\(\int \frac {x^2 (A+B x^4)}{(a+b x^4)^2} \, dx\) [20]

Optimal result

Integrand size = 20, antiderivative size = 188 \[ \int \frac {x^2 \left (A+B x^4\right )}{\left (a+b x^4\right )^2} \, dx=\frac {(A b-a B) x^3}{4 a b \left (a+b x^4\right )}-\frac {(A b+3 a B) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{5/4} b^{7/4}}+\frac {(A b+3 a B) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{5/4} b^{7/4}}-\frac {(A b+3 a B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x}{\sqrt {a}+\sqrt {b} x^2}\right )}{8 \sqrt {2} a^{5/4} b^{7/4}} \] Output:

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.14 \[ \int \frac {x^2 \left (A+B x^4\right )}{\left (a+b x^4\right )^2} \, dx=\frac {-\frac {8 \sqrt [4]{a} b^{3/4} (-A b+a B) x^3}{a+b x^4}-2 \sqrt {2} (A b+3 a B) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )+2 \sqrt {2} (A b+3 a B) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )+\sqrt {2} (A b+3 a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )-\sqrt {2} (A b+3 a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{32 a^{5/4} b^{7/4}} \] Input:

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

Output:

((-8*a^(1/4)*b^(3/4)*(-(A*b) + a*B)*x^3)/(a + b*x^4) - 2*Sqrt[2]*(A*b + 3* 
a*B)*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)] + 2*Sqrt[2]*(A*b + 3*a*B)*Arc 
Tan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)] + Sqrt[2]*(A*b + 3*a*B)*Log[Sqrt[a] - 
 Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2] - Sqrt[2]*(A*b + 3*a*B)*Log[Sqrt 
[a] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2])/(32*a^(5/4)*b^(7/4))
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.32, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {957, 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.

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

\(\Big \downarrow \) 957

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

((A*b - a*B)*x^3)/(4*a*b*(a + b*x^4)) + ((A*b + 3*a*B)*((-(ArcTan[1 - (Sqr 
t[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[Sqr 
t[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[b])))/(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 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 [C] (verified)

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

Time = 0.08 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.36

method	result	size
risch	\(\frac {\left (A b -B a \right ) x^{3}}{4 a b \left (b \,x^{4}+a \right )}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{4}+a \right )}{\sum }\frac {\left (A b +3 B a \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}}}{16 a \,b^{2}}\)	\(67\)
default	\(\frac {\left (A b -B a \right ) x^{3}}{4 a b \left (b \,x^{4}+a \right )}+\frac {\left (A b +3 B a \right ) \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}{x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{32 a \,b^{2} \left (\frac {a}{b}\right )^{\frac {1}{4}}}\)	\(142\)

Input:

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

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 768, normalized size of antiderivative = 4.09 \[ \int \frac {x^2 \left (A+B x^4\right )}{\left (a+b x^4\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/16*(4*(B*a - A*b)*x^3 - (a*b^2*x^4 + a^2*b)*(-(81*B^4*a^4 + 108*A*B^3*a 
^3*b + 54*A^2*B^2*a^2*b^2 + 12*A^3*B*a*b^3 + A^4*b^4)/(a^5*b^7))^(1/4)*log 
(a^4*b^5*(-(81*B^4*a^4 + 108*A*B^3*a^3*b + 54*A^2*B^2*a^2*b^2 + 12*A^3*B*a 
*b^3 + A^4*b^4)/(a^5*b^7))^(3/4) + (27*B^3*a^3 + 27*A*B^2*a^2*b + 9*A^2*B* 
a*b^2 + A^3*b^3)*x) + (I*a*b^2*x^4 + I*a^2*b)*(-(81*B^4*a^4 + 108*A*B^3*a^ 
3*b + 54*A^2*B^2*a^2*b^2 + 12*A^3*B*a*b^3 + A^4*b^4)/(a^5*b^7))^(1/4)*log( 
I*a^4*b^5*(-(81*B^4*a^4 + 108*A*B^3*a^3*b + 54*A^2*B^2*a^2*b^2 + 12*A^3*B* 
a*b^3 + A^4*b^4)/(a^5*b^7))^(3/4) + (27*B^3*a^3 + 27*A*B^2*a^2*b + 9*A^2*B 
*a*b^2 + A^3*b^3)*x) + (-I*a*b^2*x^4 - I*a^2*b)*(-(81*B^4*a^4 + 108*A*B^3* 
a^3*b + 54*A^2*B^2*a^2*b^2 + 12*A^3*B*a*b^3 + A^4*b^4)/(a^5*b^7))^(1/4)*lo 
g(-I*a^4*b^5*(-(81*B^4*a^4 + 108*A*B^3*a^3*b + 54*A^2*B^2*a^2*b^2 + 12*A^3 
*B*a*b^3 + A^4*b^4)/(a^5*b^7))^(3/4) + (27*B^3*a^3 + 27*A*B^2*a^2*b + 9*A^ 
2*B*a*b^2 + A^3*b^3)*x) + (a*b^2*x^4 + a^2*b)*(-(81*B^4*a^4 + 108*A*B^3*a^ 
3*b + 54*A^2*B^2*a^2*b^2 + 12*A^3*B*a*b^3 + A^4*b^4)/(a^5*b^7))^(1/4)*log( 
-a^4*b^5*(-(81*B^4*a^4 + 108*A*B^3*a^3*b + 54*A^2*B^2*a^2*b^2 + 12*A^3*B*a 
*b^3 + A^4*b^4)/(a^5*b^7))^(3/4) + (27*B^3*a^3 + 27*A*B^2*a^2*b + 9*A^2*B* 
a*b^2 + A^3*b^3)*x))/(a*b^2*x^4 + a^2*b)
 

Sympy [A] (verification not implemented)

Time = 0.64 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.79 \[ \int \frac {x^2 \left (A+B x^4\right )}{\left (a+b x^4\right )^2} \, dx=\frac {x^{3} \left (A b - B a\right )}{4 a^{2} b + 4 a b^{2} x^{4}} + \operatorname {RootSum} {\left (65536 t^{4} a^{5} b^{7} + A^{4} b^{4} + 12 A^{3} B a b^{3} + 54 A^{2} B^{2} a^{2} b^{2} + 108 A B^{3} a^{3} b + 81 B^{4} a^{4}, \left ( t \mapsto t \log {\left (\frac {4096 t^{3} a^{4} b^{5}}{A^{3} b^{3} + 9 A^{2} B a b^{2} + 27 A B^{2} a^{2} b + 27 B^{3} a^{3}} + x \right )} \right )\right )} \] Input:

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

Output:

x**3*(A*b - B*a)/(4*a**2*b + 4*a*b**2*x**4) + RootSum(65536*_t**4*a**5*b** 
7 + A**4*b**4 + 12*A**3*B*a*b**3 + 54*A**2*B**2*a**2*b**2 + 108*A*B**3*a** 
3*b + 81*B**4*a**4, Lambda(_t, _t*log(4096*_t**3*a**4*b**5/(A**3*b**3 + 9* 
A**2*B*a*b**2 + 27*A*B**2*a**2*b + 27*B**3*a**3) + x)))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.14 \[ \int \frac {x^2 \left (A+B x^4\right )}{\left (a+b x^4\right )^2} \, dx=-\frac {{\left (B a - A b\right )} x^{3}}{4 \, {\left (a b^{2} x^{4} + a^{2} b\right )}} + \frac {{\left (3 \, B a + A b\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {b} x^{2} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {b} x^{2} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{32 \, a b} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.44 \[ \int \frac {x^2 \left (A+B x^4\right )}{\left (a+b x^4\right )^2} \, dx=-\frac {B a x^{3} - A b x^{3}}{4 \, {\left (b x^{4} + a\right )} a b} + \frac {\sqrt {2} {\left (3 \, \left (a b^{3}\right )^{\frac {3}{4}} B a + \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{16 \, a^{2} b^{4}} + \frac {\sqrt {2} {\left (3 \, \left (a b^{3}\right )^{\frac {3}{4}} B a + \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{16 \, a^{2} b^{4}} - \frac {\sqrt {2} {\left (3 \, \left (a b^{3}\right )^{\frac {3}{4}} B a + \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{32 \, a^{2} b^{4}} + \frac {\sqrt {2} {\left (3 \, \left (a b^{3}\right )^{\frac {3}{4}} B a + \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{32 \, a^{2} b^{4}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 3.52 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.46 \[ \int \frac {x^2 \left (A+B x^4\right )}{\left (a+b x^4\right )^2} \, dx=\frac {\mathrm {atanh}\left (\frac {b^{1/4}\,x}{{\left (-a\right )}^{1/4}}\right )\,\left (A\,b+3\,B\,a\right )}{8\,{\left (-a\right )}^{5/4}\,b^{7/4}}-\frac {\mathrm {atan}\left (\frac {b^{1/4}\,x}{{\left (-a\right )}^{1/4}}\right )\,\left (A\,b+3\,B\,a\right )}{8\,{\left (-a\right )}^{5/4}\,b^{7/4}}+\frac {x^3\,\left (A\,b-B\,a\right )}{4\,a\,b\,\left (b\,x^4+a\right )} \] Input:

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

Output:

(atanh((b^(1/4)*x)/(-a)^(1/4))*(A*b + 3*B*a))/(8*(-a)^(5/4)*b^(7/4)) - (at 
an((b^(1/4)*x)/(-a)^(1/4))*(A*b + 3*B*a))/(8*(-a)^(5/4)*b^(7/4)) + (x^3*(A 
*b - B*a))/(4*a*b*(a + b*x^4))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.60 \[ \int \frac {x^2 \left (A+B x^4\right )}{\left (a+b x^4\right )^2} \, dx=\frac {\sqrt {2}\, \left (-2 \mathit {atan} \left (\frac {b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}-2 \sqrt {b}\, x}{b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right )+2 \mathit {atan} \left (\frac {b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+2 \sqrt {b}\, x}{b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right )+\mathrm {log}\left (-b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {a}+\sqrt {b}\, x^{2}\right )-\mathrm {log}\left (b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {a}+\sqrt {b}\, x^{2}\right )\right )}{8 b^{\frac {3}{4}} a^{\frac {1}{4}}} \] Input:

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

Output:

(b**(1/4)*a**(3/4)*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)