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 = 0.12, antiderivative size = 16, normalized size of antiderivative = 1.00, 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 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]

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 6688

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

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*}

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Mathematica [B]  time = 0.02, 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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fricas [A]  time = 0.43, size = 26, normalized size = 1.62 \[ -B \arctan \relax (w) - \frac {1}{2} \, A \log \left (A w + B\right ) + \frac {1}{2} \, A \log \left (A w - B\right ) \]

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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giac [B]  time = 1.16, size = 82, normalized size = 5.12 \[ -\frac {{\left (\frac {A^{3} B \log \left ({\left | A w + B \right |}\right )}{A^{4} + A^{2} B^{2}} - \frac {A^{3} B \log \left ({\left | A w - B \right |}\right )}{A^{4} + A^{2} B^{2}} + \frac {2 \, B^{2} \arctan \relax (w)}{A^{2} + B^{2}}\right )} {\left (A^{2} + B^{2}\right )}}{2 \, B} \]

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

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maple [B]  time = 0.01, size = 121, normalized size = 7.56 \[ \frac {A^{3} \ln \left (A w -B \right )}{2 A^{2}+2 B^{2}}-\frac {A^{3} \ln \left (A w +B \right )}{2 \left (A^{2}+B^{2}\right )}-\frac {A^{2} B \arctan \relax (w )}{A^{2}+B^{2}}+\frac {A \,B^{2} \ln \left (A w -B \right )}{2 A^{2}+2 B^{2}}-\frac {A \,B^{2} \ln \left (A w +B \right )}{2 \left (A^{2}+B^{2}\right )}-\frac {B^{3} \arctan \relax (w )}{A^{2}+B^{2}} \]

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)+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

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maxima [B]  time = 1.41, size = 68, normalized size = 4.25 \[ -\frac {{\left (A^{2} + B^{2}\right )} {\left (\frac {2 \, B^{2} \arctan \relax (w)}{A^{2} + B^{2}} + \frac {A B \log \left (A w + B\right )}{A^{2} + B^{2}} - \frac {A B \log \left (A w - B\right )}{A^{2} + B^{2}}\right )}}{2 \, B} \]

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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mupad [B]  time = 0.24, size = 352, normalized size = 22.00 \[ -A\,\mathrm {atanh}\left (\frac {2\,A^{13}\,w}{2\,A^{12}\,B+6\,A^{10}\,B^3+6\,A^8\,B^5+2\,A^6\,B^7}+\frac {2\,A^7\,B^6\,w}{2\,A^{12}\,B+6\,A^{10}\,B^3+6\,A^8\,B^5+2\,A^6\,B^7}+\frac {6\,A^9\,B^4\,w}{2\,A^{12}\,B+6\,A^{10}\,B^3+6\,A^8\,B^5+2\,A^6\,B^7}+\frac {6\,A^{11}\,B^2\,w}{2\,A^{12}\,B+6\,A^{10}\,B^3+6\,A^8\,B^5+2\,A^6\,B^7}\right )-B\,\mathrm {atan}\left (\frac {2\,A^4\,B^9\,w}{2\,A^{10}\,B^3+6\,A^8\,B^5+6\,A^6\,B^7+2\,A^4\,B^9}+\frac {6\,A^6\,B^7\,w}{2\,A^{10}\,B^3+6\,A^8\,B^5+6\,A^6\,B^7+2\,A^4\,B^9}+\frac {6\,A^8\,B^5\,w}{2\,A^{10}\,B^3+6\,A^8\,B^5+6\,A^6\,B^7+2\,A^4\,B^9}+\frac {2\,A^{10}\,B^3\,w}{2\,A^{10}\,B^3+6\,A^8\,B^5+6\,A^6\,B^7+2\,A^4\,B^9}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- A*atanh((2*A^13*w)/(2*A^12*B + 2*A^6*B^7 + 6*A^8*B^5 + 6*A^10*B^3) + (2*A^7*B^6*w)/(2*A^12*B + 2*A^6*B^7 + 6
*A^8*B^5 + 6*A^10*B^3) + (6*A^9*B^4*w)/(2*A^12*B + 2*A^6*B^7 + 6*A^8*B^5 + 6*A^10*B^3) + (6*A^11*B^2*w)/(2*A^1
2*B + 2*A^6*B^7 + 6*A^8*B^5 + 6*A^10*B^3)) - B*atan((2*A^4*B^9*w)/(2*A^4*B^9 + 6*A^6*B^7 + 6*A^8*B^5 + 2*A^10*
B^3) + (6*A^6*B^7*w)/(2*A^4*B^9 + 6*A^6*B^7 + 6*A^8*B^5 + 2*A^10*B^3) + (6*A^8*B^5*w)/(2*A^4*B^9 + 6*A^6*B^7 +
6*A^8*B^5 + 2*A^10*B^3) + (2*A^10*B^3*w)/(2*A^4*B^9 + 6*A^6*B^7 + 6*A^8*B^5 + 2*A^10*B^3))

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sympy [C]  time = 1.93, size = 422, normalized size = 26.38 \[ \left (A^{2} B + B^{3}\right ) \left (- \frac {A \log {\left (w + \frac {- \frac {A^{9}}{B \left (A^{2} + B^{2}\right )^{3}} - \frac {A^{7} B}{\left (A^{2} + B^{2}\right )^{3}} + \frac {A^{5} B^{3}}{\left (A^{2} + B^{2}\right )^{3}} + \frac {A^{5}}{B \left (A^{2} + B^{2}\right )} + \frac {A^{3} B^{5}}{\left (A^{2} + B^{2}\right )^{3}} + \frac {A B^{3}}{A^{2} + B^{2}}}{A^{2}} \right )}}{2 B \left (A^{2} + B^{2}\right )} + \frac {A \log {\left (w + \frac {\frac {A^{9}}{B \left (A^{2} + B^{2}\right )^{3}} + \frac {A^{7} B}{\left (A^{2} + B^{2}\right )^{3}} - \frac {A^{5} B^{3}}{\left (A^{2} + B^{2}\right )^{3}} - \frac {A^{5}}{B \left (A^{2} + B^{2}\right )} - \frac {A^{3} B^{5}}{\left (A^{2} + B^{2}\right )^{3}} - \frac {A B^{3}}{A^{2} + B^{2}}}{A^{2}} \right )}}{2 B \left (A^{2} + B^{2}\right )} + \frac {i \log {\left (w + \frac {- \frac {i A^{6} B^{2}}{\left (A^{2} + B^{2}\right )^{3}} - \frac {i A^{4} B^{4}}{\left (A^{2} + B^{2}\right )^{3}} - \frac {i A^{4}}{A^{2} + B^{2}} + \frac {i A^{2} B^{6}}{\left (A^{2} + B^{2}\right )^{3}} + \frac {i B^{8}}{\left (A^{2} + B^{2}\right )^{3}} - \frac {i B^{4}}{A^{2} + B^{2}}}{A^{2}} \right )}}{2 \left (A^{2} + B^{2}\right )} - \frac {i \log {\left (w + \frac {\frac {i A^{6} B^{2}}{\left (A^{2} + B^{2}\right )^{3}} + \frac {i A^{4} B^{4}}{\left (A^{2} + B^{2}\right )^{3}} + \frac {i A^{4}}{A^{2} + B^{2}} - \frac {i A^{2} B^{6}}{\left (A^{2} + B^{2}\right )^{3}} - \frac {i B^{8}}{\left (A^{2} + B^{2}\right )^{3}} + \frac {i B^{4}}{A^{2} + B^{2}}}{A^{2}} \right )}}{2 \left (A^{2} + B^{2}\right )}\right ) \]

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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