Integrand size = 25, antiderivative size = 103 \[ \int \sec (c+d x) (a+b \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {3 a \text {arctanh}(\sin (c+d x))}{8 d}-\frac {b \log (\cos (c+d x))}{d}-\frac {3 a \sec (c+d x) \tan (c+d x)}{8 d}-\frac {b \tan ^2(c+d x)}{2 d}+\frac {a \sec (c+d x) \tan ^3(c+d x)}{4 d}+\frac {b \tan ^4(c+d x)}{4 d} \]
3/8*a*arctanh(sin(d*x+c))/d-b*ln(cos(d*x+c))/d-3/8*a*sec(d*x+c)*tan(d*x+c) /d-1/2*b*tan(d*x+c)^2/d+1/4*a*sec(d*x+c)*tan(d*x+c)^3/d+1/4*b*tan(d*x+c)^4 /d
Time = 0.24 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.12 \[ \int \sec (c+d x) (a+b \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {3 a \text {arctanh}(\sin (c+d x))}{8 d}+\frac {3 a \sec (c+d x) \tan (c+d x)}{8 d}-\frac {3 a \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a \sec (c+d x) \tan ^3(c+d x)}{d}-\frac {b \left (4 \log (\cos (c+d x))+2 \tan ^2(c+d x)-\tan ^4(c+d x)\right )}{4 d} \]
(3*a*ArcTanh[Sin[c + d*x]])/(8*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 - (b*(4*Log[Cos[c + d*x]] + 2*Tan[c + d*x]^2 - Tan[c + d*x]^4))/(4*d)
Time = 0.74 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.05, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3042, 3313, 3042, 3091, 3042, 3091, 3042, 3954, 3042, 3954, 3042, 3956, 4257}
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.
| \(\displaystyle \int \tan ^4(c+d x) \sec (c+d x) (a+b \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^4 (a+b \sin (c+d x))}{\cos (c+d x)^5}dx\) |
| \(\Big \downarrow \) 3313 |
| \(\displaystyle a \int \sec (c+d x) \tan ^4(c+d x)dx+b \int \tan ^5(c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \int \sec (c+d x) \tan (c+d x)^4dx+b \int \tan (c+d x)^5dx\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle a \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \int \sec (c+d x) \tan ^2(c+d x)dx\right )+b \int \tan (c+d x)^5dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \int \sec (c+d x) \tan (c+d x)^2dx\right )+b \int \tan (c+d x)^5dx\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle a \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int \sec (c+d x)dx\right )\right )+b \int \tan (c+d x)^5dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx\right )\right )+b \int \tan (c+d x)^5dx\) |
| \(\Big \downarrow \) 3954 |
| \(\displaystyle a \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx\right )\right )+b \left (\frac {\tan ^4(c+d x)}{4 d}-\int \tan ^3(c+d x)dx\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx\right )\right )+b \left (\frac {\tan ^4(c+d x)}{4 d}-\int \tan (c+d x)^3dx\right )\) |
| \(\Big \downarrow \) 3954 |
| \(\displaystyle a \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx\right )\right )+b \left (\int \tan (c+d x)dx+\frac {\tan ^4(c+d x)}{4 d}-\frac {\tan ^2(c+d x)}{2 d}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx\right )\right )+b \left (\int \tan (c+d x)dx+\frac {\tan ^4(c+d x)}{4 d}-\frac {\tan ^2(c+d x)}{2 d}\right )\) |
| \(\Big \downarrow \) 3956 |
| \(\displaystyle a \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx\right )\right )+b \left (\frac {\tan ^4(c+d x)}{4 d}-\frac {\tan ^2(c+d x)}{2 d}-\frac {\log (\cos (c+d x))}{d}\right )\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle a \left (\frac {\tan ^3(c+d x) \sec (c+d x)}{4 d}-\frac {3}{4} \left (\frac {\tan (c+d x) \sec (c+d x)}{2 d}-\frac {\text {arctanh}(\sin (c+d x))}{2 d}\right )\right )+b \left (\frac {\tan ^4(c+d x)}{4 d}-\frac {\tan ^2(c+d x)}{2 d}-\frac {\log (\cos (c+d x))}{d}\right )\) |
b*(-(Log[Cos[c + d*x]]/d) - Tan[c + d*x]^2/(2*d) + Tan[c + d*x]^4/(4*d)) + a*((Sec[c + d*x]*Tan[c + d*x]^3)/(4*d) - (3*(-1/2*ArcTanh[Sin[c + d*x]]/d + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/4)
3.15.84.3.1 Defintions of rubi rules used
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[b*(a*Sec[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Simp[b^2*((n - 1)/(m + n - 1)) Int[(a*Sec[e + f*x])^m*( b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] & & NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_ ) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a Int[Cos[e + f*x]^ p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d Int[Cos[e + f*x]^p*(d*Sin[e + f*x ])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2 ] && IntegerQ[n] && ((LtQ[p, 0] && NeQ[a^2 - b^2, 0]) || LtQ[0, n, p - 1] | | LtQ[p + 1, -n, 2*p + 1])
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d *x])^(n - 1)/(d*(n - 1))), x] - Simp[b^2 Int[(b*Tan[c + d*x])^(n - 2), x] , x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]
Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d *x], x]]/d, x] /; FreeQ[{c, d}, x]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.59 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.08
| method | result | size |
| derivativedivides | \(\frac {a \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(111\) |
| default | \(\frac {a \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+b \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(111\) |
| parallelrisch | \(\frac {16 b \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 \left (a +\frac {8 b}{3}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+6 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a -\frac {8 b}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+3 a \sin \left (d x +c \right )-5 a \sin \left (3 d x +3 c \right )-4 b \cos \left (2 d x +2 c \right )+3 \cos \left (4 d x +4 c \right ) b +b}{4 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(196\) |
| risch | \(i x b +\frac {2 i b c}{d}+\frac {i \left (5 a \,{\mathrm e}^{7 i \left (d x +c \right )}-3 a \,{\mathrm e}^{5 i \left (d x +c \right )}+16 i b \,{\mathrm e}^{6 i \left (d x +c \right )}+3 a \,{\mathrm e}^{3 i \left (d x +c \right )}+16 i b \,{\mathrm e}^{4 i \left (d x +c \right )}-5 a \,{\mathrm e}^{i \left (d x +c \right )}+16 i b \,{\mathrm e}^{2 i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{d}\) | \(198\) |
| norman | \(\frac {-\frac {3 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {2 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {11 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {2 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {6 b \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {6 b \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 b \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {b \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (3 a -8 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}-\frac {\left (3 a +8 b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}\) | \(252\) |
1/d*(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*si n(d*x+c)^3-3/8*sin(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))+b*(1/4*tan(d*x+c) ^4-1/2*tan(d*x+c)^2-ln(cos(d*x+c))))
Time = 0.29 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.01 \[ \int \sec (c+d x) (a+b \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {{\left (3 \, a - 8 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, a + 8 \, b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 16 \, b \cos \left (d x + c\right )^{2} - 2 \, {\left (5 \, a \cos \left (d x + c\right )^{2} - 2 \, a\right )} \sin \left (d x + c\right ) + 4 \, b}{16 \, d \cos \left (d x + c\right )^{4}} \]
1/16*((3*a - 8*b)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (3*a + 8*b)*cos(d *x + c)^4*log(-sin(d*x + c) + 1) - 16*b*cos(d*x + c)^2 - 2*(5*a*cos(d*x + c)^2 - 2*a)*sin(d*x + c) + 4*b)/(d*cos(d*x + c)^4)
Timed out. \[ \int \sec (c+d x) (a+b \sin (c+d x)) \tan ^4(c+d x) \, dx=\text {Timed out} \]
Time = 0.20 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.97 \[ \int \sec (c+d x) (a+b \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {{\left (3 \, a - 8 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, a + 8 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + \frac {2 \, {\left (5 \, a \sin \left (d x + c\right )^{3} + 8 \, b \sin \left (d x + c\right )^{2} - 3 \, a \sin \left (d x + c\right ) - 6 \, b\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \]
1/16*((3*a - 8*b)*log(sin(d*x + c) + 1) - (3*a + 8*b)*log(sin(d*x + c) - 1 ) + 2*(5*a*sin(d*x + c)^3 + 8*b*sin(d*x + c)^2 - 3*a*sin(d*x + c) - 6*b)/( sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d
Time = 0.35 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.97 \[ \int \sec (c+d x) (a+b \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {{\left (3 \, a - 8 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (3 \, a + 8 \, b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (6 \, b \sin \left (d x + c\right )^{4} + 5 \, a \sin \left (d x + c\right )^{3} - 4 \, b \sin \left (d x + c\right )^{2} - 3 \, a \sin \left (d x + c\right )\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]
1/16*((3*a - 8*b)*log(abs(sin(d*x + c) + 1)) - (3*a + 8*b)*log(abs(sin(d*x + c) - 1)) + 2*(6*b*sin(d*x + c)^4 + 5*a*sin(d*x + c)^3 - 4*b*sin(d*x + c )^2 - 3*a*sin(d*x + c))/(sin(d*x + c)^2 - 1)^2)/d
Time = 12.12 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.15 \[ \int \sec (c+d x) (a+b \sin (c+d x)) \tan ^4(c+d x) \, dx=\frac {b\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-8\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {3\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (\frac {3\,a}{8}+b\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {3\,a}{8}-b\right )}{d} \]
(b*log(tan(c/2 + (d*x)/2)^2 + 1))/d - ((3*a*tan(c/2 + (d*x)/2))/4 - (11*a* tan(c/2 + (d*x)/2)^3)/4 - (11*a*tan(c/2 + (d*x)/2)^5)/4 + (3*a*tan(c/2 + ( d*x)/2)^7)/4 + 2*b*tan(c/2 + (d*x)/2)^2 - 8*b*tan(c/2 + (d*x)/2)^4 + 2*b*t an(c/2 + (d*x)/2)^6)/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) - (log(tan(c/2 + (d*x )/2) - 1)*((3*a)/8 + b))/d + (log(tan(c/2 + (d*x)/2) + 1)*((3*a)/8 - b))/d