\(\int \frac {A+B x^2}{\sqrt {x} (b x^2+c x^4)^3} \, dx\) [153]

Optimal result

Integrand size = 26, antiderivative size = 288 \[ \int \frac {A+B x^2}{\sqrt {x} \left (b x^2+c x^4\right )^3} \, dx=-\frac {2 A}{11 b^3 x^{11/2}}-\frac {2 (b B-3 A c)}{7 b^4 x^{7/2}}+\frac {2 c (b B-2 A c)}{b^5 x^{3/2}}+\frac {c^2 (b B-A c) \sqrt {x}}{4 b^4 \left (b+c x^2\right )^2}+\frac {c^2 (23 b B-31 A c) \sqrt {x}}{16 b^5 \left (b+c x^2\right )}-\frac {15 c^{7/4} (11 b B-19 A c) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{23/4}}+\frac {15 c^{7/4} (11 b B-19 A c) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{23/4}}+\frac {15 c^{7/4} (11 b B-19 A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{32 \sqrt {2} b^{23/4}} \] Output:

-2/11*A/b^3/x^(11/2)-2/7*(-3*A*c+B*b)/b^4/x^(7/2)+2*c*(-2*A*c+B*b)/b^5/x^( 
3/2)+1/4*c^2*(-A*c+B*b)*x^(1/2)/b^4/(c*x^2+b)^2+1/16*c^2*(-31*A*c+23*B*b)* 
x^(1/2)/b^5/(c*x^2+b)-15/64*c^(7/4)*(-19*A*c+11*B*b)*arctan(1-2^(1/2)*c^(1 
/4)*x^(1/2)/b^(1/4))*2^(1/2)/b^(23/4)+15/64*c^(7/4)*(-19*A*c+11*B*b)*arcta 
n(1+2^(1/2)*c^(1/4)*x^(1/2)/b^(1/4))*2^(1/2)/b^(23/4)+15/64*c^(7/4)*(-19*A 
*c+11*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^(23/4)
 

Mathematica [A] (verified)

Time = 0.69 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.80 \[ \int \frac {A+B x^2}{\sqrt {x} \left (b x^2+c x^4\right )^3} \, dx=\frac {-\frac {4 b^{3/4} \left (-11 b B x^2 \left (-32 b^3+160 b^2 c x^2+605 b c^2 x^4+385 c^3 x^6\right )+A \left (224 b^4-608 b^3 c x^2+3040 b^2 c^2 x^4+11495 b c^3 x^6+7315 c^4 x^8\right )\right )}{x^{11/2} \left (b+c x^2\right )^2}+1155 \sqrt {2} c^{7/4} (-11 b B+19 A c) \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )+1155 \sqrt {2} c^{7/4} (11 b B-19 A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{4928 b^{23/4}} \] Input:

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

Output:

((-4*b^(3/4)*(-11*b*B*x^2*(-32*b^3 + 160*b^2*c*x^2 + 605*b*c^2*x^4 + 385*c 
^3*x^6) + A*(224*b^4 - 608*b^3*c*x^2 + 3040*b^2*c^2*x^4 + 11495*b*c^3*x^6 
+ 7315*c^4*x^8)))/(x^(11/2)*(b + c*x^2)^2) + 1155*Sqrt[2]*c^(7/4)*(-11*b*B 
 + 19*A*c)*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])] 
 + 1155*Sqrt[2]*c^(7/4)*(11*b*B - 19*A*c)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4) 
*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(4928*b^(23/4))
 

Rubi [A] (verified)

Time = 0.96 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.22, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.577, Rules used = {9, 362, 253, 264, 264, 264, 266, 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.

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

\(\Big \downarrow \) 9

\(\displaystyle \int \frac {A+B x^2}{x^{13/2} \left (b+c x^2\right )^3}dx\)

\(\Big \downarrow \) 362

\(\displaystyle -\frac {(11 b B-19 A c) \int \frac {1}{x^{13/2} \left (c x^2+b\right )^2}dx}{8 b c}-\frac {b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 253

\(\displaystyle -\frac {(11 b B-19 A c) \left (\frac {15 \int \frac {1}{x^{13/2} \left (c x^2+b\right )}dx}{4 b}+\frac {1}{2 b x^{11/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 264

\(\displaystyle -\frac {(11 b B-19 A c) \left (\frac {15 \left (-\frac {c \int \frac {1}{x^{9/2} \left (c x^2+b\right )}dx}{b}-\frac {2}{11 b x^{11/2}}\right )}{4 b}+\frac {1}{2 b x^{11/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 264

\(\displaystyle -\frac {(11 b B-19 A c) \left (\frac {15 \left (-\frac {c \left (-\frac {c \int \frac {1}{x^{5/2} \left (c x^2+b\right )}dx}{b}-\frac {2}{7 b x^{7/2}}\right )}{b}-\frac {2}{11 b x^{11/2}}\right )}{4 b}+\frac {1}{2 b x^{11/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 264

\(\displaystyle -\frac {(11 b B-19 A c) \left (\frac {15 \left (-\frac {c \left (-\frac {c \left (-\frac {c \int \frac {1}{\sqrt {x} \left (c x^2+b\right )}dx}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{b}-\frac {2}{11 b x^{11/2}}\right )}{4 b}+\frac {1}{2 b x^{11/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 266

\(\displaystyle -\frac {(11 b B-19 A c) \left (\frac {15 \left (-\frac {c \left (-\frac {c \left (-\frac {2 c \int \frac {1}{c x^2+b}d\sqrt {x}}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{b}-\frac {2}{11 b x^{11/2}}\right )}{4 b}+\frac {1}{2 b x^{11/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 755

\(\displaystyle -\frac {(11 b B-19 A c) \left (\frac {15 \left (-\frac {c \left (-\frac {c \left (-\frac {2 c \left (\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\int \frac {\sqrt {c} x+\sqrt {b}}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{b}-\frac {2}{11 b x^{11/2}}\right )}{4 b}+\frac {1}{2 b x^{11/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 1476

\(\displaystyle -\frac {(11 b B-19 A c) \left (\frac {15 \left (-\frac {c \left (-\frac {c \left (-\frac {2 c \left (\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\frac {\int \frac {1}{x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {c}}+\frac {\int \frac {1}{x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{b}-\frac {2}{11 b x^{11/2}}\right )}{4 b}+\frac {1}{2 b x^{11/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 1082

\(\displaystyle -\frac {(11 b B-19 A c) \left (\frac {15 \left (-\frac {c \left (-\frac {c \left (-\frac {2 c \left (\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\frac {\int \frac {1}{-x-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\int \frac {1}{-x-1}d\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{b}-\frac {2}{11 b x^{11/2}}\right )}{4 b}+\frac {1}{2 b x^{11/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 217

\(\displaystyle -\frac {(11 b B-19 A c) \left (\frac {15 \left (-\frac {c \left (-\frac {c \left (-\frac {2 c \left (\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{b}-\frac {2}{11 b x^{11/2}}\right )}{4 b}+\frac {1}{2 b x^{11/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 1479

\(\displaystyle -\frac {(11 b B-19 A c) \left (\frac {15 \left (-\frac {c \left (-\frac {c \left (-\frac {2 c \left (\frac {-\frac {\int -\frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{c} \left (x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}\right )}{\sqrt [4]{c} \left (x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{b}-\frac {2}{11 b x^{11/2}}\right )}{4 b}+\frac {1}{2 b x^{11/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {(11 b B-19 A c) \left (\frac {15 \left (-\frac {c \left (-\frac {c \left (-\frac {2 c \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{c} \left (x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}\right )}{\sqrt [4]{c} \left (x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{b}-\frac {2}{11 b x^{11/2}}\right )}{4 b}+\frac {1}{2 b x^{11/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {(11 b B-19 A c) \left (\frac {15 \left (-\frac {c \left (-\frac {c \left (-\frac {2 c \left (\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt {c}}+\frac {\int \frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}}{x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt [4]{b} \sqrt {c}}}{2 \sqrt {b}}+\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{b}-\frac {2}{11 b x^{11/2}}\right )}{4 b}+\frac {1}{2 b x^{11/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}\)

\(\Big \downarrow \) 1103

\(\displaystyle -\frac {(11 b B-19 A c) \left (\frac {15 \left (-\frac {c \left (-\frac {c \left (-\frac {2 c \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}+\frac {\frac {\log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {b}}\right )}{b}-\frac {2}{3 b x^{3/2}}\right )}{b}-\frac {2}{7 b x^{7/2}}\right )}{b}-\frac {2}{11 b x^{11/2}}\right )}{4 b}+\frac {1}{2 b x^{11/2} \left (b+c x^2\right )}\right )}{8 b c}-\frac {b B-A c}{4 b c x^{11/2} \left (b+c x^2\right )^2}\)

Input:

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

Output:

-1/4*(b*B - A*c)/(b*c*x^(11/2)*(b + c*x^2)^2) - ((11*b*B - 19*A*c)*(1/(2*b 
*x^(11/2)*(b + c*x^2)) + (15*(-2/(11*b*x^(11/2)) - (c*(-2/(7*b*x^(7/2)) - 
(c*(-2/(3*b*x^(3/2)) - (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[b]) + (-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)*Sqrt[x] + Sqrt[c]*x]/(2*Sqrt[2]*b^(1/4)*c^(1/ 
4)))/(2*Sqrt[b])))/b))/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 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 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 = 210, normalized size of antiderivative = 0.73

method	result	size
derivativedivides	\(-\frac {2 c^{2} \left (\frac {\left (\frac {31}{32} A \,c^{2}-\frac {23}{32} B b c \right ) x^{\frac {5}{2}}+\frac {b \left (35 A c -27 B b \right ) \sqrt {x}}{32}}{\left (c \,x^{2}+b \right )^{2}}+\frac {15 \left (19 A c -11 B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{256 b}\right )}{b^{5}}-\frac {2 A}{11 b^{3} x^{\frac {11}{2}}}-\frac {2 \left (-3 A c +B b \right )}{7 b^{4} x^{\frac {7}{2}}}-\frac {2 c \left (2 A c -B b \right )}{b^{5} x^{\frac {3}{2}}}\)	\(210\)
default	\(-\frac {2 c^{2} \left (\frac {\left (\frac {31}{32} A \,c^{2}-\frac {23}{32} B b c \right ) x^{\frac {5}{2}}+\frac {b \left (35 A c -27 B b \right ) \sqrt {x}}{32}}{\left (c \,x^{2}+b \right )^{2}}+\frac {15 \left (19 A c -11 B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{256 b}\right )}{b^{5}}-\frac {2 A}{11 b^{3} x^{\frac {11}{2}}}-\frac {2 \left (-3 A c +B b \right )}{7 b^{4} x^{\frac {7}{2}}}-\frac {2 c \left (2 A c -B b \right )}{b^{5} x^{\frac {3}{2}}}\)	\(210\)
risch	\(-\frac {2 \left (154 x^{4} A \,c^{2}-77 x^{4} B b c -33 A b c \,x^{2}+11 x^{2} B \,b^{2}+7 b^{2} A \right )}{77 b^{5} x^{\frac {11}{2}}}-\frac {c^{2} \left (\frac {2 \left (\frac {31}{32} A \,c^{2}-\frac {23}{32} B b c \right ) x^{\frac {5}{2}}+\frac {b \left (35 A c -27 B b \right ) \sqrt {x}}{16}}{\left (c \,x^{2}+b \right )^{2}}+\frac {15 \left (19 A c -11 B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{128 b}\right )}{b^{5}}\)	\(217\)

Input:

int((B*x^2+A)/x^(1/2)/(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)
 

Output:

-2/b^5*c^2*(((31/32*A*c^2-23/32*B*b*c)*x^(5/2)+1/32*b*(35*A*c-27*B*b)*x^(1 
/2))/(c*x^2+b)^2+15/256*(19*A*c-11*B*b)*(b/c)^(1/4)/b*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)))-2/11*A/b^3/x^(11/2)-2/7*(-3*A*c+B*b)/b^4/x^(7/2)-2*c*(2*A*c 
-B*b)/b^5/x^(3/2)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 854, normalized size of antiderivative = 2.97 \[ \int \frac {A+B x^2}{\sqrt {x} \left (b x^2+c x^4\right )^3} \, dx =\text {Too large to display} \] Input:

integrate((B*x^2+A)/x^(1/2)/(c*x^4+b*x^2)^3,x, algorithm="fricas")
 

Output:

-1/4928*(1155*(b^5*c^2*x^10 + 2*b^6*c*x^8 + b^7*x^6)*(-(14641*B^4*b^4*c^7 
- 101156*A*B^3*b^3*c^8 + 262086*A^2*B^2*b^2*c^9 - 301796*A^3*B*b*c^10 + 13 
0321*A^4*c^11)/b^23)^(1/4)*log(15*b^6*(-(14641*B^4*b^4*c^7 - 101156*A*B^3* 
b^3*c^8 + 262086*A^2*B^2*b^2*c^9 - 301796*A^3*B*b*c^10 + 130321*A^4*c^11)/ 
b^23)^(1/4) - 15*(11*B*b*c^2 - 19*A*c^3)*sqrt(x)) + 1155*(I*b^5*c^2*x^10 + 
 2*I*b^6*c*x^8 + I*b^7*x^6)*(-(14641*B^4*b^4*c^7 - 101156*A*B^3*b^3*c^8 + 
262086*A^2*B^2*b^2*c^9 - 301796*A^3*B*b*c^10 + 130321*A^4*c^11)/b^23)^(1/4 
)*log(15*I*b^6*(-(14641*B^4*b^4*c^7 - 101156*A*B^3*b^3*c^8 + 262086*A^2*B^ 
2*b^2*c^9 - 301796*A^3*B*b*c^10 + 130321*A^4*c^11)/b^23)^(1/4) - 15*(11*B* 
b*c^2 - 19*A*c^3)*sqrt(x)) + 1155*(-I*b^5*c^2*x^10 - 2*I*b^6*c*x^8 - I*b^7 
*x^6)*(-(14641*B^4*b^4*c^7 - 101156*A*B^3*b^3*c^8 + 262086*A^2*B^2*b^2*c^9 
 - 301796*A^3*B*b*c^10 + 130321*A^4*c^11)/b^23)^(1/4)*log(-15*I*b^6*(-(146 
41*B^4*b^4*c^7 - 101156*A*B^3*b^3*c^8 + 262086*A^2*B^2*b^2*c^9 - 301796*A^ 
3*B*b*c^10 + 130321*A^4*c^11)/b^23)^(1/4) - 15*(11*B*b*c^2 - 19*A*c^3)*sqr 
t(x)) - 1155*(b^5*c^2*x^10 + 2*b^6*c*x^8 + b^7*x^6)*(-(14641*B^4*b^4*c^7 - 
 101156*A*B^3*b^3*c^8 + 262086*A^2*B^2*b^2*c^9 - 301796*A^3*B*b*c^10 + 130 
321*A^4*c^11)/b^23)^(1/4)*log(-15*b^6*(-(14641*B^4*b^4*c^7 - 101156*A*B^3* 
b^3*c^8 + 262086*A^2*B^2*b^2*c^9 - 301796*A^3*B*b*c^10 + 130321*A^4*c^11)/ 
b^23)^(1/4) - 15*(11*B*b*c^2 - 19*A*c^3)*sqrt(x)) - 4*(385*(11*B*b*c^3 - 1 
9*A*c^4)*x^8 + 605*(11*B*b^2*c^2 - 19*A*b*c^3)*x^6 - 224*A*b^4 + 160*(1...
 

Sympy [F(-1)]

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

integrate((B*x**2+A)/x**(1/2)/(c*x**4+b*x**2)**3,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.23 \[ \int \frac {A+B x^2}{\sqrt {x} \left (b x^2+c x^4\right )^3} \, dx=\frac {385 \, {\left (11 \, B b c^{3} - 19 \, A c^{4}\right )} x^{8} + 605 \, {\left (11 \, B b^{2} c^{2} - 19 \, A b c^{3}\right )} x^{6} - 224 \, A b^{4} + 160 \, {\left (11 \, B b^{3} c - 19 \, A b^{2} c^{2}\right )} x^{4} - 32 \, {\left (11 \, B b^{4} - 19 \, A b^{3} c\right )} x^{2}}{1232 \, {\left (b^{5} c^{2} x^{\frac {19}{2}} + 2 \, b^{6} c x^{\frac {15}{2}} + b^{7} x^{\frac {11}{2}}\right )}} + \frac {15 \, {\left (\frac {2 \, \sqrt {2} {\left (11 \, B b c^{2} - 19 \, A c^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {2 \, \sqrt {2} {\left (11 \, B b c^{2} - 19 \, A c^{3}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {\sqrt {2} {\left (11 \, B b c^{2} - 19 \, A c^{3}\right )} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (11 \, B b c^{2} - 19 \, A c^{3}\right )} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}}\right )}}{128 \, b^{5}} \] Input:

integrate((B*x^2+A)/x^(1/2)/(c*x^4+b*x^2)^3,x, algorithm="maxima")
 

Output:

1/1232*(385*(11*B*b*c^3 - 19*A*c^4)*x^8 + 605*(11*B*b^2*c^2 - 19*A*b*c^3)* 
x^6 - 224*A*b^4 + 160*(11*B*b^3*c - 19*A*b^2*c^2)*x^4 - 32*(11*B*b^4 - 19* 
A*b^3*c)*x^2)/(b^5*c^2*x^(19/2) + 2*b^6*c*x^(15/2) + b^7*x^(11/2)) + 15/12 
8*(2*sqrt(2)*(11*B*b*c^2 - 19*A*c^3)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c 
^(1/4) + 2*sqrt(c)*sqrt(x))/sqrt(sqrt(b)*sqrt(c)))/(sqrt(b)*sqrt(sqrt(b)*s 
qrt(c))) + 2*sqrt(2)*(11*B*b*c^2 - 19*A*c^3)*arctan(-1/2*sqrt(2)*(sqrt(2)* 
b^(1/4)*c^(1/4) - 2*sqrt(c)*sqrt(x))/sqrt(sqrt(b)*sqrt(c)))/(sqrt(b)*sqrt( 
sqrt(b)*sqrt(c))) + sqrt(2)*(11*B*b*c^2 - 19*A*c^3)*log(sqrt(2)*b^(1/4)*c^ 
(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(3/4)*c^(1/4)) - sqrt(2)*(11*B*b*c 
^2 - 19*A*c^3)*log(-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b)) 
/(b^(3/4)*c^(1/4)))/b^5
 

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.22 \[ \int \frac {A+B x^2}{\sqrt {x} \left (b x^2+c x^4\right )^3} \, dx=\frac {15 \, \sqrt {2} {\left (11 \, \left (b c^{3}\right )^{\frac {1}{4}} B b c - 19 \, \left (b c^{3}\right )^{\frac {1}{4}} A c^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{64 \, b^{6}} + \frac {15 \, \sqrt {2} {\left (11 \, \left (b c^{3}\right )^{\frac {1}{4}} B b c - 19 \, \left (b c^{3}\right )^{\frac {1}{4}} A c^{2}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{64 \, b^{6}} + \frac {15 \, \sqrt {2} {\left (11 \, \left (b c^{3}\right )^{\frac {1}{4}} B b c - 19 \, \left (b c^{3}\right )^{\frac {1}{4}} A c^{2}\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b^{6}} - \frac {15 \, \sqrt {2} {\left (11 \, \left (b c^{3}\right )^{\frac {1}{4}} B b c - 19 \, \left (b c^{3}\right )^{\frac {1}{4}} A c^{2}\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b^{6}} + \frac {23 \, B b c^{3} x^{\frac {5}{2}} - 31 \, A c^{4} x^{\frac {5}{2}} + 27 \, B b^{2} c^{2} \sqrt {x} - 35 \, A b c^{3} \sqrt {x}}{16 \, {\left (c x^{2} + b\right )}^{2} b^{5}} + \frac {2 \, {\left (77 \, B b c x^{4} - 154 \, A c^{2} x^{4} - 11 \, B b^{2} x^{2} + 33 \, A b c x^{2} - 7 \, A b^{2}\right )}}{77 \, b^{5} x^{\frac {11}{2}}} \] Input:

integrate((B*x^2+A)/x^(1/2)/(c*x^4+b*x^2)^3,x, algorithm="giac")
 

Output:

15/64*sqrt(2)*(11*(b*c^3)^(1/4)*B*b*c - 19*(b*c^3)^(1/4)*A*c^2)*arctan(1/2 
*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))/(b/c)^(1/4))/b^6 + 15/64*sqrt(2 
)*(11*(b*c^3)^(1/4)*B*b*c - 19*(b*c^3)^(1/4)*A*c^2)*arctan(-1/2*sqrt(2)*(s 
qrt(2)*(b/c)^(1/4) - 2*sqrt(x))/(b/c)^(1/4))/b^6 + 15/128*sqrt(2)*(11*(b*c 
^3)^(1/4)*B*b*c - 19*(b*c^3)^(1/4)*A*c^2)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) 
+ x + sqrt(b/c))/b^6 - 15/128*sqrt(2)*(11*(b*c^3)^(1/4)*B*b*c - 19*(b*c^3) 
^(1/4)*A*c^2)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/b^6 + 1/16 
*(23*B*b*c^3*x^(5/2) - 31*A*c^4*x^(5/2) + 27*B*b^2*c^2*sqrt(x) - 35*A*b*c^ 
3*sqrt(x))/((c*x^2 + b)^2*b^5) + 2/77*(77*B*b*c*x^4 - 154*A*c^2*x^4 - 11*B 
*b^2*x^2 + 33*A*b*c*x^2 - 7*A*b^2)/(b^5*x^(11/2))
 

Mupad [B] (verification not implemented)

Time = 9.60 (sec) , antiderivative size = 639, normalized size of antiderivative = 2.22 \[ \int \frac {A+B x^2}{\sqrt {x} \left (b x^2+c x^4\right )^3} \, dx =\text {Too large to display} \] Input:

int((A + B*x^2)/(x^(1/2)*(b*x^2 + c*x^4)^3),x)
 

Output:

(15*(-c)^(7/4)*atan((6859*A^3*c^10*x^(1/2) - 1331*B^3*b^3*c^7*x^(1/2) - 11 
913*A^2*B*b*c^9*x^(1/2) + 6897*A*B^2*b^2*c^8*x^(1/2))/(b^(1/4)*(-c)^(27/4) 
*(c*(c*(6859*A^3*c - 11913*A^2*B*b) + 6897*A*B^2*b^2) - 1331*B^3*b^3)))*(1 
9*A*c - 11*B*b))/(32*b^(23/4)) - ((2*A)/(11*b) - (2*x^2*(19*A*c - 11*B*b)) 
/(77*b^2) + (55*c^2*x^6*(19*A*c - 11*B*b))/(112*b^4) + (5*c^3*x^8*(19*A*c 
- 11*B*b))/(16*b^5) + (10*c*x^4*(19*A*c - 11*B*b))/(77*b^3))/(b^2*x^(11/2) 
 + c^2*x^(19/2) + 2*b*c*x^(15/2)) - ((-c)^(7/4)*atan((((-c)^(7/4)*(19*A*c 
- 11*B*b)*(x^(1/2)*(1330790400*A^2*b^15*c^9 + 446054400*B^2*b^17*c^7 - 154 
0915200*A*B*b^16*c^8) - (15*(-c)^(7/4)*(19*A*c - 11*B*b)*(298844160*A*b^21 
*c^6 - 173015040*B*b^22*c^5))/(64*b^(23/4)))*15i)/(64*b^(23/4)) + ((-c)^(7 
/4)*(19*A*c - 11*B*b)*(x^(1/2)*(1330790400*A^2*b^15*c^9 + 446054400*B^2*b^ 
17*c^7 - 1540915200*A*B*b^16*c^8) + (15*(-c)^(7/4)*(19*A*c - 11*B*b)*(2988 
44160*A*b^21*c^6 - 173015040*B*b^22*c^5))/(64*b^(23/4)))*15i)/(64*b^(23/4) 
))/((15*(-c)^(7/4)*(19*A*c - 11*B*b)*(x^(1/2)*(1330790400*A^2*b^15*c^9 + 4 
46054400*B^2*b^17*c^7 - 1540915200*A*B*b^16*c^8) - (15*(-c)^(7/4)*(19*A*c 
- 11*B*b)*(298844160*A*b^21*c^6 - 173015040*B*b^22*c^5))/(64*b^(23/4))))/( 
64*b^(23/4)) - (15*(-c)^(7/4)*(19*A*c - 11*B*b)*(x^(1/2)*(1330790400*A^2*b 
^15*c^9 + 446054400*B^2*b^17*c^7 - 1540915200*A*B*b^16*c^8) + (15*(-c)^(7/ 
4)*(19*A*c - 11*B*b)*(298844160*A*b^21*c^6 - 173015040*B*b^22*c^5))/(64*b^ 
(23/4))))/(64*b^(23/4))))*(19*A*c - 11*B*b)*15i)/(32*b^(23/4))
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 1064, normalized size of antiderivative = 3.69 \[ \int \frac {A+B x^2}{\sqrt {x} \left (b x^2+c x^4\right )^3} \, dx =\text {Too large to display} \] Input:

int((B*x^2+A)/x^(1/2)/(c*x^4+b*x^2)^3,x)
 

Output:

(43890*sqrt(x)*c**(3/4)*b**(1/4)*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**2*x**5 + 87780* 
sqrt(x)*c**(3/4)*b**(1/4)*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**3*x**7 + 43890*sqrt(x)*c* 
*(3/4)*b**(1/4)*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**4*x**9 - 25410*sqrt(x)*c**(3/4)*b**(1 
/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**4*c*x**5 - 50820*sqrt(x)*c**(3/4)*b**(1/4)*sqrt(2)* 
atan((c**(1/4)*b**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sq 
rt(2)))*b**3*c**2*x**7 - 25410*sqrt(x)*c**(3/4)*b**(1/4)*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**3*x**9 - 43890*sqrt(x)*c**(3/4)*b**(1/4)*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** 
2*x**5 - 87780*sqrt(x)*c**(3/4)*b**(1/4)*sqrt(2)*atan((c**(1/4)*b**(1/4)*s 
qrt(2) + 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*a*b*c**3*x**7 - 4 
3890*sqrt(x)*c**(3/4)*b**(1/4)*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**4*x**9 + 25410*sqrt(x) 
*c**(3/4)*b**(1/4)*sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) + 2*sqrt(x)*sqr 
t(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*b**4*c*x**5 + 50820*sqrt(x)*c**(3/4)*b* 
*(1/4)*sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(c))/(c*...