\(\int \frac {x^{5/2} (A+B x^2)}{(a+b x^2)^2} \, dx\) [170]

Optimal result

Integrand size = 22, antiderivative size = 213 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {2 B x^{3/2}}{3 b^2}-\frac {(A b-a B) x^{3/2}}{2 b^2 \left (a+b x^2\right )}-\frac {(3 A b-7 a B) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} \sqrt [4]{a} b^{11/4}}+\frac {(3 A b-7 a B) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} \sqrt [4]{a} b^{11/4}}-\frac {(3 A b-7 a B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{4 \sqrt {2} \sqrt [4]{a} b^{11/4}} \] Output:

2/3*B*x^(3/2)/b^2-1/2*(A*b-B*a)*x^(3/2)/b^2/(b*x^2+a)-1/8*(3*A*b-7*B*a)*ar 
ctan(1-2^(1/2)*b^(1/4)*x^(1/2)/a^(1/4))*2^(1/2)/a^(1/4)/b^(11/4)+1/8*(3*A* 
b-7*B*a)*arctan(1+2^(1/2)*b^(1/4)*x^(1/2)/a^(1/4))*2^(1/2)/a^(1/4)/b^(11/4 
)-1/8*(3*A*b-7*B*a)*arctanh(2^(1/2)*a^(1/4)*b^(1/4)*x^(1/2)/(a^(1/2)+b^(1/ 
2)*x))*2^(1/2)/a^(1/4)/b^(11/4)
 

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.76 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {\frac {4 b^{3/4} x^{3/2} \left (-3 A b+7 a B+4 b B x^2\right )}{a+b x^2}+\frac {3 \sqrt {2} (-3 A b+7 a B) \arctan \left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )}{\sqrt [4]{a}}+\frac {3 \sqrt {2} (-3 A b+7 a B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{\sqrt [4]{a}}}{24 b^{11/4}} \] Input:

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

Output:

((4*b^(3/4)*x^(3/2)*(-3*A*b + 7*a*B + 4*b*B*x^2))/(a + b*x^2) + (3*Sqrt[2] 
*(-3*A*b + 7*a*B)*ArcTan[(Sqrt[a] - Sqrt[b]*x)/(Sqrt[2]*a^(1/4)*b^(1/4)*Sq 
rt[x])])/a^(1/4) + (3*Sqrt[2]*(-3*A*b + 7*a*B)*ArcTanh[(Sqrt[2]*a^(1/4)*b^ 
(1/4)*Sqrt[x])/(Sqrt[a] + Sqrt[b]*x)])/a^(1/4))/(24*b^(11/4))
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.33, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {362, 262, 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.

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

\(\Big \downarrow \) 362

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

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

((A*b - a*B)*x^(7/2))/(2*a*b*(a + b*x^2)) - ((3*A*b - 7*a*B)*((2*x^(3/2))/ 
(3*b) - (2*a*((-(ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)]/(Sqrt[2]*a^ 
(1/4)*b^(1/4))) + ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)]/(Sqrt[2]*a 
^(1/4)*b^(1/4)))/(2*Sqrt[b]) - (-1/2*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4) 
*Sqrt[x] + Sqrt[b]*x]/(Sqrt[2]*a^(1/4)*b^(1/4)) + Log[Sqrt[a] + Sqrt[2]*a^ 
(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x]/(2*Sqrt[2]*a^(1/4)*b^(1/4)))/(2*Sqrt[b] 
)))/b))/(4*a*b)
 

Defintions of rubi rules used

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 262

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

rule 266

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

rule 362

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

rule 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 = 0.42 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.72

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

Input:

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

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 793, normalized size of antiderivative = 3.72 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

1/24*(3*(b^3*x^2 + a*b^2)*(-(2401*B^4*a^4 - 4116*A*B^3*a^3*b + 2646*A^2*B^ 
2*a^2*b^2 - 756*A^3*B*a*b^3 + 81*A^4*b^4)/(a*b^11))^(1/4)*log(a*b^8*(-(240 
1*B^4*a^4 - 4116*A*B^3*a^3*b + 2646*A^2*B^2*a^2*b^2 - 756*A^3*B*a*b^3 + 81 
*A^4*b^4)/(a*b^11))^(3/4) - (343*B^3*a^3 - 441*A*B^2*a^2*b + 189*A^2*B*a*b 
^2 - 27*A^3*b^3)*sqrt(x)) - 3*(I*b^3*x^2 + I*a*b^2)*(-(2401*B^4*a^4 - 4116 
*A*B^3*a^3*b + 2646*A^2*B^2*a^2*b^2 - 756*A^3*B*a*b^3 + 81*A^4*b^4)/(a*b^1 
1))^(1/4)*log(I*a*b^8*(-(2401*B^4*a^4 - 4116*A*B^3*a^3*b + 2646*A^2*B^2*a^ 
2*b^2 - 756*A^3*B*a*b^3 + 81*A^4*b^4)/(a*b^11))^(3/4) - (343*B^3*a^3 - 441 
*A*B^2*a^2*b + 189*A^2*B*a*b^2 - 27*A^3*b^3)*sqrt(x)) - 3*(-I*b^3*x^2 - I* 
a*b^2)*(-(2401*B^4*a^4 - 4116*A*B^3*a^3*b + 2646*A^2*B^2*a^2*b^2 - 756*A^3 
*B*a*b^3 + 81*A^4*b^4)/(a*b^11))^(1/4)*log(-I*a*b^8*(-(2401*B^4*a^4 - 4116 
*A*B^3*a^3*b + 2646*A^2*B^2*a^2*b^2 - 756*A^3*B*a*b^3 + 81*A^4*b^4)/(a*b^1 
1))^(3/4) - (343*B^3*a^3 - 441*A*B^2*a^2*b + 189*A^2*B*a*b^2 - 27*A^3*b^3) 
*sqrt(x)) - 3*(b^3*x^2 + a*b^2)*(-(2401*B^4*a^4 - 4116*A*B^3*a^3*b + 2646* 
A^2*B^2*a^2*b^2 - 756*A^3*B*a*b^3 + 81*A^4*b^4)/(a*b^11))^(1/4)*log(-a*b^8 
*(-(2401*B^4*a^4 - 4116*A*B^3*a^3*b + 2646*A^2*B^2*a^2*b^2 - 756*A^3*B*a*b 
^3 + 81*A^4*b^4)/(a*b^11))^(3/4) - (343*B^3*a^3 - 441*A*B^2*a^2*b + 189*A^ 
2*B*a*b^2 - 27*A^3*b^3)*sqrt(x)) + 4*(4*B*b*x^3 + (7*B*a - 3*A*b)*x)*sqrt( 
x))/(b^3*x^2 + a*b^2)
 

Sympy [F(-1)]

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

integrate(x**(5/2)*(B*x**2+A)/(b*x**2+a)**2,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.05 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {{\left (B a - A b\right )} x^{\frac {3}{2}}}{2 \, {\left (b^{3} x^{2} + a b^{2}\right )}} + \frac {2 \, B x^{\frac {3}{2}}}{3 \, b^{2}} - \frac {{\left (7 \, B a - 3 \, A b\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{16 \, b^{2}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.33 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {2 \, B x^{\frac {3}{2}}}{3 \, b^{2}} + \frac {B a x^{\frac {3}{2}} - A b x^{\frac {3}{2}}}{2 \, {\left (b x^{2} + a\right )} b^{2}} - \frac {\sqrt {2} {\left (7 \, \left (a b^{3}\right )^{\frac {3}{4}} B a - 3 \, \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{8 \, a b^{5}} - \frac {\sqrt {2} {\left (7 \, \left (a b^{3}\right )^{\frac {3}{4}} B a - 3 \, \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{8 \, a b^{5}} + \frac {\sqrt {2} {\left (7 \, \left (a b^{3}\right )^{\frac {3}{4}} B a - 3 \, \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{16 \, a b^{5}} - \frac {\sqrt {2} {\left (7 \, \left (a b^{3}\right )^{\frac {3}{4}} B a - 3 \, \left (a b^{3}\right )^{\frac {3}{4}} A b\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{16 \, a b^{5}} \] Input:

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

Output:

2/3*B*x^(3/2)/b^2 + 1/2*(B*a*x^(3/2) - A*b*x^(3/2))/((b*x^2 + a)*b^2) - 1/ 
8*sqrt(2)*(7*(a*b^3)^(3/4)*B*a - 3*(a*b^3)^(3/4)*A*b)*arctan(1/2*sqrt(2)*( 
sqrt(2)*(a/b)^(1/4) + 2*sqrt(x))/(a/b)^(1/4))/(a*b^5) - 1/8*sqrt(2)*(7*(a* 
b^3)^(3/4)*B*a - 3*(a*b^3)^(3/4)*A*b)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a/b)^( 
1/4) - 2*sqrt(x))/(a/b)^(1/4))/(a*b^5) + 1/16*sqrt(2)*(7*(a*b^3)^(3/4)*B*a 
 - 3*(a*b^3)^(3/4)*A*b)*log(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/( 
a*b^5) - 1/16*sqrt(2)*(7*(a*b^3)^(3/4)*B*a - 3*(a*b^3)^(3/4)*A*b)*log(-sqr 
t(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a*b^5)
 

Mupad [B] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.50 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {2\,B\,x^{3/2}}{3\,b^2}-\frac {x^{3/2}\,\left (\frac {A\,b}{2}-\frac {B\,a}{2}\right )}{b^3\,x^2+a\,b^2}+\frac {\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )\,\left (3\,A\,b-7\,B\,a\right )}{4\,{\left (-a\right )}^{1/4}\,b^{11/4}}+\frac {\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {x}\,1{}\mathrm {i}}{{\left (-a\right )}^{1/4}}\right )\,\left (3\,A\,b-7\,B\,a\right )\,1{}\mathrm {i}}{4\,{\left (-a\right )}^{1/4}\,b^{11/4}} \] Input:

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

Output:

(2*B*x^(3/2))/(3*b^2) - (x^(3/2)*((A*b)/2 - (B*a)/2))/(a*b^2 + b^3*x^2) + 
(atan((b^(1/4)*x^(1/2))/(-a)^(1/4))*(3*A*b - 7*B*a))/(4*(-a)^(1/4)*b^(11/4 
)) + (atan((b^(1/4)*x^(1/2)*1i)/(-a)^(1/4))*(3*A*b - 7*B*a)*1i)/(4*(-a)^(1 
/4)*b^(11/4))
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.69 \[ \int \frac {x^{5/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {6 b^{\frac {1}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}-2 \sqrt {x}\, \sqrt {b}}{b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right )-6 b^{\frac {1}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+2 \sqrt {x}\, \sqrt {b}}{b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right )-3 b^{\frac {1}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (-\sqrt {x}\, b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+\sqrt {a}+\sqrt {b}\, x \right )+3 b^{\frac {1}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (\sqrt {x}\, b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+\sqrt {a}+\sqrt {b}\, x \right )+8 \sqrt {x}\, b x}{12 b^{2}} \] Input:

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

Output:

(6*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(x)*s 
qrt(b))/(b**(1/4)*a**(1/4)*sqrt(2))) - 6*b**(1/4)*a**(3/4)*sqrt(2)*atan((b 
**(1/4)*a**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(b))/(b**(1/4)*a**(1/4)*sqrt(2))) 
 - 3*b**(1/4)*a**(3/4)*sqrt(2)*log( - sqrt(x)*b**(1/4)*a**(1/4)*sqrt(2) + 
sqrt(a) + sqrt(b)*x) + 3*b**(1/4)*a**(3/4)*sqrt(2)*log(sqrt(x)*b**(1/4)*a* 
*(1/4)*sqrt(2) + sqrt(a) + sqrt(b)*x) + 8*sqrt(x)*b*x)/(12*b**2)