\(\int \frac {x^4 (A+B x^4)}{(a+b x^4)^2} \, dx\) [19]

Optimal result

Integrand size = 20, antiderivative size = 189 \[ \int \frac {x^4 \left (A+B x^4\right )}{\left (a+b x^4\right )^2} \, dx=\frac {B x}{b^2}-\frac {(A b-a B) x}{4 b^2 \left (a+b x^4\right )}-\frac {(A b-5 a B) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{3/4} b^{9/4}}+\frac {(A b-5 a B) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} a^{3/4} b^{9/4}}+\frac {(A b-5 a B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x}{\sqrt {a}+\sqrt {b} x^2}\right )}{8 \sqrt {2} a^{3/4} b^{9/4}} \] Output:

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.23 \[ \int \frac {x^4 \left (A+B x^4\right )}{\left (a+b x^4\right )^2} \, dx=\frac {32 \sqrt [4]{b} B x-\frac {8 \sqrt [4]{b} (A b-a B) x}{a+b x^4}+\frac {2 \sqrt {2} (-A b+5 a B) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{a^{3/4}}+\frac {2 \sqrt {2} (A b-5 a B) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{a^{3/4}}+\frac {\sqrt {2} (-A b+5 a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{a^{3/4}}+\frac {\sqrt {2} (A b-5 a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{a^{3/4}}}{32 b^{9/4}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.38, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {957, 843, 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.

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

\(\Big \downarrow \) 957

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

\(\Big \downarrow \) 843

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

\(\Big \downarrow \) 755

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

((A*b - a*B)*x^5)/(4*a*b*(a + b*x^4)) - ((A*b - 5*a*B)*(x/b - (a*((-(ArcTa 
n[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])))/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 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 843

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n 
 - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ 
a*c^n*((m - n + 1)/(b*(m + n*p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^p, x] 
, x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* 
p + 1, 0] && 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 [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 = 0.34

method	result	size
risch	\(\frac {B x}{b^{2}}+\frac {\left (-\frac {A b}{4}+\frac {B a}{4}\right ) x}{b^{2} \left (b \,x^{4}+a \right )}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{4}+a \right )}{\sum }\frac {\left (A b -5 B a \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}}{16 b^{3}}\)	\(65\)
default	\(\frac {B x}{b^{2}}+\frac {\frac {\left (-\frac {A b}{4}+\frac {B a}{4}\right ) x}{b \,x^{4}+a}+\frac {\left (A b -5 B a \right ) \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}{x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{32 a}}{b^{2}}\)	\(142\)

Input:

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

Output:

B*x/b^2+(-1/4*A*b+1/4*B*a)*x/b^2/(b*x^4+a)+1/16/b^3*sum((A*b-5*B*a)/_R^3*l 
n(x-_R),_R=RootOf(_Z^4*b+a))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.08 (sec) , antiderivative size = 659, normalized size of antiderivative = 3.49 \[ \int \frac {x^4 \left (A+B x^4\right )}{\left (a+b x^4\right )^2} \, dx=\frac {16 \, B b x^{5} + {\left (b^{3} x^{4} + a b^{2}\right )} \left (-\frac {625 \, B^{4} a^{4} - 500 \, A B^{3} a^{3} b + 150 \, A^{2} B^{2} a^{2} b^{2} - 20 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{9}}\right )^{\frac {1}{4}} \log \left (a b^{2} \left (-\frac {625 \, B^{4} a^{4} - 500 \, A B^{3} a^{3} b + 150 \, A^{2} B^{2} a^{2} b^{2} - 20 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{9}}\right )^{\frac {1}{4}} - {\left (5 \, B a - A b\right )} x\right ) - {\left (-i \, b^{3} x^{4} - i \, a b^{2}\right )} \left (-\frac {625 \, B^{4} a^{4} - 500 \, A B^{3} a^{3} b + 150 \, A^{2} B^{2} a^{2} b^{2} - 20 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{9}}\right )^{\frac {1}{4}} \log \left (i \, a b^{2} \left (-\frac {625 \, B^{4} a^{4} - 500 \, A B^{3} a^{3} b + 150 \, A^{2} B^{2} a^{2} b^{2} - 20 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{9}}\right )^{\frac {1}{4}} - {\left (5 \, B a - A b\right )} x\right ) - {\left (i \, b^{3} x^{4} + i \, a b^{2}\right )} \left (-\frac {625 \, B^{4} a^{4} - 500 \, A B^{3} a^{3} b + 150 \, A^{2} B^{2} a^{2} b^{2} - 20 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{9}}\right )^{\frac {1}{4}} \log \left (-i \, a b^{2} \left (-\frac {625 \, B^{4} a^{4} - 500 \, A B^{3} a^{3} b + 150 \, A^{2} B^{2} a^{2} b^{2} - 20 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{9}}\right )^{\frac {1}{4}} - {\left (5 \, B a - A b\right )} x\right ) - {\left (b^{3} x^{4} + a b^{2}\right )} \left (-\frac {625 \, B^{4} a^{4} - 500 \, A B^{3} a^{3} b + 150 \, A^{2} B^{2} a^{2} b^{2} - 20 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{9}}\right )^{\frac {1}{4}} \log \left (-a b^{2} \left (-\frac {625 \, B^{4} a^{4} - 500 \, A B^{3} a^{3} b + 150 \, A^{2} B^{2} a^{2} b^{2} - 20 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{9}}\right )^{\frac {1}{4}} - {\left (5 \, B a - A b\right )} x\right ) + 4 \, {\left (5 \, B a - A b\right )} x}{16 \, {\left (b^{3} x^{4} + a b^{2}\right )}} \] Input:

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

Output:

1/16*(16*B*b*x^5 + (b^3*x^4 + a*b^2)*(-(625*B^4*a^4 - 500*A*B^3*a^3*b + 15 
0*A^2*B^2*a^2*b^2 - 20*A^3*B*a*b^3 + A^4*b^4)/(a^3*b^9))^(1/4)*log(a*b^2*( 
-(625*B^4*a^4 - 500*A*B^3*a^3*b + 150*A^2*B^2*a^2*b^2 - 20*A^3*B*a*b^3 + A 
^4*b^4)/(a^3*b^9))^(1/4) - (5*B*a - A*b)*x) - (-I*b^3*x^4 - I*a*b^2)*(-(62 
5*B^4*a^4 - 500*A*B^3*a^3*b + 150*A^2*B^2*a^2*b^2 - 20*A^3*B*a*b^3 + A^4*b 
^4)/(a^3*b^9))^(1/4)*log(I*a*b^2*(-(625*B^4*a^4 - 500*A*B^3*a^3*b + 150*A^ 
2*B^2*a^2*b^2 - 20*A^3*B*a*b^3 + A^4*b^4)/(a^3*b^9))^(1/4) - (5*B*a - A*b) 
*x) - (I*b^3*x^4 + I*a*b^2)*(-(625*B^4*a^4 - 500*A*B^3*a^3*b + 150*A^2*B^2 
*a^2*b^2 - 20*A^3*B*a*b^3 + A^4*b^4)/(a^3*b^9))^(1/4)*log(-I*a*b^2*(-(625* 
B^4*a^4 - 500*A*B^3*a^3*b + 150*A^2*B^2*a^2*b^2 - 20*A^3*B*a*b^3 + A^4*b^4 
)/(a^3*b^9))^(1/4) - (5*B*a - A*b)*x) - (b^3*x^4 + a*b^2)*(-(625*B^4*a^4 - 
 500*A*B^3*a^3*b + 150*A^2*B^2*a^2*b^2 - 20*A^3*B*a*b^3 + A^4*b^4)/(a^3*b^ 
9))^(1/4)*log(-a*b^2*(-(625*B^4*a^4 - 500*A*B^3*a^3*b + 150*A^2*B^2*a^2*b^ 
2 - 20*A^3*B*a*b^3 + A^4*b^4)/(a^3*b^9))^(1/4) - (5*B*a - A*b)*x) + 4*(5*B 
*a - A*b)*x)/(b^3*x^4 + a*b^2)
 

Sympy [A] (verification not implemented)

Time = 0.66 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.62 \[ \int \frac {x^4 \left (A+B x^4\right )}{\left (a+b x^4\right )^2} \, dx=\frac {B x}{b^{2}} + \frac {x \left (- A b + B a\right )}{4 a b^{2} + 4 b^{3} x^{4}} + \operatorname {RootSum} {\left (65536 t^{4} a^{3} b^{9} + A^{4} b^{4} - 20 A^{3} B a b^{3} + 150 A^{2} B^{2} a^{2} b^{2} - 500 A B^{3} a^{3} b + 625 B^{4} a^{4}, \left ( t \mapsto t \log {\left (- \frac {16 t a b^{2}}{- A b + 5 B a} + x \right )} \right )\right )} \] Input:

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

Output:

B*x/b**2 + x*(-A*b + B*a)/(4*a*b**2 + 4*b**3*x**4) + RootSum(65536*_t**4*a 
**3*b**9 + A**4*b**4 - 20*A**3*B*a*b**3 + 150*A**2*B**2*a**2*b**2 - 500*A* 
B**3*a**3*b + 625*B**4*a**4, Lambda(_t, _t*log(-16*_t*a*b**2/(-A*b + 5*B*a 
) + x)))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.28 \[ \int \frac {x^4 \left (A+B x^4\right )}{\left (a+b x^4\right )^2} \, dx=\frac {{\left (B a - A b\right )} x}{4 \, {\left (b^{3} x^{4} + a b^{2}\right )}} + \frac {B x}{b^{2}} - \frac {\frac {2 \, \sqrt {2} {\left (5 \, B a - A b\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {2 \, \sqrt {2} {\left (5 \, B a - A b\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {\sqrt {2} {\left (5 \, B a - A b\right )} \log \left (\sqrt {b} x^{2} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (5 \, B a - A b\right )} \log \left (\sqrt {b} x^{2} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}}}{32 \, b^{2}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.44 \[ \int \frac {x^4 \left (A+B x^4\right )}{\left (a+b x^4\right )^2} \, dx=\frac {B x}{b^{2}} - \frac {\sqrt {2} {\left (5 \, \left (a b^{3}\right )^{\frac {1}{4}} B a - \left (a b^{3}\right )^{\frac {1}{4}} A b\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{16 \, a b^{3}} - \frac {\sqrt {2} {\left (5 \, \left (a b^{3}\right )^{\frac {1}{4}} B a - \left (a b^{3}\right )^{\frac {1}{4}} A b\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{16 \, a b^{3}} - \frac {\sqrt {2} {\left (5 \, \left (a b^{3}\right )^{\frac {1}{4}} B a - \left (a b^{3}\right )^{\frac {1}{4}} A b\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{32 \, a b^{3}} + \frac {\sqrt {2} {\left (5 \, \left (a b^{3}\right )^{\frac {1}{4}} B a - \left (a b^{3}\right )^{\frac {1}{4}} A b\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{32 \, a b^{3}} + \frac {B a x - A b x}{4 \, {\left (b x^{4} + a\right )} b^{2}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 731, normalized size of antiderivative = 3.87 \[ \int \frac {x^4 \left (A+B x^4\right )}{\left (a+b x^4\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(B*x)/b^2 - (x*((A*b)/4 - (B*a)/4))/(a*b^2 + b^3*x^4) + (atan(((((x*(A^2*b 
^2 + 25*B^2*a^2 - 10*A*B*a*b))/(4*b) - ((A*b - 5*B*a)*(4*A*a*b^2 - 20*B*a^ 
2*b))/(16*(-a)^(3/4)*b^(9/4)))*(A*b - 5*B*a)*1i)/(16*(-a)^(3/4)*b^(9/4)) + 
 (((x*(A^2*b^2 + 25*B^2*a^2 - 10*A*B*a*b))/(4*b) + ((A*b - 5*B*a)*(4*A*a*b 
^2 - 20*B*a^2*b))/(16*(-a)^(3/4)*b^(9/4)))*(A*b - 5*B*a)*1i)/(16*(-a)^(3/4 
)*b^(9/4)))/((((x*(A^2*b^2 + 25*B^2*a^2 - 10*A*B*a*b))/(4*b) - ((A*b - 5*B 
*a)*(4*A*a*b^2 - 20*B*a^2*b))/(16*(-a)^(3/4)*b^(9/4)))*(A*b - 5*B*a))/(16* 
(-a)^(3/4)*b^(9/4)) - (((x*(A^2*b^2 + 25*B^2*a^2 - 10*A*B*a*b))/(4*b) + (( 
A*b - 5*B*a)*(4*A*a*b^2 - 20*B*a^2*b))/(16*(-a)^(3/4)*b^(9/4)))*(A*b - 5*B 
*a))/(16*(-a)^(3/4)*b^(9/4))))*(A*b - 5*B*a)*1i)/(8*(-a)^(3/4)*b^(9/4)) + 
(atan(((((x*(A^2*b^2 + 25*B^2*a^2 - 10*A*B*a*b))/(4*b) - ((A*b - 5*B*a)*(4 
*A*a*b^2 - 20*B*a^2*b)*1i)/(16*(-a)^(3/4)*b^(9/4)))*(A*b - 5*B*a))/(16*(-a 
)^(3/4)*b^(9/4)) + (((x*(A^2*b^2 + 25*B^2*a^2 - 10*A*B*a*b))/(4*b) + ((A*b 
 - 5*B*a)*(4*A*a*b^2 - 20*B*a^2*b)*1i)/(16*(-a)^(3/4)*b^(9/4)))*(A*b - 5*B 
*a))/(16*(-a)^(3/4)*b^(9/4)))/((((x*(A^2*b^2 + 25*B^2*a^2 - 10*A*B*a*b))/( 
4*b) - ((A*b - 5*B*a)*(4*A*a*b^2 - 20*B*a^2*b)*1i)/(16*(-a)^(3/4)*b^(9/4)) 
)*(A*b - 5*B*a)*1i)/(16*(-a)^(3/4)*b^(9/4)) - (((x*(A^2*b^2 + 25*B^2*a^2 - 
 10*A*B*a*b))/(4*b) + ((A*b - 5*B*a)*(4*A*a*b^2 - 20*B*a^2*b)*1i)/(16*(-a) 
^(3/4)*b^(9/4)))*(A*b - 5*B*a)*1i)/(16*(-a)^(3/4)*b^(9/4))))*(A*b - 5*B*a) 
)/(8*(-a)^(3/4)*b^(9/4))
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.76 \[ \int \frac {x^4 \left (A+B x^4\right )}{\left (a+b x^4\right )^2} \, dx=\frac {2 b^{\frac {3}{4}} a^{\frac {1}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}-2 \sqrt {b}\, x}{b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right )-2 b^{\frac {3}{4}} a^{\frac {1}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+2 \sqrt {b}\, x}{b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right )+b^{\frac {3}{4}} a^{\frac {1}{4}} \sqrt {2}\, \mathrm {log}\left (-b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {a}+\sqrt {b}\, x^{2}\right )-b^{\frac {3}{4}} a^{\frac {1}{4}} \sqrt {2}\, \mathrm {log}\left (b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {a}+\sqrt {b}\, x^{2}\right )+8 b x}{8 b^{2}} \] Input:

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

Output:

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