\(\int \frac {A+B x+C x^2+D x^3}{x^6 \sqrt {a+b x^2}} \, dx\) [99]

Optimal result

Integrand size = 30, antiderivative size = 173 \[ \int \frac {A+B x+C x^2+D x^3}{x^6 \sqrt {a+b x^2}} \, dx=-\frac {A \sqrt {a+b x^2}}{5 a x^5}-\frac {B \sqrt {a+b x^2}}{4 a x^4}+\frac {(4 A b-5 a C) \sqrt {a+b x^2}}{15 a^2 x^3}+\frac {(3 b B-4 a D) \sqrt {a+b x^2}}{8 a^2 x^2}-\frac {2 b (4 A b-5 a C) \sqrt {a+b x^2}}{15 a^3 x}-\frac {b (3 b B-4 a D) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{8 a^{5/2}} \] Output:

-1/5*A*(b*x^2+a)^(1/2)/a/x^5-1/4*B*(b*x^2+a)^(1/2)/a/x^4+1/15*(4*A*b-5*C*a 
)*(b*x^2+a)^(1/2)/a^2/x^3+1/8*(3*B*b-4*D*a)*(b*x^2+a)^(1/2)/a^2/x^2-2/15*b 
*(4*A*b-5*C*a)*(b*x^2+a)^(1/2)/a^3/x-1/8*b*(3*B*b-4*D*a)*arctanh((b*x^2+a) 
^(1/2)/a^(1/2))/a^(5/2)
 

Mathematica [A] (verified)

Time = 0.93 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.72 \[ \int \frac {A+B x+C x^2+D x^3}{x^6 \sqrt {a+b x^2}} \, dx=\frac {\frac {\sqrt {a+b x^2} \left (-64 A b^2 x^4+a b x^2 (32 A+5 x (9 B+16 C x))-2 a^2 \left (12 A+5 x \left (3 B+4 C x+6 D x^2\right )\right )\right )}{x^5}-30 \sqrt {a} b (-3 b B+4 a D) \text {arctanh}\left (\frac {\sqrt {b} x-\sqrt {a+b x^2}}{\sqrt {a}}\right )}{120 a^3} \] Input:

Integrate[(A + B*x + C*x^2 + D*x^3)/(x^6*Sqrt[a + b*x^2]),x]
 

Output:

((Sqrt[a + b*x^2]*(-64*A*b^2*x^4 + a*b*x^2*(32*A + 5*x*(9*B + 16*C*x)) - 2 
*a^2*(12*A + 5*x*(3*B + 4*C*x + 6*D*x^2))))/x^5 - 30*Sqrt[a]*b*(-3*b*B + 4 
*a*D)*ArcTanh[(Sqrt[b]*x - Sqrt[a + b*x^2])/Sqrt[a]])/(120*a^3)
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.09, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {2338, 25, 2338, 27, 539, 25, 539, 27, 534, 243, 73, 221}

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.

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

\(\Big \downarrow \) 2338

\(\displaystyle -\frac {\int -\frac {5 a D x^2-(4 A b-5 a C) x+5 a B}{x^5 \sqrt {b x^2+a}}dx}{5 a}-\frac {A \sqrt {a+b x^2}}{5 a x^5}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {5 a D x^2-(4 A b-5 a C) x+5 a B}{x^5 \sqrt {b x^2+a}}dx}{5 a}-\frac {A \sqrt {a+b x^2}}{5 a x^5}\)

\(\Big \downarrow \) 2338

\(\displaystyle \frac {-\frac {\int \frac {a (4 (4 A b-5 a C)+5 (3 b B-4 a D) x)}{x^4 \sqrt {b x^2+a}}dx}{4 a}-\frac {5 B \sqrt {a+b x^2}}{4 x^4}}{5 a}-\frac {A \sqrt {a+b x^2}}{5 a x^5}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {1}{4} \int \frac {4 (4 A b-5 a C)+5 (3 b B-4 a D) x}{x^4 \sqrt {b x^2+a}}dx-\frac {5 B \sqrt {a+b x^2}}{4 x^4}}{5 a}-\frac {A \sqrt {a+b x^2}}{5 a x^5}\)

\(\Big \downarrow \) 539

\(\displaystyle \frac {\frac {1}{4} \left (\frac {\int -\frac {15 a (3 b B-4 a D)-8 b (4 A b-5 a C) x}{x^3 \sqrt {b x^2+a}}dx}{3 a}+\frac {4 \sqrt {a+b x^2} (4 A b-5 a C)}{3 a x^3}\right )-\frac {5 B \sqrt {a+b x^2}}{4 x^4}}{5 a}-\frac {A \sqrt {a+b x^2}}{5 a x^5}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {1}{4} \left (\frac {4 \sqrt {a+b x^2} (4 A b-5 a C)}{3 a x^3}-\frac {\int \frac {15 a (3 b B-4 a D)-8 b (4 A b-5 a C) x}{x^3 \sqrt {b x^2+a}}dx}{3 a}\right )-\frac {5 B \sqrt {a+b x^2}}{4 x^4}}{5 a}-\frac {A \sqrt {a+b x^2}}{5 a x^5}\)

\(\Big \downarrow \) 539

\(\displaystyle \frac {\frac {1}{4} \left (\frac {4 \sqrt {a+b x^2} (4 A b-5 a C)}{3 a x^3}-\frac {-\frac {\int \frac {a b (16 (4 A b-5 a C)+15 (3 b B-4 a D) x)}{x^2 \sqrt {b x^2+a}}dx}{2 a}-\frac {15 \sqrt {a+b x^2} (3 b B-4 a D)}{2 x^2}}{3 a}\right )-\frac {5 B \sqrt {a+b x^2}}{4 x^4}}{5 a}-\frac {A \sqrt {a+b x^2}}{5 a x^5}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{4} \left (\frac {4 \sqrt {a+b x^2} (4 A b-5 a C)}{3 a x^3}-\frac {-\frac {1}{2} b \int \frac {16 (4 A b-5 a C)+15 (3 b B-4 a D) x}{x^2 \sqrt {b x^2+a}}dx-\frac {15 \sqrt {a+b x^2} (3 b B-4 a D)}{2 x^2}}{3 a}\right )-\frac {5 B \sqrt {a+b x^2}}{4 x^4}}{5 a}-\frac {A \sqrt {a+b x^2}}{5 a x^5}\)

\(\Big \downarrow \) 534

\(\displaystyle \frac {\frac {1}{4} \left (\frac {4 \sqrt {a+b x^2} (4 A b-5 a C)}{3 a x^3}-\frac {-\frac {1}{2} b \left (15 (3 b B-4 a D) \int \frac {1}{x \sqrt {b x^2+a}}dx-\frac {16 \sqrt {a+b x^2} (4 A b-5 a C)}{a x}\right )-\frac {15 \sqrt {a+b x^2} (3 b B-4 a D)}{2 x^2}}{3 a}\right )-\frac {5 B \sqrt {a+b x^2}}{4 x^4}}{5 a}-\frac {A \sqrt {a+b x^2}}{5 a x^5}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {\frac {1}{4} \left (\frac {4 \sqrt {a+b x^2} (4 A b-5 a C)}{3 a x^3}-\frac {-\frac {1}{2} b \left (\frac {15}{2} (3 b B-4 a D) \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2-\frac {16 \sqrt {a+b x^2} (4 A b-5 a C)}{a x}\right )-\frac {15 \sqrt {a+b x^2} (3 b B-4 a D)}{2 x^2}}{3 a}\right )-\frac {5 B \sqrt {a+b x^2}}{4 x^4}}{5 a}-\frac {A \sqrt {a+b x^2}}{5 a x^5}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {1}{4} \left (\frac {4 \sqrt {a+b x^2} (4 A b-5 a C)}{3 a x^3}-\frac {-\frac {1}{2} b \left (\frac {15 (3 b B-4 a D) \int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{b}-\frac {16 \sqrt {a+b x^2} (4 A b-5 a C)}{a x}\right )-\frac {15 \sqrt {a+b x^2} (3 b B-4 a D)}{2 x^2}}{3 a}\right )-\frac {5 B \sqrt {a+b x^2}}{4 x^4}}{5 a}-\frac {A \sqrt {a+b x^2}}{5 a x^5}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {1}{4} \left (\frac {4 \sqrt {a+b x^2} (4 A b-5 a C)}{3 a x^3}-\frac {-\frac {1}{2} b \left (-\frac {16 \sqrt {a+b x^2} (4 A b-5 a C)}{a x}-\frac {15 \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) (3 b B-4 a D)}{\sqrt {a}}\right )-\frac {15 \sqrt {a+b x^2} (3 b B-4 a D)}{2 x^2}}{3 a}\right )-\frac {5 B \sqrt {a+b x^2}}{4 x^4}}{5 a}-\frac {A \sqrt {a+b x^2}}{5 a x^5}\)

Input:

Int[(A + B*x + C*x^2 + D*x^3)/(x^6*Sqrt[a + b*x^2]),x]
 

Output:

-1/5*(A*Sqrt[a + b*x^2])/(a*x^5) + ((-5*B*Sqrt[a + b*x^2])/(4*x^4) + ((4*( 
4*A*b - 5*a*C)*Sqrt[a + b*x^2])/(3*a*x^3) - ((-15*(3*b*B - 4*a*D)*Sqrt[a + 
 b*x^2])/(2*x^2) - (b*((-16*(4*A*b - 5*a*C)*Sqrt[a + b*x^2])/(a*x) - (15*( 
3*b*B - 4*a*D)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/Sqrt[a]))/2)/(3*a))/4)/(5 
*a)
 

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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

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

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 534

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

rule 539

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

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 = 0.57 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.29

method	result	size
default	\(A \left (-\frac {\sqrt {b \,x^{2}+a}}{5 a \,x^{5}}-\frac {4 b \left (-\frac {\sqrt {b \,x^{2}+a}}{3 a \,x^{3}}+\frac {2 b \sqrt {b \,x^{2}+a}}{3 a^{2} x}\right )}{5 a}\right )+B \left (-\frac {\sqrt {b \,x^{2}+a}}{4 a \,x^{4}}-\frac {3 b \left (-\frac {\sqrt {b \,x^{2}+a}}{2 a \,x^{2}}+\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{2 a^{\frac {3}{2}}}\right )}{4 a}\right )+C \left (-\frac {\sqrt {b \,x^{2}+a}}{3 a \,x^{3}}+\frac {2 b \sqrt {b \,x^{2}+a}}{3 a^{2} x}\right )+D \left (-\frac {\sqrt {b \,x^{2}+a}}{2 a \,x^{2}}+\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{2 a^{\frac {3}{2}}}\right )\)	\(224\)

Input:

int((D*x^3+C*x^2+B*x+A)/x^6/(b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.51 \[ \int \frac {A+B x+C x^2+D x^3}{x^6 \sqrt {a+b x^2}} \, dx=\left [-\frac {15 \, {\left (4 \, D a b - 3 \, B b^{2}\right )} \sqrt {a} x^{5} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (16 \, {\left (5 \, C a b - 4 \, A b^{2}\right )} x^{4} - 30 \, B a^{2} x - 15 \, {\left (4 \, D a^{2} - 3 \, B a b\right )} x^{3} - 24 \, A a^{2} - 8 \, {\left (5 \, C a^{2} - 4 \, A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{240 \, a^{3} x^{5}}, -\frac {15 \, {\left (4 \, D a b - 3 \, B b^{2}\right )} \sqrt {-a} x^{5} \arctan \left (\frac {\sqrt {b x^{2} + a} \sqrt {-a}}{a}\right ) - {\left (16 \, {\left (5 \, C a b - 4 \, A b^{2}\right )} x^{4} - 30 \, B a^{2} x - 15 \, {\left (4 \, D a^{2} - 3 \, B a b\right )} x^{3} - 24 \, A a^{2} - 8 \, {\left (5 \, C a^{2} - 4 \, A a b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{120 \, a^{3} x^{5}}\right ] \] Input:

integrate((D*x^3+C*x^2+B*x+A)/x^6/(b*x^2+a)^(1/2),x, algorithm="fricas")
 

Output:

[-1/240*(15*(4*D*a*b - 3*B*b^2)*sqrt(a)*x^5*log(-(b*x^2 - 2*sqrt(b*x^2 + a 
)*sqrt(a) + 2*a)/x^2) - 2*(16*(5*C*a*b - 4*A*b^2)*x^4 - 30*B*a^2*x - 15*(4 
*D*a^2 - 3*B*a*b)*x^3 - 24*A*a^2 - 8*(5*C*a^2 - 4*A*a*b)*x^2)*sqrt(b*x^2 + 
 a))/(a^3*x^5), -1/120*(15*(4*D*a*b - 3*B*b^2)*sqrt(-a)*x^5*arctan(sqrt(b* 
x^2 + a)*sqrt(-a)/a) - (16*(5*C*a*b - 4*A*b^2)*x^4 - 30*B*a^2*x - 15*(4*D* 
a^2 - 3*B*a*b)*x^3 - 24*A*a^2 - 8*(5*C*a^2 - 4*A*a*b)*x^2)*sqrt(b*x^2 + a) 
)/(a^3*x^5)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 507 vs. \(2 (160) = 320\).

Time = 5.15 (sec) , antiderivative size = 507, normalized size of antiderivative = 2.93 \[ \int \frac {A+B x+C x^2+D x^3}{x^6 \sqrt {a+b x^2}} \, dx=- \frac {3 A a^{4} b^{\frac {9}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{5} b^{4} x^{4} + 30 a^{4} b^{5} x^{6} + 15 a^{3} b^{6} x^{8}} - \frac {2 A a^{3} b^{\frac {11}{2}} x^{2} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{5} b^{4} x^{4} + 30 a^{4} b^{5} x^{6} + 15 a^{3} b^{6} x^{8}} - \frac {3 A a^{2} b^{\frac {13}{2}} x^{4} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{5} b^{4} x^{4} + 30 a^{4} b^{5} x^{6} + 15 a^{3} b^{6} x^{8}} - \frac {12 A a b^{\frac {15}{2}} x^{6} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{5} b^{4} x^{4} + 30 a^{4} b^{5} x^{6} + 15 a^{3} b^{6} x^{8}} - \frac {8 A b^{\frac {17}{2}} x^{8} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{5} b^{4} x^{4} + 30 a^{4} b^{5} x^{6} + 15 a^{3} b^{6} x^{8}} - \frac {B}{4 \sqrt {b} x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {B \sqrt {b}}{8 a x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {3 B b^{\frac {3}{2}}}{8 a^{2} x \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {3 B b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{8 a^{\frac {5}{2}}} - \frac {C \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a x^{2}} + \frac {2 C b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a^{2}} - \frac {D \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{2 a x} + \frac {D b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{2 a^{\frac {3}{2}}} \] Input:

integrate((D*x**3+C*x**2+B*x+A)/x**6/(b*x**2+a)**(1/2),x)
 

Output:

-3*A*a**4*b**(9/2)*sqrt(a/(b*x**2) + 1)/(15*a**5*b**4*x**4 + 30*a**4*b**5* 
x**6 + 15*a**3*b**6*x**8) - 2*A*a**3*b**(11/2)*x**2*sqrt(a/(b*x**2) + 1)/( 
15*a**5*b**4*x**4 + 30*a**4*b**5*x**6 + 15*a**3*b**6*x**8) - 3*A*a**2*b**( 
13/2)*x**4*sqrt(a/(b*x**2) + 1)/(15*a**5*b**4*x**4 + 30*a**4*b**5*x**6 + 1 
5*a**3*b**6*x**8) - 12*A*a*b**(15/2)*x**6*sqrt(a/(b*x**2) + 1)/(15*a**5*b* 
*4*x**4 + 30*a**4*b**5*x**6 + 15*a**3*b**6*x**8) - 8*A*b**(17/2)*x**8*sqrt 
(a/(b*x**2) + 1)/(15*a**5*b**4*x**4 + 30*a**4*b**5*x**6 + 15*a**3*b**6*x** 
8) - B/(4*sqrt(b)*x**5*sqrt(a/(b*x**2) + 1)) + B*sqrt(b)/(8*a*x**3*sqrt(a/ 
(b*x**2) + 1)) + 3*B*b**(3/2)/(8*a**2*x*sqrt(a/(b*x**2) + 1)) - 3*B*b**2*a 
sinh(sqrt(a)/(sqrt(b)*x))/(8*a**(5/2)) - C*sqrt(b)*sqrt(a/(b*x**2) + 1)/(3 
*a*x**2) + 2*C*b**(3/2)*sqrt(a/(b*x**2) + 1)/(3*a**2) - D*sqrt(b)*sqrt(a/( 
b*x**2) + 1)/(2*a*x) + D*b*asinh(sqrt(a)/(sqrt(b)*x))/(2*a**(3/2))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.10 \[ \int \frac {A+B x+C x^2+D x^3}{x^6 \sqrt {a+b x^2}} \, dx=\frac {D b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, a^{\frac {3}{2}}} - \frac {3 \, B b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{8 \, a^{\frac {5}{2}}} + \frac {2 \, \sqrt {b x^{2} + a} C b}{3 \, a^{2} x} - \frac {8 \, \sqrt {b x^{2} + a} A b^{2}}{15 \, a^{3} x} - \frac {\sqrt {b x^{2} + a} D}{2 \, a x^{2}} + \frac {3 \, \sqrt {b x^{2} + a} B b}{8 \, a^{2} x^{2}} - \frac {\sqrt {b x^{2} + a} C}{3 \, a x^{3}} + \frac {4 \, \sqrt {b x^{2} + a} A b}{15 \, a^{2} x^{3}} - \frac {\sqrt {b x^{2} + a} B}{4 \, a x^{4}} - \frac {\sqrt {b x^{2} + a} A}{5 \, a x^{5}} \] Input:

integrate((D*x^3+C*x^2+B*x+A)/x^6/(b*x^2+a)^(1/2),x, algorithm="maxima")
 

Output:

1/2*D*b*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(3/2) - 3/8*B*b^2*arcsinh(a/(sqrt( 
a*b)*abs(x)))/a^(5/2) + 2/3*sqrt(b*x^2 + a)*C*b/(a^2*x) - 8/15*sqrt(b*x^2 
+ a)*A*b^2/(a^3*x) - 1/2*sqrt(b*x^2 + a)*D/(a*x^2) + 3/8*sqrt(b*x^2 + a)*B 
*b/(a^2*x^2) - 1/3*sqrt(b*x^2 + a)*C/(a*x^3) + 4/15*sqrt(b*x^2 + a)*A*b/(a 
^2*x^3) - 1/4*sqrt(b*x^2 + a)*B/(a*x^4) - 1/5*sqrt(b*x^2 + a)*A/(a*x^5)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 442 vs. \(2 (145) = 290\).

Time = 0.13 (sec) , antiderivative size = 442, normalized size of antiderivative = 2.55 \[ \int \frac {A+B x+C x^2+D x^3}{x^6 \sqrt {a+b x^2}} \, dx=-\frac {{\left (4 \, D a b - 3 \, B b^{2}\right )} \arctan \left (-\frac {\sqrt {b} x - \sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a^{2}} + \frac {60 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{9} D a b - 45 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{9} B b^{2} - 120 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{7} D a^{2} b + 210 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{7} B a b^{2} + 240 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} C a^{2} b^{\frac {3}{2}} - 560 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} C a^{3} b^{\frac {3}{2}} + 640 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} A a^{2} b^{\frac {5}{2}} + 120 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{3} D a^{4} b - 210 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{3} B a^{3} b^{2} + 400 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} C a^{4} b^{\frac {3}{2}} - 320 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} A a^{3} b^{\frac {5}{2}} - 60 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )} D a^{5} b + 45 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )} B a^{4} b^{2} - 80 \, C a^{5} b^{\frac {3}{2}} + 64 \, A a^{4} b^{\frac {5}{2}}}{60 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{5} a^{2}} \] Input:

integrate((D*x^3+C*x^2+B*x+A)/x^6/(b*x^2+a)^(1/2),x, algorithm="giac")
 

Output:

-1/4*(4*D*a*b - 3*B*b^2)*arctan(-(sqrt(b)*x - sqrt(b*x^2 + a))/sqrt(-a))/( 
sqrt(-a)*a^2) + 1/60*(60*(sqrt(b)*x - sqrt(b*x^2 + a))^9*D*a*b - 45*(sqrt( 
b)*x - sqrt(b*x^2 + a))^9*B*b^2 - 120*(sqrt(b)*x - sqrt(b*x^2 + a))^7*D*a^ 
2*b + 210*(sqrt(b)*x - sqrt(b*x^2 + a))^7*B*a*b^2 + 240*(sqrt(b)*x - sqrt( 
b*x^2 + a))^6*C*a^2*b^(3/2) - 560*(sqrt(b)*x - sqrt(b*x^2 + a))^4*C*a^3*b^ 
(3/2) + 640*(sqrt(b)*x - sqrt(b*x^2 + a))^4*A*a^2*b^(5/2) + 120*(sqrt(b)*x 
 - sqrt(b*x^2 + a))^3*D*a^4*b - 210*(sqrt(b)*x - sqrt(b*x^2 + a))^3*B*a^3* 
b^2 + 400*(sqrt(b)*x - sqrt(b*x^2 + a))^2*C*a^4*b^(3/2) - 320*(sqrt(b)*x - 
 sqrt(b*x^2 + a))^2*A*a^3*b^(5/2) - 60*(sqrt(b)*x - sqrt(b*x^2 + a))*D*a^5 
*b + 45*(sqrt(b)*x - sqrt(b*x^2 + a))*B*a^4*b^2 - 80*C*a^5*b^(3/2) + 64*A* 
a^4*b^(5/2))/(((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)^5*a^2)
 

Mupad [B] (verification not implemented)

Time = 3.05 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.95 \[ \int \frac {A+B x+C x^2+D x^3}{x^6 \sqrt {a+b x^2}} \, dx=\frac {b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )\,D}{2\,a^{3/2}}-\frac {\sqrt {b\,x^2+a}\,D}{2\,a\,x^2}-\frac {3\,B\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{8\,a^{5/2}}-\frac {5\,B\,\sqrt {b\,x^2+a}}{8\,a\,x^4}+\frac {3\,B\,{\left (b\,x^2+a\right )}^{3/2}}{8\,a^2\,x^4}-\frac {C\,\sqrt {b\,x^2+a}\,\left (a-2\,b\,x^2\right )}{3\,a^2\,x^3}-\frac {A\,\sqrt {b\,x^2+a}\,\left (3\,a^2-4\,a\,b\,x^2+8\,b^2\,x^4\right )}{15\,a^3\,x^5} \] Input:

int((A + B*x + C*x^2 + x^3*D)/(x^6*(a + b*x^2)^(1/2)),x)
 

Output:

(b*atanh((a + b*x^2)^(1/2)/a^(1/2))*D)/(2*a^(3/2)) - ((a + b*x^2)^(1/2)*D) 
/(2*a*x^2) - (3*B*b^2*atanh((a + b*x^2)^(1/2)/a^(1/2)))/(8*a^(5/2)) - (5*B 
*(a + b*x^2)^(1/2))/(8*a*x^4) + (3*B*(a + b*x^2)^(3/2))/(8*a^2*x^4) - (C*( 
a + b*x^2)^(1/2)*(a - 2*b*x^2))/(3*a^2*x^3) - (A*(a + b*x^2)^(1/2)*(3*a^2 
+ 8*b^2*x^4 - 4*a*b*x^2))/(15*a^3*x^5)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.66 \[ \int \frac {A+B x+C x^2+D x^3}{x^6 \sqrt {a+b x^2}} \, dx=\frac {-24 \sqrt {b \,x^{2}+a}\, a^{3}+32 \sqrt {b \,x^{2}+a}\, a^{2} b \,x^{2}-30 \sqrt {b \,x^{2}+a}\, a^{2} b x -40 \sqrt {b \,x^{2}+a}\, a^{2} c \,x^{2}-60 \sqrt {b \,x^{2}+a}\, a^{2} d \,x^{3}-64 \sqrt {b \,x^{2}+a}\, a \,b^{2} x^{4}+45 \sqrt {b \,x^{2}+a}\, a \,b^{2} x^{3}+80 \sqrt {b \,x^{2}+a}\, a b c \,x^{4}-60 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b d \,x^{5}+45 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}-\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{3} x^{5}+60 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b d \,x^{5}-45 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{3} x^{5}+64 \sqrt {b}\, a \,b^{2} x^{5}-80 \sqrt {b}\, a b c \,x^{5}}{120 a^{3} x^{5}} \] Input:

int((D*x^3+C*x^2+B*x+A)/x^6/(b*x^2+a)^(1/2),x)
 

Output:

( - 24*sqrt(a + b*x**2)*a**3 + 32*sqrt(a + b*x**2)*a**2*b*x**2 - 30*sqrt(a 
 + b*x**2)*a**2*b*x - 40*sqrt(a + b*x**2)*a**2*c*x**2 - 60*sqrt(a + b*x**2 
)*a**2*d*x**3 - 64*sqrt(a + b*x**2)*a*b**2*x**4 + 45*sqrt(a + b*x**2)*a*b* 
*2*x**3 + 80*sqrt(a + b*x**2)*a*b*c*x**4 - 60*sqrt(a)*log((sqrt(a + b*x**2 
) - sqrt(a) + sqrt(b)*x)/sqrt(a))*a*b*d*x**5 + 45*sqrt(a)*log((sqrt(a + b* 
x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*b**3*x**5 + 60*sqrt(a)*log((sqrt(a + 
 b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*a*b*d*x**5 - 45*sqrt(a)*log((sqrt 
(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*b**3*x**5 + 64*sqrt(b)*a*b**2 
*x**5 - 80*sqrt(b)*a*b*c*x**5)/(120*a**3*x**5)