\(\int \frac {x^6 (A+B x^4)}{(a+b x^4)^2} \, dx\) [18]

Optimal result

Integrand size = 20, antiderivative size = 199 \[ \int \frac {x^6 \left (A+B x^4\right )}{\left (a+b x^4\right )^2} \, dx=\frac {B x^3}{3 b^2}-\frac {(A b-a B) x^3}{4 b^2 \left (a+b x^4\right )}-\frac {(3 A b-7 a B) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} \sqrt [4]{a} b^{11/4}}+\frac {(3 A b-7 a B) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{8 \sqrt {2} \sqrt [4]{a} b^{11/4}}-\frac {(3 A b-7 a B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x}{\sqrt {a}+\sqrt {b} x^2}\right )}{8 \sqrt {2} \sqrt [4]{a} b^{11/4}} \] Output:

1/3*B*x^3/b^2-1/4*(A*b-B*a)*x^3/b^2/(b*x^4+a)+1/16*(3*A*b-7*B*a)*arctan(-1 
+2^(1/2)*b^(1/4)*x/a^(1/4))*2^(1/2)/a^(1/4)/b^(11/4)+1/16*(3*A*b-7*B*a)*ar 
ctan(1+2^(1/2)*b^(1/4)*x/a^(1/4))*2^(1/2)/a^(1/4)/b^(11/4)-1/16*(3*A*b-7*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^(1/ 
4)/b^(11/4)
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.21 \[ \int \frac {x^6 \left (A+B x^4\right )}{\left (a+b x^4\right )^2} \, dx=\frac {32 b^{3/4} B x^3-\frac {24 b^{3/4} (A b-a B) x^3}{a+b x^4}+\frac {6 \sqrt {2} (-3 A b+7 a B) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{\sqrt [4]{a}}+\frac {6 \sqrt {2} (3 A b-7 a B) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{\sqrt [4]{a}}+\frac {3 \sqrt {2} (3 A b-7 a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{\sqrt [4]{a}}+\frac {3 \sqrt {2} (-3 A b+7 a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {b} x^2\right )}{\sqrt [4]{a}}}{96 b^{11/4}} \] Input:

Integrate[(x^6*(A + B*x^4))/(a + b*x^4)^2,x]
 

Output:

(32*b^(3/4)*B*x^3 - (24*b^(3/4)*(A*b - a*B)*x^3)/(a + b*x^4) + (6*Sqrt[2]* 
(-3*A*b + 7*a*B)*ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/a^(1/4) + (6*Sqr 
t[2]*(3*A*b - 7*a*B)*ArcTan[1 + (Sqrt[2]*b^(1/4)*x)/a^(1/4)])/a^(1/4) + (3 
*Sqrt[2]*(3*A*b - 7*a*B)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*x + Sqrt[b] 
*x^2])/a^(1/4) + (3*Sqrt[2]*(-3*A*b + 7*a*B)*Log[Sqrt[a] + Sqrt[2]*a^(1/4) 
*b^(1/4)*x + Sqrt[b]*x^2])/a^(1/4))/(96*b^(11/4))
 

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.34, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {957, 843, 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.

\(\displaystyle \int \frac {x^6 \left (A+B x^4\right )}{\left (a+b x^4\right )^2} \, dx\)

\(\Big \downarrow \) 957

\(\displaystyle \frac {x^7 (A b-a B)}{4 a b \left (a+b x^4\right )}-\frac {(3 A b-7 a B) \int \frac {x^6}{b x^4+a}dx}{4 a b}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {x^7 (A b-a B)}{4 a b \left (a+b x^4\right )}-\frac {(3 A b-7 a B) \left (\frac {x^3}{3 b}-\frac {a \int \frac {x^2}{b x^4+a}dx}{b}\right )}{4 a b}\)

\(\Big \downarrow \) 826

\(\displaystyle \frac {x^7 (A b-a B)}{4 a b \left (a+b x^4\right )}-\frac {(3 A b-7 a B) \left (\frac {x^3}{3 b}-\frac {a \left (\frac {\int \frac {\sqrt {b} x^2+\sqrt {a}}{b x^4+a}dx}{2 \sqrt {b}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x^2}{b x^4+a}dx}{2 \sqrt {b}}\right )}{b}\right )}{4 a b}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {x^7 (A b-a B)}{4 a b \left (a+b x^4\right )}-\frac {(3 A b-7 a B) \left (\frac {x^3}{3 b}-\frac {a \left (\frac {\frac {\int \frac {1}{x^2-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}}dx}{2 \sqrt {b}}+\frac {\int \frac {1}{x^2+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}}dx}{2 \sqrt {b}}}{2 \sqrt {b}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x^2}{b x^4+a}dx}{2 \sqrt {b}}\right )}{b}\right )}{4 a b}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {x^7 (A b-a B)}{4 a b \left (a+b x^4\right )}-\frac {(3 A b-7 a B) \left (\frac {x^3}{3 b}-\frac {a \left (\frac {\frac {\int \frac {1}{-\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )^2-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\int \frac {1}{-\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )^2-1}d\left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {b}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x^2}{b x^4+a}dx}{2 \sqrt {b}}\right )}{b}\right )}{4 a b}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {x^7 (A b-a B)}{4 a b \left (a+b x^4\right )}-\frac {(3 A b-7 a B) \left (\frac {x^3}{3 b}-\frac {a \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {b}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x^2}{b x^4+a}dx}{2 \sqrt {b}}\right )}{b}\right )}{4 a b}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {x^7 (A b-a B)}{4 a b \left (a+b x^4\right )}-\frac {(3 A b-7 a B) \left (\frac {x^3}{3 b}-\frac {a \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {b}}-\frac {-\frac {\int -\frac {\sqrt {2} \sqrt [4]{a}-2 \sqrt [4]{b} x}{\sqrt [4]{b} \left (x^2-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}\right )}dx}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{b} x+\sqrt [4]{a}\right )}{\sqrt [4]{b} \left (x^2+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}\right )}dx}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {b}}\right )}{b}\right )}{4 a b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {x^7 (A b-a B)}{4 a b \left (a+b x^4\right )}-\frac {(3 A b-7 a B) \left (\frac {x^3}{3 b}-\frac {a \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {b}}-\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{a}-2 \sqrt [4]{b} x}{\sqrt [4]{b} \left (x^2-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}\right )}dx}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{b} x+\sqrt [4]{a}\right )}{\sqrt [4]{b} \left (x^2+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}\right )}dx}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {b}}\right )}{b}\right )}{4 a b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x^7 (A b-a B)}{4 a b \left (a+b x^4\right )}-\frac {(3 A b-7 a B) \left (\frac {x^3}{3 b}-\frac {a \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {b}}-\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{a}-2 \sqrt [4]{b} x}{x^2-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}}dx}{2 \sqrt {2} \sqrt [4]{a} \sqrt {b}}+\frac {\int \frac {\sqrt {2} \sqrt [4]{b} x+\sqrt [4]{a}}{x^2+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+\frac {\sqrt {a}}{\sqrt {b}}}dx}{2 \sqrt [4]{a} \sqrt {b}}}{2 \sqrt {b}}\right )}{b}\right )}{4 a b}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {x^7 (A b-a B)}{4 a b \left (a+b x^4\right )}-\frac {(3 A b-7 a B) \left (\frac {x^3}{3 b}-\frac {a \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}+1\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {b}}-\frac {\frac {\log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}-\frac {\log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} x+\sqrt {a}+\sqrt {b} x^2\right )}{2 \sqrt {2} \sqrt [4]{a} \sqrt [4]{b}}}{2 \sqrt {b}}\right )}{b}\right )}{4 a b}\)

Input:

Int[(x^6*(A + B*x^4))/(a + b*x^4)^2,x]
 

Output:

((A*b - a*B)*x^7)/(4*a*b*(a + b*x^4)) - ((3*A*b - 7*a*B)*(x^3/(3*b) - (a*( 
(-(ArcTan[1 - (Sqrt[2]*b^(1/4)*x)/a^(1/4)]/(Sqrt[2]*a^(1/4)*b^(1/4))) + Ar 
cTan[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*Sqrt[2]*a^(1/4)*b^(1/4)))/(2*Sqrt[b])))/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 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 = 0.09 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.36

method	result	size
risch	\(\frac {B \,x^{3}}{3 b^{2}}+\frac {\left (-\frac {A b}{4}+\frac {B a}{4}\right ) x^{3}}{b^{2} \left (b \,x^{4}+a \right )}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{4}+a \right )}{\sum }\frac {\left (3 A b -7 B a \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}}}{16 b^{3}}\)	\(71\)
default	\(\frac {B \,x^{3}}{3 b^{2}}+\frac {\frac {\left (-\frac {A b}{4}+\frac {B a}{4}\right ) x^{3}}{b \,x^{4}+a}+\frac {\left (-\frac {7 B a}{4}+\frac {3 A b}{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}{x^{2}+\left (\frac {a}{b}\right )^{\frac {1}{4}} x \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{8 b \left (\frac {a}{b}\right )^{\frac {1}{4}}}}{b^{2}}\)	\(148\)

Input:

int(x^6*(B*x^4+A)/(b*x^4+a)^2,x,method=_RETURNVERBOSE)
 

Output:

1/3*B*x^3/b^2+(-1/4*A*b+1/4*B*a)*x^3/b^2/(b*x^4+a)+1/16/b^3*sum((3*A*b-7*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 = 0.09 (sec) , antiderivative size = 782, normalized size of antiderivative = 3.93 \[ \int \frac {x^6 \left (A+B x^4\right )}{\left (a+b x^4\right )^2} \, dx =\text {Too large to display} \] Input:

integrate(x^6*(B*x^4+A)/(b*x^4+a)^2,x, algorithm="fricas")
 

Output:

1/48*(16*B*b*x^7 + 4*(7*B*a - 3*A*b)*x^3 + 3*(b^3*x^4 + a*b^2)*(-(2401*B^4 
*a^4 - 4116*A*B^3*a^3*b + 2646*A^2*B^2*a^2*b^2 - 756*A^3*B*a*b^3 + 81*A^4* 
b^4)/(a*b^11))^(1/4)*log(a*b^8*(-(2401*B^4*a^4 - 4116*A*B^3*a^3*b + 2646*A 
^2*B^2*a^2*b^2 - 756*A^3*B*a*b^3 + 81*A^4*b^4)/(a*b^11))^(3/4) - (343*B^3* 
a^3 - 441*A*B^2*a^2*b + 189*A^2*B*a*b^2 - 27*A^3*b^3)*x) - 3*(I*b^3*x^4 + 
I*a*b^2)*(-(2401*B^4*a^4 - 4116*A*B^3*a^3*b + 2646*A^2*B^2*a^2*b^2 - 756*A 
^3*B*a*b^3 + 81*A^4*b^4)/(a*b^11))^(1/4)*log(I*a*b^8*(-(2401*B^4*a^4 - 411 
6*A*B^3*a^3*b + 2646*A^2*B^2*a^2*b^2 - 756*A^3*B*a*b^3 + 81*A^4*b^4)/(a*b^ 
11))^(3/4) - (343*B^3*a^3 - 441*A*B^2*a^2*b + 189*A^2*B*a*b^2 - 27*A^3*b^3 
)*x) - 3*(-I*b^3*x^4 - I*a*b^2)*(-(2401*B^4*a^4 - 4116*A*B^3*a^3*b + 2646* 
A^2*B^2*a^2*b^2 - 756*A^3*B*a*b^3 + 81*A^4*b^4)/(a*b^11))^(1/4)*log(-I*a*b 
^8*(-(2401*B^4*a^4 - 4116*A*B^3*a^3*b + 2646*A^2*B^2*a^2*b^2 - 756*A^3*B*a 
*b^3 + 81*A^4*b^4)/(a*b^11))^(3/4) - (343*B^3*a^3 - 441*A*B^2*a^2*b + 189* 
A^2*B*a*b^2 - 27*A^3*b^3)*x) - 3*(b^3*x^4 + a*b^2)*(-(2401*B^4*a^4 - 4116* 
A*B^3*a^3*b + 2646*A^2*B^2*a^2*b^2 - 756*A^3*B*a*b^3 + 81*A^4*b^4)/(a*b^11 
))^(1/4)*log(-a*b^8*(-(2401*B^4*a^4 - 4116*A*B^3*a^3*b + 2646*A^2*B^2*a^2* 
b^2 - 756*A^3*B*a*b^3 + 81*A^4*b^4)/(a*b^11))^(3/4) - (343*B^3*a^3 - 441*A 
*B^2*a^2*b + 189*A^2*B*a*b^2 - 27*A^3*b^3)*x))/(b^3*x^4 + a*b^2)
 

Sympy [A] (verification not implemented)

Time = 0.91 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.78 \[ \int \frac {x^6 \left (A+B x^4\right )}{\left (a+b x^4\right )^2} \, dx=\frac {B x^{3}}{3 b^{2}} + \frac {x^{3} \left (- A b + B a\right )}{4 a b^{2} + 4 b^{3} x^{4}} + \operatorname {RootSum} {\left (65536 t^{4} a b^{11} + 81 A^{4} b^{4} - 756 A^{3} B a b^{3} + 2646 A^{2} B^{2} a^{2} b^{2} - 4116 A B^{3} a^{3} b + 2401 B^{4} a^{4}, \left ( t \mapsto t \log {\left (- \frac {4096 t^{3} a b^{8}}{- 27 A^{3} b^{3} + 189 A^{2} B a b^{2} - 441 A B^{2} a^{2} b + 343 B^{3} a^{3}} + x \right )} \right )\right )} \] Input:

integrate(x**6*(B*x**4+A)/(b*x**4+a)**2,x)
                                                                                    
                                                                                    
 

Output:

B*x**3/(3*b**2) + x**3*(-A*b + B*a)/(4*a*b**2 + 4*b**3*x**4) + RootSum(655 
36*_t**4*a*b**11 + 81*A**4*b**4 - 756*A**3*B*a*b**3 + 2646*A**2*B**2*a**2* 
b**2 - 4116*A*B**3*a**3*b + 2401*B**4*a**4, Lambda(_t, _t*log(-4096*_t**3* 
a*b**8/(-27*A**3*b**3 + 189*A**2*B*a*b**2 - 441*A*B**2*a**2*b + 343*B**3*a 
**3) + x)))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.11 \[ \int \frac {x^6 \left (A+B x^4\right )}{\left (a+b x^4\right )^2} \, dx=\frac {{\left (B a - A b\right )} x^{3}}{4 \, {\left (b^{3} x^{4} + a b^{2}\right )}} + \frac {B x^{3}}{3 \, b^{2}} - \frac {{\left (7 \, B a - 3 \, A b\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (2 \, \sqrt {b} x - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {b} x^{2} + \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (\sqrt {b} x^{2} - \sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{32 \, b^{2}} \] Input:

integrate(x^6*(B*x^4+A)/(b*x^4+a)^2,x, algorithm="maxima")
 

Output:

1/4*(B*a - A*b)*x^3/(b^3*x^4 + a*b^2) + 1/3*B*x^3/b^2 - 1/32*(7*B*a - 3*A* 
b)*(2*sqrt(2)*arctan(1/2*sqrt(2)*(2*sqrt(b)*x + sqrt(2)*a^(1/4)*b^(1/4))/s 
qrt(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(sqrt(b)*x^2 - sqrt( 
2)*a^(1/4)*b^(1/4)*x + sqrt(a))/(a^(1/4)*b^(3/4)))/b^2
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.41 \[ \int \frac {x^6 \left (A+B x^4\right )}{\left (a+b x^4\right )^2} \, dx=\frac {B x^{3}}{3 \, b^{2}} + \frac {B a x^{3} - A b x^{3}}{4 \, {\left (b x^{4} + a\right )} b^{2}} - \frac {\sqrt {2} {\left (7 \, \left (a b^{3}\right )^{\frac {3}{4}} B a - 3 \, \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{16 \, a b^{5}} - \frac {\sqrt {2} {\left (7 \, \left (a b^{3}\right )^{\frac {3}{4}} B a - 3 \, \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{16 \, a b^{5}} + \frac {\sqrt {2} {\left (7 \, \left (a b^{3}\right )^{\frac {3}{4}} B a - 3 \, \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \log \left (x^{2} + \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{32 \, a b^{5}} - \frac {\sqrt {2} {\left (7 \, \left (a b^{3}\right )^{\frac {3}{4}} B a - 3 \, \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \log \left (x^{2} - \sqrt {2} x \left (\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {\frac {a}{b}}\right )}{32 \, a b^{5}} \] Input:

integrate(x^6*(B*x^4+A)/(b*x^4+a)^2,x, algorithm="giac")
 

Output:

1/3*B*x^3/b^2 + 1/4*(B*a*x^3 - A*b*x^3)/((b*x^4 + a)*b^2) - 1/16*sqrt(2)*( 
7*(a*b^3)^(3/4)*B*a - 3*(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*b^5) - 1/16*sqrt(2)*(7*(a*b^3)^(3/4)*B*a - 
 3*(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*b^5) + 1/32*sqrt(2)*(7*(a*b^3)^(3/4)*B*a - 3*(a*b^3)^(3/4)*A*b) 
*log(x^2 + sqrt(2)*x*(a/b)^(1/4) + sqrt(a/b))/(a*b^5) - 1/32*sqrt(2)*(7*(a 
*b^3)^(3/4)*B*a - 3*(a*b^3)^(3/4)*A*b)*log(x^2 - sqrt(2)*x*(a/b)^(1/4) + s 
qrt(a/b))/(a*b^5)
 

Mupad [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.51 \[ \int \frac {x^6 \left (A+B x^4\right )}{\left (a+b x^4\right )^2} \, dx=\frac {B\,x^3}{3\,b^2}-\frac {x^3\,\left (\frac {A\,b}{4}-\frac {B\,a}{4}\right )}{b^3\,x^4+a\,b^2}+\frac {\mathrm {atan}\left (\frac {b^{1/4}\,x}{{\left (-a\right )}^{1/4}}\right )\,\left (3\,A\,b-7\,B\,a\right )}{8\,{\left (-a\right )}^{1/4}\,b^{11/4}}+\frac {\mathrm {atan}\left (\frac {b^{1/4}\,x\,1{}\mathrm {i}}{{\left (-a\right )}^{1/4}}\right )\,\left (3\,A\,b-7\,B\,a\right )\,1{}\mathrm {i}}{8\,{\left (-a\right )}^{1/4}\,b^{11/4}} \] Input:

int((x^6*(A + B*x^4))/(a + b*x^4)^2,x)
 

Output:

(B*x^3)/(3*b^2) - (x^3*((A*b)/4 - (B*a)/4))/(a*b^2 + b^3*x^4) + (atan((b^( 
1/4)*x)/(-a)^(1/4))*(3*A*b - 7*B*a))/(8*(-a)^(1/4)*b^(11/4)) + (atan((b^(1 
/4)*x*1i)/(-a)^(1/4))*(3*A*b - 7*B*a)*1i)/(8*(-a)^(1/4)*b^(11/4))
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.74 \[ \int \frac {x^6 \left (A+B x^4\right )}{\left (a+b x^4\right )^2} \, dx=\frac {6 b^{\frac {1}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}-2 \sqrt {b}\, x}{b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right )-6 b^{\frac {1}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+2 \sqrt {b}\, x}{b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right )-3 b^{\frac {1}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (-b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {a}+\sqrt {b}\, x^{2}\right )+3 b^{\frac {1}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}\, x +\sqrt {a}+\sqrt {b}\, x^{2}\right )+8 b \,x^{3}}{24 b^{2}} \] Input:

int(x^6*(B*x^4+A)/(b*x^4+a)^2,x)
 

Output:

(6*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))) - 6*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))) - 3*b**(1/4 
)*a**(3/4)*sqrt(2)*log( - b**(1/4)*a**(1/4)*sqrt(2)*x + sqrt(a) + sqrt(b)* 
x**2) + 3*b**(1/4)*a**(3/4)*sqrt(2)*log(b**(1/4)*a**(1/4)*sqrt(2)*x + sqrt 
(a) + sqrt(b)*x**2) + 8*b*x**3)/(24*b**2)