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

Optimal. Leaf size=135 \[ -\frac {(3 b B-a D) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{5/2} \sqrt {b}}+\frac {(2 A b-a C) \log \left (a+b x^2\right )}{2 a^3}-\frac {\log (x) (2 A b-a C)}{a^3}-\frac {A}{2 a^2 x^2}-\frac {B}{a^2 x}-\frac {\frac {A b}{a}+x \left (\frac {b B}{a}-D\right )-C}{2 a \left (a+b x^2\right )} \]

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Rubi [A]  time = 0.20, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \begin {gather*} \frac {(2 A b-a C) \log \left (a+b x^2\right )}{2 a^3}-\frac {\log (x) (2 A b-a C)}{a^3}-\frac {A}{2 a^2 x^2}-\frac {(3 b B-a D) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{5/2} \sqrt {b}}-\frac {B}{a^2 x}-\frac {\frac {A b}{a}+x \left (\frac {b B}{a}-D\right )-C}{2 a \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.10, size = 112, normalized size = 0.83 \begin {gather*} \frac {\frac {a (a (C+D x)-A b-b B x)}{a+b x^2}+(2 A b-a C) \log \left (a+b x^2\right )+2 \log (x) (a C-2 A b)-\frac {a A}{x^2}+\frac {\sqrt {a} (a D-3 b B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {b}}-\frac {2 a B}{x}}{2 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A+B x+C x^2+D x^3}{x^3 \left (a+b x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.02, size = 441, normalized size = 3.27 \begin {gather*} \left [-\frac {4 \, B a^{2} b x + 2 \, A a^{2} b - 2 \, {\left (D a^{2} b - 3 \, B a b^{2}\right )} x^{3} - 2 \, {\left (C a^{2} b - 2 \, A a b^{2}\right )} x^{2} + {\left ({\left (D a b - 3 \, B b^{2}\right )} x^{4} + {\left (D a^{2} - 3 \, B a b\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 ({\left (C a b^{2} - 2 \, A b^{3}\right )} x^{4} + {\left (C a^{2} b - 2 \, A a b^{2}\right )} x^{2}\right )} \log \left (b x^{2} + a\right ) - 4 \, {\left ({\left (C a b^{2} - 2 \, A b^{3}\right )} x^{4} + {\left (C a^{2} b - 2 \, A a b^{2}\right )} x^{2}\right )} \log \relax (x)}{4 \, {\left (a^{3} b^{2} x^{4} + a^{4} b x^{2}\right )}}, -\frac {2 \, B a^{2} b x + A a^{2} b - {\left (D a^{2} b - 3 \, B a b^{2}\right )} x^{3} - {\left (C a^{2} b - 2 \, A a b^{2}\right )} x^{2} - {\left ({\left (D a b - 3 \, B b^{2}\right )} x^{4} + {\left (D a^{2} - 3 \, B a b\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + {\left ({\left (C a b^{2} - 2 \, A b^{3}\right )} x^{4} + {\left (C a^{2} b - 2 \, A a b^{2}\right )} x^{2}\right )} \log \left (b x^{2} + a\right ) - 2 \, {\left ({\left (C a b^{2} - 2 \, A b^{3}\right )} x^{4} + {\left (C a^{2} b - 2 \, A a b^{2}\right )} x^{2}\right )} \log \relax (x)}{2 \, {\left (a^{3} b^{2} x^{4} + a^{4} b x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.42, size = 126, normalized size = 0.93 \begin {gather*} \frac {{\left (D a - 3 \, B b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{2}} - \frac {{\left (C a - 2 \, A b\right )} \log \left (b x^{2} + a\right )}{2 \, a^{3}} + \frac {{\left (C a - 2 \, A b\right )} \log \left ({\left | x \right |}\right )}{a^{3}} - \frac {2 \, B a^{2} x - {\left (D a^{2} - 3 \, B a b\right )} x^{3} + A a^{2} - {\left (C a^{2} - 2 \, A a b\right )} x^{2}}{2 \, {\left (b x^{2} + a\right )} a^{3} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 169, normalized size = 1.25 \begin {gather*} -\frac {B b x}{2 \left (b \,x^{2}+a \right ) a^{2}}-\frac {3 B b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, a^{2}}+\frac {D x}{2 \left (b \,x^{2}+a \right ) a}+\frac {D \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, a}-\frac {A b}{2 \left (b \,x^{2}+a \right ) a^{2}}-\frac {2 A b \ln \relax (x )}{a^{3}}+\frac {A b \ln \left (b \,x^{2}+a \right )}{a^{3}}+\frac {C}{2 \left (b \,x^{2}+a \right ) a}+\frac {C \ln \relax (x )}{a^{2}}-\frac {C \ln \left (b \,x^{2}+a \right )}{2 a^{2}}-\frac {B}{a^{2} x}-\frac {A}{2 a^{2} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.94, size = 117, normalized size = 0.87 \begin {gather*} \frac {{\left (D a - 3 \, B b\right )} x^{3} - 2 \, B a x + {\left (C a - 2 \, A b\right )} x^{2} - A a}{2 \, {\left (a^{2} b x^{4} + a^{3} x^{2}\right )}} + \frac {{\left (D a - 3 \, B b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{2}} - \frac {{\left (C a - 2 \, A b\right )} \log \left (b x^{2} + a\right )}{2 \, a^{3}} + \frac {{\left (C a - 2 \, A b\right )} \log \relax (x)}{a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.35, size = 158, normalized size = 1.17 \begin {gather*} \frac {C}{2\,a\,\left (b\,x^2+a\right )}-\frac {\frac {A}{2\,a}+\frac {A\,b\,x^2}{a^2}}{b\,x^4+a\,x^2}-\frac {\frac {B}{a}+\frac {3\,B\,b\,x^2}{2\,a^2}}{b\,x^3+a\,x}-\frac {C\,\ln \left (b\,x^2+a\right )}{2\,a^2}+\frac {C\,\ln \relax (x)}{a^2}+\frac {A\,b\,\ln \left (b\,x^2+a\right )}{a^3}-\frac {2\,A\,b\,\ln \relax (x)}{a^3}+\frac {x\,D\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},2;\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{a^2}-\frac {3\,B\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{2\,a^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

C/(2*a*(a + b*x^2)) - (A/(2*a) + (A*b*x^2)/a^2)/(a*x^2 + b*x^4) - (B/a + (3*B*b*x^2)/(2*a^2))/(a*x + b*x^3) -
(C*log(a + b*x^2))/(2*a^2) + (C*log(x))/a^2 + (A*b*log(a + b*x^2))/a^3 - (2*A*b*log(x))/a^3 + (x*D*hypergeom([
1/2, 2], 3/2, -(b*x^2)/a))/a^2 - (3*B*b^(1/2)*atan((b^(1/2)*x)/a^(1/2)))/(2*a^(5/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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