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

Optimal. Leaf size=144 \[ -\frac {3 (5 A b-a C) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {b}}-\frac {A}{a^3 x}-\frac {B \log \left (a+b x^2\right )}{2 a^3}+\frac {B \log (x)}{a^3}+\frac {4 B-x \left (\frac {7 A b}{a}-3 C\right )}{8 a^2 \left (a+b x^2\right )}+\frac {-b x \left (\frac {A b}{a}-C\right )-a D+b B}{4 a b \left (a+b x^2\right )^2} \]

[Out]

-A/a^3/x+1/4*(b*B-a*D-b*(A*b/a-C)*x)/a/b/(b*x^2+a)^2+1/8*(4*B-(7*A*b/a-3*C)*x)/a^2/(b*x^2+a)+B*ln(x)/a^3-1/2*B
*ln(b*x^2+a)/a^3-3/8*(5*A*b-C*a)*arctan(x*b^(1/2)/a^(1/2))/a^(7/2)/b^(1/2)

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Rubi [A]  time = 0.23, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {1805, 1802, 635, 205, 260} \[ \frac {4 B-x \left (\frac {7 A b}{a}-3 C\right )}{8 a^2 \left (a+b x^2\right )}-\frac {3 (5 A b-a C) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {b}}-\frac {A}{a^3 x}-\frac {B \log \left (a+b x^2\right )}{2 a^3}+\frac {B \log (x)}{a^3}+\frac {-b x \left (\frac {A b}{a}-C\right )-a D+b B}{4 a b \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(A/(a^3*x)) + (b*B - a*D - b*((A*b)/a - C)*x)/(4*a*b*(a + b*x^2)^2) + (4*B - ((7*A*b)/a - 3*C)*x)/(8*a^2*(a +
b*x^2)) - (3*(5*A*b - a*C)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(7/2)*Sqrt[b]) + (B*Log[x])/a^3 - (B*Log[a + b*x
^2])/(2*a^3)

Rule 205

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

Rule 260

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 635

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 1802

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 1805

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*} \int \frac {A+B x+C x^2+D x^3}{x^2 \left (a+b x^2\right )^3} \, dx &=\frac {b B-a D-b \left (\frac {A b}{a}-C\right ) x}{4 a b \left (a+b x^2\right )^2}-\frac {\int \frac {-4 A-4 B x+3 \left (\frac {A b}{a}-C\right ) x^2}{x^2 \left (a+b x^2\right )^2} \, dx}{4 a}\\ &=\frac {b B-a D-b \left (\frac {A b}{a}-C\right ) x}{4 a b \left (a+b x^2\right )^2}+\frac {4 B-\left (\frac {7 A b}{a}-3 C\right ) x}{8 a^2 \left (a+b x^2\right )}+\frac {\int \frac {8 A+8 B x-\left (\frac {7 A b}{a}-3 C\right ) x^2}{x^2 \left (a+b x^2\right )} \, dx}{8 a^2}\\ &=\frac {b B-a D-b \left (\frac {A b}{a}-C\right ) x}{4 a b \left (a+b x^2\right )^2}+\frac {4 B-\left (\frac {7 A b}{a}-3 C\right ) x}{8 a^2 \left (a+b x^2\right )}+\frac {\int \left (\frac {8 A}{a x^2}+\frac {8 B}{a x}+\frac {-15 A b+3 a C-8 b B x}{a \left (a+b x^2\right )}\right ) \, dx}{8 a^2}\\ &=-\frac {A}{a^3 x}+\frac {b B-a D-b \left (\frac {A b}{a}-C\right ) x}{4 a b \left (a+b x^2\right )^2}+\frac {4 B-\left (\frac {7 A b}{a}-3 C\right ) x}{8 a^2 \left (a+b x^2\right )}+\frac {B \log (x)}{a^3}+\frac {\int \frac {-15 A b+3 a C-8 b B x}{a+b x^2} \, dx}{8 a^3}\\ &=-\frac {A}{a^3 x}+\frac {b B-a D-b \left (\frac {A b}{a}-C\right ) x}{4 a b \left (a+b x^2\right )^2}+\frac {4 B-\left (\frac {7 A b}{a}-3 C\right ) x}{8 a^2 \left (a+b x^2\right )}+\frac {B \log (x)}{a^3}-\frac {(b B) \int \frac {x}{a+b x^2} \, dx}{a^3}-\frac {(3 (5 A b-a C)) \int \frac {1}{a+b x^2} \, dx}{8 a^3}\\ &=-\frac {A}{a^3 x}+\frac {b B-a D-b \left (\frac {A b}{a}-C\right ) x}{4 a b \left (a+b x^2\right )^2}+\frac {4 B-\left (\frac {7 A b}{a}-3 C\right ) x}{8 a^2 \left (a+b x^2\right )}-\frac {3 (5 A b-a C) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {b}}+\frac {B \log (x)}{a^3}-\frac {B \log \left (a+b x^2\right )}{2 a^3}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 141, normalized size = 0.98 \[ \frac {3 (a C-5 A b) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {b}}+\frac {4 a B+3 a C x-7 A b x}{8 a^3 \left (a+b x^2\right )}-\frac {A}{a^3 x}-\frac {B \log \left (a+b x^2\right )}{2 a^3}+\frac {B \log (x)}{a^3}+\frac {a^2 (-D)+a b B+a b C x-A b^2 x}{4 a^2 b \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(A/(a^3*x)) + (a*b*B - a^2*D - A*b^2*x + a*b*C*x)/(4*a^2*b*(a + b*x^2)^2) + (4*a*B - 7*A*b*x + 3*a*C*x)/(8*a^
3*(a + b*x^2)) + (3*(-5*A*b + a*C)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(7/2)*Sqrt[b]) + (B*Log[x])/a^3 - (B*Log[
a + b*x^2])/(2*a^3)

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fricas [B]  time = 0.70, size = 524, normalized size = 3.64 \[ \left [\frac {8 \, B a^{2} b^{2} x^{3} - 16 \, A a^{3} b + 6 \, {\left (C a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{4} + 10 \, {\left (C a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2} + 3 \, {\left ({\left (C a b^{2} - 5 \, A b^{3}\right )} x^{5} + 2 \, {\left (C a^{2} b - 5 \, A a b^{2}\right )} x^{3} + {\left (C a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x^{2} + 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) - 4 \, {\left (D a^{4} - 3 \, B a^{3} b\right )} x - 8 \, {\left (B a b^{3} x^{5} + 2 \, B a^{2} b^{2} x^{3} + B a^{3} b x\right )} \log \left (b x^{2} + a\right ) + 16 \, {\left (B a b^{3} x^{5} + 2 \, B a^{2} b^{2} x^{3} + B a^{3} b x\right )} \log \relax (x)}{16 \, {\left (a^{4} b^{3} x^{5} + 2 \, a^{5} b^{2} x^{3} + a^{6} b x\right )}}, \frac {4 \, B a^{2} b^{2} x^{3} - 8 \, A a^{3} b + 3 \, {\left (C a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{4} + 5 \, {\left (C a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2} + 3 \, {\left ({\left (C a b^{2} - 5 \, A b^{3}\right )} x^{5} + 2 \, {\left (C a^{2} b - 5 \, A a b^{2}\right )} x^{3} + {\left (C a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) - 2 \, {\left (D a^{4} - 3 \, B a^{3} b\right )} x - 4 \, {\left (B a b^{3} x^{5} + 2 \, B a^{2} b^{2} x^{3} + B a^{3} b x\right )} \log \left (b x^{2} + a\right ) + 8 \, {\left (B a b^{3} x^{5} + 2 \, B a^{2} b^{2} x^{3} + B a^{3} b x\right )} \log \relax (x)}{8 \, {\left (a^{4} b^{3} x^{5} + 2 \, a^{5} b^{2} x^{3} + a^{6} b x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(8*B*a^2*b^2*x^3 - 16*A*a^3*b + 6*(C*a^2*b^2 - 5*A*a*b^3)*x^4 + 10*(C*a^3*b - 5*A*a^2*b^2)*x^2 + 3*((C*a
*b^2 - 5*A*b^3)*x^5 + 2*(C*a^2*b - 5*A*a*b^2)*x^3 + (C*a^3 - 5*A*a^2*b)*x)*sqrt(-a*b)*log((b*x^2 + 2*sqrt(-a*b
)*x - a)/(b*x^2 + a)) - 4*(D*a^4 - 3*B*a^3*b)*x - 8*(B*a*b^3*x^5 + 2*B*a^2*b^2*x^3 + B*a^3*b*x)*log(b*x^2 + a)
+ 16*(B*a*b^3*x^5 + 2*B*a^2*b^2*x^3 + B*a^3*b*x)*log(x))/(a^4*b^3*x^5 + 2*a^5*b^2*x^3 + a^6*b*x), 1/8*(4*B*a^
2*b^2*x^3 - 8*A*a^3*b + 3*(C*a^2*b^2 - 5*A*a*b^3)*x^4 + 5*(C*a^3*b - 5*A*a^2*b^2)*x^2 + 3*((C*a*b^2 - 5*A*b^3)
*x^5 + 2*(C*a^2*b - 5*A*a*b^2)*x^3 + (C*a^3 - 5*A*a^2*b)*x)*sqrt(a*b)*arctan(sqrt(a*b)*x/a) - 2*(D*a^4 - 3*B*a
^3*b)*x - 4*(B*a*b^3*x^5 + 2*B*a^2*b^2*x^3 + B*a^3*b*x)*log(b*x^2 + a) + 8*(B*a*b^3*x^5 + 2*B*a^2*b^2*x^3 + B*
a^3*b*x)*log(x))/(a^4*b^3*x^5 + 2*a^5*b^2*x^3 + a^6*b*x)]

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giac [A]  time = 0.49, size = 141, normalized size = 0.98 \[ -\frac {B \log \left (b x^{2} + a\right )}{2 \, a^{3}} + \frac {B \log \left ({\left | x \right |}\right )}{a^{3}} + \frac {3 \, {\left (C a - 5 \, A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3}} + \frac {4 \, B a b^{2} x^{3} + 3 \, {\left (C a b^{2} - 5 \, A b^{3}\right )} x^{4} - 8 \, A a^{2} b + 5 \, {\left (C a^{2} b - 5 \, A a b^{2}\right )} x^{2} - 2 \, {\left (D a^{3} - 3 \, B a^{2} b\right )} x}{8 \, {\left (b x^{2} + a\right )}^{2} a^{3} b x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 195, normalized size = 1.35 \[ -\frac {7 A \,b^{2} x^{3}}{8 \left (b \,x^{2}+a \right )^{2} a^{3}}+\frac {3 C b \,x^{3}}{8 \left (b \,x^{2}+a \right )^{2} a^{2}}+\frac {B b \,x^{2}}{2 \left (b \,x^{2}+a \right )^{2} a^{2}}-\frac {9 A b x}{8 \left (b \,x^{2}+a \right )^{2} a^{2}}+\frac {5 C x}{8 \left (b \,x^{2}+a \right )^{2} a}-\frac {15 A b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, a^{3}}+\frac {3 B}{4 \left (b \,x^{2}+a \right )^{2} a}+\frac {3 C \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, a^{2}}-\frac {D}{4 \left (b \,x^{2}+a \right )^{2} b}+\frac {B \ln \relax (x )}{a^{3}}-\frac {B \ln \left (b \,x^{2}+a \right )}{2 a^{3}}-\frac {A}{a^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7/8/(b*x^2+a)^2*A/a^3*b^2*x^3+3/8/a^2/(b*x^2+a)^2*C*x^3*b+1/2/a^2/(b*x^2+a)^2*B*x^2*b-9/8/(b*x^2+a)^2*A/a^2*b
*x+5/8/a/(b*x^2+a)^2*C*x+3/4/(b*x^2+a)^2*B/a-1/4/(b*x^2+a)^2/b*D-1/2*B/a^3*ln(b*x^2+a)-15/8/(a*b)^(1/2)*A/a^3*
b*arctan(1/(a*b)^(1/2)*b*x)+3/8/a^2/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x)*C-A/a^3/x+B/a^3*ln(x)

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maxima [A]  time = 2.99, size = 152, normalized size = 1.06 \[ \frac {4 \, B a b^{2} x^{3} + 3 \, {\left (C a b^{2} - 5 \, A b^{3}\right )} x^{4} - 8 \, A a^{2} b + 5 \, {\left (C a^{2} b - 5 \, A a b^{2}\right )} x^{2} - 2 \, {\left (D a^{3} - 3 \, B a^{2} b\right )} x}{8 \, {\left (a^{3} b^{3} x^{5} + 2 \, a^{4} b^{2} x^{3} + a^{5} b x\right )}} - \frac {B \log \left (b x^{2} + a\right )}{2 \, a^{3}} + \frac {B \log \relax (x)}{a^{3}} + \frac {3 \, {\left (C a - 5 \, A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.40, size = 202, normalized size = 1.40 \[ \frac {\frac {3\,B}{4\,a}+\frac {B\,b\,x^2}{2\,a^2}}{a^2+2\,a\,b\,x^2+b^2\,x^4}+\frac {\frac {5\,C\,x}{8\,a}+\frac {3\,C\,b\,x^3}{8\,a^2}}{a^2+2\,a\,b\,x^2+b^2\,x^4}-\frac {\frac {A}{a}+\frac {25\,A\,b\,x^2}{8\,a^2}+\frac {15\,A\,b^2\,x^4}{8\,a^3}}{a^2\,x+2\,a\,b\,x^3+b^2\,x^5}-\frac {D}{4\,b\,{\left (b\,x^2+a\right )}^2}-\frac {B\,\ln \left (b\,x^2+a\right )}{2\,a^3}+\frac {B\,\ln \relax (x)}{a^3}-\frac {15\,A\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{8\,a^{7/2}}+\frac {3\,C\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{8\,a^{5/2}\,\sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((3*B)/(4*a) + (B*b*x^2)/(2*a^2))/(a^2 + b^2*x^4 + 2*a*b*x^2) + ((5*C*x)/(8*a) + (3*C*b*x^3)/(8*a^2))/(a^2 + b
^2*x^4 + 2*a*b*x^2) - (A/a + (25*A*b*x^2)/(8*a^2) + (15*A*b^2*x^4)/(8*a^3))/(a^2*x + b^2*x^5 + 2*a*b*x^3) - D/
(4*b*(a + b*x^2)^2) - (B*log(a + b*x^2))/(2*a^3) + (B*log(x))/a^3 - (15*A*b^(1/2)*atan((b^(1/2)*x)/a^(1/2)))/(
8*a^(7/2)) + (3*C*atan((b^(1/2)*x)/a^(1/2)))/(8*a^(5/2)*b^(1/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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