\(\int \frac {A+B x^4}{x^6 (a+b x^4)} \, dx\) [16]

Optimal result

Integrand size = 20, antiderivative size = 184 \[ \int \frac {A+B x^4}{x^6 \left (a+b x^4\right )} \, dx=-\frac {A}{5 a x^5}+\frac {A b-a B}{a^2 x}-\frac {\sqrt [4]{b} (A b-a B) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{9/4}}+\frac {\sqrt [4]{b} (A b-a B) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 \sqrt {2} a^{9/4}}-\frac {\sqrt [4]{b} (A b-a B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x}{\sqrt {a}+\sqrt {b} x^2}\right )}{2 \sqrt {2} a^{9/4}} \] Output:

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.23 \[ \int \frac {A+B x^4}{x^6 \left (a+b x^4\right )} \, dx=\frac {-\frac {8 a^{5/4} A}{x^5}+\frac {40 \sqrt [4]{a} (A b-a B)}{x}-10 \sqrt {2} \sqrt [4]{b} (A b-a B) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )+10 \sqrt {2} \sqrt [4]{b} (A b-a B) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )+5 \sqrt {2} \sqrt [4]{b} (A b-a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )+5 \sqrt {2} \sqrt [4]{b} (-A b+a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{40 a^{9/4}} \] Input:

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

Output:

((-8*a^(5/4)*A)/x^5 + (40*a^(1/4)*(A*b - a*B))/x - 10*Sqrt[2]*b^(1/4)*(A*b 
 - a*B)*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)] + 10*Sqrt[2]*b^(1/4)*(A*b 
- a*B)*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)] + 5*Sqrt[2]*b^(1/4)*(A*b - 
a*B)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2] + 5*Sqrt[2]*b^ 
(1/4)*(-(A*b) + a*B)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2 
])/(40*a^(9/4))
 

Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.30, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {955, 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.

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

\(\Big \downarrow \) 955

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

\(\Big \downarrow \) 847

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

-1/5*A/(a*x^5) - ((A*b - a*B)*(-(1/(a*x)) - (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[b]) - (-1/2*Log[Sqrt[a] - S 
qrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b]*x^2]/(Sqrt[2]*a^(1/4)*b^(1/4)) + Log[Sq 
rt[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])))/a))/a
 

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 955

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_)), x_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), 
 x] + Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1))   Int[(e 
*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b* 
c - a*d, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || 
(LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[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 = 136, normalized size of antiderivative = 0.74

method	result	size
default	\(\frac {\left (A b -B a \right ) \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}{x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{8 a^{2} \left (\frac {a}{b}\right )^{\frac {1}{4}}}-\frac {A}{5 a \,x^{5}}-\frac {-A b +B a}{a^{2} x}\)	\(136\)
risch	\(\frac {\frac {\left (A b -B a \right ) x^{4}}{a^{2}}-\frac {A}{5 a}}{x^{5}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{9} \textit {\_Z}^{4}+A^{4} b^{5}-4 A^{3} B a \,b^{4}+6 A^{2} B^{2} a^{2} b^{3}-4 A \,B^{3} a^{3} b^{2}+B^{4} a^{4} b \right )}{\sum }\textit {\_R} \ln \left (\left (5 \textit {\_R}^{4} a^{9}+4 A^{4} b^{5}-16 A^{3} B a \,b^{4}+24 A^{2} B^{2} a^{2} b^{3}-16 A \,B^{3} a^{3} b^{2}+4 B^{4} a^{4} b \right ) x +\left (-A \,a^{7} b +B \,a^{8}\right ) \textit {\_R}^{3}\right )\right )}{4}\)	\(178\)

Input:

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

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 725, normalized size of antiderivative = 3.94 \[ \int \frac {A+B x^4}{x^6 \left (a+b x^4\right )} \, dx=\frac {5 \, a^{2} x^{5} \left (-\frac {B^{4} a^{4} b - 4 \, A B^{3} a^{3} b^{2} + 6 \, A^{2} B^{2} a^{2} b^{3} - 4 \, A^{3} B a b^{4} + A^{4} b^{5}}{a^{9}}\right )^{\frac {1}{4}} \log \left (a^{7} \left (-\frac {B^{4} a^{4} b - 4 \, A B^{3} a^{3} b^{2} + 6 \, A^{2} B^{2} a^{2} b^{3} - 4 \, A^{3} B a b^{4} + A^{4} b^{5}}{a^{9}}\right )^{\frac {3}{4}} - {\left (B^{3} a^{3} b - 3 \, A B^{2} a^{2} b^{2} + 3 \, A^{2} B a b^{3} - A^{3} b^{4}\right )} x\right ) - 5 i \, a^{2} x^{5} \left (-\frac {B^{4} a^{4} b - 4 \, A B^{3} a^{3} b^{2} + 6 \, A^{2} B^{2} a^{2} b^{3} - 4 \, A^{3} B a b^{4} + A^{4} b^{5}}{a^{9}}\right )^{\frac {1}{4}} \log \left (i \, a^{7} \left (-\frac {B^{4} a^{4} b - 4 \, A B^{3} a^{3} b^{2} + 6 \, A^{2} B^{2} a^{2} b^{3} - 4 \, A^{3} B a b^{4} + A^{4} b^{5}}{a^{9}}\right )^{\frac {3}{4}} - {\left (B^{3} a^{3} b - 3 \, A B^{2} a^{2} b^{2} + 3 \, A^{2} B a b^{3} - A^{3} b^{4}\right )} x\right ) + 5 i \, a^{2} x^{5} \left (-\frac {B^{4} a^{4} b - 4 \, A B^{3} a^{3} b^{2} + 6 \, A^{2} B^{2} a^{2} b^{3} - 4 \, A^{3} B a b^{4} + A^{4} b^{5}}{a^{9}}\right )^{\frac {1}{4}} \log \left (-i \, a^{7} \left (-\frac {B^{4} a^{4} b - 4 \, A B^{3} a^{3} b^{2} + 6 \, A^{2} B^{2} a^{2} b^{3} - 4 \, A^{3} B a b^{4} + A^{4} b^{5}}{a^{9}}\right )^{\frac {3}{4}} - {\left (B^{3} a^{3} b - 3 \, A B^{2} a^{2} b^{2} + 3 \, A^{2} B a b^{3} - A^{3} b^{4}\right )} x\right ) - 5 \, a^{2} x^{5} \left (-\frac {B^{4} a^{4} b - 4 \, A B^{3} a^{3} b^{2} + 6 \, A^{2} B^{2} a^{2} b^{3} - 4 \, A^{3} B a b^{4} + A^{4} b^{5}}{a^{9}}\right )^{\frac {1}{4}} \log \left (-a^{7} \left (-\frac {B^{4} a^{4} b - 4 \, A B^{3} a^{3} b^{2} + 6 \, A^{2} B^{2} a^{2} b^{3} - 4 \, A^{3} B a b^{4} + A^{4} b^{5}}{a^{9}}\right )^{\frac {3}{4}} - {\left (B^{3} a^{3} b - 3 \, A B^{2} a^{2} b^{2} + 3 \, A^{2} B a b^{3} - A^{3} b^{4}\right )} x\right ) - 20 \, {\left (B a - A b\right )} x^{4} - 4 \, A a}{20 \, a^{2} x^{5}} \] Input:

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

Output:

1/20*(5*a^2*x^5*(-(B^4*a^4*b - 4*A*B^3*a^3*b^2 + 6*A^2*B^2*a^2*b^3 - 4*A^3 
*B*a*b^4 + A^4*b^5)/a^9)^(1/4)*log(a^7*(-(B^4*a^4*b - 4*A*B^3*a^3*b^2 + 6* 
A^2*B^2*a^2*b^3 - 4*A^3*B*a*b^4 + A^4*b^5)/a^9)^(3/4) - (B^3*a^3*b - 3*A*B 
^2*a^2*b^2 + 3*A^2*B*a*b^3 - A^3*b^4)*x) - 5*I*a^2*x^5*(-(B^4*a^4*b - 4*A* 
B^3*a^3*b^2 + 6*A^2*B^2*a^2*b^3 - 4*A^3*B*a*b^4 + A^4*b^5)/a^9)^(1/4)*log( 
I*a^7*(-(B^4*a^4*b - 4*A*B^3*a^3*b^2 + 6*A^2*B^2*a^2*b^3 - 4*A^3*B*a*b^4 + 
 A^4*b^5)/a^9)^(3/4) - (B^3*a^3*b - 3*A*B^2*a^2*b^2 + 3*A^2*B*a*b^3 - A^3* 
b^4)*x) + 5*I*a^2*x^5*(-(B^4*a^4*b - 4*A*B^3*a^3*b^2 + 6*A^2*B^2*a^2*b^3 - 
 4*A^3*B*a*b^4 + A^4*b^5)/a^9)^(1/4)*log(-I*a^7*(-(B^4*a^4*b - 4*A*B^3*a^3 
*b^2 + 6*A^2*B^2*a^2*b^3 - 4*A^3*B*a*b^4 + A^4*b^5)/a^9)^(3/4) - (B^3*a^3* 
b - 3*A*B^2*a^2*b^2 + 3*A^2*B*a*b^3 - A^3*b^4)*x) - 5*a^2*x^5*(-(B^4*a^4*b 
 - 4*A*B^3*a^3*b^2 + 6*A^2*B^2*a^2*b^3 - 4*A^3*B*a*b^4 + A^4*b^5)/a^9)^(1/ 
4)*log(-a^7*(-(B^4*a^4*b - 4*A*B^3*a^3*b^2 + 6*A^2*B^2*a^2*b^3 - 4*A^3*B*a 
*b^4 + A^4*b^5)/a^9)^(3/4) - (B^3*a^3*b - 3*A*B^2*a^2*b^2 + 3*A^2*B*a*b^3 
- A^3*b^4)*x) - 20*(B*a - A*b)*x^4 - 4*A*a)/(a^2*x^5)
 

Sympy [A] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.78 \[ \int \frac {A+B x^4}{x^6 \left (a+b x^4\right )} \, dx=\operatorname {RootSum} {\left (256 t^{4} a^{9} + A^{4} b^{5} - 4 A^{3} B a b^{4} + 6 A^{2} B^{2} a^{2} b^{3} - 4 A B^{3} a^{3} b^{2} + B^{4} a^{4} b, \left ( t \mapsto t \log {\left (- \frac {64 t^{3} a^{7}}{- A^{3} b^{4} + 3 A^{2} B a b^{3} - 3 A B^{2} a^{2} b^{2} + B^{3} a^{3} b} + x \right )} \right )\right )} + \frac {- A a + x^{4} \cdot \left (5 A b - 5 B a\right )}{5 a^{2} x^{5}} \] Input:

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

Output:

RootSum(256*_t**4*a**9 + A**4*b**5 - 4*A**3*B*a*b**4 + 6*A**2*B**2*a**2*b* 
*3 - 4*A*B**3*a**3*b**2 + B**4*a**4*b, Lambda(_t, _t*log(-64*_t**3*a**7/(- 
A**3*b**4 + 3*A**2*B*a*b**3 - 3*A*B**2*a**2*b**2 + B**3*a**3*b) + x))) + ( 
-A*a + x**4*(5*A*b - 5*B*a))/(5*a**2*x**5)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.14 \[ \int \frac {A+B x^4}{x^6 \left (a+b x^4\right )} \, dx=-\frac {{\left (B a b - A b^{2}\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {b} x^{2} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {b} x^{2} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{8 \, a^{2}} - \frac {5 \, {\left (B a - A b\right )} x^{4} + A a}{5 \, a^{2} x^{5}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.44 \[ \int \frac {A+B x^4}{x^6 \left (a+b x^4\right )} \, dx=-\frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} B a - \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{4 \, a^{3} b^{2}} - \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} B a - \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{4 \, a^{3} b^{2}} + \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} B a - \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{8 \, a^{3} b^{2}} - \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} B a - \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{8 \, a^{3} b^{2}} - \frac {5 \, B a x^{4} - 5 \, A b x^{4} + A a}{5 \, a^{2} x^{5}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.47 \[ \int \frac {A+B x^4}{x^6 \left (a+b x^4\right )} \, dx=\frac {{\left (-b\right )}^{1/4}\,\mathrm {atan}\left (\frac {{\left (-b\right )}^{1/4}\,x}{a^{1/4}}\right )\,\left (A\,b-B\,a\right )}{2\,a^{9/4}}-\frac {\frac {A}{5\,a}-\frac {x^4\,\left (A\,b-B\,a\right )}{a^2}}{x^5}-\frac {{\left (-b\right )}^{1/4}\,\mathrm {atanh}\left (\frac {{\left (-b\right )}^{1/4}\,x}{a^{1/4}}\right )\,\left (A\,b-B\,a\right )}{2\,a^{9/4}} \] Input:

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

Output:

((-b)^(1/4)*atan(((-b)^(1/4)*x)/a^(1/4))*(A*b - B*a))/(2*a^(9/4)) - (A/(5* 
a) - (x^4*(A*b - B*a))/a^2)/x^5 - ((-b)^(1/4)*atanh(((-b)^(1/4)*x)/a^(1/4) 
)*(A*b - B*a))/(2*a^(9/4))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.03 \[ \int \frac {A+B x^4}{x^6 \left (a+b x^4\right )} \, dx=-\frac {1}{5 x^{5}} \] Input:

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

Output:

( - 1)/(5*x**5)