\(\int \frac {A+B x^4}{x^2 (a+b x^4)^2} \, dx\) [22]

Optimal result

Integrand size = 20, antiderivative size = 197 \[ \int \frac {A+B x^4}{x^2 \left (a+b x^4\right )^2} \, dx=-\frac {A}{a^2 x}-\frac {(A b-a B) x^3}{4 a^2 \left (a+b x^4\right )}+\frac {(5 A b-a B) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{9/4} b^{3/4}}-\frac {(5 A b-a B) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{9/4} b^{3/4}}+\frac {(5 A b-a B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x}{\sqrt {a}+\sqrt {b} x^2}\right )}{8 \sqrt {2} a^{9/4} b^{3/4}} \] Output:

-A/a^2/x-1/4*(A*b-B*a)*x^3/a^2/(b*x^4+a)-1/16*(5*A*b-B*a)*arctan(-1+2^(1/2 
)*b^(1/4)*x/a^(1/4))*2^(1/2)/a^(9/4)/b^(3/4)-1/16*(5*A*b-B*a)*arctan(1+2^( 
1/2)*b^(1/4)*x/a^(1/4))*2^(1/2)/a^(9/4)/b^(3/4)+1/16*(5*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)/b^(3/4)
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.20 \[ \int \frac {A+B x^4}{x^2 \left (a+b x^4\right )^2} \, dx=\frac {-\frac {32 \sqrt [4]{a} A}{x}+\frac {8 \sqrt [4]{a} (-A b+a B) x^3}{a+b x^4}+\frac {2 \sqrt {2} (5 A b-a B) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{b^{3/4}}-\frac {2 \sqrt {2} (5 A b-a B) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{b^{3/4}}+\frac {\sqrt {2} (-5 A b+a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{b^{3/4}}+\frac {\sqrt {2} (5 A b-a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{b^{3/4}}}{32 a^{9/4}} \] Input:

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

Output:

((-32*a^(1/4)*A)/x + (8*a^(1/4)*(-(A*b) + a*B)*x^3)/(a + b*x^4) + (2*Sqrt[ 
2]*(5*A*b - a*B)*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/b^(3/4) - (2*Sqr 
t[2]*(5*A*b - a*B)*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/b^(3/4) + (Sqr 
t[2]*(-5*A*b + a*B)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2] 
)/b^(3/4) + (Sqrt[2]*(5*A*b - a*B)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*x 
 + Sqrt[b]*x^2])/b^(3/4))/(32*a^(9/4))
 

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.35, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {957, 847, 826, 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.

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

\(\Big \downarrow \) 957

\(\displaystyle \frac {(5 A b-a B) \int \frac {1}{x^2 \left (b x^4+a\right )}dx}{4 a b}+\frac {A b-a B}{4 a b x \left (a+b x^4\right )}\)

\(\Big \downarrow \) 847

\(\displaystyle \frac {(5 A b-a B) \left (-\frac {b \int \frac {x^2}{b x^4+a}dx}{a}-\frac {1}{a x}\right )}{4 a b}+\frac {A b-a B}{4 a b x \left (a+b x^4\right )}\)

\(\Big \downarrow \) 826

\(\displaystyle \frac {(5 A b-a B) \left (-\frac {b \left (\frac {\int \frac {\sqrt {b} x^2+\sqrt {a}}{b x^4+a}dx}{2 \sqrt {b}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x^2}{b x^4+a}dx}{2 \sqrt {b}}\right )}{a}-\frac {1}{a x}\right )}{4 a b}+\frac {A b-a B}{4 a b x \left (a+b x^4\right )}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {(5 A b-a B) \left (-\frac {b \left (\frac {\frac {\int \frac {1}{x^2-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}}dx}{2 \sqrt {b}}+\frac {\int \frac {1}{x^2+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}}dx}{2 \sqrt {b}}}{2 \sqrt {b}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x^2}{b x^4+a}dx}{2 \sqrt {b}}\right )}{a}-\frac {1}{a x}\right )}{4 a b}+\frac {A b-a B}{4 a b x \left (a+b x^4\right )}\)

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

(A*b - a*B)/(4*a*b*x*(a + b*x^4)) + ((5*A*b - a*B)*(-(1/(a*x)) - (b*((-(Ar 
cTan[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[b]) - 
(-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*S 
qrt[2]*a^(1/4)*b^(1/4)))/(2*Sqrt[b])))/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 826

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 
2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s)   Int[(r + s*x^2)/(a + b*x^ 
4), x], x] - Simp[1/(2*s)   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 847

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x 
)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + n*(p + 1) 
+ 1)/(a*c^n*(m + 1)))   Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a 
, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p 
, x]
 

rule 957

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_)), x_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a 
*b*e*n*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b*n* 
(p + 1))   Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, 
 m, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && N 
eQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] && LeQ[-1 
, m, (-n)*(p + 1)]))
 

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 [A] (verified)

Time = 0.09 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.76

method	result	size
default	\(-\frac {\frac {\left (\frac {A b}{4}-\frac {B a}{4}\right ) x^{3}}{b \,x^{4}+a}+\frac {\left (\frac {5 A b}{4}-\frac {B a}{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}{x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{8 b \left (\frac {a}{b}\right )^{\frac {1}{4}}}}{a^{2}}-\frac {A}{a^{2} x}\)	\(149\)
risch	\(\frac {-\frac {\left (5 A b -B a \right ) x^{4}}{4 a^{2}}-\frac {A}{a}}{x \left (b \,x^{4}+a \right )}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{9} b^{3} \textit {\_Z}^{4}+625 A^{4} b^{4}-500 A^{3} B a \,b^{3}+150 A^{2} B^{2} a^{2} b^{2}-20 A \,B^{3} a^{3} b +B^{4} a^{4}\right )}{\sum }\textit {\_R} \ln \left (\left (5 \textit {\_R}^{4} a^{9} b^{3}+2500 A^{4} b^{4}-2000 A^{3} B a \,b^{3}+600 A^{2} B^{2} a^{2} b^{2}-80 A \,B^{3} a^{3} b +4 B^{4} a^{4}\right ) x +\left (5 A \,a^{7} b^{3}-B \,a^{8} b^{2}\right ) \textit {\_R}^{3}\right )\right )}{16}\)	\(196\)

Input:

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

Output:

-1/a^2*((1/4*A*b-1/4*B*a)*x^3/(b*x^4+a)+1/8*(5/4*A*b-1/4*B*a)/b/(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^2/x
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 778, normalized size of antiderivative = 3.95 \[ \int \frac {A+B x^4}{x^2 \left (a+b x^4\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

1/16*(4*(B*a - 5*A*b)*x^4 - (a^2*b*x^5 + a^3*x)*(-(B^4*a^4 - 20*A*B^3*a^3* 
b + 150*A^2*B^2*a^2*b^2 - 500*A^3*B*a*b^3 + 625*A^4*b^4)/(a^9*b^3))^(1/4)* 
log(a^7*b^2*(-(B^4*a^4 - 20*A*B^3*a^3*b + 150*A^2*B^2*a^2*b^2 - 500*A^3*B* 
a*b^3 + 625*A^4*b^4)/(a^9*b^3))^(3/4) - (B^3*a^3 - 15*A*B^2*a^2*b + 75*A^2 
*B*a*b^2 - 125*A^3*b^3)*x) - (-I*a^2*b*x^5 - I*a^3*x)*(-(B^4*a^4 - 20*A*B^ 
3*a^3*b + 150*A^2*B^2*a^2*b^2 - 500*A^3*B*a*b^3 + 625*A^4*b^4)/(a^9*b^3))^ 
(1/4)*log(I*a^7*b^2*(-(B^4*a^4 - 20*A*B^3*a^3*b + 150*A^2*B^2*a^2*b^2 - 50 
0*A^3*B*a*b^3 + 625*A^4*b^4)/(a^9*b^3))^(3/4) - (B^3*a^3 - 15*A*B^2*a^2*b 
+ 75*A^2*B*a*b^2 - 125*A^3*b^3)*x) - (I*a^2*b*x^5 + I*a^3*x)*(-(B^4*a^4 - 
20*A*B^3*a^3*b + 150*A^2*B^2*a^2*b^2 - 500*A^3*B*a*b^3 + 625*A^4*b^4)/(a^9 
*b^3))^(1/4)*log(-I*a^7*b^2*(-(B^4*a^4 - 20*A*B^3*a^3*b + 150*A^2*B^2*a^2* 
b^2 - 500*A^3*B*a*b^3 + 625*A^4*b^4)/(a^9*b^3))^(3/4) - (B^3*a^3 - 15*A*B^ 
2*a^2*b + 75*A^2*B*a*b^2 - 125*A^3*b^3)*x) + (a^2*b*x^5 + a^3*x)*(-(B^4*a^ 
4 - 20*A*B^3*a^3*b + 150*A^2*B^2*a^2*b^2 - 500*A^3*B*a*b^3 + 625*A^4*b^4)/ 
(a^9*b^3))^(1/4)*log(-a^7*b^2*(-(B^4*a^4 - 20*A*B^3*a^3*b + 150*A^2*B^2*a^ 
2*b^2 - 500*A^3*B*a*b^3 + 625*A^4*b^4)/(a^9*b^3))^(3/4) - (B^3*a^3 - 15*A* 
B^2*a^2*b + 75*A^2*B*a*b^2 - 125*A^3*b^3)*x) - 16*A*a)/(a^2*b*x^5 + a^3*x)
 

Sympy [A] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.79 \[ \int \frac {A+B x^4}{x^2 \left (a+b x^4\right )^2} \, dx=\frac {- 4 A a + x^{4} \left (- 5 A b + B a\right )}{4 a^{3} x + 4 a^{2} b x^{5}} + \operatorname {RootSum} {\left (65536 t^{4} a^{9} b^{3} + 625 A^{4} b^{4} - 500 A^{3} B a b^{3} + 150 A^{2} B^{2} a^{2} b^{2} - 20 A B^{3} a^{3} b + B^{4} a^{4}, \left ( t \mapsto t \log {\left (\frac {4096 t^{3} a^{7} b^{2}}{- 125 A^{3} b^{3} + 75 A^{2} B a b^{2} - 15 A B^{2} a^{2} b + B^{3} a^{3}} + x \right )} \right )\right )} \] Input:

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

Output:

(-4*A*a + x**4*(-5*A*b + B*a))/(4*a**3*x + 4*a**2*b*x**5) + RootSum(65536* 
_t**4*a**9*b**3 + 625*A**4*b**4 - 500*A**3*B*a*b**3 + 150*A**2*B**2*a**2*b 
**2 - 20*A*B**3*a**3*b + B**4*a**4, Lambda(_t, _t*log(4096*_t**3*a**7*b**2 
/(-125*A**3*b**3 + 75*A**2*B*a*b**2 - 15*A*B**2*a**2*b + B**3*a**3) + x)))
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.10 \[ \int \frac {A+B x^4}{x^2 \left (a+b x^4\right )^2} \, dx=\frac {{\left (B a - 5 \, A b\right )} x^{4} - 4 \, A a}{4 \, {\left (a^{2} b x^{5} + a^{3} x\right )}} + \frac {{\left (B a - 5 \, A b\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {b} x^{2} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {b} x^{2} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{32 \, a^{2}} \] Input:

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

Output:

1/4*((B*a - 5*A*b)*x^4 - 4*A*a)/(a^2*b*x^5 + a^3*x) + 1/32*(B*a - 5*A*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)) + 2*sqrt(2)*arctan(1/2*s 
qrt(2)*(2*sqrt(b)*x - sqrt(2)*a^(1/4)*b^(1/4))/sqrt(sqrt(a)*sqrt(b)))/(sqr 
t(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
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.39 \[ \int \frac {A+B x^4}{x^2 \left (a+b x^4\right )^2} \, dx=\frac {B a x^{4} - 5 \, A b x^{4} - 4 \, A a}{4 \, {\left (b x^{5} + a x\right )} a^{2}} + \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} B a - 5 \, \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{16 \, a^{3} b^{3}} + \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} B a - 5 \, \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{16 \, a^{3} b^{3}} - \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} B a - 5 \, \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{32 \, a^{3} b^{3}} + \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} B a - 5 \, \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{32 \, a^{3} b^{3}} \] Input:

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

Output:

1/4*(B*a*x^4 - 5*A*b*x^4 - 4*A*a)/((b*x^5 + a*x)*a^2) + 1/16*sqrt(2)*((a*b 
^3)^(3/4)*B*a - 5*(a*b^3)^(3/4)*A*b)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)*(a/ 
b)^(1/4))/(a/b)^(1/4))/(a^3*b^3) + 1/16*sqrt(2)*((a*b^3)^(3/4)*B*a - 5*(a* 
b^3)^(3/4)*A*b)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*(a/b)^(1/4))/(a/b)^(1/4) 
)/(a^3*b^3) - 1/32*sqrt(2)*((a*b^3)^(3/4)*B*a - 5*(a*b^3)^(3/4)*A*b)*log(x 
^2 + sqrt(2)*x*(a/b)^(1/4) + sqrt(a/b))/(a^3*b^3) + 1/32*sqrt(2)*((a*b^3)^ 
(3/4)*B*a - 5*(a*b^3)^(3/4)*A*b)*log(x^2 - sqrt(2)*x*(a/b)^(1/4) + sqrt(a/ 
b))/(a^3*b^3)
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 3.59 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.49 \[ \int \frac {A+B x^4}{x^2 \left (a+b x^4\right )^2} \, dx=\frac {\mathrm {atanh}\left (\frac {b^{1/4}\,x}{{\left (-a\right )}^{1/4}}\right )\,\left (5\,A\,b-B\,a\right )}{8\,{\left (-a\right )}^{9/4}\,b^{3/4}}-\frac {\mathrm {atan}\left (\frac {b^{1/4}\,x}{{\left (-a\right )}^{1/4}}\right )\,\left (5\,A\,b-B\,a\right )}{8\,{\left (-a\right )}^{9/4}\,b^{3/4}}-\frac {\frac {A}{a}+\frac {x^4\,\left (5\,A\,b-B\,a\right )}{4\,a^2}}{b\,x^5+a\,x} \] Input:

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

Output:

(atanh((b^(1/4)*x)/(-a)^(1/4))*(5*A*b - B*a))/(8*(-a)^(9/4)*b^(3/4)) - (at 
an((b^(1/4)*x)/(-a)^(1/4))*(5*A*b - B*a))/(8*(-a)^(9/4)*b^(3/4)) - (A/a + 
(x^4*(5*A*b - B*a))/(4*a^2))/(a*x + b*x^5)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.76 \[ \int \frac {A+B x^4}{x^2 \left (a+b x^4\right )^2} \, dx=\frac {2 b^{\frac {1}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}-2 \sqrt {b}\, x}{b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) x -2 b^{\frac {1}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+2 \sqrt {b}\, x}{b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right ) x -b^{\frac {1}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (-b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {a}+\sqrt {b}\, x^{2}\right ) x +b^{\frac {1}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {a}+\sqrt {b}\, x^{2}\right ) x -8 a}{8 a^{2} x} \] Input:

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

Output:

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