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

Optimal. Leaf size=130 \[ \frac{x (a D+3 b B)+4 A b}{8 a^2 b \left (a+b x^2\right )}-\frac{A \log \left (a+b x^2\right )}{2 a^3}+\frac{A \log (x)}{a^3}+\frac{(a D+3 b B) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} b^{3/2}}+\frac{x (b B-a D)-a C+A b}{4 a b \left (a+b x^2\right )^2} \]

[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)

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Rubi [A]  time = 0.134627, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {1805, 823, 801, 635, 205, 260} \[ \frac{x (a D+3 b B)+4 A b}{8 a^2 b \left (a+b x^2\right )}-\frac{A \log \left (a+b x^2\right )}{2 a^3}+\frac{A \log (x)}{a^3}+\frac{(a D+3 b B) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{5/2} b^{3/2}}+\frac{x (b B-a D)-a C+A b}{4 a b \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x + C*x^2 + D*x^3)/(x*(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 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 823

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] + Di
st[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 801

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 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 \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{\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]  time = 0.102998, size = 117, normalized size = 0.9 \[ \frac{\frac{2 a^2 (-a (C+D x)+A b+b B x)}{b \left (a+b x^2\right )^2}+\frac{a (a D x+4 A b+3 b B x)}{b \left (a+b x^2\right )}-4 A \log \left (a+b x^2\right )+\frac{\sqrt{a} (a D+3 b B) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{b^{3/2}}+8 A \log (x)}{8 a^3} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.013, size = 184, normalized size = 1.4 \begin{align*}{\frac{A\ln \left ( x \right ) }{{a}^{3}}}+{\frac{3\,bB{x}^{3}}{8\,{a}^{2} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{D{x}^{3}}{8\,a \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{A{x}^{2}b}{2\,{a}^{2} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{5\,Bx}{8\,a \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{xD}{8\, \left ( b{x}^{2}+a \right ) ^{2}b}}+{\frac{3\,A}{4\,a \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{C}{4\, \left ( b{x}^{2}+a \right ) ^{2}b}}-{\frac{A\ln \left ( b{x}^{2}+a \right ) }{2\,{a}^{3}}}+{\frac{3\,B}{8\,{a}^{2}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{D}{8\,ab}\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/(b*x^2+a)^3,x)

[Out]

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

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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/(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/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [B]  time = 16.3593, size = 872, normalized size = 6.71 \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/(b*x**2+a)**3,x)

[Out]

A*log(x)/a**3 + (-A/(2*a**3) - sqrt(-a**7*b**3)*(3*B*b + D*a)/(16*a**6*b**3))*log(x + (3072*A**3*b**4 + 3072*A
**2*a**3*b**4*(-A/(2*a**3) - sqrt(-a**7*b**3)*(3*B*b + D*a)/(16*a**6*b**3)) - 144*A*B**2*a*b**3 - 96*A*B*D*a**
2*b**2 - 16*A*D**2*a**3*b - 6144*A*a**6*b**4*(-A/(2*a**3) - sqrt(-a**7*b**3)*(3*B*b + D*a)/(16*a**6*b**3))**2
+ 144*B**2*a**4*b**3*(-A/(2*a**3) - sqrt(-a**7*b**3)*(3*B*b + D*a)/(16*a**6*b**3)) + 96*B*D*a**5*b**2*(-A/(2*a
**3) - sqrt(-a**7*b**3)*(3*B*b + D*a)/(16*a**6*b**3)) + 16*D**2*a**6*b*(-A/(2*a**3) - sqrt(-a**7*b**3)*(3*B*b
+ D*a)/(16*a**6*b**3)))/(1728*A**2*B*b**4 + 576*A**2*D*a*b**3 + 27*B**3*a*b**3 + 27*B**2*D*a**2*b**2 + 9*B*D**
2*a**3*b + D**3*a**4)) + (-A/(2*a**3) + sqrt(-a**7*b**3)*(3*B*b + D*a)/(16*a**6*b**3))*log(x + (3072*A**3*b**4
+ 3072*A**2*a**3*b**4*(-A/(2*a**3) + sqrt(-a**7*b**3)*(3*B*b + D*a)/(16*a**6*b**3)) - 144*A*B**2*a*b**3 - 96*
A*B*D*a**2*b**2 - 16*A*D**2*a**3*b - 6144*A*a**6*b**4*(-A/(2*a**3) + sqrt(-a**7*b**3)*(3*B*b + D*a)/(16*a**6*b
**3))**2 + 144*B**2*a**4*b**3*(-A/(2*a**3) + sqrt(-a**7*b**3)*(3*B*b + D*a)/(16*a**6*b**3)) + 96*B*D*a**5*b**2
*(-A/(2*a**3) + sqrt(-a**7*b**3)*(3*B*b + D*a)/(16*a**6*b**3)) + 16*D**2*a**6*b*(-A/(2*a**3) + sqrt(-a**7*b**3
)*(3*B*b + D*a)/(16*a**6*b**3)))/(1728*A**2*B*b**4 + 576*A**2*D*a*b**3 + 27*B**3*a*b**3 + 27*B**2*D*a**2*b**2
+ 9*B*D**2*a**3*b + D**3*a**4)) + (6*A*a*b + 4*A*b**2*x**2 - 2*C*a**2 + x**3*(3*B*b**2 + D*a*b) + x*(5*B*a*b -
D*a**2))/(8*a**4*b + 16*a**3*b**2*x**2 + 8*a**2*b**3*x**4)

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Giac [A]  time = 1.67072, size = 173, normalized size = 1.33 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)