\(\int \frac {A+B x^2+C x^4}{(a+b x^2)^3} \, dx\) [114]

Optimal result

Integrand size = 22, antiderivative size = 115 \[ \int \frac {A+B x^2+C x^4}{\left (a+b x^2\right )^3} \, dx=\frac {\left (\frac {A}{a}-\frac {b B-a C}{b^2}\right ) x}{4 \left (a+b x^2\right )^2}+\frac {\left (3 A b^2+a (b B-5 a C)\right ) x}{8 a^2 b^2 \left (a+b x^2\right )}+\frac {\left (3 A b^2+a (b B+3 a C)\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{5/2} b^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.95 \[ \int \frac {A+B x^2+C x^4}{\left (a+b x^2\right )^3} \, dx=\frac {x \left (-3 a^3 C+3 A b^3 x^2+a b^2 \left (5 A+B x^2\right )-a^2 b \left (B+5 C x^2\right )\right )}{8 a^2 b^2 \left (a+b x^2\right )^2}+\frac {\left (3 A b^2+a (b B+3 a C)\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{5/2} b^{5/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1471, 25, 27, 298, 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.

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

\(\Big \downarrow \) 1471

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 298

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

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

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 298

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

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.93

method	result	size
default	\(\frac {\frac {\left (3 b^{2} A +a b B -5 a^{2} C \right ) x^{3}}{8 a^{2} b}+\frac {\left (5 b^{2} A -a b B -3 a^{2} C \right ) x}{8 a \,b^{2}}}{\left (b \,x^{2}+a \right )^{2}}+\frac {\left (3 b^{2} A +a b B +3 a^{2} C \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 a^{2} b^{2} \sqrt {a b}}\)	\(107\)
risch	\(\frac {\frac {\left (3 b^{2} A +a b B -5 a^{2} C \right ) x^{3}}{8 a^{2} b}+\frac {\left (5 b^{2} A -a b B -3 a^{2} C \right ) x}{8 a \,b^{2}}}{\left (b \,x^{2}+a \right )^{2}}-\frac {3 A \ln \left (b x +\sqrt {-a b}\right )}{16 \sqrt {-a b}\, a^{2}}-\frac {\ln \left (b x +\sqrt {-a b}\right ) B}{16 \sqrt {-a b}\, b a}-\frac {3 \ln \left (b x +\sqrt {-a b}\right ) C}{16 \sqrt {-a b}\, b^{2}}+\frac {3 A \ln \left (-b x +\sqrt {-a b}\right )}{16 \sqrt {-a b}\, a^{2}}+\frac {\ln \left (-b x +\sqrt {-a b}\right ) B}{16 \sqrt {-a b}\, b a}+\frac {3 \ln \left (-b x +\sqrt {-a b}\right ) C}{16 \sqrt {-a b}\, b^{2}}\)	\(215\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 391, normalized size of antiderivative = 3.40 \[ \int \frac {A+B x^2+C x^4}{\left (a+b x^2\right )^3} \, dx=\left [-\frac {2 \, {\left (5 \, C a^{3} b^{2} - B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{3} + {\left (3 \, C a^{4} + B a^{3} b + 3 \, A a^{2} b^{2} + {\left (3 \, C a^{2} b^{2} + B a b^{3} + 3 \, A b^{4}\right )} x^{4} + 2 \, {\left (3 \, C a^{3} b + B a^{2} b^{2} + 3 \, A a b^{3}\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) + 2 \, {\left (3 \, C a^{4} b + B a^{3} b^{2} - 5 \, A a^{2} b^{3}\right )} x}{16 \, {\left (a^{3} b^{5} x^{4} + 2 \, a^{4} b^{4} x^{2} + a^{5} b^{3}\right )}}, -\frac {{\left (5 \, C a^{3} b^{2} - B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{3} - {\left (3 \, C a^{4} + B a^{3} b + 3 \, A a^{2} b^{2} + {\left (3 \, C a^{2} b^{2} + B a b^{3} + 3 \, A b^{4}\right )} x^{4} + 2 \, {\left (3 \, C a^{3} b + B a^{2} b^{2} + 3 \, A a b^{3}\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + {\left (3 \, C a^{4} b + B a^{3} b^{2} - 5 \, A a^{2} b^{3}\right )} x}{8 \, {\left (a^{3} b^{5} x^{4} + 2 \, a^{4} b^{4} x^{2} + a^{5} b^{3}\right )}}\right ] \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 0.72 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.70 \[ \int \frac {A+B x^2+C x^4}{\left (a+b x^2\right )^3} \, dx=- \frac {\sqrt {- \frac {1}{a^{5} b^{5}}} \cdot \left (3 A b^{2} + B a b + 3 C a^{2}\right ) \log {\left (- a^{3} b^{2} \sqrt {- \frac {1}{a^{5} b^{5}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{a^{5} b^{5}}} \cdot \left (3 A b^{2} + B a b + 3 C a^{2}\right ) \log {\left (a^{3} b^{2} \sqrt {- \frac {1}{a^{5} b^{5}}} + x \right )}}{16} + \frac {x^{3} \cdot \left (3 A b^{3} + B a b^{2} - 5 C a^{2} b\right ) + x \left (5 A a b^{2} - B a^{2} b - 3 C a^{3}\right )}{8 a^{4} b^{2} + 16 a^{3} b^{3} x^{2} + 8 a^{2} b^{4} x^{4}} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.05 \[ \int \frac {A+B x^2+C x^4}{\left (a+b x^2\right )^3} \, dx=-\frac {{\left (5 \, C a^{2} b - B a b^{2} - 3 \, A b^{3}\right )} x^{3} + {\left (3 \, C a^{3} + B a^{2} b - 5 \, A a b^{2}\right )} x}{8 \, {\left (a^{2} b^{4} x^{4} + 2 \, a^{3} b^{3} x^{2} + a^{4} b^{2}\right )}} + \frac {{\left (3 \, C a^{2} + B a b + 3 \, A b^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2} b^{2}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.95 \[ \int \frac {A+B x^2+C x^4}{\left (a+b x^2\right )^3} \, dx=\frac {{\left (3 \, C a^{2} + B a b + 3 \, A b^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2} b^{2}} - \frac {5 \, C a^{2} b x^{3} - B a b^{2} x^{3} - 3 \, A b^{3} x^{3} + 3 \, C a^{3} x + B a^{2} b x - 5 \, A a b^{2} x}{8 \, {\left (b x^{2} + a\right )}^{2} a^{2} b^{2}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.97 \[ \int \frac {A+B x^2+C x^4}{\left (a+b x^2\right )^3} \, dx=\frac {\frac {x^3\,\left (-5\,C\,a^2+B\,a\,b+3\,A\,b^2\right )}{8\,a^2\,b}-\frac {x\,\left (3\,C\,a^2+B\,a\,b-5\,A\,b^2\right )}{8\,a\,b^2}}{a^2+2\,a\,b\,x^2+b^2\,x^4}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (3\,C\,a^2+B\,a\,b+3\,A\,b^2\right )}{8\,a^{5/2}\,b^{5/2}} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.86 \[ \int \frac {A+B x^2+C x^4}{\left (a+b x^2\right )^3} \, dx=\frac {3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{3} c +4 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{2}+6 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b c \,x^{2}+8 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{3} x^{2}+3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{2} c \,x^{4}+4 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{4} x^{4}-3 a^{3} b c x +4 a^{2} b^{3} x -5 a^{2} b^{2} c \,x^{3}+4 a \,b^{4} x^{3}}{8 a^{2} b^{3} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:

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

Output:

(3*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3*c + 4*sqrt(b)*sqrt(a 
)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*b**2 + 6*sqrt(b)*sqrt(a)*atan((b*x)/( 
sqrt(b)*sqrt(a)))*a**2*b*c*x**2 + 8*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sq 
rt(a)))*a*b**3*x**2 + 3*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b* 
*2*c*x**4 + 4*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b**4*x**4 - 3* 
a**3*b*c*x + 4*a**2*b**3*x - 5*a**2*b**2*c*x**3 + 4*a*b**4*x**3)/(8*a**2*b 
**3*(a**2 + 2*a*b*x**2 + b**2*x**4))