\(\int \frac {x^{5/2} (A+B x^2)}{(b x^2+c x^4)^2} \, dx\) [136]

Optimal result

Integrand size = 26, antiderivative size = 208 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {2 A}{b^2 \sqrt {x}}+\frac {(b B-A c) x^{3/2}}{2 b^2 \left (b+c x^2\right )}-\frac {(b B-5 A c) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{9/4} c^{3/4}}+\frac {(b B-5 A c) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{9/4} c^{3/4}}-\frac {(b B-5 A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{4 \sqrt {2} b^{9/4} c^{3/4}} \] Output:

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

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.78 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {\frac {4 \sqrt [4]{b} \left (-4 A b+b B x^2-5 A c x^2\right )}{\sqrt {x} \left (b+c x^2\right )}+\frac {\sqrt {2} (-b B+5 A c) \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )}{c^{3/4}}+\frac {\sqrt {2} (-b B+5 A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{c^{3/4}}}{8 b^{9/4}} \] Input:

Integrate[(x^(5/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.35, 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, 826, 1476, 1082, 217, 1479, 25, 27, 1103}

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

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

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 362

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

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

Int[(x^(5/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]
 

Output:

-1/2*(b*B - A*c)/(b*c*Sqrt[x]*(b + c*x^2)) - ((b*B - 5*A*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)*Sqr 
t[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))/(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 = 153, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(-\frac {2 A}{b^{2} \sqrt {x}}-\frac {2 \left (\frac {\left (\frac {A c}{4}-\frac {B b}{4}\right ) x^{\frac {3}{2}}}{c \,x^{2}+b}+\frac {\left (\frac {5 A c}{4}-\frac {B b}{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{8 c \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{b^{2}}\)	\(153\)
default	\(-\frac {2 A}{b^{2} \sqrt {x}}-\frac {2 \left (\frac {\left (\frac {A c}{4}-\frac {B b}{4}\right ) x^{\frac {3}{2}}}{c \,x^{2}+b}+\frac {\left (\frac {5 A c}{4}-\frac {B b}{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{8 c \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{b^{2}}\)	\(153\)
risch	\(-\frac {2 A}{b^{2} \sqrt {x}}-\frac {\frac {2 \left (\frac {A c}{4}-\frac {B b}{4}\right ) x^{\frac {3}{2}}}{c \,x^{2}+b}+\frac {\left (\frac {5 A c}{4}-\frac {B b}{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 c \left (\frac {b}{c}\right )^{\frac {1}{4}}}}{b^{2}}\)	\(154\)

Input:

int(x^(5/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x,method=_RETURNVERBOSE)
 

Output:

-2*A/b^2/x^(1/2)-2/b^2*((1/4*A*c-1/4*B*b)*x^(3/2)/(c*x^2+b)+1/8*(5/4*A*c-1 
/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 = 0.12 (sec) , antiderivative size = 789, normalized size of antiderivative = 3.79 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx =\text {Too large to display} \] Input:

integrate(x^(5/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="fricas")
 

Output:

-1/8*((b^2*c*x^3 + b^3*x)*(-(B^4*b^4 - 20*A*B^3*b^3*c + 150*A^2*B^2*b^2*c^ 
2 - 500*A^3*B*b*c^3 + 625*A^4*c^4)/(b^9*c^3))^(1/4)*log(b^7*c^2*(-(B^4*b^4 
 - 20*A*B^3*b^3*c + 150*A^2*B^2*b^2*c^2 - 500*A^3*B*b*c^3 + 625*A^4*c^4)/( 
b^9*c^3))^(3/4) - (B^3*b^3 - 15*A*B^2*b^2*c + 75*A^2*B*b*c^2 - 125*A^3*c^3 
)*sqrt(x)) + (-I*b^2*c*x^3 - I*b^3*x)*(-(B^4*b^4 - 20*A*B^3*b^3*c + 150*A^ 
2*B^2*b^2*c^2 - 500*A^3*B*b*c^3 + 625*A^4*c^4)/(b^9*c^3))^(1/4)*log(I*b^7* 
c^2*(-(B^4*b^4 - 20*A*B^3*b^3*c + 150*A^2*B^2*b^2*c^2 - 500*A^3*B*b*c^3 + 
625*A^4*c^4)/(b^9*c^3))^(3/4) - (B^3*b^3 - 15*A*B^2*b^2*c + 75*A^2*B*b*c^2 
 - 125*A^3*c^3)*sqrt(x)) + (I*b^2*c*x^3 + I*b^3*x)*(-(B^4*b^4 - 20*A*B^3*b 
^3*c + 150*A^2*B^2*b^2*c^2 - 500*A^3*B*b*c^3 + 625*A^4*c^4)/(b^9*c^3))^(1/ 
4)*log(-I*b^7*c^2*(-(B^4*b^4 - 20*A*B^3*b^3*c + 150*A^2*B^2*b^2*c^2 - 500* 
A^3*B*b*c^3 + 625*A^4*c^4)/(b^9*c^3))^(3/4) - (B^3*b^3 - 15*A*B^2*b^2*c + 
75*A^2*B*b*c^2 - 125*A^3*c^3)*sqrt(x)) - (b^2*c*x^3 + b^3*x)*(-(B^4*b^4 - 
20*A*B^3*b^3*c + 150*A^2*B^2*b^2*c^2 - 500*A^3*B*b*c^3 + 625*A^4*c^4)/(b^9 
*c^3))^(1/4)*log(-b^7*c^2*(-(B^4*b^4 - 20*A*B^3*b^3*c + 150*A^2*B^2*b^2*c^ 
2 - 500*A^3*B*b*c^3 + 625*A^4*c^4)/(b^9*c^3))^(3/4) - (B^3*b^3 - 15*A*B^2* 
b^2*c + 75*A^2*B*b*c^2 - 125*A^3*c^3)*sqrt(x)) - 4*((B*b - 5*A*c)*x^2 - 4* 
A*b)*sqrt(x))/(b^2*c*x^3 + b^3*x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\text {Timed out} \] Input:

integrate(x**(5/2)*(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 = 222, normalized size of antiderivative = 1.07 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {{\left (B b - 5 \, A c\right )} x^{2} - 4 \, A b}{2 \, {\left (b^{2} c x^{\frac {5}{2}} + b^{3} \sqrt {x}\right )}} + \frac {{\left (B b - 5 \, A c\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}}\right )}}{16 \, b^{2}} \] Input:

integrate(x^(5/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.34 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {B b x^{2} - 5 \, A c x^{2} - 4 \, A b}{2 \, {\left (c x^{\frac {5}{2}} + b \sqrt {x}\right )} b^{2}} + \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {3}{4}} B b - 5 \, \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, b^{3} c^{3}} + \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {3}{4}} B b - 5 \, \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, b^{3} c^{3}} - \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {3}{4}} B b - 5 \, \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, b^{3} c^{3}} + \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {3}{4}} B b - 5 \, \left (b c^{3}\right )^{\frac {3}{4}} A c\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, b^{3} c^{3}} \] Input:

integrate(x^(5/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="giac")
 

Output:

1/2*(B*b*x^2 - 5*A*c*x^2 - 4*A*b)/((c*x^(5/2) + b*sqrt(x))*b^2) + 1/8*sqrt 
(2)*((b*c^3)^(3/4)*B*b - 5*(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^3*c^3) + 1/8*sqrt(2)*((b*c^3)^(3/ 
4)*B*b - 5*(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^3*c^3) - 1/16*sqrt(2)*((b*c^3)^(3/4)*B*b - 5*(b* 
c^3)^(3/4)*A*c)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^3*c^3) 
 + 1/16*sqrt(2)*((b*c^3)^(3/4)*B*b - 5*(b*c^3)^(3/4)*A*c)*log(-sqrt(2)*sqr 
t(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^3*c^3)
 

Mupad [B] (verification not implemented)

Time = 8.75 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.50 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {\mathrm {atanh}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )\,\left (5\,A\,c-B\,b\right )}{4\,{\left (-b\right )}^{9/4}\,c^{3/4}}-\frac {\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )\,\left (5\,A\,c-B\,b\right )}{4\,{\left (-b\right )}^{9/4}\,c^{3/4}}-\frac {\frac {2\,A}{b}+\frac {x^2\,\left (5\,A\,c-B\,b\right )}{2\,b^2}}{b\,\sqrt {x}+c\,x^{5/2}} \] Input:

int((x^(5/2)*(A + B*x^2))/(b*x^2 + c*x^4)^2,x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 650, normalized size of antiderivative = 3.12 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx =\text {Too large to display} \] Input:

int(x^(5/2)*(B*x^2+A)/(c*x^4+b*x^2)^2,x)
 

Output:

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