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

[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)

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Rubi [A]  time = 0.228149, antiderivative size = 144, normalized size of antiderivative = 1., 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 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]

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 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 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]

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*}

Mathematica [A]  time = 0.0961463, size = 141, normalized size = 0.98 \[ \frac{a^2 (-D)+a b B+a b C x-A b^2 x}{4 a^2 b \left (a+b x^2\right )^2}+\frac{4 a B+3 a C x-7 A b x}{8 a^3 \left (a+b x^2\right )}+\frac{3 (a C-5 A b) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{7/2} \sqrt{b}}-\frac{A}{a^3 x}-\frac{B \log \left (a+b x^2\right )}{2 a^3}+\frac{B \log (x)}{a^3} \]

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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Maple [A]  time = 0.013, size = 195, normalized size = 1.4 \begin{align*} -{\frac{A}{{a}^{3}x}}+{\frac{B\ln \left ( x \right ) }{{a}^{3}}}-{\frac{7\,A{x}^{3}{b}^{2}}{8\,{a}^{3} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{3\,bC{x}^{3}}{8\,{a}^{2} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{B{x}^{2}b}{2\,{a}^{2} \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{9\,Abx}{8\,{a}^{2} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{5\,Cx}{8\,a \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{3\,B}{4\,a \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{D}{4\, \left ( b{x}^{2}+a \right ) ^{2}b}}-{\frac{B\ln \left ( b{x}^{2}+a \right ) }{2\,{a}^{3}}}-{\frac{15\,Ab}{8\,{a}^{3}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{3\,C}{8\,{a}^{2}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

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]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

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]

Exception raised: UnboundLocalError

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Sympy [B]  time = 20.804, size = 860, normalized size = 5.97 \begin{align*} \text{result too large to display} \end{align*}

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]

B*log(x)/a**3 + (-B/(2*a**3) - 3*sqrt(-a**7*b)*(-5*A*b + C*a)/(16*a**7*b))*log(x + (-1200*A**2*B*a*b**2 + 1200
*A**2*a**4*b**2*(-B/(2*a**3) - 3*sqrt(-a**7*b)*(-5*A*b + C*a)/(16*a**7*b)) + 480*A*B*C*a**2*b - 480*A*C*a**5*b
*(-B/(2*a**3) - 3*sqrt(-a**7*b)*(-5*A*b + C*a)/(16*a**7*b)) + 1024*B**3*a**2*b + 1024*B**2*a**5*b*(-B/(2*a**3)
- 3*sqrt(-a**7*b)*(-5*A*b + C*a)/(16*a**7*b)) - 48*B*C**2*a**3 - 2048*B*a**8*b*(-B/(2*a**3) - 3*sqrt(-a**7*b)
*(-5*A*b + C*a)/(16*a**7*b))**2 + 48*C**2*a**6*(-B/(2*a**3) - 3*sqrt(-a**7*b)*(-5*A*b + C*a)/(16*a**7*b)))/(-1
125*A**3*b**3 + 675*A**2*C*a*b**2 - 2880*A*B**2*a*b**2 - 135*A*C**2*a**2*b + 576*B**2*C*a**2*b + 9*C**3*a**3))
+ (-B/(2*a**3) + 3*sqrt(-a**7*b)*(-5*A*b + C*a)/(16*a**7*b))*log(x + (-1200*A**2*B*a*b**2 + 1200*A**2*a**4*b*
*2*(-B/(2*a**3) + 3*sqrt(-a**7*b)*(-5*A*b + C*a)/(16*a**7*b)) + 480*A*B*C*a**2*b - 480*A*C*a**5*b*(-B/(2*a**3)
+ 3*sqrt(-a**7*b)*(-5*A*b + C*a)/(16*a**7*b)) + 1024*B**3*a**2*b + 1024*B**2*a**5*b*(-B/(2*a**3) + 3*sqrt(-a*
*7*b)*(-5*A*b + C*a)/(16*a**7*b)) - 48*B*C**2*a**3 - 2048*B*a**8*b*(-B/(2*a**3) + 3*sqrt(-a**7*b)*(-5*A*b + C*
a)/(16*a**7*b))**2 + 48*C**2*a**6*(-B/(2*a**3) + 3*sqrt(-a**7*b)*(-5*A*b + C*a)/(16*a**7*b)))/(-1125*A**3*b**3
+ 675*A**2*C*a*b**2 - 2880*A*B**2*a*b**2 - 135*A*C**2*a**2*b + 576*B**2*C*a**2*b + 9*C**3*a**3)) + (-8*A*a**2
*b + 4*B*a*b**2*x**3 + x**4*(-15*A*b**3 + 3*C*a*b**2) + x**2*(-25*A*a*b**2 + 5*C*a**2*b) + x*(6*B*a**2*b - 2*D
*a**3))/(8*a**5*b*x + 16*a**4*b**2*x**3 + 8*a**3*b**3*x**5)

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Giac [A]  time = 1.70337, size = 190, normalized size = 1.32 \begin{align*} -\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} \end{align*}

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)