\(\int \frac {x^{5/2} (A+B x^2)}{(b x^2+c x^4)^3} \, dx\) [150]

Optimal result

Integrand size = 26, antiderivative size = 263 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {2 A}{5 b^3 x^{5/2}}-\frac {2 (b B-3 A c)}{b^4 \sqrt {x}}-\frac {c (b B-A c) x^{3/2}}{4 b^3 \left (b+c x^2\right )^2}-\frac {c (13 b B-21 A c) x^{3/2}}{16 b^4 \left (b+c x^2\right )}+\frac {9 \sqrt [4]{c} (5 b B-13 A c) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{17/4}}-\frac {9 \sqrt [4]{c} (5 b B-13 A c) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{17/4}}+\frac {9 \sqrt [4]{c} (5 b B-13 A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{32 \sqrt {2} b^{17/4}} \] Output:

-2/5*A/b^3/x^(5/2)-2*(-3*A*c+B*b)/b^4/x^(1/2)-1/4*c*(-A*c+B*b)*x^(3/2)/b^3 
/(c*x^2+b)^2-1/16*c*(-21*A*c+13*B*b)*x^(3/2)/b^4/(c*x^2+b)+9/64*c^(1/4)*(- 
13*A*c+5*B*b)*arctan(1-2^(1/2)*c^(1/4)*x^(1/2)/b^(1/4))*2^(1/2)/b^(17/4)-9 
/64*c^(1/4)*(-13*A*c+5*B*b)*arctan(1+2^(1/2)*c^(1/4)*x^(1/2)/b^(1/4))*2^(1 
/2)/b^(17/4)+9/64*c^(1/4)*(-13*A*c+5*B*b)*arctanh(2^(1/2)*b^(1/4)*c^(1/4)* 
x^(1/2)/(b^(1/2)+c^(1/2)*x))*2^(1/2)/b^(17/4)
 

Mathematica [A] (verified)

Time = 0.67 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.79 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {-\frac {4 \sqrt [4]{b} \left (5 b B x^2 \left (32 b^2+81 b c x^2+45 c^2 x^4\right )+A \left (32 b^3-416 b^2 c x^2-1053 b c^2 x^4-585 c^3 x^6\right )\right )}{x^{5/2} \left (b+c x^2\right )^2}+45 \sqrt {2} \sqrt [4]{c} (5 b B-13 A c) \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )+45 \sqrt {2} \sqrt [4]{c} (5 b B-13 A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{320 b^{17/4}} \] Input:

Integrate[(x^(5/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]
 

Output:

((-4*b^(1/4)*(5*b*B*x^2*(32*b^2 + 81*b*c*x^2 + 45*c^2*x^4) + A*(32*b^3 - 4 
16*b^2*c*x^2 - 1053*b*c^2*x^4 - 585*c^3*x^6)))/(x^(5/2)*(b + c*x^2)^2) + 4 
5*Sqrt[2]*c^(1/4)*(5*b*B - 13*A*c)*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b 
^(1/4)*c^(1/4)*Sqrt[x])] + 45*Sqrt[2]*c^(1/4)*(5*b*B - 13*A*c)*ArcTanh[(Sq 
rt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(320*b^(17/4))
 

Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 329, normalized size of antiderivative = 1.25, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {9, 362, 253, 264, 264, 266, 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.

\(\displaystyle \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx\)

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 362

\(\displaystyle -\frac {(5 b B-13 A c) \int \frac {1}{x^{7/2} \left (c x^2+b\right )^2}dx}{8 b c}-\frac {b B-A c}{4 b c x^{5/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 253

\(\displaystyle -\frac {(5 b B-13 A c) \left (\frac {9 \int \frac {1}{x^{7/2} \left (c x^2+b\right )}dx}{4 b}+\frac {1}{2 b x^{5/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{5/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 264

\(\displaystyle -\frac {(5 b B-13 A c) \left (\frac {9 \left (-\frac {c \int \frac {1}{x^{3/2} \left (c x^2+b\right )}dx}{b}-\frac {2}{5 b x^{5/2}}\right )}{4 b}+\frac {1}{2 b x^{5/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{5/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 264

\(\displaystyle -\frac {(5 b B-13 A c) \left (\frac {9 \left (-\frac {c \left (-\frac {c \int \frac {\sqrt {x}}{c x^2+b}dx}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{4 b}+\frac {1}{2 b x^{5/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{5/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 266

\(\displaystyle -\frac {(5 b B-13 A c) \left (\frac {9 \left (-\frac {c \left (-\frac {2 c \int \frac {x}{c x^2+b}d\sqrt {x}}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{4 b}+\frac {1}{2 b x^{5/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{5/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 826

\(\displaystyle -\frac {(5 b B-13 A c) \left (\frac {9 \left (-\frac {c \left (-\frac {2 c \left (\frac {\int \frac {\sqrt {c} x+\sqrt {b}}{c x^2+b}d\sqrt {x}}{2 \sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{4 b}+\frac {1}{2 b x^{5/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{5/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 1476

\(\displaystyle -\frac {(5 b B-13 A c) \left (\frac {9 \left (-\frac {c \left (-\frac {2 c \left (\frac {\frac {\int \frac {1}{x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {c}}+\frac {\int \frac {1}{x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {c}}}{2 \sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{4 b}+\frac {1}{2 b x^{5/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{5/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 1082

\(\displaystyle -\frac {(5 b B-13 A c) \left (\frac {9 \left (-\frac {c \left (-\frac {2 c \left (\frac {\frac {\int \frac {1}{-x-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\int \frac {1}{-x-1}d\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{4 b}+\frac {1}{2 b x^{5/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{5/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 217

\(\displaystyle -\frac {(5 b B-13 A c) \left (\frac {9 \left (-\frac {c \left (-\frac {2 c \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{4 b}+\frac {1}{2 b x^{5/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{5/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 1479

\(\displaystyle -\frac {(5 b B-13 A c) \left (\frac {9 \left (-\frac {c \left (-\frac {2 c \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {-\frac {\int -\frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{c} \left (x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}\right )}{\sqrt [4]{c} \left (x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{4 b}+\frac {1}{2 b x^{5/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{5/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {(5 b B-13 A c) \left (\frac {9 \left (-\frac {c \left (-\frac {2 c \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{c} \left (x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}\right )}{\sqrt [4]{c} \left (x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{4 b}+\frac {1}{2 b x^{5/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{5/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {(5 b B-13 A c) \left (\frac {9 \left (-\frac {c \left (-\frac {2 c \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt {c}}+\frac {\int \frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}}{x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt [4]{b} \sqrt {c}}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{4 b}+\frac {1}{2 b x^{5/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{5/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 1103

\(\displaystyle -\frac {(5 b B-13 A c) \left (\frac {9 \left (-\frac {c \left (-\frac {2 c \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\frac {\log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{4 b}+\frac {1}{2 b x^{5/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{5/2} \left (b+c x^2\right )^2}\)

Input:

Int[(x^(5/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]
 

Output:

-1/4*(b*B - A*c)/(b*c*x^(5/2)*(b + c*x^2)^2) - ((5*b*B - 13*A*c)*(1/(2*b*x 
^(5/2)*(b + c*x^2)) + (9*(-2/(5*b*x^(5/2)) - (c*(-2/(b*Sqrt[x]) - (2*c*((- 
(ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)]/(Sqrt[2]*b^(1/4)*c^(1/4))) 
+ ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)]/(Sqrt[2]*b^(1/4)*c^(1/4))) 
/(2*Sqrt[c]) - (-1/2*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[ 
c]*x]/(Sqrt[2]*b^(1/4)*c^(1/4)) + Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sq 
rt[x] + Sqrt[c]*x]/(2*Sqrt[2]*b^(1/4)*c^(1/4)))/(2*Sqrt[c])))/b))/b))/(4*b 
)))/(8*b*c)
 

Defintions of rubi rules used

rule 9

Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, 
x, Min]}, Simp[1/e^(p*r)   Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, 
x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && 
  !MonomialQ[Px, x]
 

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 253

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(c*x 
)^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 
2*a*(p + 1))   Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m 
}, x] && LtQ[p, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 264

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 362

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*b*e 
*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(2*a*b*(p + 1))   I 
nt[(e*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && N 
eQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) || 
  !RationalQ[m] || (ILtQ[p + 1/2, 0] && LeQ[-1, m, -2*(p + 1)]))
 

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 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 = 0.44 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.72

method	result	size
derivativedivides	\(\frac {2 c \left (\frac {\left (\frac {21}{32} A \,c^{2}-\frac {13}{32} B b c \right ) x^{\frac {7}{2}}+\frac {b \left (25 A c -17 B b \right ) x^{\frac {3}{2}}}{32}}{\left (c \,x^{2}+b \right )^{2}}+\frac {\left (\frac {117 A c}{32}-\frac {45 B b}{32}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{8 c \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{b^{4}}-\frac {2 A}{5 b^{3} x^{\frac {5}{2}}}-\frac {2 \left (-3 A c +B b \right )}{b^{4} \sqrt {x}}\)	\(190\)
default	\(\frac {2 c \left (\frac {\left (\frac {21}{32} A \,c^{2}-\frac {13}{32} B b c \right ) x^{\frac {7}{2}}+\frac {b \left (25 A c -17 B b \right ) x^{\frac {3}{2}}}{32}}{\left (c \,x^{2}+b \right )^{2}}+\frac {\left (\frac {117 A c}{32}-\frac {45 B b}{32}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{8 c \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{b^{4}}-\frac {2 A}{5 b^{3} x^{\frac {5}{2}}}-\frac {2 \left (-3 A c +B b \right )}{b^{4} \sqrt {x}}\)	\(190\)
risch	\(-\frac {2 \left (-15 A c \,x^{2}+5 B b \,x^{2}+A b \right )}{5 b^{4} x^{\frac {5}{2}}}+\frac {c \left (\frac {2 \left (\frac {21}{32} A \,c^{2}-\frac {13}{32} B b c \right ) x^{\frac {7}{2}}+\frac {b \left (25 A c -17 B b \right ) x^{\frac {3}{2}}}{16}}{\left (c \,x^{2}+b \right )^{2}}+\frac {\left (\frac {117 A c}{32}-\frac {45 B b}{32}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 c \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{b^{4}}\)	\(191\)

Input:

int(x^(5/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)
 

Output:

2/b^4*c*(((21/32*A*c^2-13/32*B*b*c)*x^(7/2)+1/32*b*(25*A*c-17*B*b)*x^(3/2) 
)/(c*x^2+b)^2+1/8*(117/32*A*c-45/32*B*b)/c/(b/c)^(1/4)*2^(1/2)*(ln((x-(b/c 
)^(1/4)*x^(1/2)*2^(1/2)+(b/c)^(1/2))/(x+(b/c)^(1/4)*x^(1/2)*2^(1/2)+(b/c)^ 
(1/2)))+2*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+2*arctan(2^(1/2)/(b/c)^(1/ 
4)*x^(1/2)-1)))-2/5*A/b^3/x^(5/2)-2*(-3*A*c+B*b)/b^4/x^(1/2)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 918, normalized size of antiderivative = 3.49 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx =\text {Too large to display} \] Input:

integrate(x^(5/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="fricas")
 

Output:

1/320*(45*(b^4*c^2*x^7 + 2*b^5*c*x^5 + b^6*x^3)*(-(625*B^4*b^4*c - 6500*A* 
B^3*b^3*c^2 + 25350*A^2*B^2*b^2*c^3 - 43940*A^3*B*b*c^4 + 28561*A^4*c^5)/b 
^17)^(1/4)*log(729*b^13*(-(625*B^4*b^4*c - 6500*A*B^3*b^3*c^2 + 25350*A^2* 
B^2*b^2*c^3 - 43940*A^3*B*b*c^4 + 28561*A^4*c^5)/b^17)^(3/4) - 729*(125*B^ 
3*b^3*c - 975*A*B^2*b^2*c^2 + 2535*A^2*B*b*c^3 - 2197*A^3*c^4)*sqrt(x)) - 
45*(I*b^4*c^2*x^7 + 2*I*b^5*c*x^5 + I*b^6*x^3)*(-(625*B^4*b^4*c - 6500*A*B 
^3*b^3*c^2 + 25350*A^2*B^2*b^2*c^3 - 43940*A^3*B*b*c^4 + 28561*A^4*c^5)/b^ 
17)^(1/4)*log(729*I*b^13*(-(625*B^4*b^4*c - 6500*A*B^3*b^3*c^2 + 25350*A^2 
*B^2*b^2*c^3 - 43940*A^3*B*b*c^4 + 28561*A^4*c^5)/b^17)^(3/4) - 729*(125*B 
^3*b^3*c - 975*A*B^2*b^2*c^2 + 2535*A^2*B*b*c^3 - 2197*A^3*c^4)*sqrt(x)) - 
 45*(-I*b^4*c^2*x^7 - 2*I*b^5*c*x^5 - I*b^6*x^3)*(-(625*B^4*b^4*c - 6500*A 
*B^3*b^3*c^2 + 25350*A^2*B^2*b^2*c^3 - 43940*A^3*B*b*c^4 + 28561*A^4*c^5)/ 
b^17)^(1/4)*log(-729*I*b^13*(-(625*B^4*b^4*c - 6500*A*B^3*b^3*c^2 + 25350* 
A^2*B^2*b^2*c^3 - 43940*A^3*B*b*c^4 + 28561*A^4*c^5)/b^17)^(3/4) - 729*(12 
5*B^3*b^3*c - 975*A*B^2*b^2*c^2 + 2535*A^2*B*b*c^3 - 2197*A^3*c^4)*sqrt(x) 
) - 45*(b^4*c^2*x^7 + 2*b^5*c*x^5 + b^6*x^3)*(-(625*B^4*b^4*c - 6500*A*B^3 
*b^3*c^2 + 25350*A^2*B^2*b^2*c^3 - 43940*A^3*B*b*c^4 + 28561*A^4*c^5)/b^17 
)^(1/4)*log(-729*b^13*(-(625*B^4*b^4*c - 6500*A*B^3*b^3*c^2 + 25350*A^2*B^ 
2*b^2*c^3 - 43940*A^3*B*b*c^4 + 28561*A^4*c^5)/b^17)^(3/4) - 729*(125*B^3* 
b^3*c - 975*A*B^2*b^2*c^2 + 2535*A^2*B*b*c^3 - 2197*A^3*c^4)*sqrt(x)) -...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\text {Timed out} \] Input:

integrate(x**(5/2)*(B*x**2+A)/(c*x**4+b*x**2)**3,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.08 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {45 \, {\left (5 \, B b c^{2} - 13 \, A c^{3}\right )} x^{6} + 81 \, {\left (5 \, B b^{2} c - 13 \, A b c^{2}\right )} x^{4} + 32 \, A b^{3} + 32 \, {\left (5 \, B b^{3} - 13 \, A b^{2} c\right )} x^{2}}{80 \, {\left (b^{4} c^{2} x^{\frac {13}{2}} + 2 \, b^{5} c x^{\frac {9}{2}} + b^{6} x^{\frac {5}{2}}\right )}} - \frac {9 \, {\left (5 \, B b c - 13 \, A c^{2}\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}}\right )}}{128 \, b^{4}} \] Input:

integrate(x^(5/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="maxima")
 

Output:

-1/80*(45*(5*B*b*c^2 - 13*A*c^3)*x^6 + 81*(5*B*b^2*c - 13*A*b*c^2)*x^4 + 3 
2*A*b^3 + 32*(5*B*b^3 - 13*A*b^2*c)*x^2)/(b^4*c^2*x^(13/2) + 2*b^5*c*x^(9/ 
2) + b^6*x^(5/2)) - 9/128*(5*B*b*c - 13*A*c^2)*(2*sqrt(2)*arctan(1/2*sqrt( 
2)*(sqrt(2)*b^(1/4)*c^(1/4) + 2*sqrt(c)*sqrt(x))/sqrt(sqrt(b)*sqrt(c)))/(s 
qrt(sqrt(b)*sqrt(c))*sqrt(c)) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*b^( 
1/4)*c^(1/4) - 2*sqrt(c)*sqrt(x))/sqrt(sqrt(b)*sqrt(c)))/(sqrt(sqrt(b)*sqr 
t(c))*sqrt(c)) - sqrt(2)*log(sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + 
 sqrt(b))/(b^(1/4)*c^(3/4)) + sqrt(2)*log(-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) 
 + sqrt(c)*x + sqrt(b))/(b^(1/4)*c^(3/4)))/b^4
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.24 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {9 \, \sqrt {2} {\left (5 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 13 \, \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{64 \, b^{5} c^{2}} - \frac {9 \, \sqrt {2} {\left (5 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 13 \, \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{64 \, b^{5} c^{2}} + \frac {9 \, \sqrt {2} {\left (5 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 13 \, \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b^{5} c^{2}} - \frac {9 \, \sqrt {2} {\left (5 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 13 \, \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b^{5} c^{2}} - \frac {13 \, B b c^{2} x^{\frac {7}{2}} - 21 \, A c^{3} x^{\frac {7}{2}} + 17 \, B b^{2} c x^{\frac {3}{2}} - 25 \, A b c^{2} x^{\frac {3}{2}}}{16 \, {\left (c x^{2} + b\right )}^{2} b^{4}} - \frac {2 \, {\left (5 \, B b x^{2} - 15 \, A c x^{2} + A b\right )}}{5 \, b^{4} x^{\frac {5}{2}}} \] Input:

integrate(x^(5/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="giac")
 

Output:

-9/64*sqrt(2)*(5*(b*c^3)^(3/4)*B*b - 13*(b*c^3)^(3/4)*A*c)*arctan(1/2*sqrt 
(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))/(b/c)^(1/4))/(b^5*c^2) - 9/64*sqrt(2 
)*(5*(b*c^3)^(3/4)*B*b - 13*(b*c^3)^(3/4)*A*c)*arctan(-1/2*sqrt(2)*(sqrt(2 
)*(b/c)^(1/4) - 2*sqrt(x))/(b/c)^(1/4))/(b^5*c^2) + 9/128*sqrt(2)*(5*(b*c^ 
3)^(3/4)*B*b - 13*(b*c^3)^(3/4)*A*c)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + 
 sqrt(b/c))/(b^5*c^2) - 9/128*sqrt(2)*(5*(b*c^3)^(3/4)*B*b - 13*(b*c^3)^(3 
/4)*A*c)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^5*c^2) - 1/1 
6*(13*B*b*c^2*x^(7/2) - 21*A*c^3*x^(7/2) + 17*B*b^2*c*x^(3/2) - 25*A*b*c^2 
*x^(3/2))/((c*x^2 + b)^2*b^4) - 2/5*(5*B*b*x^2 - 15*A*c*x^2 + A*b)/(b^4*x^ 
(5/2))
 

Mupad [B] (verification not implemented)

Time = 9.28 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.58 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {\frac {2\,x^2\,\left (13\,A\,c-5\,B\,b\right )}{5\,b^2}-\frac {2\,A}{5\,b}+\frac {9\,c^2\,x^6\,\left (13\,A\,c-5\,B\,b\right )}{16\,b^4}+\frac {81\,c\,x^4\,\left (13\,A\,c-5\,B\,b\right )}{80\,b^3}}{b^2\,x^{5/2}+c^2\,x^{13/2}+2\,b\,c\,x^{9/2}}+\frac {9\,{\left (-c\right )}^{1/4}\,\mathrm {atan}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )\,\left (13\,A\,c-5\,B\,b\right )}{32\,b^{17/4}}-\frac {9\,{\left (-c\right )}^{1/4}\,\mathrm {atanh}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )\,\left (13\,A\,c-5\,B\,b\right )}{32\,b^{17/4}} \] Input:

int((x^(5/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x)
 

Output:

((2*x^2*(13*A*c - 5*B*b))/(5*b^2) - (2*A)/(5*b) + (9*c^2*x^6*(13*A*c - 5*B 
*b))/(16*b^4) + (81*c*x^4*(13*A*c - 5*B*b))/(80*b^3))/(b^2*x^(5/2) + c^2*x 
^(13/2) + 2*b*c*x^(9/2)) + (9*(-c)^(1/4)*atan(((-c)^(1/4)*x^(1/2))/b^(1/4) 
)*(13*A*c - 5*B*b))/(32*b^(17/4)) - (9*(-c)^(1/4)*atanh(((-c)^(1/4)*x^(1/2 
))/b^(1/4))*(13*A*c - 5*B*b))/(32*b^(17/4))
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 1041, normalized size of antiderivative = 3.96 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx =\text {Too large to display} \] Input:

int(x^(5/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x)
 

Output:

( - 1170*sqrt(x)*c**(1/4)*b**(3/4)*sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) 
 - 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*a*b**2*c*x**2 - 2340*sq 
rt(x)*c**(1/4)*b**(3/4)*sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) - 2*sqrt(x 
)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*a*b*c**2*x**4 - 1170*sqrt(x)*c**(1 
/4)*b**(3/4)*sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(c))/ 
(c**(1/4)*b**(1/4)*sqrt(2)))*a*c**3*x**6 + 450*sqrt(x)*c**(1/4)*b**(3/4)*s 
qrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**( 
1/4)*sqrt(2)))*b**4*x**2 + 900*sqrt(x)*c**(1/4)*b**(3/4)*sqrt(2)*atan((c** 
(1/4)*b**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*b 
**3*c*x**4 + 450*sqrt(x)*c**(1/4)*b**(3/4)*sqrt(2)*atan((c**(1/4)*b**(1/4) 
*sqrt(2) - 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*b**2*c**2*x**6 
+ 1170*sqrt(x)*c**(1/4)*b**(3/4)*sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) + 
 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*a*b**2*c*x**2 + 2340*sqrt 
(x)*c**(1/4)*b**(3/4)*sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) + 2*sqrt(x)* 
sqrt(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*a*b*c**2*x**4 + 1170*sqrt(x)*c**(1/4 
)*b**(3/4)*sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(c))/(c 
**(1/4)*b**(1/4)*sqrt(2)))*a*c**3*x**6 - 450*sqrt(x)*c**(1/4)*b**(3/4)*sqr 
t(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/ 
4)*sqrt(2)))*b**4*x**2 - 900*sqrt(x)*c**(1/4)*b**(3/4)*sqrt(2)*atan((c**(1 
/4)*b**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*...