Integrand size = 9, antiderivative size = 74 \[ \int \frac {1}{(\cos (x)+\cos (3 x))^4} \, dx=-\frac {5}{8} \text {arctanh}(2 \cos (x) \sin (x))+\frac {9 \tan (x)}{16}+\frac {\tan ^3(x)}{48}+\frac {\tan (x)}{3 \left (1-\tan ^2(x)\right )^3}-\frac {5 \tan (x)}{6 \left (1-\tan ^2(x)\right )^2}+\frac {5 \tan (x)}{4 \left (1-\tan ^2(x)\right )} \] Output:
-5/8*arctanh(2*cos(x)*sin(x))+9/16*tan(x)+1/48*tan(x)^3+1/3*tan(x)/(1-tan( x)^2)^3-5/6*tan(x)/(1-tan(x)^2)^2+5*tan(x)/(4-4*tan(x)^2)
Time = 0.38 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.08 \[ \int \frac {1}{(\cos (x)+\cos (3 x))^4} \, dx=\frac {1}{96} \left (60 \log (\cos (x)-\sin (x))-60 \log (\cos (x)+\sin (x))-\frac {5}{(\cos (x)-\sin (x))^2}+2 (25 \cos (x)+10 \cos (3 x)+11 \cos (5 x)) \sec ^3(2 x) \sin (x)+\frac {5}{(\cos (x)+\sin (x))^2}+2 \left (26+\sec ^2(x)\right ) \tan (x)\right ) \] Input:
Integrate[(Cos[x] + Cos[3*x])^(-4),x]
Output:
(60*Log[Cos[x] - Sin[x]] - 60*Log[Cos[x] + Sin[x]] - 5/(Cos[x] - Sin[x])^2 + 2*(25*Cos[x] + 10*Cos[3*x] + 11*Cos[5*x])*Sec[2*x]^3*Sin[x] + 5/(Cos[x] + Sin[x])^2 + 2*(26 + Sec[x]^2)*Tan[x])/96
Time = 0.26 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.92, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {3042, 4823, 27, 300, 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.
\(\displaystyle \int \frac {1}{(\cos (x)+\cos (3 x))^4} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(\cos (x)+\cos (3 x))^4}dx\) |
\(\Big \downarrow \) 4823 |
\(\displaystyle \int \frac {\left (\tan ^2(x)+1\right )^5}{16 \left (1-\tan ^2(x)\right )^4}d\tan (x)\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{16} \int \frac {\left (\tan ^2(x)+1\right )^5}{\left (1-\tan ^2(x)\right )^4}d\tan (x)\) |
\(\Big \downarrow \) 300 |
\(\displaystyle \frac {1}{16} \int \left (\tan ^2(x)-\frac {8 \left (-5 \tan ^6(x)+5 \tan ^4(x)-5 \tan ^2(x)+1\right )}{\left (1-\tan ^2(x)\right )^4}+9\right )d\tan (x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{16} \left (-20 \text {arctanh}(\tan (x))+\frac {\tan ^3(x)}{3}+\frac {20 \tan (x)}{1-\tan ^2(x)}-\frac {40 \tan (x)}{3 \left (1-\tan ^2(x)\right )^2}+\frac {16 \tan (x)}{3 \left (1-\tan ^2(x)\right )^3}+9 \tan (x)\right )\) |
Input:
Int[(Cos[x] + Cos[3*x])^(-4),x]
Output:
(-20*ArcTanh[Tan[x]] + 9*Tan[x] + Tan[x]^3/3 + (16*Tan[x])/(3*(1 - Tan[x]^ 2)^3) - (40*Tan[x])/(3*(1 - Tan[x]^2)^2) + (20*Tan[x])/(1 - Tan[x]^2))/16
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)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int [PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c , d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
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 = 13.26 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00
method | result | size |
default | \(\frac {\tan \left (x \right )^{3}}{48}+\frac {9 \tan \left (x \right )}{16}-\frac {1}{24 \left (\tan \left (x \right )-1\right )^{3}}-\frac {3}{16 \left (\tan \left (x \right )-1\right )^{2}}-\frac {5}{8 \left (\tan \left (x \right )-1\right )}+\frac {5 \ln \left (\tan \left (x \right )-1\right )}{8}-\frac {1}{24 \left (\tan \left (x \right )+1\right )^{3}}+\frac {3}{16 \left (\tan \left (x \right )+1\right )^{2}}-\frac {5}{8 \left (\tan \left (x \right )+1\right )}-\frac {5 \ln \left (\tan \left (x \right )+1\right )}{8}\) | \(74\) |
risch | \(\frac {i \left (15 \,{\mathrm e}^{16 i x}+45 \,{\mathrm e}^{14 i x}+85 \,{\mathrm e}^{12 i x}+135 \,{\mathrm e}^{10 i x}+153 \,{\mathrm e}^{8 i x}+155 \,{\mathrm e}^{6 i x}+99 \,{\mathrm e}^{4 i x}+57 \,{\mathrm e}^{2 i x}+24\right )}{12 \left ({\mathrm e}^{6 i x}+{\mathrm e}^{4 i x}+{\mathrm e}^{2 i x}+1\right )^{3}}+\frac {5 \ln \left ({\mathrm e}^{2 i x}-i\right )}{8}-\frac {5 \ln \left ({\mathrm e}^{2 i x}+i\right )}{8}\) | \(104\) |
Input:
int(1/(cos(x)+cos(3*x))^4,x,method=_RETURNVERBOSE)
Output:
1/48*tan(x)^3+9/16*tan(x)-1/24/(tan(x)-1)^3-3/16/(tan(x)-1)^2-5/8/(tan(x)- 1)+5/8*ln(tan(x)-1)-1/24/(tan(x)+1)^3+3/16/(tan(x)+1)^2-5/8/(tan(x)+1)-5/8 *ln(tan(x)+1)
Leaf count of result is larger than twice the leaf count of optimal. 132 vs. \(2 (56) = 112\).
Time = 0.09 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.78 \[ \int \frac {1}{(\cos (x)+\cos (3 x))^4} \, dx=-\frac {15 \, {\left (8 \, \cos \left (x\right )^{9} - 12 \, \cos \left (x\right )^{7} + 6 \, \cos \left (x\right )^{5} - \cos \left (x\right )^{3}\right )} \log \left (2 \, \cos \left (x\right ) \sin \left (x\right ) + 1\right ) - 15 \, {\left (8 \, \cos \left (x\right )^{9} - 12 \, \cos \left (x\right )^{7} + 6 \, \cos \left (x\right )^{5} - \cos \left (x\right )^{3}\right )} \log \left (-2 \, \cos \left (x\right ) \sin \left (x\right ) + 1\right ) - {\left (384 \, \cos \left (x\right )^{8} - 504 \, \cos \left (x\right )^{6} + 204 \, \cos \left (x\right )^{4} - 20 \, \cos \left (x\right )^{2} - 1\right )} \sin \left (x\right )}{48 \, {\left (8 \, \cos \left (x\right )^{9} - 12 \, \cos \left (x\right )^{7} + 6 \, \cos \left (x\right )^{5} - \cos \left (x\right )^{3}\right )}} \] Input:
integrate(1/(cos(x)+cos(3*x))^4,x, algorithm="fricas")
Output:
-1/48*(15*(8*cos(x)^9 - 12*cos(x)^7 + 6*cos(x)^5 - cos(x)^3)*log(2*cos(x)* sin(x) + 1) - 15*(8*cos(x)^9 - 12*cos(x)^7 + 6*cos(x)^5 - cos(x)^3)*log(-2 *cos(x)*sin(x) + 1) - (384*cos(x)^8 - 504*cos(x)^6 + 204*cos(x)^4 - 20*cos (x)^2 - 1)*sin(x))/(8*cos(x)^9 - 12*cos(x)^7 + 6*cos(x)^5 - cos(x)^3)
\[ \int \frac {1}{(\cos (x)+\cos (3 x))^4} \, dx=\int \frac {1}{\left (\cos {\left (x \right )} + \cos {\left (3 x \right )}\right )^{4}}\, dx \] Input:
integrate(1/(cos(x)+cos(3*x))**4,x)
Output:
Integral((cos(x) + cos(3*x))**(-4), x)
Leaf count of result is larger than twice the leaf count of optimal. 4157 vs. \(2 (56) = 112\).
Time = 0.25 (sec) , antiderivative size = 4157, normalized size of antiderivative = 56.18 \[ \int \frac {1}{(\cos (x)+\cos (3 x))^4} \, dx=\text {Too large to display} \] Input:
integrate(1/(cos(x)+cos(3*x))^4,x, algorithm="maxima")
Output:
-1/48*(4*(15*sin(16*x) + 45*sin(14*x) + 85*sin(12*x) + 135*sin(10*x) + 153 *sin(8*x) + 155*sin(6*x) + 99*sin(4*x) + 57*sin(2*x))*cos(18*x) + 12*(15*s in(14*x) + 35*sin(12*x) + 75*sin(10*x) + 93*sin(8*x) + 105*sin(6*x) + 69*s in(4*x) + 42*sin(2*x))*cos(16*x) + 12*(20*sin(12*x) + 90*sin(10*x) + 126*s in(8*x) + 160*sin(6*x) + 108*sin(4*x) + 69*sin(2*x))*cos(14*x) + 20*(66*si n(10*x) + 102*sin(8*x) + 140*sin(6*x) + 96*sin(4*x) + 63*sin(2*x))*cos(12* x) + 12*(72*sin(8*x) + 170*sin(6*x) + 126*sin(4*x) + 93*sin(2*x))*cos(10*x ) + 60*(22*sin(6*x) + 18*sin(4*x) + 15*sin(2*x))*cos(8*x) + 60*(4*sin(4*x) + 7*sin(2*x))*cos(6*x) - 15*(2*(3*cos(16*x) + 6*cos(14*x) + 10*cos(12*x) + 12*cos(10*x) + 12*cos(8*x) + 10*cos(6*x) + 6*cos(4*x) + 3*cos(2*x) + 1)* cos(18*x) + cos(18*x)^2 + 6*(6*cos(14*x) + 10*cos(12*x) + 12*cos(10*x) + 1 2*cos(8*x) + 10*cos(6*x) + 6*cos(4*x) + 3*cos(2*x) + 1)*cos(16*x) + 9*cos( 16*x)^2 + 12*(10*cos(12*x) + 12*cos(10*x) + 12*cos(8*x) + 10*cos(6*x) + 6* cos(4*x) + 3*cos(2*x) + 1)*cos(14*x) + 36*cos(14*x)^2 + 20*(12*cos(10*x) + 12*cos(8*x) + 10*cos(6*x) + 6*cos(4*x) + 3*cos(2*x) + 1)*cos(12*x) + 100* cos(12*x)^2 + 24*(12*cos(8*x) + 10*cos(6*x) + 6*cos(4*x) + 3*cos(2*x) + 1) *cos(10*x) + 144*cos(10*x)^2 + 24*(10*cos(6*x) + 6*cos(4*x) + 3*cos(2*x) + 1)*cos(8*x) + 144*cos(8*x)^2 + 20*(6*cos(4*x) + 3*cos(2*x) + 1)*cos(6*x) + 100*cos(6*x)^2 + 12*(3*cos(2*x) + 1)*cos(4*x) + 36*cos(4*x)^2 + 9*cos(2* x)^2 + 2*(3*sin(16*x) + 6*sin(14*x) + 10*sin(12*x) + 12*sin(10*x) + 12*...
Time = 0.12 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.73 \[ \int \frac {1}{(\cos (x)+\cos (3 x))^4} \, dx=\frac {1}{48} \, \tan \left (x\right )^{3} - \frac {15 \, \tan \left (x\right )^{5} - 20 \, \tan \left (x\right )^{3} + 9 \, \tan \left (x\right )}{12 \, {\left (\tan \left (x\right )^{2} - 1\right )}^{3}} - \frac {5}{8} \, \log \left ({\left | \tan \left (x\right ) + 1 \right |}\right ) + \frac {5}{8} \, \log \left ({\left | \tan \left (x\right ) - 1 \right |}\right ) + \frac {9}{16} \, \tan \left (x\right ) \] Input:
integrate(1/(cos(x)+cos(3*x))^4,x, algorithm="giac")
Output:
1/48*tan(x)^3 - 1/12*(15*tan(x)^5 - 20*tan(x)^3 + 9*tan(x))/(tan(x)^2 - 1) ^3 - 5/8*log(abs(tan(x) + 1)) + 5/8*log(abs(tan(x) - 1)) + 9/16*tan(x)
Time = 21.25 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(\cos (x)+\cos (3 x))^4} \, dx=\frac {-384\,\sin \left (x\right )\,{\cos \left (x\right )}^8+504\,\sin \left (x\right )\,{\cos \left (x\right )}^6-204\,\sin \left (x\right )\,{\cos \left (x\right )}^4+20\,\sin \left (x\right )\,{\cos \left (x\right )}^2+\sin \left (x\right )}{-384\,{\cos \left (x\right )}^9+576\,{\cos \left (x\right )}^7-288\,{\cos \left (x\right )}^5+48\,{\cos \left (x\right )}^3}-\frac {5\,\mathrm {atanh}\left (\frac {\sin \left (x\right )}{\cos \left (x\right )}\right )}{4} \] Input:
int(1/(cos(3*x) + cos(x))^4,x)
Output:
(sin(x) + 20*cos(x)^2*sin(x) - 204*cos(x)^4*sin(x) + 504*cos(x)^6*sin(x) - 384*cos(x)^8*sin(x))/(48*cos(x)^3 - 288*cos(x)^5 + 576*cos(x)^7 - 384*cos (x)^9) - (5*atanh(sin(x)/cos(x)))/4
\[ \int \frac {1}{(\cos (x)+\cos (3 x))^4} \, dx=\int \frac {1}{\cos \left (3 x \right )^{4}+4 \cos \left (3 x \right )^{3} \cos \left (x \right )+6 \cos \left (3 x \right )^{2} \cos \left (x \right )^{2}+4 \cos \left (3 x \right ) \cos \left (x \right )^{3}+\cos \left (x \right )^{4}}d x \] Input:
int(1/(cos(x)+cos(3*x))^4,x)
Output:
int(1/(cos(3*x)**4 + 4*cos(3*x)**3*cos(x) + 6*cos(3*x)**2*cos(x)**2 + 4*co s(3*x)*cos(x)**3 + cos(x)**4),x)