\(\int \frac {A+B x^2+C x^4+D x^6}{(a+b x^2)^3} \, dx\) [132]

Optimal result

Integrand size = 27, antiderivative size = 147 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^3} \, dx=\frac {D x}{b^3}+\frac {\left (\frac {A}{a}-\frac {b^2 B-a b C+a^2 D}{b^3}\right ) x}{4 \left (a+b x^2\right )^2}+\frac {\left (3 A b^3+a \left (b^2 B-5 a b C+9 a^2 D\right )\right ) x}{8 a^2 b^3 \left (a+b x^2\right )}+\frac {\left (3 A b^3+a \left (b^2 B+3 a b C-15 a^2 D\right )\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{5/2} b^{7/2}} \] Output:

D*x/b^3+1/4*(A/a-(B*b^2-C*a*b+D*a^2)/b^3)*x/(b*x^2+a)^2+1/8*(3*A*b^3+a*(B* 
b^2-5*C*a*b+9*D*a^2))*x/a^2/b^3/(b*x^2+a)+1/8*(3*A*b^3+a*(B*b^2+3*C*a*b-15 
*D*a^2))*arctan(b^(1/2)*x/a^(1/2))/a^(5/2)/b^(7/2)
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.96 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^3} \, dx=\frac {x \left (15 a^4 D+3 A b^4 x^2+a b^3 \left (5 A+B x^2\right )+a^3 b \left (-3 C+25 D x^2\right )-a^2 b^2 \left (B+5 C x^2-8 D x^4\right )\right )}{8 a^2 b^3 \left (a+b x^2\right )^2}+\frac {\left (3 A b^3+a \left (b^2 B+3 a b C-15 a^2 D\right )\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{5/2} b^{7/2}} \] Input:

Integrate[(A + B*x^2 + C*x^4 + D*x^6)/(a + b*x^2)^3,x]
 

Output:

(x*(15*a^4*D + 3*A*b^4*x^2 + a*b^3*(5*A + B*x^2) + a^3*b*(-3*C + 25*D*x^2) 
 - a^2*b^2*(B + 5*C*x^2 - 8*D*x^4)))/(8*a^2*b^3*(a + b*x^2)^2) + ((3*A*b^3 
 + a*(b^2*B + 3*a*b*C - 15*a^2*D))*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(5/2) 
*b^(7/2))
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2345, 25, 1471, 25, 27, 299, 218}

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.

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

\(\Big \downarrow \) 2345

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1471

\(\displaystyle \frac {\frac {x \left (\frac {9 a^2 D-5 a b C+b^2 B}{b^3}+\frac {3 A}{a}\right )}{2 \left (a+b x^2\right )}-\frac {\int -\frac {-7 D a^3+8 b D x^2 a^2+3 b C a^2+b^2 B a+3 A b^3}{b^3 \left (b x^2+a\right )}dx}{2 a}}{4 a}+\frac {x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {3 A b^3+8 a^2 D x^2 b+a \left (-7 D a^2+3 b C a+b^2 B\right )}{b^3 \left (b x^2+a\right )}dx}{2 a}+\frac {x \left (\frac {9 a^2 D-5 a b C+b^2 B}{b^3}+\frac {3 A}{a}\right )}{2 \left (a+b x^2\right )}}{4 a}+\frac {x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {3 A b^3+8 a^2 D x^2 b+a \left (-7 D a^2+3 b C a+b^2 B\right )}{b x^2+a}dx}{2 a b^3}+\frac {x \left (\frac {9 a^2 D-5 a b C+b^2 B}{b^3}+\frac {3 A}{a}\right )}{2 \left (a+b x^2\right )}}{4 a}+\frac {x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {\left (a \left (-15 a^2 D+3 a b C+b^2 B\right )+3 A b^3\right ) \int \frac {1}{b x^2+a}dx+8 a^2 D x}{2 a b^3}+\frac {x \left (\frac {9 a^2 D-5 a b C+b^2 B}{b^3}+\frac {3 A}{a}\right )}{2 \left (a+b x^2\right )}}{4 a}+\frac {x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (a \left (-15 a^2 D+3 a b C+b^2 B\right )+3 A b^3\right )}{\sqrt {a} \sqrt {b}}+8 a^2 D x}{2 a b^3}+\frac {x \left (\frac {9 a^2 D-5 a b C+b^2 B}{b^3}+\frac {3 A}{a}\right )}{2 \left (a+b x^2\right )}}{4 a}+\frac {x \left (A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}\)

Input:

Int[(A + B*x^2 + C*x^4 + D*x^6)/(a + b*x^2)^3,x]
 

Output:

((A - (a*(b^2*B - a*b*C + a^2*D))/b^3)*x)/(4*a*(a + b*x^2)^2) + ((((3*A)/a 
 + (b^2*B - 5*a*b*C + 9*a^2*D)/b^3)*x)/(2*(a + b*x^2)) + (8*a^2*D*x + ((3* 
A*b^3 + a*(b^2*B + 3*a*b*C - 15*a^2*D))*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(Sqrt 
[a]*Sqrt[b]))/(2*a*b^3))/(4*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 218

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

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 1471

Int[((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, d + e*x^2 
, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], x 
, 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Simp[1/(2*d*(q 
 + 1))   Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), 
x], x], x]] /; FreeQ[{a, b, c, d, e}, 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 2345

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot 
ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 
 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b 
*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   In 
t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] 
/; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
 

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.93

method	result	size
default	\(\frac {D x}{b^{3}}+\frac {\frac {\frac {b \left (3 b^{3} A +a \,b^{2} B -5 a^{2} b C +9 a^{3} D\right ) x^{3}}{8 a^{2}}+\frac {\left (5 b^{3} A -a \,b^{2} B -3 a^{2} b C +7 a^{3} D\right ) x}{8 a}}{\left (b \,x^{2}+a \right )^{2}}+\frac {\left (3 b^{3} A +a \,b^{2} B +3 a^{2} b C -15 a^{3} D\right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 a^{2} \sqrt {a b}}}{b^{3}}\)	\(137\)

Input:

int((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^3,x,method=_RETURNVERBOSE)
 

Output:

D*x/b^3+1/b^3*((1/8*b*(3*A*b^3+B*a*b^2-5*C*a^2*b+9*D*a^3)/a^2*x^3+1/8*(5*A 
*b^3-B*a*b^2-3*C*a^2*b+7*D*a^3)/a*x)/(b*x^2+a)^2+1/8*(3*A*b^3+B*a*b^2+3*C* 
a^2*b-15*D*a^3)/a^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 511, normalized size of antiderivative = 3.48 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^3} \, dx=\left [\frac {16 \, D a^{3} b^{3} x^{5} + 2 \, {\left (25 \, D a^{4} b^{2} - 5 \, C a^{3} b^{3} + B a^{2} b^{4} + 3 \, A a b^{5}\right )} x^{3} + {\left (15 \, D a^{5} - 3 \, C a^{4} b - B a^{3} b^{2} - 3 \, A a^{2} b^{3} + {\left (15 \, D a^{3} b^{2} - 3 \, C a^{2} b^{3} - B a b^{4} - 3 \, A b^{5}\right )} x^{4} + 2 \, {\left (15 \, D a^{4} b - 3 \, C a^{3} b^{2} - B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) + 2 \, {\left (15 \, D a^{5} b - 3 \, C a^{4} b^{2} - B a^{3} b^{3} + 5 \, A a^{2} b^{4}\right )} x}{16 \, {\left (a^{3} b^{6} x^{4} + 2 \, a^{4} b^{5} x^{2} + a^{5} b^{4}\right )}}, \frac {8 \, D a^{3} b^{3} x^{5} + {\left (25 \, D a^{4} b^{2} - 5 \, C a^{3} b^{3} + B a^{2} b^{4} + 3 \, A a b^{5}\right )} x^{3} - {\left (15 \, D a^{5} - 3 \, C a^{4} b - B a^{3} b^{2} - 3 \, A a^{2} b^{3} + {\left (15 \, D a^{3} b^{2} - 3 \, C a^{2} b^{3} - B a b^{4} - 3 \, A b^{5}\right )} x^{4} + 2 \, {\left (15 \, D a^{4} b - 3 \, C a^{3} b^{2} - B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + {\left (15 \, D a^{5} b - 3 \, C a^{4} b^{2} - B a^{3} b^{3} + 5 \, A a^{2} b^{4}\right )} x}{8 \, {\left (a^{3} b^{6} x^{4} + 2 \, a^{4} b^{5} x^{2} + a^{5} b^{4}\right )}}\right ] \] Input:

integrate((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^3,x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

[1/16*(16*D*a^3*b^3*x^5 + 2*(25*D*a^4*b^2 - 5*C*a^3*b^3 + B*a^2*b^4 + 3*A* 
a*b^5)*x^3 + (15*D*a^5 - 3*C*a^4*b - B*a^3*b^2 - 3*A*a^2*b^3 + (15*D*a^3*b 
^2 - 3*C*a^2*b^3 - B*a*b^4 - 3*A*b^5)*x^4 + 2*(15*D*a^4*b - 3*C*a^3*b^2 - 
B*a^2*b^3 - 3*A*a*b^4)*x^2)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b 
*x^2 + a)) + 2*(15*D*a^5*b - 3*C*a^4*b^2 - B*a^3*b^3 + 5*A*a^2*b^4)*x)/(a^ 
3*b^6*x^4 + 2*a^4*b^5*x^2 + a^5*b^4), 1/8*(8*D*a^3*b^3*x^5 + (25*D*a^4*b^2 
 - 5*C*a^3*b^3 + B*a^2*b^4 + 3*A*a*b^5)*x^3 - (15*D*a^5 - 3*C*a^4*b - B*a^ 
3*b^2 - 3*A*a^2*b^3 + (15*D*a^3*b^2 - 3*C*a^2*b^3 - B*a*b^4 - 3*A*b^5)*x^4 
 + 2*(15*D*a^4*b - 3*C*a^3*b^2 - B*a^2*b^3 - 3*A*a*b^4)*x^2)*sqrt(a*b)*arc 
tan(sqrt(a*b)*x/a) + (15*D*a^5*b - 3*C*a^4*b^2 - B*a^3*b^3 + 5*A*a^2*b^4)* 
x)/(a^3*b^6*x^4 + 2*a^4*b^5*x^2 + a^5*b^4)]
 

Sympy [A] (verification not implemented)

Time = 3.16 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.65 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^3} \, dx=\frac {D x}{b^{3}} + \frac {\sqrt {- \frac {1}{a^{5} b^{7}}} \left (- 3 A b^{3} - B a b^{2} - 3 C a^{2} b + 15 D a^{3}\right ) \log {\left (- a^{3} b^{3} \sqrt {- \frac {1}{a^{5} b^{7}}} + x \right )}}{16} - \frac {\sqrt {- \frac {1}{a^{5} b^{7}}} \left (- 3 A b^{3} - B a b^{2} - 3 C a^{2} b + 15 D a^{3}\right ) \log {\left (a^{3} b^{3} \sqrt {- \frac {1}{a^{5} b^{7}}} + x \right )}}{16} + \frac {x^{3} \cdot \left (3 A b^{4} + B a b^{3} - 5 C a^{2} b^{2} + 9 D a^{3} b\right ) + x \left (5 A a b^{3} - B a^{2} b^{2} - 3 C a^{3} b + 7 D a^{4}\right )}{8 a^{4} b^{3} + 16 a^{3} b^{4} x^{2} + 8 a^{2} b^{5} x^{4}} \] Input:

integrate((D*x**6+C*x**4+B*x**2+A)/(b*x**2+a)**3,x)
 

Output:

D*x/b**3 + sqrt(-1/(a**5*b**7))*(-3*A*b**3 - B*a*b**2 - 3*C*a**2*b + 15*D* 
a**3)*log(-a**3*b**3*sqrt(-1/(a**5*b**7)) + x)/16 - sqrt(-1/(a**5*b**7))*( 
-3*A*b**3 - B*a*b**2 - 3*C*a**2*b + 15*D*a**3)*log(a**3*b**3*sqrt(-1/(a**5 
*b**7)) + x)/16 + (x**3*(3*A*b**4 + B*a*b**3 - 5*C*a**2*b**2 + 9*D*a**3*b) 
 + x*(5*A*a*b**3 - B*a**2*b**2 - 3*C*a**3*b + 7*D*a**4))/(8*a**4*b**3 + 16 
*a**3*b**4*x**2 + 8*a**2*b**5*x**4)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.05 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^3} \, dx=\frac {{\left (9 \, D a^{3} b - 5 \, C a^{2} b^{2} + B a b^{3} + 3 \, A b^{4}\right )} x^{3} + {\left (7 \, D a^{4} - 3 \, C a^{3} b - B a^{2} b^{2} + 5 \, A a b^{3}\right )} x}{8 \, {\left (a^{2} b^{5} x^{4} + 2 \, a^{3} b^{4} x^{2} + a^{4} b^{3}\right )}} + \frac {D x}{b^{3}} - \frac {{\left (15 \, D a^{3} - 3 \, C a^{2} b - B a b^{2} - 3 \, A b^{3}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2} b^{3}} \] Input:

integrate((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^3,x, algorithm="maxima")
 

Output:

1/8*((9*D*a^3*b - 5*C*a^2*b^2 + B*a*b^3 + 3*A*b^4)*x^3 + (7*D*a^4 - 3*C*a^ 
3*b - B*a^2*b^2 + 5*A*a*b^3)*x)/(a^2*b^5*x^4 + 2*a^3*b^4*x^2 + a^4*b^3) + 
D*x/b^3 - 1/8*(15*D*a^3 - 3*C*a^2*b - B*a*b^2 - 3*A*b^3)*arctan(b*x/sqrt(a 
*b))/(sqrt(a*b)*a^2*b^3)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^3} \, dx=\frac {D x}{b^{3}} - \frac {{\left (15 \, D a^{3} - 3 \, C a^{2} b - B a b^{2} - 3 \, A b^{3}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2} b^{3}} + \frac {9 \, D a^{3} b x^{3} - 5 \, C a^{2} b^{2} x^{3} + B a b^{3} x^{3} + 3 \, A b^{4} x^{3} + 7 \, D a^{4} x - 3 \, C a^{3} b x - B a^{2} b^{2} x + 5 \, A a b^{3} x}{8 \, {\left (b x^{2} + a\right )}^{2} a^{2} b^{3}} \] Input:

integrate((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^3,x, algorithm="giac")
 

Output:

D*x/b^3 - 1/8*(15*D*a^3 - 3*C*a^2*b - B*a*b^2 - 3*A*b^3)*arctan(b*x/sqrt(a 
*b))/(sqrt(a*b)*a^2*b^3) + 1/8*(9*D*a^3*b*x^3 - 5*C*a^2*b^2*x^3 + B*a*b^3* 
x^3 + 3*A*b^4*x^3 + 7*D*a^4*x - 3*C*a^3*b*x - B*a^2*b^2*x + 5*A*a*b^3*x)/( 
(b*x^2 + a)^2*a^2*b^3)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^3} \, dx=\int \frac {A+B\,x^2+C\,x^4+x^6\,D}{{\left (b\,x^2+a\right )}^3} \,d x \] Input:

int((A + B*x^2 + C*x^4 + x^6*D)/(a + b*x^2)^3,x)
 

Output:

int((A + B*x^2 + C*x^4 + x^6*D)/(a + b*x^2)^3, x)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 327, normalized size of antiderivative = 2.22 \[ \int \frac {A+B x^2+C x^4+D x^6}{\left (a+b x^2\right )^3} \, dx=\frac {-15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{4} d +3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{3} b c -30 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{3} b d \,x^{2}+4 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{3}+6 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{2} c \,x^{2}-15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{2} d \,x^{4}+8 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{4} x^{2}+3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{3} c \,x^{4}+4 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{5} x^{4}+15 a^{4} b d x -3 a^{3} b^{2} c x +25 a^{3} b^{2} d \,x^{3}+4 a^{2} b^{4} x -5 a^{2} b^{3} c \,x^{3}+8 a^{2} b^{3} d \,x^{5}+4 a \,b^{5} x^{3}}{8 a^{2} b^{4} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:

int((D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^3,x)
 

Output:

( - 15*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**4*d + 3*sqrt(b)*sq 
rt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3*b*c - 30*sqrt(b)*sqrt(a)*atan((b* 
x)/(sqrt(b)*sqrt(a)))*a**3*b*d*x**2 + 4*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b 
)*sqrt(a)))*a**2*b**3 + 6*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a* 
*2*b**2*c*x**2 - 15*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*b** 
2*d*x**4 + 8*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b**4*x**2 + 3 
*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b**3*c*x**4 + 4*sqrt(b)*s 
qrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b**5*x**4 + 15*a**4*b*d*x - 3*a**3*b* 
*2*c*x + 25*a**3*b**2*d*x**3 + 4*a**2*b**4*x - 5*a**2*b**3*c*x**3 + 8*a**2 
*b**3*d*x**5 + 4*a*b**5*x**3)/(8*a**2*b**4*(a**2 + 2*a*b*x**2 + b**2*x**4) 
)