\(\int \frac {A+B x^2}{x^{3/2} (b x^2+c x^4)^2} \, dx\) [140]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [C] (verification not implemented)
Sympy [F(-1)]
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 26, antiderivative size = 256 \[ \int \frac {A+B x^2}{x^{3/2} \left (b x^2+c x^4\right )^2} \, dx=-\frac {2 A}{9 b^2 x^{9/2}}-\frac {2 (b B-2 A c)}{5 b^3 x^{5/2}}+\frac {2 c (2 b B-3 A c)}{b^4 \sqrt {x}}+\frac {c^2 (b B-A c) x^{3/2}}{2 b^4 \left (b+c x^2\right )}-\frac {c^{5/4} (9 b B-13 A c) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{17/4}}+\frac {c^{5/4} (9 b B-13 A c) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{17/4}}-\frac {c^{5/4} (9 b B-13 A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{4 \sqrt {2} b^{17/4}} \] Output:

-2/9*A/b^2/x^(9/2)-2/5*(-2*A*c+B*b)/b^3/x^(5/2)+2*c*(-3*A*c+2*B*b)/b^4/x^( 
1/2)+1/2*c^2*(-A*c+B*b)*x^(3/2)/b^4/(c*x^2+b)-1/8*c^(5/4)*(-13*A*c+9*B*b)* 
arctan(1-2^(1/2)*c^(1/4)*x^(1/2)/b^(1/4))*2^(1/2)/b^(17/4)+1/8*c^(5/4)*(-1 
3*A*c+9*B*b)*arctan(1+2^(1/2)*c^(1/4)*x^(1/2)/b^(1/4))*2^(1/2)/b^(17/4)-1/ 
8*c^(5/4)*(-13*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^(17/4)
 

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.82 \[ \int \frac {A+B x^2}{x^{3/2} \left (b x^2+c x^4\right )^2} \, dx=\frac {-\frac {4 \sqrt [4]{b} \left (9 b B x^2 \left (4 b^2-36 b c x^2-45 c^2 x^4\right )+A \left (20 b^3-52 b^2 c x^2+468 b c^2 x^4+585 c^3 x^6\right )\right )}{x^{9/2} \left (b+c x^2\right )}+45 \sqrt {2} c^{5/4} (-9 b B+13 A c) \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )+45 \sqrt {2} c^{5/4} (-9 b B+13 A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{360 b^{17/4}} \] Input:

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

Output:

((-4*b^(1/4)*(9*b*B*x^2*(4*b^2 - 36*b*c*x^2 - 45*c^2*x^4) + A*(20*b^3 - 52 
*b^2*c*x^2 + 468*b*c^2*x^4 + 585*c^3*x^6)))/(x^(9/2)*(b + c*x^2)) + 45*Sqr 
t[2]*c^(5/4)*(-9*b*B + 13*A*c)*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/ 
4)*c^(1/4)*Sqrt[x])] + 45*Sqrt[2]*c^(5/4)*(-9*b*B + 13*A*c)*ArcTanh[(Sqrt[ 
2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(360*b^(17/4))
 

Rubi [A] (verified)

Time = 0.91 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.25, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {9, 362, 264, 264, 264, 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.

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

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 362

\(\displaystyle -\frac {(9 b B-13 A c) \int \frac {1}{x^{11/2} \left (c x^2+b\right )}dx}{4 b c}-\frac {b B-A c}{2 b c x^{9/2} \left (b+c x^2\right )}\)

\(\Big \downarrow \) 264

\(\displaystyle -\frac {(9 b B-13 A c) \left (-\frac {c \int \frac {1}{x^{7/2} \left (c x^2+b\right )}dx}{b}-\frac {2}{9 b x^{9/2}}\right )}{4 b c}-\frac {b B-A c}{2 b c x^{9/2} \left (b+c x^2\right )}\)

\(\Big \downarrow \) 264

\(\displaystyle -\frac {(9 b B-13 A c) \left (-\frac {c \left (-\frac {c \int \frac {1}{x^{3/2} \left (c x^2+b\right )}dx}{b}-\frac {2}{5 b x^{5/2}}\right )}{b}-\frac {2}{9 b x^{9/2}}\right )}{4 b c}-\frac {b B-A c}{2 b c x^{9/2} \left (b+c x^2\right )}\)

\(\Big \downarrow \) 264

\(\displaystyle -\frac {(9 b B-13 A c) \left (-\frac {c \left (-\frac {c \left (-\frac {c \int \frac {\sqrt {x}}{c x^2+b}dx}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{b}-\frac {2}{9 b x^{9/2}}\right )}{4 b c}-\frac {b B-A c}{2 b c x^{9/2} \left (b+c x^2\right )}\)

\(\Big \downarrow \) 266

\(\displaystyle -\frac {(9 b B-13 A c) \left (-\frac {c \left (-\frac {c \left (-\frac {2 c \int \frac {x}{c x^2+b}d\sqrt {x}}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{b}-\frac {2}{9 b x^{9/2}}\right )}{4 b c}-\frac {b B-A c}{2 b c x^{9/2} \left (b+c x^2\right )}\)

\(\Big \downarrow \) 826

\(\displaystyle -\frac {(9 b B-13 A c) \left (-\frac {c \left (-\frac {c \left (-\frac {2 c \left (\frac {\int \frac {\sqrt {c} x+\sqrt {b}}{c x^2+b}d\sqrt {x}}{2 \sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{b}-\frac {2}{9 b x^{9/2}}\right )}{4 b c}-\frac {b B-A c}{2 b c x^{9/2} \left (b+c x^2\right )}\)

\(\Big \downarrow \) 1476

\(\displaystyle -\frac {(9 b B-13 A c) \left (-\frac {c \left (-\frac {c \left (-\frac {2 c \left (\frac {\frac {\int \frac {1}{x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {c}}+\frac {\int \frac {1}{x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {c}}}{2 \sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{b}-\frac {2}{9 b x^{9/2}}\right )}{4 b c}-\frac {b B-A c}{2 b c x^{9/2} \left (b+c x^2\right )}\)

\(\Big \downarrow \) 1082

\(\displaystyle -\frac {(9 b B-13 A c) \left (-\frac {c \left (-\frac {c \left (-\frac {2 c \left (\frac {\frac {\int \frac {1}{-x-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\int \frac {1}{-x-1}d\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{b}-\frac {2}{9 b x^{9/2}}\right )}{4 b c}-\frac {b B-A c}{2 b c x^{9/2} \left (b+c x^2\right )}\)

\(\Big \downarrow \) 217

\(\displaystyle -\frac {(9 b B-13 A c) \left (-\frac {c \left (-\frac {c \left (-\frac {2 c \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\int \frac {\sqrt {b}-\sqrt {c} x}{c x^2+b}d\sqrt {x}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{b}-\frac {2}{9 b x^{9/2}}\right )}{4 b c}-\frac {b B-A c}{2 b c x^{9/2} \left (b+c x^2\right )}\)

\(\Big \downarrow \) 1479

\(\displaystyle -\frac {(9 b B-13 A c) \left (-\frac {c \left (-\frac {c \left (-\frac {2 c \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {-\frac {\int -\frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{c} \left (x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}\right )}{\sqrt [4]{c} \left (x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{b}-\frac {2}{9 b x^{9/2}}\right )}{4 b c}-\frac {b B-A c}{2 b c x^{9/2} \left (b+c x^2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {(9 b B-13 A c) \left (-\frac {c \left (-\frac {c \left (-\frac {2 c \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{c} \left (x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}\right )}{\sqrt [4]{c} \left (x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}\right )}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{b}-\frac {2}{9 b x^{9/2}}\right )}{4 b c}-\frac {b B-A c}{2 b c x^{9/2} \left (b+c x^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {(9 b B-13 A c) \left (-\frac {c \left (-\frac {c \left (-\frac {2 c \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{c} \sqrt {x}}{x-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt {2} \sqrt [4]{b} \sqrt {c}}+\frac {\int \frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}+\sqrt [4]{b}}{x+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{c}}+\frac {\sqrt {b}}{\sqrt {c}}}d\sqrt {x}}{2 \sqrt [4]{b} \sqrt {c}}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{b}-\frac {2}{9 b x^{9/2}}\right )}{4 b c}-\frac {b B-A c}{2 b c x^{9/2} \left (b+c x^2\right )}\)

\(\Big \downarrow \) 1103

\(\displaystyle -\frac {(9 b B-13 A c) \left (-\frac {c \left (-\frac {c \left (-\frac {2 c \left (\frac {\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}-\frac {\frac {\log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}-\frac {\log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} \sqrt [4]{b} \sqrt [4]{c}}}{2 \sqrt {c}}\right )}{b}-\frac {2}{b \sqrt {x}}\right )}{b}-\frac {2}{5 b x^{5/2}}\right )}{b}-\frac {2}{9 b x^{9/2}}\right )}{4 b c}-\frac {b B-A c}{2 b c x^{9/2} \left (b+c x^2\right )}\)

Input:

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

Output:

-1/2*(b*B - A*c)/(b*c*x^(9/2)*(b + c*x^2)) - ((9*b*B - 13*A*c)*(-2/(9*b*x^ 
(9/2)) - (c*(-2/(5*b*x^(5/2)) - (c*(-2/(b*Sqrt[x]) - (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[c]) 
- (-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[c])))/b))/b))/b))/(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 264
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 266
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 362
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*b*e 
*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(2*a*b*(p + 1))   I 
nt[(e*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && N 
eQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) || 
  !RationalQ[m] || (ILtQ[p + 1/2, 0] && LeQ[-1, m, -2*(p + 1)]))
 

rule 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 = 190, normalized size of antiderivative = 0.74

method result size
derivativedivides \(-\frac {2 A}{9 b^{2} x^{\frac {9}{2}}}-\frac {2 \left (-2 A c +B b \right )}{5 b^{3} x^{\frac {5}{2}}}-\frac {2 c \left (3 A c -2 B b \right )}{b^{4} \sqrt {x}}-\frac {2 c^{2} \left (\frac {\left (\frac {A c}{4}-\frac {B b}{4}\right ) x^{\frac {3}{2}}}{c \,x^{2}+b}+\frac {\left (\frac {13 A c}{4}-\frac {9 B b}{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{8 c \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{b^{4}}\) \(190\)
default \(-\frac {2 A}{9 b^{2} x^{\frac {9}{2}}}-\frac {2 \left (-2 A c +B b \right )}{5 b^{3} x^{\frac {5}{2}}}-\frac {2 c \left (3 A c -2 B b \right )}{b^{4} \sqrt {x}}-\frac {2 c^{2} \left (\frac {\left (\frac {A c}{4}-\frac {B b}{4}\right ) x^{\frac {3}{2}}}{c \,x^{2}+b}+\frac {\left (\frac {13 A c}{4}-\frac {9 B b}{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{8 c \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{b^{4}}\) \(190\)
risch \(-\frac {2 \left (135 x^{4} A \,c^{2}-90 x^{4} B b c -18 A b c \,x^{2}+9 x^{2} B \,b^{2}+5 b^{2} A \right )}{45 b^{4} x^{\frac {9}{2}}}-\frac {c^{2} \left (\frac {2 \left (\frac {A c}{4}-\frac {B b}{4}\right ) x^{\frac {3}{2}}}{c \,x^{2}+b}+\frac {\left (\frac {13 A c}{4}-\frac {9 B b}{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 c \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{b^{4}}\) \(197\)

Input:

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

Output:

-2/9*A/b^2/x^(9/2)-2/5*(-2*A*c+B*b)/b^3/x^(5/2)-2*c*(3*A*c-2*B*b)/b^4/x^(1 
/2)-2/b^4*c^2*((1/4*A*c-1/4*B*b)*x^(3/2)/(c*x^2+b)+1/8*(13/4*A*c-9/4*B*b)/ 
c/(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 = 886, normalized size of antiderivative = 3.46 \[ \int \frac {A+B x^2}{x^{3/2} \left (b x^2+c x^4\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/360*(45*(b^4*c*x^7 + b^5*x^5)*(-(6561*B^4*b^4*c^5 - 37908*A*B^3*b^3*c^6 
 + 82134*A^2*B^2*b^2*c^7 - 79092*A^3*B*b*c^8 + 28561*A^4*c^9)/b^17)^(1/4)* 
log(b^13*(-(6561*B^4*b^4*c^5 - 37908*A*B^3*b^3*c^6 + 82134*A^2*B^2*b^2*c^7 
 - 79092*A^3*B*b*c^8 + 28561*A^4*c^9)/b^17)^(3/4) - (729*B^3*b^3*c^4 - 315 
9*A*B^2*b^2*c^5 + 4563*A^2*B*b*c^6 - 2197*A^3*c^7)*sqrt(x)) + 45*(-I*b^4*c 
*x^7 - I*b^5*x^5)*(-(6561*B^4*b^4*c^5 - 37908*A*B^3*b^3*c^6 + 82134*A^2*B^ 
2*b^2*c^7 - 79092*A^3*B*b*c^8 + 28561*A^4*c^9)/b^17)^(1/4)*log(I*b^13*(-(6 
561*B^4*b^4*c^5 - 37908*A*B^3*b^3*c^6 + 82134*A^2*B^2*b^2*c^7 - 79092*A^3* 
B*b*c^8 + 28561*A^4*c^9)/b^17)^(3/4) - (729*B^3*b^3*c^4 - 3159*A*B^2*b^2*c 
^5 + 4563*A^2*B*b*c^6 - 2197*A^3*c^7)*sqrt(x)) + 45*(I*b^4*c*x^7 + I*b^5*x 
^5)*(-(6561*B^4*b^4*c^5 - 37908*A*B^3*b^3*c^6 + 82134*A^2*B^2*b^2*c^7 - 79 
092*A^3*B*b*c^8 + 28561*A^4*c^9)/b^17)^(1/4)*log(-I*b^13*(-(6561*B^4*b^4*c 
^5 - 37908*A*B^3*b^3*c^6 + 82134*A^2*B^2*b^2*c^7 - 79092*A^3*B*b*c^8 + 285 
61*A^4*c^9)/b^17)^(3/4) - (729*B^3*b^3*c^4 - 3159*A*B^2*b^2*c^5 + 4563*A^2 
*B*b*c^6 - 2197*A^3*c^7)*sqrt(x)) - 45*(b^4*c*x^7 + b^5*x^5)*(-(6561*B^4*b 
^4*c^5 - 37908*A*B^3*b^3*c^6 + 82134*A^2*B^2*b^2*c^7 - 79092*A^3*B*b*c^8 + 
 28561*A^4*c^9)/b^17)^(1/4)*log(-b^13*(-(6561*B^4*b^4*c^5 - 37908*A*B^3*b^ 
3*c^6 + 82134*A^2*B^2*b^2*c^7 - 79092*A^3*B*b*c^8 + 28561*A^4*c^9)/b^17)^( 
3/4) - (729*B^3*b^3*c^4 - 3159*A*B^2*b^2*c^5 + 4563*A^2*B*b*c^6 - 2197*A^3 
*c^7)*sqrt(x)) - 4*(45*(9*B*b*c^2 - 13*A*c^3)*x^6 + 36*(9*B*b^2*c - 13*...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.08 \[ \int \frac {A+B x^2}{x^{3/2} \left (b x^2+c x^4\right )^2} \, dx=\frac {45 \, {\left (9 \, B b c^{2} - 13 \, A c^{3}\right )} x^{6} + 36 \, {\left (9 \, B b^{2} c - 13 \, A b c^{2}\right )} x^{4} - 20 \, A b^{3} - 4 \, {\left (9 \, B b^{3} - 13 \, A b^{2} c\right )} x^{2}}{90 \, {\left (b^{4} c x^{\frac {13}{2}} + b^{5} x^{\frac {9}{2}}\right )}} + \frac {{\left (9 \, B b c^{2} - 13 \, A c^{3}\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}}\right )}}{16 \, b^{4}} \] Input:

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

Output:

1/90*(45*(9*B*b*c^2 - 13*A*c^3)*x^6 + 36*(9*B*b^2*c - 13*A*b*c^2)*x^4 - 20 
*A*b^3 - 4*(9*B*b^3 - 13*A*b^2*c)*x^2)/(b^4*c*x^(13/2) + b^5*x^(9/2)) + 1/ 
16*(9*B*b*c^2 - 13*A*c^3)*(2*sqrt(2)*arctan(1/2*sqrt(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)) + 2*sqrt(2)*arctan(-1/2*sqrt(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)*sqrt(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^4
 

Giac [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.28 \[ \int \frac {A+B x^2}{x^{3/2} \left (b x^2+c x^4\right )^2} \, dx=\frac {\sqrt {2} {\left (9 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 13 \, \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, b^{5} c} + \frac {\sqrt {2} {\left (9 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 13 \, \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, b^{5} c} - \frac {\sqrt {2} {\left (9 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 13 \, \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, b^{5} c} + \frac {\sqrt {2} {\left (9 \, \left (b c^{3}\right )^{\frac {3}{4}} B b - 13 \, \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, b^{5} c} + \frac {B b c^{2} x^{\frac {3}{2}} - A c^{3} x^{\frac {3}{2}}}{2 \, {\left (c x^{2} + b\right )} b^{4}} + \frac {2 \, {\left (90 \, B b c x^{4} - 135 \, A c^{2} x^{4} - 9 \, B b^{2} x^{2} + 18 \, A b c x^{2} - 5 \, A b^{2}\right )}}{45 \, b^{4} x^{\frac {9}{2}}} \] Input:

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

Output:

1/8*sqrt(2)*(9*(b*c^3)^(3/4)*B*b - 13*(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^5*c) + 1/8*sqrt(2)*(9* 
(b*c^3)^(3/4)*B*b - 13*(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^5*c) - 1/16*sqrt(2)*(9*(b*c^3)^(3/4) 
*B*b - 13*(b*c^3)^(3/4)*A*c)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/ 
c))/(b^5*c) + 1/16*sqrt(2)*(9*(b*c^3)^(3/4)*B*b - 13*(b*c^3)^(3/4)*A*c)*lo 
g(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^5*c) + 1/2*(B*b*c^2*x^( 
3/2) - A*c^3*x^(3/2))/((c*x^2 + b)*b^4) + 2/45*(90*B*b*c*x^4 - 135*A*c^2*x 
^4 - 9*B*b^2*x^2 + 18*A*b*c*x^2 - 5*A*b^2)/(b^4*x^(9/2))
 

Mupad [B] (verification not implemented)

Time = 9.62 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.55 \[ \int \frac {A+B x^2}{x^{3/2} \left (b x^2+c x^4\right )^2} \, dx=\frac {{\left (-c\right )}^{5/4}\,\mathrm {atan}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )\,\left (13\,A\,c-9\,B\,b\right )}{4\,b^{17/4}}-\frac {\frac {2\,A}{9\,b}-\frac {2\,x^2\,\left (13\,A\,c-9\,B\,b\right )}{45\,b^2}+\frac {c^2\,x^6\,\left (13\,A\,c-9\,B\,b\right )}{2\,b^4}+\frac {2\,c\,x^4\,\left (13\,A\,c-9\,B\,b\right )}{5\,b^3}}{b\,x^{9/2}+c\,x^{13/2}}-\frac {{\left (-c\right )}^{5/4}\,\mathrm {atanh}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )\,\left (13\,A\,c-9\,B\,b\right )}{4\,b^{17/4}} \] Input:

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

Output:

((-c)^(5/4)*atan(((-c)^(1/4)*x^(1/2))/b^(1/4))*(13*A*c - 9*B*b))/(4*b^(17/ 
4)) - ((2*A)/(9*b) - (2*x^2*(13*A*c - 9*B*b))/(45*b^2) + (c^2*x^6*(13*A*c 
- 9*B*b))/(2*b^4) + (2*c*x^4*(13*A*c - 9*B*b))/(5*b^3))/(b*x^(9/2) + c*x^( 
13/2)) - ((-c)^(5/4)*atanh(((-c)^(1/4)*x^(1/2))/b^(1/4))*(13*A*c - 9*B*b)) 
/(4*b^(17/4))
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 716, normalized size of antiderivative = 2.80 \[ \int \frac {A+B x^2}{x^{3/2} \left (b x^2+c x^4\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(1170*sqrt(x)*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*x**4 + 1170*sqrt( 
x)*c**(1/4)*b**(3/4)*sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) - 2*sqrt(x)*s 
qrt(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*a*c**3*x**6 - 810*sqrt(x)*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*c*x**4 - 810*sqrt(x)*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)*s 
qrt(2)))*b**2*c**2*x**6 - 1170*sqrt(x)*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*x**4 - 1170*sqrt(x)*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**3*x**6 
+ 810*sqrt(x)*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*c*x**4 + 810*sqrt(x)* 
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**2*x**6 - 585*sqrt(x)*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**2*x**4 - 585*sqrt(x)*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**3*x**6 + 405*sqrt(x) 
*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*c*x**4 + 405*sqrt(x)*c**(1/4)*b**(3/4)*sqrt(2)*lo...