Integrand size = 21, antiderivative size = 69 \[ \int \frac {\tan ^4(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\sec (c+d x)}{a d}+\frac {2 \sec ^3(c+d x)}{3 a d}-\frac {\sec ^5(c+d x)}{5 a d}+\frac {\tan ^5(c+d x)}{5 a d} \]
Time = 0.17 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.54 \[ \int \frac {\tan ^4(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\sec ^3(c+d x) (200-534 \cos (c+d x)+288 \cos (2 (c+d x))-178 \cos (3 (c+d x))+24 \cos (4 (c+d x))-64 \sin (c+d x)-178 \sin (2 (c+d x))+192 \sin (3 (c+d x))-89 \sin (4 (c+d x)))}{960 a d (1+\sin (c+d x))} \]
-1/960*(Sec[c + d*x]^3*(200 - 534*Cos[c + d*x] + 288*Cos[2*(c + d*x)] - 17 8*Cos[3*(c + d*x)] + 24*Cos[4*(c + d*x)] - 64*Sin[c + d*x] - 178*Sin[2*(c + d*x)] + 192*Sin[3*(c + d*x)] - 89*Sin[4*(c + d*x)]))/(a*d*(1 + Sin[c + d *x]))
Time = 0.36 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.84, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3185, 3042, 3086, 210, 2009, 3087, 15}
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.
\(\displaystyle \int \frac {\tan ^4(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)^4}{a \sin (c+d x)+a}dx\) |
\(\Big \downarrow \) 3185 |
\(\displaystyle \frac {\int \sec ^2(c+d x) \tan ^4(c+d x)dx}{a}-\frac {\int \sec (c+d x) \tan ^5(c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \sec (c+d x)^2 \tan (c+d x)^4dx}{a}-\frac {\int \sec (c+d x) \tan (c+d x)^5dx}{a}\) |
\(\Big \downarrow \) 3086 |
\(\displaystyle \frac {\int \sec (c+d x)^2 \tan (c+d x)^4dx}{a}-\frac {\int \left (\sec ^2(c+d x)-1\right )^2d\sec (c+d x)}{a d}\) |
\(\Big \downarrow \) 210 |
\(\displaystyle \frac {\int \sec (c+d x)^2 \tan (c+d x)^4dx}{a}-\frac {\int \left (\sec ^4(c+d x)-2 \sec ^2(c+d x)+1\right )d\sec (c+d x)}{a d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\int \sec (c+d x)^2 \tan (c+d x)^4dx}{a}-\frac {\frac {1}{5} \sec ^5(c+d x)-\frac {2}{3} \sec ^3(c+d x)+\sec (c+d x)}{a d}\) |
\(\Big \downarrow \) 3087 |
\(\displaystyle \frac {\int \tan ^4(c+d x)d\tan (c+d x)}{a d}-\frac {\frac {1}{5} \sec ^5(c+d x)-\frac {2}{3} \sec ^3(c+d x)+\sec (c+d x)}{a d}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle \frac {\tan ^5(c+d x)}{5 a d}-\frac {\frac {1}{5} \sec ^5(c+d x)-\frac {2}{3} \sec ^3(c+d x)+\sec (c+d x)}{a d}\) |
3.9.23.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^2 )^p, x], x] /; FreeQ[{a, b}, x] && IGtQ[p, 0]
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_.), x_Symbol] :> Simp[a/f Subst[Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2 ), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2 ] && !(IntegerQ[m/2] && LtQ[0, m, n + 1])
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S ymbol] :> Simp[1/f Subst[Int[(b*x)^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] && !(IntegerQ[(n - 1) /2] && LtQ[0, n, m - 1])
Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*( x_)]), x_Symbol] :> Simp[1/a Int[Sec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x ] - Simp[1/(b*g) Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /; Fre eQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]
Time = 0.29 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.07
method | result | size |
parallelrisch | \(\frac {\frac {16}{15}-\frac {32 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {32 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15}+\frac {32 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{15}}{d a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\) | \(74\) |
norman | \(\frac {\frac {16}{15 a d}+\frac {32 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{15 d a}-\frac {32 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d a}-\frac {32 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}\) | \(92\) |
risch | \(-\frac {2 \left (25 i {\mathrm e}^{4 i \left (d x +c \right )}+5 \,{\mathrm e}^{5 i \left (d x +c \right )}+21 i {\mathrm e}^{2 i \left (d x +c \right )}+13 \,{\mathrm e}^{3 i \left (d x +c \right )}+15 i {\mathrm e}^{6 i \left (d x +c \right )}+15 \,{\mathrm e}^{7 i \left (d x +c \right )}-9 \,{\mathrm e}^{i \left (d x +c \right )}+3 i\right )}{15 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{5} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} d a}\) | \(120\) |
derivativedivides | \(\frac {-\frac {1}{6 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {3}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {1}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {3}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d a}\) | \(130\) |
default | \(\frac {-\frac {1}{6 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {3}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {1}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {3}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d a}\) | \(130\) |
16/15*(1-6*tan(1/2*d*x+1/2*c)^3-2*tan(1/2*d*x+1/2*c)^2+2*tan(1/2*d*x+1/2*c ))/d/a/(tan(1/2*d*x+1/2*c)-1)^3/(tan(1/2*d*x+1/2*c)+1)^5
Time = 0.25 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.09 \[ \int \frac {\tan ^4(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {3 \, \cos \left (d x + c\right )^{4} + 6 \, \cos \left (d x + c\right )^{2} + 4 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - 1}{15 \, {\left (a d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{3}\right )}} \]
-1/15*(3*cos(d*x + c)^4 + 6*cos(d*x + c)^2 + 4*(3*cos(d*x + c)^2 - 1)*sin( d*x + c) - 1)/(a*d*cos(d*x + c)^3*sin(d*x + c) + a*d*cos(d*x + c)^3)
\[ \int \frac {\tan ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\sin ^{4}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]
Leaf count of result is larger than twice the leaf count of optimal. 214 vs. \(2 (63) = 126\).
Time = 0.23 (sec) , antiderivative size = 214, normalized size of antiderivative = 3.10 \[ \int \frac {\tan ^4(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {16 \, {\left (\frac {2 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {2 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {6 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + 1\right )}}{15 \, {\left (a + \frac {2 \, a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {6 \, a \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {6 \, a \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {2 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {2 \, a \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )} d} \]
-16/15*(2*sin(d*x + c)/(cos(d*x + c) + 1) - 2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 6*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 1)/((a + 2*a*sin(d*x + c )/(cos(d*x + c) + 1) - 2*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 6*a*sin(d *x + c)^3/(cos(d*x + c) + 1)^3 + 6*a*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 2*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 2*a*sin(d*x + c)^7/(cos(d*x + c ) + 1)^7 - a*sin(d*x + c)^8/(cos(d*x + c) + 1)^8)*d)
Time = 0.33 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.74 \[ \int \frac {\tan ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {5 \, {\left (9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 11\right )}}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} - \frac {45 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 240 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 490 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 320 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 73}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{120 \, d} \]
1/120*(5*(9*tan(1/2*d*x + 1/2*c)^2 - 24*tan(1/2*d*x + 1/2*c) + 11)/(a*(tan (1/2*d*x + 1/2*c) - 1)^3) - (45*tan(1/2*d*x + 1/2*c)^4 + 240*tan(1/2*d*x + 1/2*c)^3 + 490*tan(1/2*d*x + 1/2*c)^2 + 320*tan(1/2*d*x + 1/2*c) + 73)/(a *(tan(1/2*d*x + 1/2*c) + 1)^5))/d
Time = 10.11 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.06 \[ \int \frac {\tan ^4(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {16\,\left (-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{15\,a\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^5} \]