Integrand size = 21, antiderivative size = 126 \[ \int \frac {\tan ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\text {arctanh}(\sin (c+d x))}{32 a^3 d}+\frac {a}{16 d (a+a \sin (c+d x))^4}-\frac {1}{6 d (a+a \sin (c+d x))^3}+\frac {3}{32 a d (a+a \sin (c+d x))^2}+\frac {1}{32 d \left (a^3-a^3 \sin (c+d x)\right )}+\frac {1}{16 d \left (a^3+a^3 \sin (c+d x)\right )} \]
-1/32*arctanh(sin(d*x+c))/a^3/d+1/16*a/d/(a+a*sin(d*x+c))^4-1/6/d/(a+a*sin (d*x+c))^3+3/32/a/d/(a+a*sin(d*x+c))^2+1/32/d/(a^3-a^3*sin(d*x+c))+1/16/d/ (a^3+a^3*sin(d*x+c))
Time = 0.24 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.65 \[ \int \frac {\tan ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {3 \text {arctanh}(\sin (c+d x))-\frac {3}{1-\sin (c+d x)}-\frac {6}{(1+\sin (c+d x))^4}+\frac {16}{(1+\sin (c+d x))^3}-\frac {9}{(1+\sin (c+d x))^2}-\frac {6}{1+\sin (c+d x)}}{96 a^3 d} \]
-1/96*(3*ArcTanh[Sin[c + d*x]] - 3/(1 - Sin[c + d*x]) - 6/(1 + Sin[c + d*x ])^4 + 16/(1 + Sin[c + d*x])^3 - 9/(1 + Sin[c + d*x])^2 - 6/(1 + Sin[c + d *x]))/(a^3*d)
Time = 0.28 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.87, 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.
| \(\displaystyle \int \frac {\tan ^3(c+d x)}{(a \sin (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^3}{(a \sin (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 3186 |
| \(\displaystyle \frac {\int \frac {a^3 \sin ^3(c+d x)}{(a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^5}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {\int \left (-\frac {a}{4 (\sin (c+d x) a+a)^5}+\frac {1}{2 (\sin (c+d x) a+a)^4}-\frac {3}{16 (\sin (c+d x) a+a)^3 a}-\frac {1}{32 \left (a^2-a^2 \sin ^2(c+d x)\right ) a^2}+\frac {1}{32 (a-a \sin (c+d x))^2 a^2}-\frac {1}{16 (\sin (c+d x) a+a)^2 a^2}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {\text {arctanh}(\sin (c+d x))}{32 a^3}+\frac {1}{32 a^2 (a-a \sin (c+d x))}+\frac {1}{16 a^2 (a \sin (c+d x)+a)}+\frac {a}{16 (a \sin (c+d x)+a)^4}-\frac {1}{6 (a \sin (c+d x)+a)^3}+\frac {3}{32 a (a \sin (c+d x)+a)^2}}{d}\) |
(-1/32*ArcTanh[Sin[c + d*x]]/a^3 + 1/(32*a^2*(a - a*Sin[c + d*x])) + a/(16 *(a + a*Sin[c + d*x])^4) - 1/(6*(a + a*Sin[c + d*x])^3) + 3/(32*a*(a + a*S in[c + d*x])^2) + 1/(16*a^2*(a + a*Sin[c + d*x])))/d
3.1.73.3.1 Defintions of rubi rules used
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 = 2.14 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.72
| method | result | size |
| derivativedivides | \(\frac {\frac {1}{16 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{6 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {3}{32 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {1}{16+16 \sin \left (d x +c \right )}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{64}-\frac {1}{32 \left (\sin \left (d x +c \right )-1\right )}+\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{64}}{d \,a^{3}}\) | \(91\) |
| default | \(\frac {\frac {1}{16 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{6 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {3}{32 \left (1+\sin \left (d x +c \right )\right )^{2}}+\frac {1}{16+16 \sin \left (d x +c \right )}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{64}-\frac {1}{32 \left (\sin \left (d x +c \right )-1\right )}+\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{64}}{d \,a^{3}}\) | \(91\) |
| risch | \(\frac {i \left (-310 \,{\mathrm e}^{5 i \left (d x +c \right )}-162 i {\mathrm e}^{4 i \left (d x +c \right )}+88 \,{\mathrm e}^{3 i \left (d x +c \right )}-18 i {\mathrm e}^{2 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )}+88 \,{\mathrm e}^{7 i \left (d x +c \right )}+162 i {\mathrm e}^{6 i \left (d x +c \right )}+18 i {\mathrm e}^{8 i \left (d x +c \right )}+3 \,{\mathrm e}^{9 i \left (d x +c \right )}\right )}{48 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{8} \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )^{2} d \,a^{3}}+\frac {\ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )}{32 a^{3} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{32 a^{3} d}\) | \(185\) |
1/d/a^3*(1/16/(1+sin(d*x+c))^4-1/6/(1+sin(d*x+c))^3+3/32/(1+sin(d*x+c))^2+ 1/16/(1+sin(d*x+c))-1/64*ln(1+sin(d*x+c))-1/32/(sin(d*x+c)-1)+1/64*ln(sin( d*x+c)-1))
Time = 0.31 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.79 \[ \int \frac {\tan ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {6 \, \cos \left (d x + c\right )^{4} + 38 \, \cos \left (d x + c\right )^{2} - 3 \, {\left (3 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (3 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 18 \, {\left (\cos \left (d x + c\right )^{2} + 2\right )} \sin \left (d x + c\right ) - 60}{192 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{4} - 4 \, a^{3} d \cos \left (d x + c\right )^{2} + {\left (a^{3} d \cos \left (d x + c\right )^{4} - 4 \, a^{3} d \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )}} \]
1/192*(6*cos(d*x + c)^4 + 38*cos(d*x + c)^2 - 3*(3*cos(d*x + c)^4 - 4*cos( d*x + c)^2 + (cos(d*x + c)^4 - 4*cos(d*x + c)^2)*sin(d*x + c))*log(sin(d*x + c) + 1) + 3*(3*cos(d*x + c)^4 - 4*cos(d*x + c)^2 + (cos(d*x + c)^4 - 4* cos(d*x + c)^2)*sin(d*x + c))*log(-sin(d*x + c) + 1) - 18*(cos(d*x + c)^2 + 2)*sin(d*x + c) - 60)/(3*a^3*d*cos(d*x + c)^4 - 4*a^3*d*cos(d*x + c)^2 + (a^3*d*cos(d*x + c)^4 - 4*a^3*d*cos(d*x + c)^2)*sin(d*x + c))
\[ \int \frac {\tan ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\int \frac {\tan ^{3}{\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
Time = 0.20 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.16 \[ \int \frac {\tan ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{4} + 9 \, \sin \left (d x + c\right )^{3} - 25 \, \sin \left (d x + c\right )^{2} - 27 \, \sin \left (d x + c\right ) - 8\right )}}{a^{3} \sin \left (d x + c\right )^{5} + 3 \, a^{3} \sin \left (d x + c\right )^{4} + 2 \, a^{3} \sin \left (d x + c\right )^{3} - 2 \, a^{3} \sin \left (d x + c\right )^{2} - 3 \, a^{3} \sin \left (d x + c\right ) - a^{3}} - \frac {3 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} + \frac {3 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3}}}{192 \, d} \]
1/192*(2*(3*sin(d*x + c)^4 + 9*sin(d*x + c)^3 - 25*sin(d*x + c)^2 - 27*sin (d*x + c) - 8)/(a^3*sin(d*x + c)^5 + 3*a^3*sin(d*x + c)^4 + 2*a^3*sin(d*x + c)^3 - 2*a^3*sin(d*x + c)^2 - 3*a^3*sin(d*x + c) - a^3) - 3*log(sin(d*x + c) + 1)/a^3 + 3*log(sin(d*x + c) - 1)/a^3)/d
Time = 1.00 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.90 \[ \int \frac {\tan ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {12 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3}} - \frac {12 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3}} + \frac {12 \, {\left (\sin \left (d x + c\right ) + 1\right )}}{a^{3} {\left (\sin \left (d x + c\right ) - 1\right )}} - \frac {25 \, \sin \left (d x + c\right )^{4} + 148 \, \sin \left (d x + c\right )^{3} + 366 \, \sin \left (d x + c\right )^{2} + 260 \, \sin \left (d x + c\right ) + 65}{a^{3} {\left (\sin \left (d x + c\right ) + 1\right )}^{4}}}{768 \, d} \]
-1/768*(12*log(abs(sin(d*x + c) + 1))/a^3 - 12*log(abs(sin(d*x + c) - 1))/ a^3 + 12*(sin(d*x + c) + 1)/(a^3*(sin(d*x + c) - 1)) - (25*sin(d*x + c)^4 + 148*sin(d*x + c)^3 + 366*sin(d*x + c)^2 + 260*sin(d*x + c) + 65)/(a^3*(s in(d*x + c) + 1)^4))/d
Time = 9.69 (sec) , antiderivative size = 302, normalized size of antiderivative = 2.40 \[ \int \frac {\tan ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{16}+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{8}+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{6}+\frac {37\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{8}+\frac {101\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{24}+\frac {37\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{8}+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+6\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+13\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+8\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-14\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-28\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-14\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+8\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+13\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+6\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a^3\right )}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{16\,a^3\,d} \]
(tan(c/2 + (d*x)/2)/16 + (3*tan(c/2 + (d*x)/2)^2)/8 + (5*tan(c/2 + (d*x)/2 )^3)/6 + (37*tan(c/2 + (d*x)/2)^4)/8 + (101*tan(c/2 + (d*x)/2)^5)/24 + (37 *tan(c/2 + (d*x)/2)^6)/8 + (5*tan(c/2 + (d*x)/2)^7)/6 + (3*tan(c/2 + (d*x) /2)^8)/8 + tan(c/2 + (d*x)/2)^9/16)/(d*(13*a^3*tan(c/2 + (d*x)/2)^2 + 8*a^ 3*tan(c/2 + (d*x)/2)^3 - 14*a^3*tan(c/2 + (d*x)/2)^4 - 28*a^3*tan(c/2 + (d *x)/2)^5 - 14*a^3*tan(c/2 + (d*x)/2)^6 + 8*a^3*tan(c/2 + (d*x)/2)^7 + 13*a ^3*tan(c/2 + (d*x)/2)^8 + 6*a^3*tan(c/2 + (d*x)/2)^9 + a^3*tan(c/2 + (d*x) /2)^10 + a^3 + 6*a^3*tan(c/2 + (d*x)/2))) - atanh(tan(c/2 + (d*x)/2))/(16* a^3*d)