Integrand size = 39, antiderivative size = 16 \[ \int \frac {\left (-A^2-B^2\right ) \cos ^2(z)}{B \left (1-\frac {\left (A^2+B^2\right ) \sin ^2(z)}{B^2}\right )} \, dz=-B z-A \text {arctanh}\left (\frac {A \tan (z)}{B}\right ) \] Output:
-B*z-A*arctanh(A*tan(z)/B)
Leaf count is larger than twice the leaf count of optimal. \(35\) vs. \(2(16)=32\).
Time = 0.17 (sec) , antiderivative size = 35, normalized size of antiderivative = 2.19 \[ \int \frac {\left (-A^2-B^2\right ) \cos ^2(z)}{B \left (1-\frac {\left (A^2+B^2\right ) \sin ^2(z)}{B^2}\right )} \, dz=-\frac {B \left (A^2+B^2\right ) \left (B z+A \text {arctanh}\left (\frac {A \tan (z)}{B}\right )\right )}{A^2 B+B^3} \] Input:
Integrate[((-A^2 - B^2)*Cos[z]^2)/(B*(1 - ((A^2 + B^2)*Sin[z]^2)/B^2)),z]
Output:
-((B*(A^2 + B^2)*(B*z + A*ArcTanh[(A*Tan[z])/B]))/(A^2*B + B^3))
Leaf count is larger than twice the leaf count of optimal. \(46\) vs. \(2(16)=32\).
Time = 0.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.88, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {27, 3042, 3670, 27, 303, 216, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (-A^2-B^2\right ) \cos ^2(z)}{B \left (1-\frac {\left (A^2+B^2\right ) \sin ^2(z)}{B^2}\right )} \, dz\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\left (A^2+B^2\right ) \int \frac {\cos ^2(z)}{1-\frac {\left (A^2+B^2\right ) \sin ^2(z)}{B^2}}dz}{B}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\left (A^2+B^2\right ) \int \frac {\cos (z)^2}{1-\frac {\left (A^2+B^2\right ) \sin (z)^2}{B^2}}dz}{B}\) |
\(\Big \downarrow \) 3670 |
\(\displaystyle -\frac {\left (A^2+B^2\right ) \int \frac {B^2}{\left (\tan ^2(z)+1\right ) \left (B^2-A^2 \tan ^2(z)\right )}d\tan (z)}{B}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -B \left (A^2+B^2\right ) \int \frac {1}{\left (\tan ^2(z)+1\right ) \left (B^2-A^2 \tan ^2(z)\right )}d\tan (z)\) |
\(\Big \downarrow \) 303 |
\(\displaystyle -B \left (A^2+B^2\right ) \left (\frac {A^2 \int \frac {1}{B^2-A^2 \tan ^2(z)}d\tan (z)}{A^2+B^2}+\frac {\int \frac {1}{\tan ^2(z)+1}d\tan (z)}{A^2+B^2}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -B \left (A^2+B^2\right ) \left (\frac {A^2 \int \frac {1}{B^2-A^2 \tan ^2(z)}d\tan (z)}{A^2+B^2}+\frac {\arctan (\tan (z))}{A^2+B^2}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -B \left (A^2+B^2\right ) \left (\frac {\arctan (\tan (z))}{A^2+B^2}+\frac {A \text {arctanh}\left (\frac {A \tan (z)}{B}\right )}{B \left (A^2+B^2\right )}\right )\) |
Input:
Int[((-A^2 - B^2)*Cos[z]^2)/(B*(1 - ((A^2 + B^2)*Sin[z]^2)/B^2)),z]
Output:
-(B*(A^2 + B^2)*(ArcTan[Tan[z]]/(A^2 + B^2) + (A*ArcTanh[(A*Tan[z])/B])/(B *(A^2 + B^2))))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[1/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Simp[b/(b *c - a*d) Int[1/(a + b*x^2), x], x] - Simp[d/(b*c - a*d) Int[1/(c + d*x ^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^( p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Su bst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]
Leaf count of result is larger than twice the leaf count of optimal. \(73\) vs. \(2(16)=32\).
Time = 0.55 (sec) , antiderivative size = 74, normalized size of antiderivative = 4.62
method | result | size |
default | \(\left (-A^{2}-B^{2}\right ) B \left (\frac {\arctan \left (\tan \left (z \right )\right )}{A^{2}+B^{2}}+\frac {A \ln \left (A \tan \left (z \right )+B \right )}{2 B \left (A^{2}+B^{2}\right )}-\frac {A \ln \left (A \tan \left (z \right )-B \right )}{2 B \left (A^{2}+B^{2}\right )}\right )\) | \(74\) |
parallelrisch | \(\frac {\left (-A^{2}-B^{2}\right ) \left (2 B z +A \left (\ln \left (2 A \tan \left (\frac {z}{2}\right )+2 B -\sec \left (\frac {z}{2}\right )^{2} B \right )-\ln \left (2 A \tan \left (\frac {z}{2}\right )-2 B +\sec \left (\frac {z}{2}\right )^{2} B \right )\right )\right )}{2 A^{2}+2 B^{2}}\) | \(77\) |
norman | \(\frac {-B z -2 B z \tan \left (\frac {z}{2}\right )^{2}-B z \tan \left (\frac {z}{2}\right )^{4}}{\left (1+\tan \left (\frac {z}{2}\right )^{2}\right )^{2}}-\frac {A \ln \left (-B \tan \left (\frac {z}{2}\right )^{2}+2 A \tan \left (\frac {z}{2}\right )+B \right )}{2}+\frac {A \ln \left (B \tan \left (\frac {z}{2}\right )^{2}+2 A \tan \left (\frac {z}{2}\right )-B \right )}{2}\) | \(83\) |
risch | \(-\frac {B z \,A^{2}}{A^{2}+B^{2}}-\frac {B^{3} z}{A^{2}+B^{2}}+\frac {A^{3} \ln \left ({\mathrm e}^{2 i z}-\frac {i B +A}{-i B +A}\right )}{2 A^{2}+2 B^{2}}+\frac {A \ln \left ({\mathrm e}^{2 i z}-\frac {i B +A}{-i B +A}\right ) B^{2}}{2 A^{2}+2 B^{2}}-\frac {A^{3} \ln \left ({\mathrm e}^{2 i z}-\frac {-i B +A}{i B +A}\right )}{2 \left (A^{2}+B^{2}\right )}-\frac {A \ln \left ({\mathrm e}^{2 i z}-\frac {-i B +A}{i B +A}\right ) B^{2}}{2 \left (A^{2}+B^{2}\right )}\) | \(183\) |
Input:
int((-A^2-B^2)*cos(z)^2/B/(1-(A^2+B^2)*sin(z)^2/B^2),z,method=_RETURNVERBO SE)
Output:
(-A^2-B^2)*B*(1/(A^2+B^2)*arctan(tan(z))+1/2*A/B/(A^2+B^2)*ln(A*tan(z)+B)- 1/2*A/B/(A^2+B^2)*ln(A*tan(z)-B))
Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (16) = 32\).
Time = 0.09 (sec) , antiderivative size = 67, normalized size of antiderivative = 4.19 \[ \int \frac {\left (-A^2-B^2\right ) \cos ^2(z)}{B \left (1-\frac {\left (A^2+B^2\right ) \sin ^2(z)}{B^2}\right )} \, dz=-B z - \frac {1}{4} \, A \log \left (2 \, A B \cos \left (z\right ) \sin \left (z\right ) - {\left (A^{2} - B^{2}\right )} \cos \left (z\right )^{2} + A^{2}\right ) + \frac {1}{4} \, A \log \left (-2 \, A B \cos \left (z\right ) \sin \left (z\right ) - {\left (A^{2} - B^{2}\right )} \cos \left (z\right )^{2} + A^{2}\right ) \] Input:
integrate((-A^2-B^2)*cos(z)^2/B/(1-(A^2+B^2)*sin(z)^2/B^2),z, algorithm="f ricas")
Output:
-B*z - 1/4*A*log(2*A*B*cos(z)*sin(z) - (A^2 - B^2)*cos(z)^2 + A^2) + 1/4*A *log(-2*A*B*cos(z)*sin(z) - (A^2 - B^2)*cos(z)^2 + A^2)
Time = 101.86 (sec) , antiderivative size = 202, normalized size of antiderivative = 12.62 \[ \int \frac {\left (-A^2-B^2\right ) \cos ^2(z)}{B \left (1-\frac {\left (A^2+B^2\right ) \sin ^2(z)}{B^2}\right )} \, dz=\frac {\left (- A^{2} - B^{2}\right ) \left (\begin {cases} z & \text {for}\: A = 0 \wedge B = 0 \\\frac {z \sin ^{2}{\left (z \right )}}{2} + \frac {z \cos ^{2}{\left (z \right )}}{2} + \frac {\sin {\left (z \right )} \cos {\left (z \right )}}{2} & \text {for}\: A = - i B \vee A = i B \\\frac {A B \log {\left (- \frac {A}{B} + \tan {\left (\frac {z}{2} \right )} - \frac {\sqrt {A^{2} + B^{2}}}{B} \right )}}{2 A^{2} + 2 B^{2}} + \frac {A B \log {\left (- \frac {A}{B} + \tan {\left (\frac {z}{2} \right )} + \frac {\sqrt {A^{2} + B^{2}}}{B} \right )}}{2 A^{2} + 2 B^{2}} - \frac {A B \log {\left (\frac {A}{B} + \tan {\left (\frac {z}{2} \right )} - \frac {\sqrt {A^{2} + B^{2}}}{B} \right )}}{2 A^{2} + 2 B^{2}} - \frac {A B \log {\left (\frac {A}{B} + \tan {\left (\frac {z}{2} \right )} + \frac {\sqrt {A^{2} + B^{2}}}{B} \right )}}{2 A^{2} + 2 B^{2}} + \frac {2 B^{2} z}{2 A^{2} + 2 B^{2}} & \text {otherwise} \end {cases}\right )}{B} \] Input:
integrate((-A**2-B**2)*cos(z)**2/B/(1-(A**2+B**2)*sin(z)**2/B**2),z)
Output:
(-A**2 - B**2)*Piecewise((z, Eq(A, 0) & Eq(B, 0)), (z*sin(z)**2/2 + z*cos( z)**2/2 + sin(z)*cos(z)/2, Eq(A, I*B) | Eq(A, -I*B)), (A*B*log(-A/B + tan( z/2) - sqrt(A**2 + B**2)/B)/(2*A**2 + 2*B**2) + A*B*log(-A/B + tan(z/2) + sqrt(A**2 + B**2)/B)/(2*A**2 + 2*B**2) - A*B*log(A/B + tan(z/2) - sqrt(A** 2 + B**2)/B)/(2*A**2 + 2*B**2) - A*B*log(A/B + tan(z/2) + sqrt(A**2 + B**2 )/B)/(2*A**2 + 2*B**2) + 2*B**2*z/(2*A**2 + 2*B**2), True))/B
Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (16) = 32\).
Time = 0.10 (sec) , antiderivative size = 69, normalized size of antiderivative = 4.31 \[ \int \frac {\left (-A^2-B^2\right ) \cos ^2(z)}{B \left (1-\frac {\left (A^2+B^2\right ) \sin ^2(z)}{B^2}\right )} \, dz=-\frac {{\left (A^{2} + B^{2}\right )} {\left (\frac {2 \, B^{2} z}{A^{2} + B^{2}} + \frac {A B \log \left (A \tan \left (z\right ) + B\right )}{A^{2} + B^{2}} - \frac {A B \log \left (A \tan \left (z\right ) - B\right )}{A^{2} + B^{2}}\right )}}{2 \, B} \] Input:
integrate((-A^2-B^2)*cos(z)^2/B/(1-(A^2+B^2)*sin(z)^2/B^2),z, algorithm="m axima")
Output:
-1/2*(A^2 + B^2)*(2*B^2*z/(A^2 + B^2) + A*B*log(A*tan(z) + B)/(A^2 + B^2) - A*B*log(A*tan(z) - B)/(A^2 + B^2))/B
Leaf count of result is larger than twice the leaf count of optimal. 83 vs. \(2 (16) = 32\).
Time = 0.13 (sec) , antiderivative size = 83, normalized size of antiderivative = 5.19 \[ \int \frac {\left (-A^2-B^2\right ) \cos ^2(z)}{B \left (1-\frac {\left (A^2+B^2\right ) \sin ^2(z)}{B^2}\right )} \, dz=-\frac {{\left (\frac {A^{3} B \log \left ({\left | A \tan \left (z\right ) + B \right |}\right )}{A^{4} + A^{2} B^{2}} - \frac {A^{3} B \log \left ({\left | A \tan \left (z\right ) - B \right |}\right )}{A^{4} + A^{2} B^{2}} + \frac {2 \, B^{2} z}{A^{2} + B^{2}}\right )} {\left (A^{2} + B^{2}\right )}}{2 \, B} \] Input:
integrate((-A^2-B^2)*cos(z)^2/B/(1-(A^2+B^2)*sin(z)^2/B^2),z, algorithm="g iac")
Output:
-1/2*(A^3*B*log(abs(A*tan(z) + B))/(A^4 + A^2*B^2) - A^3*B*log(abs(A*tan(z ) - B))/(A^4 + A^2*B^2) + 2*B^2*z/(A^2 + B^2))*(A^2 + B^2)/B
Time = 0.27 (sec) , antiderivative size = 360, normalized size of antiderivative = 22.50 \[ \int \frac {\left (-A^2-B^2\right ) \cos ^2(z)}{B \left (1-\frac {\left (A^2+B^2\right ) \sin ^2(z)}{B^2}\right )} \, dz=-A\,\mathrm {atanh}\left (\frac {2\,A^{13}\,\mathrm {tan}\left (z\right )}{2\,A^{12}\,B+6\,A^{10}\,B^3+6\,A^8\,B^5+2\,A^6\,B^7}+\frac {2\,A^7\,B^6\,\mathrm {tan}\left (z\right )}{2\,A^{12}\,B+6\,A^{10}\,B^3+6\,A^8\,B^5+2\,A^6\,B^7}+\frac {6\,A^9\,B^4\,\mathrm {tan}\left (z\right )}{2\,A^{12}\,B+6\,A^{10}\,B^3+6\,A^8\,B^5+2\,A^6\,B^7}+\frac {6\,A^{11}\,B^2\,\mathrm {tan}\left (z\right )}{2\,A^{12}\,B+6\,A^{10}\,B^3+6\,A^8\,B^5+2\,A^6\,B^7}\right )-B\,\mathrm {atan}\left (\frac {2\,A^4\,B^9\,\mathrm {tan}\left (z\right )}{2\,A^{10}\,B^3+6\,A^8\,B^5+6\,A^6\,B^7+2\,A^4\,B^9}+\frac {6\,A^6\,B^7\,\mathrm {tan}\left (z\right )}{2\,A^{10}\,B^3+6\,A^8\,B^5+6\,A^6\,B^7+2\,A^4\,B^9}+\frac {6\,A^8\,B^5\,\mathrm {tan}\left (z\right )}{2\,A^{10}\,B^3+6\,A^8\,B^5+6\,A^6\,B^7+2\,A^4\,B^9}+\frac {2\,A^{10}\,B^3\,\mathrm {tan}\left (z\right )}{2\,A^{10}\,B^3+6\,A^8\,B^5+6\,A^6\,B^7+2\,A^4\,B^9}\right ) \] Input:
int((cos(z)^2*(A^2 + B^2))/(B*((sin(z)^2*(A^2 + B^2))/B^2 - 1)),z)
Output:
- A*atanh((2*A^13*tan(z))/(2*A^12*B + 2*A^6*B^7 + 6*A^8*B^5 + 6*A^10*B^3) + (2*A^7*B^6*tan(z))/(2*A^12*B + 2*A^6*B^7 + 6*A^8*B^5 + 6*A^10*B^3) + (6* A^9*B^4*tan(z))/(2*A^12*B + 2*A^6*B^7 + 6*A^8*B^5 + 6*A^10*B^3) + (6*A^11* B^2*tan(z))/(2*A^12*B + 2*A^6*B^7 + 6*A^8*B^5 + 6*A^10*B^3)) - B*atan((2*A ^4*B^9*tan(z))/(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*tan(z))/(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*t an(z))/(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*tan( z))/(2*A^4*B^9 + 6*A^6*B^7 + 6*A^8*B^5 + 2*A^10*B^3))
Time = 0.16 (sec) , antiderivative size = 74, normalized size of antiderivative = 4.62 \[ \int \frac {\left (-A^2-B^2\right ) \cos ^2(z)}{B \left (1-\frac {\left (A^2+B^2\right ) \sin ^2(z)}{B^2}\right )} \, dz=\mathit {atan} \left (\frac {\tan \left (\frac {z}{2}\right ) b i}{\sqrt {a^{2}+b^{2}}-a}\right ) a i -\frac {\mathrm {log}\left (-\sqrt {a^{2}+b^{2}}+\tan \left (\frac {z}{2}\right ) b -a \right ) a}{2}+\frac {\mathrm {log}\left (\sqrt {a^{2}+b^{2}}+\tan \left (\frac {z}{2}\right ) b +a \right ) a}{2}-b z \] Input:
int((-A^2-B^2)*cos(z)^2/B/(1-(A^2+B^2)*sin(z)^2/B^2),z)
Output:
(2*atan((tan(z/2)*b*i)/(sqrt(a**2 + b**2) - a))*a*i - log( - sqrt(a**2 + b **2) + tan(z/2)*b - a)*a + log(sqrt(a**2 + b**2) + tan(z/2)*b + a)*a - 2*b *z)/2