\(\int \frac {A+B x+C x^2+D x^3}{x^2 (a+b x^2)^2} \, dx\) [100]

Optimal result

Integrand size = 28, antiderivative size = 110 \[ \int \frac {A+B x+C x^2+D x^3}{x^2 \left (a+b x^2\right )^2} \, dx=-\frac {A}{a^2 x}+\frac {b B-a D-b \left (\frac {A b}{a}-C\right ) x}{2 a b \left (a+b x^2\right )}-\frac {(3 A b-a C) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{5/2} \sqrt {b}}+\frac {B \log (x)}{a^2}-\frac {B \log \left (a+b x^2\right )}{2 a^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {1819, 1816, 649, 211, 266} \[ \int \frac {A+B x+C x^2+D x^3}{x^2 \left (a+b x^2\right )^2} \, dx=-\frac {(3 A b-a C) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{5/2} \sqrt {b}}-\frac {A}{a^2 x}-\frac {B \log \left (a+b x^2\right )}{2 a^2}+\frac {B \log (x)}{a^2}+\frac {-b x \left (\frac {A b}{a}-C\right )-a D+b B}{2 a b \left (a+b x^2\right )} \]

[In]

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

[Out]

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

Rule 211

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

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 649

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[(-a)*c]

Rule 1816

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 1819

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*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] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {b B-a D-b \left (\frac {A b}{a}-C\right ) x}{2 a b \left (a+b x^2\right )}-\frac {\int \frac {-2 A-2 B x+\left (\frac {A b}{a}-C\right ) x^2}{x^2 \left (a+b x^2\right )} \, dx}{2 a} \\ & = \frac {b B-a D-b \left (\frac {A b}{a}-C\right ) x}{2 a b \left (a+b x^2\right )}-\frac {\int \left (-\frac {2 A}{a x^2}-\frac {2 B}{a x}+\frac {3 A b-a C+2 b B x}{a \left (a+b x^2\right )}\right ) \, dx}{2 a} \\ & = -\frac {A}{a^2 x}+\frac {b B-a D-b \left (\frac {A b}{a}-C\right ) x}{2 a b \left (a+b x^2\right )}+\frac {B \log (x)}{a^2}-\frac {\int \frac {3 A b-a C+2 b B x}{a+b x^2} \, dx}{2 a^2} \\ & = -\frac {A}{a^2 x}+\frac {b B-a D-b \left (\frac {A b}{a}-C\right ) x}{2 a b \left (a+b x^2\right )}+\frac {B \log (x)}{a^2}-\frac {(b B) \int \frac {x}{a+b x^2} \, dx}{a^2}-\frac {(3 A b-a C) \int \frac {1}{a+b x^2} \, dx}{2 a^2} \\ & = -\frac {A}{a^2 x}+\frac {b B-a D-b \left (\frac {A b}{a}-C\right ) x}{2 a b \left (a+b x^2\right )}-\frac {(3 A b-a C) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{5/2} \sqrt {b}}+\frac {B \log (x)}{a^2}-\frac {B \log \left (a+b x^2\right )}{2 a^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x+C x^2+D x^3}{x^2 \left (a+b x^2\right )^2} \, dx=-\frac {A}{a^2 x}+\frac {a b B-a^2 D-A b^2 x+a b C x}{2 a^2 b \left (a+b x^2\right )}+\frac {(-3 A b+a C) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{5/2} \sqrt {b}}+\frac {B \log (x)}{a^2}-\frac {B \log \left (a+b x^2\right )}{2 a^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.55 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.87

method	result	size
default	\(-\frac {A}{a^{2} x}+\frac {B \ln \left (x \right )}{a^{2}}-\frac {\frac {\left (\frac {A b}{2}-\frac {C a}{2}\right ) x -\frac {a \left (B b -D a \right )}{2 b}}{b \,x^{2}+a}+\frac {B \ln \left (b \,x^{2}+a \right )}{2}+\frac {\left (3 A b -C a \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}}}{a^{2}}\)	\(96\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 336, normalized size of antiderivative = 3.05 \[ \int \frac {A+B x+C x^2+D x^3}{x^2 \left (a+b x^2\right )^2} \, dx=\left [-\frac {4 \, A a^{2} b - 2 \, {\left (C a^{2} b - 3 \, A a b^{2}\right )} x^{2} - {\left ({\left (C a b - 3 \, A b^{2}\right )} x^{3} + {\left (C a^{2} - 3 \, A a b\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x^{2} + 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) + 2 \, {\left (D a^{3} - B a^{2} b\right )} x + 2 \, {\left (B a b^{2} x^{3} + B a^{2} b x\right )} \log \left (b x^{2} + a\right ) - 4 \, {\left (B a b^{2} x^{3} + B a^{2} b x\right )} \log \left (x\right )}{4 \, {\left (a^{3} b^{2} x^{3} + a^{4} b x\right )}}, -\frac {2 \, A a^{2} b - {\left (C a^{2} b - 3 \, A a b^{2}\right )} x^{2} - {\left ({\left (C a b - 3 \, A b^{2}\right )} x^{3} + {\left (C a^{2} - 3 \, A a b\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + {\left (D a^{3} - B a^{2} b\right )} x + {\left (B a b^{2} x^{3} + B a^{2} b x\right )} \log \left (b x^{2} + a\right ) - 2 \, {\left (B a b^{2} x^{3} + B a^{2} b x\right )} \log \left (x\right )}{2 \, {\left (a^{3} b^{2} x^{3} + a^{4} b x\right )}}\right ] \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.95 \[ \int \frac {A+B x+C x^2+D x^3}{x^2 \left (a+b x^2\right )^2} \, dx=-\frac {2 \, A a b - {\left (C a b - 3 \, A b^{2}\right )} x^{2} + {\left (D a^{2} - B a b\right )} x}{2 \, {\left (a^{2} b^{2} x^{3} + a^{3} b x\right )}} - \frac {B \log \left (b x^{2} + a\right )}{2 \, a^{2}} + \frac {B \log \left (x\right )}{a^{2}} + \frac {{\left (C a - 3 \, A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.94 \[ \int \frac {A+B x+C x^2+D x^3}{x^2 \left (a+b x^2\right )^2} \, dx=-\frac {B \log \left (b x^{2} + a\right )}{2 \, a^{2}} + \frac {B \log \left ({\left | x \right |}\right )}{a^{2}} + \frac {{\left (C a - 3 \, A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{2}} + \frac {C a b x^{2} - 3 \, A b^{2} x^{2} - D a^{2} x + B a b x - 2 \, A a b}{2 \, {\left (b x^{3} + a x\right )} a^{2} b} \]

[In]

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

[Out]

-1/2*B*log(b*x^2 + a)/a^2 + B*log(abs(x))/a^2 + 1/2*(C*a - 3*A*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^2) + 1/2*
(C*a*b*x^2 - 3*A*b^2*x^2 - D*a^2*x + B*a*b*x - 2*A*a*b)/((b*x^3 + a*x)*a^2*b)

Mupad [B] (verification not implemented)

Time = 6.19 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.21 \[ \int \frac {A+B x+C x^2+D x^3}{x^2 \left (a+b x^2\right )^2} \, dx=\frac {B}{2\,a\,\left (b\,x^2+a\right )}-\frac {\frac {A}{a}+\frac {3\,A\,b\,x^2}{2\,a^2}}{b\,x^3+a\,x}-\frac {B\,\ln \left (b\,x^2+a\right )}{2\,a^2}+\frac {B\,\ln \left (x\right )}{a^2}-\frac {D}{2\,b\,\left (b\,x^2+a\right )}+\frac {C\,x}{2\,a\,\left (b\,x^2+a\right )}-\frac {3\,A\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{2\,a^{5/2}}+\frac {C\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{2\,a^{3/2}\,\sqrt {b}} \]

[In]

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

[Out]

B/(2*a*(a + b*x^2)) - (A/a + (3*A*b*x^2)/(2*a^2))/(a*x + b*x^3) - (B*log(a + b*x^2))/(2*a^2) + (B*log(x))/a^2
- D/(2*b*(a + b*x^2)) + (C*x)/(2*a*(a + b*x^2)) - (3*A*b^(1/2)*atan((b^(1/2)*x)/a^(1/2)))/(2*a^(5/2)) + (C*ata
n((b^(1/2)*x)/a^(1/2)))/(2*a^(3/2)*b^(1/2))