Integrand size = 19, antiderivative size = 115 \[ \int (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx=-\frac {23 a \log (1-\sin (c+d x))}{16 d}+\frac {7 a \log (1+\sin (c+d x))}{16 d}-\frac {a \sin (c+d x)}{d}+\frac {a^3}{8 d (a-a \sin (c+d x))^2}-\frac {a^2}{d (a-a \sin (c+d x))}+\frac {a^2}{8 d (a+a \sin (c+d x))} \] Output:
-23/16*a*ln(1-sin(d*x+c))/d+7/16*a*ln(1+sin(d*x+c))/d-a*sin(d*x+c)/d+1/8*a ^3/d/(a-a*sin(d*x+c))^2-a^2/d/(a-a*sin(d*x+c))+1/8*a^2/d/(a+a*sin(d*x+c))
Time = 0.08 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.23 \[ \int (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {15 a \text {arctanh}(\sin (c+d x))}{8 d}-\frac {a \log (\cos (c+d x))}{d}-\frac {a \sec ^2(c+d x)}{d}+\frac {a \sec ^4(c+d x)}{4 d}+\frac {15 a \sec (c+d x) \tan (c+d x)}{8 d}-\frac {15 a \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {5 a \sec (c+d x) \tan ^3(c+d x)}{d}-\frac {a \sin (c+d x) \tan ^4(c+d x)}{d} \] Input:
Integrate[(a + a*Sin[c + d*x])*Tan[c + d*x]^5,x]
Output:
(15*a*ArcTanh[Sin[c + d*x]])/(8*d) - (a*Log[Cos[c + d*x]])/d - (a*Sec[c + d*x]^2)/d + (a*Sec[c + d*x]^4)/(4*d) + (15*a*Sec[c + d*x]*Tan[c + d*x])/(8 *d) - (15*a*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (5*a*Sec[c + d*x]*Tan[c + d*x]^3)/d - (a*Sin[c + d*x]*Tan[c + d*x]^4)/d
Time = 0.27 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, 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.
| \(\displaystyle \int \tan ^5(c+d x) (a \sin (c+d x)+a) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan (c+d x)^5 (a \sin (c+d x)+a)dx\) |
| \(\Big \downarrow \) 3186 |
| \(\displaystyle \frac {\int \frac {a^5 \sin ^5(c+d x)}{(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {\int \left (\frac {a^3}{4 (a-a \sin (c+d x))^3}-\frac {a^2}{(a-a \sin (c+d x))^2}-\frac {a^2}{8 (\sin (c+d x) a+a)^2}+\frac {23 a}{16 (a-a \sin (c+d x))}+\frac {7 a}{16 (\sin (c+d x) a+a)}-1\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {a^3}{8 (a-a \sin (c+d x))^2}-\frac {a^2}{a-a \sin (c+d x)}+\frac {a^2}{8 (a \sin (c+d x)+a)}-a \sin (c+d x)-\frac {23}{16} a \log (a-a \sin (c+d x))+\frac {7}{16} a \log (a \sin (c+d x)+a)}{d}\) |
Input:
Int[(a + a*Sin[c + d*x])*Tan[c + d*x]^5,x]
Output:
((-23*a*Log[a - a*Sin[c + d*x]])/16 + (7*a*Log[a + a*Sin[c + d*x]])/16 - a *Sin[c + d*x] + a^3/(8*(a - a*Sin[c + d*x])^2) - a^2/(a - a*Sin[c + d*x]) + a^2/(8*(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 = 1.21 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.05
| method | result | size |
| derivativedivides | \(\frac {a \left (\frac {\sin \left (d x +c \right )^{7}}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \sin \left (d x +c \right )^{7}}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \sin \left (d x +c \right )^{5}}{8}-\frac {5 \sin \left (d x +c \right )^{3}}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a \left (\frac {\tan \left (d x +c \right )^{4}}{4}-\frac {\tan \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(121\) |
| default | \(\frac {a \left (\frac {\sin \left (d x +c \right )^{7}}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \sin \left (d x +c \right )^{7}}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \sin \left (d x +c \right )^{5}}{8}-\frac {5 \sin \left (d x +c \right )^{3}}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a \left (\frac {\tan \left (d x +c \right )^{4}}{4}-\frac {\tan \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(121\) |
| parts | \(\frac {a \left (\frac {\tan \left (d x +c \right )^{4}}{4}-\frac {\tan \left (d x +c \right )^{2}}{2}+\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}+\frac {a \left (\frac {\sin \left (d x +c \right )^{7}}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \sin \left (d x +c \right )^{7}}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \sin \left (d x +c \right )^{5}}{8}-\frac {5 \sin \left (d x +c \right )^{3}}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(127\) |
| risch | \(i a x +\frac {i a \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {2 i a c}{d}+\frac {i \left (2 i a \,{\mathrm e}^{2 i \left (d x +c \right )}+9 a \,{\mathrm e}^{i \left (d x +c \right )}-2 i a \,{\mathrm e}^{4 i \left (d x +c \right )}+6 a \,{\mathrm e}^{3 i \left (d x +c \right )}+9 a \,{\mathrm e}^{5 i \left (d x +c \right )}\right )}{4 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2} d}+\frac {7 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {23 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}\) | \(182\) |
Input:
int((a+a*sin(d*x+c))*tan(d*x+c)^5,x,method=_RETURNVERBOSE)
Output:
1/d*(a*(1/4*sin(d*x+c)^7/cos(d*x+c)^4-3/8*sin(d*x+c)^7/cos(d*x+c)^2-3/8*si n(d*x+c)^5-5/8*sin(d*x+c)^3-15/8*sin(d*x+c)+15/8*ln(sec(d*x+c)+tan(d*x+c)) )+a*(1/4*tan(d*x+c)^4-1/2*tan(d*x+c)^2-ln(cos(d*x+c))))
Time = 0.14 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.38 \[ \int (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {16 \, a \cos \left (d x + c\right )^{4} + 2 \, a \cos \left (d x + c\right )^{2} + 7 \, {\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 23 \, {\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, a \cos \left (d x + c\right )^{2} + a\right )} \sin \left (d x + c\right ) - 6 \, a}{16 \, {\left (d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{2}\right )}} \] Input:
integrate((a+a*sin(d*x+c))*tan(d*x+c)^5,x, algorithm="fricas")
Output:
1/16*(16*a*cos(d*x + c)^4 + 2*a*cos(d*x + c)^2 + 7*(a*cos(d*x + c)^2*sin(d *x + c) - a*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 23*(a*cos(d*x + c)^2*s in(d*x + c) - a*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) + 2*(8*a*cos(d*x + c)^2 + a)*sin(d*x + c) - 6*a)/(d*cos(d*x + c)^2*sin(d*x + c) - d*cos(d*x + c)^2)
\[ \int (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx=a \left (\int \sin {\left (c + d x \right )} \tan ^{5}{\left (c + d x \right )}\, dx + \int \tan ^{5}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate((a+a*sin(d*x+c))*tan(d*x+c)**5,x)
Output:
a*(Integral(sin(c + d*x)*tan(c + d*x)**5, x) + Integral(tan(c + d*x)**5, x ))
Time = 0.03 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.83 \[ \int (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {7 \, a \log \left (\sin \left (d x + c\right ) + 1\right ) - 23 \, a \log \left (\sin \left (d x + c\right ) - 1\right ) - 16 \, a \sin \left (d x + c\right ) + \frac {2 \, {\left (9 \, a \sin \left (d x + c\right )^{2} - a \sin \left (d x + c\right ) - 6 \, a\right )}}{\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )^{2} - \sin \left (d x + c\right ) + 1}}{16 \, d} \] Input:
integrate((a+a*sin(d*x+c))*tan(d*x+c)^5,x, algorithm="maxima")
Output:
1/16*(7*a*log(sin(d*x + c) + 1) - 23*a*log(sin(d*x + c) - 1) - 16*a*sin(d* x + c) + 2*(9*a*sin(d*x + c)^2 - a*sin(d*x + c) - 6*a)/(sin(d*x + c)^3 - s in(d*x + c)^2 - sin(d*x + c) + 1))/d
Time = 0.19 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.78 \[ \int (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {1}{16} \, a {\left (\frac {7 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{d} - \frac {23 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{d} - \frac {16 \, \sin \left (d x + c\right )}{d} + \frac {2 \, {\left (9 \, \sin \left (d x + c\right )^{2} - \sin \left (d x + c\right ) - 6\right )}}{d {\left (\sin \left (d x + c\right ) + 1\right )} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}}\right )} \] Input:
integrate((a+a*sin(d*x+c))*tan(d*x+c)^5,x, algorithm="giac")
Output:
1/16*a*(7*log(abs(sin(d*x + c) + 1))/d - 23*log(abs(sin(d*x + c) - 1))/d - 16*sin(d*x + c)/d + 2*(9*sin(d*x + c)^2 - sin(d*x + c) - 6)/(d*(sin(d*x + c) + 1)*(sin(d*x + c) - 1)^2))
Time = 17.28 (sec) , antiderivative size = 235, normalized size of antiderivative = 2.04 \[ \int (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {-\frac {15\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{2}+\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}-\frac {15\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}-\frac {23\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{8\,d}+\frac {7\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{8\,d}+\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \] Input:
int(tan(c + d*x)^5*(a + a*sin(c + d*x)),x)
Output:
((11*a*tan(c/2 + (d*x)/2)^2)/2 - (15*a*tan(c/2 + (d*x)/2))/4 + (11*a*tan(c /2 + (d*x)/2)^3)/4 - 5*a*tan(c/2 + (d*x)/2)^4 + (11*a*tan(c/2 + (d*x)/2)^5 )/4 + (11*a*tan(c/2 + (d*x)/2)^6)/2 - (15*a*tan(c/2 + (d*x)/2)^7)/4)/(d*(2 *tan(c/2 + (d*x)/2)^3 - 2*tan(c/2 + (d*x)/2) - 2*tan(c/2 + (d*x)/2)^4 + 2* tan(c/2 + (d*x)/2)^5 - 2*tan(c/2 + (d*x)/2)^7 + tan(c/2 + (d*x)/2)^8 + 1)) - (23*a*log(tan(c/2 + (d*x)/2) - 1))/(8*d) + (7*a*log(tan(c/2 + (d*x)/2) + 1))/(8*d) + (a*log(tan(c/2 + (d*x)/2)^2 + 1))/d
Time = 0.18 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.20 \[ \int (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx=\frac {a \left (-2 \cos \left (d x +c \right ) \tan \left (d x +c \right )^{3}+19 \cos \left (d x +c \right ) \tan \left (d x +c \right )+12 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right )-45 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+45 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+6 \sin \left (d x +c \right ) \tan \left (d x +c \right )^{4}-13 \sin \left (d x +c \right ) \tan \left (d x +c \right )^{2}-64 \sin \left (d x +c \right )+6 \tan \left (d x +c \right )^{4}-12 \tan \left (d x +c \right )^{2}\right )}{24 d} \] Input:
int((a+a*sin(d*x+c))*tan(d*x+c)^5,x)
Output:
(a*( - 2*cos(c + d*x)*tan(c + d*x)**3 + 19*cos(c + d*x)*tan(c + d*x) + 12* log(tan(c + d*x)**2 + 1) - 45*log(tan((c + d*x)/2) - 1) + 45*log(tan((c + d*x)/2) + 1) + 6*sin(c + d*x)*tan(c + d*x)**4 - 13*sin(c + d*x)*tan(c + d* x)**2 - 64*sin(c + d*x) + 6*tan(c + d*x)**4 - 12*tan(c + d*x)**2))/(24*d)