\(\int \frac {A+B x+C x^2+D x^3}{x (a+b x^2)^3} \, dx\) [107]

Optimal result

Integrand size = 28, antiderivative size = 130 \[ \int \frac {A+B x+C x^2+D x^3}{x \left (a+b x^2\right )^3} \, dx=\frac {A b-a C+(b B-a D) x}{4 a b \left (a+b x^2\right )^2}+\frac {4 A b+(3 b B+a D) x}{8 a^2 b \left (a+b x^2\right )}+\frac {(3 b B+a D) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{5/2} b^{3/2}}+\frac {A \log (x)}{a^3}-\frac {A \log \left (a+b x^2\right )}{2 a^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {1819, 837, 815, 649, 211, 266} \[ \int \frac {A+B x+C x^2+D x^3}{x \left (a+b x^2\right )^3} \, dx=\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) (a D+3 b B)}{8 a^{5/2} b^{3/2}}-\frac {A \log \left (a+b x^2\right )}{2 a^3}+\frac {A \log (x)}{a^3}+\frac {x (a D+3 b B)+4 A b}{8 a^2 b \left (a+b x^2\right )}+\frac {x (b B-a D)-a C+A b}{4 a b \left (a+b x^2\right )^2} \]

[In]

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

[Out]

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

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 815

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m*((f + g*x)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 837

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^(
m + 1))*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)*(c*d^2 + a*e^2))), x] +
Dist[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^
2*(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g},
x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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 {A b-a C+(b B-a D) x}{4 a b \left (a+b x^2\right )^2}-\frac {\int \frac {-4 A-\frac {(3 b B+a D) x}{b}}{x \left (a+b x^2\right )^2} \, dx}{4 a} \\ & = \frac {A b-a C+(b B-a D) x}{4 a b \left (a+b x^2\right )^2}+\frac {4 A b+(3 b B+a D) x}{8 a^2 b \left (a+b x^2\right )}+\frac {\int \frac {8 a A b+a (3 b B+a D) x}{x \left (a+b x^2\right )} \, dx}{8 a^3 b} \\ & = \frac {A b-a C+(b B-a D) x}{4 a b \left (a+b x^2\right )^2}+\frac {4 A b+(3 b B+a D) x}{8 a^2 b \left (a+b x^2\right )}+\frac {\int \left (\frac {8 A b}{x}+\frac {3 a b B+a^2 D-8 A b^2 x}{a+b x^2}\right ) \, dx}{8 a^3 b} \\ & = \frac {A b-a C+(b B-a D) x}{4 a b \left (a+b x^2\right )^2}+\frac {4 A b+(3 b B+a D) x}{8 a^2 b \left (a+b x^2\right )}+\frac {A \log (x)}{a^3}+\frac {\int \frac {3 a b B+a^2 D-8 A b^2 x}{a+b x^2} \, dx}{8 a^3 b} \\ & = \frac {A b-a C+(b B-a D) x}{4 a b \left (a+b x^2\right )^2}+\frac {4 A b+(3 b B+a D) x}{8 a^2 b \left (a+b x^2\right )}+\frac {A \log (x)}{a^3}-\frac {(A b) \int \frac {x}{a+b x^2} \, dx}{a^3}+\frac {(3 b B+a D) \int \frac {1}{a+b x^2} \, dx}{8 a^2 b} \\ & = \frac {A b-a C+(b B-a D) x}{4 a b \left (a+b x^2\right )^2}+\frac {4 A b+(3 b B+a D) x}{8 a^2 b \left (a+b x^2\right )}+\frac {(3 b B+a D) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{5/2} b^{3/2}}+\frac {A \log (x)}{a^3}-\frac {A \log \left (a+b x^2\right )}{2 a^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.90 \[ \int \frac {A+B x+C x^2+D x^3}{x \left (a+b x^2\right )^3} \, dx=\frac {\frac {a (4 A b+3 b B x+a D x)}{b \left (a+b x^2\right )}+\frac {2 a^2 (A b+b B x-a (C+D x))}{b \left (a+b x^2\right )^2}+\frac {\sqrt {a} (3 b B+a D) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{3/2}}+8 A \log (x)-4 A \log \left (a+b x^2\right )}{8 a^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.45 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.00

method	result	size
default	\(\frac {A \ln \left (x \right )}{a^{3}}-\frac {\frac {\left (-\frac {3}{8} a b B -\frac {1}{8} D a^{2}\right ) x^{3}-\frac {a A b \,x^{2}}{2}-\frac {a^{2} \left (5 B b -D a \right ) x}{8 b}-\frac {a^{2} \left (3 A b -C a \right )}{4 b}}{\left (b \,x^{2}+a \right )^{2}}+\frac {4 b A \ln \left (b \,x^{2}+a \right )+\frac {\left (-3 a b B -D a^{2}\right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}}}{8 b}}{a^{3}}\)	\(130\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 233 vs. \(2 (114) = 228\).

Time = 0.32 (sec) , antiderivative size = 488, normalized size of antiderivative = 3.75 \[ \int \frac {A+B x+C x^2+D x^3}{x \left (a+b x^2\right )^3} \, dx=\left [\frac {8 \, A a b^{3} x^{2} - 4 \, C a^{3} b + 12 \, A a^{2} b^{2} + 2 \, {\left (D a^{2} b^{2} + 3 \, B a b^{3}\right )} x^{3} - {\left ({\left (D a b^{2} + 3 \, B b^{3}\right )} x^{4} + D a^{3} + 3 \, B a^{2} b + 2 \, {\left (D a^{2} b + 3 \, B a b^{2}\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) - 2 \, {\left (D a^{3} b - 5 \, B a^{2} b^{2}\right )} x - 8 \, {\left (A b^{4} x^{4} + 2 \, A a b^{3} x^{2} + A a^{2} b^{2}\right )} \log \left (b x^{2} + a\right ) + 16 \, {\left (A b^{4} x^{4} + 2 \, A a b^{3} x^{2} + A a^{2} b^{2}\right )} \log \left (x\right )}{16 \, {\left (a^{3} b^{4} x^{4} + 2 \, a^{4} b^{3} x^{2} + a^{5} b^{2}\right )}}, \frac {4 \, A a b^{3} x^{2} - 2 \, C a^{3} b + 6 \, A a^{2} b^{2} + {\left (D a^{2} b^{2} + 3 \, B a b^{3}\right )} x^{3} + {\left ({\left (D a b^{2} + 3 \, B b^{3}\right )} x^{4} + D a^{3} + 3 \, B a^{2} b + 2 \, {\left (D a^{2} b + 3 \, B a b^{2}\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) - {\left (D a^{3} b - 5 \, B a^{2} b^{2}\right )} x - 4 \, {\left (A b^{4} x^{4} + 2 \, A a b^{3} x^{2} + A a^{2} b^{2}\right )} \log \left (b x^{2} + a\right ) + 8 \, {\left (A b^{4} x^{4} + 2 \, A a b^{3} x^{2} + A a^{2} b^{2}\right )} \log \left (x\right )}{8 \, {\left (a^{3} b^{4} x^{4} + 2 \, a^{4} b^{3} x^{2} + a^{5} b^{2}\right )}}\right ] \]

[In]

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

[Out]

[1/16*(8*A*a*b^3*x^2 - 4*C*a^3*b + 12*A*a^2*b^2 + 2*(D*a^2*b^2 + 3*B*a*b^3)*x^3 - ((D*a*b^2 + 3*B*b^3)*x^4 + D
*a^3 + 3*B*a^2*b + 2*(D*a^2*b + 3*B*a*b^2)*x^2)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)) - 2*(
D*a^3*b - 5*B*a^2*b^2)*x - 8*(A*b^4*x^4 + 2*A*a*b^3*x^2 + A*a^2*b^2)*log(b*x^2 + a) + 16*(A*b^4*x^4 + 2*A*a*b^
3*x^2 + A*a^2*b^2)*log(x))/(a^3*b^4*x^4 + 2*a^4*b^3*x^2 + a^5*b^2), 1/8*(4*A*a*b^3*x^2 - 2*C*a^3*b + 6*A*a^2*b
^2 + (D*a^2*b^2 + 3*B*a*b^3)*x^3 + ((D*a*b^2 + 3*B*b^3)*x^4 + D*a^3 + 3*B*a^2*b + 2*(D*a^2*b + 3*B*a*b^2)*x^2)
*sqrt(a*b)*arctan(sqrt(a*b)*x/a) - (D*a^3*b - 5*B*a^2*b^2)*x - 4*(A*b^4*x^4 + 2*A*a*b^3*x^2 + A*a^2*b^2)*log(b
*x^2 + a) + 8*(A*b^4*x^4 + 2*A*a*b^3*x^2 + A*a^2*b^2)*log(x))/(a^3*b^4*x^4 + 2*a^4*b^3*x^2 + a^5*b^2)]

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.02 \[ \int \frac {A+B x+C x^2+D x^3}{x \left (a+b x^2\right )^3} \, dx=\frac {4 \, A b^{2} x^{2} + {\left (D a b + 3 \, B b^{2}\right )} x^{3} - 2 \, C a^{2} + 6 \, A a b - {\left (D a^{2} - 5 \, B a b\right )} x}{8 \, {\left (a^{2} b^{3} x^{4} + 2 \, a^{3} b^{2} x^{2} + a^{4} b\right )}} - \frac {A \log \left (b x^{2} + a\right )}{2 \, a^{3}} + \frac {A \log \left (x\right )}{a^{3}} + \frac {{\left (D a + 3 \, B b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2} b} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.98 \[ \int \frac {A+B x+C x^2+D x^3}{x \left (a+b x^2\right )^3} \, dx=-\frac {A \log \left (b x^{2} + a\right )}{2 \, a^{3}} + \frac {A \log \left ({\left | x \right |}\right )}{a^{3}} + \frac {{\left (D a + 3 \, B b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2} b} + \frac {4 \, A a b^{2} x^{2} - 2 \, C a^{3} + 6 \, A a^{2} b + {\left (D a^{2} b + 3 \, B a b^{2}\right )} x^{3} - {\left (D a^{3} - 5 \, B a^{2} b\right )} x}{8 \, {\left (b x^{2} + a\right )}^{2} a^{3} b} \]

[In]

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

[Out]

-1/2*A*log(b*x^2 + a)/a^3 + A*log(abs(x))/a^3 + 1/8*(D*a + 3*B*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^2*b) + 1/
8*(4*A*a*b^2*x^2 - 2*C*a^3 + 6*A*a^2*b + (D*a^2*b + 3*B*a*b^2)*x^3 - (D*a^3 - 5*B*a^2*b)*x)/((b*x^2 + a)^2*a^3
*b)

Mupad [F(-1)]

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

[In]

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

[Out]

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