Integrand size = 17, antiderivative size = 55 \[ \int \sec ^3(a+b x) \tan ^2(a+b x) \, dx=-\frac {\text {arctanh}(\sin (a+b x))}{8 b}-\frac {\sec (a+b x) \tan (a+b x)}{8 b}+\frac {\sec ^3(a+b x) \tan (a+b x)}{4 b} \] Output:
-1/8*arctanh(sin(b*x+a))/b-1/8*sec(b*x+a)*tan(b*x+a)/b+1/4*sec(b*x+a)^3*ta n(b*x+a)/b
Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00 \[ \int \sec ^3(a+b x) \tan ^2(a+b x) \, dx=-\frac {\text {arctanh}(\sin (a+b x))}{8 b}-\frac {\sec (a+b x) \tan (a+b x)}{8 b}+\frac {\sec ^3(a+b x) \tan (a+b x)}{4 b} \] Input:
Integrate[Sec[a + b*x]^3*Tan[a + b*x]^2,x]
Output:
-1/8*ArcTanh[Sin[a + b*x]]/b - (Sec[a + b*x]*Tan[a + b*x])/(8*b) + (Sec[a + b*x]^3*Tan[a + b*x])/(4*b)
Time = 0.56 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {3042, 3091, 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.
\(\displaystyle \int \tan ^2(a+b x) \sec ^3(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (a+b x)^2 \sec (a+b x)^3dx\) |
\(\Big \downarrow \) 3091 |
\(\displaystyle \frac {\tan (a+b x) \sec ^3(a+b x)}{4 b}-\frac {1}{4} \int \sec ^3(a+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tan (a+b x) \sec ^3(a+b x)}{4 b}-\frac {1}{4} \int \csc \left (a+b x+\frac {\pi }{2}\right )^3dx\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {1}{4} \left (-\frac {1}{2} \int \sec (a+b x)dx-\frac {\tan (a+b x) \sec (a+b x)}{2 b}\right )+\frac {\tan (a+b x) \sec ^3(a+b x)}{4 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} \left (-\frac {1}{2} \int \csc \left (a+b x+\frac {\pi }{2}\right )dx-\frac {\tan (a+b x) \sec (a+b x)}{2 b}\right )+\frac {\tan (a+b x) \sec ^3(a+b x)}{4 b}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {1}{4} \left (-\frac {\text {arctanh}(\sin (a+b x))}{2 b}-\frac {\tan (a+b x) \sec (a+b x)}{2 b}\right )+\frac {\tan (a+b x) \sec ^3(a+b x)}{4 b}\) |
Input:
Int[Sec[a + b*x]^3*Tan[a + b*x]^2,x]
Output:
(Sec[a + b*x]^3*Tan[a + b*x])/(4*b) + (-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_)]*(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 = 3.05 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.20
method | result | size |
derivativedivides | \(\frac {\frac {\sin \left (b x +a \right )^{3}}{4 \cos \left (b x +a \right )^{4}}+\frac {\sin \left (b x +a \right )^{3}}{8 \cos \left (b x +a \right )^{2}}+\frac {\sin \left (b x +a \right )}{8}-\frac {\ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{8}}{b}\) | \(66\) |
default | \(\frac {\frac {\sin \left (b x +a \right )^{3}}{4 \cos \left (b x +a \right )^{4}}+\frac {\sin \left (b x +a \right )^{3}}{8 \cos \left (b x +a \right )^{2}}+\frac {\sin \left (b x +a \right )}{8}-\frac {\ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{8}}{b}\) | \(66\) |
risch | \(\frac {i \left ({\mathrm e}^{7 i \left (b x +a \right )}-7 \,{\mathrm e}^{5 i \left (b x +a \right )}+7 \,{\mathrm e}^{3 i \left (b x +a \right )}-{\mathrm e}^{i \left (b x +a \right )}\right )}{4 b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{4}}+\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}-i\right )}{8 b}-\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}{8 b}\) | \(100\) |
Input:
int(sec(b*x+a)^3*tan(b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
1/b*(1/4*sin(b*x+a)^3/cos(b*x+a)^4+1/8*sin(b*x+a)^3/cos(b*x+a)^2+1/8*sin(b *x+a)-1/8*ln(sec(b*x+a)+tan(b*x+a)))
Time = 0.09 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.29 \[ \int \sec ^3(a+b x) \tan ^2(a+b x) \, dx=-\frac {\cos \left (b x + a\right )^{4} \log \left (\sin \left (b x + a\right ) + 1\right ) - \cos \left (b x + a\right )^{4} \log \left (-\sin \left (b x + a\right ) + 1\right ) + 2 \, {\left (\cos \left (b x + a\right )^{2} - 2\right )} \sin \left (b x + a\right )}{16 \, b \cos \left (b x + a\right )^{4}} \] Input:
integrate(sec(b*x+a)^3*tan(b*x+a)^2,x, algorithm="fricas")
Output:
-1/16*(cos(b*x + a)^4*log(sin(b*x + a) + 1) - cos(b*x + a)^4*log(-sin(b*x + a) + 1) + 2*(cos(b*x + a)^2 - 2)*sin(b*x + a))/(b*cos(b*x + a)^4)
\[ \int \sec ^3(a+b x) \tan ^2(a+b x) \, dx=\int \tan ^{2}{\left (a + b x \right )} \sec ^{3}{\left (a + b x \right )}\, dx \] Input:
integrate(sec(b*x+a)**3*tan(b*x+a)**2,x)
Output:
Integral(tan(a + b*x)**2*sec(a + b*x)**3, x)
Time = 0.05 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.18 \[ \int \sec ^3(a+b x) \tan ^2(a+b x) \, dx=\frac {\frac {2 \, {\left (\sin \left (b x + a\right )^{3} + \sin \left (b x + a\right )\right )}}{\sin \left (b x + a\right )^{4} - 2 \, \sin \left (b x + a\right )^{2} + 1} - \log \left (\sin \left (b x + a\right ) + 1\right ) + \log \left (\sin \left (b x + a\right ) - 1\right )}{16 \, b} \] Input:
integrate(sec(b*x+a)^3*tan(b*x+a)^2,x, algorithm="maxima")
Output:
1/16*(2*(sin(b*x + a)^3 + sin(b*x + a))/(sin(b*x + a)^4 - 2*sin(b*x + a)^2 + 1) - log(sin(b*x + a) + 1) + log(sin(b*x + a) - 1))/b
Time = 0.21 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.49 \[ \int \sec ^3(a+b x) \tan ^2(a+b x) \, dx=\frac {\frac {4 \, {\left (\frac {1}{\sin \left (b x + a\right )} + \sin \left (b x + a\right )\right )}}{{\left (\frac {1}{\sin \left (b x + a\right )} + \sin \left (b x + a\right )\right )}^{2} - 4} - \log \left ({\left | \frac {1}{\sin \left (b x + a\right )} + \sin \left (b x + a\right ) + 2 \right |}\right ) + \log \left ({\left | \frac {1}{\sin \left (b x + a\right )} + \sin \left (b x + a\right ) - 2 \right |}\right )}{32 \, b} \] Input:
integrate(sec(b*x+a)^3*tan(b*x+a)^2,x, algorithm="giac")
Output:
1/32*(4*(1/sin(b*x + a) + sin(b*x + a))/((1/sin(b*x + a) + sin(b*x + a))^2 - 4) - log(abs(1/sin(b*x + a) + sin(b*x + a) + 2)) + log(abs(1/sin(b*x + a) + sin(b*x + a) - 2)))/b
Time = 28.55 (sec) , antiderivative size = 125, normalized size of antiderivative = 2.27 \[ \int \sec ^3(a+b x) \tan ^2(a+b x) \, dx=\frac {\frac {{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^7}{4}+\frac {7\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^5}{4}+\frac {7\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^3}{4}+\frac {\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}{4}}{b\,\left ({\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2+1\right )}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )\right )}{4\,b} \] Input:
int(tan(a + b*x)^2/cos(a + b*x)^3,x)
Output:
(tan(a/2 + (b*x)/2)/4 + (7*tan(a/2 + (b*x)/2)^3)/4 + (7*tan(a/2 + (b*x)/2) ^5)/4 + 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)) - atanh(ta n(a/2 + (b*x)/2))/(4*b)
Time = 0.28 (sec) , antiderivative size = 155, normalized size of antiderivative = 2.82 \[ \int \sec ^3(a+b x) \tan ^2(a+b x) \, dx=\frac {\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b x +a \right )^{4}-2 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b x +a \right )^{2}+\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )-\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b x +a \right )^{4}+2 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b x +a \right )^{2}-\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )+\sin \left (b x +a \right )^{3}+\sin \left (b x +a \right )}{8 b \left (\sin \left (b x +a \right )^{4}-2 \sin \left (b x +a \right )^{2}+1\right )} \] Input:
int(sec(b*x+a)^3*tan(b*x+a)^2,x)
Output:
(log(tan((a + b*x)/2) - 1)*sin(a + b*x)**4 - 2*log(tan((a + b*x)/2) - 1)*s in(a + b*x)**2 + log(tan((a + b*x)/2) - 1) - log(tan((a + b*x)/2) + 1)*sin (a + b*x)**4 + 2*log(tan((a + b*x)/2) + 1)*sin(a + b*x)**2 - log(tan((a + b*x)/2) + 1) + sin(a + b*x)**3 + sin(a + b*x))/(8*b*(sin(a + b*x)**4 - 2*s in(a + b*x)**2 + 1))