\(\int \frac {x^{3/2} (A+B x^2)}{b x^2+c x^4} \, dx\) [122]

Optimal result

Integrand size = 26, antiderivative size = 172 \[ \int \frac {x^{3/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {2 B \sqrt {x}}{c}+\frac {(b B-A c) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{3/4} c^{5/4}}-\frac {(b B-A c) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{3/4} c^{5/4}}-\frac {(b B-A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{\sqrt {2} b^{3/4} c^{5/4}} \] Output:

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.78 \[ \int \frac {x^{3/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {2 B \sqrt {x}}{c}+\frac {(b B-A c) \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )}{\sqrt {2} b^{3/4} c^{5/4}}-\frac {(b B-A c) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{\sqrt {2} b^{3/4} c^{5/4}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.38, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {9, 363, 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.

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

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 363

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 755

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

(2*B*Sqrt[x])/c - (2*(b*B - A*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] - Sqr 
t[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])))/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 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 363

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), 
 x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3))   Int[(e*x)^ 
m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d 
, 0] && NeQ[m + 2*p + 3, 0]
 

rule 755

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] 
], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r)   Int[(r - s*x^2)/(a + b*x^4) 
, x], x] + Simp[1/(2*r)   Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{a, 
 b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] & 
& AtomQ[SplitProduct[SumBaseQ, b]]))
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(\frac {2 B \sqrt {x}}{c}+\frac {\left (A c -B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 c b}\)	\(127\)
default	\(\frac {2 B \sqrt {x}}{c}+\frac {\left (A c -B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 c b}\)	\(127\)
risch	\(\frac {2 B \sqrt {x}}{c}+\frac {\left (A c -B b \right ) \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 c b}\)	\(127\)

Input:

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

Output:

2*B*x^(1/2)/c+1/4*(A*c-B*b)/c*(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*a 
rctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+2*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)- 
1))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 570, normalized size of antiderivative = 3.31 \[ \int \frac {x^{3/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {c \left (-\frac {B^{4} b^{4} - 4 \, A B^{3} b^{3} c + 6 \, A^{2} B^{2} b^{2} c^{2} - 4 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{5}}\right )^{\frac {1}{4}} \log \left (b c \left (-\frac {B^{4} b^{4} - 4 \, A B^{3} b^{3} c + 6 \, A^{2} B^{2} b^{2} c^{2} - 4 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{5}}\right )^{\frac {1}{4}} - {\left (B b - A c\right )} \sqrt {x}\right ) + i \, c \left (-\frac {B^{4} b^{4} - 4 \, A B^{3} b^{3} c + 6 \, A^{2} B^{2} b^{2} c^{2} - 4 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{5}}\right )^{\frac {1}{4}} \log \left (i \, b c \left (-\frac {B^{4} b^{4} - 4 \, A B^{3} b^{3} c + 6 \, A^{2} B^{2} b^{2} c^{2} - 4 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{5}}\right )^{\frac {1}{4}} - {\left (B b - A c\right )} \sqrt {x}\right ) - i \, c \left (-\frac {B^{4} b^{4} - 4 \, A B^{3} b^{3} c + 6 \, A^{2} B^{2} b^{2} c^{2} - 4 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{5}}\right )^{\frac {1}{4}} \log \left (-i \, b c \left (-\frac {B^{4} b^{4} - 4 \, A B^{3} b^{3} c + 6 \, A^{2} B^{2} b^{2} c^{2} - 4 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{5}}\right )^{\frac {1}{4}} - {\left (B b - A c\right )} \sqrt {x}\right ) - c \left (-\frac {B^{4} b^{4} - 4 \, A B^{3} b^{3} c + 6 \, A^{2} B^{2} b^{2} c^{2} - 4 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{5}}\right )^{\frac {1}{4}} \log \left (-b c \left (-\frac {B^{4} b^{4} - 4 \, A B^{3} b^{3} c + 6 \, A^{2} B^{2} b^{2} c^{2} - 4 \, A^{3} B b c^{3} + A^{4} c^{4}}{b^{3} c^{5}}\right )^{\frac {1}{4}} - {\left (B b - A c\right )} \sqrt {x}\right ) + 4 \, B \sqrt {x}}{2 \, c} \] Input:

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

Output:

1/2*(c*(-(B^4*b^4 - 4*A*B^3*b^3*c + 6*A^2*B^2*b^2*c^2 - 4*A^3*B*b*c^3 + A^ 
4*c^4)/(b^3*c^5))^(1/4)*log(b*c*(-(B^4*b^4 - 4*A*B^3*b^3*c + 6*A^2*B^2*b^2 
*c^2 - 4*A^3*B*b*c^3 + A^4*c^4)/(b^3*c^5))^(1/4) - (B*b - A*c)*sqrt(x)) + 
I*c*(-(B^4*b^4 - 4*A*B^3*b^3*c + 6*A^2*B^2*b^2*c^2 - 4*A^3*B*b*c^3 + A^4*c 
^4)/(b^3*c^5))^(1/4)*log(I*b*c*(-(B^4*b^4 - 4*A*B^3*b^3*c + 6*A^2*B^2*b^2* 
c^2 - 4*A^3*B*b*c^3 + A^4*c^4)/(b^3*c^5))^(1/4) - (B*b - A*c)*sqrt(x)) - I 
*c*(-(B^4*b^4 - 4*A*B^3*b^3*c + 6*A^2*B^2*b^2*c^2 - 4*A^3*B*b*c^3 + A^4*c^ 
4)/(b^3*c^5))^(1/4)*log(-I*b*c*(-(B^4*b^4 - 4*A*B^3*b^3*c + 6*A^2*B^2*b^2* 
c^2 - 4*A^3*B*b*c^3 + A^4*c^4)/(b^3*c^5))^(1/4) - (B*b - A*c)*sqrt(x)) - c 
*(-(B^4*b^4 - 4*A*B^3*b^3*c + 6*A^2*B^2*b^2*c^2 - 4*A^3*B*b*c^3 + A^4*c^4) 
/(b^3*c^5))^(1/4)*log(-b*c*(-(B^4*b^4 - 4*A*B^3*b^3*c + 6*A^2*B^2*b^2*c^2 
- 4*A^3*B*b*c^3 + A^4*c^4)/(b^3*c^5))^(1/4) - (B*b - A*c)*sqrt(x)) + 4*B*s 
qrt(x))/c
 

Sympy [A] (verification not implemented)

Time = 8.09 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.38 \[ \int \frac {x^{3/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\begin {cases} \tilde {\infty } \left (- \frac {2 A}{3 x^{\frac {3}{2}}} + 2 B \sqrt {x}\right ) & \text {for}\: b = 0 \wedge c = 0 \\\frac {- \frac {2 A}{3 x^{\frac {3}{2}}} + 2 B \sqrt {x}}{c} & \text {for}\: b = 0 \\\frac {2 A \sqrt {x} + \frac {2 B x^{\frac {5}{2}}}{5}}{b} & \text {for}\: c = 0 \\- \frac {A \sqrt [4]{- \frac {b}{c}} \log {\left (\sqrt {x} - \sqrt [4]{- \frac {b}{c}} \right )}}{2 b} + \frac {A \sqrt [4]{- \frac {b}{c}} \log {\left (\sqrt {x} + \sqrt [4]{- \frac {b}{c}} \right )}}{2 b} + \frac {A \sqrt [4]{- \frac {b}{c}} \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {b}{c}}} \right )}}{b} + \frac {2 B \sqrt {x}}{c} + \frac {B \sqrt [4]{- \frac {b}{c}} \log {\left (\sqrt {x} - \sqrt [4]{- \frac {b}{c}} \right )}}{2 c} - \frac {B \sqrt [4]{- \frac {b}{c}} \log {\left (\sqrt {x} + \sqrt [4]{- \frac {b}{c}} \right )}}{2 c} - \frac {B \sqrt [4]{- \frac {b}{c}} \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {b}{c}}} \right )}}{c} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((zoo*(-2*A/(3*x**(3/2)) + 2*B*sqrt(x)), Eq(b, 0) & Eq(c, 0)), (( 
-2*A/(3*x**(3/2)) + 2*B*sqrt(x))/c, Eq(b, 0)), ((2*A*sqrt(x) + 2*B*x**(5/2 
)/5)/b, Eq(c, 0)), (-A*(-b/c)**(1/4)*log(sqrt(x) - (-b/c)**(1/4))/(2*b) + 
A*(-b/c)**(1/4)*log(sqrt(x) + (-b/c)**(1/4))/(2*b) + A*(-b/c)**(1/4)*atan( 
sqrt(x)/(-b/c)**(1/4))/b + 2*B*sqrt(x)/c + B*(-b/c)**(1/4)*log(sqrt(x) - ( 
-b/c)**(1/4))/(2*c) - B*(-b/c)**(1/4)*log(sqrt(x) + (-b/c)**(1/4))/(2*c) - 
 B*(-b/c)**(1/4)*atan(sqrt(x)/(-b/c)**(1/4))/c, True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.27 \[ \int \frac {x^{3/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {2 \, B \sqrt {x}}{c} - \frac {\frac {2 \, \sqrt {2} {\left (B b - A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {2 \, \sqrt {2} {\left (B b - A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {\sqrt {2} {\left (B b - A c\right )} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (B b - A c\right )} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}} c^{\frac {1}{4}}}}{4 \, c} \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 251 vs. \(2 (125) = 250\).

Time = 0.23 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.46 \[ \int \frac {x^{3/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {2 \, B \sqrt {x}}{c} - \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b - \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{2 \, b c^{2}} - \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b - \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{2 \, b c^{2}} - \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b - \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, b c^{2}} + \frac {\sqrt {2} {\left (\left (b c^{3}\right )^{\frac {1}{4}} B b - \left (b c^{3}\right )^{\frac {1}{4}} A c\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, b c^{2}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 739, normalized size of antiderivative = 4.30 \[ \int \frac {x^{3/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx =\text {Too large to display} \] Input:

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

Output:

(2*B*x^(1/2))/c - (atan((((A*c - B*b)*(x^(1/2)*(16*A^2*c^3 + 16*B^2*b^2*c 
- 32*A*B*b*c^2) - ((32*B*b^2*c^2 - 32*A*b*c^3)*(A*c - B*b))/(2*(-b)^(3/4)* 
c^(5/4)))*1i)/(2*(-b)^(3/4)*c^(5/4)) + ((A*c - B*b)*(x^(1/2)*(16*A^2*c^3 + 
 16*B^2*b^2*c - 32*A*B*b*c^2) + ((32*B*b^2*c^2 - 32*A*b*c^3)*(A*c - B*b))/ 
(2*(-b)^(3/4)*c^(5/4)))*1i)/(2*(-b)^(3/4)*c^(5/4)))/(((A*c - B*b)*(x^(1/2) 
*(16*A^2*c^3 + 16*B^2*b^2*c - 32*A*B*b*c^2) - ((32*B*b^2*c^2 - 32*A*b*c^3) 
*(A*c - B*b))/(2*(-b)^(3/4)*c^(5/4))))/(2*(-b)^(3/4)*c^(5/4)) - ((A*c - B* 
b)*(x^(1/2)*(16*A^2*c^3 + 16*B^2*b^2*c - 32*A*B*b*c^2) + ((32*B*b^2*c^2 - 
32*A*b*c^3)*(A*c - B*b))/(2*(-b)^(3/4)*c^(5/4))))/(2*(-b)^(3/4)*c^(5/4)))) 
*(A*c - B*b)*1i)/((-b)^(3/4)*c^(5/4)) - (atan((((A*c - B*b)*(x^(1/2)*(16*A 
^2*c^3 + 16*B^2*b^2*c - 32*A*B*b*c^2) - ((32*B*b^2*c^2 - 32*A*b*c^3)*(A*c 
- B*b)*1i)/(2*(-b)^(3/4)*c^(5/4))))/(2*(-b)^(3/4)*c^(5/4)) + ((A*c - B*b)* 
(x^(1/2)*(16*A^2*c^3 + 16*B^2*b^2*c - 32*A*B*b*c^2) + ((32*B*b^2*c^2 - 32* 
A*b*c^3)*(A*c - B*b)*1i)/(2*(-b)^(3/4)*c^(5/4))))/(2*(-b)^(3/4)*c^(5/4)))/ 
(((A*c - B*b)*(x^(1/2)*(16*A^2*c^3 + 16*B^2*b^2*c - 32*A*B*b*c^2) - ((32*B 
*b^2*c^2 - 32*A*b*c^3)*(A*c - B*b)*1i)/(2*(-b)^(3/4)*c^(5/4)))*1i)/(2*(-b) 
^(3/4)*c^(5/4)) - ((A*c - B*b)*(x^(1/2)*(16*A^2*c^3 + 16*B^2*b^2*c - 32*A* 
B*b*c^2) + ((32*B*b^2*c^2 - 32*A*b*c^3)*(A*c - B*b)*1i)/(2*(-b)^(3/4)*c^(5 
/4)))*1i)/(2*(-b)^(3/4)*c^(5/4))))*(A*c - B*b))/((-b)^(3/4)*c^(5/4))
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.68 \[ \int \frac {x^{3/2} \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {-2 c^{\frac {7}{4}} b^{\frac {1}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}-2 \sqrt {x}\, \sqrt {c}}{c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}}\right ) a +2 c^{\frac {3}{4}} b^{\frac {9}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}-2 \sqrt {x}\, \sqrt {c}}{c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}}\right )+2 c^{\frac {7}{4}} b^{\frac {1}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+2 \sqrt {x}\, \sqrt {c}}{c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}}\right ) a -2 c^{\frac {3}{4}} b^{\frac {9}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+2 \sqrt {x}\, \sqrt {c}}{c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}}\right )-c^{\frac {7}{4}} b^{\frac {1}{4}} \sqrt {2}\, \mathrm {log}\left (-\sqrt {x}\, c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+\sqrt {b}+\sqrt {c}\, x \right ) a +c^{\frac {3}{4}} b^{\frac {9}{4}} \sqrt {2}\, \mathrm {log}\left (-\sqrt {x}\, c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+\sqrt {b}+\sqrt {c}\, x \right )+c^{\frac {7}{4}} b^{\frac {1}{4}} \sqrt {2}\, \mathrm {log}\left (\sqrt {x}\, c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+\sqrt {b}+\sqrt {c}\, x \right ) a -c^{\frac {3}{4}} b^{\frac {9}{4}} \sqrt {2}\, \mathrm {log}\left (\sqrt {x}\, c^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {2}+\sqrt {b}+\sqrt {c}\, x \right )+8 \sqrt {x}\, b^{2} c}{4 b \,c^{2}} \] Input:

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

Output:

( - 2*c**(3/4)*b**(1/4)*sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) - 2*sqrt(x 
)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*a*c + 2*c**(3/4)*b**(1/4)*sqrt(2)* 
atan((c**(1/4)*b**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sq 
rt(2)))*b**2 + 2*c**(3/4)*b**(1/4)*sqrt(2)*atan((c**(1/4)*b**(1/4)*sqrt(2) 
 + 2*sqrt(x)*sqrt(c))/(c**(1/4)*b**(1/4)*sqrt(2)))*a*c - 2*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**(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 + c**(3/4)*b**(1/4)*sqrt(2)* 
log( - sqrt(x)*c**(1/4)*b**(1/4)*sqrt(2) + sqrt(b) + sqrt(c)*x)*b**2 + c** 
(3/4)*b**(1/4)*sqrt(2)*log(sqrt(x)*c**(1/4)*b**(1/4)*sqrt(2) + sqrt(b) + s 
qrt(c)*x)*a*c - c**(3/4)*b**(1/4)*sqrt(2)*log(sqrt(x)*c**(1/4)*b**(1/4)*sq 
rt(2) + sqrt(b) + sqrt(c)*x)*b**2 + 8*sqrt(x)*b**2*c)/(4*b*c**2)