3.2.37 \(\int \frac {c+d x^2+e x^4+f x^6}{(a+b x^2)^3} \, dx\) [137]

3.2.37.1 Optimal result

Integrand size = 27, antiderivative size = 147 \[ \int \frac {c+d x^2+e x^4+f x^6}{\left (a+b x^2\right )^3} \, dx=\frac {f x}{b^3}+\frac {\left (c-\frac {a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x}{4 a \left (a+b x^2\right )^2}+\frac {\left (3 b^3 c+a b^2 d-5 a^2 b e+9 a^3 f\right ) x}{8 a^2 b^3 \left (a+b x^2\right )}+\frac {\left (3 b^3 c+a b^2 d+3 a^2 b e-15 a^3 f\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{5/2} b^{7/2}} \]

output

f*x/b^3+1/4*(c-a*(a^2*f-a*b*e+b^2*d)/b^3)*x/a/(b*x^2+a)^2+1/8*(9*a^3*f-5*a 
^2*b*e+a*b^2*d+3*b^3*c)*x/a^2/b^3/(b*x^2+a)+1/8*(-15*a^3*f+3*a^2*b*e+a*b^2 
*d+3*b^3*c)*arctan(x*b^(1/2)/a^(1/2))/a^(5/2)/b^(7/2)
 

3.2.37.2 Mathematica [A] (verified)

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

input

Integrate[(c + d*x^2 + e*x^4 + f*x^6)/(a + b*x^2)^3,x]
 

output

(x*(15*a^4*f + 3*b^4*c*x^2 + a*b^3*(5*c + d*x^2) + a^3*b*(-3*e + 25*f*x^2) 
 - a^2*b^2*(d + 5*e*x^2 - 8*f*x^4)))/(8*a^2*b^3*(a + b*x^2)^2) + ((3*b^3*c 
 + a*b^2*d + 3*a^2*b*e - 15*a^3*f)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(5/2) 
*b^(7/2))
 

3.2.37.3 Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.12, 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.

\(\displaystyle \int \frac {c+d x^2+e x^4+f x^6}{\left (a+b x^2\right )^3} \, dx\)

\(\Big \downarrow \) 2345

\(\displaystyle \frac {x \left (c-\frac {a \left (a^2 f-a b e+b^2 d\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}-\frac {\int -\frac {\frac {4 a f x^4}{b}+\frac {4 a (b e-a f) x^2}{b^2}+\frac {f a^3-b e a^2+b^2 d a+3 b^3 c}{b^3}}{\left (b x^2+a\right )^2}dx}{4 a}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {\frac {4 a f x^4}{b}+\frac {4 a (b e-a f) x^2}{b^2}+\frac {f a^3-b e a^2+b^2 d a+3 b^3 c}{b^3}}{\left (b x^2+a\right )^2}dx}{4 a}+\frac {x \left (c-\frac {a \left (a^2 f-a b e+b^2 d\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}\)

\(\Big \downarrow \) 1471

\(\displaystyle \frac {\frac {x \left (9 a^3 f-5 a^2 b e+a b^2 d+3 b^3 c\right )}{2 a b^3 \left (a+b x^2\right )}-\frac {\int -\frac {-7 f a^3+8 b f x^2 a^2+3 b e a^2+b^2 d a+3 b^3 c}{b^3 \left (b x^2+a\right )}dx}{2 a}}{4 a}+\frac {x \left (c-\frac {a \left (a^2 f-a b e+b^2 d\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {-7 f a^3+8 b f x^2 a^2+3 b e a^2+b^2 d a+3 b^3 c}{b^3 \left (b x^2+a\right )}dx}{2 a}+\frac {x \left (9 a^3 f-5 a^2 b e+a b^2 d+3 b^3 c\right )}{2 a b^3 \left (a+b x^2\right )}}{4 a}+\frac {x \left (c-\frac {a \left (a^2 f-a b e+b^2 d\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {-7 f a^3+8 b f x^2 a^2+3 b e a^2+b^2 d a+3 b^3 c}{b x^2+a}dx}{2 a b^3}+\frac {x \left (9 a^3 f-5 a^2 b e+a b^2 d+3 b^3 c\right )}{2 a b^3 \left (a+b x^2\right )}}{4 a}+\frac {x \left (c-\frac {a \left (a^2 f-a b e+b^2 d\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {\left (-15 a^3 f+3 a^2 b e+a b^2 d+3 b^3 c\right ) \int \frac {1}{b x^2+a}dx+8 a^2 f x}{2 a b^3}+\frac {x \left (9 a^3 f-5 a^2 b e+a b^2 d+3 b^3 c\right )}{2 a b^3 \left (a+b x^2\right )}}{4 a}+\frac {x \left (c-\frac {a \left (a^2 f-a b e+b^2 d\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {x \left (c-\frac {a \left (a^2 f-a b e+b^2 d\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}+\frac {\frac {8 a^2 f x+\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (-15 a^3 f+3 a^2 b e+a b^2 d+3 b^3 c\right )}{\sqrt {a} \sqrt {b}}}{2 a b^3}+\frac {x \left (9 a^3 f-5 a^2 b e+a b^2 d+3 b^3 c\right )}{2 a b^3 \left (a+b x^2\right )}}{4 a}\)

input

Int[(c + d*x^2 + e*x^4 + f*x^6)/(a + b*x^2)^3,x]
 

output

((c - (a*(b^2*d - a*b*e + a^2*f))/b^3)*x)/(4*a*(a + b*x^2)^2) + (((3*b^3*c 
 + a*b^2*d - 5*a^2*b*e + 9*a^3*f)*x)/(2*a*b^3*(a + b*x^2)) + (8*a^2*f*x + 
((3*b^3*c + a*b^2*d + 3*a^2*b*e - 15*a^3*f)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/( 
Sqrt[a]*Sqrt[b]))/(2*a*b^3))/(4*a)
 

3.2.37.3.1 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]
 

3.2.37.4 Maple [A] (verified)

Time = 3.50 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.95

method	result	size
default	\(\frac {f x}{b^{3}}-\frac {\frac {-\frac {b \left (9 f \,a^{3}-5 a^{2} b e +a \,b^{2} d +3 b^{3} c \right ) x^{3}}{8 a^{2}}-\frac {\left (7 f \,a^{3}-3 a^{2} b e -a \,b^{2} d +5 b^{3} c \right ) x}{8 a}}{\left (b \,x^{2}+a \right )^{2}}+\frac {\left (15 f \,a^{3}-3 a^{2} b e -a \,b^{2} d -3 b^{3} c \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 a^{2} \sqrt {a b}}}{b^{3}}\)	\(139\)
risch	\(\frac {f x}{b^{3}}+\frac {\frac {b \left (9 f \,a^{3}-5 a^{2} b e +a \,b^{2} d +3 b^{3} c \right ) x^{3}}{8 a^{2}}+\frac {\left (7 f \,a^{3}-3 a^{2} b e -a \,b^{2} d +5 b^{3} c \right ) x}{8 a}}{b^{3} \left (b \,x^{2}+a \right )^{2}}-\frac {15 a \ln \left (b x -\sqrt {-a b}\right ) f}{16 b^{3} \sqrt {-a b}}+\frac {3 \ln \left (b x -\sqrt {-a b}\right ) e}{16 b^{2} \sqrt {-a b}}+\frac {\ln \left (b x -\sqrt {-a b}\right ) d}{16 b \sqrt {-a b}\, a}+\frac {3 \ln \left (b x -\sqrt {-a b}\right ) c}{16 \sqrt {-a b}\, a^{2}}+\frac {15 a \ln \left (-b x -\sqrt {-a b}\right ) f}{16 b^{3} \sqrt {-a b}}-\frac {3 \ln \left (-b x -\sqrt {-a b}\right ) e}{16 b^{2} \sqrt {-a b}}-\frac {\ln \left (-b x -\sqrt {-a b}\right ) d}{16 b \sqrt {-a b}\, a}-\frac {3 \ln \left (-b x -\sqrt {-a b}\right ) c}{16 \sqrt {-a b}\, a^{2}}\)	\(302\)

input

int((f*x^6+e*x^4+d*x^2+c)/(b*x^2+a)^3,x,method=_RETURNVERBOSE)
 

output

f*x/b^3-1/b^3*((-1/8*b*(9*a^3*f-5*a^2*b*e+a*b^2*d+3*b^3*c)/a^2*x^3-1/8*(7* 
a^3*f-3*a^2*b*e-a*b^2*d+5*b^3*c)/a*x)/(b*x^2+a)^2+1/8*(15*a^3*f-3*a^2*b*e- 
a*b^2*d-3*b^3*c)/a^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))
 

3.2.37.5 Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 504, normalized size of antiderivative = 3.43 \[ \int \frac {c+d x^2+e x^4+f x^6}{\left (a+b x^2\right )^3} \, dx=\left [\frac {16 \, a^{3} b^{3} f x^{5} + 2 \, {\left (3 \, a b^{5} c + a^{2} b^{4} d - 5 \, a^{3} b^{3} e + 25 \, a^{4} b^{2} f\right )} x^{3} + {\left (3 \, a^{2} b^{3} c + a^{3} b^{2} d + 3 \, a^{4} b e - 15 \, a^{5} f + {\left (3 \, b^{5} c + a b^{4} d + 3 \, a^{2} b^{3} e - 15 \, a^{3} b^{2} f\right )} x^{4} + 2 \, {\left (3 \, a b^{4} c + a^{2} b^{3} d + 3 \, a^{3} b^{2} e - 15 \, a^{4} b f\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} + 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) + 2 \, {\left (5 \, a^{2} b^{4} c - a^{3} b^{3} d - 3 \, a^{4} b^{2} e + 15 \, a^{5} b f\right )} x}{16 \, {\left (a^{3} b^{6} x^{4} + 2 \, a^{4} b^{5} x^{2} + a^{5} b^{4}\right )}}, \frac {8 \, a^{3} b^{3} f x^{5} + {\left (3 \, a b^{5} c + a^{2} b^{4} d - 5 \, a^{3} b^{3} e + 25 \, a^{4} b^{2} f\right )} x^{3} + {\left (3 \, a^{2} b^{3} c + a^{3} b^{2} d + 3 \, a^{4} b e - 15 \, a^{5} f + {\left (3 \, b^{5} c + a b^{4} d + 3 \, a^{2} b^{3} e - 15 \, a^{3} b^{2} f\right )} x^{4} + 2 \, {\left (3 \, a b^{4} c + a^{2} b^{3} d + 3 \, a^{3} b^{2} e - 15 \, a^{4} b f\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + {\left (5 \, a^{2} b^{4} c - a^{3} b^{3} d - 3 \, a^{4} b^{2} e + 15 \, a^{5} b f\right )} x}{8 \, {\left (a^{3} b^{6} x^{4} + 2 \, a^{4} b^{5} x^{2} + a^{5} b^{4}\right )}}\right ] \]

input

integrate((f*x^6+e*x^4+d*x^2+c)/(b*x^2+a)^3,x, algorithm="fricas")
 

output

[1/16*(16*a^3*b^3*f*x^5 + 2*(3*a*b^5*c + a^2*b^4*d - 5*a^3*b^3*e + 25*a^4* 
b^2*f)*x^3 + (3*a^2*b^3*c + a^3*b^2*d + 3*a^4*b*e - 15*a^5*f + (3*b^5*c + 
a*b^4*d + 3*a^2*b^3*e - 15*a^3*b^2*f)*x^4 + 2*(3*a*b^4*c + a^2*b^3*d + 3*a 
^3*b^2*e - 15*a^4*b*f)*x^2)*sqrt(-a*b)*log((b*x^2 + 2*sqrt(-a*b)*x - a)/(b 
*x^2 + a)) + 2*(5*a^2*b^4*c - a^3*b^3*d - 3*a^4*b^2*e + 15*a^5*b*f)*x)/(a^ 
3*b^6*x^4 + 2*a^4*b^5*x^2 + a^5*b^4), 1/8*(8*a^3*b^3*f*x^5 + (3*a*b^5*c + 
a^2*b^4*d - 5*a^3*b^3*e + 25*a^4*b^2*f)*x^3 + (3*a^2*b^3*c + a^3*b^2*d + 3 
*a^4*b*e - 15*a^5*f + (3*b^5*c + a*b^4*d + 3*a^2*b^3*e - 15*a^3*b^2*f)*x^4 
 + 2*(3*a*b^4*c + a^2*b^3*d + 3*a^3*b^2*e - 15*a^4*b*f)*x^2)*sqrt(a*b)*arc 
tan(sqrt(a*b)*x/a) + (5*a^2*b^4*c - a^3*b^3*d - 3*a^4*b^2*e + 15*a^5*b*f)* 
x)/(a^3*b^6*x^4 + 2*a^4*b^5*x^2 + a^5*b^4)]
 

3.2.37.6 Sympy [A] (verification not implemented)

Time = 3.40 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.65 \[ \int \frac {c+d x^2+e x^4+f x^6}{\left (a+b x^2\right )^3} \, dx=\frac {\sqrt {- \frac {1}{a^{5} b^{7}}} \cdot \left (15 a^{3} f - 3 a^{2} b e - a b^{2} d - 3 b^{3} c\right ) \log {\left (- a^{3} b^{3} \sqrt {- \frac {1}{a^{5} b^{7}}} + x \right )}}{16} - \frac {\sqrt {- \frac {1}{a^{5} b^{7}}} \cdot \left (15 a^{3} f - 3 a^{2} b e - a b^{2} d - 3 b^{3} c\right ) \log {\left (a^{3} b^{3} \sqrt {- \frac {1}{a^{5} b^{7}}} + x \right )}}{16} + \frac {x^{3} \cdot \left (9 a^{3} b f - 5 a^{2} b^{2} e + a b^{3} d + 3 b^{4} c\right ) + x \left (7 a^{4} f - 3 a^{3} b e - a^{2} b^{2} d + 5 a b^{3} c\right )}{8 a^{4} b^{3} + 16 a^{3} b^{4} x^{2} + 8 a^{2} b^{5} x^{4}} + \frac {f x}{b^{3}} \]

input

integrate((f*x**6+e*x**4+d*x**2+c)/(b*x**2+a)**3,x)
 

output

sqrt(-1/(a**5*b**7))*(15*a**3*f - 3*a**2*b*e - a*b**2*d - 3*b**3*c)*log(-a 
**3*b**3*sqrt(-1/(a**5*b**7)) + x)/16 - sqrt(-1/(a**5*b**7))*(15*a**3*f - 
3*a**2*b*e - a*b**2*d - 3*b**3*c)*log(a**3*b**3*sqrt(-1/(a**5*b**7)) + x)/ 
16 + (x**3*(9*a**3*b*f - 5*a**2*b**2*e + a*b**3*d + 3*b**4*c) + x*(7*a**4* 
f - 3*a**3*b*e - a**2*b**2*d + 5*a*b**3*c))/(8*a**4*b**3 + 16*a**3*b**4*x* 
*2 + 8*a**2*b**5*x**4) + f*x/b**3
 

3.2.37.7 Maxima [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.05 \[ \int \frac {c+d x^2+e x^4+f x^6}{\left (a+b x^2\right )^3} \, dx=\frac {{\left (3 \, b^{4} c + a b^{3} d - 5 \, a^{2} b^{2} e + 9 \, a^{3} b f\right )} x^{3} + {\left (5 \, a b^{3} c - a^{2} b^{2} d - 3 \, a^{3} b e + 7 \, a^{4} f\right )} x}{8 \, {\left (a^{2} b^{5} x^{4} + 2 \, a^{3} b^{4} x^{2} + a^{4} b^{3}\right )}} + \frac {f x}{b^{3}} + \frac {{\left (3 \, b^{3} c + a b^{2} d + 3 \, a^{2} b e - 15 \, a^{3} f\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2} b^{3}} \]

input

integrate((f*x^6+e*x^4+d*x^2+c)/(b*x^2+a)^3,x, algorithm="maxima")
 

output

1/8*((3*b^4*c + a*b^3*d - 5*a^2*b^2*e + 9*a^3*b*f)*x^3 + (5*a*b^3*c - a^2* 
b^2*d - 3*a^3*b*e + 7*a^4*f)*x)/(a^2*b^5*x^4 + 2*a^3*b^4*x^2 + a^4*b^3) + 
f*x/b^3 + 1/8*(3*b^3*c + a*b^2*d + 3*a^2*b*e - 15*a^3*f)*arctan(b*x/sqrt(a 
*b))/(sqrt(a*b)*a^2*b^3)
 

3.2.37.8 Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.99 \[ \int \frac {c+d x^2+e x^4+f x^6}{\left (a+b x^2\right )^3} \, dx=\frac {f x}{b^{3}} + \frac {{\left (3 \, b^{3} c + a b^{2} d + 3 \, a^{2} b e - 15 \, a^{3} f\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2} b^{3}} + \frac {3 \, b^{4} c x^{3} + a b^{3} d x^{3} - 5 \, a^{2} b^{2} e x^{3} + 9 \, a^{3} b f x^{3} + 5 \, a b^{3} c x - a^{2} b^{2} d x - 3 \, a^{3} b e x + 7 \, a^{4} f x}{8 \, {\left (b x^{2} + a\right )}^{2} a^{2} b^{3}} \]

input

integrate((f*x^6+e*x^4+d*x^2+c)/(b*x^2+a)^3,x, algorithm="giac")
 

output

f*x/b^3 + 1/8*(3*b^3*c + a*b^2*d + 3*a^2*b*e - 15*a^3*f)*arctan(b*x/sqrt(a 
*b))/(sqrt(a*b)*a^2*b^3) + 1/8*(3*b^4*c*x^3 + a*b^3*d*x^3 - 5*a^2*b^2*e*x^ 
3 + 9*a^3*b*f*x^3 + 5*a*b^3*c*x - a^2*b^2*d*x - 3*a^3*b*e*x + 7*a^4*f*x)/( 
(b*x^2 + a)^2*a^2*b^3)
 

3.2.37.9 Mupad [B] (verification not implemented)

Time = 5.54 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.01 \[ \int \frac {c+d x^2+e x^4+f x^6}{\left (a+b x^2\right )^3} \, dx=\frac {\frac {x\,\left (7\,f\,a^3-3\,e\,a^2\,b-d\,a\,b^2+5\,c\,b^3\right )}{8\,a}+\frac {x^3\,\left (9\,f\,a^3\,b-5\,e\,a^2\,b^2+d\,a\,b^3+3\,c\,b^4\right )}{8\,a^2}}{a^2\,b^3+2\,a\,b^4\,x^2+b^5\,x^4}+\frac {f\,x}{b^3}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (-15\,f\,a^3+3\,e\,a^2\,b+d\,a\,b^2+3\,c\,b^3\right )}{8\,a^{5/2}\,b^{7/2}} \]

input

int((c + d*x^2 + e*x^4 + f*x^6)/(a + b*x^2)^3,x)
 

output

((x*(5*b^3*c + 7*a^3*f - a*b^2*d - 3*a^2*b*e))/(8*a) + (x^3*(3*b^4*c - 5*a 
^2*b^2*e + a*b^3*d + 9*a^3*b*f))/(8*a^2))/(a^2*b^3 + b^5*x^4 + 2*a*b^4*x^2 
) + (f*x)/b^3 + (atan((b^(1/2)*x)/a^(1/2))*(3*b^3*c - 15*a^3*f + a*b^2*d + 
 3*a^2*b*e))/(8*a^(5/2)*b^(7/2))