Integrand size = 11, antiderivative size = 324 \[ \int \frac {1}{(\cos (3 x)+\cos (5 x))^3} \, dx=\frac {49}{16} \text {arctanh}(\sin (x))+\frac {1}{128} \sqrt {842-391 \sqrt {2}} \text {arctanh}\left (\frac {2 \sin (x)}{\sqrt {2-\sqrt {2}}}\right )+\frac {1}{4} \sqrt {122-71 \sqrt {2}} \text {arctanh}\left (\frac {2 \sin (x)}{\sqrt {2-\sqrt {2}}}\right )-\frac {3}{16} \sqrt {34+7 \sqrt {2}} \text {arctanh}\left (\frac {2 \sin (x)}{\sqrt {2-\sqrt {2}}}\right )+\frac {3}{16} \sqrt {34-7 \sqrt {2}} \text {arctanh}\left (\frac {2 \sin (x)}{\sqrt {2+\sqrt {2}}}\right )-\frac {1}{4} \sqrt {122+71 \sqrt {2}} \text {arctanh}\left (\frac {2 \sin (x)}{\sqrt {2+\sqrt {2}}}\right )-\frac {1}{128} \sqrt {842+391 \sqrt {2}} \text {arctanh}\left (\frac {2 \sin (x)}{\sqrt {2+\sqrt {2}}}\right )+\frac {1}{32 (1-\sin (x))}-\frac {1}{32 (1+\sin (x))}+\frac {\sin (x) \left (1-4 \sin ^2(x)\right )}{8 \left (1-8 \sin ^2(x)+8 \sin ^4(x)\right )^2}-\frac {\sin (x) \left (37-56 \sin ^2(x)\right )}{32 \left (1-8 \sin ^2(x)+8 \sin ^4(x)\right )}+\frac {\sin (x) \left (5-12 \sin ^2(x)\right )}{4 \left (1-8 \sin ^2(x)+8 \sin ^4(x)\right )} \] Output:
49/16*arctanh(sin(x))+1/128*(842-391*2^(1/2))^(1/2)*arctanh(2*sin(x)/(2-2^ (1/2))^(1/2))+1/4*(122-71*2^(1/2))^(1/2)*arctanh(2*sin(x)/(2-2^(1/2))^(1/2 ))-3/16*(34+7*2^(1/2))^(1/2)*arctanh(2*sin(x)/(2-2^(1/2))^(1/2))+3/16*(34- 7*2^(1/2))^(1/2)*arctanh(2*sin(x)/(2+2^(1/2))^(1/2))-1/4*(122+71*2^(1/2))^ (1/2)*arctanh(2*sin(x)/(2+2^(1/2))^(1/2))-1/128*(842+391*2^(1/2))^(1/2)*ar ctanh(2*sin(x)/(2+2^(1/2))^(1/2))+1/(32-32*sin(x))-1/(32+32*sin(x))+1/8*si n(x)*(1-4*sin(x)^2)/(1-8*sin(x)^2+8*sin(x)^4)^2-sin(x)*(37-56*sin(x)^2)/(3 2-256*sin(x)^2+256*sin(x)^4)+sin(x)*(5-12*sin(x)^2)/(4-32*sin(x)^2+32*sin( x)^4)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.61 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.02 \[ \int \frac {1}{(\cos (3 x)+\cos (5 x))^3} \, dx=\frac {1}{512} \left (-1568 \log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+1568 \log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )-\text {RootSum}\left [1+\text {$\#$1}^8\&,\frac {364 \arctan \left (\frac {\sin (x)}{\cos (x)-\text {$\#$1}}\right )-182 i \log \left (1-2 \cos (x) \text {$\#$1}+\text {$\#$1}^2\right )-158 \arctan \left (\frac {\sin (x)}{\cos (x)-\text {$\#$1}}\right ) \text {$\#$1}^2+79 i \log \left (1-2 \cos (x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2-158 \arctan \left (\frac {\sin (x)}{\cos (x)-\text {$\#$1}}\right ) \text {$\#$1}^4+79 i \log \left (1-2 \cos (x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^4+364 \arctan \left (\frac {\sin (x)}{\cos (x)-\text {$\#$1}}\right ) \text {$\#$1}^6-182 i \log \left (1-2 \cos (x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^6}{\text {$\#$1}^7}\&\right ]+\frac {16}{\left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )^2}-\frac {16}{\left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^2}+\frac {32 (\cos (x)-2 \sin (x))}{(\cos (2 x)-\sin (2 x))^2}+\frac {8 (-21 \cos (x)+41 \sin (x))}{\cos (2 x)-\sin (2 x)}-\frac {32 (\cos (x)+2 \sin (x))}{(\cos (2 x)+\sin (2 x))^2}+\frac {8 (21 \cos (x)+41 \sin (x))}{\cos (2 x)+\sin (2 x)}\right ) \] Input:
Integrate[(Cos[3*x] + Cos[5*x])^(-3),x]
Output:
(-1568*Log[Cos[x/2] - Sin[x/2]] + 1568*Log[Cos[x/2] + Sin[x/2]] - RootSum[ 1 + #1^8 & , (364*ArcTan[Sin[x]/(Cos[x] - #1)] - (182*I)*Log[1 - 2*Cos[x]* #1 + #1^2] - 158*ArcTan[Sin[x]/(Cos[x] - #1)]*#1^2 + (79*I)*Log[1 - 2*Cos[ x]*#1 + #1^2]*#1^2 - 158*ArcTan[Sin[x]/(Cos[x] - #1)]*#1^4 + (79*I)*Log[1 - 2*Cos[x]*#1 + #1^2]*#1^4 + 364*ArcTan[Sin[x]/(Cos[x] - #1)]*#1^6 - (182* I)*Log[1 - 2*Cos[x]*#1 + #1^2]*#1^6)/#1^7 & ] + 16/(Cos[x/2] - Sin[x/2])^2 - 16/(Cos[x/2] + Sin[x/2])^2 + (32*(Cos[x] - 2*Sin[x]))/(Cos[2*x] - Sin[2 *x])^2 + (8*(-21*Cos[x] + 41*Sin[x]))/(Cos[2*x] - Sin[2*x]) - (32*(Cos[x] + 2*Sin[x]))/(Cos[2*x] + Sin[2*x])^2 + (8*(21*Cos[x] + 41*Sin[x]))/(Cos[2* x] + Sin[2*x]))/512
Time = 0.55 (sec) , antiderivative size = 319, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 4825, 27, 1567, 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.
| \(\displaystyle \int \frac {1}{(\cos (3 x)+\cos (5 x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(\cos (3 x)+\cos (5 x))^3}dx\) |
| \(\Big \downarrow \) 4825 |
| \(\displaystyle \int \frac {1}{8 \left (1-\sin ^2(x)\right )^2 \left (8 \sin ^4(x)-8 \sin ^2(x)+1\right )^3}d\sin (x)\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{8} \int \frac {1}{\left (1-\sin ^2(x)\right )^2 \left (8 \sin ^4(x)-8 \sin ^2(x)+1\right )^3}d\sin (x)\) |
| \(\Big \downarrow \) 1567 |
| \(\displaystyle \frac {1}{8} \int \left (\frac {8 \left (8 \sin ^2(x)-1\right )}{\left (8 \sin ^4(x)-8 \sin ^2(x)+1\right )^3}-\frac {49}{2 \left (\sin ^2(x)-1\right )}+\frac {8 \left (24 \sin ^2(x)-1\right )}{8 \sin ^4(x)-8 \sin ^2(x)+1}+\frac {1}{4 (\sin (x)-1)^2}+\frac {1}{4 (\sin (x)+1)^2}+\frac {8 \left (16 \sin ^2(x)-1\right )}{\left (8 \sin ^4(x)-8 \sin ^2(x)+1\right )^2}\right )d\sin (x)\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{8} \left (\frac {49}{2} \text {arctanh}(\sin (x))-\frac {3}{2} \sqrt {34+7 \sqrt {2}} \text {arctanh}\left (\frac {2 \sin (x)}{\sqrt {2-\sqrt {2}}}\right )+2 \sqrt {122-71 \sqrt {2}} \text {arctanh}\left (\frac {2 \sin (x)}{\sqrt {2-\sqrt {2}}}\right )+\frac {1}{16} \sqrt {842-391 \sqrt {2}} \text {arctanh}\left (\frac {2 \sin (x)}{\sqrt {2-\sqrt {2}}}\right )-\frac {1}{16} \sqrt {842+391 \sqrt {2}} \text {arctanh}\left (\frac {2 \sin (x)}{\sqrt {2+\sqrt {2}}}\right )-2 \sqrt {122+71 \sqrt {2}} \text {arctanh}\left (\frac {2 \sin (x)}{\sqrt {2+\sqrt {2}}}\right )+\frac {3}{2} \sqrt {34-7 \sqrt {2}} \text {arctanh}\left (\frac {2 \sin (x)}{\sqrt {2+\sqrt {2}}}\right )-\frac {\sin (x) \left (37-56 \sin ^2(x)\right )}{4 \left (8 \sin ^4(x)-8 \sin ^2(x)+1\right )}+\frac {2 \sin (x) \left (5-12 \sin ^2(x)\right )}{8 \sin ^4(x)-8 \sin ^2(x)+1}+\frac {\sin (x) \left (1-4 \sin ^2(x)\right )}{\left (8 \sin ^4(x)-8 \sin ^2(x)+1\right )^2}+\frac {1}{4 (1-\sin (x))}-\frac {1}{4 (\sin (x)+1)}\right )\) |
Input:
Int[(Cos[3*x] + Cos[5*x])^(-3),x]
Output:
((49*ArcTanh[Sin[x]])/2 + (Sqrt[842 - 391*Sqrt[2]]*ArcTanh[(2*Sin[x])/Sqrt [2 - Sqrt[2]]])/16 + 2*Sqrt[122 - 71*Sqrt[2]]*ArcTanh[(2*Sin[x])/Sqrt[2 - Sqrt[2]]] - (3*Sqrt[34 + 7*Sqrt[2]]*ArcTanh[(2*Sin[x])/Sqrt[2 - Sqrt[2]]]) /2 + (3*Sqrt[34 - 7*Sqrt[2]]*ArcTanh[(2*Sin[x])/Sqrt[2 + Sqrt[2]]])/2 - 2* Sqrt[122 + 71*Sqrt[2]]*ArcTanh[(2*Sin[x])/Sqrt[2 + Sqrt[2]]] - (Sqrt[842 + 391*Sqrt[2]]*ArcTanh[(2*Sin[x])/Sqrt[2 + Sqrt[2]]])/16 + 1/(4*(1 - Sin[x] )) - 1/(4*(1 + Sin[x])) + (Sin[x]*(1 - 4*Sin[x]^2))/(1 - 8*Sin[x]^2 + 8*Si n[x]^4)^2 - (Sin[x]*(37 - 56*Sin[x]^2))/(4*(1 - 8*Sin[x]^2 + 8*Sin[x]^4)) + (2*Sin[x]*(5 - 12*Sin[x]^2))/(1 - 8*Sin[x]^2 + 8*Sin[x]^4))/8
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] :> Int[ExpandIntegrand[(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b^2 - 4*a*c, 0] && ((IntegerQ[p] && IntegerQ[q]) || IGtQ[p, 0] || IGtQ[q, 0])
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*ArcSin[x]] + b*Cos[n*ArcSin[x]]]]^p/Sqrt[1 - x^2], x], x, Sin[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[(p - 1)/2, 0] && IntegerQ[(m - 1)/2] & & IntegerQ[(n - 1)/2]
Time = 4.16 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.43
| method | result | size |
| default | \(-\frac {1}{32 \left (\sin \left (x \right )-1\right )}-\frac {49 \ln \left (\sin \left (x \right )-1\right )}{32}+\frac {-10 \sin \left (x \right )^{7}+\frac {43 \sin \left (x \right )^{5}}{4}-\frac {5 \sin \left (x \right )^{3}}{2}+\frac {7 \sin \left (x \right )}{32}}{\left (1-8 \sin \left (x \right )^{2}+8 \sin \left (x \right )^{4}\right )^{2}}-\frac {\left (261+182 \sqrt {2}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {2 \sin \left (x \right )}{\sqrt {2+\sqrt {2}}}\right )}{128 \sqrt {2+\sqrt {2}}}-\frac {\left (-261+182 \sqrt {2}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {2 \sin \left (x \right )}{\sqrt {2-\sqrt {2}}}\right )}{128 \sqrt {2-\sqrt {2}}}-\frac {1}{32 \left (1+\sin \left (x \right )\right )}+\frac {49 \ln \left (1+\sin \left (x \right )\right )}{32}\) | \(139\) |
| risch | \(-\frac {i \left (14 \,{\mathrm e}^{19 i x}-11 \,{\mathrm e}^{17 i x}-9 \,{\mathrm e}^{15 i x}+17 \,{\mathrm e}^{13 i x}+17 \,{\mathrm e}^{11 i x}-17 \,{\mathrm e}^{9 i x}-17 \,{\mathrm e}^{7 i x}+9 \,{\mathrm e}^{5 i x}+11 \,{\mathrm e}^{3 i x}-14 \,{\mathrm e}^{i x}\right )}{32 \left ({\mathrm e}^{10 i x}+{\mathrm e}^{8 i x}+{\mathrm e}^{2 i x}+1\right )^{2}}+\frac {49 \ln \left ({\mathrm e}^{i x}+i\right )}{16}-\frac {49 \ln \left ({\mathrm e}^{i x}-i\right )}{16}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (2147483648 \textit {\_Z}^{4}-5159649280 \textit {\_Z}^{2}+3508129\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i x}+\left (-\frac {1325400064}{104211847} i \textit {\_R}^{3}+\frac {3059324672}{104211847} i \textit {\_R} \right ) {\mathrm e}^{i x}-1\right )\right )\) | \(160\) |
Input:
int(1/(cos(3*x)+cos(5*x))^3,x,method=_RETURNVERBOSE)
Output:
-1/32/(sin(x)-1)-49/32*ln(sin(x)-1)+64*(-5/32*sin(x)^7+43/256*sin(x)^5-5/1 28*sin(x)^3+7/2048*sin(x))/(1-8*sin(x)^2+8*sin(x)^4)^2-1/128*(261+182*2^(1 /2))*2^(1/2)/(2+2^(1/2))^(1/2)*arctanh(2*sin(x)/(2+2^(1/2))^(1/2))-1/128*( -261+182*2^(1/2))*2^(1/2)/(2-2^(1/2))^(1/2)*arctanh(2*sin(x)/(2-2^(1/2))^( 1/2))-1/32/(1+sin(x))+49/32*ln(1+sin(x))
Time = 0.12 (sec) , antiderivative size = 388, normalized size of antiderivative = 1.20 \[ \int \frac {1}{(\cos (3 x)+\cos (5 x))^3} \, dx =\text {Too large to display} \] Input:
integrate(1/(cos(3*x)+cos(5*x))^3,x, algorithm="fricas")
Output:
-1/256*((64*cos(x)^10 - 128*cos(x)^8 + 80*cos(x)^6 - 16*cos(x)^4 + cos(x)^ 2)*sqrt(55639*sqrt(2) + 78730)*log(sqrt(55639*sqrt(2) + 78730)*(79*sqrt(2) - 103) + 3746*sin(x)) - (64*cos(x)^10 - 128*cos(x)^8 + 80*cos(x)^6 - 16*c os(x)^4 + cos(x)^2)*sqrt(55639*sqrt(2) + 78730)*log(sqrt(55639*sqrt(2) + 7 8730)*(79*sqrt(2) - 103) - 3746*sin(x)) - (64*cos(x)^10 - 128*cos(x)^8 + 8 0*cos(x)^6 - 16*cos(x)^4 + cos(x)^2)*sqrt(-55639*sqrt(2) + 78730)*log((79* sqrt(2) + 103)*sqrt(-55639*sqrt(2) + 78730) + 3746*sin(x)) + (64*cos(x)^10 - 128*cos(x)^8 + 80*cos(x)^6 - 16*cos(x)^4 + cos(x)^2)*sqrt(-55639*sqrt(2 ) + 78730)*log((79*sqrt(2) + 103)*sqrt(-55639*sqrt(2) + 78730) - 3746*sin( x)) - 392*(64*cos(x)^10 - 128*cos(x)^8 + 80*cos(x)^6 - 16*cos(x)^4 + cos(x )^2)*log(sin(x) + 1) + 392*(64*cos(x)^10 - 128*cos(x)^8 + 80*cos(x)^6 - 16 *cos(x)^4 + cos(x)^2)*log(-sin(x) + 1) - 8*(448*cos(x)^8 - 872*cos(x)^6 + 512*cos(x)^4 - 81*cos(x)^2 + 2)*sin(x))/(64*cos(x)^10 - 128*cos(x)^8 + 80* cos(x)^6 - 16*cos(x)^4 + cos(x)^2)
\[ \int \frac {1}{(\cos (3 x)+\cos (5 x))^3} \, dx=\int \frac {1}{\left (\cos {\left (3 x \right )} + \cos {\left (5 x \right )}\right )^{3}}\, dx \] Input:
integrate(1/(cos(3*x)+cos(5*x))**3,x)
Output:
Integral((cos(3*x) + cos(5*x))**(-3), x)
\[ \int \frac {1}{(\cos (3 x)+\cos (5 x))^3} \, dx=\int { \frac {1}{{\left (\cos \left (5 \, x\right ) + \cos \left (3 \, x\right )\right )}^{3}} \,d x } \] Input:
integrate(1/(cos(3*x)+cos(5*x))^3,x, algorithm="maxima")
Output:
1/32*((14*sin(19*x) - 11*sin(17*x) - 9*sin(15*x) + 17*sin(13*x) + 17*sin(1 1*x) - 17*sin(9*x) - 17*sin(7*x) + 9*sin(5*x) + 11*sin(3*x) - 14*sin(x))*c os(20*x) - 14*(2*sin(18*x) + sin(16*x) + 2*sin(12*x) + 4*sin(10*x) + 2*sin (8*x) + sin(4*x) + 2*sin(2*x))*cos(19*x) - 2*(11*sin(17*x) + 9*sin(15*x) - 17*sin(13*x) - 17*sin(11*x) + 17*sin(9*x) + 17*sin(7*x) - 9*sin(5*x) - 11 *sin(3*x) + 14*sin(x))*cos(18*x) + 11*(sin(16*x) + 2*sin(12*x) + 4*sin(10* x) + 2*sin(8*x) + sin(4*x) + 2*sin(2*x))*cos(17*x) - (9*sin(15*x) - 17*sin (13*x) - 17*sin(11*x) + 17*sin(9*x) + 17*sin(7*x) - 9*sin(5*x) - 11*sin(3* x) + 14*sin(x))*cos(16*x) + 9*(2*sin(12*x) + 4*sin(10*x) + 2*sin(8*x) + si n(4*x) + 2*sin(2*x))*cos(15*x) - 17*(2*sin(12*x) + 4*sin(10*x) + 2*sin(8*x ) + sin(4*x) + 2*sin(2*x))*cos(13*x) + 2*(17*sin(11*x) - 17*sin(9*x) - 17* sin(7*x) + 9*sin(5*x) + 11*sin(3*x) - 14*sin(x))*cos(12*x) - 17*(4*sin(10* x) + 2*sin(8*x) + sin(4*x) + 2*sin(2*x))*cos(11*x) - 4*(17*sin(9*x) + 17*s in(7*x) - 9*sin(5*x) - 11*sin(3*x) + 14*sin(x))*cos(10*x) + 17*(2*sin(8*x) + sin(4*x) + 2*sin(2*x))*cos(9*x) - 2*(17*sin(7*x) - 9*sin(5*x) - 11*sin( 3*x) + 14*sin(x))*cos(8*x) + 17*(sin(4*x) + 2*sin(2*x))*cos(7*x) - 9*(sin( 4*x) + 2*sin(2*x))*cos(5*x) + (11*sin(3*x) - 14*sin(x))*cos(4*x) - 32*(2*( 2*cos(18*x) + cos(16*x) + 2*cos(12*x) + 4*cos(10*x) + 2*cos(8*x) + cos(4*x ) + 2*cos(2*x) + 1)*cos(20*x) + cos(20*x)^2 + 4*(cos(16*x) + 2*cos(12*x) + 4*cos(10*x) + 2*cos(8*x) + cos(4*x) + 2*cos(2*x) + 1)*cos(18*x) + 4*co...
Time = 0.16 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.53 \[ \int \frac {1}{(\cos (3 x)+\cos (5 x))^3} \, dx=-\frac {1}{256} \, \sqrt {55639 \, \sqrt {2} + 78730} \log \left ({\left | \frac {1}{2} \, \sqrt {\sqrt {2} + 2} + \sin \left (x\right ) \right |}\right ) + \frac {1}{256} \, \sqrt {55639 \, \sqrt {2} + 78730} \log \left ({\left | -\frac {1}{2} \, \sqrt {\sqrt {2} + 2} + \sin \left (x\right ) \right |}\right ) + \frac {1}{256} \, \sqrt {-55639 \, \sqrt {2} + 78730} \log \left ({\left | \sqrt {-\frac {1}{4} \, \sqrt {2} + \frac {1}{2}} + \sin \left (x\right ) \right |}\right ) - \frac {1}{256} \, \sqrt {-55639 \, \sqrt {2} + 78730} \log \left ({\left | -\sqrt {-\frac {1}{4} \, \sqrt {2} + \frac {1}{2}} + \sin \left (x\right ) \right |}\right ) - \frac {\sin \left (x\right )}{16 \, {\left (\sin \left (x\right )^{2} - 1\right )}} - \frac {320 \, \sin \left (x\right )^{7} - 344 \, \sin \left (x\right )^{5} + 80 \, \sin \left (x\right )^{3} - 7 \, \sin \left (x\right )}{32 \, {\left (8 \, \sin \left (x\right )^{4} - 8 \, \sin \left (x\right )^{2} + 1\right )}^{2}} + \frac {49}{32} \, \log \left (\sin \left (x\right ) + 1\right ) - \frac {49}{32} \, \log \left (-\sin \left (x\right ) + 1\right ) \] Input:
integrate(1/(cos(3*x)+cos(5*x))^3,x, algorithm="giac")
Output:
-1/256*sqrt(55639*sqrt(2) + 78730)*log(abs(1/2*sqrt(sqrt(2) + 2) + sin(x)) ) + 1/256*sqrt(55639*sqrt(2) + 78730)*log(abs(-1/2*sqrt(sqrt(2) + 2) + sin (x))) + 1/256*sqrt(-55639*sqrt(2) + 78730)*log(abs(sqrt(-1/4*sqrt(2) + 1/2 ) + sin(x))) - 1/256*sqrt(-55639*sqrt(2) + 78730)*log(abs(-sqrt(-1/4*sqrt( 2) + 1/2) + sin(x))) - 1/16*sin(x)/(sin(x)^2 - 1) - 1/32*(320*sin(x)^7 - 3 44*sin(x)^5 + 80*sin(x)^3 - 7*sin(x))/(8*sin(x)^4 - 8*sin(x)^2 + 1)^2 + 49 /32*log(sin(x) + 1) - 49/32*log(-sin(x) + 1)
Time = 21.63 (sec) , antiderivative size = 374, normalized size of antiderivative = 1.15 \[ \int \frac {1}{(\cos (3 x)+\cos (5 x))^3} \, dx =\text {Too large to display} \] Input:
int(1/(cos(3*x) + cos(5*x))^3,x)
Output:
(49*atanh(tan(x/2)))/8 + (atanh((21921344091656122463*tan(x/2)*(78730 - 55 639*2^(1/2))^(1/2))/(16*((22727474752897861007033*2^(1/2))/128 + (22727474 752897861007033*2^(1/2)*tan(x/2)^2)/128 - (64283006104824152616365*tan(x/2 )^2)/256 - 64283006104824152616365/256)) - (124005858648938077817*2^(1/2)* tan(x/2)*(78730 - 55639*2^(1/2))^(1/2))/(128*((22727474752897861007033*2^( 1/2))/128 + (22727474752897861007033*2^(1/2)*tan(x/2)^2)/128 - (6428300610 4824152616365*tan(x/2)^2)/256 - 64283006104824152616365/256)))*(78730 - 55 639*2^(1/2))^(1/2))/128 - (atanh((21921344091656122463*tan(x/2)*(55639*2^( 1/2) + 78730)^(1/2))/(16*((22727474752897861007033*2^(1/2))/128 + (2272747 4752897861007033*2^(1/2)*tan(x/2)^2)/128 + (64283006104824152616365*tan(x/ 2)^2)/256 + 64283006104824152616365/256)) + (124005858648938077817*2^(1/2) *tan(x/2)*(55639*2^(1/2) + 78730)^(1/2))/(128*((22727474752897861007033*2^ (1/2))/128 + (22727474752897861007033*2^(1/2)*tan(x/2)^2)/128 + (642830061 04824152616365*tan(x/2)^2)/256 + 64283006104824152616365/256)))*(55639*2^( 1/2) + 78730)^(1/2))/128 + ((9*tan(x/2))/16 - (395*tan(x/2)^3)/16 + 396*ta n(x/2)^5 - (2675*tan(x/2)^7)/2 + (7981*tan(x/2)^9)/8 + (7981*tan(x/2)^11)/ 8 - (2675*tan(x/2)^13)/2 + 396*tan(x/2)^15 - (395*tan(x/2)^17)/16 + (9*tan (x/2)^19)/16)/(1037*tan(x/2)^4 - 58*tan(x/2)^2 - 5880*tan(x/2)^6 + 15346*t an(x/2)^8 - 20892*tan(x/2)^10 + 15346*tan(x/2)^12 - 5880*tan(x/2)^14 + 103 7*tan(x/2)^16 - 58*tan(x/2)^18 + tan(x/2)^20 + 1)
\[ \int \frac {1}{(\cos (3 x)+\cos (5 x))^3} \, dx=\int \frac {1}{\cos \left (5 x \right )^{3}+3 \cos \left (5 x \right )^{2} \cos \left (3 x \right )+3 \cos \left (5 x \right ) \cos \left (3 x \right )^{2}+\cos \left (3 x \right )^{3}}d x \] Input:
int(1/(cos(3*x)+cos(5*x))^3,x)
Output:
int(1/(cos(5*x)**3 + 3*cos(5*x)**2*cos(3*x) + 3*cos(5*x)*cos(3*x)**2 + cos (3*x)**3),x)