\(\int \frac {x^6 (A+B x^4)}{(a+b x^4) (c+d x^4)} \, dx\) [12]

Optimal result

Integrand size = 29, antiderivative size = 389 \[ \int \frac {x^6 \left (A+B x^4\right )}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {B x^3}{3 b d}+\frac {a^{3/4} (A b-a B) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} b^{7/4} (b c-a d)}-\frac {a^{3/4} (A b-a B) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} b^{7/4} (b c-a d)}+\frac {c^{3/4} (B c-A d) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{2 \sqrt {2} d^{7/4} (b c-a d)}-\frac {c^{3/4} (B c-A d) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )}{2 \sqrt {2} d^{7/4} (b c-a d)}+\frac {a^{3/4} (A b-a B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x}{\sqrt {a}+\sqrt {b} x^2}\right )}{2 \sqrt {2} b^{7/4} (b c-a d)}+\frac {c^{3/4} (B c-A d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x}{\sqrt {c}+\sqrt {d} x^2}\right )}{2 \sqrt {2} d^{7/4} (b c-a d)} \] Output:

1/3*B*x^3/b/d-1/4*a^(3/4)*(A*b-B*a)*arctan(-1+2^(1/2)*b^(1/4)*x/a^(1/4))*2 
^(1/2)/b^(7/4)/(-a*d+b*c)-1/4*a^(3/4)*(A*b-B*a)*arctan(1+2^(1/2)*b^(1/4)*x 
/a^(1/4))*2^(1/2)/b^(7/4)/(-a*d+b*c)-1/4*c^(3/4)*(-A*d+B*c)*arctan(-1+2^(1 
/2)*d^(1/4)*x/c^(1/4))*2^(1/2)/d^(7/4)/(-a*d+b*c)-1/4*c^(3/4)*(-A*d+B*c)*a 
rctan(1+2^(1/2)*d^(1/4)*x/c^(1/4))*2^(1/2)/d^(7/4)/(-a*d+b*c)+1/4*a^(3/4)* 
(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) 
/b^(7/4)/(-a*d+b*c)+1/4*c^(3/4)*(-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)/d^(7/4)/(-a*d+b*c)
 

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 465, normalized size of antiderivative = 1.20 \[ \int \frac {x^6 \left (A+B x^4\right )}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {8 b^{3/4} B d^{3/4} (b c-a d) x^3-6 \sqrt {2} a^{3/4} (-A b+a B) d^{7/4} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )+6 \sqrt {2} a^{3/4} (-A b+a B) d^{7/4} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )+6 \sqrt {2} b^{7/4} c^{3/4} (B c-A d) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )-6 \sqrt {2} b^{7/4} c^{3/4} (B c-A d) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{d} x}{\sqrt [4]{c}}\right )+3 \sqrt {2} a^{3/4} (-A b+a B) d^{7/4} \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )+3 \sqrt {2} a^{3/4} (A b-a B) d^{7/4} \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )-3 \sqrt {2} b^{7/4} c^{3/4} (B c-A d) \log \left (\sqrt {c}-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {d} x^2\right )+3 \sqrt {2} b^{7/4} c^{3/4} (B c-A d) \log \left (\sqrt {c}+\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} x+\sqrt {d} x^2\right )}{24 b^{7/4} d^{7/4} (b c-a d)} \] Input:

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

Output:

(8*b^(3/4)*B*d^(3/4)*(b*c - a*d)*x^3 - 6*Sqrt[2]*a^(3/4)*(-(A*b) + a*B)*d^ 
(7/4)*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)] + 6*Sqrt[2]*a^(3/4)*(-(A*b) 
+ a*B)*d^(7/4)*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)] + 6*Sqrt[2]*b^(7/4) 
*c^(3/4)*(B*c - A*d)*ArcTan[1 - (Sqrt[2]*d^(1/4)*x)/c^(1/4)] - 6*Sqrt[2]*b 
^(7/4)*c^(3/4)*(B*c - A*d)*ArcTan[1 + (Sqrt[2]*d^(1/4)*x)/c^(1/4)] + 3*Sqr 
t[2]*a^(3/4)*(-(A*b) + a*B)*d^(7/4)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)* 
x + Sqrt[b]*x^2] + 3*Sqrt[2]*a^(3/4)*(A*b - a*B)*d^(7/4)*Log[Sqrt[a] + Sqr 
t[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2] - 3*Sqrt[2]*b^(7/4)*c^(3/4)*(B*c - A 
*d)*Log[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*x + Sqrt[d]*x^2] + 3*Sqrt[2]*b^( 
7/4)*c^(3/4)*(B*c - A*d)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*x + Sqrt[d] 
*x^2])/(24*b^(7/4)*d^(7/4)*(b*c - a*d))
 

Rubi [A] (verified)

Time = 1.19 (sec) , antiderivative size = 544, normalized size of antiderivative = 1.40, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1052, 27, 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.

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

\(\Big \downarrow \) 1052

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1054

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(B*x^3)/(3*b*d) - (-1/2*(a^(3/4)*(A*b - a*B)*d*ArcTan[1 - (Sqrt[2]*b^(1/4) 
*x)/a^(1/4)])/(Sqrt[2]*b^(3/4)*(b*c - a*d)) + (a^(3/4)*(A*b - a*B)*d*ArcTa 
n[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/(2*Sqrt[2]*b^(3/4)*(b*c - a*d)) - (b*c 
^(3/4)*(B*c - A*d)*ArcTan[1 - (Sqrt[2]*d^(1/4)*x)/c^(1/4)])/(2*Sqrt[2]*d^( 
3/4)*(b*c - a*d)) + (b*c^(3/4)*(B*c - A*d)*ArcTan[1 + (Sqrt[2]*d^(1/4)*x)/ 
c^(1/4)])/(2*Sqrt[2]*d^(3/4)*(b*c - a*d)) + (a^(3/4)*(A*b - a*B)*d*Log[Sqr 
t[a] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(4*Sqrt[2]*b^(3/4)*(b*c - 
 a*d)) - (a^(3/4)*(A*b - a*B)*d*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*x + 
Sqrt[b]*x^2])/(4*Sqrt[2]*b^(3/4)*(b*c - a*d)) + (b*c^(3/4)*(B*c - A*d)*Log 
[Sqrt[c] - Sqrt[2]*c^(1/4)*d^(1/4)*x + Sqrt[d]*x^2])/(4*Sqrt[2]*d^(3/4)*(b 
*c - a*d)) - (b*c^(3/4)*(B*c - A*d)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)* 
x + Sqrt[d]*x^2])/(4*Sqrt[2]*d^(3/4)*(b*c - a*d)))/(b*d)
 

Defintions of rubi rules used

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 1052

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[f*g^(n - 1)*(g*x)^(m 
- n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*d*(m + n*(p + q + 1) + 
 1))), x] - Simp[g^n/(b*d*(m + n*(p + q + 1) + 1))   Int[(g*x)^(m - n)*(a + 
 b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m - n + 1) + (a*f*d*(m + n*q + 1) + b*( 
f*c*(m + n*p + 1) - e*d*(m + n*(p + q + 1) + 1)))*x^n, x], x], x] /; FreeQ[ 
{a, b, c, d, e, f, g, p, q}, x] && IGtQ[n, 0] && GtQ[m, n - 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 = 0.19 (sec) , antiderivative size = 254, normalized size of antiderivative = 0.65

method	result	size
default	\(\frac {B \,x^{3}}{3 b d}+\frac {a \left (A b -B a \right ) \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}{x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{8 \left (a d -c b \right ) b^{2} \left (\frac {a}{b}\right )^{\frac {1}{4}}}-\frac {c \left (A d -B c \right ) \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {c}{d}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {c}{d}}}{x^{2}+\left (\frac {c}{d}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {c}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )\right )}{8 d^{2} \left (a d -c b \right ) \left (\frac {c}{d}\right )^{\frac {1}{4}}}\)	\(254\)
risch	\(\text {Expression too large to display}\)	\(3768\)

Input:

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

Output:

1/3*B*x^3/b/d+1/8*a*(A*b-B*a)/(a*d-b*c)/b^2/(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))-1/8* 
c*(A*d-B*c)/d^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. \[ \int \frac {x^6 \left (A+B x^4\right )}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 401, normalized size of antiderivative = 1.03 \[ \int \frac {x^6 \left (A+B x^4\right )}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {B x^{3}}{3 \, b d} + \frac {{\left (B a^{2} - A a b\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {b} x^{2} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {b} x^{2} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{8 \, {\left (b^{2} c - a b d\right )}} - \frac {{\left (B c^{2} - A c d\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {d} x + \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {\sqrt {c} \sqrt {d}} \sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {d} x - \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {c} \sqrt {d}}}\right )}{\sqrt {\sqrt {c} \sqrt {d}} \sqrt {d}} - \frac {\sqrt {2} \log \left (\sqrt {d} x^{2} + \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {d} x^{2} - \sqrt {2} c^{\frac {1}{4}} d^{\frac {1}{4}} x + \sqrt {c}\right )}{c^{\frac {1}{4}} d^{\frac {3}{4}}}\right )}}{8 \, {\left (b c d - a d^{2}\right )}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 585, normalized size of antiderivative = 1.50 \[ \int \frac {x^6 \left (A+B x^4\right )}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx =\text {Too large to display} \] Input:

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

Output:

1/3*B*x^3/(b*d) + 1/2*((a*b^3)^(3/4)*B*a - (a*b^3)^(3/4)*A*b)*arctan(1/2*s 
qrt(2)*(2*x + sqrt(2)*(a/b)^(1/4))/(a/b)^(1/4))/(sqrt(2)*b^5*c - sqrt(2)*a 
*b^4*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)*b^5*c - sqrt(2)*a*b^4*d) 
- 1/2*((c*d^3)^(3/4)*B*c - (c*d^3)^(3/4)*A*d)*arctan(1/2*sqrt(2)*(2*x + sq 
rt(2)*(c/d)^(1/4))/(c/d)^(1/4))/(sqrt(2)*b*c*d^4 - sqrt(2)*a*d^5) - 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*d^4 - sqrt(2)*a*d^5) - 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)*b^5*c - sqrt(2)*a*b^4*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)*b^5*c - sqrt( 
2)*a*b^4*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*d^4 - sqrt(2)*a*d^5) - 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*d^4 - sqrt(2)*a*d^5)
 

Mupad [B] (verification not implemented)

Time = 11.58 (sec) , antiderivative size = 29549, normalized size of antiderivative = 75.96 \[ \int \frac {x^6 \left (A+B x^4\right )}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

(B*x^3)/(3*b*d) - 2*atan(((((256*A^3*a^3*b^10*c^7*d^6 - 1024*A^3*a^4*b^9*c 
^6*d^7 + 1536*A^3*a^5*b^8*c^5*d^8 - 1024*A^3*a^6*b^7*c^4*d^9 + 256*A^3*a^7 
*b^6*c^3*d^10 - 256*B^3*a^3*b^10*c^10*d^3 + 768*B^3*a^4*b^9*c^9*d^4 - 768* 
B^3*a^5*b^8*c^8*d^5 + 256*B^3*a^6*b^7*c^7*d^6 + 256*B^3*a^7*b^6*c^6*d^7 - 
768*B^3*a^8*b^5*c^5*d^8 + 768*B^3*a^9*b^4*c^4*d^9 - 256*B^3*a^10*b^3*c^3*d 
^10 + 768*A*B^2*a^3*b^10*c^9*d^4 - 2304*A*B^2*a^4*b^9*c^8*d^5 + 2304*A*B^2 
*a^5*b^8*c^7*d^6 - 1536*A*B^2*a^6*b^7*c^6*d^7 + 2304*A*B^2*a^7*b^6*c^5*d^8 
 - 2304*A*B^2*a^8*b^5*c^4*d^9 + 768*A*B^2*a^9*b^4*c^3*d^10 - 768*A^2*B*a^3 
*b^10*c^8*d^5 + 2304*A^2*B*a^4*b^9*c^7*d^6 - 1536*A^2*B*a^5*b^8*c^6*d^7 - 
1536*A^2*B*a^6*b^7*c^5*d^8 + 2304*A^2*B*a^7*b^6*c^4*d^9 - 768*A^2*B*a^8*b^ 
5*c^3*d^10)/(b^3*d^3) - (x*(-(B^4*a^7 + A^4*a^3*b^4 + 6*A^2*B^2*a^5*b^2 - 
4*A*B^3*a^6*b - 4*A^3*B*a^4*b^3)/(256*b^11*c^4 + 256*a^4*b^7*d^4 - 1024*a^ 
3*b^8*c*d^3 + 1536*a^2*b^9*c^2*d^2 - 1024*a*b^10*c^3*d))^(1/4)*(512*A^2*a^ 
3*b^11*c^7*d^7 - 2048*A^2*a^4*b^10*c^6*d^8 + 3072*A^2*a^5*b^9*c^5*d^9 - 20 
48*A^2*a^6*b^8*c^4*d^10 + 512*A^2*a^7*b^7*c^3*d^11 + 256*B^2*a^3*b^11*c^9* 
d^5 - 1024*B^2*a^4*b^10*c^8*d^6 + 1792*B^2*a^5*b^9*c^7*d^7 - 2048*B^2*a^6* 
b^8*c^6*d^8 + 1792*B^2*a^7*b^7*c^5*d^9 - 1024*B^2*a^8*b^6*c^4*d^10 + 256*B 
^2*a^9*b^5*c^3*d^11 - 512*A*B*a^3*b^11*c^8*d^6 + 1536*A*B*a^4*b^10*c^7*d^7 
 - 1024*A*B*a^5*b^9*c^6*d^8 - 1024*A*B*a^6*b^8*c^5*d^9 + 1536*A*B*a^7*b^7* 
c^4*d^10 - 512*A*B*a^8*b^6*c^3*d^11)*4i)/(b^3*d^3))*(-(B^4*a^7 + A^4*a^...
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.38 \[ \int \frac {x^6 \left (A+B x^4\right )}{\left (a+b x^4\right ) \left (c+d x^4\right )} \, dx=\frac {6 d^{\frac {1}{4}} c^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}-2 \sqrt {d}\, x}{d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}}\right )-6 d^{\frac {1}{4}} c^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}+2 \sqrt {d}\, x}{d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}}\right )-3 d^{\frac {1}{4}} c^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (-d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {c}+\sqrt {d}\, x^{2}\right )+3 d^{\frac {1}{4}} c^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (d^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {c}+\sqrt {d}\, x^{2}\right )+8 d \,x^{3}}{24 d^{2}} \] Input:

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

Output:

(6*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))) - 6*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))) - 3*d**(1/4 
)*c**(3/4)*sqrt(2)*log( - d**(1/4)*c**(1/4)*sqrt(2)*x + sqrt(c) + sqrt(d)* 
x**2) + 3*d**(1/4)*c**(3/4)*sqrt(2)*log(d**(1/4)*c**(1/4)*sqrt(2)*x + sqrt 
(c) + sqrt(d)*x**2) + 8*d*x**3)/(24*d**2)