Integrand size = 7, antiderivative size = 60 \[ \int \frac {1}{(\sin (x)+\tan (x))^3} \, dx=\frac {1}{32} \text {arctanh}(\cos (x))-\frac {1}{32 (1-\cos (x))}-\frac {1}{16 (1+\cos (x))^4}+\frac {1}{6 (1+\cos (x))^3}-\frac {3}{32 (1+\cos (x))^2}-\frac {1}{16 (1+\cos (x))} \] Output:
1/32*arctanh(cos(x))-1/(32-32*cos(x))-1/16/(1+cos(x))^4+1/6/(1+cos(x))^3-3 /32/(1+cos(x))^2-1/(16+16*cos(x))
Time = 0.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.38 \[ \int \frac {1}{(\sin (x)+\tan (x))^3} \, dx=-\frac {1}{64} \csc ^2\left (\frac {x}{2}\right )+\frac {1}{32} \log \left (\cos \left (\frac {x}{2}\right )\right )-\frac {1}{32} \log \left (\sin \left (\frac {x}{2}\right )\right )-\frac {1}{32} \sec ^2\left (\frac {x}{2}\right )-\frac {3}{128} \sec ^4\left (\frac {x}{2}\right )+\frac {1}{48} \sec ^6\left (\frac {x}{2}\right )-\frac {1}{256} \sec ^8\left (\frac {x}{2}\right ) \] Input:
Integrate[(Sin[x] + Tan[x])^(-3),x]
Output:
-1/64*Csc[x/2]^2 + Log[Cos[x/2]]/32 - Log[Sin[x/2]]/32 - Sec[x/2]^2/32 - ( 3*Sec[x/2]^4)/128 + Sec[x/2]^6/48 - Sec[x/2]^8/256
Time = 0.29 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {3042, 4897, 3042, 25, 3186, 99, 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.
\(\displaystyle \int \frac {1}{(\sin (x)+\tan (x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(\sin (x)+\tan (x))^3}dx\) |
\(\Big \downarrow \) 4897 |
\(\displaystyle \int \frac {\cot ^3(x)}{(\cos (x)+1)^3}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\tan \left (x-\frac {\pi }{2}\right )^3}{\left (1-\sin \left (x-\frac {\pi }{2}\right )\right )^3}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\tan \left (x-\frac {\pi }{2}\right )^3}{\left (1-\sin \left (x-\frac {\pi }{2}\right )\right )^3}dx\) |
\(\Big \downarrow \) 3186 |
\(\displaystyle -\int \frac {\cos ^3(x)}{(1-\cos (x))^2 (\cos (x)+1)^5}d\cos (x)\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\int \left (\frac {1}{32 (\cos (x)-1)^2}-\frac {1}{16 (\cos (x)+1)^2}-\frac {3}{16 (\cos (x)+1)^3}+\frac {1}{2 (\cos (x)+1)^4}-\frac {1}{4 (\cos (x)+1)^5}+\frac {1}{32 \left (\cos ^2(x)-1\right )}\right )d\cos (x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{32} \text {arctanh}(\cos (x))-\frac {1}{32 (1-\cos (x))}-\frac {1}{16 (\cos (x)+1)}-\frac {3}{32 (\cos (x)+1)^2}+\frac {1}{6 (\cos (x)+1)^3}-\frac {1}{16 (\cos (x)+1)^4}\) |
Input:
Int[(Sin[x] + Tan[x])^(-3),x]
Output:
ArcTanh[Cos[x]]/32 - 1/(32*(1 - Cos[x])) - 1/(16*(1 + Cos[x])^4) + 1/(6*(1 + Cos[x])^3) - 3/(32*(1 + Cos[x])^2) - 1/(16*(1 + Cos[x]))
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p _.), x_Symbol] :> Simp[1/f Subst[Int[x^p*((a + x)^(m - (p + 1)/2)/(a - x) ^((p + 1)/2)), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && E qQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]
Time = 0.54 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.93
method | result | size |
default | \(\frac {1}{32 \cos \left (x \right )-32}-\frac {\ln \left (\cos \left (x \right )-1\right )}{64}-\frac {1}{16 \left (\cos \left (x \right )+1\right )^{4}}+\frac {1}{6 \left (\cos \left (x \right )+1\right )^{3}}-\frac {3}{32 \left (\cos \left (x \right )+1\right )^{2}}-\frac {1}{16 \left (\cos \left (x \right )+1\right )}+\frac {\ln \left (\cos \left (x \right )+1\right )}{64}\) | \(56\) |
risch | \(-\frac {3 \,{\mathrm e}^{9 i x}+18 \,{\mathrm e}^{8 i x}-88 \,{\mathrm e}^{7 i x}-162 \,{\mathrm e}^{6 i x}-310 \,{\mathrm e}^{5 i x}-162 \,{\mathrm e}^{4 i x}-88 \,{\mathrm e}^{3 i x}+18 \,{\mathrm e}^{2 i x}+3 \,{\mathrm e}^{i x}}{48 \left ({\mathrm e}^{i x}+1\right )^{8} \left ({\mathrm e}^{i x}-1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i x}-1\right )}{32}+\frac {\ln \left ({\mathrm e}^{i x}+1\right )}{32}\) | \(106\) |
Input:
int(1/(sin(x)+tan(x))^3,x,method=_RETURNVERBOSE)
Output:
1/32/(cos(x)-1)-1/64*ln(cos(x)-1)-1/16/(cos(x)+1)^4+1/6/(cos(x)+1)^3-3/32/ (cos(x)+1)^2-1/16/(cos(x)+1)+1/64*ln(cos(x)+1)
Leaf count of result is larger than twice the leaf count of optimal. 130 vs. \(2 (46) = 92\).
Time = 0.08 (sec) , antiderivative size = 130, normalized size of antiderivative = 2.17 \[ \int \frac {1}{(\sin (x)+\tan (x))^3} \, dx=-\frac {6 \, \cos \left (x\right )^{4} + 18 \, \cos \left (x\right )^{3} - 50 \, \cos \left (x\right )^{2} - 3 \, {\left (\cos \left (x\right )^{5} + 3 \, \cos \left (x\right )^{4} + 2 \, \cos \left (x\right )^{3} - 2 \, \cos \left (x\right )^{2} - 3 \, \cos \left (x\right ) - 1\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) + 3 \, {\left (\cos \left (x\right )^{5} + 3 \, \cos \left (x\right )^{4} + 2 \, \cos \left (x\right )^{3} - 2 \, \cos \left (x\right )^{2} - 3 \, \cos \left (x\right ) - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) - 54 \, \cos \left (x\right ) - 16}{192 \, {\left (\cos \left (x\right )^{5} + 3 \, \cos \left (x\right )^{4} + 2 \, \cos \left (x\right )^{3} - 2 \, \cos \left (x\right )^{2} - 3 \, \cos \left (x\right ) - 1\right )}} \] Input:
integrate(1/(sin(x)+tan(x))^3,x, algorithm="fricas")
Output:
-1/192*(6*cos(x)^4 + 18*cos(x)^3 - 50*cos(x)^2 - 3*(cos(x)^5 + 3*cos(x)^4 + 2*cos(x)^3 - 2*cos(x)^2 - 3*cos(x) - 1)*log(1/2*cos(x) + 1/2) + 3*(cos(x )^5 + 3*cos(x)^4 + 2*cos(x)^3 - 2*cos(x)^2 - 3*cos(x) - 1)*log(-1/2*cos(x) + 1/2) - 54*cos(x) - 16)/(cos(x)^5 + 3*cos(x)^4 + 2*cos(x)^3 - 2*cos(x)^2 - 3*cos(x) - 1)
\[ \int \frac {1}{(\sin (x)+\tan (x))^3} \, dx=\int \frac {1}{\left (\sin {\left (x \right )} + \tan {\left (x \right )}\right )^{3}}\, dx \] Input:
integrate(1/(sin(x)+tan(x))**3,x)
Output:
Integral((sin(x) + tan(x))**(-3), x)
Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.22 \[ \int \frac {1}{(\sin (x)+\tan (x))^3} \, dx=-\frac {{\left (\cos \left (x\right ) + 1\right )}^{2}}{64 \, \sin \left (x\right )^{2}} - \frac {\sin \left (x\right )^{2}}{32 \, {\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac {\sin \left (x\right )^{4}}{64 \, {\left (\cos \left (x\right ) + 1\right )}^{4}} + \frac {\sin \left (x\right )^{6}}{192 \, {\left (\cos \left (x\right ) + 1\right )}^{6}} - \frac {\sin \left (x\right )^{8}}{256 \, {\left (\cos \left (x\right ) + 1\right )}^{8}} - \frac {1}{32} \, \log \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1}\right ) \] Input:
integrate(1/(sin(x)+tan(x))^3,x, algorithm="maxima")
Output:
-1/64*(cos(x) + 1)^2/sin(x)^2 - 1/32*sin(x)^2/(cos(x) + 1)^2 + 1/64*sin(x) ^4/(cos(x) + 1)^4 + 1/192*sin(x)^6/(cos(x) + 1)^6 - 1/256*sin(x)^8/(cos(x) + 1)^8 - 1/32*log(sin(x)/(cos(x) + 1))
Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (46) = 92\).
Time = 0.13 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.58 \[ \int \frac {1}{(\sin (x)+\tan (x))^3} \, dx=\frac {{\left (\frac {\cos \left (x\right ) - 1}{\cos \left (x\right ) + 1} + 1\right )} {\left (\cos \left (x\right ) + 1\right )}}{64 \, {\left (\cos \left (x\right ) - 1\right )}} + \frac {\cos \left (x\right ) - 1}{32 \, {\left (\cos \left (x\right ) + 1\right )}} + \frac {{\left (\cos \left (x\right ) - 1\right )}^{2}}{64 \, {\left (\cos \left (x\right ) + 1\right )}^{2}} - \frac {{\left (\cos \left (x\right ) - 1\right )}^{3}}{192 \, {\left (\cos \left (x\right ) + 1\right )}^{3}} - \frac {{\left (\cos \left (x\right ) - 1\right )}^{4}}{256 \, {\left (\cos \left (x\right ) + 1\right )}^{4}} - \frac {1}{64} \, \log \left (-\frac {\cos \left (x\right ) - 1}{\cos \left (x\right ) + 1}\right ) \] Input:
integrate(1/(sin(x)+tan(x))^3,x, algorithm="giac")
Output:
1/64*((cos(x) - 1)/(cos(x) + 1) + 1)*(cos(x) + 1)/(cos(x) - 1) + 1/32*(cos (x) - 1)/(cos(x) + 1) + 1/64*(cos(x) - 1)^2/(cos(x) + 1)^2 - 1/192*(cos(x) - 1)^3/(cos(x) + 1)^3 - 1/256*(cos(x) - 1)^4/(cos(x) + 1)^4 - 1/64*log(-( cos(x) - 1)/(cos(x) + 1))
Time = 15.30 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.80 \[ \int \frac {1}{(\sin (x)+\tan (x))^3} \, dx=\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^4}{64}-\frac {1}{64\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{32}-\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{32}+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^6}{192}-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^8}{256} \] Input:
int(1/(sin(x) + tan(x))^3,x)
Output:
tan(x/2)^4/64 - 1/(64*tan(x/2)^2) - tan(x/2)^2/32 - log(tan(x/2))/32 + tan (x/2)^6/192 - tan(x/2)^8/256
Time = 0.16 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.92 \[ \int \frac {1}{(\sin (x)+\tan (x))^3} \, dx=\frac {-24 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )\right ) \tan \left (\frac {x}{2}\right )^{2}-3 \tan \left (\frac {x}{2}\right )^{10}+4 \tan \left (\frac {x}{2}\right )^{8}+12 \tan \left (\frac {x}{2}\right )^{6}-24 \tan \left (\frac {x}{2}\right )^{4}-12}{768 \tan \left (\frac {x}{2}\right )^{2}} \] Input:
int(1/(sin(x)+tan(x))^3,x)
Output:
( - 24*log(tan(x/2))*tan(x/2)**2 - 3*tan(x/2)**10 + 4*tan(x/2)**8 + 12*tan (x/2)**6 - 24*tan(x/2)**4 - 12)/(768*tan(x/2)**2)