\(\int \frac {(c x)^{5/2}}{(a+b x^2)^{5/2}} \, dx\) [643]

Optimal result

Integrand size = 19, antiderivative size = 304 \[ \int \frac {(c x)^{5/2}}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {c (c x)^{3/2}}{3 b \left (a+b x^2\right )^{3/2}}+\frac {c (c x)^{3/2}}{2 a b \sqrt {a+b x^2}}-\frac {c^2 \sqrt {c x} \sqrt {a+b x^2}}{2 a b^{3/2} \left (\sqrt {a}+\sqrt {b} x\right )}+\frac {c^{5/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{2 a^{3/4} b^{7/4} \sqrt {a+b x^2}}-\frac {c^{5/2} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{4 a^{3/4} b^{7/4} \sqrt {a+b x^2}} \] Output:

-1/3*c*(c*x)^(3/2)/b/(b*x^2+a)^(3/2)+1/2*c*(c*x)^(3/2)/a/b/(b*x^2+a)^(1/2) 
-1/2*c^2*(c*x)^(1/2)*(b*x^2+a)^(1/2)/a/b^(3/2)/(a^(1/2)+b^(1/2)*x)+1/2*c^( 
5/2)*(a^(1/2)+b^(1/2)*x)*((b*x^2+a)/(a^(1/2)+b^(1/2)*x)^2)^(1/2)*EllipticE 
(sin(2*arctan(b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2))),1/2*2^(1/2))/a^(3/4)/b 
^(7/4)/(b*x^2+a)^(1/2)-1/4*c^(5/2)*(a^(1/2)+b^(1/2)*x)*((b*x^2+a)/(a^(1/2) 
+b^(1/2)*x)^2)^(1/2)*InverseJacobiAM(2*arctan(b^(1/4)*(c*x)^(1/2)/a^(1/4)/ 
c^(1/2)),1/2*2^(1/2))/a^(3/4)/b^(7/4)/(b*x^2+a)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.24 \[ \int \frac {(c x)^{5/2}}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {2 c (c x)^{3/2} \left (-a+\left (a+b x^2\right ) \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{2},\frac {7}{4},-\frac {b x^2}{a}\right )\right )}{3 a b \left (a+b x^2\right )^{3/2}} \] Input:

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

Output:

(2*c*(c*x)^(3/2)*(-a + (a + b*x^2)*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[3 
/4, 5/2, 7/4, -((b*x^2)/a)]))/(3*a*b*(a + b*x^2)^(3/2))
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.13, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {252, 253, 266, 834, 27, 761, 1510}

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.

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

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 253

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 834

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 761

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

\(\Big \downarrow \) 1510

\(\displaystyle \frac {c^2 \left (\frac {(c x)^{3/2}}{a c \sqrt {a+b x^2}}-\frac {\frac {\sqrt [4]{a} \sqrt {c} \left (\sqrt {a} c+\sqrt {b} c x\right ) \sqrt {\frac {a c^2+b c^2 x^2}{\left (\sqrt {a} c+\sqrt {b} c x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^2}}-\frac {\frac {\sqrt [4]{a} \sqrt {c} \left (\sqrt {a} c+\sqrt {b} c x\right ) \sqrt {\frac {a c^2+b c^2 x^2}{\left (\sqrt {a} c+\sqrt {b} c x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{\sqrt [4]{b} \sqrt {a+b x^2}}-\frac {c^2 \sqrt {c x} \sqrt {a+b x^2}}{\sqrt {a} c+\sqrt {b} c x}}{\sqrt {b}}}{a c}\right )}{2 b}-\frac {c (c x)^{3/2}}{3 b \left (a+b x^2\right )^{3/2}}\)

Input:

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

Output:

-1/3*(c*(c*x)^(3/2))/(b*(a + b*x^2)^(3/2)) + (c^2*((c*x)^(3/2)/(a*c*Sqrt[a 
 + b*x^2]) - (-((-((c^2*Sqrt[c*x]*Sqrt[a + b*x^2])/(Sqrt[a]*c + Sqrt[b]*c* 
x)) + (a^(1/4)*Sqrt[c]*(Sqrt[a]*c + Sqrt[b]*c*x)*Sqrt[(a*c^2 + b*c^2*x^2)/ 
(Sqrt[a]*c + Sqrt[b]*c*x)^2]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[c*x])/(a^(1/ 
4)*Sqrt[c])], 1/2])/(b^(1/4)*Sqrt[a + b*x^2]))/Sqrt[b]) + (a^(1/4)*Sqrt[c] 
*(Sqrt[a]*c + Sqrt[b]*c*x)*Sqrt[(a*c^2 + b*c^2*x^2)/(Sqrt[a]*c + Sqrt[b]*c 
*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[c*x])/(a^(1/4)*Sqrt[c])], 1/2])/(2 
*b^(3/4)*Sqrt[a + b*x^2]))/(a*c)))/(2*b)
 

Defintions of rubi rules used

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 252

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x 
)^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* 
(p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c 
}, x] && LtQ[p, -1] && GtQ[m, 1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi 
alQ[a, b, c, 2, m, p, x]
 

rule 253

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(c*x 
)^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 
2*a*(p + 1))   Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m 
}, x] && LtQ[p, -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 761

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 
1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* 
EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 834

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S 
imp[1/q   Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q   Int[(1 - q*x^2)/Sqrt[a 
 + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 1510

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

Maple [A] (verified)

Time = 1.64 (sec) , antiderivative size = 263, normalized size of antiderivative = 0.87

method	result	size
elliptic	\(\frac {\sqrt {c x}\, \sqrt {c x \left (b \,x^{2}+a \right )}\, \left (-\frac {c^{2} x \sqrt {b c \,x^{3}+a c x}}{3 b^{3} \left (x^{2}+\frac {a}{b}\right )^{2}}+\frac {c^{3} x^{2}}{2 b a \sqrt {\left (x^{2}+\frac {a}{b}\right ) b c x}}-\frac {c^{3} \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \operatorname {EllipticE}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right )}{4 b^{2} a \sqrt {b c \,x^{3}+a c x}}\right )}{c x \sqrt {b \,x^{2}+a}}\)	\(263\)
default	\(-\frac {\left (6 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticE}\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a b \,x^{2}-3 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a b \,x^{2}+6 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticE}\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a^{2}-3 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) a^{2}-6 b^{2} x^{4}-2 a b \,x^{2}\right ) c^{2} \sqrt {c x}}{12 x \,b^{2} a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}\)	\(385\)

Input:

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

Output:

1/c/x*(c*x)^(1/2)/(b*x^2+a)^(1/2)*(c*x*(b*x^2+a))^(1/2)*(-1/3*c^2/b^3*x*(b 
*c*x^3+a*c*x)^(1/2)/(x^2+a/b)^2+1/2/b*c^3*x^2/a/((x^2+a/b)*b*c*x)^(1/2)-1/ 
4/b^2/a*c^3*(-a*b)^(1/2)*((x+1/b*(-a*b)^(1/2))*b/(-a*b)^(1/2))^(1/2)*(-2*( 
x-1/b*(-a*b)^(1/2))*b/(-a*b)^(1/2))^(1/2)*(-b/(-a*b)^(1/2)*x)^(1/2)/(b*c*x 
^3+a*c*x)^(1/2)*(-2/b*(-a*b)^(1/2)*EllipticE(((x+1/b*(-a*b)^(1/2))*b/(-a*b 
)^(1/2))^(1/2),1/2*2^(1/2))+1/b*(-a*b)^(1/2)*EllipticF(((x+1/b*(-a*b)^(1/2 
))*b/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.39 \[ \int \frac {(c x)^{5/2}}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {3 \, {\left (b^{2} c^{2} x^{4} + 2 \, a b c^{2} x^{2} + a^{2} c^{2}\right )} \sqrt {b c} {\rm weierstrassZeta}\left (-\frac {4 \, a}{b}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right )\right ) + {\left (3 \, b^{2} c^{2} x^{3} + a b c^{2} x\right )} \sqrt {b x^{2} + a} \sqrt {c x}}{6 \, {\left (a b^{4} x^{4} + 2 \, a^{2} b^{3} x^{2} + a^{3} b^{2}\right )}} \] Input:

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

Output:

1/6*(3*(b^2*c^2*x^4 + 2*a*b*c^2*x^2 + a^2*c^2)*sqrt(b*c)*weierstrassZeta(- 
4*a/b, 0, weierstrassPInverse(-4*a/b, 0, x)) + (3*b^2*c^2*x^3 + a*b*c^2*x) 
*sqrt(b*x^2 + a)*sqrt(c*x))/(a*b^4*x^4 + 2*a^2*b^3*x^2 + a^3*b^2)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.65 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.14 \[ \int \frac {(c x)^{5/2}}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {c^{\frac {5}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{4}, \frac {5}{2} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{2}} \Gamma \left (\frac {11}{4}\right )} \] Input:

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

Output:

c**(5/2)*x**(7/2)*gamma(7/4)*hyper((7/4, 5/2), (11/4,), b*x**2*exp_polar(I 
*pi)/a)/(2*a**(5/2)*gamma(11/4))
 

Maxima [F]

\[ \int \frac {(c x)^{5/2}}{\left (a+b x^2\right )^{5/2}} \, dx=\int { \frac {\left (c x\right )^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {(c x)^{5/2}}{\left (a+b x^2\right )^{5/2}} \, dx=\int { \frac {\left (c x\right )^{\frac {5}{2}}}{{\left (b x^{2} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(c x)^{5/2}}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {\sqrt {c}\, c^{2} \left (-2 \sqrt {x}\, \sqrt {b \,x^{2}+a}\, x +3 \left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+a}}{b^{3} x^{6}+3 a \,b^{2} x^{4}+3 a^{2} b \,x^{2}+a^{3}}d x \right ) a^{3}+6 \left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+a}}{b^{3} x^{6}+3 a \,b^{2} x^{4}+3 a^{2} b \,x^{2}+a^{3}}d x \right ) a^{2} b \,x^{2}+3 \left (\int \frac {\sqrt {x}\, \sqrt {b \,x^{2}+a}}{b^{3} x^{6}+3 a \,b^{2} x^{4}+3 a^{2} b \,x^{2}+a^{3}}d x \right ) a \,b^{2} x^{4}\right )}{3 b \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:

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

Output:

(sqrt(c)*c**2*( - 2*sqrt(x)*sqrt(a + b*x**2)*x + 3*int((sqrt(x)*sqrt(a + b 
*x**2))/(a**3 + 3*a**2*b*x**2 + 3*a*b**2*x**4 + b**3*x**6),x)*a**3 + 6*int 
((sqrt(x)*sqrt(a + b*x**2))/(a**3 + 3*a**2*b*x**2 + 3*a*b**2*x**4 + b**3*x 
**6),x)*a**2*b*x**2 + 3*int((sqrt(x)*sqrt(a + b*x**2))/(a**3 + 3*a**2*b*x* 
*2 + 3*a*b**2*x**4 + b**3*x**6),x)*a*b**2*x**4))/(3*b*(a**2 + 2*a*b*x**2 + 
 b**2*x**4))