\(\int \frac {1}{\cos (3 x)+\sin (2 x)} \, dx\) [1]

Optimal result

Integrand size = 11, antiderivative size = 71 \[ \int \frac {1}{\cos (3 x)+\sin (2 x)} \, dx=-\frac {1}{5} \left (1-\sqrt {5}\right ) \log \left (1-\sqrt {5}-4 \sin (x)\right )-\frac {1}{5} \left (1+\sqrt {5}\right ) \log \left (1+\sqrt {5}-4 \sin (x)\right )+\frac {1}{2} \log (1-\sin (x))-\frac {1}{10} \log (1+\sin (x)) \] Output:

-1/5*(-5^(1/2)+1)*ln(1-5^(1/2)-4*sin(x))-1/5*(5^(1/2)+1)*ln(1+5^(1/2)-4*si 
n(x))+1/2*ln(1-sin(x))-1/10*ln(1+sin(x))
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.31 \[ \int \frac {1}{\cos (3 x)+\sin (2 x)} \, dx=\frac {1}{25} \left (25 \log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )-5 \log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )-\sqrt {5} \left (-5+\sqrt {5}\right ) \log \left (1-\sqrt {5}-4 \sin (x)\right )-\sqrt {5} \left (5+\sqrt {5}\right ) \log \left (1+\sqrt {5}-4 \sin (x)\right )\right ) \] Input:

Integrate[(Cos[3*x] + Sin[2*x])^(-1),x]
 

Output:

(25*Log[Cos[x/2] - Sin[x/2]] - 5*Log[Cos[x/2] + Sin[x/2]] - Sqrt[5]*(-5 + 
Sqrt[5])*Log[1 - Sqrt[5] - 4*Sin[x]] - Sqrt[5]*(5 + Sqrt[5])*Log[1 + Sqrt[ 
5] - 4*Sin[x]])/25
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.11, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 4829, 1300, 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.

\(\displaystyle \int \frac {1}{\sin (2 x)+\cos (3 x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (2 x)+\cos (3 x)}dx\)

\(\Big \downarrow \) 4829

\(\displaystyle \int \frac {1}{\left (-4 \sin ^2(x)+2 \sin (x)+1\right ) \left (1-\sin ^2(x)\right )}d\sin (x)\)

\(\Big \downarrow \) 1300

\(\displaystyle -4 \int \left (\frac {1}{40 (\sin (x)+1)}-\frac {1}{5 \left (\left (1-\sqrt {5}\right ) \sin (x)+1\right )}-\frac {1}{5 \left (\left (1+\sqrt {5}\right ) \sin (x)+1\right )}+\frac {1}{8 (1-\sin (x))}\right )d\sin (x)\)

\(\Big \downarrow \) 2009

\(\displaystyle -4 \left (-\frac {1}{8} \log (1-\sin (x))+\frac {1}{40} \log (\sin (x)+1)-\frac {\log \left (\left (1-\sqrt {5}\right ) \sin (x)+1\right )}{5 \left (1-\sqrt {5}\right )}-\frac {\log \left (\left (1+\sqrt {5}\right ) \sin (x)+1\right )}{5 \left (1+\sqrt {5}\right )}\right )\)

Input:

Int[(Cos[3*x] + Sin[2*x])^(-1),x]
 

Output:

-4*(-1/8*Log[1 - Sin[x]] + Log[1 + Sin[x]]/40 - Log[1 + (1 - Sqrt[5])*Sin[ 
x]]/(5*(1 - Sqrt[5])) - Log[1 + (1 + Sqrt[5])*Sin[x]]/(5*(1 + Sqrt[5])))
 

Defintions of rubi rules used

rule 1300

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_.) + (f_.)*(x_)^2)^(q_), x 
_Symbol] :> With[{r = Rt[b^2 - 4*a*c, 2]}, Simp[1/c^p   Int[ExpandIntegrand 
[(b/2 - r/2 + c*x)^p*(b/2 + r/2 + c*x)^p*(d + f*x^2)^q, x], x], x] /; EqQ[p 
, -1] ||  !FractionalPowerFactorQ[r]] /; FreeQ[{a, b, c, d, f}, x] && ILtQ[ 
p, 0] && IntegerQ[q] && NiceSqrtQ[b^2 - 4*a*c]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4829

Int[(cos[(n_.)*((c_.) + (d_.)*(x_))]*(b_.) + (a_.)*sin[(m_.)*((c_.) + (d_.) 
*(x_))])^(p_), x_Symbol] :> Simp[1/d   Subst[Int[Simplify[TrigExpand[a*Sin[ 
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/2] && Inte 
gerQ[(n - 1)/2]
 

Maple [A] (verified)

Time = 0.72 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.68

method	result	size
default	\(\frac {\ln \left (\sin \left (x \right )-1\right )}{2}-\frac {\ln \left (1+\sin \left (x \right )\right )}{10}-\frac {\ln \left (4 \sin \left (x \right )^{2}-2 \sin \left (x \right )-1\right )}{5}+\frac {2 \sqrt {5}\, \operatorname {arctanh}\left (\frac {\left (-2+8 \sin \left (x \right )\right ) \sqrt {5}}{10}\right )}{5}\)	\(48\)
risch	\(\ln \left ({\mathrm e}^{i x}-i\right )-\frac {\ln \left ({\mathrm e}^{i x}+i\right )}{5}-\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {i \left (\sqrt {5}-1\right ) {\mathrm e}^{i x}}{2}-1\right )}{5}+\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {i \left (\sqrt {5}-1\right ) {\mathrm e}^{i x}}{2}-1\right ) \sqrt {5}}{5}-\frac {\ln \left ({\mathrm e}^{2 i x}-\frac {i \left (\sqrt {5}+1\right ) {\mathrm e}^{i x}}{2}-1\right )}{5}-\frac {\ln \left ({\mathrm e}^{2 i x}-\frac {i \left (\sqrt {5}+1\right ) {\mathrm e}^{i x}}{2}-1\right ) \sqrt {5}}{5}\)	\(120\)

Input:

int(1/(cos(3*x)+sin(2*x)),x,method=_RETURNVERBOSE)
 

Output:

1/2*ln(sin(x)-1)-1/10*ln(1+sin(x))-1/5*ln(4*sin(x)^2-2*sin(x)-1)+2/5*5^(1/ 
2)*arctanh(1/10*(-2+8*sin(x))*5^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.03 \[ \int \frac {1}{\cos (3 x)+\sin (2 x)} \, dx=\frac {1}{5} \, \sqrt {5} \log \left (\frac {8 \, \cos \left (x\right )^{2} - 4 \, {\left (\sqrt {5} - 1\right )} \sin \left (x\right ) + \sqrt {5} - 11}{4 \, \cos \left (x\right )^{2} + 2 \, \sin \left (x\right ) - 3}\right ) - \frac {1}{5} \, \log \left (4 \, \cos \left (x\right )^{2} + 2 \, \sin \left (x\right ) - 3\right ) - \frac {1}{10} \, \log \left (\sin \left (x\right ) + 1\right ) + \frac {1}{2} \, \log \left (-\sin \left (x\right ) + 1\right ) \] Input:

integrate(1/(cos(3*x)+sin(2*x)),x, algorithm="fricas")
 

Output:

1/5*sqrt(5)*log((8*cos(x)^2 - 4*(sqrt(5) - 1)*sin(x) + sqrt(5) - 11)/(4*co 
s(x)^2 + 2*sin(x) - 3)) - 1/5*log(4*cos(x)^2 + 2*sin(x) - 3) - 1/10*log(si 
n(x) + 1) + 1/2*log(-sin(x) + 1)
 

Sympy [F]

\[ \int \frac {1}{\cos (3 x)+\sin (2 x)} \, dx=\int \frac {1}{\sin {\left (2 x \right )} + \cos {\left (3 x \right )}}\, dx \] Input:

integrate(1/(cos(3*x)+sin(2*x)),x)
 

Output:

Integral(1/(sin(2*x) + cos(3*x)), x)
 

Maxima [F]

\[ \int \frac {1}{\cos (3 x)+\sin (2 x)} \, dx=\int { \frac {1}{\cos \left (3 \, x\right ) + \sin \left (2 \, x\right )} \,d x } \] Input:

integrate(1/(cos(3*x)+sin(2*x)),x, algorithm="maxima")
 

Output:

2/5*integrate(-(cos(4*x)*cos(x) - cos(2*x)*cos(x) + cos(x)*sin(3*x) - cos( 
3*x)*sin(x) + sin(4*x)*sin(x) - sin(2*x)*sin(x) + cos(x))/(2*(cos(2*x) - s 
in(3*x) + sin(x) - 1)*cos(4*x) - cos(4*x)^2 + 2*(cos(x) - sin(2*x))*cos(3* 
x) - cos(3*x)^2 - 2*(sin(x) - 1)*cos(2*x) - cos(2*x)^2 - cos(x)^2 + 2*(cos 
(3*x) - cos(x) + sin(2*x))*sin(4*x) - sin(4*x)^2 + 2*(cos(2*x) + sin(x) - 
1)*sin(3*x) - sin(3*x)^2 + 2*cos(x)*sin(2*x) - sin(2*x)^2 - sin(x)^2 + 2*s 
in(x) - 1), x) + 6/5*integrate((cos(3*x)*cos(4/3*arctan2(sin(3*x), cos(3*x 
))) - cos(3*x)*cos(2/3*arctan2(sin(3*x), cos(3*x))) + cos(1/3*arctan2(sin( 
3*x), cos(3*x)))*sin(3*x) + sin(3*x)*sin(4/3*arctan2(sin(3*x), cos(3*x))) 
- sin(3*x)*sin(2/3*arctan2(sin(3*x), cos(3*x))) - cos(3*x)*sin(1/3*arctan2 
(sin(3*x), cos(3*x))) + cos(3*x))/(cos(3*x)^2 - 2*(cos(2/3*arctan2(sin(3*x 
), cos(3*x))) - sin(3*x) + sin(1/3*arctan2(sin(3*x), cos(3*x))) - 1)*cos(4 
/3*arctan2(sin(3*x), cos(3*x))) + cos(4/3*arctan2(sin(3*x), cos(3*x)))^2 - 
 2*(sin(3*x) - sin(1/3*arctan2(sin(3*x), cos(3*x))) + 1)*cos(2/3*arctan2(s 
in(3*x), cos(3*x))) + cos(2/3*arctan2(sin(3*x), cos(3*x)))^2 - 2*cos(3*x)* 
cos(1/3*arctan2(sin(3*x), cos(3*x))) + cos(1/3*arctan2(sin(3*x), cos(3*x)) 
)^2 + sin(3*x)^2 - 2*(cos(3*x) - cos(1/3*arctan2(sin(3*x), cos(3*x))) + si 
n(2/3*arctan2(sin(3*x), cos(3*x))))*sin(4/3*arctan2(sin(3*x), cos(3*x))) + 
 sin(4/3*arctan2(sin(3*x), cos(3*x)))^2 + 2*(cos(3*x) - cos(1/3*arctan2(si 
n(3*x), cos(3*x))))*sin(2/3*arctan2(sin(3*x), cos(3*x))) + sin(2/3*arct...
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.93 \[ \int \frac {1}{\cos (3 x)+\sin (2 x)} \, dx=-\frac {1}{5} \, \sqrt {5} \log \left (\frac {{\left | -2 \, \sqrt {5} + 8 \, \sin \left (x\right ) - 2 \right |}}{{\left | 2 \, \sqrt {5} + 8 \, \sin \left (x\right ) - 2 \right |}}\right ) - \frac {1}{10} \, \log \left (\sin \left (x\right ) + 1\right ) + \frac {1}{2} \, \log \left (-\sin \left (x\right ) + 1\right ) - \frac {1}{5} \, \log \left ({\left | 4 \, \sin \left (x\right )^{2} - 2 \, \sin \left (x\right ) - 1 \right |}\right ) \] Input:

integrate(1/(cos(3*x)+sin(2*x)),x, algorithm="giac")
 

Output:

-1/5*sqrt(5)*log(abs(-2*sqrt(5) + 8*sin(x) - 2)/abs(2*sqrt(5) + 8*sin(x) - 
 2)) - 1/10*log(sin(x) + 1) + 1/2*log(-sin(x) + 1) - 1/5*log(abs(4*sin(x)^ 
2 - 2*sin(x) - 1))
 

Mupad [B] (verification not implemented)

Time = 20.93 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.15 \[ \int \frac {1}{\cos (3 x)+\sin (2 x)} \, dx=\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )-1\right )-\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )+1\right )}{5}-\ln \left (2\,\mathrm {tan}\left (\frac {x}{2}\right )-2\,\sqrt {5}\,\mathrm {tan}\left (\frac {x}{2}\right )+{\mathrm {tan}\left (\frac {x}{2}\right )}^2+1\right )\,\left (\frac {\sqrt {5}}{5}+\frac {1}{5}\right )+\ln \left (2\,\mathrm {tan}\left (\frac {x}{2}\right )+2\,\sqrt {5}\,\mathrm {tan}\left (\frac {x}{2}\right )+{\mathrm {tan}\left (\frac {x}{2}\right )}^2+1\right )\,\left (\frac {\sqrt {5}}{5}-\frac {1}{5}\right ) \] Input:

int(1/(cos(3*x) + sin(2*x)),x)
 

Output:

log(tan(x/2) - 1) - log(tan(x/2) + 1)/5 - log(2*tan(x/2) - 2*5^(1/2)*tan(x 
/2) + tan(x/2)^2 + 1)*(5^(1/2)/5 + 1/5) + log(2*tan(x/2) + 2*5^(1/2)*tan(x 
/2) + tan(x/2)^2 + 1)*(5^(1/2)/5 - 1/5)
 

Reduce [F]

\[ \int \frac {1}{\cos (3 x)+\sin (2 x)} \, dx=\int \frac {1}{\cos \left (3 x \right )+\sin \left (2 x \right )}d x \] Input:

int(1/(cos(3*x)+sin(2*x)),x)
 

Output:

int(1/(cos(3*x) + sin(2*x)),x)