\(\int \frac {x^{11/2} (A+B x^2)}{(b x^2+c x^4)^2} \, dx\) [133]

Optimal result

Integrand size = 26, antiderivative size = 211 \[ \int \frac {x^{11/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {2 B \sqrt {x}}{c^2}+\frac {(b B-A c) \sqrt {x}}{2 c^2 \left (b+c x^2\right )}+\frac {(5 b B-A c) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{3/4} c^{9/4}}-\frac {(5 b B-A c) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{3/4} c^{9/4}}-\frac {(5 b B-A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{4 \sqrt {2} b^{3/4} c^{9/4}} \] Output:

2*B*x^(1/2)/c^2+1/2*(-A*c+B*b)*x^(1/2)/c^2/(c*x^2+b)+1/8*(-A*c+5*B*b)*arct 
an(1-2^(1/2)*c^(1/4)*x^(1/2)/b^(1/4))*2^(1/2)/b^(3/4)/c^(9/4)-1/8*(-A*c+5* 
B*b)*arctan(1+2^(1/2)*c^(1/4)*x^(1/2)/b^(1/4))*2^(1/2)/b^(3/4)/c^(9/4)-1/8 
*(-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^(3/4)/c^(9/4)
 

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.76 \[ \int \frac {x^{11/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {\frac {4 \sqrt [4]{c} \sqrt {x} \left (5 b B-A c+4 B c x^2\right )}{b+c x^2}+\frac {\sqrt {2} (5 b B-A c) \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )}{b^{3/4}}-\frac {\sqrt {2} (5 b B-A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{b^{3/4}}}{8 c^{9/4}} \] Input:

Integrate[(x^(11/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]
 

Output:

((4*c^(1/4)*Sqrt[x]*(5*b*B - A*c + 4*B*c*x^2))/(b + c*x^2) + (Sqrt[2]*(5*b 
*B - A*c)*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])]) 
/b^(3/4) - (Sqrt[2]*(5*b*B - A*c)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] 
)/(Sqrt[b] + Sqrt[c]*x)])/b^(3/4))/(8*c^(9/4))
 

Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.33, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {9, 362, 262, 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.

\(\displaystyle \int \frac {x^{11/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx\)

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 362

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

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 755

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

Int[(x^(11/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]
 

Output:

-1/2*((b*B - A*c)*x^(5/2))/(b*c*(b + c*x^2)) + ((5*b*B - A*c)*((2*Sqrt[x]) 
/c - (2*b*((-(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)*Sq 
rt[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]))) 
/c))/(4*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 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && 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 = 0.45 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.72

method	result	size
derivativedivides	\(\frac {2 B \sqrt {x}}{c^{2}}+\frac {\frac {2 \left (-\frac {A c}{4}+\frac {B b}{4}\right ) \sqrt {x}}{c \,x^{2}+b}+\frac {\left (A c -5 B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{16 b}}{c^{2}}\)	\(152\)
default	\(\frac {2 B \sqrt {x}}{c^{2}}+\frac {\frac {2 \left (-\frac {A c}{4}+\frac {B b}{4}\right ) \sqrt {x}}{c \,x^{2}+b}+\frac {\left (A c -5 B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{16 b}}{c^{2}}\)	\(152\)
risch	\(\frac {2 B \sqrt {x}}{c^{2}}+\frac {\frac {2 \left (-\frac {A c}{4}+\frac {B b}{4}\right ) \sqrt {x}}{c \,x^{2}+b}+\frac {\left (A c -5 B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{16 b}}{c^{2}}\)	\(152\)

Input:

int(x^(11/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x,method=_RETURNVERBOSE)
 

Output:

2*B*x^(1/2)/c^2+2/c^2*((-1/4*A*c+1/4*B*b)*x^(1/2)/(c*x^2+b)+1/32*(A*c-5*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)))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 669, normalized size of antiderivative = 3.17 \[ \int \frac {x^{11/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {{\left (c^{3} x^{2} + b c^{2}\right )} \left (-\frac {625 \, B^{4} b^{4} - 500 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 20 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{9}}\right )^{\frac {1}{4}} \log \left (b c^{2} \left (-\frac {625 \, B^{4} b^{4} - 500 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 20 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{9}}\right )^{\frac {1}{4}} - {\left (5 \, B b - A c\right )} \sqrt {x}\right ) - {\left (-i \, c^{3} x^{2} - i \, b c^{2}\right )} \left (-\frac {625 \, B^{4} b^{4} - 500 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 20 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{9}}\right )^{\frac {1}{4}} \log \left (i \, b c^{2} \left (-\frac {625 \, B^{4} b^{4} - 500 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 20 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{9}}\right )^{\frac {1}{4}} - {\left (5 \, B b - A c\right )} \sqrt {x}\right ) - {\left (i \, c^{3} x^{2} + i \, b c^{2}\right )} \left (-\frac {625 \, B^{4} b^{4} - 500 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 20 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{9}}\right )^{\frac {1}{4}} \log \left (-i \, b c^{2} \left (-\frac {625 \, B^{4} b^{4} - 500 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 20 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{9}}\right )^{\frac {1}{4}} - {\left (5 \, B b - A c\right )} \sqrt {x}\right ) - {\left (c^{3} x^{2} + b c^{2}\right )} \left (-\frac {625 \, B^{4} b^{4} - 500 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 20 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{9}}\right )^{\frac {1}{4}} \log \left (-b c^{2} \left (-\frac {625 \, B^{4} b^{4} - 500 \, A B^{3} b^{3} c + 150 \, A^{2} B^{2} b^{2} c^{2} - 20 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{9}}\right )^{\frac {1}{4}} - {\left (5 \, B b - A c\right )} \sqrt {x}\right ) + 4 \, {\left (4 \, B c x^{2} + 5 \, B b - A c\right )} \sqrt {x}}{8 \, {\left (c^{3} x^{2} + b c^{2}\right )}} \] Input:

integrate(x^(11/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="fricas")
 

Output:

1/8*((c^3*x^2 + b*c^2)*(-(625*B^4*b^4 - 500*A*B^3*b^3*c + 150*A^2*B^2*b^2* 
c^2 - 20*A^3*B*b*c^3 + A^4*c^4)/(b^3*c^9))^(1/4)*log(b*c^2*(-(625*B^4*b^4 
- 500*A*B^3*b^3*c + 150*A^2*B^2*b^2*c^2 - 20*A^3*B*b*c^3 + A^4*c^4)/(b^3*c 
^9))^(1/4) - (5*B*b - A*c)*sqrt(x)) - (-I*c^3*x^2 - I*b*c^2)*(-(625*B^4*b^ 
4 - 500*A*B^3*b^3*c + 150*A^2*B^2*b^2*c^2 - 20*A^3*B*b*c^3 + A^4*c^4)/(b^3 
*c^9))^(1/4)*log(I*b*c^2*(-(625*B^4*b^4 - 500*A*B^3*b^3*c + 150*A^2*B^2*b^ 
2*c^2 - 20*A^3*B*b*c^3 + A^4*c^4)/(b^3*c^9))^(1/4) - (5*B*b - A*c)*sqrt(x) 
) - (I*c^3*x^2 + I*b*c^2)*(-(625*B^4*b^4 - 500*A*B^3*b^3*c + 150*A^2*B^2*b 
^2*c^2 - 20*A^3*B*b*c^3 + A^4*c^4)/(b^3*c^9))^(1/4)*log(-I*b*c^2*(-(625*B^ 
4*b^4 - 500*A*B^3*b^3*c + 150*A^2*B^2*b^2*c^2 - 20*A^3*B*b*c^3 + A^4*c^4)/ 
(b^3*c^9))^(1/4) - (5*B*b - A*c)*sqrt(x)) - (c^3*x^2 + b*c^2)*(-(625*B^4*b 
^4 - 500*A*B^3*b^3*c + 150*A^2*B^2*b^2*c^2 - 20*A^3*B*b*c^3 + A^4*c^4)/(b^ 
3*c^9))^(1/4)*log(-b*c^2*(-(625*B^4*b^4 - 500*A*B^3*b^3*c + 150*A^2*B^2*b^ 
2*c^2 - 20*A^3*B*b*c^3 + A^4*c^4)/(b^3*c^9))^(1/4) - (5*B*b - A*c)*sqrt(x) 
) + 4*(4*B*c*x^2 + 5*B*b - A*c)*sqrt(x))/(c^3*x^2 + b*c^2)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{11/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\text {Timed out} \] Input:

integrate(x**(11/2)*(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 = 250, normalized size of antiderivative = 1.18 \[ \int \frac {x^{11/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {{\left (B b - A c\right )} \sqrt {x}}{2 \, {\left (c^{3} x^{2} + b c^{2}\right )}} + \frac {2 \, B \sqrt {x}}{c^{2}} - \frac {\frac {2 \, \sqrt {2} {\left (5 \, B b - A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {2 \, \sqrt {2} {\left (5 \, B b - A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {\sqrt {2} {\left (5 \, B b - A c\right )} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (5 \, B b - A c\right )} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}}}{16 \, c^{2}} \] Input:

integrate(x^(11/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.34 \[ \int \frac {x^{11/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {2 \, B \sqrt {x}}{c^{2}} - \frac {\sqrt {2} {\left (5 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, b c^{3}} - \frac {\sqrt {2} {\left (5 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, b c^{3}} - \frac {\sqrt {2} {\left (5 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, b c^{3}} + \frac {\sqrt {2} {\left (5 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, b c^{3}} + \frac {B b \sqrt {x} - A c \sqrt {x}}{2 \, {\left (c x^{2} + b\right )} c^{2}} \] Input:

integrate(x^(11/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="giac")
 

Output:

2*B*sqrt(x)/c^2 - 1/8*sqrt(2)*(5*(b*c^3)^(1/4)*B*b - (b*c^3)^(1/4)*A*c)*ar 
ctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))/(b/c)^(1/4))/(b*c^3) - 
1/8*sqrt(2)*(5*(b*c^3)^(1/4)*B*b - (b*c^3)^(1/4)*A*c)*arctan(-1/2*sqrt(2)* 
(sqrt(2)*(b/c)^(1/4) - 2*sqrt(x))/(b/c)^(1/4))/(b*c^3) - 1/16*sqrt(2)*(5*( 
b*c^3)^(1/4)*B*b - (b*c^3)^(1/4)*A*c)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x 
+ sqrt(b/c))/(b*c^3) + 1/16*sqrt(2)*(5*(b*c^3)^(1/4)*B*b - (b*c^3)^(1/4)*A 
*c)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b*c^3) + 1/2*(B*b*s 
qrt(x) - A*c*sqrt(x))/((c*x^2 + b)*c^2)
 

Mupad [B] (verification not implemented)

Time = 9.30 (sec) , antiderivative size = 744, normalized size of antiderivative = 3.53 \[ \int \frac {x^{11/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx =\text {Too large to display} \] Input:

int((x^(11/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x)
 

Output:

(2*B*x^(1/2))/c^2 - (x^(1/2)*((A*c)/2 - (B*b)/2))/(b*c^2 + c^3*x^2) + (ata 
n((((A*c - 5*B*b)*((x^(1/2)*(A^2*c^2 + 25*B^2*b^2 - 10*A*B*b*c))/c - ((A*c 
 - 5*B*b)*(8*A*b*c^2 - 40*B*b^2*c))/(8*(-b)^(3/4)*c^(9/4)))*1i)/(8*(-b)^(3 
/4)*c^(9/4)) + ((A*c - 5*B*b)*((x^(1/2)*(A^2*c^2 + 25*B^2*b^2 - 10*A*B*b*c 
))/c + ((A*c - 5*B*b)*(8*A*b*c^2 - 40*B*b^2*c))/(8*(-b)^(3/4)*c^(9/4)))*1i 
)/(8*(-b)^(3/4)*c^(9/4)))/(((A*c - 5*B*b)*((x^(1/2)*(A^2*c^2 + 25*B^2*b^2 
- 10*A*B*b*c))/c - ((A*c - 5*B*b)*(8*A*b*c^2 - 40*B*b^2*c))/(8*(-b)^(3/4)* 
c^(9/4))))/(8*(-b)^(3/4)*c^(9/4)) - ((A*c - 5*B*b)*((x^(1/2)*(A^2*c^2 + 25 
*B^2*b^2 - 10*A*B*b*c))/c + ((A*c - 5*B*b)*(8*A*b*c^2 - 40*B*b^2*c))/(8*(- 
b)^(3/4)*c^(9/4))))/(8*(-b)^(3/4)*c^(9/4))))*(A*c - 5*B*b)*1i)/(4*(-b)^(3/ 
4)*c^(9/4)) + (atan((((A*c - 5*B*b)*((x^(1/2)*(A^2*c^2 + 25*B^2*b^2 - 10*A 
*B*b*c))/c - ((A*c - 5*B*b)*(8*A*b*c^2 - 40*B*b^2*c)*1i)/(8*(-b)^(3/4)*c^( 
9/4))))/(8*(-b)^(3/4)*c^(9/4)) + ((A*c - 5*B*b)*((x^(1/2)*(A^2*c^2 + 25*B^ 
2*b^2 - 10*A*B*b*c))/c + ((A*c - 5*B*b)*(8*A*b*c^2 - 40*B*b^2*c)*1i)/(8*(- 
b)^(3/4)*c^(9/4))))/(8*(-b)^(3/4)*c^(9/4)))/(((A*c - 5*B*b)*((x^(1/2)*(A^2 
*c^2 + 25*B^2*b^2 - 10*A*B*b*c))/c - ((A*c - 5*B*b)*(8*A*b*c^2 - 40*B*b^2* 
c)*1i)/(8*(-b)^(3/4)*c^(9/4)))*1i)/(8*(-b)^(3/4)*c^(9/4)) - ((A*c - 5*B*b) 
*((x^(1/2)*(A^2*c^2 + 25*B^2*b^2 - 10*A*B*b*c))/c + ((A*c - 5*B*b)*(8*A*b* 
c^2 - 40*B*b^2*c)*1i)/(8*(-b)^(3/4)*c^(9/4)))*1i)/(8*(-b)^(3/4)*c^(9/4)))) 
*(A*c - 5*B*b))/(4*(-b)^(3/4)*c^(9/4))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 618, normalized size of antiderivative = 2.93 \[ \int \frac {x^{11/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx =\text {Too large to display} \] Input:

int(x^(11/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x)
 

Output:

( - 2*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 - 2*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**2 + 10*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**3 + 10*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**2 + 2*c**(3/4)*b**(1/4)*sqrt(2)*a 
tan((c**(1/4)*b**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sqr 
t(2)))*a*b*c + 2*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**2 - 10*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**3 - 10*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**2 - c**(3/4)*b**(1/4)*sqrt(2)*log( - sqrt(x)*c**(1/4)*b**(1/4)*sqr 
t(2) + sqrt(b) + sqrt(c)*x)*a*b*c - 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**2 + 5*c**(3/ 
4)*b**(1/4)*sqrt(2)*log( - sqrt(x)*c**(1/4)*b**(1/4)*sqrt(2) + sqrt(b) + s 
qrt(c)*x)*b**3 + 5*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**2 + 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*b*c + c...