\(\int \frac {A+B x^4}{(a+b x^4)^2} \, dx\) [21]

Optimal result

Integrand size = 17, antiderivative size = 186 \[ \int \frac {A+B x^4}{\left (a+b x^4\right )^2} \, dx=\frac {(A b-a B) x}{4 a b \left (a+b x^4\right )}-\frac {(3 A b+a B) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} b^{5/4}}+\frac {(3 A b+a B) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{7/4} b^{5/4}}+\frac {(3 A b+a B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x}{\sqrt {a}+\sqrt {b} x^2}\right )}{8 \sqrt {2} a^{7/4} b^{5/4}} \] Output:

1/4*(A*b-B*a)*x/a/b/(b*x^4+a)+1/16*(3*A*b+B*a)*arctan(-1+2^(1/2)*b^(1/4)*x 
/a^(1/4))*2^(1/2)/a^(7/4)/b^(5/4)+1/16*(3*A*b+B*a)*arctan(1+2^(1/2)*b^(1/4 
)*x/a^(1/4))*2^(1/2)/a^(7/4)/b^(5/4)+1/16*(3*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^(7/4)/b^(5/4)
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.14 \[ \int \frac {A+B x^4}{\left (a+b x^4\right )^2} \, dx=\frac {-\frac {8 a^{3/4} \sqrt [4]{b} (-A b+a B) x}{a+b x^4}-2 \sqrt {2} (3 A b+a B) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )+2 \sqrt {2} (3 A b+a B) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )-\sqrt {2} (3 A b+a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )+\sqrt {2} (3 A b+a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{32 a^{7/4} b^{5/4}} \] Input:

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

Output:

((-8*a^(3/4)*b^(1/4)*(-(A*b) + a*B)*x)/(a + b*x^4) - 2*Sqrt[2]*(3*A*b + a* 
B)*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)] + 2*Sqrt[2]*(3*A*b + a*B)*ArcTa 
n[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)] - Sqrt[2]*(3*A*b + a*B)*Log[Sqrt[a] - S 
qrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2] + Sqrt[2]*(3*A*b + a*B)*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 = 0.71 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.33, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {910, 755, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

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

\(\Big \downarrow \) 910

\(\displaystyle \frac {(a B+3 A b) \int \frac {1}{b x^4+a}dx}{4 a b}+\frac {x (A b-a B)}{4 a b \left (a+b x^4\right )}\)

\(\Big \downarrow \) 755

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

\(\Big \downarrow \) 1476

\(\displaystyle \frac {(a B+3 A b) \left (\frac {\frac {\int \frac {1}{x^2-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}}dx}{2 \sqrt {b}}+\frac {\int \frac {1}{x^2+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}}dx}{2 \sqrt {b}}}{2 \sqrt {a}}+\frac {\int \frac {\sqrt {a}-\sqrt {b} x^2}{b x^4+a}dx}{2 \sqrt {a}}\right )}{4 a b}+\frac {x (A b-a B)}{4 a b \left (a+b x^4\right )}\)

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

((A*b - a*B)*x)/(4*a*b*(a + b*x^4)) + ((3*A*b + a*B)*((-(ArcTan[1 - (Sqrt[ 
2]*b^(1/4)*x)/a^(1/4)]/(Sqrt[2]*a^(1/4)*b^(1/4))) + ArcTan[1 + (Sqrt[2]*b^ 
(1/4)*x)/a^(1/4)]/(Sqrt[2]*a^(1/4)*b^(1/4)))/(2*Sqrt[a]) + (-1/2*Log[Sqrt[ 
a] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2]/(Sqrt[2]*a^(1/4)*b^(1/4)) + 
Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2]/(2*Sqrt[2]*a^(1/4)* 
b^(1/4)))/(2*Sqrt[a])))/(4*a*b)
 

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 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 755

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] 
], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r)   Int[(r - s*x^2)/(a + b*x^4) 
, x], x] + Simp[1/(2*r)   Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{a, 
 b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] & 
& AtomQ[SplitProduct[SumBaseQ, b]]))
 

rule 910

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Si 
mp[(-(b*c - a*d))*x*((a + b*x^n)^(p + 1)/(a*b*n*(p + 1))), x] - Simp[(a*d - 
 b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1))   Int[(a + b*x^n)^(p + 1), x], x] /; 
FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/ 
n + p, 0])
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

Maple [C] (verified)

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

Time = 0.08 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.35

method	result	size
risch	\(\frac {\left (A b -B a \right ) x}{4 a b \left (b \,x^{4}+a \right )}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{4}+a \right )}{\sum }\frac {\left (3 A b +B a \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}}{16 a \,b^{2}}\)	\(65\)
default	\(\frac {\left (A b -B a \right ) x}{4 a b \left (b \,x^{4}+a \right )}+\frac {\left (3 A b +B a \right ) \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}{x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{32 a^{2} b}\)	\(140\)

Input:

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

Output:

1/4*(A*b-B*a)*x/a/b/(b*x^4+a)+1/16/a/b^2*sum((3*A*b+B*a)/_R^3*ln(x-_R),_R= 
RootOf(_Z^4*b+a))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 648, normalized size of antiderivative = 3.48 \[ \int \frac {A+B x^4}{\left (a+b x^4\right )^2} \, dx=\frac {{\left (a b^{2} x^{4} + a^{2} b\right )} \left (-\frac {B^{4} a^{4} + 12 \, A B^{3} a^{3} b + 54 \, A^{2} B^{2} a^{2} b^{2} + 108 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a^{7} b^{5}}\right )^{\frac {1}{4}} \log \left (a^{2} b \left (-\frac {B^{4} a^{4} + 12 \, A B^{3} a^{3} b + 54 \, A^{2} B^{2} a^{2} b^{2} + 108 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a^{7} b^{5}}\right )^{\frac {1}{4}} + {\left (B a + 3 \, A b\right )} x\right ) - {\left (-i \, a b^{2} x^{4} - i \, a^{2} b\right )} \left (-\frac {B^{4} a^{4} + 12 \, A B^{3} a^{3} b + 54 \, A^{2} B^{2} a^{2} b^{2} + 108 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a^{7} b^{5}}\right )^{\frac {1}{4}} \log \left (i \, a^{2} b \left (-\frac {B^{4} a^{4} + 12 \, A B^{3} a^{3} b + 54 \, A^{2} B^{2} a^{2} b^{2} + 108 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a^{7} b^{5}}\right )^{\frac {1}{4}} + {\left (B a + 3 \, A b\right )} x\right ) - {\left (i \, a b^{2} x^{4} + i \, a^{2} b\right )} \left (-\frac {B^{4} a^{4} + 12 \, A B^{3} a^{3} b + 54 \, A^{2} B^{2} a^{2} b^{2} + 108 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a^{7} b^{5}}\right )^{\frac {1}{4}} \log \left (-i \, a^{2} b \left (-\frac {B^{4} a^{4} + 12 \, A B^{3} a^{3} b + 54 \, A^{2} B^{2} a^{2} b^{2} + 108 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a^{7} b^{5}}\right )^{\frac {1}{4}} + {\left (B a + 3 \, A b\right )} x\right ) - {\left (a b^{2} x^{4} + a^{2} b\right )} \left (-\frac {B^{4} a^{4} + 12 \, A B^{3} a^{3} b + 54 \, A^{2} B^{2} a^{2} b^{2} + 108 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a^{7} b^{5}}\right )^{\frac {1}{4}} \log \left (-a^{2} b \left (-\frac {B^{4} a^{4} + 12 \, A B^{3} a^{3} b + 54 \, A^{2} B^{2} a^{2} b^{2} + 108 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a^{7} b^{5}}\right )^{\frac {1}{4}} + {\left (B a + 3 \, A b\right )} x\right ) - 4 \, {\left (B a - A b\right )} x}{16 \, {\left (a b^{2} x^{4} + a^{2} b\right )}} \] Input:

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

Output:

1/16*((a*b^2*x^4 + a^2*b)*(-(B^4*a^4 + 12*A*B^3*a^3*b + 54*A^2*B^2*a^2*b^2 
 + 108*A^3*B*a*b^3 + 81*A^4*b^4)/(a^7*b^5))^(1/4)*log(a^2*b*(-(B^4*a^4 + 1 
2*A*B^3*a^3*b + 54*A^2*B^2*a^2*b^2 + 108*A^3*B*a*b^3 + 81*A^4*b^4)/(a^7*b^ 
5))^(1/4) + (B*a + 3*A*b)*x) - (-I*a*b^2*x^4 - I*a^2*b)*(-(B^4*a^4 + 12*A* 
B^3*a^3*b + 54*A^2*B^2*a^2*b^2 + 108*A^3*B*a*b^3 + 81*A^4*b^4)/(a^7*b^5))^ 
(1/4)*log(I*a^2*b*(-(B^4*a^4 + 12*A*B^3*a^3*b + 54*A^2*B^2*a^2*b^2 + 108*A 
^3*B*a*b^3 + 81*A^4*b^4)/(a^7*b^5))^(1/4) + (B*a + 3*A*b)*x) - (I*a*b^2*x^ 
4 + I*a^2*b)*(-(B^4*a^4 + 12*A*B^3*a^3*b + 54*A^2*B^2*a^2*b^2 + 108*A^3*B* 
a*b^3 + 81*A^4*b^4)/(a^7*b^5))^(1/4)*log(-I*a^2*b*(-(B^4*a^4 + 12*A*B^3*a^ 
3*b + 54*A^2*B^2*a^2*b^2 + 108*A^3*B*a*b^3 + 81*A^4*b^4)/(a^7*b^5))^(1/4) 
+ (B*a + 3*A*b)*x) - (a*b^2*x^4 + a^2*b)*(-(B^4*a^4 + 12*A*B^3*a^3*b + 54* 
A^2*B^2*a^2*b^2 + 108*A^3*B*a*b^3 + 81*A^4*b^4)/(a^7*b^5))^(1/4)*log(-a^2* 
b*(-(B^4*a^4 + 12*A*B^3*a^3*b + 54*A^2*B^2*a^2*b^2 + 108*A^3*B*a*b^3 + 81* 
A^4*b^4)/(a^7*b^5))^(1/4) + (B*a + 3*A*b)*x) - 4*(B*a - A*b)*x)/(a*b^2*x^4 
 + a^2*b)
 

Sympy [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.60 \[ \int \frac {A+B x^4}{\left (a+b x^4\right )^2} \, dx=\frac {x \left (A b - B a\right )}{4 a^{2} b + 4 a b^{2} x^{4}} + \operatorname {RootSum} {\left (65536 t^{4} a^{7} b^{5} + 81 A^{4} b^{4} + 108 A^{3} B a b^{3} + 54 A^{2} B^{2} a^{2} b^{2} + 12 A B^{3} a^{3} b + B^{4} a^{4}, \left ( t \mapsto t \log {\left (\frac {16 t a^{2} b}{3 A b + B a} + x \right )} \right )\right )} \] Input:

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

Output:

x*(A*b - B*a)/(4*a**2*b + 4*a*b**2*x**4) + RootSum(65536*_t**4*a**7*b**5 + 
 81*A**4*b**4 + 108*A**3*B*a*b**3 + 54*A**2*B**2*a**2*b**2 + 12*A*B**3*a** 
3*b + B**4*a**4, Lambda(_t, _t*log(16*_t*a**2*b/(3*A*b + B*a) + x)))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.27 \[ \int \frac {A+B x^4}{\left (a+b x^4\right )^2} \, dx=-\frac {{\left (B a - A b\right )} x}{4 \, {\left (a b^{2} x^{4} + a^{2} b\right )}} + \frac {\frac {2 \, \sqrt {2} {\left (B a + 3 \, A b\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {2 \, \sqrt {2} {\left (B a + 3 \, A b\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {\sqrt {2} {\left (B a + 3 \, A b\right )} \log \left (\sqrt {b} x^{2} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (B a + 3 \, A b\right )} \log \left (\sqrt {b} x^{2} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}}}{32 \, a b} \] Input:

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

Output:

-1/4*(B*a - A*b)*x/(a*b^2*x^4 + a^2*b) + 1/32*(2*sqrt(2)*(B*a + 3*A*b)*arc 
tan(1/2*sqrt(2)*(2*sqrt(b)*x + sqrt(2)*a^(1/4)*b^(1/4))/sqrt(sqrt(a)*sqrt( 
b)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(b))) + 2*sqrt(2)*(B*a + 3*A*b)*arctan(1/2* 
sqrt(2)*(2*sqrt(b)*x - sqrt(2)*a^(1/4)*b^(1/4))/sqrt(sqrt(a)*sqrt(b)))/(sq 
rt(a)*sqrt(sqrt(a)*sqrt(b))) + sqrt(2)*(B*a + 3*A*b)*log(sqrt(b)*x^2 + sqr 
t(2)*a^(1/4)*b^(1/4)*x + sqrt(a))/(a^(3/4)*b^(1/4)) - sqrt(2)*(B*a + 3*A*b 
)*log(sqrt(b)*x^2 - sqrt(2)*a^(1/4)*b^(1/4)*x + sqrt(a))/(a^(3/4)*b^(1/4)) 
)/(a*b)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.43 \[ \int \frac {A+B x^4}{\left (a+b x^4\right )^2} \, dx=\frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {1}{4}} B a + 3 \, \left (a b^{3}\right )^{\frac {1}{4}} A b\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{16 \, a^{2} b^{2}} + \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {1}{4}} B a + 3 \, \left (a b^{3}\right )^{\frac {1}{4}} A b\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{16 \, a^{2} b^{2}} + \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {1}{4}} B a + 3 \, \left (a b^{3}\right )^{\frac {1}{4}} A b\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{32 \, a^{2} b^{2}} - \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {1}{4}} B a + 3 \, \left (a b^{3}\right )^{\frac {1}{4}} A b\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{32 \, a^{2} b^{2}} - \frac {B a x - A b x}{4 \, {\left (b x^{4} + a\right )} a b} \] Input:

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

Output:

1/16*sqrt(2)*((a*b^3)^(1/4)*B*a + 3*(a*b^3)^(1/4)*A*b)*arctan(1/2*sqrt(2)* 
(2*x + sqrt(2)*(a/b)^(1/4))/(a/b)^(1/4))/(a^2*b^2) + 1/16*sqrt(2)*((a*b^3) 
^(1/4)*B*a + 3*(a*b^3)^(1/4)*A*b)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*(a/b)^ 
(1/4))/(a/b)^(1/4))/(a^2*b^2) + 1/32*sqrt(2)*((a*b^3)^(1/4)*B*a + 3*(a*b^3 
)^(1/4)*A*b)*log(x^2 + sqrt(2)*x*(a/b)^(1/4) + sqrt(a/b))/(a^2*b^2) - 1/32 
*sqrt(2)*((a*b^3)^(1/4)*B*a + 3*(a*b^3)^(1/4)*A*b)*log(x^2 - sqrt(2)*x*(a/ 
b)^(1/4) + sqrt(a/b))/(a^2*b^2) - 1/4*(B*a*x - A*b*x)/((b*x^4 + a)*a*b)
 

Mupad [B] (verification not implemented)

Time = 4.06 (sec) , antiderivative size = 740, normalized size of antiderivative = 3.98 \[ \int \frac {A+B x^4}{\left (a+b x^4\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(atan((((3*A*b + B*a)*((x*(9*A^2*b^3 + B^2*a^2*b + 6*A*B*a*b^2))/(4*a^2) - 
 ((3*A*b + B*a)*(12*A*b^3 + 4*B*a*b^2))/(16*(-a)^(7/4)*b^(5/4)))*1i)/(16*( 
-a)^(7/4)*b^(5/4)) + ((3*A*b + B*a)*((x*(9*A^2*b^3 + B^2*a^2*b + 6*A*B*a*b 
^2))/(4*a^2) + ((3*A*b + B*a)*(12*A*b^3 + 4*B*a*b^2))/(16*(-a)^(7/4)*b^(5/ 
4)))*1i)/(16*(-a)^(7/4)*b^(5/4)))/(((3*A*b + B*a)*((x*(9*A^2*b^3 + B^2*a^2 
*b + 6*A*B*a*b^2))/(4*a^2) - ((3*A*b + B*a)*(12*A*b^3 + 4*B*a*b^2))/(16*(- 
a)^(7/4)*b^(5/4))))/(16*(-a)^(7/4)*b^(5/4)) - ((3*A*b + B*a)*((x*(9*A^2*b^ 
3 + B^2*a^2*b + 6*A*B*a*b^2))/(4*a^2) + ((3*A*b + B*a)*(12*A*b^3 + 4*B*a*b 
^2))/(16*(-a)^(7/4)*b^(5/4))))/(16*(-a)^(7/4)*b^(5/4))))*(3*A*b + B*a)*1i) 
/(8*(-a)^(7/4)*b^(5/4)) + (atan((((3*A*b + B*a)*((x*(9*A^2*b^3 + B^2*a^2*b 
 + 6*A*B*a*b^2))/(4*a^2) - ((3*A*b + B*a)*(12*A*b^3 + 4*B*a*b^2)*1i)/(16*( 
-a)^(7/4)*b^(5/4))))/(16*(-a)^(7/4)*b^(5/4)) + ((3*A*b + B*a)*((x*(9*A^2*b 
^3 + B^2*a^2*b + 6*A*B*a*b^2))/(4*a^2) + ((3*A*b + B*a)*(12*A*b^3 + 4*B*a* 
b^2)*1i)/(16*(-a)^(7/4)*b^(5/4))))/(16*(-a)^(7/4)*b^(5/4)))/(((3*A*b + B*a 
)*((x*(9*A^2*b^3 + B^2*a^2*b + 6*A*B*a*b^2))/(4*a^2) - ((3*A*b + B*a)*(12* 
A*b^3 + 4*B*a*b^2)*1i)/(16*(-a)^(7/4)*b^(5/4)))*1i)/(16*(-a)^(7/4)*b^(5/4) 
) - ((3*A*b + B*a)*((x*(9*A^2*b^3 + B^2*a^2*b + 6*A*B*a*b^2))/(4*a^2) + (( 
3*A*b + B*a)*(12*A*b^3 + 4*B*a*b^2)*1i)/(16*(-a)^(7/4)*b^(5/4)))*1i)/(16*( 
-a)^(7/4)*b^(5/4))))*(3*A*b + B*a))/(8*(-a)^(7/4)*b^(5/4)) + (x*(A*b - B*a 
))/(4*a*b*(a + b*x^4))
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.60 \[ \int \frac {A+B x^4}{\left (a+b x^4\right )^2} \, dx=\frac {\sqrt {2}\, \left (-2 \mathit {atan} \left (\frac {b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}-2 \sqrt {b}\, x}{b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right )+2 \mathit {atan} \left (\frac {b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+2 \sqrt {b}\, x}{b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right )-\mathrm {log}\left (-b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {a}+\sqrt {b}\, x^{2}\right )+\mathrm {log}\left (b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {a}+\sqrt {b}\, x^{2}\right )\right )}{8 b^{\frac {1}{4}} a^{\frac {3}{4}}} \] Input:

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

Output:

(b**(3/4)*a**(1/4)*sqrt(2)*( - 2*atan((b**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt( 
b)*x)/(b**(1/4)*a**(1/4)*sqrt(2))) + 2*atan((b**(1/4)*a**(1/4)*sqrt(2) + 2 
*sqrt(b)*x)/(b**(1/4)*a**(1/4)*sqrt(2))) - log( - b**(1/4)*a**(1/4)*sqrt(2 
)*x + sqrt(a) + sqrt(b)*x**2) + log(b**(1/4)*a**(1/4)*sqrt(2)*x + sqrt(a) 
+ sqrt(b)*x**2)))/(8*a*b)