Integrand size = 17, antiderivative size = 99 \[ \int \sec ^5(a+b x) \tan ^4(a+b x) \, dx=\frac {3 \text {arctanh}(\sin (a+b x))}{128 b}+\frac {3 \sec (a+b x) \tan (a+b x)}{128 b}+\frac {\sec ^3(a+b x) \tan (a+b x)}{64 b}-\frac {\sec ^5(a+b x) \tan (a+b x)}{16 b}+\frac {\sec ^5(a+b x) \tan ^3(a+b x)}{8 b} \] Output:
3/128*arctanh(sin(b*x+a))/b+3/128*sec(b*x+a)*tan(b*x+a)/b+1/64*sec(b*x+a)^ 3*tan(b*x+a)/b-1/16*sec(b*x+a)^5*tan(b*x+a)/b+1/8*sec(b*x+a)^5*tan(b*x+a)^ 3/b
Time = 0.03 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.21 \[ \int \sec ^5(a+b x) \tan ^4(a+b x) \, dx=\frac {3 \text {arctanh}(\sin (a+b x))}{128 b}+\frac {3 \sec (a+b x) \tan (a+b x)}{128 b}+\frac {\sec ^3(a+b x) \tan (a+b x)}{64 b}+\frac {\sec ^5(a+b x) \tan (a+b x)}{80 b}-\frac {3 \sec ^7(a+b x) \tan (a+b x)}{40 b}+\frac {\sec ^5(a+b x) \tan ^3(a+b x)}{5 b} \] Input:
Integrate[Sec[a + b*x]^5*Tan[a + b*x]^4,x]
Output:
(3*ArcTanh[Sin[a + b*x]])/(128*b) + (3*Sec[a + b*x]*Tan[a + b*x])/(128*b) + (Sec[a + b*x]^3*Tan[a + b*x])/(64*b) + (Sec[a + b*x]^5*Tan[a + b*x])/(80 *b) - (3*Sec[a + b*x]^7*Tan[a + b*x])/(40*b) + (Sec[a + b*x]^5*Tan[a + b*x ]^3)/(5*b)
Time = 0.95 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.15, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.588, Rules used = {3042, 3091, 3042, 3091, 3042, 4255, 3042, 4255, 3042, 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.
| \(\displaystyle \int \tan ^4(a+b x) \sec ^5(a+b x) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan (a+b x)^4 \sec (a+b x)^5dx\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {\tan ^3(a+b x) \sec ^5(a+b x)}{8 b}-\frac {3}{8} \int \sec ^5(a+b x) \tan ^2(a+b x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan ^3(a+b x) \sec ^5(a+b x)}{8 b}-\frac {3}{8} \int \sec (a+b x)^5 \tan (a+b x)^2dx\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {\tan ^3(a+b x) \sec ^5(a+b x)}{8 b}-\frac {3}{8} \left (\frac {\tan (a+b x) \sec ^5(a+b x)}{6 b}-\frac {1}{6} \int \sec ^5(a+b x)dx\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan ^3(a+b x) \sec ^5(a+b x)}{8 b}-\frac {3}{8} \left (\frac {\tan (a+b x) \sec ^5(a+b x)}{6 b}-\frac {1}{6} \int \csc \left (a+b x+\frac {\pi }{2}\right )^5dx\right )\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {\tan ^3(a+b x) \sec ^5(a+b x)}{8 b}-\frac {3}{8} \left (\frac {1}{6} \left (-\frac {3}{4} \int \sec ^3(a+b x)dx-\frac {\tan (a+b x) \sec ^3(a+b x)}{4 b}\right )+\frac {\tan (a+b x) \sec ^5(a+b x)}{6 b}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan ^3(a+b x) \sec ^5(a+b x)}{8 b}-\frac {3}{8} \left (\frac {1}{6} \left (-\frac {3}{4} \int \csc \left (a+b x+\frac {\pi }{2}\right )^3dx-\frac {\tan (a+b x) \sec ^3(a+b x)}{4 b}\right )+\frac {\tan (a+b x) \sec ^5(a+b x)}{6 b}\right )\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {\tan ^3(a+b x) \sec ^5(a+b x)}{8 b}-\frac {3}{8} \left (\frac {1}{6} \left (-\frac {3}{4} \left (\frac {1}{2} \int \sec (a+b x)dx+\frac {\tan (a+b x) \sec (a+b x)}{2 b}\right )-\frac {\tan (a+b x) \sec ^3(a+b x)}{4 b}\right )+\frac {\tan (a+b x) \sec ^5(a+b x)}{6 b}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan ^3(a+b x) \sec ^5(a+b x)}{8 b}-\frac {3}{8} \left (\frac {1}{6} \left (-\frac {3}{4} \left (\frac {1}{2} \int \csc \left (a+b x+\frac {\pi }{2}\right )dx+\frac {\tan (a+b x) \sec (a+b x)}{2 b}\right )-\frac {\tan (a+b x) \sec ^3(a+b x)}{4 b}\right )+\frac {\tan (a+b x) \sec ^5(a+b x)}{6 b}\right )\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\tan ^3(a+b x) \sec ^5(a+b x)}{8 b}-\frac {3}{8} \left (\frac {1}{6} \left (-\frac {3}{4} \left (\frac {\text {arctanh}(\sin (a+b x))}{2 b}+\frac {\tan (a+b x) \sec (a+b x)}{2 b}\right )-\frac {\tan (a+b x) \sec ^3(a+b x)}{4 b}\right )+\frac {\tan (a+b x) \sec ^5(a+b x)}{6 b}\right )\) |
Input:
Int[Sec[a + b*x]^5*Tan[a + b*x]^4,x]
Output:
(Sec[a + b*x]^5*Tan[a + b*x]^3)/(8*b) - (3*((Sec[a + b*x]^5*Tan[a + b*x])/ (6*b) + (-1/4*(Sec[a + b*x]^3*Tan[a + b*x])/b - (3*(ArcTanh[Sin[a + b*x]]/ (2*b) + (Sec[a + b*x]*Tan[a + b*x])/(2*b)))/4)/6))/8
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[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 9.27 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.13
| method | result | size |
| derivativedivides | \(\frac {\frac {\sin \left (b x +a \right )^{5}}{8 \cos \left (b x +a \right )^{8}}+\frac {\sin \left (b x +a \right )^{5}}{16 \cos \left (b x +a \right )^{6}}+\frac {\sin \left (b x +a \right )^{5}}{64 \cos \left (b x +a \right )^{4}}-\frac {\sin \left (b x +a \right )^{5}}{128 \cos \left (b x +a \right )^{2}}-\frac {\sin \left (b x +a \right )^{3}}{128}-\frac {3 \sin \left (b x +a \right )}{128}+\frac {3 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{128}}{b}\) | \(112\) |
| default | \(\frac {\frac {\sin \left (b x +a \right )^{5}}{8 \cos \left (b x +a \right )^{8}}+\frac {\sin \left (b x +a \right )^{5}}{16 \cos \left (b x +a \right )^{6}}+\frac {\sin \left (b x +a \right )^{5}}{64 \cos \left (b x +a \right )^{4}}-\frac {\sin \left (b x +a \right )^{5}}{128 \cos \left (b x +a \right )^{2}}-\frac {\sin \left (b x +a \right )^{3}}{128}-\frac {3 \sin \left (b x +a \right )}{128}+\frac {3 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{128}}{b}\) | \(112\) |
| risch | \(-\frac {i \left (3 \,{\mathrm e}^{15 i \left (b x +a \right )}+23 \,{\mathrm e}^{13 i \left (b x +a \right )}-333 \,{\mathrm e}^{11 i \left (b x +a \right )}+671 \,{\mathrm e}^{9 i \left (b x +a \right )}-671 \,{\mathrm e}^{7 i \left (b x +a \right )}+333 \,{\mathrm e}^{5 i \left (b x +a \right )}-23 \,{\mathrm e}^{3 i \left (b x +a \right )}-3 \,{\mathrm e}^{i \left (b x +a \right )}\right )}{64 b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{8}}-\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-i\right )}{128 b}+\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}{128 b}\) | \(146\) |
Input:
int(sec(b*x+a)^5*tan(b*x+a)^4,x,method=_RETURNVERBOSE)
Output:
1/b*(1/8*sin(b*x+a)^5/cos(b*x+a)^8+1/16*sin(b*x+a)^5/cos(b*x+a)^6+1/64*sin (b*x+a)^5/cos(b*x+a)^4-1/128*sin(b*x+a)^5/cos(b*x+a)^2-1/128*sin(b*x+a)^3- 3/128*sin(b*x+a)+3/128*ln(sec(b*x+a)+tan(b*x+a)))
Time = 0.10 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.95 \[ \int \sec ^5(a+b x) \tan ^4(a+b x) \, dx=\frac {3 \, \cos \left (b x + a\right )^{8} \log \left (\sin \left (b x + a\right ) + 1\right ) - 3 \, \cos \left (b x + a\right )^{8} \log \left (-\sin \left (b x + a\right ) + 1\right ) + 2 \, {\left (3 \, \cos \left (b x + a\right )^{6} + 2 \, \cos \left (b x + a\right )^{4} - 24 \, \cos \left (b x + a\right )^{2} + 16\right )} \sin \left (b x + a\right )}{256 \, b \cos \left (b x + a\right )^{8}} \] Input:
integrate(sec(b*x+a)^5*tan(b*x+a)^4,x, algorithm="fricas")
Output:
1/256*(3*cos(b*x + a)^8*log(sin(b*x + a) + 1) - 3*cos(b*x + a)^8*log(-sin( b*x + a) + 1) + 2*(3*cos(b*x + a)^6 + 2*cos(b*x + a)^4 - 24*cos(b*x + a)^2 + 16)*sin(b*x + a))/(b*cos(b*x + a)^8)
\[ \int \sec ^5(a+b x) \tan ^4(a+b x) \, dx=\int \tan ^{4}{\left (a + b x \right )} \sec ^{5}{\left (a + b x \right )}\, dx \] Input:
integrate(sec(b*x+a)**5*tan(b*x+a)**4,x)
Output:
Integral(tan(a + b*x)**4*sec(a + b*x)**5, x)
Time = 0.05 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.12 \[ \int \sec ^5(a+b x) \tan ^4(a+b x) \, dx=-\frac {\frac {2 \, {\left (3 \, \sin \left (b x + a\right )^{7} - 11 \, \sin \left (b x + a\right )^{5} - 11 \, \sin \left (b x + a\right )^{3} + 3 \, \sin \left (b x + a\right )\right )}}{\sin \left (b x + a\right )^{8} - 4 \, \sin \left (b x + a\right )^{6} + 6 \, \sin \left (b x + a\right )^{4} - 4 \, \sin \left (b x + a\right )^{2} + 1} - 3 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 3 \, \log \left (\sin \left (b x + a\right ) - 1\right )}{256 \, b} \] Input:
integrate(sec(b*x+a)^5*tan(b*x+a)^4,x, algorithm="maxima")
Output:
-1/256*(2*(3*sin(b*x + a)^7 - 11*sin(b*x + a)^5 - 11*sin(b*x + a)^3 + 3*si n(b*x + a))/(sin(b*x + a)^8 - 4*sin(b*x + a)^6 + 6*sin(b*x + a)^4 - 4*sin( b*x + a)^2 + 1) - 3*log(sin(b*x + a) + 1) + 3*log(sin(b*x + a) - 1))/b
Time = 0.25 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.08 \[ \int \sec ^5(a+b x) \tan ^4(a+b x) \, dx=-\frac {\frac {4 \, {\left (3 \, {\left (\frac {1}{\sin \left (b x + a\right )} + \sin \left (b x + a\right )\right )}^{3} - \frac {20}{\sin \left (b x + a\right )} - 20 \, \sin \left (b x + a\right )\right )}}{{\left ({\left (\frac {1}{\sin \left (b x + a\right )} + \sin \left (b x + a\right )\right )}^{2} - 4\right )}^{2}} - 3 \, \log \left ({\left | \frac {1}{\sin \left (b x + a\right )} + \sin \left (b x + a\right ) + 2 \right |}\right ) + 3 \, \log \left ({\left | \frac {1}{\sin \left (b x + a\right )} + \sin \left (b x + a\right ) - 2 \right |}\right )}{512 \, b} \] Input:
integrate(sec(b*x+a)^5*tan(b*x+a)^4,x, algorithm="giac")
Output:
-1/512*(4*(3*(1/sin(b*x + a) + sin(b*x + a))^3 - 20/sin(b*x + a) - 20*sin( b*x + a))/((1/sin(b*x + a) + sin(b*x + a))^2 - 4)^2 - 3*log(abs(1/sin(b*x + a) + sin(b*x + a) + 2)) + 3*log(abs(1/sin(b*x + a) + sin(b*x + a) - 2))) /b
Time = 30.40 (sec) , antiderivative size = 229, normalized size of antiderivative = 2.31 \[ \int \sec ^5(a+b x) \tan ^4(a+b x) \, dx=\frac {3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )\right )}{64\,b}+\frac {-\frac {3\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^{15}}{64}+\frac {23\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^{13}}{64}+\frac {333\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^{11}}{64}+\frac {671\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^9}{64}+\frac {671\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^7}{64}+\frac {333\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^5}{64}+\frac {23\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^3}{64}-\frac {3\,\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}{64}}{b\,\left ({\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^{16}-8\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^{14}+28\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^{12}-56\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^{10}+70\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^8-56\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^6+28\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4-8\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2+1\right )} \] Input:
int(tan(a + b*x)^4/cos(a + b*x)^5,x)
Output:
(3*atanh(tan(a/2 + (b*x)/2)))/(64*b) + ((23*tan(a/2 + (b*x)/2)^3)/64 - (3* tan(a/2 + (b*x)/2))/64 + (333*tan(a/2 + (b*x)/2)^5)/64 + (671*tan(a/2 + (b *x)/2)^7)/64 + (671*tan(a/2 + (b*x)/2)^9)/64 + (333*tan(a/2 + (b*x)/2)^11) /64 + (23*tan(a/2 + (b*x)/2)^13)/64 - (3*tan(a/2 + (b*x)/2)^15)/64)/(b*(28 *tan(a/2 + (b*x)/2)^4 - 8*tan(a/2 + (b*x)/2)^2 - 56*tan(a/2 + (b*x)/2)^6 + 70*tan(a/2 + (b*x)/2)^8 - 56*tan(a/2 + (b*x)/2)^10 + 28*tan(a/2 + (b*x)/2 )^12 - 8*tan(a/2 + (b*x)/2)^14 + tan(a/2 + (b*x)/2)^16 + 1))
Time = 0.18 (sec) , antiderivative size = 290, normalized size of antiderivative = 2.93 \[ \int \sec ^5(a+b x) \tan ^4(a+b x) \, dx=\frac {-3 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b x +a \right )^{8}+12 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b x +a \right )^{6}-18 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b x +a \right )^{4}+12 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b x +a \right )^{2}-3 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )+3 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b x +a \right )^{8}-12 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b x +a \right )^{6}+18 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b x +a \right )^{4}-12 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b x +a \right )^{2}+3 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )-3 \sin \left (b x +a \right )^{7}+11 \sin \left (b x +a \right )^{5}+11 \sin \left (b x +a \right )^{3}-3 \sin \left (b x +a \right )}{128 b \left (\sin \left (b x +a \right )^{8}-4 \sin \left (b x +a \right )^{6}+6 \sin \left (b x +a \right )^{4}-4 \sin \left (b x +a \right )^{2}+1\right )} \] Input:
int(sec(b*x+a)^5*tan(b*x+a)^4,x)
Output:
( - 3*log(tan((a + b*x)/2) - 1)*sin(a + b*x)**8 + 12*log(tan((a + b*x)/2) - 1)*sin(a + b*x)**6 - 18*log(tan((a + b*x)/2) - 1)*sin(a + b*x)**4 + 12*l og(tan((a + b*x)/2) - 1)*sin(a + b*x)**2 - 3*log(tan((a + b*x)/2) - 1) + 3 *log(tan((a + b*x)/2) + 1)*sin(a + b*x)**8 - 12*log(tan((a + b*x)/2) + 1)* sin(a + b*x)**6 + 18*log(tan((a + b*x)/2) + 1)*sin(a + b*x)**4 - 12*log(ta n((a + b*x)/2) + 1)*sin(a + b*x)**2 + 3*log(tan((a + b*x)/2) + 1) - 3*sin( a + b*x)**7 + 11*sin(a + b*x)**5 + 11*sin(a + b*x)**3 - 3*sin(a + b*x))/(1 28*b*(sin(a + b*x)**8 - 4*sin(a + b*x)**6 + 6*sin(a + b*x)**4 - 4*sin(a + b*x)**2 + 1))