Integrand size = 17, antiderivative size = 76 \[ \int \sec ^5(a+b x) \tan ^2(a+b x) \, dx=-\frac {\text {arctanh}(\sin (a+b x))}{16 b}-\frac {\sec (a+b x) \tan (a+b x)}{16 b}-\frac {\sec ^3(a+b x) \tan (a+b x)}{24 b}+\frac {\sec ^5(a+b x) \tan (a+b x)}{6 b} \] Output:
-1/16*arctanh(sin(b*x+a))/b-1/16*sec(b*x+a)*tan(b*x+a)/b-1/24*sec(b*x+a)^3 *tan(b*x+a)/b+1/6*sec(b*x+a)^5*tan(b*x+a)/b
Time = 0.02 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00 \[ \int \sec ^5(a+b x) \tan ^2(a+b x) \, dx=-\frac {\text {arctanh}(\sin (a+b x))}{16 b}-\frac {\sec (a+b x) \tan (a+b x)}{16 b}-\frac {\sec ^3(a+b x) \tan (a+b x)}{24 b}+\frac {\sec ^5(a+b x) \tan (a+b x)}{6 b} \] Input:
Integrate[Sec[a + b*x]^5*Tan[a + b*x]^2,x]
Output:
-1/16*ArcTanh[Sin[a + b*x]]/b - (Sec[a + b*x]*Tan[a + b*x])/(16*b) - (Sec[ a + b*x]^3*Tan[a + b*x])/(24*b) + (Sec[a + b*x]^5*Tan[a + b*x])/(6*b)
Time = 0.72 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.13, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {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.
| \(\displaystyle \int \tan ^2(a+b x) \sec ^5(a+b x) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan (a+b x)^2 \sec (a+b x)^5dx\) |
| \(\Big \downarrow \) 3091 |
| \(\displaystyle \frac {\tan (a+b x) \sec ^5(a+b x)}{6 b}-\frac {1}{6} \int \sec ^5(a+b x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\tan (a+b x) \sec ^5(a+b x)}{6 b}-\frac {1}{6} \int \csc \left (a+b x+\frac {\pi }{2}\right )^5dx\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{6} \left (-\frac {3}{4} \int \sec ^3(a+b x)dx-\frac {\tan (a+b x) \sec ^3(a+b x)}{4 b}\right )+\frac {\tan (a+b x) \sec ^5(a+b x)}{6 b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{6} \left (-\frac {3}{4} \int \csc \left (a+b x+\frac {\pi }{2}\right )^3dx-\frac {\tan (a+b x) \sec ^3(a+b x)}{4 b}\right )+\frac {\tan (a+b x) \sec ^5(a+b x)}{6 b}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{6} \left (-\frac {3}{4} \left (\frac {1}{2} \int \sec (a+b x)dx+\frac {\tan (a+b x) \sec (a+b x)}{2 b}\right )-\frac {\tan (a+b x) \sec ^3(a+b x)}{4 b}\right )+\frac {\tan (a+b x) \sec ^5(a+b x)}{6 b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{6} \left (-\frac {3}{4} \left (\frac {1}{2} \int \csc \left (a+b x+\frac {\pi }{2}\right )dx+\frac {\tan (a+b x) \sec (a+b x)}{2 b}\right )-\frac {\tan (a+b x) \sec ^3(a+b x)}{4 b}\right )+\frac {\tan (a+b x) \sec ^5(a+b x)}{6 b}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{6} \left (-\frac {3}{4} \left (\frac {\text {arctanh}(\sin (a+b x))}{2 b}+\frac {\tan (a+b x) \sec (a+b x)}{2 b}\right )-\frac {\tan (a+b x) \sec ^3(a+b x)}{4 b}\right )+\frac {\tan (a+b x) \sec ^5(a+b x)}{6 b}\) |
Input:
Int[Sec[a + b*x]^5*Tan[a + b*x]^2,x]
Output:
(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
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 = 8.76 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.11
| method | result | size |
| derivativedivides | \(\frac {\frac {\sin \left (b x +a \right )^{3}}{6 \cos \left (b x +a \right )^{6}}+\frac {\sin \left (b x +a \right )^{3}}{8 \cos \left (b x +a \right )^{4}}+\frac {\sin \left (b x +a \right )^{3}}{16 \cos \left (b x +a \right )^{2}}+\frac {\sin \left (b x +a \right )}{16}-\frac {\ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{16}}{b}\) | \(84\) |
| default | \(\frac {\frac {\sin \left (b x +a \right )^{3}}{6 \cos \left (b x +a \right )^{6}}+\frac {\sin \left (b x +a \right )^{3}}{8 \cos \left (b x +a \right )^{4}}+\frac {\sin \left (b x +a \right )^{3}}{16 \cos \left (b x +a \right )^{2}}+\frac {\sin \left (b x +a \right )}{16}-\frac {\ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{16}}{b}\) | \(84\) |
| risch | \(\frac {i \left (3 \,{\mathrm e}^{11 i \left (b x +a \right )}+17 \,{\mathrm e}^{9 i \left (b x +a \right )}-114 \,{\mathrm e}^{7 i \left (b x +a \right )}+114 \,{\mathrm e}^{5 i \left (b x +a \right )}-17 \,{\mathrm e}^{3 i \left (b x +a \right )}-3 \,{\mathrm e}^{i \left (b x +a \right )}\right )}{24 b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{6}}+\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}-i\right )}{16 b}-\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}{16 b}\) | \(124\) |
Input:
int(sec(b*x+a)^5*tan(b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
1/b*(1/6*sin(b*x+a)^3/cos(b*x+a)^6+1/8*sin(b*x+a)^3/cos(b*x+a)^4+1/16*sin( b*x+a)^3/cos(b*x+a)^2+1/16*sin(b*x+a)-1/16*ln(sec(b*x+a)+tan(b*x+a)))
Time = 0.12 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.11 \[ \int \sec ^5(a+b x) \tan ^2(a+b x) \, dx=-\frac {3 \, \cos \left (b x + a\right )^{6} \log \left (\sin \left (b x + a\right ) + 1\right ) - 3 \, \cos \left (b x + a\right )^{6} \log \left (-\sin \left (b x + a\right ) + 1\right ) + 2 \, {\left (3 \, \cos \left (b x + a\right )^{4} + 2 \, \cos \left (b x + a\right )^{2} - 8\right )} \sin \left (b x + a\right )}{96 \, b \cos \left (b x + a\right )^{6}} \] Input:
integrate(sec(b*x+a)^5*tan(b*x+a)^2,x, algorithm="fricas")
Output:
-1/96*(3*cos(b*x + a)^6*log(sin(b*x + a) + 1) - 3*cos(b*x + a)^6*log(-sin( b*x + a) + 1) + 2*(3*cos(b*x + a)^4 + 2*cos(b*x + a)^2 - 8)*sin(b*x + a))/ (b*cos(b*x + a)^6)
\[ \int \sec ^5(a+b x) \tan ^2(a+b x) \, dx=\int \tan ^{2}{\left (a + b x \right )} \sec ^{5}{\left (a + b x \right )}\, dx \] Input:
integrate(sec(b*x+a)**5*tan(b*x+a)**2,x)
Output:
Integral(tan(a + b*x)**2*sec(a + b*x)**5, x)
Time = 0.04 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.20 \[ \int \sec ^5(a+b x) \tan ^2(a+b x) \, dx=\frac {\frac {2 \, {\left (3 \, \sin \left (b x + a\right )^{5} - 8 \, \sin \left (b x + a\right )^{3} - 3 \, \sin \left (b x + a\right )\right )}}{\sin \left (b x + a\right )^{6} - 3 \, \sin \left (b x + a\right )^{4} + 3 \, \sin \left (b x + a\right )^{2} - 1} - 3 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 3 \, \log \left (\sin \left (b x + a\right ) - 1\right )}{96 \, b} \] Input:
integrate(sec(b*x+a)^5*tan(b*x+a)^2,x, algorithm="maxima")
Output:
1/96*(2*(3*sin(b*x + a)^5 - 8*sin(b*x + a)^3 - 3*sin(b*x + a))/(sin(b*x + a)^6 - 3*sin(b*x + a)^4 + 3*sin(b*x + a)^2 - 1) - 3*log(sin(b*x + a) + 1) + 3*log(sin(b*x + a) - 1))/b
Time = 0.22 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.96 \[ \int \sec ^5(a+b x) \tan ^2(a+b x) \, dx=\frac {\frac {2 \, {\left (3 \, \sin \left (b x + a\right )^{5} - 8 \, \sin \left (b x + a\right )^{3} - 3 \, \sin \left (b x + a\right )\right )}}{{\left (\sin \left (b x + a\right )^{2} - 1\right )}^{3}} - 3 \, \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right ) + 3 \, \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right )}{96 \, b} \] Input:
integrate(sec(b*x+a)^5*tan(b*x+a)^2,x, algorithm="giac")
Output:
1/96*(2*(3*sin(b*x + a)^5 - 8*sin(b*x + a)^3 - 3*sin(b*x + a))/(sin(b*x + a)^2 - 1)^3 - 3*log(abs(sin(b*x + a) + 1)) + 3*log(abs(sin(b*x + a) - 1))) /b
Time = 29.28 (sec) , antiderivative size = 177, normalized size of antiderivative = 2.33 \[ \int \sec ^5(a+b x) \tan ^2(a+b x) \, dx=\frac {\frac {{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^{11}}{8}+\frac {47\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^9}{24}+\frac {13\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^7}{4}+\frac {13\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^5}{4}+\frac {47\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^3}{24}+\frac {\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}{8}}{b\,\left ({\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2+1\right )}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )\right )}{8\,b} \] Input:
int(tan(a + b*x)^2/cos(a + b*x)^5,x)
Output:
(tan(a/2 + (b*x)/2)/8 + (47*tan(a/2 + (b*x)/2)^3)/24 + (13*tan(a/2 + (b*x) /2)^5)/4 + (13*tan(a/2 + (b*x)/2)^7)/4 + (47*tan(a/2 + (b*x)/2)^9)/24 + ta n(a/2 + (b*x)/2)^11/8)/(b*(15*tan(a/2 + (b*x)/2)^4 - 6*tan(a/2 + (b*x)/2)^ 2 - 20*tan(a/2 + (b*x)/2)^6 + 15*tan(a/2 + (b*x)/2)^8 - 6*tan(a/2 + (b*x)/ 2)^10 + tan(a/2 + (b*x)/2)^12 + 1)) - atanh(tan(a/2 + (b*x)/2))/(8*b)
Time = 0.25 (sec) , antiderivative size = 226, normalized size of antiderivative = 2.97 \[ \int \sec ^5(a+b x) \tan ^2(a+b x) \, dx=\frac {3 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b x +a \right )^{6}-9 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b x +a \right )^{4}+9 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b x +a \right )^{2}-3 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )-3 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b x +a \right )^{6}+9 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b x +a \right )^{4}-9 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b x +a \right )^{2}+3 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )+3 \sin \left (b x +a \right )^{5}-8 \sin \left (b x +a \right )^{3}-3 \sin \left (b x +a \right )}{48 b \left (\sin \left (b x +a \right )^{6}-3 \sin \left (b x +a \right )^{4}+3 \sin \left (b x +a \right )^{2}-1\right )} \] Input:
int(sec(b*x+a)^5*tan(b*x+a)^2,x)
Output:
(3*log(tan((a + b*x)/2) - 1)*sin(a + b*x)**6 - 9*log(tan((a + b*x)/2) - 1) *sin(a + b*x)**4 + 9*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)**6 + 9*log(t an((a + b*x)/2) + 1)*sin(a + b*x)**4 - 9*log(tan((a + b*x)/2) + 1)*sin(a + b*x)**2 + 3*log(tan((a + b*x)/2) + 1) + 3*sin(a + b*x)**5 - 8*sin(a + b*x )**3 - 3*sin(a + b*x))/(48*b*(sin(a + b*x)**6 - 3*sin(a + b*x)**4 + 3*sin( a + b*x)**2 - 1))