3.71 \(\int -\frac{A^2+B^2}{B (1+w^2)^2 (1-\frac{(A^2+B^2) w^2}{B^2 (1+w^2)})} \, dw\)

Optimal. Leaf size=16 \[ -A \tanh ^{-1}\left (\frac{A w}{B}\right )-B \tan ^{-1}(w) \]

[Out]

-(B*ArcTan[w]) - A*ArcTanh[(A*w)/B]

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Rubi [A]  time = 0.124371, antiderivative size = 16, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {12, 6688, 391, 203, 208} \[ -A \tanh ^{-1}\left (\frac{A w}{B}\right )-B \tan ^{-1}(w) \]

Antiderivative was successfully verified.

[In]

Int[-((A^2 + B^2)/(B*(1 + w^2)^2*(1 - ((A^2 + B^2)*w^2)/(B^2*(1 + w^2))))),w]

[Out]

-(B*ArcTan[w]) - A*ArcTanh[(A*w)/B]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 391

Int[1/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x^n),
x], x] - Dist[d/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 208

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

Rubi steps

\begin{align*} \int -\frac{A^2+B^2}{B \left (1+w^2\right )^2 \left (1-\frac{\left (A^2+B^2\right ) w^2}{B^2 \left (1+w^2\right )}\right )} \, dw &=-\frac{\left (A^2+B^2\right ) \int \frac{1}{\left (1+w^2\right )^2 \left (1-\frac{\left (A^2+B^2\right ) w^2}{B^2 \left (1+w^2\right )}\right )} \, dw}{B}\\ &=-\frac{\left (A^2+B^2\right ) \int \frac{B^2}{\left (1+w^2\right ) \left (B^2-A^2 w^2\right )} \, dw}{B}\\ &=-\left (\left (B \left (A^2+B^2\right )\right ) \int \frac{1}{\left (1+w^2\right ) \left (B^2-A^2 w^2\right )} \, dw\right )\\ &=-\left (B \int \frac{1}{1+w^2} \, dw\right )-\left (A^2 B\right ) \int \frac{1}{B^2-A^2 w^2} \, dw\\ &=-B \tan ^{-1}(w)-A \tanh ^{-1}\left (\frac{A w}{B}\right )\\ \end{align*}

Mathematica [B]  time = 0.01813, size = 35, normalized size = 2.19 \[ -\frac{B \left (A^2+B^2\right ) \left (A \tanh ^{-1}\left (\frac{A w}{B}\right )+B \tan ^{-1}(w)\right )}{A^2 B+B^3} \]

Antiderivative was successfully verified.

[In]

Integrate[-((A^2 + B^2)/(B*(1 + w^2)^2*(1 - ((A^2 + B^2)*w^2)/(B^2*(1 + w^2))))),w]

[Out]

-((B*(A^2 + B^2)*(B*ArcTan[w] + A*ArcTanh[(A*w)/B]))/(A^2*B + B^3))

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Maple [B]  time = 0.007, size = 121, normalized size = 7.6 \begin{align*}{\frac{{A}^{3}\ln \left ( Aw-B \right ) }{2\,{A}^{2}+2\,{B}^{2}}}+{\frac{A{B}^{2}\ln \left ( Aw-B \right ) }{2\,{A}^{2}+2\,{B}^{2}}}-{\frac{B\arctan \left ( w \right ){A}^{2}}{{A}^{2}+{B}^{2}}}-{\frac{\arctan \left ( w \right ){B}^{3}}{{A}^{2}+{B}^{2}}}-{\frac{{A}^{3}\ln \left ( Aw+B \right ) }{2\,{A}^{2}+2\,{B}^{2}}}-{\frac{A{B}^{2}\ln \left ( Aw+B \right ) }{2\,{A}^{2}+2\,{B}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-A^2-B^2)/B/(w^2+1)^2/(1-(A^2+B^2)*w^2/B^2/(w^2+1)),w)

[Out]

1/2*A^3/(A^2+B^2)*ln(A*w-B)+1/2*A*B^2/(A^2+B^2)*ln(A*w-B)-B/(A^2+B^2)*arctan(w)*A^2-1/(A^2+B^2)*arctan(w)*B^3-
1/2*A^3/(A^2+B^2)*ln(A*w+B)-1/2*A*B^2/(A^2+B^2)*ln(A*w+B)

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Maxima [B]  time = 1.41031, size = 92, normalized size = 5.75 \begin{align*} -\frac{{\left (A^{2} + B^{2}\right )}{\left (\frac{2 \, B^{2} \arctan \left (w\right )}{A^{2} + B^{2}} + \frac{A B \log \left (A w + B\right )}{A^{2} + B^{2}} - \frac{A B \log \left (A w - B\right )}{A^{2} + B^{2}}\right )}}{2 \, B} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-A^2-B^2)/B/(w^2+1)^2/(1-(A^2+B^2)*w^2/B^2/(w^2+1)),w, algorithm="maxima")

[Out]

-1/2*(A^2 + B^2)*(2*B^2*arctan(w)/(A^2 + B^2) + A*B*log(A*w + B)/(A^2 + B^2) - A*B*log(A*w - B)/(A^2 + B^2))/B

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Fricas [A]  time = 2.16152, size = 76, normalized size = 4.75 \begin{align*} -B \arctan \left (w\right ) - \frac{1}{2} \, A \log \left (A w + B\right ) + \frac{1}{2} \, A \log \left (A w - B\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-A^2-B^2)/B/(w^2+1)^2/(1-(A^2+B^2)*w^2/B^2/(w^2+1)),w, algorithm="fricas")

[Out]

-B*arctan(w) - 1/2*A*log(A*w + B) + 1/2*A*log(A*w - B)

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Sympy [C]  time = 1.30127, size = 422, normalized size = 26.38 \begin{align*} \left (A^{2} B + B^{3}\right ) \left (- \frac{A \log{\left (w + \frac{- \frac{A^{9}}{B \left (A^{2} + B^{2}\right )^{3}} - \frac{A^{7} B}{\left (A^{2} + B^{2}\right )^{3}} + \frac{A^{5} B^{3}}{\left (A^{2} + B^{2}\right )^{3}} + \frac{A^{5}}{B \left (A^{2} + B^{2}\right )} + \frac{A^{3} B^{5}}{\left (A^{2} + B^{2}\right )^{3}} + \frac{A B^{3}}{A^{2} + B^{2}}}{A^{2}} \right )}}{2 B \left (A^{2} + B^{2}\right )} + \frac{A \log{\left (w + \frac{\frac{A^{9}}{B \left (A^{2} + B^{2}\right )^{3}} + \frac{A^{7} B}{\left (A^{2} + B^{2}\right )^{3}} - \frac{A^{5} B^{3}}{\left (A^{2} + B^{2}\right )^{3}} - \frac{A^{5}}{B \left (A^{2} + B^{2}\right )} - \frac{A^{3} B^{5}}{\left (A^{2} + B^{2}\right )^{3}} - \frac{A B^{3}}{A^{2} + B^{2}}}{A^{2}} \right )}}{2 B \left (A^{2} + B^{2}\right )} + \frac{i \log{\left (w + \frac{- \frac{i A^{6} B^{2}}{\left (A^{2} + B^{2}\right )^{3}} - \frac{i A^{4} B^{4}}{\left (A^{2} + B^{2}\right )^{3}} - \frac{i A^{4}}{A^{2} + B^{2}} + \frac{i A^{2} B^{6}}{\left (A^{2} + B^{2}\right )^{3}} + \frac{i B^{8}}{\left (A^{2} + B^{2}\right )^{3}} - \frac{i B^{4}}{A^{2} + B^{2}}}{A^{2}} \right )}}{2 \left (A^{2} + B^{2}\right )} - \frac{i \log{\left (w + \frac{\frac{i A^{6} B^{2}}{\left (A^{2} + B^{2}\right )^{3}} + \frac{i A^{4} B^{4}}{\left (A^{2} + B^{2}\right )^{3}} + \frac{i A^{4}}{A^{2} + B^{2}} - \frac{i A^{2} B^{6}}{\left (A^{2} + B^{2}\right )^{3}} - \frac{i B^{8}}{\left (A^{2} + B^{2}\right )^{3}} + \frac{i B^{4}}{A^{2} + B^{2}}}{A^{2}} \right )}}{2 \left (A^{2} + B^{2}\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-A**2-B**2)/B/(w**2+1)**2/(1-(A**2+B**2)*w**2/B**2/(w**2+1)),w)

[Out]

(A**2*B + B**3)*(-A*log(w + (-A**9/(B*(A**2 + B**2)**3) - A**7*B/(A**2 + B**2)**3 + A**5*B**3/(A**2 + B**2)**3
+ A**5/(B*(A**2 + B**2)) + A**3*B**5/(A**2 + B**2)**3 + A*B**3/(A**2 + B**2))/A**2)/(2*B*(A**2 + B**2)) + A*l
og(w + (A**9/(B*(A**2 + B**2)**3) + A**7*B/(A**2 + B**2)**3 - A**5*B**3/(A**2 + B**2)**3 - A**5/(B*(A**2 + B**
2)) - A**3*B**5/(A**2 + B**2)**3 - A*B**3/(A**2 + B**2))/A**2)/(2*B*(A**2 + B**2)) + I*log(w + (-I*A**6*B**2/(
A**2 + B**2)**3 - I*A**4*B**4/(A**2 + B**2)**3 - I*A**4/(A**2 + B**2) + I*A**2*B**6/(A**2 + B**2)**3 + I*B**8/
(A**2 + B**2)**3 - I*B**4/(A**2 + B**2))/A**2)/(2*(A**2 + B**2)) - I*log(w + (I*A**6*B**2/(A**2 + B**2)**3 + I
*A**4*B**4/(A**2 + B**2)**3 + I*A**4/(A**2 + B**2) - I*A**2*B**6/(A**2 + B**2)**3 - I*B**8/(A**2 + B**2)**3 +
I*B**4/(A**2 + B**2))/A**2)/(2*(A**2 + B**2)))

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Giac [B]  time = 1.09746, size = 111, normalized size = 6.94 \begin{align*} -\frac{{\left (\frac{A^{3} B \log \left ({\left | A w + B \right |}\right )}{A^{4} + A^{2} B^{2}} - \frac{A^{3} B \log \left ({\left | A w - B \right |}\right )}{A^{4} + A^{2} B^{2}} + \frac{2 \, B^{2} \arctan \left (w\right )}{A^{2} + B^{2}}\right )}{\left (A^{2} + B^{2}\right )}}{2 \, B} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-A^2-B^2)/B/(w^2+1)^2/(1-(A^2+B^2)*w^2/B^2/(w^2+1)),w, algorithm="giac")

[Out]

-1/2*(A^3*B*log(abs(A*w + B))/(A^4 + A^2*B^2) - A^3*B*log(abs(A*w - B))/(A^4 + A^2*B^2) + 2*B^2*arctan(w)/(A^2
+ B^2))*(A^2 + B^2)/B