Integrand size = 11, antiderivative size = 203 \[ \int \frac {1}{(\cos (3 x)+\cos (5 x))^2} \, dx=\frac {1}{8} \sqrt {\frac {1}{2} \left (17-12 \sqrt {2}\right )} \log \left (\sqrt {3-2 \sqrt {2}} \cos (x)-\sin (x)\right )+\frac {1}{8} \sqrt {\frac {1}{2} \left (17+12 \sqrt {2}\right )} \log \left (\sqrt {3+2 \sqrt {2}} \cos (x)-\sin (x)\right )-\frac {1}{8} \sqrt {\frac {1}{2} \left (17-12 \sqrt {2}\right )} \log \left (\sqrt {3-2 \sqrt {2}} \cos (x)+\sin (x)\right )-\frac {1}{8} \sqrt {\frac {1}{2} \left (17+12 \sqrt {2}\right )} \log \left (\sqrt {3+2 \sqrt {2}} \cos (x)+\sin (x)\right )+\frac {\tan (x)}{4}+\frac {\tan (x) \left (1-3 \tan ^2(x)\right )}{2 \left (1-6 \tan ^2(x)+\tan ^4(x)\right )} \] Output:
1/8*(3/2*2^(1/2)-2)*ln((2^(1/2)-1)*cos(x)-sin(x))+1/8*(3/2*2^(1/2)+2)*ln(( 1+2^(1/2))*cos(x)-sin(x))-1/8*(3/2*2^(1/2)-2)*ln((2^(1/2)-1)*cos(x)+sin(x) )-1/8*(3/2*2^(1/2)+2)*ln((1+2^(1/2))*cos(x)+sin(x))+1/4*tan(x)+tan(x)*(1-3 *tan(x)^2)/(2-12*tan(x)^2+2*tan(x)^4)
Result contains complex when optimal does not.
Time = 0.19 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.81 \[ \int \frac {1}{(\cos (3 x)+\cos (5 x))^2} \, dx=-\frac {1}{2} i \arctan \left (\frac {\sec (x) \left (-2 i \cos (x)-i \sqrt {2} \sin (x)\right )}{\sqrt {2}}\right )+\frac {1}{2} i \arctan \left (\frac {\sec (x) \left (-2 i \cos (x)+i \sqrt {2} \sin (x)\right )}{\sqrt {2}}\right )+\frac {3 \log \left (\sqrt {2}-2 \sin (2 x)\right )}{8 \sqrt {2}}-\frac {3 \log \left (\sqrt {2}+2 \sin (2 x)\right )}{8 \sqrt {2}}+\frac {-1+2 \sin (2 x)}{8 (\cos (2 x)-\sin (2 x))}+\frac {1+2 \sin (2 x)}{8 (\cos (2 x)+\sin (2 x))}+\frac {\tan (x)}{4} \] Input:
Integrate[(Cos[3*x] + Cos[5*x])^(-2),x]
Output:
(-1/2*I)*ArcTan[(Sec[x]*((-2*I)*Cos[x] - I*Sqrt[2]*Sin[x]))/Sqrt[2]] + (I/ 2)*ArcTan[(Sec[x]*((-2*I)*Cos[x] + I*Sqrt[2]*Sin[x]))/Sqrt[2]] + (3*Log[Sq rt[2] - 2*Sin[2*x]])/(8*Sqrt[2]) - (3*Log[Sqrt[2] + 2*Sin[2*x]])/(8*Sqrt[2 ]) + (-1 + 2*Sin[2*x])/(8*(Cos[2*x] - Sin[2*x])) + (1 + 2*Sin[2*x])/(8*(Co s[2*x] + Sin[2*x])) + Tan[x]/4
Time = 0.37 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.52, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {3042, 4823, 27, 1517, 27, 2205, 2009}
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.
| \(\displaystyle \int \frac {1}{(\cos (3 x)+\cos (5 x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(\cos (3 x)+\cos (5 x))^2}dx\) |
| \(\Big \downarrow \) 4823 |
| \(\displaystyle \int \frac {\left (\tan ^2(x)+1\right )^4}{4 \left (\tan ^4(x)-6 \tan ^2(x)+1\right )^2}d\tan (x)\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \int \frac {\left (\tan ^2(x)+1\right )^4}{\left (\tan ^4(x)-6 \tan ^2(x)+1\right )^2}d\tan (x)\) |
| \(\Big \downarrow \) 1517 |
| \(\displaystyle \frac {1}{4} \left (\frac {2 \tan (x) \left (1-3 \tan ^2(x)\right )}{\tan ^4(x)-6 \tan ^2(x)+1}-\frac {1}{64} \int \frac {64 \left (-\tan ^4(x)-4 \tan ^2(x)+1\right )}{\tan ^4(x)-6 \tan ^2(x)+1}d\tan (x)\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \left (\frac {2 \tan (x) \left (1-3 \tan ^2(x)\right )}{\tan ^4(x)-6 \tan ^2(x)+1}-\int \frac {-\tan ^4(x)-4 \tan ^2(x)+1}{\tan ^4(x)-6 \tan ^2(x)+1}d\tan (x)\right )\) |
| \(\Big \downarrow \) 2205 |
| \(\displaystyle \frac {1}{4} \left (\frac {2 \tan (x) \left (1-3 \tan ^2(x)\right )}{\tan ^4(x)-6 \tan ^2(x)+1}-\int \left (\frac {2 \left (1-5 \tan ^2(x)\right )}{\tan ^4(x)-6 \tan ^2(x)+1}-1\right )d\tan (x)\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{4} \left (-\sqrt {\frac {1}{2} \left (17-12 \sqrt {2}\right )} \text {arctanh}\left (\frac {\tan (x)}{\sqrt {3-2 \sqrt {2}}}\right )-\sqrt {\frac {1}{2} \left (17+12 \sqrt {2}\right )} \text {arctanh}\left (\frac {\tan (x)}{\sqrt {3+2 \sqrt {2}}}\right )+\frac {2 \tan (x) \left (1-3 \tan ^2(x)\right )}{\tan ^4(x)-6 \tan ^2(x)+1}+\tan (x)\right )\) |
Input:
Int[(Cos[3*x] + Cos[5*x])^(-2),x]
Output:
(-(Sqrt[(17 - 12*Sqrt[2])/2]*ArcTanh[Tan[x]/Sqrt[3 - 2*Sqrt[2]]]) - Sqrt[( 17 + 12*Sqrt[2])/2]*ArcTanh[Tan[x]/Sqrt[3 + 2*Sqrt[2]]] + Tan[x] + (2*Tan[ x]*(1 - 3*Tan[x]^2))/(1 - 6*Tan[x]^2 + Tan[x]^4))/4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x _Symbol] :> With[{f = Coeff[PolynomialRemainder[(d + e*x^2)^q, a + b*x^2 + c*x^4, x], x, 0], g = Coeff[PolynomialRemainder[(d + e*x^2)^q, a + b*x^2 + c*x^4, x], x, 2]}, Simp[x*(a + b*x^2 + c*x^4)^(p + 1)*((a*b*g - f*(b^2 - 2* a*c) - c*(b*f - 2*a*g)*x^2)/(2*a*(p + 1)*(b^2 - 4*a*c))), x] + Simp[1/(2*a* (p + 1)*(b^2 - 4*a*c)) Int[(a + b*x^2 + c*x^4)^(p + 1)*ExpandToSum[2*a*(p + 1)*(b^2 - 4*a*c)*PolynomialQuotient[(d + e*x^2)^q, a + b*x^2 + c*x^4, x] + b^2*f*(2*p + 3) - 2*a*c*f*(4*p + 5) - a*b*g + c*(4*p + 7)*(b*f - 2*a*g)* x^2, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ [c*d^2 - b*d*e + a*e^2, 0] && IGtQ[q, 1] && LtQ[p, -1]
Int[(Px_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[ExpandInte grand[Px/(a + b*x^2 + c*x^4), x], x] /; FreeQ[{a, b, c}, x] && PolyQ[Px, x^ 2] && Expon[Px, x^2] > 1
Int[(cos[(m_.)*((c_.) + (d_.)*(x_))]*(a_.) + cos[(n_.)*((c_.) + (d_.)*(x_)) ]*(b_.))^(p_), x_Symbol] :> Simp[1/d Subst[Int[Simplify[TrigExpand[a*Cos[ m*ArcTan[x]] + b*Cos[n*ArcTan[x]]]]^p/(1 + x^2), x], x, Tan[c + d*x]], x] / ; FreeQ[{a, b, c, d}, x] && ILtQ[p/2, 0] && IntegerQ[(m - 1)/2] && IntegerQ [(n - 1)/2]
Time = 1.11 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.52
| method | result | size |
| default | \(\frac {\tan \left (x \right )}{4}+\frac {-\frac {3 \tan \left (x \right )}{2}-\frac {1}{2}}{2 \tan \left (x \right )^{2}-4 \tan \left (x \right )-2}+\frac {\ln \left (\tan \left (x \right )^{2}-2 \tan \left (x \right )-1\right )}{4}-\frac {3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (2 \tan \left (x \right )-2\right ) \sqrt {2}}{4}\right )}{8}-\frac {\frac {3 \tan \left (x \right )}{2}-\frac {1}{2}}{2 \left (\tan \left (x \right )^{2}+2 \tan \left (x \right )-1\right )}-\frac {\ln \left (\tan \left (x \right )^{2}+2 \tan \left (x \right )-1\right )}{4}-\frac {3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (2 \tan \left (x \right )+2\right ) \sqrt {2}}{4}\right )}{8}\) | \(106\) |
| risch | \(\frac {i \left (3 \,{\mathrm e}^{8 i x}+{\mathrm e}^{6 i x}-{\mathrm e}^{4 i x}+{\mathrm e}^{2 i x}+4\right )}{4 \,{\mathrm e}^{10 i x}+4 \,{\mathrm e}^{8 i x}+4 \,{\mathrm e}^{2 i x}+4}-\frac {\ln \left ({\mathrm e}^{2 i x}-\frac {i \sqrt {2}}{2}-\frac {\sqrt {2}}{2}\right )}{4}+\frac {3 \ln \left ({\mathrm e}^{2 i x}-\frac {i \sqrt {2}}{2}-\frac {\sqrt {2}}{2}\right ) \sqrt {2}}{16}-\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {i \sqrt {2}}{2}+\frac {\sqrt {2}}{2}\right )}{4}-\frac {3 \ln \left ({\mathrm e}^{2 i x}+\frac {i \sqrt {2}}{2}+\frac {\sqrt {2}}{2}\right ) \sqrt {2}}{16}+\frac {\ln \left ({\mathrm e}^{2 i x}-\frac {i \sqrt {2}}{2}+\frac {\sqrt {2}}{2}\right )}{4}+\frac {3 \ln \left ({\mathrm e}^{2 i x}-\frac {i \sqrt {2}}{2}+\frac {\sqrt {2}}{2}\right ) \sqrt {2}}{16}+\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {i \sqrt {2}}{2}-\frac {\sqrt {2}}{2}\right )}{4}-\frac {3 \ln \left ({\mathrm e}^{2 i x}+\frac {i \sqrt {2}}{2}-\frac {\sqrt {2}}{2}\right ) \sqrt {2}}{16}\) | \(222\) |
Input:
int(1/(cos(3*x)+cos(5*x))^2,x,method=_RETURNVERBOSE)
Output:
1/4*tan(x)+1/2*(-3/2*tan(x)-1/2)/(tan(x)^2-2*tan(x)-1)+1/4*ln(tan(x)^2-2*t an(x)-1)-3/8*2^(1/2)*arctanh(1/4*(2*tan(x)-2)*2^(1/2))-1/2*(3/2*tan(x)-1/2 )/(tan(x)^2+2*tan(x)-1)-1/4*ln(tan(x)^2+2*tan(x)-1)-3/8*2^(1/2)*arctanh(1/ 4*(2*tan(x)+2)*2^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 276 vs. \(2 (119) = 238\).
Time = 0.10 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.36 \[ \int \frac {1}{(\cos (3 x)+\cos (5 x))^2} \, dx=\frac {4 \, {\left (8 \, \cos \left (x\right )^{5} - 8 \, \cos \left (x\right )^{3} + \cos \left (x\right )\right )} \log \left (4 \, {\left (2 \, \cos \left (x\right )^{3} - \cos \left (x\right )\right )} \sin \left (x\right ) + 1\right ) - 4 \, {\left (8 \, \cos \left (x\right )^{5} - 8 \, \cos \left (x\right )^{3} + \cos \left (x\right )\right )} \log \left (-4 \, {\left (2 \, \cos \left (x\right )^{3} - \cos \left (x\right )\right )} \sin \left (x\right ) + 1\right ) + 3 \, {\left (8 \, \sqrt {2} \cos \left (x\right )^{5} - 8 \, \sqrt {2} \cos \left (x\right )^{3} + \sqrt {2} \cos \left (x\right )\right )} \log \left (\frac {4 \, \sqrt {2} \cos \left (x\right )^{2} - 4 \, {\left (2 \, \cos \left (x\right )^{3} + {\left (\sqrt {2} - 1\right )} \cos \left (x\right )\right )} \sin \left (x\right ) - 2 \, \sqrt {2} + 3}{4 \, {\left (2 \, \cos \left (x\right )^{3} - \cos \left (x\right )\right )} \sin \left (x\right ) + 1}\right ) + 3 \, {\left (8 \, \sqrt {2} \cos \left (x\right )^{5} - 8 \, \sqrt {2} \cos \left (x\right )^{3} + \sqrt {2} \cos \left (x\right )\right )} \log \left (-\frac {4 \, \sqrt {2} \cos \left (x\right )^{2} - 4 \, {\left (2 \, \cos \left (x\right )^{3} - {\left (\sqrt {2} + 1\right )} \cos \left (x\right )\right )} \sin \left (x\right ) - 2 \, \sqrt {2} - 3}{4 \, {\left (2 \, \cos \left (x\right )^{3} - \cos \left (x\right )\right )} \sin \left (x\right ) - 1}\right ) + 8 \, {\left (16 \, \cos \left (x\right )^{4} - 14 \, \cos \left (x\right )^{2} + 1\right )} \sin \left (x\right )}{32 \, {\left (8 \, \cos \left (x\right )^{5} - 8 \, \cos \left (x\right )^{3} + \cos \left (x\right )\right )}} \] Input:
integrate(1/(cos(3*x)+cos(5*x))^2,x, algorithm="fricas")
Output:
1/32*(4*(8*cos(x)^5 - 8*cos(x)^3 + cos(x))*log(4*(2*cos(x)^3 - cos(x))*sin (x) + 1) - 4*(8*cos(x)^5 - 8*cos(x)^3 + cos(x))*log(-4*(2*cos(x)^3 - cos(x ))*sin(x) + 1) + 3*(8*sqrt(2)*cos(x)^5 - 8*sqrt(2)*cos(x)^3 + sqrt(2)*cos( x))*log((4*sqrt(2)*cos(x)^2 - 4*(2*cos(x)^3 + (sqrt(2) - 1)*cos(x))*sin(x) - 2*sqrt(2) + 3)/(4*(2*cos(x)^3 - cos(x))*sin(x) + 1)) + 3*(8*sqrt(2)*cos (x)^5 - 8*sqrt(2)*cos(x)^3 + sqrt(2)*cos(x))*log(-(4*sqrt(2)*cos(x)^2 - 4* (2*cos(x)^3 - (sqrt(2) + 1)*cos(x))*sin(x) - 2*sqrt(2) - 3)/(4*(2*cos(x)^3 - cos(x))*sin(x) - 1)) + 8*(16*cos(x)^4 - 14*cos(x)^2 + 1)*sin(x))/(8*cos (x)^5 - 8*cos(x)^3 + cos(x))
\[ \int \frac {1}{(\cos (3 x)+\cos (5 x))^2} \, dx=\int \frac {1}{\left (\cos {\left (3 x \right )} + \cos {\left (5 x \right )}\right )^{2}}\, dx \] Input:
integrate(1/(cos(3*x)+cos(5*x))**2,x)
Output:
Integral((cos(3*x) + cos(5*x))**(-2), x)
Leaf count of result is larger than twice the leaf count of optimal. 1264 vs. \(2 (119) = 238\).
Time = 0.15 (sec) , antiderivative size = 1264, normalized size of antiderivative = 6.23 \[ \int \frac {1}{(\cos (3 x)+\cos (5 x))^2} \, dx=\text {Too large to display} \] Input:
integrate(1/(cos(3*x)+cos(5*x))^2,x, algorithm="maxima")
Output:
-1/32*(8*(3*sin(8*x) + sin(6*x) - sin(4*x) + sin(2*x))*cos(10*x) + 8*(sin( 6*x) - sin(4*x) - 2*sin(2*x))*cos(8*x) + ((3*sqrt(2) + 4)*cos(10*x)^2 + (3 *sqrt(2) + 4)*cos(8*x)^2 + (3*sqrt(2) + 4)*cos(2*x)^2 + (3*sqrt(2) + 4)*si n(10*x)^2 + (3*sqrt(2) + 4)*sin(8*x)^2 + 2*(3*sqrt(2) + 4)*sin(8*x)*sin(2* x) + (3*sqrt(2) + 4)*sin(2*x)^2 + 2*((3*sqrt(2) + 4)*cos(8*x) + (3*sqrt(2) + 4)*cos(2*x) + 3*sqrt(2) + 4)*cos(10*x) + 2*((3*sqrt(2) + 4)*cos(2*x) + 3*sqrt(2) + 4)*cos(8*x) + 2*(3*sqrt(2) + 4)*cos(2*x) + 2*((3*sqrt(2) + 4)* sin(8*x) + (3*sqrt(2) + 4)*sin(2*x))*sin(10*x) + 3*sqrt(2) + 4)*log(2*cos( 2*x)^2 + 2*sin(2*x)^2 + 2*sqrt(2)*cos(2*x) + 2*sqrt(2)*sin(2*x) + 2) - ((3 *sqrt(2) + 4)*cos(10*x)^2 + (3*sqrt(2) + 4)*cos(8*x)^2 + (3*sqrt(2) + 4)*c os(2*x)^2 + (3*sqrt(2) + 4)*sin(10*x)^2 + (3*sqrt(2) + 4)*sin(8*x)^2 + 2*( 3*sqrt(2) + 4)*sin(8*x)*sin(2*x) + (3*sqrt(2) + 4)*sin(2*x)^2 + 2*((3*sqrt (2) + 4)*cos(8*x) + (3*sqrt(2) + 4)*cos(2*x) + 3*sqrt(2) + 4)*cos(10*x) + 2*((3*sqrt(2) + 4)*cos(2*x) + 3*sqrt(2) + 4)*cos(8*x) + 2*(3*sqrt(2) + 4)* cos(2*x) + 2*((3*sqrt(2) + 4)*sin(8*x) + (3*sqrt(2) + 4)*sin(2*x))*sin(10* x) + 3*sqrt(2) + 4)*log(2*cos(2*x)^2 + 2*sin(2*x)^2 + 2*sqrt(2)*cos(2*x) - 2*sqrt(2)*sin(2*x) + 2) + ((3*sqrt(2) - 4)*cos(10*x)^2 + (3*sqrt(2) - 4)* cos(8*x)^2 + (3*sqrt(2) - 4)*cos(2*x)^2 + (3*sqrt(2) - 4)*sin(10*x)^2 + (3 *sqrt(2) - 4)*sin(8*x)^2 + 2*(3*sqrt(2) - 4)*sin(8*x)*sin(2*x) + (3*sqrt(2 ) - 4)*sin(2*x)^2 + 2*((3*sqrt(2) - 4)*cos(8*x) + (3*sqrt(2) - 4)*cos(2...
Time = 0.15 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.62 \[ \int \frac {1}{(\cos (3 x)+\cos (5 x))^2} \, dx=\frac {3}{16} \, \sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 2 \, \tan \left (x\right ) + 2 \right |}}{{\left | 2 \, \sqrt {2} + 2 \, \tan \left (x\right ) + 2 \right |}}\right ) + \frac {3}{16} \, \sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 2 \, \tan \left (x\right ) - 2 \right |}}{{\left | 2 \, \sqrt {2} + 2 \, \tan \left (x\right ) - 2 \right |}}\right ) - \frac {3 \, \tan \left (x\right )^{3} - \tan \left (x\right )}{2 \, {\left (\tan \left (x\right )^{4} - 6 \, \tan \left (x\right )^{2} + 1\right )}} - \frac {1}{4} \, \log \left ({\left | \tan \left (x\right )^{2} + 2 \, \tan \left (x\right ) - 1 \right |}\right ) + \frac {1}{4} \, \log \left ({\left | \tan \left (x\right )^{2} - 2 \, \tan \left (x\right ) - 1 \right |}\right ) + \frac {1}{4} \, \tan \left (x\right ) \] Input:
integrate(1/(cos(3*x)+cos(5*x))^2,x, algorithm="giac")
Output:
3/16*sqrt(2)*log(abs(-2*sqrt(2) + 2*tan(x) + 2)/abs(2*sqrt(2) + 2*tan(x) + 2)) + 3/16*sqrt(2)*log(abs(-2*sqrt(2) + 2*tan(x) - 2)/abs(2*sqrt(2) + 2*t an(x) - 2)) - 1/2*(3*tan(x)^3 - tan(x))/(tan(x)^4 - 6*tan(x)^2 + 1) - 1/4* log(abs(tan(x)^2 + 2*tan(x) - 1)) + 1/4*log(abs(tan(x)^2 - 2*tan(x) - 1)) + 1/4*tan(x)
Time = 21.07 (sec) , antiderivative size = 645, normalized size of antiderivative = 3.18 \[ \int \frac {1}{(\cos (3 x)+\cos (5 x))^2} \, dx=\text {Too large to display} \] Input:
int(1/(cos(3*x) + cos(5*x))^2,x)
Output:
atanh(- (492847497216*tan(x/2))/(420671913984*2^(1/2) - 420671913984*2^(1/ 2)*tan(x/2)^2 - 594920079360*tan(x/2)^2 + 594920079360) - (348496330752*2^ (1/2)*tan(x/2))/(420671913984*2^(1/2) - 420671913984*2^(1/2)*tan(x/2)^2 - 594920079360*tan(x/2)^2 + 594920079360))/2 + atanh((16384*tan(x/2))/(8192* tan(x/2)^4 - 16384*tan(x/2)^2 + 8192) - (16384*tan(x/2)^3)/(8192*tan(x/2)^ 4 - 16384*tan(x/2)^2 + 8192) - (16384*2^(1/2)*tan(x/2)^3)/(8192*tan(x/2)^4 - 16384*tan(x/2)^2 + 8192) + (16384*2^(1/2)*tan(x/2))/(8192*tan(x/2)^4 - 16384*tan(x/2)^2 + 8192))/2 - (3*tan(x/2))/(2*(29*tan(x/2)^2 - 98*tan(x/2) ^4 + 98*tan(x/2)^6 - 29*tan(x/2)^8 + tan(x/2)^10 - 1)) + (30*tan(x/2)^3)/( 29*tan(x/2)^2 - 98*tan(x/2)^4 + 98*tan(x/2)^6 - 29*tan(x/2)^8 + tan(x/2)^1 0 - 1) - (65*tan(x/2)^5)/(29*tan(x/2)^2 - 98*tan(x/2)^4 + 98*tan(x/2)^6 - 29*tan(x/2)^8 + tan(x/2)^10 - 1) + (30*tan(x/2)^7)/(29*tan(x/2)^2 - 98*tan (x/2)^4 + 98*tan(x/2)^6 - 29*tan(x/2)^8 + tan(x/2)^10 - 1) - (3*tan(x/2)^9 )/(2*(29*tan(x/2)^2 - 98*tan(x/2)^4 + 98*tan(x/2)^6 - 29*tan(x/2)^8 + tan( x/2)^10 - 1)) + (3*2^(1/2)*atanh(- (492847497216*tan(x/2))/(420671913984*2 ^(1/2) - 420671913984*2^(1/2)*tan(x/2)^2 - 594920079360*tan(x/2)^2 + 59492 0079360) - (348496330752*2^(1/2)*tan(x/2))/(420671913984*2^(1/2) - 4206719 13984*2^(1/2)*tan(x/2)^2 - 594920079360*tan(x/2)^2 + 594920079360)))/8 - ( 3*2^(1/2)*atanh((16384*tan(x/2))/(8192*tan(x/2)^4 - 16384*tan(x/2)^2 + 819 2) - (16384*tan(x/2)^3)/(8192*tan(x/2)^4 - 16384*tan(x/2)^2 + 8192) - (...
\[ \int \frac {1}{(\cos (3 x)+\cos (5 x))^2} \, dx=\int \frac {1}{\cos \left (5 x \right )^{2}+2 \cos \left (5 x \right ) \cos \left (3 x \right )+\cos \left (3 x \right )^{2}}d x \] Input:
int(1/(cos(3*x)+cos(5*x))^2,x)
Output:
int(1/(cos(5*x)**2 + 2*cos(5*x)*cos(3*x) + cos(3*x)**2),x)