Integrand size = 7, antiderivative size = 65 \[ \int \frac {1}{(\sin (x)+\tan (x))^4} \, dx=-\frac {1}{5} \cot ^5(x)-\frac {9 \cot ^7(x)}{7}-\frac {16 \cot ^9(x)}{9}-\frac {8 \cot ^{11}(x)}{11}-\frac {4 \csc ^5(x)}{5}+\frac {16 \csc ^7(x)}{7}-\frac {20 \csc ^9(x)}{9}+\frac {8 \csc ^{11}(x)}{11} \]
-1/5*cot(x)^5-9/7*cot(x)^7-16/9*cot(x)^9-8/11*cot(x)^11-4/5*csc(x)^5+16/7* csc(x)^7-20/9*csc(x)^9+8/11*csc(x)^11
Time = 0.02 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.98 \[ \int \frac {1}{(\sin (x)+\tan (x))^4} \, dx=\frac {1}{96} \cot \left (\frac {x}{2}\right )-\frac {1}{384} \cot \left (\frac {x}{2}\right ) \csc ^2\left (\frac {x}{2}\right )-\frac {2749 \tan \left (\frac {x}{2}\right )}{110880}-\frac {2033 \sec ^2\left (\frac {x}{2}\right ) \tan \left (\frac {x}{2}\right )}{443520}+\frac {179 \sec ^4\left (\frac {x}{2}\right ) \tan \left (\frac {x}{2}\right )}{73920}+\frac {641 \sec ^6\left (\frac {x}{2}\right ) \tan \left (\frac {x}{2}\right )}{88704}-\frac {7 \sec ^8\left (\frac {x}{2}\right ) \tan \left (\frac {x}{2}\right )}{1584}+\frac {\sec ^{10}\left (\frac {x}{2}\right ) \tan \left (\frac {x}{2}\right )}{1408} \]
Cot[x/2]/96 - (Cot[x/2]*Csc[x/2]^2)/384 - (2749*Tan[x/2])/110880 - (2033*S ec[x/2]^2*Tan[x/2])/443520 + (179*Sec[x/2]^4*Tan[x/2])/73920 + (641*Sec[x/ 2]^6*Tan[x/2])/88704 - (7*Sec[x/2]^8*Tan[x/2])/1584 + (Sec[x/2]^10*Tan[x/2 ])/1408
Time = 0.41 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {3042, 4897, 3042, 3190, 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.
\(\displaystyle \int \frac {1}{(\sin (x)+\tan (x))^4} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(\sin (x)+\tan (x))^4}dx\) |
\(\Big \downarrow \) 4897 |
\(\displaystyle \int \frac {\cot ^4(x)}{(\cos (x)+1)^4}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan \left (x-\frac {\pi }{2}\right )^4}{\left (1-\sin \left (x-\frac {\pi }{2}\right )\right )^4}dx\) |
\(\Big \downarrow \) 3190 |
\(\displaystyle \int \left (\cot ^8(x) \csc ^4(x)-4 \cot ^7(x) \csc ^5(x)+6 \cot ^6(x) \csc ^6(x)-4 \cot ^5(x) \csc ^7(x)+\cot ^4(x) \csc ^8(x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {8}{11} \cot ^{11}(x)-\frac {16 \cot ^9(x)}{9}-\frac {9 \cot ^7(x)}{7}-\frac {\cot ^5(x)}{5}+\frac {8 \csc ^{11}(x)}{11}-\frac {20 \csc ^9(x)}{9}+\frac {16 \csc ^7(x)}{7}-\frac {4 \csc ^5(x)}{5}\) |
-1/5*Cot[x]^5 - (9*Cot[x]^7)/7 - (16*Cot[x]^9)/9 - (8*Cot[x]^11)/11 - (4*C sc[x]^5)/5 + (16*Csc[x]^7)/7 - (20*Csc[x]^9)/9 + (8*Csc[x]^11)/11
3.4.48.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((g_.)*tan[(e_.) + (f_.)*(x _)])^(p_.), x_Symbol] :> Simp[a^(2*m) Int[ExpandIntegrand[(g*Tan[e + f*x] )^p/Sec[e + f*x]^m, (a*Sec[e + f*x] - b*Tan[e + f*x])^(-m), x], x], x] /; F reeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 0]
Time = 1.66 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.98
method | result | size |
default | \(\frac {\tan \left (\frac {x}{2}\right )^{11}}{1408}-\frac {\tan \left (\frac {x}{2}\right )^{9}}{1152}-\frac {3 \tan \left (\frac {x}{2}\right )^{7}}{896}+\frac {3 \tan \left (\frac {x}{2}\right )^{5}}{640}+\frac {\tan \left (\frac {x}{2}\right )^{3}}{128}-\frac {3 \tan \left (\frac {x}{2}\right )}{128}+\frac {1}{128 \tan \left (\frac {x}{2}\right )}-\frac {1}{384 \tan \left (\frac {x}{2}\right )^{3}}\) | \(64\) |
risch | \(\frac {4 i \left (3465 \,{\mathrm e}^{10 i x}+5544 \,{\mathrm e}^{9 i x}+10857 \,{\mathrm e}^{8 i x}+5280 \,{\mathrm e}^{7 i x}+4818 \,{\mathrm e}^{6 i x}+176 \,{\mathrm e}^{5 i x}+2794 \,{\mathrm e}^{4 i x}+1952 \,{\mathrm e}^{3 i x}+1525 \,{\mathrm e}^{2 i x}+488 \,{\mathrm e}^{i x}+61\right )}{3465 \left ({\mathrm e}^{i x}+1\right )^{11} \left ({\mathrm e}^{i x}-1\right )^{3}}\) | \(94\) |
1/1408*tan(1/2*x)^11-1/1152*tan(1/2*x)^9-3/896*tan(1/2*x)^7+3/640*tan(1/2* x)^5+1/128*tan(1/2*x)^3-3/128*tan(1/2*x)+1/128/tan(1/2*x)-1/384/tan(1/2*x) ^3
Time = 0.24 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.20 \[ \int \frac {1}{(\sin (x)+\tan (x))^4} \, dx=\frac {122 \, \cos \left (x\right )^{7} + 488 \, \cos \left (x\right )^{6} + 549 \, \cos \left (x\right )^{5} - 244 \, \cos \left (x\right )^{4} - 64 \, \cos \left (x\right )^{3} + 144 \, \cos \left (x\right )^{2} + 128 \, \cos \left (x\right ) + 32}{3465 \, {\left (\cos \left (x\right )^{6} + 4 \, \cos \left (x\right )^{5} + 5 \, \cos \left (x\right )^{4} - 5 \, \cos \left (x\right )^{2} - 4 \, \cos \left (x\right ) - 1\right )} \sin \left (x\right )} \]
1/3465*(122*cos(x)^7 + 488*cos(x)^6 + 549*cos(x)^5 - 244*cos(x)^4 - 64*cos (x)^3 + 144*cos(x)^2 + 128*cos(x) + 32)/((cos(x)^6 + 4*cos(x)^5 + 5*cos(x) ^4 - 5*cos(x)^2 - 4*cos(x) - 1)*sin(x))
\[ \int \frac {1}{(\sin (x)+\tan (x))^4} \, dx=\int \frac {1}{\left (\sin {\left (x \right )} + \tan {\left (x \right )}\right )^{4}}\, dx \]
Time = 0.24 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.49 \[ \int \frac {1}{(\sin (x)+\tan (x))^4} \, dx=\frac {{\left (\frac {3 \, \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} - 1\right )} {\left (\cos \left (x\right ) + 1\right )}^{3}}{384 \, \sin \left (x\right )^{3}} - \frac {3 \, \sin \left (x\right )}{128 \, {\left (\cos \left (x\right ) + 1\right )}} + \frac {\sin \left (x\right )^{3}}{128 \, {\left (\cos \left (x\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (x\right )^{5}}{640 \, {\left (\cos \left (x\right ) + 1\right )}^{5}} - \frac {3 \, \sin \left (x\right )^{7}}{896 \, {\left (\cos \left (x\right ) + 1\right )}^{7}} - \frac {\sin \left (x\right )^{9}}{1152 \, {\left (\cos \left (x\right ) + 1\right )}^{9}} + \frac {\sin \left (x\right )^{11}}{1408 \, {\left (\cos \left (x\right ) + 1\right )}^{11}} \]
1/384*(3*sin(x)^2/(cos(x) + 1)^2 - 1)*(cos(x) + 1)^3/sin(x)^3 - 3/128*sin( x)/(cos(x) + 1) + 1/128*sin(x)^3/(cos(x) + 1)^3 + 3/640*sin(x)^5/(cos(x) + 1)^5 - 3/896*sin(x)^7/(cos(x) + 1)^7 - 1/1152*sin(x)^9/(cos(x) + 1)^9 + 1 /1408*sin(x)^11/(cos(x) + 1)^11
Time = 0.28 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(\sin (x)+\tan (x))^4} \, dx=\frac {1}{1408} \, \tan \left (\frac {1}{2} \, x\right )^{11} - \frac {1}{1152} \, \tan \left (\frac {1}{2} \, x\right )^{9} - \frac {3}{896} \, \tan \left (\frac {1}{2} \, x\right )^{7} + \frac {3}{640} \, \tan \left (\frac {1}{2} \, x\right )^{5} + \frac {1}{128} \, \tan \left (\frac {1}{2} \, x\right )^{3} + \frac {3 \, \tan \left (\frac {1}{2} \, x\right )^{2} - 1}{384 \, \tan \left (\frac {1}{2} \, x\right )^{3}} - \frac {3}{128} \, \tan \left (\frac {1}{2} \, x\right ) \]
1/1408*tan(1/2*x)^11 - 1/1152*tan(1/2*x)^9 - 3/896*tan(1/2*x)^7 + 3/640*ta n(1/2*x)^5 + 1/128*tan(1/2*x)^3 + 1/384*(3*tan(1/2*x)^2 - 1)/tan(1/2*x)^3 - 3/128*tan(1/2*x)
Time = 29.11 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.34 \[ \int \frac {1}{(\sin (x)+\tan (x))^4} \, dx=-\frac {15616\,{\cos \left (\frac {x}{2}\right )}^{14}-23424\,{\cos \left (\frac {x}{2}\right )}^{12}+5856\,{\cos \left (\frac {x}{2}\right )}^{10}+976\,{\cos \left (\frac {x}{2}\right )}^8+7296\,{\cos \left (\frac {x}{2}\right )}^6-7440\,{\cos \left (\frac {x}{2}\right )}^4+2590\,{\cos \left (\frac {x}{2}\right )}^2-315}{443520\,\left ({\cos \left (\frac {x}{2}\right )}^{11}\,\sin \left (\frac {x}{2}\right )-{\cos \left (\frac {x}{2}\right )}^{13}\,\sin \left (\frac {x}{2}\right )\right )} \]