\(\int \frac {A+B x+C x^2}{(a+b x^2)^4} \, dx\) [30]

Optimal result

Integrand size = 20, antiderivative size = 126 \[ \int \frac {A+B x+C x^2}{\left (a+b x^2\right )^4} \, dx=-\frac {a B-(A b-a C) x}{6 a b \left (a+b x^2\right )^3}+\frac {(5 A b+a C) x}{24 a^2 b \left (a+b x^2\right )^2}+\frac {(5 A b+a C) x}{16 a^3 b \left (a+b x^2\right )}+\frac {(5 A b+a C) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 a^{7/2} b^{3/2}} \] Output:

-1/6*(B*a-(A*b-C*a)*x)/a/b/(b*x^2+a)^3+1/24*(5*A*b+C*a)*x/a^2/b/(b*x^2+a)^ 
2+1/16*(5*A*b+C*a)*x/a^3/b/(b*x^2+a)+1/16*(5*A*b+C*a)*arctan(b^(1/2)*x/a^( 
1/2))/a^(7/2)/b^(3/2)
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.89 \[ \int \frac {A+B x+C x^2}{\left (a+b x^2\right )^4} \, dx=\frac {15 A b^3 x^5-a^3 (8 B+3 C x)+a b^2 x^3 \left (40 A+3 C x^2\right )+a^2 b x \left (33 A+8 C x^2\right )}{48 a^3 b \left (a+b x^2\right )^3}+\frac {(5 A b+a C) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 a^{7/2} b^{3/2}} \] Input:

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

Output:

(15*A*b^3*x^5 - a^3*(8*B + 3*C*x) + a*b^2*x^3*(40*A + 3*C*x^2) + a^2*b*x*( 
33*A + 8*C*x^2))/(48*a^3*b*(a + b*x^2)^3) + ((5*A*b + a*C)*ArcTan[(Sqrt[b] 
*x)/Sqrt[a]])/(16*a^(7/2)*b^(3/2))
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2345, 25, 27, 215, 215, 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.

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

\(\Big \downarrow \) 2345

\(\displaystyle -\frac {\int -\frac {5 A b+a C}{b \left (b x^2+a\right )^3}dx}{6 a}-\frac {a B-x (A b-a C)}{6 a b \left (a+b x^2\right )^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {5 A b+a C}{b \left (b x^2+a\right )^3}dx}{6 a}-\frac {a B-x (A b-a C)}{6 a b \left (a+b x^2\right )^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(a C+5 A b) \int \frac {1}{\left (b x^2+a\right )^3}dx}{6 a b}-\frac {a B-x (A b-a C)}{6 a b \left (a+b x^2\right )^3}\)

\(\Big \downarrow \) 215

\(\displaystyle \frac {(a C+5 A b) \left (\frac {3 \int \frac {1}{\left (b x^2+a\right )^2}dx}{4 a}+\frac {x}{4 a \left (a+b x^2\right )^2}\right )}{6 a b}-\frac {a B-x (A b-a C)}{6 a b \left (a+b x^2\right )^3}\)

\(\Big \downarrow \) 215

\(\displaystyle \frac {(a C+5 A b) \left (\frac {3 \left (\frac {\int \frac {1}{b x^2+a}dx}{2 a}+\frac {x}{2 a \left (a+b x^2\right )}\right )}{4 a}+\frac {x}{4 a \left (a+b x^2\right )^2}\right )}{6 a b}-\frac {a B-x (A b-a C)}{6 a b \left (a+b x^2\right )^3}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {(a C+5 A b) \left (\frac {3 \left (\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b}}+\frac {x}{2 a \left (a+b x^2\right )}\right )}{4 a}+\frac {x}{4 a \left (a+b x^2\right )^2}\right )}{6 a b}-\frac {a B-x (A b-a C)}{6 a b \left (a+b x^2\right )^3}\)

Input:

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

Output:

-1/6*(a*B - (A*b - a*C)*x)/(a*b*(a + b*x^2)^3) + ((5*A*b + a*C)*(x/(4*a*(a 
 + b*x^2)^2) + (3*(x/(2*a*(a + b*x^2)) + ArcTan[(Sqrt[b]*x)/Sqrt[a]]/(2*a^ 
(3/2)*Sqrt[b])))/(4*a)))/(6*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 215

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) 
/(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1))   Int[(a + b*x^2)^(p + 1 
), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 
*p])
 

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 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 = 0.58 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.79

method	result	size
default	\(\frac {\frac {\left (5 A b +C a \right ) b \,x^{5}}{16 a^{3}}+\frac {\left (5 A b +C a \right ) x^{3}}{6 a^{2}}+\frac {\left (11 A b -C a \right ) x}{16 a b}-\frac {B}{6 b}}{\left (b \,x^{2}+a \right )^{3}}+\frac {\left (5 A b +C a \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 a^{3} b \sqrt {a b}}\)	\(100\)
risch	\(\frac {\frac {\left (5 A b +C a \right ) b \,x^{5}}{16 a^{3}}+\frac {\left (5 A b +C a \right ) x^{3}}{6 a^{2}}+\frac {\left (11 A b -C a \right ) x}{16 a b}-\frac {B}{6 b}}{\left (b \,x^{2}+a \right )^{3}}-\frac {5 \ln \left (b x +\sqrt {-a b}\right ) A}{32 \sqrt {-a b}\, a^{3}}-\frac {\ln \left (b x +\sqrt {-a b}\right ) C}{32 \sqrt {-a b}\, b \,a^{2}}+\frac {5 \ln \left (-b x +\sqrt {-a b}\right ) A}{32 \sqrt {-a b}\, a^{3}}+\frac {\ln \left (-b x +\sqrt {-a b}\right ) C}{32 \sqrt {-a b}\, b \,a^{2}}\)	\(170\)

Input:

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

Output:

(1/16*(5*A*b+C*a)/a^3*b*x^5+1/6/a^2*(5*A*b+C*a)*x^3+1/16*(11*A*b-C*a)/a/b* 
x-1/6*B/b)/(b*x^2+a)^3+1/16*(5*A*b+C*a)/a^3/b/(a*b)^(1/2)*arctan(b*x/(a*b) 
^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 430, normalized size of antiderivative = 3.41 \[ \int \frac {A+B x+C x^2}{\left (a+b x^2\right )^4} \, dx=\left [-\frac {16 \, B a^{4} b - 6 \, {\left (C a^{2} b^{3} + 5 \, A a b^{4}\right )} x^{5} - 16 \, {\left (C a^{3} b^{2} + 5 \, A a^{2} b^{3}\right )} x^{3} + 3 \, {\left ({\left (C a b^{3} + 5 \, A b^{4}\right )} x^{6} + C a^{4} + 5 \, A a^{3} b + 3 \, {\left (C a^{2} b^{2} + 5 \, A a b^{3}\right )} x^{4} + 3 \, {\left (C a^{3} b + 5 \, A a^{2} b^{2}\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) + 6 \, {\left (C a^{4} b - 11 \, A a^{3} b^{2}\right )} x}{96 \, {\left (a^{4} b^{5} x^{6} + 3 \, a^{5} b^{4} x^{4} + 3 \, a^{6} b^{3} x^{2} + a^{7} b^{2}\right )}}, -\frac {8 \, B a^{4} b - 3 \, {\left (C a^{2} b^{3} + 5 \, A a b^{4}\right )} x^{5} - 8 \, {\left (C a^{3} b^{2} + 5 \, A a^{2} b^{3}\right )} x^{3} - 3 \, {\left ({\left (C a b^{3} + 5 \, A b^{4}\right )} x^{6} + C a^{4} + 5 \, A a^{3} b + 3 \, {\left (C a^{2} b^{2} + 5 \, A a b^{3}\right )} x^{4} + 3 \, {\left (C a^{3} b + 5 \, A a^{2} b^{2}\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + 3 \, {\left (C a^{4} b - 11 \, A a^{3} b^{2}\right )} x}{48 \, {\left (a^{4} b^{5} x^{6} + 3 \, a^{5} b^{4} x^{4} + 3 \, a^{6} b^{3} x^{2} + a^{7} b^{2}\right )}}\right ] \] Input:

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

Output:

[-1/96*(16*B*a^4*b - 6*(C*a^2*b^3 + 5*A*a*b^4)*x^5 - 16*(C*a^3*b^2 + 5*A*a 
^2*b^3)*x^3 + 3*((C*a*b^3 + 5*A*b^4)*x^6 + C*a^4 + 5*A*a^3*b + 3*(C*a^2*b^ 
2 + 5*A*a*b^3)*x^4 + 3*(C*a^3*b + 5*A*a^2*b^2)*x^2)*sqrt(-a*b)*log((b*x^2 
- 2*sqrt(-a*b)*x - a)/(b*x^2 + a)) + 6*(C*a^4*b - 11*A*a^3*b^2)*x)/(a^4*b^ 
5*x^6 + 3*a^5*b^4*x^4 + 3*a^6*b^3*x^2 + a^7*b^2), -1/48*(8*B*a^4*b - 3*(C* 
a^2*b^3 + 5*A*a*b^4)*x^5 - 8*(C*a^3*b^2 + 5*A*a^2*b^3)*x^3 - 3*((C*a*b^3 + 
 5*A*b^4)*x^6 + C*a^4 + 5*A*a^3*b + 3*(C*a^2*b^2 + 5*A*a*b^3)*x^4 + 3*(C*a 
^3*b + 5*A*a^2*b^2)*x^2)*sqrt(a*b)*arctan(sqrt(a*b)*x/a) + 3*(C*a^4*b - 11 
*A*a^3*b^2)*x)/(a^4*b^5*x^6 + 3*a^5*b^4*x^4 + 3*a^6*b^3*x^2 + a^7*b^2)]
 

Sympy [A] (verification not implemented)

Time = 0.89 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.56 \[ \int \frac {A+B x+C x^2}{\left (a+b x^2\right )^4} \, dx=- \frac {\sqrt {- \frac {1}{a^{7} b^{3}}} \cdot \left (5 A b + C a\right ) \log {\left (- a^{4} b \sqrt {- \frac {1}{a^{7} b^{3}}} + x \right )}}{32} + \frac {\sqrt {- \frac {1}{a^{7} b^{3}}} \cdot \left (5 A b + C a\right ) \log {\left (a^{4} b \sqrt {- \frac {1}{a^{7} b^{3}}} + x \right )}}{32} + \frac {- 8 B a^{3} + x^{5} \cdot \left (15 A b^{3} + 3 C a b^{2}\right ) + x^{3} \cdot \left (40 A a b^{2} + 8 C a^{2} b\right ) + x \left (33 A a^{2} b - 3 C a^{3}\right )}{48 a^{6} b + 144 a^{5} b^{2} x^{2} + 144 a^{4} b^{3} x^{4} + 48 a^{3} b^{4} x^{6}} \] Input:

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

Output:

-sqrt(-1/(a**7*b**3))*(5*A*b + C*a)*log(-a**4*b*sqrt(-1/(a**7*b**3)) + x)/ 
32 + sqrt(-1/(a**7*b**3))*(5*A*b + C*a)*log(a**4*b*sqrt(-1/(a**7*b**3)) + 
x)/32 + (-8*B*a**3 + x**5*(15*A*b**3 + 3*C*a*b**2) + x**3*(40*A*a*b**2 + 8 
*C*a**2*b) + x*(33*A*a**2*b - 3*C*a**3))/(48*a**6*b + 144*a**5*b**2*x**2 + 
 144*a**4*b**3*x**4 + 48*a**3*b**4*x**6)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.06 \[ \int \frac {A+B x+C x^2}{\left (a+b x^2\right )^4} \, dx=\frac {3 \, {\left (C a b^{2} + 5 \, A b^{3}\right )} x^{5} - 8 \, B a^{3} + 8 \, {\left (C a^{2} b + 5 \, A a b^{2}\right )} x^{3} - 3 \, {\left (C a^{3} - 11 \, A a^{2} b\right )} x}{48 \, {\left (a^{3} b^{4} x^{6} + 3 \, a^{4} b^{3} x^{4} + 3 \, a^{5} b^{2} x^{2} + a^{6} b\right )}} + \frac {{\left (C a + 5 \, A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} a^{3} b} \] Input:

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

Output:

1/48*(3*(C*a*b^2 + 5*A*b^3)*x^5 - 8*B*a^3 + 8*(C*a^2*b + 5*A*a*b^2)*x^3 - 
3*(C*a^3 - 11*A*a^2*b)*x)/(a^3*b^4*x^6 + 3*a^4*b^3*x^4 + 3*a^5*b^2*x^2 + a 
^6*b) + 1/16*(C*a + 5*A*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^3*b)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.87 \[ \int \frac {A+B x+C x^2}{\left (a+b x^2\right )^4} \, dx=\frac {{\left (C a + 5 \, A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} a^{3} b} + \frac {3 \, C a b^{2} x^{5} + 15 \, A b^{3} x^{5} + 8 \, C a^{2} b x^{3} + 40 \, A a b^{2} x^{3} - 3 \, C a^{3} x + 33 \, A a^{2} b x - 8 \, B a^{3}}{48 \, {\left (b x^{2} + a\right )}^{3} a^{3} b} \] Input:

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

Output:

1/16*(C*a + 5*A*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^3*b) + 1/48*(3*C*a*b 
^2*x^5 + 15*A*b^3*x^5 + 8*C*a^2*b*x^3 + 40*A*a*b^2*x^3 - 3*C*a^3*x + 33*A* 
a^2*b*x - 8*B*a^3)/((b*x^2 + a)^3*a^3*b)
 

Mupad [B] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.92 \[ \int \frac {A+B x+C x^2}{\left (a+b x^2\right )^4} \, dx=\frac {\frac {x^3\,\left (5\,A\,b+C\,a\right )}{6\,a^2}-\frac {B}{6\,b}+\frac {b\,x^5\,\left (5\,A\,b+C\,a\right )}{16\,a^3}+\frac {x\,\left (11\,A\,b-C\,a\right )}{16\,a\,b}}{a^3+3\,a^2\,b\,x^2+3\,a\,b^2\,x^4+b^3\,x^6}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (5\,A\,b+C\,a\right )}{16\,a^{7/2}\,b^{3/2}} \] Input:

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

Output:

((x^3*(5*A*b + C*a))/(6*a^2) - B/(6*b) + (b*x^5*(5*A*b + C*a))/(16*a^3) + 
(x*(11*A*b - C*a))/(16*a*b))/(a^3 + b^3*x^6 + 3*a^2*b*x^2 + 3*a*b^2*x^4) + 
 (atan((b^(1/2)*x)/a^(1/2))*(5*A*b + C*a))/(16*a^(7/2)*b^(3/2))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.41 \[ \int \frac {A+B x+C x^2}{\left (a+b x^2\right )^4} \, dx=\frac {15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{3} b +3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{3} c +45 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{2} x^{2}+9 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b c \,x^{2}+45 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{3} x^{4}+9 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{2} c \,x^{4}+15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{4} x^{6}+3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{3} c \,x^{6}+33 a^{3} b^{2} x -8 a^{3} b^{2}-3 a^{3} b c x +40 a^{2} b^{3} x^{3}+8 a^{2} b^{2} c \,x^{3}+15 a \,b^{4} x^{5}+3 a \,b^{3} c \,x^{5}}{48 a^{3} b^{2} \left (b^{3} x^{6}+3 a \,b^{2} x^{4}+3 a^{2} b \,x^{2}+a^{3}\right )} \] Input:

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

Output:

(15*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3*b + 3*sqrt(b)*sqrt( 
a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3*c + 45*sqrt(b)*sqrt(a)*atan((b*x)/(s 
qrt(b)*sqrt(a)))*a**2*b**2*x**2 + 9*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sq 
rt(a)))*a**2*b*c*x**2 + 45*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a 
*b**3*x**4 + 9*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b**2*c*x**4 
 + 15*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b**4*x**6 + 3*sqrt(b)* 
sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b**3*c*x**6 + 33*a**3*b**2*x - 8*a** 
3*b**2 - 3*a**3*b*c*x + 40*a**2*b**3*x**3 + 8*a**2*b**2*c*x**3 + 15*a*b**4 
*x**5 + 3*a*b**3*c*x**5)/(48*a**3*b**2*(a**3 + 3*a**2*b*x**2 + 3*a*b**2*x* 
*4 + b**3*x**6))