\(\int \frac {x^{19/2} (A+B x^2)}{(b x^2+c x^4)^3} \, dx\) [143]

Optimal result

Integrand size = 26, antiderivative size = 242 \[ \int \frac {x^{19/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {2 B \sqrt {x}}{c^3}-\frac {b (b B-A c) \sqrt {x}}{4 c^3 \left (b+c x^2\right )^2}+\frac {(17 b B-9 A c) \sqrt {x}}{16 c^3 \left (b+c x^2\right )}+\frac {5 (9 b B-A c) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{3/4} c^{13/4}}-\frac {5 (9 b B-A c) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{3/4} c^{13/4}}-\frac {5 (9 b B-A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{32 \sqrt {2} b^{3/4} c^{13/4}} \] Output:

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

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.76 \[ \int \frac {x^{19/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {\frac {4 \sqrt [4]{c} \sqrt {x} \left (45 b^2 B-5 A b c+81 b B c x^2-9 A c^2 x^2+32 B c^2 x^4\right )}{\left (b+c x^2\right )^2}+\frac {5 \sqrt {2} (9 b B-A c) \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )}{b^{3/4}}-\frac {5 \sqrt {2} (9 b B-A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{b^{3/4}}}{64 c^{13/4}} \] Input:

Integrate[(x^(19/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]
 

Output:

((4*c^(1/4)*Sqrt[x]*(45*b^2*B - 5*A*b*c + 81*b*B*c*x^2 - 9*A*c^2*x^2 + 32* 
B*c^2*x^4))/(b + c*x^2)^2 + (5*Sqrt[2]*(9*b*B - A*c)*ArcTan[(Sqrt[b] - Sqr 
t[c]*x)/(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])])/b^(3/4) - (5*Sqrt[2]*(9*b*B - 
A*c)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/b^( 
3/4))/(64*c^(13/4))
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.28, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {9, 362, 252, 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.

\(\displaystyle \int \frac {x^{19/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx\)

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 362

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

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 755

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

Int[(x^(19/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x]
 

Output:

-1/4*((b*B - A*c)*x^(9/2))/(b*c*(b + c*x^2)^2) + ((9*b*B - A*c)*(-1/2*x^(5 
/2)/(c*(b + c*x^2)) + (5*((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)*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*Sq 
rt[2]*b^(1/4)*c^(1/4)))/(2*Sqrt[b])))/c))/(4*c)))/(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 252

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x 
)^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* 
(p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c 
}, x] && LtQ[p, -1] && GtQ[m, 1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi 
alQ[a, b, c, 2, m, p, x]
 

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 = 0.48 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.71

method	result	size
derivativedivides	\(\frac {2 B \sqrt {x}}{c^{3}}+\frac {\frac {2 \left (\left (-\frac {9}{32} A \,c^{2}+\frac {17}{32} B b c \right ) x^{\frac {5}{2}}-\frac {b \left (5 A c -13 B b \right ) \sqrt {x}}{32}\right )}{\left (c \,x^{2}+b \right )^{2}}+\frac {5 \left (A c -9 B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{128 b}}{c^{3}}\)	\(172\)
default	\(\frac {2 B \sqrt {x}}{c^{3}}+\frac {\frac {2 \left (\left (-\frac {9}{32} A \,c^{2}+\frac {17}{32} B b c \right ) x^{\frac {5}{2}}-\frac {b \left (5 A c -13 B b \right ) \sqrt {x}}{32}\right )}{\left (c \,x^{2}+b \right )^{2}}+\frac {5 \left (A c -9 B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{128 b}}{c^{3}}\)	\(172\)
risch	\(\frac {2 B \sqrt {x}}{c^{3}}+\frac {\frac {2 \left (-\frac {9}{32} A \,c^{2}+\frac {17}{32} B b c \right ) x^{\frac {5}{2}}-\frac {b \left (5 A c -13 B b \right ) \sqrt {x}}{16}}{\left (c \,x^{2}+b \right )^{2}}+\frac {5 \left (A c -9 B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{128 b}}{c^{3}}\)	\(172\)

Input:

int(x^(19/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)
 

Output:

2*B*x^(1/2)/c^3+2/c^3*(((-9/32*A*c^2+17/32*B*b*c)*x^(5/2)-1/32*b*(5*A*c-13 
*B*b)*x^(1/2))/(c*x^2+b)^2+5/256*(A*c-9*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 = 0.12 (sec) , antiderivative size = 748, normalized size of antiderivative = 3.09 \[ \int \frac {x^{19/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx =\text {Too large to display} \] Input:

integrate(x^(19/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="fricas")
 

Output:

1/64*(5*(c^5*x^4 + 2*b*c^4*x^2 + b^2*c^3)*(-(6561*B^4*b^4 - 2916*A*B^3*b^3 
*c + 486*A^2*B^2*b^2*c^2 - 36*A^3*B*b*c^3 + A^4*c^4)/(b^3*c^13))^(1/4)*log 
(5*b*c^3*(-(6561*B^4*b^4 - 2916*A*B^3*b^3*c + 486*A^2*B^2*b^2*c^2 - 36*A^3 
*B*b*c^3 + A^4*c^4)/(b^3*c^13))^(1/4) - 5*(9*B*b - A*c)*sqrt(x)) - 5*(-I*c 
^5*x^4 - 2*I*b*c^4*x^2 - I*b^2*c^3)*(-(6561*B^4*b^4 - 2916*A*B^3*b^3*c + 4 
86*A^2*B^2*b^2*c^2 - 36*A^3*B*b*c^3 + A^4*c^4)/(b^3*c^13))^(1/4)*log(5*I*b 
*c^3*(-(6561*B^4*b^4 - 2916*A*B^3*b^3*c + 486*A^2*B^2*b^2*c^2 - 36*A^3*B*b 
*c^3 + A^4*c^4)/(b^3*c^13))^(1/4) - 5*(9*B*b - A*c)*sqrt(x)) - 5*(I*c^5*x^ 
4 + 2*I*b*c^4*x^2 + I*b^2*c^3)*(-(6561*B^4*b^4 - 2916*A*B^3*b^3*c + 486*A^ 
2*B^2*b^2*c^2 - 36*A^3*B*b*c^3 + A^4*c^4)/(b^3*c^13))^(1/4)*log(-5*I*b*c^3 
*(-(6561*B^4*b^4 - 2916*A*B^3*b^3*c + 486*A^2*B^2*b^2*c^2 - 36*A^3*B*b*c^3 
 + A^4*c^4)/(b^3*c^13))^(1/4) - 5*(9*B*b - A*c)*sqrt(x)) - 5*(c^5*x^4 + 2* 
b*c^4*x^2 + b^2*c^3)*(-(6561*B^4*b^4 - 2916*A*B^3*b^3*c + 486*A^2*B^2*b^2* 
c^2 - 36*A^3*B*b*c^3 + A^4*c^4)/(b^3*c^13))^(1/4)*log(-5*b*c^3*(-(6561*B^4 
*b^4 - 2916*A*B^3*b^3*c + 486*A^2*B^2*b^2*c^2 - 36*A^3*B*b*c^3 + A^4*c^4)/ 
(b^3*c^13))^(1/4) - 5*(9*B*b - A*c)*sqrt(x)) + 4*(32*B*c^2*x^4 + 45*B*b^2 
- 5*A*b*c + 9*(9*B*b*c - A*c^2)*x^2)*sqrt(x))/(c^5*x^4 + 2*b*c^4*x^2 + b^2 
*c^3)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{19/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\text {Timed out} \] Input:

integrate(x**(19/2)*(B*x**2+A)/(c*x**4+b*x**2)**3,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.17 \[ \int \frac {x^{19/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {{\left (17 \, B b c - 9 \, A c^{2}\right )} x^{\frac {5}{2}} + {\left (13 \, B b^{2} - 5 \, A b c\right )} \sqrt {x}}{16 \, {\left (c^{5} x^{4} + 2 \, b c^{4} x^{2} + b^{2} c^{3}\right )}} + \frac {2 \, B \sqrt {x}}{c^{3}} - \frac {5 \, {\left (\frac {2 \, \sqrt {2} {\left (9 \, B b - A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {2 \, \sqrt {2} {\left (9 \, B b - A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {\sqrt {2} {\left (9 \, B b - A c\right )} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (9 \, B b - A c\right )} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}}\right )}}{128 \, c^{3}} \] Input:

integrate(x^(19/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="maxima")
 

Output:

1/16*((17*B*b*c - 9*A*c^2)*x^(5/2) + (13*B*b^2 - 5*A*b*c)*sqrt(x))/(c^5*x^ 
4 + 2*b*c^4*x^2 + b^2*c^3) + 2*B*sqrt(x)/c^3 - 5/128*(2*sqrt(2)*(9*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)*(9*B*b - A*c 
)*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(b)*sqrt(sqrt(b)*sqrt(c))) + sqrt(2)*(9*B*b - A*c)*l 
og(sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(3/4)*c^(1/4) 
) - sqrt(2)*(9*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^3
 

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.26 \[ \int \frac {x^{19/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx=\frac {2 \, B \sqrt {x}}{c^{3}} - \frac {5 \, \sqrt {2} {\left (9 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{64 \, b c^{4}} - \frac {5 \, \sqrt {2} {\left (9 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{64 \, b c^{4}} - \frac {5 \, \sqrt {2} {\left (9 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b c^{4}} + \frac {5 \, \sqrt {2} {\left (9 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b c^{4}} + \frac {17 \, B b c x^{\frac {5}{2}} - 9 \, A c^{2} x^{\frac {5}{2}} + 13 \, B b^{2} \sqrt {x} - 5 \, A b c \sqrt {x}}{16 \, {\left (c x^{2} + b\right )}^{2} c^{3}} \] Input:

integrate(x^(19/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x, algorithm="giac")
 

Output:

2*B*sqrt(x)/c^3 - 5/64*sqrt(2)*(9*(b*c^3)^(1/4)*B*b - (b*c^3)^(1/4)*A*c)*a 
rctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))/(b/c)^(1/4))/(b*c^4) - 
 5/64*sqrt(2)*(9*(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^4) - 5/128*sqrt(2)*( 
9*(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^4) + 5/128*sqrt(2)*(9*(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^4) + 1/16*( 
17*B*b*c*x^(5/2) - 9*A*c^2*x^(5/2) + 13*B*b^2*sqrt(x) - 5*A*b*c*sqrt(x))/( 
(c*x^2 + b)^2*c^3)
 

Mupad [B] (verification not implemented)

Time = 9.63 (sec) , antiderivative size = 760, normalized size of antiderivative = 3.14 \[ \int \frac {x^{19/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx =\text {Too large to display} \] Input:

int((x^(19/2)*(A + B*x^2))/(b*x^2 + c*x^4)^3,x)
 

Output:

(x^(1/2)*((13*B*b^2)/16 - (5*A*b*c)/16) - x^(5/2)*((9*A*c^2)/16 - (17*B*b* 
c)/16))/(b^2*c^3 + c^5*x^4 + 2*b*c^4*x^2) + (2*B*x^(1/2))/c^3 - (atan((((A 
*c - 9*B*b)*((25*x^(1/2)*(A^2*c^2 + 81*B^2*b^2 - 18*A*B*b*c))/(64*c^3) - ( 
5*(45*B*b^2 - 5*A*b*c)*(A*c - 9*B*b))/(64*(-b)^(3/4)*c^(13/4)))*5i)/(64*(- 
b)^(3/4)*c^(13/4)) + ((A*c - 9*B*b)*((25*x^(1/2)*(A^2*c^2 + 81*B^2*b^2 - 1 
8*A*B*b*c))/(64*c^3) + (5*(45*B*b^2 - 5*A*b*c)*(A*c - 9*B*b))/(64*(-b)^(3/ 
4)*c^(13/4)))*5i)/(64*(-b)^(3/4)*c^(13/4)))/((5*(A*c - 9*B*b)*((25*x^(1/2) 
*(A^2*c^2 + 81*B^2*b^2 - 18*A*B*b*c))/(64*c^3) - (5*(45*B*b^2 - 5*A*b*c)*( 
A*c - 9*B*b))/(64*(-b)^(3/4)*c^(13/4))))/(64*(-b)^(3/4)*c^(13/4)) - (5*(A* 
c - 9*B*b)*((25*x^(1/2)*(A^2*c^2 + 81*B^2*b^2 - 18*A*B*b*c))/(64*c^3) + (5 
*(45*B*b^2 - 5*A*b*c)*(A*c - 9*B*b))/(64*(-b)^(3/4)*c^(13/4))))/(64*(-b)^( 
3/4)*c^(13/4))))*(A*c - 9*B*b)*5i)/(32*(-b)^(3/4)*c^(13/4)) - (5*atan(((5* 
(A*c - 9*B*b)*((25*x^(1/2)*(A^2*c^2 + 81*B^2*b^2 - 18*A*B*b*c))/(64*c^3) - 
 ((45*B*b^2 - 5*A*b*c)*(A*c - 9*B*b)*5i)/(64*(-b)^(3/4)*c^(13/4))))/(64*(- 
b)^(3/4)*c^(13/4)) + (5*(A*c - 9*B*b)*((25*x^(1/2)*(A^2*c^2 + 81*B^2*b^2 - 
 18*A*B*b*c))/(64*c^3) + ((45*B*b^2 - 5*A*b*c)*(A*c - 9*B*b)*5i)/(64*(-b)^ 
(3/4)*c^(13/4))))/(64*(-b)^(3/4)*c^(13/4)))/(((A*c - 9*B*b)*((25*x^(1/2)*( 
A^2*c^2 + 81*B^2*b^2 - 18*A*B*b*c))/(64*c^3) - ((45*B*b^2 - 5*A*b*c)*(A*c 
- 9*B*b)*5i)/(64*(-b)^(3/4)*c^(13/4)))*5i)/(64*(-b)^(3/4)*c^(13/4)) - ((A* 
c - 9*B*b)*((25*x^(1/2)*(A^2*c^2 + 81*B^2*b^2 - 18*A*B*b*c))/(64*c^3) +...
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 956, normalized size of antiderivative = 3.95 \[ \int \frac {x^{19/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^3} \, dx =\text {Too large to display} \] Input:

int(x^(19/2)*(B*x^2+A)/(c*x^4+b*x^2)^3,x)
 

Output:

( - 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)))*a*b**2*c - 20*c**(3/4)*b**(1/4)*s 
qrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**( 
1/4)*sqrt(2)))*a*b*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)))*a*c**3 
*x**4 + 90*c**(3/4)*b**(1/4)*sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) - 2*s 
qrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*b**4 + 180*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*c*x**2 + 90*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* 
*2*x**4 + 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)))*a*b**2*c + 20*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*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)))* 
a*c**3*x**4 - 90*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 - 180*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*c*x**2 - 90*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))...