Integrand size = 21, antiderivative size = 195 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {\text {arctanh}(\sin (c+d x))}{128 a^4 d}+\frac {a^2}{48 d (a+a \sin (c+d x))^6}-\frac {7 a}{80 d (a+a \sin (c+d x))^5}+\frac {1}{8 d (a+a \sin (c+d x))^4}-\frac {5}{96 a d (a+a \sin (c+d x))^3}+\frac {1}{256 d \left (a^2-a^2 \sin (c+d x)\right )^2}-\frac {5}{256 d \left (a^2+a^2 \sin (c+d x)\right )^2}-\frac {3}{256 d \left (a^4-a^4 \sin (c+d x)\right )}-\frac {1}{256 d \left (a^4+a^4 \sin (c+d x)\right )} \] Output:
-1/128*arctanh(sin(d*x+c))/a^4/d+1/48*a^2/d/(a+a*sin(d*x+c))^6-7/80*a/d/(a +a*sin(d*x+c))^5+1/8/d/(a+a*sin(d*x+c))^4-5/96/a/d/(a+a*sin(d*x+c))^3+1/25 6/d/(a^2-a^2*sin(d*x+c))^2-5/256/d/(a^2+a^2*sin(d*x+c))^2-3/256/d/(a^4-a^4 *sin(d*x+c))-1/256/d/(a^4+a^4*sin(d*x+c))
Time = 1.59 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.57 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {30 \text {arctanh}(\sin (c+d x))-\frac {2 \left (-48-177 \sin (c+d x)-132 \sin ^2(c+d x)+257 \sin ^3(c+d x)+440 \sin ^4(c+d x)+65 \sin ^5(c+d x)+60 \sin ^6(c+d x)+15 \sin ^7(c+d x)\right )}{(-1+\sin (c+d x))^2 (1+\sin (c+d x))^6}}{3840 a^4 d} \] Input:
Integrate[Tan[c + d*x]^5/(a + a*Sin[c + d*x])^4,x]
Output:
-1/3840*(30*ArcTanh[Sin[c + d*x]] - (2*(-48 - 177*Sin[c + d*x] - 132*Sin[c + d*x]^2 + 257*Sin[c + d*x]^3 + 440*Sin[c + d*x]^4 + 65*Sin[c + d*x]^5 + 60*Sin[c + d*x]^6 + 15*Sin[c + d*x]^7))/((-1 + Sin[c + d*x])^2*(1 + Sin[c + d*x])^6))/(a^4*d)
Time = 0.35 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.86, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 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.
| \(\displaystyle \int \frac {\tan ^5(c+d x)}{(a \sin (c+d x)+a)^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^5}{(a \sin (c+d x)+a)^4}dx\) |
| \(\Big \downarrow \) 3186 |
| \(\displaystyle \frac {\int \frac {a^5 \sin ^5(c+d x)}{(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^7}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {\int \left (-\frac {a^2}{8 (\sin (c+d x) a+a)^7}+\frac {7 a}{16 (\sin (c+d x) a+a)^6}-\frac {1}{2 (\sin (c+d x) a+a)^5}+\frac {5}{32 (\sin (c+d x) a+a)^4 a}+\frac {1}{128 (a-a \sin (c+d x))^3 a^2}+\frac {5}{128 (\sin (c+d x) a+a)^3 a^2}-\frac {1}{128 \left (a^2-a^2 \sin ^2(c+d x)\right ) a^3}-\frac {3}{256 (a-a \sin (c+d x))^2 a^3}+\frac {1}{256 (\sin (c+d x) a+a)^2 a^3}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {\text {arctanh}(\sin (c+d x))}{128 a^4}-\frac {3}{256 a^3 (a-a \sin (c+d x))}-\frac {1}{256 a^3 (a \sin (c+d x)+a)}+\frac {a^2}{48 (a \sin (c+d x)+a)^6}+\frac {1}{256 a^2 (a-a \sin (c+d x))^2}-\frac {5}{256 a^2 (a \sin (c+d x)+a)^2}-\frac {7 a}{80 (a \sin (c+d x)+a)^5}+\frac {1}{8 (a \sin (c+d x)+a)^4}-\frac {5}{96 a (a \sin (c+d x)+a)^3}}{d}\) |
Input:
Int[Tan[c + d*x]^5/(a + a*Sin[c + d*x])^4,x]
Output:
(-1/128*ArcTanh[Sin[c + d*x]]/a^4 + 1/(256*a^2*(a - a*Sin[c + d*x])^2) - 3 /(256*a^3*(a - a*Sin[c + d*x])) + a^2/(48*(a + a*Sin[c + d*x])^6) - (7*a)/ (80*(a + a*Sin[c + d*x])^5) + 1/(8*(a + a*Sin[c + d*x])^4) - 5/(96*a*(a + a*Sin[c + d*x])^3) - 5/(256*a^2*(a + a*Sin[c + d*x])^2) - 1/(256*a^3*(a + a*Sin[c + d*x])))/d
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 = 7.73 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.65
| method | result | size |
| derivativedivides | \(\frac {\frac {1}{48 \left (1+\sin \left (d x +c \right )\right )^{6}}-\frac {7}{80 \left (1+\sin \left (d x +c \right )\right )^{5}}+\frac {1}{8 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {5}{96 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {5}{256 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{256 \left (1+\sin \left (d x +c \right )\right )}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{256}+\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {3}{256 \left (\sin \left (d x +c \right )-1\right )}+\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{256}}{d \,a^{4}}\) | \(127\) |
| default | \(\frac {\frac {1}{48 \left (1+\sin \left (d x +c \right )\right )^{6}}-\frac {7}{80 \left (1+\sin \left (d x +c \right )\right )^{5}}+\frac {1}{8 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {5}{96 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {5}{256 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{256 \left (1+\sin \left (d x +c \right )\right )}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{256}+\frac {1}{256 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {3}{256 \left (\sin \left (d x +c \right )-1\right )}+\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{256}}{d \,a^{4}}\) | \(127\) |
| risch | \(\frac {i \left (-4240 i {\mathrm e}^{12 i \left (d x +c \right )}-4133 \,{\mathrm e}^{9 i \left (d x +c \right )}+4133 \,{\mathrm e}^{7 i \left (d x +c \right )}+5727 \,{\mathrm e}^{11 i \left (d x +c \right )}-365 \,{\mathrm e}^{13 i \left (d x +c \right )}+15 \,{\mathrm e}^{15 i \left (d x +c \right )}-15 \,{\mathrm e}^{i \left (d x +c \right )}+365 \,{\mathrm e}^{3 i \left (d x +c \right )}-5727 \,{\mathrm e}^{5 i \left (d x +c \right )}+120 i {\mathrm e}^{2 i \left (d x +c \right )}+11656 i {\mathrm e}^{6 i \left (d x +c \right )}-8928 i {\mathrm e}^{8 i \left (d x +c \right )}+11656 i {\mathrm e}^{10 i \left (d x +c \right )}+120 i {\mathrm e}^{14 i \left (d x +c \right )}-4240 i {\mathrm e}^{4 i \left (d x +c \right )}\right )}{960 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{12} d \,a^{4}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{128 a^{4} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{128 a^{4} d}\) | \(254\) |
Input:
int(tan(d*x+c)^5/(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)
Output:
1/d/a^4*(1/48/(1+sin(d*x+c))^6-7/80/(1+sin(d*x+c))^5+1/8/(1+sin(d*x+c))^4- 5/96/(1+sin(d*x+c))^3-5/256/(1+sin(d*x+c))^2-1/256/(1+sin(d*x+c))-1/256*ln (1+sin(d*x+c))+1/256/(sin(d*x+c)-1)^2+3/256/(sin(d*x+c)-1)+1/256*ln(sin(d* x+c)-1))
Time = 0.11 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.49 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {120 \, \cos \left (d x + c\right )^{6} - 1240 \, \cos \left (d x + c\right )^{4} + 1856 \, \cos \left (d x + c\right )^{2} + 15 \, {\left (\cos \left (d x + c\right )^{8} - 8 \, \cos \left (d x + c\right )^{6} + 8 \, \cos \left (d x + c\right )^{4} - 4 \, {\left (\cos \left (d x + c\right )^{6} - 2 \, \cos \left (d x + c\right )^{4}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (\cos \left (d x + c\right )^{8} - 8 \, \cos \left (d x + c\right )^{6} + 8 \, \cos \left (d x + c\right )^{4} - 4 \, {\left (\cos \left (d x + c\right )^{6} - 2 \, \cos \left (d x + c\right )^{4}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (15 \, \cos \left (d x + c\right )^{6} - 110 \, \cos \left (d x + c\right )^{4} + 432 \, \cos \left (d x + c\right )^{2} - 160\right )} \sin \left (d x + c\right ) - 640}{3840 \, {\left (a^{4} d \cos \left (d x + c\right )^{8} - 8 \, a^{4} d \cos \left (d x + c\right )^{6} + 8 \, a^{4} d \cos \left (d x + c\right )^{4} - 4 \, {\left (a^{4} d \cos \left (d x + c\right )^{6} - 2 \, a^{4} d \cos \left (d x + c\right )^{4}\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(tan(d*x+c)^5/(a+a*sin(d*x+c))^4,x, algorithm="fricas")
Output:
-1/3840*(120*cos(d*x + c)^6 - 1240*cos(d*x + c)^4 + 1856*cos(d*x + c)^2 + 15*(cos(d*x + c)^8 - 8*cos(d*x + c)^6 + 8*cos(d*x + c)^4 - 4*(cos(d*x + c) ^6 - 2*cos(d*x + c)^4)*sin(d*x + c))*log(sin(d*x + c) + 1) - 15*(cos(d*x + c)^8 - 8*cos(d*x + c)^6 + 8*cos(d*x + c)^4 - 4*(cos(d*x + c)^6 - 2*cos(d* x + c)^4)*sin(d*x + c))*log(-sin(d*x + c) + 1) + 2*(15*cos(d*x + c)^6 - 11 0*cos(d*x + c)^4 + 432*cos(d*x + c)^2 - 160)*sin(d*x + c) - 640)/(a^4*d*co s(d*x + c)^8 - 8*a^4*d*cos(d*x + c)^6 + 8*a^4*d*cos(d*x + c)^4 - 4*(a^4*d* cos(d*x + c)^6 - 2*a^4*d*cos(d*x + c)^4)*sin(d*x + c))
\[ \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\int \frac {\tan ^{5}{\left (c + d x \right )}}{\sin ^{4}{\left (c + d x \right )} + 4 \sin ^{3}{\left (c + d x \right )} + 6 \sin ^{2}{\left (c + d x \right )} + 4 \sin {\left (c + d x \right )} + 1}\, dx}{a^{4}} \] Input:
integrate(tan(d*x+c)**5/(a+a*sin(d*x+c))**4,x)
Output:
Integral(tan(c + d*x)**5/(sin(c + d*x)**4 + 4*sin(c + d*x)**3 + 6*sin(c + d*x)**2 + 4*sin(c + d*x) + 1), x)/a**4
Time = 0.04 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.09 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{7} + 60 \, \sin \left (d x + c\right )^{6} + 65 \, \sin \left (d x + c\right )^{5} + 440 \, \sin \left (d x + c\right )^{4} + 257 \, \sin \left (d x + c\right )^{3} - 132 \, \sin \left (d x + c\right )^{2} - 177 \, \sin \left (d x + c\right ) - 48\right )}}{a^{4} \sin \left (d x + c\right )^{8} + 4 \, a^{4} \sin \left (d x + c\right )^{7} + 4 \, a^{4} \sin \left (d x + c\right )^{6} - 4 \, a^{4} \sin \left (d x + c\right )^{5} - 10 \, a^{4} \sin \left (d x + c\right )^{4} - 4 \, a^{4} \sin \left (d x + c\right )^{3} + 4 \, a^{4} \sin \left (d x + c\right )^{2} + 4 \, a^{4} \sin \left (d x + c\right ) + a^{4}} - \frac {15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{4}} + \frac {15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{4}}}{3840 \, d} \] Input:
integrate(tan(d*x+c)^5/(a+a*sin(d*x+c))^4,x, algorithm="maxima")
Output:
1/3840*(2*(15*sin(d*x + c)^7 + 60*sin(d*x + c)^6 + 65*sin(d*x + c)^5 + 440 *sin(d*x + c)^4 + 257*sin(d*x + c)^3 - 132*sin(d*x + c)^2 - 177*sin(d*x + c) - 48)/(a^4*sin(d*x + c)^8 + 4*a^4*sin(d*x + c)^7 + 4*a^4*sin(d*x + c)^6 - 4*a^4*sin(d*x + c)^5 - 10*a^4*sin(d*x + c)^4 - 4*a^4*sin(d*x + c)^3 + 4 *a^4*sin(d*x + c)^2 + 4*a^4*sin(d*x + c) + a^4) - 15*log(sin(d*x + c) + 1) /a^4 + 15*log(sin(d*x + c) - 1)/a^4)/d
Time = 0.16 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.69 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{256 \, a^{4} d} + \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{256 \, a^{4} d} + \frac {15 \, \sin \left (d x + c\right )^{7} + 60 \, \sin \left (d x + c\right )^{6} + 65 \, \sin \left (d x + c\right )^{5} + 440 \, \sin \left (d x + c\right )^{4} + 257 \, \sin \left (d x + c\right )^{3} - 132 \, \sin \left (d x + c\right )^{2} - 177 \, \sin \left (d x + c\right ) - 48}{1920 \, a^{4} d {\left (\sin \left (d x + c\right ) + 1\right )}^{6} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} \] Input:
integrate(tan(d*x+c)^5/(a+a*sin(d*x+c))^4,x, algorithm="giac")
Output:
-1/256*log(abs(sin(d*x + c) + 1))/(a^4*d) + 1/256*log(abs(sin(d*x + c) - 1 ))/(a^4*d) + 1/1920*(15*sin(d*x + c)^7 + 60*sin(d*x + c)^6 + 65*sin(d*x + c)^5 + 440*sin(d*x + c)^4 + 257*sin(d*x + c)^3 - 132*sin(d*x + c)^2 - 177* sin(d*x + c) - 48)/(a^4*d*(sin(d*x + c) + 1)^6*(sin(d*x + c) - 1)^2)
Time = 20.49 (sec) , antiderivative size = 476, normalized size of antiderivative = 2.44 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx =\text {Too large to display} \] Input:
int(tan(c + d*x)^5/(a + a*sin(c + d*x))^4,x)
Output:
(tan(c/2 + (d*x)/2)/64 + tan(c/2 + (d*x)/2)^2/8 + (73*tan(c/2 + (d*x)/2)^3 )/192 + (5*tan(c/2 + (d*x)/2)^4)/12 - (139*tan(c/2 + (d*x)/2)^5)/320 + (10 73*tan(c/2 + (d*x)/2)^6)/120 + (10277*tan(c/2 + (d*x)/2)^7)/960 + (237*tan (c/2 + (d*x)/2)^8)/10 + (10277*tan(c/2 + (d*x)/2)^9)/960 + (1073*tan(c/2 + (d*x)/2)^10)/120 - (139*tan(c/2 + (d*x)/2)^11)/320 + (5*tan(c/2 + (d*x)/2 )^12)/12 + (73*tan(c/2 + (d*x)/2)^13)/192 + tan(c/2 + (d*x)/2)^14/8 + tan( c/2 + (d*x)/2)^15/64)/(d*(24*a^4*tan(c/2 + (d*x)/2)^2 + 24*a^4*tan(c/2 + ( d*x)/2)^3 - 36*a^4*tan(c/2 + (d*x)/2)^4 - 120*a^4*tan(c/2 + (d*x)/2)^5 - 8 8*a^4*tan(c/2 + (d*x)/2)^6 + 88*a^4*tan(c/2 + (d*x)/2)^7 + 198*a^4*tan(c/2 + (d*x)/2)^8 + 88*a^4*tan(c/2 + (d*x)/2)^9 - 88*a^4*tan(c/2 + (d*x)/2)^10 - 120*a^4*tan(c/2 + (d*x)/2)^11 - 36*a^4*tan(c/2 + (d*x)/2)^12 + 24*a^4*t an(c/2 + (d*x)/2)^13 + 24*a^4*tan(c/2 + (d*x)/2)^14 + 8*a^4*tan(c/2 + (d*x )/2)^15 + a^4*tan(c/2 + (d*x)/2)^16 + a^4 + 8*a^4*tan(c/2 + (d*x)/2))) - a tanh(tan(c/2 + (d*x)/2))/(64*a^4*d)
Time = 0.24 (sec) , antiderivative size = 536, normalized size of antiderivative = 2.75 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^4} \, dx =\text {Too large to display} \] Input:
int(tan(d*x+c)^5/(a+a*sin(d*x+c))^4,x)
Output:
(60*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**8 + 240*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**7 + 240*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**6 - 240* log(tan((c + d*x)/2) - 1)*sin(c + d*x)**5 - 600*log(tan((c + d*x)/2) - 1)* sin(c + d*x)**4 - 240*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3 + 240*log( tan((c + d*x)/2) - 1)*sin(c + d*x)**2 + 240*log(tan((c + d*x)/2) - 1)*sin( c + d*x) + 60*log(tan((c + d*x)/2) - 1) - 60*log(tan((c + d*x)/2) + 1)*sin (c + d*x)**8 - 240*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**7 - 240*log(tan ((c + d*x)/2) + 1)*sin(c + d*x)**6 + 240*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**5 + 600*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4 + 240*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3 - 240*log(tan((c + d*x)/2) + 1)*sin(c + d*x )**2 - 240*log(tan((c + d*x)/2) + 1)*sin(c + d*x) - 60*log(tan((c + d*x)/2 ) + 1) + 177*sin(c + d*x)**8 + 768*sin(c + d*x)**7 + 948*sin(c + d*x)**6 - 448*sin(c + d*x)**5 - 10*sin(c + d*x)**4 + 320*sin(c + d*x)**3 + 180*sin( c + d*x)**2 - 15)/(7680*a**4*d*(sin(c + d*x)**8 + 4*sin(c + d*x)**7 + 4*si n(c + d*x)**6 - 4*sin(c + d*x)**5 - 10*sin(c + d*x)**4 - 4*sin(c + d*x)**3 + 4*sin(c + d*x)**2 + 4*sin(c + d*x) + 1))