\(\int \frac {A+B x^2}{x^{9/2} (b x^2+c x^4)} \, dx\) [128]

Optimal result

Integrand size = 26, antiderivative size = 214 \[ \int \frac {A+B x^2}{x^{9/2} \left (b x^2+c x^4\right )} \, dx=-\frac {2 A}{11 b x^{11/2}}-\frac {2 (b B-A c)}{7 b^2 x^{7/2}}+\frac {2 c (b B-A c)}{3 b^3 x^{3/2}}-\frac {c^{7/4} (b B-A c) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{15/4}}+\frac {c^{7/4} (b B-A c) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{15/4}}+\frac {c^{7/4} (b B-A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{\sqrt {2} b^{15/4}} \] Output:

-2/11*A/b/x^(11/2)-2/7*(-A*c+B*b)/b^2/x^(7/2)+2/3*c*(-A*c+B*b)/b^3/x^(3/2) 
-1/2*c^(7/4)*(-A*c+B*b)*arctan(1-2^(1/2)*c^(1/4)*x^(1/2)/b^(1/4))*2^(1/2)/ 
b^(15/4)+1/2*c^(7/4)*(-A*c+B*b)*arctan(1+2^(1/2)*c^(1/4)*x^(1/2)/b^(1/4))* 
2^(1/2)/b^(15/4)+1/2*c^(7/4)*(-A*c+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^(15/4)
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.82 \[ \int \frac {A+B x^2}{x^{9/2} \left (b x^2+c x^4\right )} \, dx=-\frac {2 \left (21 A b^2+33 b^2 B x^2-33 A b c x^2-77 b B c x^4+77 A c^2 x^4\right )}{231 b^3 x^{11/2}}-\frac {c^{7/4} (b B-A c) \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )}{\sqrt {2} b^{15/4}}+\frac {c^{7/4} (b B-A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{\sqrt {2} b^{15/4}} \] Input:

Integrate[(A + B*x^2)/(x^(9/2)*(b*x^2 + c*x^4)),x]
 

Output:

(-2*(21*A*b^2 + 33*b^2*B*x^2 - 33*A*b*c*x^2 - 77*b*B*c*x^4 + 77*A*c^2*x^4) 
)/(231*b^3*x^(11/2)) - (c^(7/4)*(b*B - A*c)*ArcTan[(Sqrt[b] - Sqrt[c]*x)/( 
Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])])/(Sqrt[2]*b^(15/4)) + (c^(7/4)*(b*B - A* 
c)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(Sqrt 
[2]*b^(15/4))
 

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.29, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {9, 359, 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.

\(\displaystyle \int \frac {A+B x^2}{x^{9/2} \left (b x^2+c x^4\right )} \, dx\)

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 359

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

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 755

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

Int[(A + B*x^2)/(x^(9/2)*(b*x^2 + c*x^4)),x]
 

Output:

(-2*A)/(11*b*x^(11/2)) + ((b*B - A*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
 

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 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 359

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + 
Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1))   Int[(e*x)^(m + 2)* 
(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] 
&& LtQ[m, -1] &&  !ILtQ[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 = 0.45 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.75

method	result	size
derivativedivides	\(-\frac {2 A}{11 b \,x^{\frac {11}{2}}}-\frac {2 \left (-A c +B b \right )}{7 b^{2} x^{\frac {7}{2}}}-\frac {2 \left (A c -B b \right ) c}{3 b^{3} x^{\frac {3}{2}}}-\frac {\left (A c -B b \right ) c^{2} \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 b^{4}}\)	\(160\)
default	\(-\frac {2 A}{11 b \,x^{\frac {11}{2}}}-\frac {2 \left (-A c +B b \right )}{7 b^{2} x^{\frac {7}{2}}}-\frac {2 \left (A c -B b \right ) c}{3 b^{3} x^{\frac {3}{2}}}-\frac {\left (A c -B b \right ) c^{2} \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 b^{4}}\)	\(160\)
risch	\(-\frac {2 \left (77 x^{4} A \,c^{2}-77 x^{4} B b c -33 A b c \,x^{2}+33 x^{2} B \,b^{2}+21 b^{2} A \right )}{231 b^{3} x^{\frac {11}{2}}}-\frac {\left (A c -B b \right ) c^{2} \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 b^{4}}\)	\(167\)

Input:

int((B*x^2+A)/x^(9/2)/(c*x^4+b*x^2),x,method=_RETURNVERBOSE)
 

Output:

-2/11*A/b/x^(11/2)-2/7*(-A*c+B*b)/b^2/x^(7/2)-2/3*(A*c-B*b)/b^3*c/x^(3/2)- 
1/4*(A*c-B*b)/b^4*c^2*(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))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.14 (sec) , antiderivative size = 672, normalized size of antiderivative = 3.14 \[ \int \frac {A+B x^2}{x^{9/2} \left (b x^2+c x^4\right )} \, dx=-\frac {231 \, b^{3} x^{6} \left (-\frac {B^{4} b^{4} c^{7} - 4 \, A B^{3} b^{3} c^{8} + 6 \, A^{2} B^{2} b^{2} c^{9} - 4 \, A^{3} B b c^{10} + A^{4} c^{11}}{b^{15}}\right )^{\frac {1}{4}} \log \left (b^{4} \left (-\frac {B^{4} b^{4} c^{7} - 4 \, A B^{3} b^{3} c^{8} + 6 \, A^{2} B^{2} b^{2} c^{9} - 4 \, A^{3} B b c^{10} + A^{4} c^{11}}{b^{15}}\right )^{\frac {1}{4}} - {\left (B b c^{2} - A c^{3}\right )} \sqrt {x}\right ) + 231 i \, b^{3} x^{6} \left (-\frac {B^{4} b^{4} c^{7} - 4 \, A B^{3} b^{3} c^{8} + 6 \, A^{2} B^{2} b^{2} c^{9} - 4 \, A^{3} B b c^{10} + A^{4} c^{11}}{b^{15}}\right )^{\frac {1}{4}} \log \left (i \, b^{4} \left (-\frac {B^{4} b^{4} c^{7} - 4 \, A B^{3} b^{3} c^{8} + 6 \, A^{2} B^{2} b^{2} c^{9} - 4 \, A^{3} B b c^{10} + A^{4} c^{11}}{b^{15}}\right )^{\frac {1}{4}} - {\left (B b c^{2} - A c^{3}\right )} \sqrt {x}\right ) - 231 i \, b^{3} x^{6} \left (-\frac {B^{4} b^{4} c^{7} - 4 \, A B^{3} b^{3} c^{8} + 6 \, A^{2} B^{2} b^{2} c^{9} - 4 \, A^{3} B b c^{10} + A^{4} c^{11}}{b^{15}}\right )^{\frac {1}{4}} \log \left (-i \, b^{4} \left (-\frac {B^{4} b^{4} c^{7} - 4 \, A B^{3} b^{3} c^{8} + 6 \, A^{2} B^{2} b^{2} c^{9} - 4 \, A^{3} B b c^{10} + A^{4} c^{11}}{b^{15}}\right )^{\frac {1}{4}} - {\left (B b c^{2} - A c^{3}\right )} \sqrt {x}\right ) - 231 \, b^{3} x^{6} \left (-\frac {B^{4} b^{4} c^{7} - 4 \, A B^{3} b^{3} c^{8} + 6 \, A^{2} B^{2} b^{2} c^{9} - 4 \, A^{3} B b c^{10} + A^{4} c^{11}}{b^{15}}\right )^{\frac {1}{4}} \log \left (-b^{4} \left (-\frac {B^{4} b^{4} c^{7} - 4 \, A B^{3} b^{3} c^{8} + 6 \, A^{2} B^{2} b^{2} c^{9} - 4 \, A^{3} B b c^{10} + A^{4} c^{11}}{b^{15}}\right )^{\frac {1}{4}} - {\left (B b c^{2} - A c^{3}\right )} \sqrt {x}\right ) - 4 \, {\left (77 \, {\left (B b c - A c^{2}\right )} x^{4} - 21 \, A b^{2} - 33 \, {\left (B b^{2} - A b c\right )} x^{2}\right )} \sqrt {x}}{462 \, b^{3} x^{6}} \] Input:

integrate((B*x^2+A)/x^(9/2)/(c*x^4+b*x^2),x, algorithm="fricas")
 

Output:

-1/462*(231*b^3*x^6*(-(B^4*b^4*c^7 - 4*A*B^3*b^3*c^8 + 6*A^2*B^2*b^2*c^9 - 
 4*A^3*B*b*c^10 + A^4*c^11)/b^15)^(1/4)*log(b^4*(-(B^4*b^4*c^7 - 4*A*B^3*b 
^3*c^8 + 6*A^2*B^2*b^2*c^9 - 4*A^3*B*b*c^10 + A^4*c^11)/b^15)^(1/4) - (B*b 
*c^2 - A*c^3)*sqrt(x)) + 231*I*b^3*x^6*(-(B^4*b^4*c^7 - 4*A*B^3*b^3*c^8 + 
6*A^2*B^2*b^2*c^9 - 4*A^3*B*b*c^10 + A^4*c^11)/b^15)^(1/4)*log(I*b^4*(-(B^ 
4*b^4*c^7 - 4*A*B^3*b^3*c^8 + 6*A^2*B^2*b^2*c^9 - 4*A^3*B*b*c^10 + A^4*c^1 
1)/b^15)^(1/4) - (B*b*c^2 - A*c^3)*sqrt(x)) - 231*I*b^3*x^6*(-(B^4*b^4*c^7 
 - 4*A*B^3*b^3*c^8 + 6*A^2*B^2*b^2*c^9 - 4*A^3*B*b*c^10 + A^4*c^11)/b^15)^ 
(1/4)*log(-I*b^4*(-(B^4*b^4*c^7 - 4*A*B^3*b^3*c^8 + 6*A^2*B^2*b^2*c^9 - 4* 
A^3*B*b*c^10 + A^4*c^11)/b^15)^(1/4) - (B*b*c^2 - A*c^3)*sqrt(x)) - 231*b^ 
3*x^6*(-(B^4*b^4*c^7 - 4*A*B^3*b^3*c^8 + 6*A^2*B^2*b^2*c^9 - 4*A^3*B*b*c^1 
0 + A^4*c^11)/b^15)^(1/4)*log(-b^4*(-(B^4*b^4*c^7 - 4*A*B^3*b^3*c^8 + 6*A^ 
2*B^2*b^2*c^9 - 4*A^3*B*b*c^10 + A^4*c^11)/b^15)^(1/4) - (B*b*c^2 - A*c^3) 
*sqrt(x)) - 4*(77*(B*b*c - A*c^2)*x^4 - 21*A*b^2 - 33*(B*b^2 - A*b*c)*x^2) 
*sqrt(x))/(b^3*x^6)
 

Sympy [A] (verification not implemented)

Time = 131.68 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.59 \[ \int \frac {A+B x^2}{x^{9/2} \left (b x^2+c x^4\right )} \, dx=\begin {cases} \tilde {\infty } \left (- \frac {2 A}{15 x^{\frac {15}{2}}} - \frac {2 B}{11 x^{\frac {11}{2}}}\right ) & \text {for}\: b = 0 \wedge c = 0 \\\frac {- \frac {2 A}{15 x^{\frac {15}{2}}} - \frac {2 B}{11 x^{\frac {11}{2}}}}{c} & \text {for}\: b = 0 \\\frac {- \frac {2 A}{11 x^{\frac {11}{2}}} - \frac {2 B}{7 x^{\frac {7}{2}}}}{b} & \text {for}\: c = 0 \\- \frac {2 A}{11 b x^{\frac {11}{2}}} + \frac {2 A c}{7 b^{2} x^{\frac {7}{2}}} - \frac {2 A c^{2}}{3 b^{3} x^{\frac {3}{2}}} + \frac {A c^{3} \sqrt [4]{- \frac {b}{c}} \log {\left (\sqrt {x} - \sqrt [4]{- \frac {b}{c}} \right )}}{2 b^{4}} - \frac {A c^{3} \sqrt [4]{- \frac {b}{c}} \log {\left (\sqrt {x} + \sqrt [4]{- \frac {b}{c}} \right )}}{2 b^{4}} - \frac {A c^{3} \sqrt [4]{- \frac {b}{c}} \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {b}{c}}} \right )}}{b^{4}} - \frac {2 B}{7 b x^{\frac {7}{2}}} + \frac {2 B c}{3 b^{2} x^{\frac {3}{2}}} - \frac {B c^{2} \sqrt [4]{- \frac {b}{c}} \log {\left (\sqrt {x} - \sqrt [4]{- \frac {b}{c}} \right )}}{2 b^{3}} + \frac {B c^{2} \sqrt [4]{- \frac {b}{c}} \log {\left (\sqrt {x} + \sqrt [4]{- \frac {b}{c}} \right )}}{2 b^{3}} + \frac {B c^{2} \sqrt [4]{- \frac {b}{c}} \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {b}{c}}} \right )}}{b^{3}} & \text {otherwise} \end {cases} \] Input:

integrate((B*x**2+A)/x**(9/2)/(c*x**4+b*x**2),x)
 

Output:

Piecewise((zoo*(-2*A/(15*x**(15/2)) - 2*B/(11*x**(11/2))), Eq(b, 0) & Eq(c 
, 0)), ((-2*A/(15*x**(15/2)) - 2*B/(11*x**(11/2)))/c, Eq(b, 0)), ((-2*A/(1 
1*x**(11/2)) - 2*B/(7*x**(7/2)))/b, Eq(c, 0)), (-2*A/(11*b*x**(11/2)) + 2* 
A*c/(7*b**2*x**(7/2)) - 2*A*c**2/(3*b**3*x**(3/2)) + A*c**3*(-b/c)**(1/4)* 
log(sqrt(x) - (-b/c)**(1/4))/(2*b**4) - A*c**3*(-b/c)**(1/4)*log(sqrt(x) + 
 (-b/c)**(1/4))/(2*b**4) - A*c**3*(-b/c)**(1/4)*atan(sqrt(x)/(-b/c)**(1/4) 
)/b**4 - 2*B/(7*b*x**(7/2)) + 2*B*c/(3*b**2*x**(3/2)) - B*c**2*(-b/c)**(1/ 
4)*log(sqrt(x) - (-b/c)**(1/4))/(2*b**3) + B*c**2*(-b/c)**(1/4)*log(sqrt(x 
) + (-b/c)**(1/4))/(2*b**3) + B*c**2*(-b/c)**(1/4)*atan(sqrt(x)/(-b/c)**(1 
/4))/b**3, True))
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.29 \[ \int \frac {A+B x^2}{x^{9/2} \left (b x^2+c x^4\right )} \, dx=\frac {\frac {2 \, \sqrt {2} {\left (B b c^{2} - A c^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {2 \, \sqrt {2} {\left (B b c^{2} - A c^{3}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {\sqrt {2} {\left (B b c^{2} - A c^{3}\right )} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (B b c^{2} - A c^{3}\right )} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}}}{4 \, b^{3}} + \frac {2 \, {\left (77 \, {\left (B b c - A c^{2}\right )} x^{4} - 21 \, A b^{2} - 33 \, {\left (B b^{2} - A b c\right )} x^{2}\right )}}{231 \, b^{3} x^{\frac {11}{2}}} \] Input:

integrate((B*x^2+A)/x^(9/2)/(c*x^4+b*x^2),x, algorithm="maxima")
 

Output:

1/4*(2*sqrt(2)*(B*b*c^2 - 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))) + 2*sqrt(2)*(B*b*c^2 - 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)*sq 
rt(c))) + sqrt(2)*(B*b*c^2 - 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)*(B*b*c^2 - A*c^3)*log(-sq 
rt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(3/4)*c^(1/4)))/b^ 
3 + 2/231*(77*(B*b*c - A*c^2)*x^4 - 21*A*b^2 - 33*(B*b^2 - A*b*c)*x^2)/(b^ 
3*x^(11/2))
 

Giac [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.36 \[ \int \frac {A+B x^2}{x^{9/2} \left (b x^2+c x^4\right )} \, dx=\frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b c - \left (b c^{3}\right )^{\frac {1}{4}} A c^{2}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{2 \, b^{4}} + \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b c - \left (b c^{3}\right )^{\frac {1}{4}} A c^{2}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{2 \, b^{4}} + \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b c - \left (b c^{3}\right )^{\frac {1}{4}} A c^{2}\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, b^{4}} - \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b c - \left (b c^{3}\right )^{\frac {1}{4}} A c^{2}\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, b^{4}} + \frac {2 \, {\left (77 \, B b c x^{4} - 77 \, A c^{2} x^{4} - 33 \, B b^{2} x^{2} + 33 \, A b c x^{2} - 21 \, A b^{2}\right )}}{231 \, b^{3} x^{\frac {11}{2}}} \] Input:

integrate((B*x^2+A)/x^(9/2)/(c*x^4+b*x^2),x, algorithm="giac")
 

Output:

1/2*sqrt(2)*((b*c^3)^(1/4)*B*b*c - (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^4 + 1/2*sqrt(2)*((b*c^3) 
^(1/4)*B*b*c - (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^4 + 1/4*sqrt(2)*((b*c^3)^(1/4)*B*b*c - (b*c 
^3)^(1/4)*A*c^2)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/b^4 - 1/ 
4*sqrt(2)*((b*c^3)^(1/4)*B*b*c - (b*c^3)^(1/4)*A*c^2)*log(-sqrt(2)*sqrt(x) 
*(b/c)^(1/4) + x + sqrt(b/c))/b^4 + 2/231*(77*B*b*c*x^4 - 77*A*c^2*x^4 - 3 
3*B*b^2*x^2 + 33*A*b*c*x^2 - 21*A*b^2)/(b^3*x^(11/2))
 

Mupad [B] (verification not implemented)

Time = 9.29 (sec) , antiderivative size = 563, normalized size of antiderivative = 2.63 \[ \int \frac {A+B x^2}{x^{9/2} \left (b x^2+c x^4\right )} \, dx=\frac {{\left (-c\right )}^{7/4}\,\mathrm {atan}\left (\frac {A^3\,c^{10}\,\sqrt {x}-B^3\,b^3\,c^7\,\sqrt {x}-3\,A^2\,B\,b\,c^9\,\sqrt {x}+3\,A\,B^2\,b^2\,c^8\,\sqrt {x}}{b^{1/4}\,{\left (-c\right )}^{27/4}\,\left (c\,\left (c\,\left (A^3\,c-3\,A^2\,B\,b\right )+3\,A\,B^2\,b^2\right )-B^3\,b^3\right )}\right )\,\left (A\,c-B\,b\right )}{b^{15/4}}-\frac {\frac {2\,A}{11\,b}-\frac {2\,x^2\,\left (A\,c-B\,b\right )}{7\,b^2}+\frac {2\,c\,x^4\,\left (A\,c-B\,b\right )}{3\,b^3}}{x^{11/2}}-\frac {{\left (-c\right )}^{7/4}\,\mathrm {atan}\left (\frac {\frac {{\left (-c\right )}^{7/4}\,\left (A\,c-B\,b\right )\,\left (\sqrt {x}\,\left (16\,A^2\,b^9\,c^9-32\,A\,B\,b^{10}\,c^8+16\,B^2\,b^{11}\,c^7\right )-\frac {{\left (-c\right )}^{7/4}\,\left (A\,c-B\,b\right )\,\left (32\,A\,b^{13}\,c^6-32\,B\,b^{14}\,c^5\right )}{2\,b^{15/4}}\right )\,1{}\mathrm {i}}{2\,b^{15/4}}+\frac {{\left (-c\right )}^{7/4}\,\left (A\,c-B\,b\right )\,\left (\sqrt {x}\,\left (16\,A^2\,b^9\,c^9-32\,A\,B\,b^{10}\,c^8+16\,B^2\,b^{11}\,c^7\right )+\frac {{\left (-c\right )}^{7/4}\,\left (A\,c-B\,b\right )\,\left (32\,A\,b^{13}\,c^6-32\,B\,b^{14}\,c^5\right )}{2\,b^{15/4}}\right )\,1{}\mathrm {i}}{2\,b^{15/4}}}{\frac {{\left (-c\right )}^{7/4}\,\left (A\,c-B\,b\right )\,\left (\sqrt {x}\,\left (16\,A^2\,b^9\,c^9-32\,A\,B\,b^{10}\,c^8+16\,B^2\,b^{11}\,c^7\right )-\frac {{\left (-c\right )}^{7/4}\,\left (A\,c-B\,b\right )\,\left (32\,A\,b^{13}\,c^6-32\,B\,b^{14}\,c^5\right )}{2\,b^{15/4}}\right )}{2\,b^{15/4}}-\frac {{\left (-c\right )}^{7/4}\,\left (A\,c-B\,b\right )\,\left (\sqrt {x}\,\left (16\,A^2\,b^9\,c^9-32\,A\,B\,b^{10}\,c^8+16\,B^2\,b^{11}\,c^7\right )+\frac {{\left (-c\right )}^{7/4}\,\left (A\,c-B\,b\right )\,\left (32\,A\,b^{13}\,c^6-32\,B\,b^{14}\,c^5\right )}{2\,b^{15/4}}\right )}{2\,b^{15/4}}}\right )\,\left (A\,c-B\,b\right )\,1{}\mathrm {i}}{b^{15/4}} \] Input:

int((A + B*x^2)/(x^(9/2)*(b*x^2 + c*x^4)),x)
 

Output:

((-c)^(7/4)*atan((A^3*c^10*x^(1/2) - B^3*b^3*c^7*x^(1/2) - 3*A^2*B*b*c^9*x 
^(1/2) + 3*A*B^2*b^2*c^8*x^(1/2))/(b^(1/4)*(-c)^(27/4)*(c*(c*(A^3*c - 3*A^ 
2*B*b) + 3*A*B^2*b^2) - B^3*b^3)))*(A*c - B*b))/b^(15/4) - ((-c)^(7/4)*ata 
n((((-c)^(7/4)*(A*c - B*b)*(x^(1/2)*(16*A^2*b^9*c^9 + 16*B^2*b^11*c^7 - 32 
*A*B*b^10*c^8) - ((-c)^(7/4)*(A*c - B*b)*(32*A*b^13*c^6 - 32*B*b^14*c^5))/ 
(2*b^(15/4)))*1i)/(2*b^(15/4)) + ((-c)^(7/4)*(A*c - B*b)*(x^(1/2)*(16*A^2* 
b^9*c^9 + 16*B^2*b^11*c^7 - 32*A*B*b^10*c^8) + ((-c)^(7/4)*(A*c - B*b)*(32 
*A*b^13*c^6 - 32*B*b^14*c^5))/(2*b^(15/4)))*1i)/(2*b^(15/4)))/(((-c)^(7/4) 
*(A*c - B*b)*(x^(1/2)*(16*A^2*b^9*c^9 + 16*B^2*b^11*c^7 - 32*A*B*b^10*c^8) 
 - ((-c)^(7/4)*(A*c - B*b)*(32*A*b^13*c^6 - 32*B*b^14*c^5))/(2*b^(15/4)))) 
/(2*b^(15/4)) - ((-c)^(7/4)*(A*c - B*b)*(x^(1/2)*(16*A^2*b^9*c^9 + 16*B^2* 
b^11*c^7 - 32*A*B*b^10*c^8) + ((-c)^(7/4)*(A*c - B*b)*(32*A*b^13*c^6 - 32* 
B*b^14*c^5))/(2*b^(15/4))))/(2*b^(15/4))))*(A*c - B*b)*1i)/b^(15/4) - ((2* 
A)/(11*b) - (2*x^2*(A*c - B*b))/(7*b^2) + (2*c*x^4*(A*c - B*b))/(3*b^3))/x 
^(11/2)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.73 \[ \int \frac {A+B x^2}{x^{9/2} \left (b x^2+c x^4\right )} \, dx=\frac {462 \sqrt {x}\, c^{\frac {11}{4}} b^{\frac {1}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}-2 \sqrt {x}\, \sqrt {c}}{c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}}\right ) a \,x^{5}-462 \sqrt {x}\, c^{\frac {7}{4}} b^{\frac {9}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}-2 \sqrt {x}\, \sqrt {c}}{c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}}\right ) x^{5}-462 \sqrt {x}\, c^{\frac {11}{4}} b^{\frac {1}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+2 \sqrt {x}\, \sqrt {c}}{c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}}\right ) a \,x^{5}+462 \sqrt {x}\, c^{\frac {7}{4}} b^{\frac {9}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+2 \sqrt {x}\, \sqrt {c}}{c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}}\right ) x^{5}+231 \sqrt {x}\, c^{\frac {11}{4}} b^{\frac {1}{4}} \sqrt {2}\, \mathrm {log}\left (-\sqrt {x}\, c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+\sqrt {b}+\sqrt {c}\, x \right ) a \,x^{5}-231 \sqrt {x}\, c^{\frac {7}{4}} b^{\frac {9}{4}} \sqrt {2}\, \mathrm {log}\left (-\sqrt {x}\, c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+\sqrt {b}+\sqrt {c}\, x \right ) x^{5}-231 \sqrt {x}\, c^{\frac {11}{4}} b^{\frac {1}{4}} \sqrt {2}\, \mathrm {log}\left (\sqrt {x}\, c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+\sqrt {b}+\sqrt {c}\, x \right ) a \,x^{5}+231 \sqrt {x}\, c^{\frac {7}{4}} b^{\frac {9}{4}} \sqrt {2}\, \mathrm {log}\left (\sqrt {x}\, c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+\sqrt {b}+\sqrt {c}\, x \right ) x^{5}-168 a \,b^{3}+264 a \,b^{2} c \,x^{2}-616 a b \,c^{2} x^{4}-264 b^{4} x^{2}+616 b^{3} c \,x^{4}}{924 \sqrt {x}\, b^{4} x^{5}} \] Input:

int((B*x^2+A)/x^(9/2)/(c*x^4+b*x^2),x)
 

Output:

(462*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**2*x**5 - 462*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*x**5 - 462*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**2*x**5 + 462*sqrt(x)*c**(3/4)*b**(1/4)*sqrt(2)*ata 
n((c**(1/4)*b**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt( 
2)))*b**2*c*x**5 + 231*sqrt(x)*c**(3/4)*b**(1/4)*sqrt(2)*log( - sqrt(x)*c* 
*(1/4)*b**(1/4)*sqrt(2) + sqrt(b) + sqrt(c)*x)*a*c**2*x**5 - 231*sqrt(x)*c 
**(3/4)*b**(1/4)*sqrt(2)*log( - sqrt(x)*c**(1/4)*b**(1/4)*sqrt(2) + sqrt(b 
) + sqrt(c)*x)*b**2*c*x**5 - 231*sqrt(x)*c**(3/4)*b**(1/4)*sqrt(2)*log(sqr 
t(x)*c**(1/4)*b**(1/4)*sqrt(2) + sqrt(b) + sqrt(c)*x)*a*c**2*x**5 + 231*sq 
rt(x)*c**(3/4)*b**(1/4)*sqrt(2)*log(sqrt(x)*c**(1/4)*b**(1/4)*sqrt(2) + sq 
rt(b) + sqrt(c)*x)*b**2*c*x**5 - 168*a*b**3 + 264*a*b**2*c*x**2 - 616*a*b* 
c**2*x**4 - 264*b**4*x**2 + 616*b**3*c*x**4)/(924*sqrt(x)*b**4*x**5)