\(\int \frac {x^{5/2}}{a+b x^2} \, dx\) [289]

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

Optimal result

Integrand size = 15, antiderivative size = 148 \[ \int \frac {x^{5/2}}{a+b x^2} \, dx=\frac {2 x^{3/2}}{3 b}+\frac {a^{3/4} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} b^{7/4}}-\frac {a^{3/4} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} b^{7/4}}+\frac {a^{3/4} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{\sqrt {2} b^{7/4}} \] Output:

2/3*x^(3/2)/b+1/2*a^(3/4)*arctan(1-2^(1/2)*b^(1/4)*x^(1/2)/a^(1/4))*2^(1/2 
)/b^(7/4)-1/2*a^(3/4)*arctan(1+2^(1/2)*b^(1/4)*x^(1/2)/a^(1/4))*2^(1/2)/b^ 
(7/4)+1/2*a^(3/4)*arctanh(2^(1/2)*a^(1/4)*b^(1/4)*x^(1/2)/(a^(1/2)+b^(1/2) 
*x))*2^(1/2)/b^(7/4)
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.80 \[ \int \frac {x^{5/2}}{a+b x^2} \, dx=\frac {4 b^{3/4} x^{3/2}+3 \sqrt {2} a^{3/4} \arctan \left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )+3 \sqrt {2} a^{3/4} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{6 b^{7/4}} \] Input:

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

Output:

(4*b^(3/4)*x^(3/2) + 3*Sqrt[2]*a^(3/4)*ArcTan[(Sqrt[a] - Sqrt[b]*x)/(Sqrt[ 
2]*a^(1/4)*b^(1/4)*Sqrt[x])] + 3*Sqrt[2]*a^(3/4)*ArcTanh[(Sqrt[2]*a^(1/4)* 
b^(1/4)*Sqrt[x])/(Sqrt[a] + Sqrt[b]*x)])/(6*b^(7/4))
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.56, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {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.

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

\(\Big \downarrow \) 262

\(\displaystyle \frac {2 x^{3/2}}{3 b}-\frac {a \int \frac {\sqrt {x}}{b x^2+a}dx}{b}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {2 x^{3/2}}{3 b}-\frac {2 a \int \frac {x}{b x^2+a}d\sqrt {x}}{b}\)

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

(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
 

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 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 = 0.32 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.78

method result size
derivativedivides \(\frac {2 x^{\frac {3}{2}}}{3 b}-\frac {a \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{4 b^{2} \left (\frac {a}{b}\right )^{\frac {1}{4}}}\) \(116\)
default \(\frac {2 x^{\frac {3}{2}}}{3 b}-\frac {a \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{4 b^{2} \left (\frac {a}{b}\right )^{\frac {1}{4}}}\) \(116\)
risch \(\frac {2 x^{\frac {3}{2}}}{3 b}-\frac {a \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{4 b^{2} \left (\frac {a}{b}\right )^{\frac {1}{4}}}\) \(116\)

Input:

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

Output:

2/3*x^(3/2)/b-1/4*a/b^2/(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 = 0.07 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.07 \[ \int \frac {x^{5/2}}{a+b x^2} \, dx=-\frac {3 \, b \left (-\frac {a^{3}}{b^{7}}\right )^{\frac {1}{4}} \log \left (b^{5} \left (-\frac {a^{3}}{b^{7}}\right )^{\frac {3}{4}} + a^{2} \sqrt {x}\right ) - 3 i \, b \left (-\frac {a^{3}}{b^{7}}\right )^{\frac {1}{4}} \log \left (i \, b^{5} \left (-\frac {a^{3}}{b^{7}}\right )^{\frac {3}{4}} + a^{2} \sqrt {x}\right ) + 3 i \, b \left (-\frac {a^{3}}{b^{7}}\right )^{\frac {1}{4}} \log \left (-i \, b^{5} \left (-\frac {a^{3}}{b^{7}}\right )^{\frac {3}{4}} + a^{2} \sqrt {x}\right ) - 3 \, b \left (-\frac {a^{3}}{b^{7}}\right )^{\frac {1}{4}} \log \left (-b^{5} \left (-\frac {a^{3}}{b^{7}}\right )^{\frac {3}{4}} + a^{2} \sqrt {x}\right ) - 4 \, x^{\frac {3}{2}}}{6 \, b} \] Input:

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

Output:

-1/6*(3*b*(-a^3/b^7)^(1/4)*log(b^5*(-a^3/b^7)^(3/4) + a^2*sqrt(x)) - 3*I*b 
*(-a^3/b^7)^(1/4)*log(I*b^5*(-a^3/b^7)^(3/4) + a^2*sqrt(x)) + 3*I*b*(-a^3/ 
b^7)^(1/4)*log(-I*b^5*(-a^3/b^7)^(3/4) + a^2*sqrt(x)) - 3*b*(-a^3/b^7)^(1/ 
4)*log(-b^5*(-a^3/b^7)^(3/4) + a^2*sqrt(x)) - 4*x^(3/2))/b
 

Sympy [A] (verification not implemented)

Time = 5.34 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.84 \[ \int \frac {x^{5/2}}{a+b x^2} \, dx=\begin {cases} \tilde {\infty } x^{\frac {3}{2}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 x^{\frac {7}{2}}}{7 a} & \text {for}\: b = 0 \\\frac {2 x^{\frac {3}{2}}}{3 b} & \text {for}\: a = 0 \\- \frac {a \log {\left (\sqrt {x} - \sqrt [4]{- \frac {a}{b}} \right )}}{2 b^{2} \sqrt [4]{- \frac {a}{b}}} + \frac {a \log {\left (\sqrt {x} + \sqrt [4]{- \frac {a}{b}} \right )}}{2 b^{2} \sqrt [4]{- \frac {a}{b}}} - \frac {a \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {a}{b}}} \right )}}{b^{2} \sqrt [4]{- \frac {a}{b}}} + \frac {2 x^{\frac {3}{2}}}{3 b} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((zoo*x**(3/2), Eq(a, 0) & Eq(b, 0)), (2*x**(7/2)/(7*a), Eq(b, 0) 
), (2*x**(3/2)/(3*b), Eq(a, 0)), (-a*log(sqrt(x) - (-a/b)**(1/4))/(2*b**2* 
(-a/b)**(1/4)) + a*log(sqrt(x) + (-a/b)**(1/4))/(2*b**2*(-a/b)**(1/4)) - a 
*atan(sqrt(x)/(-a/b)**(1/4))/(b**2*(-a/b)**(1/4)) + 2*x**(3/2)/(3*b), True 
))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.26 \[ \int \frac {x^{5/2}}{a+b x^2} \, dx=-\frac {a {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{4 \, b} + \frac {2 \, x^{\frac {3}{2}}}{3 \, b} \] Input:

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

Output:

-1/4*a*(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*sqrt(2 
)*arctan(-1/2*sqrt(2)*(sqrt(2)*a^(1/4)*b^(1/4) - 2*sqrt(b)*sqrt(x))/sqrt(s 
qrt(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)*lo 
g(-sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a))/(a^(1/4)*b^(3/4) 
))/b + 2/3*x^(3/2)/b
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.20 \[ \int \frac {x^{5/2}}{a+b x^2} \, dx=\frac {2 \, x^{\frac {3}{2}}}{3 \, b} - \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \, b^{4}} - \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \, b^{4}} + \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {3}{4}} \log \left (\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{4 \, b^{4}} - \frac {\sqrt {2} \left (a b^{3}\right )^{\frac {3}{4}} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{4 \, b^{4}} \] Input:

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

Output:

2/3*x^(3/2)/b - 1/2*sqrt(2)*(a*b^3)^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(a/b 
)^(1/4) + 2*sqrt(x))/(a/b)^(1/4))/b^4 - 1/2*sqrt(2)*(a*b^3)^(3/4)*arctan(- 
1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) - 2*sqrt(x))/(a/b)^(1/4))/b^4 + 1/4*sqrt( 
2)*(a*b^3)^(3/4)*log(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/b^4 - 1/ 
4*sqrt(2)*(a*b^3)^(3/4)*log(-sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/ 
b^4
 

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.36 \[ \int \frac {x^{5/2}}{a+b x^2} \, dx=\frac {2\,x^{3/2}}{3\,b}+\frac {{\left (-a\right )}^{3/4}\,\mathrm {atan}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )}{b^{7/4}}-\frac {{\left (-a\right )}^{3/4}\,\mathrm {atanh}\left (\frac {b^{1/4}\,\sqrt {x}}{{\left (-a\right )}^{1/4}}\right )}{b^{7/4}} \] Input:

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

Output:

(2*x^(3/2))/(3*b) + ((-a)^(3/4)*atan((b^(1/4)*x^(1/2))/(-a)^(1/4)))/b^(7/4 
) - ((-a)^(3/4)*atanh((b^(1/4)*x^(1/2))/(-a)^(1/4)))/b^(7/4)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.99 \[ \int \frac {x^{5/2}}{a+b x^2} \, dx=\frac {6 b^{\frac {1}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}-2 \sqrt {x}\, \sqrt {b}}{b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right )-6 b^{\frac {1}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathit {atan} \left (\frac {b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+2 \sqrt {x}\, \sqrt {b}}{b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}}\right )-3 b^{\frac {1}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (-\sqrt {x}\, b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+\sqrt {a}+\sqrt {b}\, x \right )+3 b^{\frac {1}{4}} a^{\frac {3}{4}} \sqrt {2}\, \mathrm {log}\left (\sqrt {x}\, b^{\frac {1}{4}} a^{\frac {1}{4}} \sqrt {2}+\sqrt {a}+\sqrt {b}\, x \right )+8 \sqrt {x}\, b x}{12 b^{2}} \] Input:

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