\(\int \frac {\sqrt {a+b x^2} (A+B x^2+C x^4+D x^6)}{x^2} \, dx\) [191]

Optimal result

Integrand size = 32, antiderivative size = 170 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2+C x^4+D x^6\right )}{x^2} \, dx=\frac {\left (\frac {16 A b^3}{a}+8 b^2 B-2 a b C+a^2 D\right ) x \sqrt {a+b x^2}}{16 b^2}-\frac {A \left (a+b x^2\right )^{3/2}}{a x}+\frac {(2 b C-a D) x \left (a+b x^2\right )^{3/2}}{8 b^2}+\frac {D x^3 \left (a+b x^2\right )^{3/2}}{6 b}+\frac {\left (16 A b^3+a \left (8 b^2 B-2 a b C+a^2 D\right )\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{5/2}} \] Output:

1/16*(16*A*b^3/a+8*B*b^2-2*C*a*b+a^2*D)*x*(b*x^2+a)^(1/2)/b^2-A*(b*x^2+a)^ 
(3/2)/a/x+1/8*(2*C*b-D*a)*x*(b*x^2+a)^(3/2)/b^2+1/6*D*x^3*(b*x^2+a)^(3/2)/ 
b+1/16*(16*A*b^3+a*(8*B*b^2-2*C*a*b+D*a^2))*arctanh(b^(1/2)*x/(b*x^2+a)^(1 
/2))/b^(5/2)
 

Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.81 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2+C x^4+D x^6\right )}{x^2} \, dx=\frac {\sqrt {a+b x^2} \left (-48 A b^2+x^2 \left (-3 a^2 D+2 a b \left (3 C+D x^2\right )+4 b^2 \left (6 B+3 C x^2+2 D x^4\right )\right )\right )}{48 b^2 x}+\frac {\left (16 A b^3+a \left (8 b^2 B-2 a b C+a^2 D\right )\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{8 b^{5/2}} \] Input:

Integrate[(Sqrt[a + b*x^2]*(A + B*x^2 + C*x^4 + D*x^6))/x^2,x]
 

Output:

(Sqrt[a + b*x^2]*(-48*A*b^2 + x^2*(-3*a^2*D + 2*a*b*(3*C + D*x^2) + 4*b^2* 
(6*B + 3*C*x^2 + 2*D*x^4))))/(48*b^2*x) + ((16*A*b^3 + a*(8*b^2*B - 2*a*b* 
C + a^2*D))*ArcTanh[(Sqrt[b]*x)/(-Sqrt[a] + Sqrt[a + b*x^2])])/(8*b^(5/2))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.98, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.281, Rules used = {2338, 9, 25, 1473, 27, 299, 211, 224, 219}

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.

\(\displaystyle \int \frac {\sqrt {a+b x^2} \left (A+B x^2+C x^4+D x^6\right )}{x^2} \, dx\)

\(\Big \downarrow \) 2338

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

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1473

\(\displaystyle \frac {\frac {\int 3 \sqrt {b x^2+a} \left (a (2 b C-a D) x^2+2 b (2 A b+a B)\right )dx}{6 b}+\frac {a D x^3 \left (a+b x^2\right )^{3/2}}{6 b}}{a}-\frac {A \left (a+b x^2\right )^{3/2}}{a x}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \sqrt {b x^2+a} \left (a (2 b C-a D) x^2+2 b (2 A b+a B)\right )dx}{2 b}+\frac {a D x^3 \left (a+b x^2\right )^{3/2}}{6 b}}{a}-\frac {A \left (a+b x^2\right )^{3/2}}{a x}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {\frac {\left (8 b^2 (a B+2 A b)-a^2 (2 b C-a D)\right ) \int \sqrt {b x^2+a}dx}{4 b}+\frac {a x \left (a+b x^2\right )^{3/2} (2 b C-a D)}{4 b}}{2 b}+\frac {a D x^3 \left (a+b x^2\right )^{3/2}}{6 b}}{a}-\frac {A \left (a+b x^2\right )^{3/2}}{a x}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {\frac {\frac {\left (8 b^2 (a B+2 A b)-a^2 (2 b C-a D)\right ) \left (\frac {1}{2} a \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {1}{2} x \sqrt {a+b x^2}\right )}{4 b}+\frac {a x \left (a+b x^2\right )^{3/2} (2 b C-a D)}{4 b}}{2 b}+\frac {a D x^3 \left (a+b x^2\right )^{3/2}}{6 b}}{a}-\frac {A \left (a+b x^2\right )^{3/2}}{a x}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {\frac {\left (8 b^2 (a B+2 A b)-a^2 (2 b C-a D)\right ) \left (\frac {1}{2} a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {1}{2} x \sqrt {a+b x^2}\right )}{4 b}+\frac {a x \left (a+b x^2\right )^{3/2} (2 b C-a D)}{4 b}}{2 b}+\frac {a D x^3 \left (a+b x^2\right )^{3/2}}{6 b}}{a}-\frac {A \left (a+b x^2\right )^{3/2}}{a x}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {\left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {a+b x^2}\right ) \left (8 b^2 (a B+2 A b)-a^2 (2 b C-a D)\right )}{4 b}+\frac {a x \left (a+b x^2\right )^{3/2} (2 b C-a D)}{4 b}}{2 b}+\frac {a D x^3 \left (a+b x^2\right )^{3/2}}{6 b}}{a}-\frac {A \left (a+b x^2\right )^{3/2}}{a x}\)

Input:

Int[(Sqrt[a + b*x^2]*(A + B*x^2 + C*x^4 + D*x^6))/x^2,x]
 

Output:

-((A*(a + b*x^2)^(3/2))/(a*x)) + ((a*D*x^3*(a + b*x^2)^(3/2))/(6*b) + ((a* 
(2*b*C - a*D)*x*(a + b*x^2)^(3/2))/(4*b) + ((8*b^2*(2*A*b + a*B) - a^2*(2* 
b*C - a*D))*((x*Sqrt[a + b*x^2])/2 + (a*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2 
]])/(2*Sqrt[b])))/(4*b))/(2*b))/a
 

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 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 299

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

rule 1473

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
x_Symbol] :> Simp[c^p*x^(4*p - 1)*((d + e*x^2)^(q + 1)/(e*(4*p + 2*q + 1))) 
, x] + Simp[1/(e*(4*p + 2*q + 1))   Int[(d + e*x^2)^q*ExpandToSum[e*(4*p + 
2*q + 1)*(a + b*x^2 + c*x^4)^p - d*c^p*(4*p - 1)*x^(4*p - 2) - e*c^p*(4*p + 
 2*q + 1)*x^(4*p), x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b^2 - 
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] &&  !LtQ[q, -1]
 

rule 2338

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{ 
Q = PolynomialQuotient[Pq, c*x, x], R = PolynomialRemainder[Pq, c*x, x]}, S 
imp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Simp[1/(a*c*( 
m + 1))   Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*( 
m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && Lt 
Q[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])
 

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.69

method	result	size
pseudoelliptic	\(-\frac {-x \left (b^{3} A +\frac {1}{2} a \,b^{2} B -\frac {1}{8} a^{2} b C +\frac {1}{16} a^{3} D\right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )+\sqrt {b \,x^{2}+a}\, \left (\left (-\frac {1}{6} D x^{6}-\frac {1}{4} C \,x^{4}-\frac {1}{2} x^{2} B +A \right ) b^{\frac {5}{2}}+\frac {\left (\left (-\frac {2 D x^{2}}{3}-2 C \right ) b^{\frac {3}{2}}+D a \sqrt {b}\right ) x^{2} a}{16}\right )}{b^{\frac {5}{2}} x}\)	\(118\)
default	\(B \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )+A \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{a x}+\frac {2 b \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{a}\right )+C \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )+D \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )}{2 b}\right )\)	\(242\)

Input:

int((b*x^2+a)^(1/2)*(D*x^6+C*x^4+B*x^2+A)/x^2,x,method=_RETURNVERBOSE)
 

Output:

-(-x*(b^3*A+1/2*a*b^2*B-1/8*a^2*b*C+1/16*a^3*D)*arctanh((b*x^2+a)^(1/2)/x/ 
b^(1/2))+(b*x^2+a)^(1/2)*((-1/6*D*x^6-1/4*C*x^4-1/2*x^2*B+A)*b^(5/2)+1/16* 
((-2/3*D*x^2-2*C)*b^(3/2)+D*a*b^(1/2))*x^2*a))/b^(5/2)/x
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.58 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2+C x^4+D x^6\right )}{x^2} \, dx=\left [\frac {3 \, {\left (D a^{3} - 2 \, C a^{2} b + 8 \, B a b^{2} + 16 \, A b^{3}\right )} \sqrt {b} x \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (8 \, D b^{3} x^{6} + 2 \, {\left (D a b^{2} + 6 \, C b^{3}\right )} x^{4} - 48 \, A b^{3} - 3 \, {\left (D a^{2} b - 2 \, C a b^{2} - 8 \, B b^{3}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{96 \, b^{3} x}, -\frac {3 \, {\left (D a^{3} - 2 \, C a^{2} b + 8 \, B a b^{2} + 16 \, A b^{3}\right )} \sqrt {-b} x \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (8 \, D b^{3} x^{6} + 2 \, {\left (D a b^{2} + 6 \, C b^{3}\right )} x^{4} - 48 \, A b^{3} - 3 \, {\left (D a^{2} b - 2 \, C a b^{2} - 8 \, B b^{3}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{48 \, b^{3} x}\right ] \] Input:

integrate((b*x^2+a)^(1/2)*(D*x^6+C*x^4+B*x^2+A)/x^2,x, algorithm="fricas")
 

Output:

[1/96*(3*(D*a^3 - 2*C*a^2*b + 8*B*a*b^2 + 16*A*b^3)*sqrt(b)*x*log(-2*b*x^2 
 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(8*D*b^3*x^6 + 2*(D*a*b^2 + 6*C*b^ 
3)*x^4 - 48*A*b^3 - 3*(D*a^2*b - 2*C*a*b^2 - 8*B*b^3)*x^2)*sqrt(b*x^2 + a) 
)/(b^3*x), -1/48*(3*(D*a^3 - 2*C*a^2*b + 8*B*a*b^2 + 16*A*b^3)*sqrt(-b)*x* 
arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (8*D*b^3*x^6 + 2*(D*a*b^2 + 6*C*b^3)* 
x^4 - 48*A*b^3 - 3*(D*a^2*b - 2*C*a*b^2 - 8*B*b^3)*x^2)*sqrt(b*x^2 + a))/( 
b^3*x)]
 

Sympy [A] (verification not implemented)

Time = 1.55 (sec) , antiderivative size = 340, normalized size of antiderivative = 2.00 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2+C x^4+D x^6\right )}{x^2} \, dx=- \frac {A \sqrt {a}}{x \sqrt {1 + \frac {b x^{2}}{a}}} + A \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )} - \frac {A b x}{\sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} + B \left (\begin {cases} \frac {a \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {a + b x^{2}}}{2} & \text {for}\: b \neq 0 \\\sqrt {a} x & \text {otherwise} \end {cases}\right ) + C \left (\begin {cases} - \frac {a^{2} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{8 b} + \frac {a x \sqrt {a + b x^{2}}}{8 b} + \frac {x^{3} \sqrt {a + b x^{2}}}{4} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{3}}{3} & \text {otherwise} \end {cases}\right ) + D \left (\begin {cases} \frac {a^{3} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{16 b^{2}} - \frac {a^{2} x \sqrt {a + b x^{2}}}{16 b^{2}} + \frac {a x^{3} \sqrt {a + b x^{2}}}{24 b} + \frac {x^{5} \sqrt {a + b x^{2}}}{6} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{5}}{5} & \text {otherwise} \end {cases}\right ) \] Input:

integrate((b*x**2+a)**(1/2)*(D*x**6+C*x**4+B*x**2+A)/x**2,x)
 

Output:

-A*sqrt(a)/(x*sqrt(1 + b*x**2/a)) + A*sqrt(b)*asinh(sqrt(b)*x/sqrt(a)) - A 
*b*x/(sqrt(a)*sqrt(1 + b*x**2/a)) + B*Piecewise((a*Piecewise((log(2*sqrt(b 
)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*log(x)/sqrt(b*x**2), Tr 
ue))/2 + x*sqrt(a + b*x**2)/2, Ne(b, 0)), (sqrt(a)*x, True)) + C*Piecewise 
((-a**2*Piecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 
0)), (x*log(x)/sqrt(b*x**2), True))/(8*b) + a*x*sqrt(a + b*x**2)/(8*b) + x 
**3*sqrt(a + b*x**2)/4, Ne(b, 0)), (sqrt(a)*x**3/3, True)) + D*Piecewise(( 
a**3*Piecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)) 
, (x*log(x)/sqrt(b*x**2), True))/(16*b**2) - a**2*x*sqrt(a + b*x**2)/(16*b 
**2) + a*x**3*sqrt(a + b*x**2)/(24*b) + x**5*sqrt(a + b*x**2)/6, Ne(b, 0)) 
, (sqrt(a)*x**5/5, True))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.07 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2+C x^4+D x^6\right )}{x^2} \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} D x^{3}}{6 \, b} + \frac {1}{2} \, \sqrt {b x^{2} + a} B x - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} D a x}{8 \, b^{2}} + \frac {\sqrt {b x^{2} + a} D a^{2} x}{16 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} C x}{4 \, b} - \frac {\sqrt {b x^{2} + a} C a x}{8 \, b} + \frac {D a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {5}{2}}} - \frac {C a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {3}{2}}} + \frac {B a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {b}} + A \sqrt {b} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - \frac {\sqrt {b x^{2} + a} A}{x} \] Input:

integrate((b*x^2+a)^(1/2)*(D*x^6+C*x^4+B*x^2+A)/x^2,x, algorithm="maxima")
 

Output:

1/6*(b*x^2 + a)^(3/2)*D*x^3/b + 1/2*sqrt(b*x^2 + a)*B*x - 1/8*(b*x^2 + a)^ 
(3/2)*D*a*x/b^2 + 1/16*sqrt(b*x^2 + a)*D*a^2*x/b^2 + 1/4*(b*x^2 + a)^(3/2) 
*C*x/b - 1/8*sqrt(b*x^2 + a)*C*a*x/b + 1/16*D*a^3*arcsinh(b*x/sqrt(a*b))/b 
^(5/2) - 1/8*C*a^2*arcsinh(b*x/sqrt(a*b))/b^(3/2) + 1/2*B*a*arcsinh(b*x/sq 
rt(a*b))/sqrt(b) + A*sqrt(b)*arcsinh(b*x/sqrt(a*b)) - sqrt(b*x^2 + a)*A/x
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2+C x^4+D x^6\right )}{x^2} \, dx=\frac {1}{48} \, {\left (2 \, {\left (4 \, D x^{2} + \frac {D a b^{3} + 6 \, C b^{4}}{b^{4}}\right )} x^{2} - \frac {3 \, {\left (D a^{2} b^{2} - 2 \, C a b^{3} - 8 \, B b^{4}\right )}}{b^{4}}\right )} \sqrt {b x^{2} + a} x + \frac {2 \, A a \sqrt {b}}{{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a} - \frac {{\left (D a^{3} - 2 \, C a^{2} b + 8 \, B a b^{2} + 16 \, A b^{3}\right )} \log \left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2}\right )}{32 \, b^{\frac {5}{2}}} \] Input:

integrate((b*x^2+a)^(1/2)*(D*x^6+C*x^4+B*x^2+A)/x^2,x, algorithm="giac")
 

Output:

1/48*(2*(4*D*x^2 + (D*a*b^3 + 6*C*b^4)/b^4)*x^2 - 3*(D*a^2*b^2 - 2*C*a*b^3 
 - 8*B*b^4)/b^4)*sqrt(b*x^2 + a)*x + 2*A*a*sqrt(b)/((sqrt(b)*x - sqrt(b*x^ 
2 + a))^2 - a) - 1/32*(D*a^3 - 2*C*a^2*b + 8*B*a*b^2 + 16*A*b^3)*log((sqrt 
(b)*x - sqrt(b*x^2 + a))^2)/b^(5/2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2+C x^4+D x^6\right )}{x^2} \, dx=\int \frac {\sqrt {b\,x^2+a}\,\left (A+B\,x^2+C\,x^4+x^6\,D\right )}{x^2} \,d x \] Input:

int(((a + b*x^2)^(1/2)*(A + B*x^2 + C*x^4 + x^6*D))/x^2,x)
 

Output:

int(((a + b*x^2)^(1/2)*(A + B*x^2 + C*x^4 + x^6*D))/x^2, x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.35 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2+C x^4+D x^6\right )}{x^2} \, dx=\frac {-24 \sqrt {b \,x^{2}+a}\, a^{2} b d \,x^{2}-384 \sqrt {b \,x^{2}+a}\, a \,b^{3}+48 \sqrt {b \,x^{2}+a}\, a \,b^{2} c \,x^{2}+16 \sqrt {b \,x^{2}+a}\, a \,b^{2} d \,x^{4}+192 \sqrt {b \,x^{2}+a}\, b^{4} x^{2}+96 \sqrt {b \,x^{2}+a}\, b^{3} c \,x^{4}+64 \sqrt {b \,x^{2}+a}\, b^{3} d \,x^{6}+24 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} d x -48 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b c x +576 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{3} x +3 \sqrt {b}\, a^{3} d x -432 \sqrt {b}\, a \,b^{3} x}{384 b^{3} x} \] Input:

int((b*x^2+a)^(1/2)*(D*x^6+C*x^4+B*x^2+A)/x^2,x)
 

Output:

( - 24*sqrt(a + b*x**2)*a**2*b*d*x**2 - 384*sqrt(a + b*x**2)*a*b**3 + 48*s 
qrt(a + b*x**2)*a*b**2*c*x**2 + 16*sqrt(a + b*x**2)*a*b**2*d*x**4 + 192*sq 
rt(a + b*x**2)*b**4*x**2 + 96*sqrt(a + b*x**2)*b**3*c*x**4 + 64*sqrt(a + b 
*x**2)*b**3*d*x**6 + 24*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a) 
)*a**3*d*x - 48*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*b 
*c*x + 576*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a*b**3*x + 
3*sqrt(b)*a**3*d*x - 432*sqrt(b)*a*b**3*x)/(384*b**3*x)