Integrand size = 28, antiderivative size = 38 \[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x)) \, dx=\frac {a c \text {arctanh}(\sin (e+f x))}{2 f}-\frac {a c \sec (e+f x) \tan (e+f x)}{2 f} \] Output:
1/2*a*c*arctanh(sin(f*x+e))/f-1/2*a*c*sec(f*x+e)*tan(f*x+e)/f
Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00 \[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x)) \, dx=-a c \left (-\frac {\text {arctanh}(\sin (e+f x))}{2 f}+\frac {\sec (e+f x) \tan (e+f x)}{2 f}\right ) \] Input:
Integrate[Sec[e + f*x]*(a + a*Sec[e + f*x])*(c - c*Sec[e + f*x]),x]
Output:
-(a*c*(-1/2*ArcTanh[Sin[e + f*x]]/f + (Sec[e + f*x]*Tan[e + f*x])/(2*f)))
Time = 0.32 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3042, 4446, 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.
\(\displaystyle \int \sec (e+f x) (a \sec (e+f x)+a) (c-c \sec (e+f x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right ) \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )dx\) |
\(\Big \downarrow \) 4446 |
\(\displaystyle -a c \int \sec (e+f x) \tan ^2(e+f x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -a c \int \sec (e+f x) \tan (e+f x)^2dx\) |
\(\Big \downarrow \) 3091 |
\(\displaystyle -a c \left (\frac {\tan (e+f x) \sec (e+f x)}{2 f}-\frac {1}{2} \int \sec (e+f x)dx\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -a c \left (\frac {\tan (e+f x) \sec (e+f x)}{2 f}-\frac {1}{2} \int \csc \left (e+f x+\frac {\pi }{2}\right )dx\right )\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle -a c \left (\frac {\tan (e+f x) \sec (e+f x)}{2 f}-\frac {\text {arctanh}(\sin (e+f x))}{2 f}\right )\) |
Input:
Int[Sec[e + f*x]*(a + a*Sec[e + f*x])*(c - c*Sec[e + f*x]),x]
Output:
-(a*c*(-1/2*ArcTanh[Sin[e + f*x]]/f + (Sec[e + f*x]*Tan[e + f*x])/(2*f)))
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]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(c sc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Simp[((-a)*c)^m Int[ExpandTrig[csc[e + f*x]*cot[e + f*x]^(2*m), (c + d*csc[e + f*x])^(n - m ), x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[b*c + a*d, 0] && Eq Q[a^2 - b^2, 0] && IntegersQ[m, n] && GeQ[n - m, 0] && GtQ[m*n, 0]
Time = 0.17 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.53
method | result | size |
derivativedivides | \(\frac {a c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )-a c \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )}{f}\) | \(58\) |
default | \(\frac {a c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )-a c \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )}{f}\) | \(58\) |
parts | \(\frac {a c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}-\frac {a c \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )}{f}\) | \(60\) |
parallelrisch | \(\frac {a \left (\left (-1-\cos \left (2 f x +2 e \right )\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )+\left (1+\cos \left (2 f x +2 e \right )\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )-2 \sin \left (f x +e \right )\right ) c}{2 f \left (1+\cos \left (2 f x +2 e \right )\right )}\) | \(80\) |
risch | \(\frac {i a c \left ({\mathrm e}^{3 i \left (f x +e \right )}-{\mathrm e}^{i \left (f x +e \right )}\right )}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}+\frac {a c \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{2 f}-\frac {a c \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{2 f}\) | \(84\) |
norman | \(\frac {-\frac {a c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}-\frac {a c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{2}}-\frac {a c \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2 f}+\frac {a c \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2 f}\) | \(91\) |
Input:
int(sec(f*x+e)*(a+a*sec(f*x+e))*(c-c*sec(f*x+e)),x,method=_RETURNVERBOSE)
Output:
1/f*(a*c*ln(sec(f*x+e)+tan(f*x+e))-a*c*(1/2*sec(f*x+e)*tan(f*x+e)+1/2*ln(s ec(f*x+e)+tan(f*x+e))))
Time = 0.08 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.76 \[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x)) \, dx=\frac {a c \cos \left (f x + e\right )^{2} \log \left (\sin \left (f x + e\right ) + 1\right ) - a c \cos \left (f x + e\right )^{2} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \, a c \sin \left (f x + e\right )}{4 \, f \cos \left (f x + e\right )^{2}} \] Input:
integrate(sec(f*x+e)*(a+a*sec(f*x+e))*(c-c*sec(f*x+e)),x, algorithm="frica s")
Output:
1/4*(a*c*cos(f*x + e)^2*log(sin(f*x + e) + 1) - a*c*cos(f*x + e)^2*log(-si n(f*x + e) + 1) - 2*a*c*sin(f*x + e))/(f*cos(f*x + e)^2)
\[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x)) \, dx=- a c \left (\int \left (- \sec {\left (e + f x \right )}\right )\, dx + \int \sec ^{3}{\left (e + f x \right )}\, dx\right ) \] Input:
integrate(sec(f*x+e)*(a+a*sec(f*x+e))*(c-c*sec(f*x+e)),x)
Output:
-a*c*(Integral(-sec(e + f*x), x) + Integral(sec(e + f*x)**3, x))
Time = 0.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.79 \[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x)) \, dx=\frac {a c {\left (\frac {2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} + 4 \, a c \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right )}{4 \, f} \] Input:
integrate(sec(f*x+e)*(a+a*sec(f*x+e))*(c-c*sec(f*x+e)),x, algorithm="maxim a")
Output:
1/4*(a*c*(2*sin(f*x + e)/(sin(f*x + e)^2 - 1) - log(sin(f*x + e) + 1) + lo g(sin(f*x + e) - 1)) + 4*a*c*log(sec(f*x + e) + tan(f*x + e)))/f
Time = 0.13 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.45 \[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x)) \, dx=\frac {a c \log \left ({\left | \sin \left (f x + e\right ) + 1 \right |}\right ) - a c \log \left ({\left | \sin \left (f x + e\right ) - 1 \right |}\right ) + \frac {2 \, a c \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1}}{4 \, f} \] Input:
integrate(sec(f*x+e)*(a+a*sec(f*x+e))*(c-c*sec(f*x+e)),x, algorithm="giac" )
Output:
1/4*(a*c*log(abs(sin(f*x + e) + 1)) - a*c*log(abs(sin(f*x + e) - 1)) + 2*a *c*sin(f*x + e)/(sin(f*x + e)^2 - 1))/f
Time = 11.45 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.03 \[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x)) \, dx=\frac {a\,c\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{f}-\frac {a\,c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+a\,c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )} \] Input:
int(((a + a/cos(e + f*x))*(c - c/cos(e + f*x)))/cos(e + f*x),x)
Output:
(a*c*atanh(tan(e/2 + (f*x)/2)))/f - (a*c*tan(e/2 + (f*x)/2)^3 + a*c*tan(e/ 2 + (f*x)/2))/(f*(tan(e/2 + (f*x)/2)^4 - 2*tan(e/2 + (f*x)/2)^2 + 1))
Time = 0.16 (sec) , antiderivative size = 95, normalized size of antiderivative = 2.50 \[ \int \sec (e+f x) (a+a \sec (e+f x)) (c-c \sec (e+f x)) \, dx=\frac {a c \left (-\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \sin \left (f x +e \right )^{2}+\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )+\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \sin \left (f x +e \right )^{2}-\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )+\sin \left (f x +e \right )\right )}{2 f \left (\sin \left (f x +e \right )^{2}-1\right )} \] Input:
int(sec(f*x+e)*(a+a*sec(f*x+e))*(c-c*sec(f*x+e)),x)
Output:
(a*c*( - log(tan((e + f*x)/2) - 1)*sin(e + f*x)**2 + log(tan((e + f*x)/2) - 1) + log(tan((e + f*x)/2) + 1)*sin(e + f*x)**2 - log(tan((e + f*x)/2) + 1) + sin(e + f*x)))/(2*f*(sin(e + f*x)**2 - 1))