\(\int \frac {\sqrt {a+b x^2} (A+B x^2+C x^4+D x^6)}{x^4} \, dx\) [192]

Optimal result

Integrand size = 32, antiderivative size = 143 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2+C x^4+D x^6\right )}{x^4} \, dx=\frac {1}{8} \left (\frac {8 b B}{a}+4 C-\frac {a D}{b}\right ) x \sqrt {a+b x^2}-\frac {A \left (a+b x^2\right )^{3/2}}{3 a x^3}-\frac {B \left (a+b x^2\right )^{3/2}}{a x}+\frac {D x \left (a+b x^2\right )^{3/2}}{4 b}+\frac {\left (8 b^2 B+4 a b C-a^2 D\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{3/2}} \] Output:

1/8*(8*b*B/a+4*C-a*D/b)*x*(b*x^2+a)^(1/2)-1/3*A*(b*x^2+a)^(3/2)/a/x^3-B*(b 
*x^2+a)^(3/2)/a/x+1/4*D*x*(b*x^2+a)^(3/2)/b+1/8*(8*B*b^2+4*C*a*b-D*a^2)*ar 
ctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(3/2)
 

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.82 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2+C x^4+D x^6\right )}{x^4} \, dx=\frac {\sqrt {a+b x^2} \left (-8 A b^2 x^2+3 a^2 D x^4+2 a b \left (-4 A-12 B x^2+6 C x^4+3 D x^6\right )\right )}{24 a b x^3}+\frac {\left (-8 b^2 B-4 a b C+a^2 D\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{8 b^{3/2}} \] Input:

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

Output:

(Sqrt[a + b*x^2]*(-8*A*b^2*x^2 + 3*a^2*D*x^4 + 2*a*b*(-4*A - 12*B*x^2 + 6* 
C*x^4 + 3*D*x^6)))/(24*a*b*x^3) + ((-8*b^2*B - 4*a*b*C + a^2*D)*Log[-(Sqrt 
[b]*x) + Sqrt[a + b*x^2]])/(8*b^(3/2))
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.96, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {2338, 9, 27, 1588, 25, 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.

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

\(\Big \downarrow \) 2338

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

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1588

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 299

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

-1/3*(A*(a + b*x^2)^(3/2))/(a*x^3) + (-((B*(a + b*x^2)^(3/2))/x) + (a*D*x* 
(a + b*x^2)^(3/2))/(4*b) + ((4*b*(2*b*B + a*C) - a^2*D)*((x*Sqrt[a + b*x^2 
])/2 + (a*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*Sqrt[b])))/(4*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 1588

Int[((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c 
_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + 
c*x^4)^p, f*x, x], R = PolynomialRemainder[(a + b*x^2 + c*x^4)^p, f*x, x]}, 
 Simp[R*(f*x)^(m + 1)*((d + e*x^2)^(q + 1)/(d*f*(m + 1))), x] + Simp[1/(d*f 
^2*(m + 1))   Int[(f*x)^(m + 2)*(d + e*x^2)^q*ExpandToSum[d*f*(m + 1)*(Qx/x 
) - e*R*(m + 2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e, f, q}, x] && Ne 
Q[b^2 - 4*a*c, 0] && IGtQ[p, 0] && LtQ[m, -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 = 0.56 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.77

method	result	size
pseudoelliptic	\(-\frac {-3 \left (B \,b^{2}+\frac {1}{2} C a b -\frac {1}{8} D a^{2}\right ) x^{3} b a \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )+\left (-\frac {3 D b^{\frac {3}{2}} a^{2} x^{4}}{8}+b^{\frac {5}{2}} \left (a \left (-\frac {3}{4} D x^{6}-\frac {3}{2} C \,x^{4}+3 x^{2} B +A \right )+A b \,x^{2}\right )\right ) \sqrt {b \,x^{2}+a}}{3 b^{\frac {5}{2}} a \,x^{3}}\)	\(110\)
default	\(C \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )-\frac {A \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3 a \,x^{3}}+B \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{a x}+\frac {2 b \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{a}\right )+D \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )\)	\(177\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.80 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2+C x^4+D x^6\right )}{x^4} \, dx=\left [-\frac {3 \, {\left (D a^{3} - 4 \, C a^{2} b - 8 \, B a b^{2}\right )} \sqrt {b} x^{3} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (6 \, D a b^{2} x^{6} + 3 \, {\left (D a^{2} b + 4 \, C a b^{2}\right )} x^{4} - 8 \, A a b^{2} - 8 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{48 \, a b^{2} x^{3}}, \frac {3 \, {\left (D a^{3} - 4 \, C a^{2} b - 8 \, B a b^{2}\right )} \sqrt {-b} x^{3} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (6 \, D a b^{2} x^{6} + 3 \, {\left (D a^{2} b + 4 \, C a b^{2}\right )} x^{4} - 8 \, A a b^{2} - 8 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{24 \, a b^{2} x^{3}}\right ] \] Input:

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

Output:

[-1/48*(3*(D*a^3 - 4*C*a^2*b - 8*B*a*b^2)*sqrt(b)*x^3*log(-2*b*x^2 - 2*sqr 
t(b*x^2 + a)*sqrt(b)*x - a) - 2*(6*D*a*b^2*x^6 + 3*(D*a^2*b + 4*C*a*b^2)*x 
^4 - 8*A*a*b^2 - 8*(3*B*a*b^2 + A*b^3)*x^2)*sqrt(b*x^2 + a))/(a*b^2*x^3), 
1/24*(3*(D*a^3 - 4*C*a^2*b - 8*B*a*b^2)*sqrt(-b)*x^3*arctan(sqrt(-b)*x/sqr 
t(b*x^2 + a)) + (6*D*a*b^2*x^6 + 3*(D*a^2*b + 4*C*a*b^2)*x^4 - 8*A*a*b^2 - 
 8*(3*B*a*b^2 + A*b^3)*x^2)*sqrt(b*x^2 + a))/(a*b^2*x^3)]
 

Sympy [A] (verification not implemented)

Time = 1.73 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.88 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2+C x^4+D x^6\right )}{x^4} \, dx=- \frac {A \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{3 x^{2}} - \frac {A b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a} - \frac {B \sqrt {a}}{x \sqrt {1 + \frac {b x^{2}}{a}}} + B \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )} - \frac {B b x}{\sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} + C \left (\begin {cases} \frac {a \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {a + b x^{2}}}{2} & \text {for}\: b \neq 0 \\\sqrt {a} x & \text {otherwise} \end {cases}\right ) + D \left (\begin {cases} - \frac {a^{2} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{8 b} + \frac {a x \sqrt {a + b x^{2}}}{8 b} + \frac {x^{3} \sqrt {a + b x^{2}}}{4} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{3}}{3} & \text {otherwise} \end {cases}\right ) \] Input:

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

Output:

-A*sqrt(b)*sqrt(a/(b*x**2) + 1)/(3*x**2) - A*b**(3/2)*sqrt(a/(b*x**2) + 1) 
/(3*a) - B*sqrt(a)/(x*sqrt(1 + b*x**2/a)) + B*sqrt(b)*asinh(sqrt(b)*x/sqrt 
(a)) - B*b*x/(sqrt(a)*sqrt(1 + b*x**2/a)) + C*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), True))/2 + x*sqrt(a + b*x**2)/2, Ne(b, 0)), (sqrt(a)*x, True)) + D*P 
iecewise((-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))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2+C x^4+D x^6\right )}{x^4} \, dx=\frac {1}{2} \, \sqrt {b x^{2} + a} C x + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} D x}{4 \, b} - \frac {\sqrt {b x^{2} + a} D a x}{8 \, b} - \frac {D a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {3}{2}}} + \frac {C a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {b}} + B \sqrt {b} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - \frac {\sqrt {b x^{2} + a} B}{x} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A}{3 \, a x^{3}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.41 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2+C x^4+D x^6\right )}{x^4} \, dx=\frac {1}{8} \, {\left (2 \, D x^{2} + \frac {D a b + 4 \, C b^{2}}{b^{2}}\right )} \sqrt {b x^{2} + a} x + \frac {{\left (D a^{2} - 4 \, C a b - 8 \, B b^{2}\right )} \log \left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2}\right )}{16 \, b^{\frac {3}{2}}} + \frac {2 \, {\left (3 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} B a \sqrt {b} + 3 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} A b^{\frac {3}{2}} - 6 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} B a^{2} \sqrt {b} + 3 \, B a^{3} \sqrt {b} + A a^{2} b^{\frac {3}{2}}\right )}}{3 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{3}} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.38 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2+C x^4+D x^6\right )}{x^4} \, dx=\frac {-64 \sqrt {b \,x^{2}+a}\, a \,b^{2}+24 \sqrt {b \,x^{2}+a}\, a b d \,x^{4}-256 \sqrt {b \,x^{2}+a}\, b^{3} x^{2}+96 \sqrt {b \,x^{2}+a}\, b^{2} c \,x^{4}+48 \sqrt {b \,x^{2}+a}\, b^{2} d \,x^{6}-24 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} d \,x^{3}+96 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b c \,x^{3}+192 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{3} x^{3}-\sqrt {b}\, a^{2} d \,x^{3}+16 \sqrt {b}\, a b c \,x^{3}}{192 b^{2} x^{3}} \] Input:

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

Output:

( - 64*sqrt(a + b*x**2)*a*b**2 + 24*sqrt(a + b*x**2)*a*b*d*x**4 - 256*sqrt 
(a + b*x**2)*b**3*x**2 + 96*sqrt(a + b*x**2)*b**2*c*x**4 + 48*sqrt(a + b*x 
**2)*b**2*d*x**6 - 24*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))* 
a**2*d*x**3 + 96*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a*b*c 
*x**3 + 192*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*b**3*x**3 
- sqrt(b)*a**2*d*x**3 + 16*sqrt(b)*a*b*c*x**3)/(192*b**2*x**3)