\(\int \frac {x^{9/2} (A+B x^2)}{(b x^2+c x^4)^2} \, dx\) [134]

Optimal result

Integrand size = 26, antiderivative size = 200 \[ \int \frac {x^{9/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {(b B-A c) x^{3/2}}{2 b c \left (b+c x^2\right )}-\frac {(3 b B+A c) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{5/4} c^{7/4}}+\frac {(3 b B+A c) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{5/4} c^{7/4}}-\frac {(3 b B+A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{4 \sqrt {2} b^{5/4} c^{7/4}} \] Output:

-1/2*(-A*c+B*b)*x^(3/2)/b/c/(c*x^2+b)-1/8*(A*c+3*B*b)*arctan(1-2^(1/2)*c^( 
1/4)*x^(1/2)/b^(1/4))*2^(1/2)/b^(5/4)/c^(7/4)+1/8*(A*c+3*B*b)*arctan(1+2^( 
1/2)*c^(1/4)*x^(1/2)/b^(1/4))*2^(1/2)/b^(5/4)/c^(7/4)-1/8*(A*c+3*B*b)*arct 
anh(2^(1/2)*b^(1/4)*c^(1/4)*x^(1/2)/(b^(1/2)+c^(1/2)*x))*2^(1/2)/b^(5/4)/c 
^(7/4)
 

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.76 \[ \int \frac {x^{9/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {\frac {4 \sqrt [4]{b} c^{3/4} (-b B+A c) x^{3/2}}{b+c x^2}-\sqrt {2} (3 b B+A c) \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )-\sqrt {2} (3 b B+A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{8 b^{5/4} c^{7/4}} \] Input:

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

Output:

((4*b^(1/4)*c^(3/4)*(-(b*B) + A*c)*x^(3/2))/(b + c*x^2) - Sqrt[2]*(3*b*B + 
 A*c)*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])] - Sq 
rt[2]*(3*b*B + A*c)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + S 
qrt[c]*x)])/(8*b^(5/4)*c^(7/4))
 

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.32, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {9, 362, 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.

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

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 362

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

-1/2*((b*B - A*c)*x^(3/2))/(b*c*(b + c*x^2)) + ((3*b*B + A*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]/(Sq 
rt[2]*b^(1/4)*c^(1/4)) + Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + S 
qrt[c]*x]/(2*Sqrt[2]*b^(1/4)*c^(1/4)))/(2*Sqrt[c])))/(2*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 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 = 0.43 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.73

method	result	size
derivativedivides	\(\frac {\left (A c -B b \right ) x^{\frac {3}{2}}}{2 b c \left (c \,x^{2}+b \right )}+\frac {\left (A c +3 B b \right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{16 b \,c^{2} \left (\frac {b}{c}\right )^{\frac {1}{4}}}\)	\(146\)
default	\(\frac {\left (A c -B b \right ) x^{\frac {3}{2}}}{2 b c \left (c \,x^{2}+b \right )}+\frac {\left (A c +3 B b \right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{16 b \,c^{2} \left (\frac {b}{c}\right )^{\frac {1}{4}}}\)	\(146\)

Input:

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

Output:

1/2*(A*c-B*b)/b/c*x^(3/2)/(c*x^2+b)+1/16*(A*c+3*B*b)/b/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 = 0.11 (sec) , antiderivative size = 776, normalized size of antiderivative = 3.88 \[ \int \frac {x^{9/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/8*(4*(B*b - A*c)*x^(3/2) - (b*c^2*x^2 + b^2*c)*(-(81*B^4*b^4 + 108*A*B^ 
3*b^3*c + 54*A^2*B^2*b^2*c^2 + 12*A^3*B*b*c^3 + A^4*c^4)/(b^5*c^7))^(1/4)* 
log(b^4*c^5*(-(81*B^4*b^4 + 108*A*B^3*b^3*c + 54*A^2*B^2*b^2*c^2 + 12*A^3* 
B*b*c^3 + A^4*c^4)/(b^5*c^7))^(3/4) + (27*B^3*b^3 + 27*A*B^2*b^2*c + 9*A^2 
*B*b*c^2 + A^3*c^3)*sqrt(x)) + (I*b*c^2*x^2 + I*b^2*c)*(-(81*B^4*b^4 + 108 
*A*B^3*b^3*c + 54*A^2*B^2*b^2*c^2 + 12*A^3*B*b*c^3 + A^4*c^4)/(b^5*c^7))^( 
1/4)*log(I*b^4*c^5*(-(81*B^4*b^4 + 108*A*B^3*b^3*c + 54*A^2*B^2*b^2*c^2 + 
12*A^3*B*b*c^3 + A^4*c^4)/(b^5*c^7))^(3/4) + (27*B^3*b^3 + 27*A*B^2*b^2*c 
+ 9*A^2*B*b*c^2 + A^3*c^3)*sqrt(x)) + (-I*b*c^2*x^2 - I*b^2*c)*(-(81*B^4*b 
^4 + 108*A*B^3*b^3*c + 54*A^2*B^2*b^2*c^2 + 12*A^3*B*b*c^3 + A^4*c^4)/(b^5 
*c^7))^(1/4)*log(-I*b^4*c^5*(-(81*B^4*b^4 + 108*A*B^3*b^3*c + 54*A^2*B^2*b 
^2*c^2 + 12*A^3*B*b*c^3 + A^4*c^4)/(b^5*c^7))^(3/4) + (27*B^3*b^3 + 27*A*B 
^2*b^2*c + 9*A^2*B*b*c^2 + A^3*c^3)*sqrt(x)) + (b*c^2*x^2 + b^2*c)*(-(81*B 
^4*b^4 + 108*A*B^3*b^3*c + 54*A^2*B^2*b^2*c^2 + 12*A^3*B*b*c^3 + A^4*c^4)/ 
(b^5*c^7))^(1/4)*log(-b^4*c^5*(-(81*B^4*b^4 + 108*A*B^3*b^3*c + 54*A^2*B^2 
*b^2*c^2 + 12*A^3*B*b*c^3 + A^4*c^4)/(b^5*c^7))^(3/4) + (27*B^3*b^3 + 27*A 
*B^2*b^2*c + 9*A^2*B*b*c^2 + A^3*c^3)*sqrt(x)))/(b*c^2*x^2 + b^2*c)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{9/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.08 \[ \int \frac {x^{9/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {{\left (B b - A c\right )} x^{\frac {3}{2}}}{2 \, {\left (b c^{2} x^{2} + b^{2} c\right )}} + \frac {{\left (3 \, B b + A c\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}}\right )}}{16 \, b c} \] Input:

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

Output:

-1/2*(B*b - A*c)*x^(3/2)/(b*c^2*x^2 + b^2*c) + 1/16*(3*B*b + A*c)*(2*sqrt( 
2)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) + 2*sqrt(c)*sqrt(x))/sqrt(s 
qrt(b)*sqrt(c)))/(sqrt(sqrt(b)*sqrt(c))*sqrt(c)) + 2*sqrt(2)*arctan(-1/2*s 
qrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) - 2*sqrt(c)*sqrt(x))/sqrt(sqrt(b)*sqrt(c)) 
)/(sqrt(sqrt(b)*sqrt(c))*sqrt(c)) - sqrt(2)*log(sqrt(2)*b^(1/4)*c^(1/4)*sq 
rt(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*c)
 

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.36 \[ \int \frac {x^{9/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {B b x^{\frac {3}{2}} - A c x^{\frac {3}{2}}}{2 \, {\left (c x^{2} + b\right )} b c} + \frac {\sqrt {2} {\left (3 \, \left (b c^{3}\right )^{\frac {3}{4}} B b + \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, b^{2} c^{4}} + \frac {\sqrt {2} {\left (3 \, \left (b c^{3}\right )^{\frac {3}{4}} B b + \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, b^{2} c^{4}} - \frac {\sqrt {2} {\left (3 \, \left (b c^{3}\right )^{\frac {3}{4}} B b + \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, b^{2} c^{4}} + \frac {\sqrt {2} {\left (3 \, \left (b c^{3}\right )^{\frac {3}{4}} B b + \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, b^{2} c^{4}} \] Input:

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

Output:

-1/2*(B*b*x^(3/2) - A*c*x^(3/2))/((c*x^2 + b)*b*c) + 1/8*sqrt(2)*(3*(b*c^3 
)^(3/4)*B*b + (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^2*c^4) + 1/8*sqrt(2)*(3*(b*c^3)^(3/4)*B*b + (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^2*c^4) - 1/16*sqrt(2)*(3*(b*c^3)^(3/4)*B*b + (b*c^3)^(3/4)*A* 
c)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^2*c^4) + 1/16*sqrt( 
2)*(3*(b*c^3)^(3/4)*B*b + (b*c^3)^(3/4)*A*c)*log(-sqrt(2)*sqrt(x)*(b/c)^(1 
/4) + x + sqrt(b/c))/(b^2*c^4)
 

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.46 \[ \int \frac {x^{9/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {\mathrm {atanh}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )\,\left (A\,c+3\,B\,b\right )}{4\,{\left (-b\right )}^{5/4}\,c^{7/4}}-\frac {\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )\,\left (A\,c+3\,B\,b\right )}{4\,{\left (-b\right )}^{5/4}\,c^{7/4}}+\frac {x^{3/2}\,\left (A\,c-B\,b\right )}{2\,b\,c\,\left (c\,x^2+b\right )} \] Input:

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

Output:

(atanh((c^(1/4)*x^(1/2))/(-b)^(1/4))*(A*c + 3*B*b))/(4*(-b)^(5/4)*c^(7/4)) 
 - (atan((c^(1/4)*x^(1/2))/(-b)^(1/4))*(A*c + 3*B*b))/(4*(-b)^(5/4)*c^(7/4 
)) + (x^(3/2)*(A*c - B*b))/(2*b*c*(b + c*x^2))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 607, normalized size of antiderivative = 3.04 \[ \int \frac {x^{9/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

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