*c^2*d^2 - 1024*a^ 
12*b*c*d^3))^(5/4) - 8*A^5*B*a^12*b^5*d^9*x*(-(A^4*b^9 + B^4*a^4*b^5 + 6*A 
^2*B^2*a^2*b^7 - 4*A^3*B*a*b^8 - 4*A*B^3*a^3*b^6)/(256*a^13*d^4 + 256*a^9* 
b^4*c^4 - 1024*a^10*b^3*c^3*d + 1536*a^11*b^2*c^2*d^2 - 1024*a^12*b*c*d^3) 
)^(1/4) - 4096*A^2*a^12*b^9*c^12*d*x*(-(A^4*b^9 + B^4*a^4*b^5 + 6*A^2*B^2* 
a^2*b^7 - 4*A^3*B*a*b^8 - 4*A*B^3*a^3*b^6)/(256*a^13*d^4 + 256*a^9*b^4*...
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.41 \[ \int \frac {A+B x^4}{x^6 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {-10 d^{\frac {5}{4}} c^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}-2 \sqrt {d}\, x}{d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}}\right ) x^{5}+10 d^{\frac {5}{4}} c^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}+2 \sqrt {d}\, x}{d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}}\right ) x^{5}+5 d^{\frac {5}{4}} c^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (-d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {c}+\sqrt {d}\, x^{2}\right ) x^{5}-5 d^{\frac {5}{4}} c^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {c}+\sqrt {d}\, x^{2}\right ) x^{5}-8 c^{2}+40 c d \,x^{4}}{40 c^{3} x^{5}} \] Input:

int((B*x^4+A)/x^6/(b*x^4+a)/(d*x^4+c),x)
 

Output:

( - 10*d**(1/4)*c**(3/4)*sqrt(2)*atan((d**(1/4)*c**(1/4)*sqrt(2) - 2*sqrt( 
d)*x)/(d**(1/4)*c**(1/4)*sqrt(2)))*d*x**5 + 10*d**(1/4)*c**(3/4)*sqrt(2)*a 
tan((d**(1/4)*c**(1/4)*sqrt(2) + 2*sqrt(d)*x)/(d**(1/4)*c**(1/4)*sqrt(2))) 
*d*x**5 + 5*d**(1/4)*c**(3/4)*sqrt(2)*log( - d**(1/4)*c**(1/4)*sqrt(2)*x + 
 sqrt(c) + sqrt(d)*x**2)*d*x**5 - 5*d**(1/4)*c**(3/4)*sqrt(2)*log(d**(1/4) 
*c**(1/4)*sqrt(2)*x + sqrt(c) + sqrt(d)*x**2)*d*x**5 - 8*c**2 + 40*c*d*x** 
4)/(40*c**3*x**5)