\(\int \frac {x^{3/2} (A+B x^2)}{(b x^2+c x^4)^2} \, dx\) [137]

Optimal result

Integrand size = 26, antiderivative size = 213 \[ \int \frac {x^{3/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {2 A}{3 b^2 x^{3/2}}+\frac {(b B-A c) \sqrt {x}}{2 b^2 \left (b+c x^2\right )}-\frac {(3 b B-7 A c) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{11/4} \sqrt [4]{c}}+\frac {(3 b B-7 A c) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{11/4} \sqrt [4]{c}}+\frac {(3 b B-7 A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{4 \sqrt {2} b^{11/4} \sqrt [4]{c}} \] Output:

-2/3*A/b^2/x^(3/2)+1/2*(-A*c+B*b)*x^(1/2)/b^2/(c*x^2+b)-1/8*(-7*A*c+3*B*b) 
*arctan(1-2^(1/2)*c^(1/4)*x^(1/2)/b^(1/4))*2^(1/2)/b^(11/4)/c^(1/4)+1/8*(- 
7*A*c+3*B*b)*arctan(1+2^(1/2)*c^(1/4)*x^(1/2)/b^(1/4))*2^(1/2)/b^(11/4)/c^ 
(1/4)+1/8*(-7*A*c+3*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^(11/4)/c^(1/4)
 

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.77 \[ \int \frac {x^{3/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {\frac {4 b^{3/4} \left (-4 A b+3 b B x^2-7 A c x^2\right )}{x^{3/2} \left (b+c x^2\right )}+\frac {3 \sqrt {2} (-3 b B+7 A c) \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )}{\sqrt [4]{c}}+\frac {3 \sqrt {2} (3 b B-7 A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{\sqrt [4]{c}}}{24 b^{11/4}} \] Input:

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

Output:

((4*b^(3/4)*(-4*A*b + 3*b*B*x^2 - 7*A*c*x^2))/(x^(3/2)*(b + c*x^2)) + (3*S 
qrt[2]*(-3*b*B + 7*A*c)*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^(1 
/4)*Sqrt[x])])/c^(1/4) + (3*Sqrt[2]*(3*b*B - 7*A*c)*ArcTanh[(Sqrt[2]*b^(1/ 
4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/c^(1/4))/(24*b^(11/4))
 

Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 283, 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, 264, 266, 755, 1476, 1082, 217, 1479, 25, 27, 1103}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 362

\(\displaystyle -\frac {(3 b B-7 A c) \int \frac {1}{x^{5/2} \left (c x^2+b\right )}dx}{4 b c}-\frac {b B-A c}{2 b c x^{3/2} \left (b+c x^2\right )}\)

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 755

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

-1/2*(b*B - A*c)/(b*c*x^(3/2)*(b + c*x^2)) - ((3*b*B - 7*A*c)*(-2/(3*b*x^( 
3/2)) - (2*c*((-(ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)]/(Sqrt[2]*b^ 
(1/4)*c^(1/4))) + ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)]/(Sqrt[2]*b 
^(1/4)*c^(1/4)))/(2*Sqrt[b]) + (-1/2*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4) 
*Sqrt[x] + Sqrt[c]*x]/(Sqrt[2]*b^(1/4)*c^(1/4)) + Log[Sqrt[b] + Sqrt[2]*b^ 
(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x]/(2*Sqrt[2]*b^(1/4)*c^(1/4)))/(2*Sqrt[b] 
)))/b))/(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 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 = 0.43 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.72

method	result	size
derivativedivides	\(-\frac {2 \left (\frac {\left (\frac {A c}{4}-\frac {B b}{4}\right ) \sqrt {x}}{c \,x^{2}+b}+\frac {\left (7 A c -3 B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{32 b}\right )}{b^{2}}-\frac {2 A}{3 b^{2} x^{\frac {3}{2}}}\)	\(153\)
default	\(-\frac {2 \left (\frac {\left (\frac {A c}{4}-\frac {B b}{4}\right ) \sqrt {x}}{c \,x^{2}+b}+\frac {\left (7 A c -3 B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{32 b}\right )}{b^{2}}-\frac {2 A}{3 b^{2} x^{\frac {3}{2}}}\)	\(153\)
risch	\(-\frac {2 A}{3 b^{2} x^{\frac {3}{2}}}-\frac {\frac {2 \left (\frac {A c}{4}-\frac {B b}{4}\right ) \sqrt {x}}{c \,x^{2}+b}+\frac {\left (7 A c -3 B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{16 b}}{b^{2}}\)	\(154\)

Input:

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

Output:

-2/b^2*((1/4*A*c-1/4*B*b)*x^(1/2)/(c*x^2+b)+1/32*(7*A*c-3*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*ar 
ctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)))-2/3*A/b^2/x^(3/2)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 691, normalized size of antiderivative = 3.24 \[ \int \frac {x^{3/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/24*(3*(b^2*c*x^4 + b^3*x^2)*(-(81*B^4*b^4 - 756*A*B^3*b^3*c + 2646*A^2* 
B^2*b^2*c^2 - 4116*A^3*B*b*c^3 + 2401*A^4*c^4)/(b^11*c))^(1/4)*log(b^3*(-( 
81*B^4*b^4 - 756*A*B^3*b^3*c + 2646*A^2*B^2*b^2*c^2 - 4116*A^3*B*b*c^3 + 2 
401*A^4*c^4)/(b^11*c))^(1/4) - (3*B*b - 7*A*c)*sqrt(x)) + 3*(I*b^2*c*x^4 + 
 I*b^3*x^2)*(-(81*B^4*b^4 - 756*A*B^3*b^3*c + 2646*A^2*B^2*b^2*c^2 - 4116* 
A^3*B*b*c^3 + 2401*A^4*c^4)/(b^11*c))^(1/4)*log(I*b^3*(-(81*B^4*b^4 - 756* 
A*B^3*b^3*c + 2646*A^2*B^2*b^2*c^2 - 4116*A^3*B*b*c^3 + 2401*A^4*c^4)/(b^1 
1*c))^(1/4) - (3*B*b - 7*A*c)*sqrt(x)) + 3*(-I*b^2*c*x^4 - I*b^3*x^2)*(-(8 
1*B^4*b^4 - 756*A*B^3*b^3*c + 2646*A^2*B^2*b^2*c^2 - 4116*A^3*B*b*c^3 + 24 
01*A^4*c^4)/(b^11*c))^(1/4)*log(-I*b^3*(-(81*B^4*b^4 - 756*A*B^3*b^3*c + 2 
646*A^2*B^2*b^2*c^2 - 4116*A^3*B*b*c^3 + 2401*A^4*c^4)/(b^11*c))^(1/4) - ( 
3*B*b - 7*A*c)*sqrt(x)) - 3*(b^2*c*x^4 + b^3*x^2)*(-(81*B^4*b^4 - 756*A*B^ 
3*b^3*c + 2646*A^2*B^2*b^2*c^2 - 4116*A^3*B*b*c^3 + 2401*A^4*c^4)/(b^11*c) 
)^(1/4)*log(-b^3*(-(81*B^4*b^4 - 756*A*B^3*b^3*c + 2646*A^2*B^2*b^2*c^2 - 
4116*A^3*B*b*c^3 + 2401*A^4*c^4)/(b^11*c))^(1/4) - (3*B*b - 7*A*c)*sqrt(x) 
) - 4*((3*B*b - 7*A*c)*x^2 - 4*A*b)*sqrt(x))/(b^2*c*x^4 + b^3*x^2)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.18 \[ \int \frac {x^{3/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {{\left (3 \, B b - 7 \, A c\right )} x^{2} - 4 \, A b}{6 \, {\left (b^{2} c x^{\frac {7}{2}} + b^{3} x^{\frac {3}{2}}\right )}} + \frac {\frac {2 \, \sqrt {2} {\left (3 \, B b - 7 \, A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {2 \, \sqrt {2} {\left (3 \, B b - 7 \, A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {\sqrt {2} {\left (3 \, B b - 7 \, A c\right )} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (3 \, B b - 7 \, A c\right )} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}}}{16 \, b^{2}} \] Input:

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

Output:

1/6*((3*B*b - 7*A*c)*x^2 - 4*A*b)/(b^2*c*x^(7/2) + b^3*x^(3/2)) + 1/16*(2* 
sqrt(2)*(3*B*b - 7*A*c)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) + 2*sq 
rt(c)*sqrt(x))/sqrt(sqrt(b)*sqrt(c)))/(sqrt(b)*sqrt(sqrt(b)*sqrt(c))) + 2* 
sqrt(2)*(3*B*b - 7*A*c)*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) - 2*s 
qrt(c)*sqrt(x))/sqrt(sqrt(b)*sqrt(c)))/(sqrt(b)*sqrt(sqrt(b)*sqrt(c))) + s 
qrt(2)*(3*B*b - 7*A*c)*log(sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + s 
qrt(b))/(b^(3/4)*c^(1/4)) - sqrt(2)*(3*B*b - 7*A*c)*log(-sqrt(2)*b^(1/4)*c 
^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(3/4)*c^(1/4)))/b^2
 

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.33 \[ \int \frac {x^{3/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {\sqrt {2} {\left (3 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 7 \, \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, b^{3} c} + \frac {\sqrt {2} {\left (3 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 7 \, \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, b^{3} c} + \frac {\sqrt {2} {\left (3 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 7 \, \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, b^{3} c} - \frac {\sqrt {2} {\left (3 \, \left (b c^{3}\right )^{\frac {1}{4}} B b - 7 \, \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, b^{3} c} + \frac {B b \sqrt {x} - A c \sqrt {x}}{2 \, {\left (c x^{2} + b\right )} b^{2}} - \frac {2 \, A}{3 \, b^{2} x^{\frac {3}{2}}} \] Input:

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

Output:

1/8*sqrt(2)*(3*(b*c^3)^(1/4)*B*b - 7*(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^3*c) + 1/8*sqrt(2)*(3*( 
b*c^3)^(1/4)*B*b - 7*(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^3*c) + 1/16*sqrt(2)*(3*(b*c^3)^(1/4)*B 
*b - 7*(b*c^3)^(1/4)*A*c)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c)) 
/(b^3*c) - 1/16*sqrt(2)*(3*(b*c^3)^(1/4)*B*b - 7*(b*c^3)^(1/4)*A*c)*log(-s 
qrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^3*c) + 1/2*(B*b*sqrt(x) - A 
*c*sqrt(x))/((c*x^2 + b)*b^2) - 2/3*A/(b^2*x^(3/2))
 

Mupad [B] (verification not implemented)

Time = 9.02 (sec) , antiderivative size = 859, normalized size of antiderivative = 4.03 \[ \int \frac {x^{3/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

- ((2*A)/(3*b) + (x^2*(7*A*c - 3*B*b))/(6*b^2))/(b*x^(3/2) + c*x^(7/2)) - 
(atan((((7*A*c - 3*B*b)*(x^(1/2)*(1568*A^2*b^6*c^5 + 288*B^2*b^8*c^3 - 134 
4*A*B*b^7*c^4) - ((7*A*c - 3*B*b)*(1792*A*b^9*c^4 - 768*B*b^10*c^3))/(8*(- 
b)^(11/4)*c^(1/4)))*1i)/(8*(-b)^(11/4)*c^(1/4)) + ((7*A*c - 3*B*b)*(x^(1/2 
)*(1568*A^2*b^6*c^5 + 288*B^2*b^8*c^3 - 1344*A*B*b^7*c^4) + ((7*A*c - 3*B* 
b)*(1792*A*b^9*c^4 - 768*B*b^10*c^3))/(8*(-b)^(11/4)*c^(1/4)))*1i)/(8*(-b) 
^(11/4)*c^(1/4)))/(((7*A*c - 3*B*b)*(x^(1/2)*(1568*A^2*b^6*c^5 + 288*B^2*b 
^8*c^3 - 1344*A*B*b^7*c^4) - ((7*A*c - 3*B*b)*(1792*A*b^9*c^4 - 768*B*b^10 
*c^3))/(8*(-b)^(11/4)*c^(1/4))))/(8*(-b)^(11/4)*c^(1/4)) - ((7*A*c - 3*B*b 
)*(x^(1/2)*(1568*A^2*b^6*c^5 + 288*B^2*b^8*c^3 - 1344*A*B*b^7*c^4) + ((7*A 
*c - 3*B*b)*(1792*A*b^9*c^4 - 768*B*b^10*c^3))/(8*(-b)^(11/4)*c^(1/4))))/( 
8*(-b)^(11/4)*c^(1/4))))*(7*A*c - 3*B*b)*1i)/(4*(-b)^(11/4)*c^(1/4)) - (at 
an((((7*A*c - 3*B*b)*(x^(1/2)*(1568*A^2*b^6*c^5 + 288*B^2*b^8*c^3 - 1344*A 
*B*b^7*c^4) - ((7*A*c - 3*B*b)*(1792*A*b^9*c^4 - 768*B*b^10*c^3)*1i)/(8*(- 
b)^(11/4)*c^(1/4))))/(8*(-b)^(11/4)*c^(1/4)) + ((7*A*c - 3*B*b)*(x^(1/2)*( 
1568*A^2*b^6*c^5 + 288*B^2*b^8*c^3 - 1344*A*B*b^7*c^4) + ((7*A*c - 3*B*b)* 
(1792*A*b^9*c^4 - 768*B*b^10*c^3)*1i)/(8*(-b)^(11/4)*c^(1/4))))/(8*(-b)^(1 
1/4)*c^(1/4)))/(((7*A*c - 3*B*b)*(x^(1/2)*(1568*A^2*b^6*c^5 + 288*B^2*b^8* 
c^3 - 1344*A*B*b^7*c^4) - ((7*A*c - 3*B*b)*(1792*A*b^9*c^4 - 768*B*b^10*c^ 
3)*1i)/(8*(-b)^(11/4)*c^(1/4)))*1i)/(8*(-b)^(11/4)*c^(1/4)) - ((7*A*c -...
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 663, normalized size of antiderivative = 3.11 \[ \int \frac {x^{3/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(42*sqrt(x)*c**(3/4)*b**(1/4)*sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) - 2* 
sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*a*b*c*x + 42*sqrt(x)*c**(3/4 
)*b**(1/4)*sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(c))/(c 
**(1/4)*b**(1/4)*sqrt(2)))*a*c**2*x**3 - 18*sqrt(x)*c**(3/4)*b**(1/4)*sqrt 
(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4 
)*sqrt(2)))*b**3*x - 18*sqrt(x)*c**(3/4)*b**(1/4)*sqrt(2)*atan((c**(1/4)*b 
**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*b**2*c*x 
**3 - 42*sqrt(x)*c**(3/4)*b**(1/4)*sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) 
 + 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*a*b*c*x - 42*sqrt(x)*c* 
*(3/4)*b**(1/4)*sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(c 
))/(c**(1/4)*b**(1/4)*sqrt(2)))*a*c**2*x**3 + 18*sqrt(x)*c**(3/4)*b**(1/4) 
*sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(c))/(c**(1/4)*b* 
*(1/4)*sqrt(2)))*b**3*x + 18*sqrt(x)*c**(3/4)*b**(1/4)*sqrt(2)*atan((c**(1 
/4)*b**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*b** 
2*c*x**3 + 21*sqrt(x)*c**(3/4)*b**(1/4)*sqrt(2)*log( - sqrt(x)*c**(1/4)*b* 
*(1/4)*sqrt(2) + sqrt(b) + sqrt(c)*x)*a*b*c*x + 21*sqrt(x)*c**(3/4)*b**(1/ 
4)*sqrt(2)*log( - sqrt(x)*c**(1/4)*b**(1/4)*sqrt(2) + sqrt(b) + sqrt(c)*x) 
*a*c**2*x**3 - 9*sqrt(x)*c**(3/4)*b**(1/4)*sqrt(2)*log( - sqrt(x)*c**(1/4) 
*b**(1/4)*sqrt(2) + sqrt(b) + sqrt(c)*x)*b**3*x - 9*sqrt(x)*c**(3/4)*b**(1 
/4)*sqrt(2)*log( - sqrt(x)*c**(1/4)*b**(1/4)*sqrt(2) + sqrt(b) + sqrt(c...