\(\int \frac {A+B x^4}{x^2 (a+b x^4) (c+d x^4)} \, dx\) [16]

Optimal result

Integrand size = 29, antiderivative size = 387 \[ \int \frac {A+B x^4}{x^2 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=-\frac {A}{a c x}+\frac {\sqrt [4]{b} (A b-a B) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{5/4} (b c-a d)}-\frac {\sqrt [4]{b} (A b-a B) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{5/4} (b c-a d)}+\frac {\sqrt [4]{d} (B c-A d) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{2 \sqrt {2} c^{5/4} (b c-a d)}-\frac {\sqrt [4]{d} (B c-A d) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{2 \sqrt {2} c^{5/4} (b c-a d)}+\frac {\sqrt [4]{b} (A b-a B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x}{\sqrt {a}+\sqrt {b} x^2}\right )}{2 \sqrt {2} a^{5/4} (b c-a d)}+\frac {\sqrt [4]{d} (B c-A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x}{\sqrt {c}+\sqrt {d} x^2}\right )}{2 \sqrt {2} c^{5/4} (b c-a d)} \] Output:

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

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 470, normalized size of antiderivative = 1.21 \[ \int \frac {A+B x^4}{x^2 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {8 \sqrt [4]{a} A \sqrt [4]{c} (-b c+a d)+2 \sqrt {2} \sqrt [4]{b} (A b-a B) c^{5/4} x \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )-2 \sqrt {2} \sqrt [4]{b} (A b-a B) c^{5/4} x \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )-2 \sqrt {2} a^{5/4} \sqrt [4]{d} (-B c+A d) x \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )+2 \sqrt {2} a^{5/4} \sqrt [4]{d} (-B c+A d) x \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )-\sqrt {2} \sqrt [4]{b} (A b-a B) c^{5/4} x \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )+\sqrt {2} \sqrt [4]{b} (A b-a B) c^{5/4} x \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )+\sqrt {2} a^{5/4} \sqrt [4]{d} (-B c+A d) x \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {d} x^2\right )+\sqrt {2} a^{5/4} \sqrt [4]{d} (B c-A d) x \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {d} x^2\right )}{8 a^{5/4} c^{5/4} (b c-a d) x} \] Input:

Integrate[(A + B*x^4)/(x^2*(a + b*x^4)*(c + d*x^4)),x]
 

Output:

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

Rubi [A] (verified)

Time = 1.15 (sec) , antiderivative size = 542, normalized size of antiderivative = 1.40, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {1053, 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.

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

\(\Big \downarrow \) 1053

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

\(\Big \downarrow \) 1054

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

\(\Big \downarrow \) 2009

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

Input:

Int[(A + B*x^4)/(x^2*(a + b*x^4)*(c + d*x^4)),x]
 

Output:

-(A/(a*c*x)) - (-1/2*(b^(1/4)*(A*b - a*B)*c*ArcTan[1 - (Sqrt[2]*b^(1/4)*x) 
/a^(1/4)])/(Sqrt[2]*a^(1/4)*(b*c - a*d)) + (b^(1/4)*(A*b - a*B)*c*ArcTan[1 
 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]*a^(1/4)*(b*c - a*d)) - (a*d^(1 
/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)) + (a*d^(1/4)*(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^(1/4)*(A*b - a*B)*c*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^(1/4)*(A*b - a*B)*c*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqr 
t[b]*x^2])/(4*Sqrt[2]*a^(1/4)*(b*c - a*d)) + (a*d^(1/4)*(B*c - A*d)*Log[Sq 
rt[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*d^(1/4)*(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*c)
 

Defintions of rubi rules used

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 = 0.20 (sec) , antiderivative size = 252, normalized size of antiderivative = 0.65

method	result	size
default	\(\frac {\left (A b -B a \right ) \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}{x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{8 a \left (a d -c b \right ) \left (\frac {a}{b}\right )^{\frac {1}{4}}}-\frac {A}{a c x}-\frac {\left (A d -B c \right ) \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {c}{d}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {c}{d}}}{x^{2}+\left (\frac {c}{d}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {c}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )\right )}{8 c \left (a d -c b \right ) \left (\frac {c}{d}\right )^{\frac {1}{4}}}\)	\(252\)
risch	\(\text {Expression too large to display}\)	\(3956\)

Input:

int((B*x^4+A)/x^2/(b*x^4+a)/(d*x^4+c),x,method=_RETURNVERBOSE)
 

Output:

1/8*(A*b-B*a)/a/(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))-A/a/c/x-1/8*(A*d-B*c)/ 
c/(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 [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 62.51 (sec) , antiderivative size = 2543, normalized size of antiderivative = 6.57 \[ \int \frac {A+B x^4}{x^2 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\text {Too large to display} \] Input:

integrate((B*x^4+A)/x^2/(b*x^4+a)/(d*x^4+c),x, algorithm="fricas")
 

Output:

-1/4*(a*c*x*(-(B^4*a^4*b - 4*A*B^3*a^3*b^2 + 6*A^2*B^2*a^2*b^3 - 4*A^3*B*a 
*b^4 + A^4*b^5)/(a^5*b^4*c^4 - 4*a^6*b^3*c^3*d + 6*a^7*b^2*c^2*d^2 - 4*a^8 
*b*c*d^3 + a^9*d^4))^(1/4)*log(-(B^3*a^3*b - 3*A*B^2*a^2*b^2 + 3*A^2*B*a*b 
^3 - A^3*b^4)*x + (a^4*b^3*c^3 - 3*a^5*b^2*c^2*d + 3*a^6*b*c*d^2 - a^7*d^3 
)*(-(B^4*a^4*b - 4*A*B^3*a^3*b^2 + 6*A^2*B^2*a^2*b^3 - 4*A^3*B*a*b^4 + A^4 
*b^5)/(a^5*b^4*c^4 - 4*a^6*b^3*c^3*d + 6*a^7*b^2*c^2*d^2 - 4*a^8*b*c*d^3 + 
 a^9*d^4))^(3/4)) - a*c*x*(-(B^4*a^4*b - 4*A*B^3*a^3*b^2 + 6*A^2*B^2*a^2*b 
^3 - 4*A^3*B*a*b^4 + A^4*b^5)/(a^5*b^4*c^4 - 4*a^6*b^3*c^3*d + 6*a^7*b^2*c 
^2*d^2 - 4*a^8*b*c*d^3 + a^9*d^4))^(1/4)*log(-(B^3*a^3*b - 3*A*B^2*a^2*b^2 
 + 3*A^2*B*a*b^3 - A^3*b^4)*x - (a^4*b^3*c^3 - 3*a^5*b^2*c^2*d + 3*a^6*b*c 
*d^2 - a^7*d^3)*(-(B^4*a^4*b - 4*A*B^3*a^3*b^2 + 6*A^2*B^2*a^2*b^3 - 4*A^3 
*B*a*b^4 + A^4*b^5)/(a^5*b^4*c^4 - 4*a^6*b^3*c^3*d + 6*a^7*b^2*c^2*d^2 - 4 
*a^8*b*c*d^3 + a^9*d^4))^(3/4)) + I*a*c*x*(-(B^4*a^4*b - 4*A*B^3*a^3*b^2 + 
 6*A^2*B^2*a^2*b^3 - 4*A^3*B*a*b^4 + A^4*b^5)/(a^5*b^4*c^4 - 4*a^6*b^3*c^3 
*d + 6*a^7*b^2*c^2*d^2 - 4*a^8*b*c*d^3 + a^9*d^4))^(1/4)*log(-(B^3*a^3*b - 
 3*A*B^2*a^2*b^2 + 3*A^2*B*a*b^3 - A^3*b^4)*x - (I*a^4*b^3*c^3 - 3*I*a^5*b 
^2*c^2*d + 3*I*a^6*b*c*d^2 - I*a^7*d^3)*(-(B^4*a^4*b - 4*A*B^3*a^3*b^2 + 6 
*A^2*B^2*a^2*b^3 - 4*A^3*B*a*b^4 + A^4*b^5)/(a^5*b^4*c^4 - 4*a^6*b^3*c^3*d 
 + 6*a^7*b^2*c^2*d^2 - 4*a^8*b*c*d^3 + a^9*d^4))^(3/4)) - I*a*c*x*(-(B^4*a 
^4*b - 4*A*B^3*a^3*b^2 + 6*A^2*B^2*a^2*b^3 - 4*A^3*B*a*b^4 + A^4*b^5)/(...
 

Sympy [F(-1)]

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

integrate((B*x**4+A)/x**2/(b*x**4+a)/(d*x**4+c),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 401, normalized size of antiderivative = 1.04 \[ \int \frac {A+B x^4}{x^2 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {{\left (B a b - A b^{2}\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {b} x^{2} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {b} x^{2} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{8 \, {\left (a b c - a^{2} d\right )}} - \frac {{\left (B c d - A d^{2}\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {d} x + \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {\sqrt {c} \sqrt {d}} \sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {d} x - \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {\sqrt {c} \sqrt {d}} \sqrt {d}} - \frac {\sqrt {2} \log \left (\sqrt {d} x^{2} + \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {d} x^{2} - \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}}\right )}}{8 \, {\left (b c^{2} - a c d\right )}} - \frac {A}{a c x} \] Input:

integrate((B*x^4+A)/x^2/(b*x^4+a)/(d*x^4+c),x, algorithm="maxima")
 

Output:

1/8*(B*a*b - A*b^2)*(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))/sqrt(s 
qrt(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(sq 
rt(b)*x^2 - sqrt(2)*a^(1/4)*b^(1/4)*x + sqrt(a))/(a^(1/4)*b^(3/4)))/(a*b*c 
 - a^2*d) - 1/8*(B*c*d - A*d^2)*(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)) + 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)) + sq 
rt(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^2 - a*c*d) - A/(a*c*x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 625 vs. \(2 (293) = 586\).

Time = 0.13 (sec) , antiderivative size = 625, normalized size of antiderivative = 1.61 \[ \int \frac {A+B x^4}{x^2 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx =\text {Too large to display} \] Input:

integrate((B*x^4+A)/x^2/(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 + sqrt 
(2)*(a/b)^(1/4))/(a/b)^(1/4))/(sqrt(2)*a^2*b^3*c - sqrt(2)*a^3*b^2*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^2*b^3*c - sqrt(2)*a^3*b^2*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^3*d^2 - sqrt(2)*a*c^2*d^3) - 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^3*d^2 - sqrt(2)*a*c^2*d^3) - 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^2*b^3*c - sqrt(2)*a^3*b^2*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 
^2*b^3*c - sqrt(2)*a^3*b^2*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^3*d^2 - sqrt(2 
)*a*c^2*d^3) - 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^3*d^2 - sqrt(2)*a*c^2*d^3) - A/ 
(a*c*x)
 

Mupad [B] (verification not implemented)

Time = 11.10 (sec) , antiderivative size = 26209, normalized size of antiderivative = 67.72 \[ \int \frac {A+B x^4}{x^2 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\text {Too large to display} \] Input:

int((A + B*x^4)/(x^2*(a + b*x^4)*(c + d*x^4)),x)
 

Output:

2*atan((1024*A^2*a*b^8*c^14*x*(-(A^4*d^5 + B^4*c^4*d + 6*A^2*B^2*c^2*d^3 - 
 4*A^3*B*c*d^4 - 4*A*B^3*c^3*d^2)/(256*b^4*c^9 + 256*a^4*c^5*d^4 - 1024*a^ 
3*b*c^6*d^3 + 1536*a^2*b^2*c^7*d^2 - 1024*a*b^3*c^8*d))^(5/4) + 1024*B^2*a 
^3*b^6*c^14*x*(-(A^4*d^5 + B^4*c^4*d + 6*A^2*B^2*c^2*d^3 - 4*A^3*B*c*d^4 - 
 4*A*B^3*c^3*d^2)/(256*b^4*c^9 + 256*a^4*c^5*d^4 - 1024*a^3*b*c^6*d^3 + 15 
36*a^2*b^2*c^7*d^2 - 1024*a*b^3*c^8*d))^(5/4) + 1024*A^2*a^9*c^6*d^8*x*(-( 
A^4*d^5 + B^4*c^4*d + 6*A^2*B^2*c^2*d^3 - 4*A^3*B*c*d^4 - 4*A*B^3*c^3*d^2) 
/(256*b^4*c^9 + 256*a^4*c^5*d^4 - 1024*a^3*b*c^6*d^3 + 1536*a^2*b^2*c^7*d^ 
2 - 1024*a*b^3*c^8*d))^(5/4) + 4*A^6*b^5*c^6*d^4*x*(-(A^4*d^5 + B^4*c^4*d 
+ 6*A^2*B^2*c^2*d^3 - 4*A^3*B*c*d^4 - 4*A*B^3*c^3*d^2)/(256*b^4*c^9 + 256* 
a^4*c^5*d^4 - 1024*a^3*b*c^6*d^3 + 1536*a^2*b^2*c^7*d^2 - 1024*a*b^3*c^8*d 
))^(1/4) + 1024*B^2*a^9*c^8*d^6*x*(-(A^4*d^5 + B^4*c^4*d + 6*A^2*B^2*c^2*d 
^3 - 4*A^3*B*c*d^4 - 4*A*B^3*c^3*d^2)/(256*b^4*c^9 + 256*a^4*c^5*d^4 - 102 
4*a^3*b*c^6*d^3 + 1536*a^2*b^2*c^7*d^2 - 1024*a*b^3*c^8*d))^(5/4) + 4*A^4* 
B^2*b^5*c^8*d^2*x*(-(A^4*d^5 + B^4*c^4*d + 6*A^2*B^2*c^2*d^3 - 4*A^3*B*c*d 
^4 - 4*A*B^3*c^3*d^2)/(256*b^4*c^9 + 256*a^4*c^5*d^4 - 1024*a^3*b*c^6*d^3 
+ 1536*a^2*b^2*c^7*d^2 - 1024*a*b^3*c^8*d))^(1/4) + 6144*A^2*a^3*b^6*c^12* 
d^2*x*(-(A^4*d^5 + B^4*c^4*d + 6*A^2*B^2*c^2*d^3 - 4*A^3*B*c*d^4 - 4*A*B^3 
*c^3*d^2)/(256*b^4*c^9 + 256*a^4*c^5*d^4 - 1024*a^3*b*c^6*d^3 + 1536*a^2*b 
^2*c^7*d^2 - 1024*a*b^3*c^8*d))^(5/4) - 4096*A^2*a^4*b^5*c^11*d^3*x*(-(...
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.39 \[ \int \frac {A+B x^4}{x^2 \left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {2 d^{\frac {1}{4}} c^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}-2 \sqrt {d}\, x}{d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}}\right ) x -2 d^{\frac {1}{4}} c^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}+2 \sqrt {d}\, x}{d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}}\right ) x -d^{\frac {1}{4}} c^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (-d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {c}+\sqrt {d}\, x^{2}\right ) x +d^{\frac {1}{4}} c^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {c}+\sqrt {d}\, x^{2}\right ) x -8 c}{8 c^{2} x} \] Input:

int((B*x^4+A)/x^2/(b*x^4+a)/(d*x^4+c),x)
 

Output:

(2*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)))*x - 2*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)))*x - d**(1 
/4)*c**(3/4)*sqrt(2)*log( - d**(1/4)*c**(1/4)*sqrt(2)*x + sqrt(c) + sqrt(d 
)*x**2)*x + d**(1/4)*c**(3/4)*sqrt(2)*log(d**(1/4)*c**(1/4)*sqrt(2)*x + sq 
rt(c) + sqrt(d)*x**2)*x - 8*c)/(8*c**2*x)