Integrand size = 25, antiderivative size = 117 \[ \int \sec (c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=-\frac {11 a \log (1-\sin (c+d x))}{16 d}-\frac {5 a \log (1+\sin (c+d x))}{16 d}+\frac {a^5}{8 d \left (a^2-a^2 \sin (c+d x)\right )^2}-\frac {3 a^5}{4 d \left (a^4-a^4 \sin (c+d x)\right )}-\frac {a^5}{8 d \left (a^4+a^4 \sin (c+d x)\right )} \] Output:
-11/16*a*ln(1-sin(d*x+c))/d-5/16*a*ln(1+sin(d*x+c))/d+1/8*a^5/d/(a^2-a^2*s in(d*x+c))^2-3/4*a^5/d/(a^4-a^4*sin(d*x+c))-1/8*a^5/d/(a^4+a^4*sin(d*x+c))
Time = 0.04 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.03 \[ \int \sec (c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {3 a \text {arctanh}(\sin (c+d x))}{8 d}-\frac {a \log (\cos (c+d x))}{d}-\frac {a \sec ^2(c+d x)}{d}+\frac {a \sec ^4(c+d x)}{4 d}+\frac {3 a \sec (c+d x) \tan (c+d x)}{8 d}-\frac {3 a \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a \sec (c+d x) \tan ^3(c+d x)}{d} \] Input:
Integrate[Sec[c + d*x]*(a + a*Sin[c + d*x])*Tan[c + d*x]^4,x]
Output:
(3*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) + (3*a*Sec[c + d*x]*Tan[c + d*x])/(8*d ) - (3*a*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (a*Sec[c + d*x]*Tan[c + d*x] ^3)/d
Time = 0.31 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.79, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3315, 27, 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.
| \(\displaystyle \int \tan ^4(c+d x) \sec (c+d x) (a \sin (c+d x)+a) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^4 (a \sin (c+d x)+a)}{\cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3315 |
| \(\displaystyle \frac {a^5 \int \frac {\sin ^4(c+d x)}{(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a \int \frac {a^4 \sin ^4(c+d x)}{(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {a \int \left (\frac {a^2}{4 (a-a \sin (c+d x))^3}-\frac {3 a}{4 (a-a \sin (c+d x))^2}+\frac {a}{8 (\sin (c+d x) a+a)^2}+\frac {11}{16 (a-a \sin (c+d x))}-\frac {5}{16 (\sin (c+d x) a+a)}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a \left (\frac {a^2}{8 (a-a \sin (c+d x))^2}-\frac {3 a}{4 (a-a \sin (c+d x))}-\frac {a}{8 (a \sin (c+d x)+a)}-\frac {11}{16} \log (a-a \sin (c+d x))-\frac {5}{16} \log (a \sin (c+d x)+a)\right )}{d}\) |
Input:
Int[Sec[c + d*x]*(a + a*Sin[c + d*x])*Tan[c + d*x]^4,x]
Output:
(a*((-11*Log[a - a*Sin[c + d*x]])/16 - (5*Log[a + a*Sin[c + d*x]])/16 + a^ 2/(8*(a - a*Sin[c + d*x])^2) - (3*a)/(4*(a - a*Sin[c + d*x])) - a/(8*(a + a*Sin[c + d*x]))))/d
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
Time = 0.39 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.95
| method | result | size |
| derivativedivides | \(\frac {a \left (\frac {\tan \left (d x +c \right )^{4}}{4}-\frac {\tan \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+a \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(111\) |
| default | \(\frac {a \left (\frac {\tan \left (d x +c \right )^{4}}{4}-\frac {\tan \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+a \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(111\) |
| risch | \(i a x +\frac {2 i a c}{d}+\frac {i a \left (6 i {\mathrm e}^{4 i \left (d x +c \right )}+5 \,{\mathrm e}^{5 i \left (d x +c \right )}-6 i {\mathrm e}^{2 i \left (d x +c \right )}+14 \,{\mathrm e}^{3 i \left (d x +c \right )}+5 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d}-\frac {5 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {11 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}\) | \(146\) |
Input:
int(sec(d*x+c)*(a+a*sin(d*x+c))*tan(d*x+c)^4,x,method=_RETURNVERBOSE)
Output:
1/d*(a*(1/4*tan(d*x+c)^4-1/2*tan(d*x+c)^2-ln(cos(d*x+c)))+a*(1/4*sin(d*x+c )^5/cos(d*x+c)^4-1/8*sin(d*x+c)^5/cos(d*x+c)^2-1/8*sin(d*x+c)^3-3/8*sin(d* x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c))))
Time = 0.11 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.16 \[ \int \sec (c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {10 \, a \cos \left (d x + c\right )^{2} - 5 \, {\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 11 \, {\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 6 \, a \sin \left (d x + c\right ) + 2 \, a}{16 \, {\left (d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{2}\right )}} \] Input:
integrate(sec(d*x+c)*(a+a*sin(d*x+c))*tan(d*x+c)^4,x, algorithm="fricas")
Output:
1/16*(10*a*cos(d*x + c)^2 - 5*(a*cos(d*x + c)^2*sin(d*x + c) - a*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 11*(a*cos(d*x + c)^2*sin(d*x + c) - a*cos(d *x + c)^2)*log(-sin(d*x + c) + 1) - 6*a*sin(d*x + c) + 2*a)/(d*cos(d*x + c )^2*sin(d*x + c) - d*cos(d*x + c)^2)
\[ \int \sec (c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=a \left (\int \tan ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \sin {\left (c + d x \right )} \tan ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx\right ) \] Input:
integrate(sec(d*x+c)*(a+a*sin(d*x+c))*tan(d*x+c)**4,x)
Output:
a*(Integral(tan(c + d*x)**4*sec(c + d*x), x) + Integral(sin(c + d*x)*tan(c + d*x)**4*sec(c + d*x), x))
Time = 0.03 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.74 \[ \int \sec (c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=-\frac {5 \, a \log \left (\sin \left (d x + c\right ) + 1\right ) + 11 \, a \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (5 \, a \sin \left (d x + c\right )^{2} + 3 \, a \sin \left (d x + c\right ) - 6 \, a\right )}}{\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )^{2} - \sin \left (d x + c\right ) + 1}}{16 \, d} \] Input:
integrate(sec(d*x+c)*(a+a*sin(d*x+c))*tan(d*x+c)^4,x, algorithm="maxima")
Output:
-1/16*(5*a*log(sin(d*x + c) + 1) + 11*a*log(sin(d*x + c) - 1) - 2*(5*a*sin (d*x + c)^2 + 3*a*sin(d*x + c) - 6*a)/(sin(d*x + c)^3 - sin(d*x + c)^2 - s in(d*x + c) + 1))/d
Time = 0.19 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.68 \[ \int \sec (c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=-\frac {1}{16} \, a {\left (\frac {5 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{d} + \frac {11 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{d} - \frac {2 \, {\left (5 \, \sin \left (d x + c\right )^{2} + 3 \, \sin \left (d x + c\right ) - 6\right )}}{d {\left (\sin \left (d x + c\right ) + 1\right )} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}}\right )} \] Input:
integrate(sec(d*x+c)*(a+a*sin(d*x+c))*tan(d*x+c)^4,x, algorithm="giac")
Output:
-1/16*a*(5*log(abs(sin(d*x + c) + 1))/d + 11*log(abs(sin(d*x + c) - 1))/d - 2*(5*sin(d*x + c)^2 + 3*sin(d*x + c) - 6)/(d*(sin(d*x + c) + 1)*(sin(d*x + c) - 1)^2))
Time = 32.78 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.75 \[ \int \sec (c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {5\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{8\,d}-\frac {11\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{8\,d}+\frac {\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}-\frac {9\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}+\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+\frac {3\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )} \] Input:
int((tan(c + d*x)^4*(a + a*sin(c + d*x)))/cos(c + d*x),x)
Output:
(a*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (5*a*log(tan(c/2 + (d*x)/2) + 1))/(8 *d) - (11*a*log(tan(c/2 + (d*x)/2) - 1))/(8*d) + ((3*a*tan(c/2 + (d*x)/2)) /4 + (a*tan(c/2 + (d*x)/2)^2)/2 - (9*a*tan(c/2 + (d*x)/2)^3)/2 + (a*tan(c/ 2 + (d*x)/2)^4)/2 + (3*a*tan(c/2 + (d*x)/2)^5)/4)/(d*(2*tan(c/2 + (d*x)/2) + tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^3 + tan(c/2 + (d*x)/2)^4 + 2*tan(c/2 + (d*x)/2)^5 - tan(c/2 + (d*x)/2)^6 - 1))
Time = 0.19 (sec) , antiderivative size = 300, normalized size of antiderivative = 2.56 \[ \int \sec (c+d x) (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {a \left (8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{3}-8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2}-8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )+8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )-11 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{3}+11 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}+11 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )-11 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-5 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3}+5 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}+5 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )-5 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+3 \sin \left (d x +c \right )^{3}+2 \sin \left (d x +c \right )^{2}-3\right )}{8 d \left (\sin \left (d x +c \right )^{3}-\sin \left (d x +c \right )^{2}-\sin \left (d x +c \right )+1\right )} \] Input:
int(sec(d*x+c)*(a+a*sin(d*x+c))*tan(d*x+c)^4,x)
Output:
(a*(8*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**3 - 8*log(tan((c + d*x)/2 )**2 + 1)*sin(c + d*x)**2 - 8*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x) + 8*log(tan((c + d*x)/2)**2 + 1) - 11*log(tan((c + d*x)/2) - 1)*sin(c + d*x) **3 + 11*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 + 11*log(tan((c + d*x)/ 2) - 1)*sin(c + d*x) - 11*log(tan((c + d*x)/2) - 1) - 5*log(tan((c + d*x)/ 2) + 1)*sin(c + d*x)**3 + 5*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 + 5* log(tan((c + d*x)/2) + 1)*sin(c + d*x) - 5*log(tan((c + d*x)/2) + 1) + 3*s in(c + d*x)**3 + 2*sin(c + d*x)**2 - 3))/(8*d*(sin(c + d*x)**3 - sin(c + d *x)**2 - sin(c + d*x) + 1))