Integrand size = 15, antiderivative size = 55 \[ \int \sec (a+b x) \tan ^4(a+b x) \, dx=\frac {3 \text {arctanh}(\sin (a+b x))}{8 b}-\frac {3 \sec (a+b x) \tan (a+b x)}{8 b}+\frac {\sec (a+b x) \tan ^3(a+b x)}{4 b} \] Output:
3/8*arctanh(sin(b*x+a))/b-3/8*sec(b*x+a)*tan(b*x+a)/b+1/4*sec(b*x+a)*tan(b *x+a)^3/b
Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.33 \[ \int \sec (a+b x) \tan ^4(a+b x) \, dx=\frac {3 \text {arctanh}(\sin (a+b x))}{8 b}+\frac {3 \sec (a+b x) \tan (a+b x)}{8 b}-\frac {3 \sec ^3(a+b x) \tan (a+b x)}{4 b}+\frac {\sec (a+b x) \tan ^3(a+b x)}{b} \] Input:
Integrate[Sec[a + b*x]*Tan[a + b*x]^4,x]
Output:
(3*ArcTanh[Sin[a + b*x]])/(8*b) + (3*Sec[a + b*x]*Tan[a + b*x])/(8*b) - (3 *Sec[a + b*x]^3*Tan[a + b*x])/(4*b) + (Sec[a + b*x]*Tan[a + b*x]^3)/b
Time = 0.57 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3091, 3042, 3091, 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.
\(\displaystyle \int \tan ^4(a+b x) \sec (a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (a+b x)^4 \sec (a+b x)dx\) |
\(\Big \downarrow \) 3091 |
\(\displaystyle \frac {\tan ^3(a+b x) \sec (a+b x)}{4 b}-\frac {3}{4} \int \sec (a+b x) \tan ^2(a+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tan ^3(a+b x) \sec (a+b x)}{4 b}-\frac {3}{4} \int \sec (a+b x) \tan (a+b x)^2dx\) |
\(\Big \downarrow \) 3091 |
\(\displaystyle \frac {\tan ^3(a+b x) \sec (a+b x)}{4 b}-\frac {3}{4} \left (\frac {\tan (a+b x) \sec (a+b x)}{2 b}-\frac {1}{2} \int \sec (a+b x)dx\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tan ^3(a+b x) \sec (a+b x)}{4 b}-\frac {3}{4} \left (\frac {\tan (a+b x) \sec (a+b x)}{2 b}-\frac {1}{2} \int \csc \left (a+b x+\frac {\pi }{2}\right )dx\right )\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\tan ^3(a+b x) \sec (a+b x)}{4 b}-\frac {3}{4} \left (\frac {\tan (a+b x) \sec (a+b x)}{2 b}-\frac {\text {arctanh}(\sin (a+b x))}{2 b}\right )\) |
Input:
Int[Sec[a + b*x]*Tan[a + b*x]^4,x]
Output:
(Sec[a + b*x]*Tan[a + b*x]^3)/(4*b) - (3*(-1/2*ArcTanh[Sin[a + b*x]]/b + ( Sec[a + b*x]*Tan[a + b*x])/(2*b)))/4
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_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 1.24 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.38
method | result | size |
derivativedivides | \(\frac {\frac {\sin \left (b x +a \right )^{5}}{4 \cos \left (b x +a \right )^{4}}-\frac {\sin \left (b x +a \right )^{5}}{8 \cos \left (b x +a \right )^{2}}-\frac {\sin \left (b x +a \right )^{3}}{8}-\frac {3 \sin \left (b x +a \right )}{8}+\frac {3 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{8}}{b}\) | \(76\) |
default | \(\frac {\frac {\sin \left (b x +a \right )^{5}}{4 \cos \left (b x +a \right )^{4}}-\frac {\sin \left (b x +a \right )^{5}}{8 \cos \left (b x +a \right )^{2}}-\frac {\sin \left (b x +a \right )^{3}}{8}-\frac {3 \sin \left (b x +a \right )}{8}+\frac {3 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{8}}{b}\) | \(76\) |
risch | \(\frac {i \left (5 \,{\mathrm e}^{7 i \left (b x +a \right )}-3 \,{\mathrm e}^{5 i \left (b x +a \right )}+3 \,{\mathrm e}^{3 i \left (b x +a \right )}-5 \,{\mathrm e}^{i \left (b x +a \right )}\right )}{4 b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{4}}-\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-i\right )}{8 b}+\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}{8 b}\) | \(102\) |
Input:
int(sec(b*x+a)*tan(b*x+a)^4,x,method=_RETURNVERBOSE)
Output:
1/b*(1/4*sin(b*x+a)^5/cos(b*x+a)^4-1/8*sin(b*x+a)^5/cos(b*x+a)^2-1/8*sin(b *x+a)^3-3/8*sin(b*x+a)+3/8*ln(sec(b*x+a)+tan(b*x+a)))
Time = 0.10 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.35 \[ \int \sec (a+b x) \tan ^4(a+b x) \, dx=\frac {3 \, \cos \left (b x + a\right )^{4} \log \left (\sin \left (b x + a\right ) + 1\right ) - 3 \, \cos \left (b x + a\right )^{4} \log \left (-\sin \left (b x + a\right ) + 1\right ) - 2 \, {\left (5 \, \cos \left (b x + a\right )^{2} - 2\right )} \sin \left (b x + a\right )}{16 \, b \cos \left (b x + a\right )^{4}} \] Input:
integrate(sec(b*x+a)*tan(b*x+a)^4,x, algorithm="fricas")
Output:
1/16*(3*cos(b*x + a)^4*log(sin(b*x + a) + 1) - 3*cos(b*x + a)^4*log(-sin(b *x + a) + 1) - 2*(5*cos(b*x + a)^2 - 2)*sin(b*x + a))/(b*cos(b*x + a)^4)
\[ \int \sec (a+b x) \tan ^4(a+b x) \, dx=\int \tan ^{4}{\left (a + b x \right )} \sec {\left (a + b x \right )}\, dx \] Input:
integrate(sec(b*x+a)*tan(b*x+a)**4,x)
Output:
Integral(tan(a + b*x)**4*sec(a + b*x), x)
Time = 0.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.29 \[ \int \sec (a+b x) \tan ^4(a+b x) \, dx=\frac {\frac {2 \, {\left (5 \, \sin \left (b x + a\right )^{3} - 3 \, \sin \left (b x + a\right )\right )}}{\sin \left (b x + a\right )^{4} - 2 \, \sin \left (b x + a\right )^{2} + 1} + 3 \, \log \left (\sin \left (b x + a\right ) + 1\right ) - 3 \, \log \left (\sin \left (b x + a\right ) - 1\right )}{16 \, b} \] Input:
integrate(sec(b*x+a)*tan(b*x+a)^4,x, algorithm="maxima")
Output:
1/16*(2*(5*sin(b*x + a)^3 - 3*sin(b*x + a))/(sin(b*x + a)^4 - 2*sin(b*x + a)^2 + 1) + 3*log(sin(b*x + a) + 1) - 3*log(sin(b*x + a) - 1))/b
Time = 0.20 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.15 \[ \int \sec (a+b x) \tan ^4(a+b x) \, dx=\frac {\frac {2 \, {\left (5 \, \sin \left (b x + a\right )^{3} - 3 \, \sin \left (b x + a\right )\right )}}{{\left (\sin \left (b x + a\right )^{2} - 1\right )}^{2}} + 3 \, \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right ) - 3 \, \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right )}{16 \, b} \] Input:
integrate(sec(b*x+a)*tan(b*x+a)^4,x, algorithm="giac")
Output:
1/16*(2*(5*sin(b*x + a)^3 - 3*sin(b*x + a))/(sin(b*x + a)^2 - 1)^2 + 3*log (abs(sin(b*x + a) + 1)) - 3*log(abs(sin(b*x + a) - 1)))/b
Time = 29.01 (sec) , antiderivative size = 126, normalized size of antiderivative = 2.29 \[ \int \sec (a+b x) \tan ^4(a+b x) \, dx=\frac {3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )\right )}{4\,b}-\frac {\frac {3\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^7}{4}-\frac {11\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^5}{4}-\frac {11\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^3}{4}+\frac {3\,\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}{4}}{b\,\left ({\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2+1\right )} \] Input:
int(tan(a + b*x)^4/cos(a + b*x),x)
Output:
(3*atanh(tan(a/2 + (b*x)/2)))/(4*b) - ((3*tan(a/2 + (b*x)/2))/4 - (11*tan( a/2 + (b*x)/2)^3)/4 - (11*tan(a/2 + (b*x)/2)^5)/4 + (3*tan(a/2 + (b*x)/2)^ 7)/4)/(b*(6*tan(a/2 + (b*x)/2)^4 - 4*tan(a/2 + (b*x)/2)^2 - 4*tan(a/2 + (b *x)/2)^6 + tan(a/2 + (b*x)/2)^8 + 1))
Time = 0.18 (sec) , antiderivative size = 162, normalized size of antiderivative = 2.95 \[ \int \sec (a+b x) \tan ^4(a+b x) \, dx=\frac {-3 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b x +a \right )^{4}+6 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b x +a \right )^{2}-3 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )+3 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b x +a \right )^{4}-6 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b x +a \right )^{2}+3 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )+5 \sin \left (b x +a \right )^{3}-3 \sin \left (b x +a \right )}{8 b \left (\sin \left (b x +a \right )^{4}-2 \sin \left (b x +a \right )^{2}+1\right )} \] Input:
int(sec(b*x+a)*tan(b*x+a)^4,x)
Output:
( - 3*log(tan((a + b*x)/2) - 1)*sin(a + b*x)**4 + 6*log(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)**4 - 6*log(tan((a + b*x)/2) + 1)*sin(a + b*x)**2 + 3*lo g(tan((a + b*x)/2) + 1) + 5*sin(a + b*x)**3 - 3*sin(a + b*x))/(8*b*(sin(a + b*x)**4 - 2*sin(a + b*x)**2 + 1))