Integrand size = 30, antiderivative size = 86 \[ \int \sec (e+f x) (a+a \sec (e+f x))^3 (c-c \sec (e+f x)) \, dx=\frac {5 a^3 c \text {arctanh}(\sin (e+f x))}{8 f}-\frac {3 a^3 c \sec (e+f x) \tan (e+f x)}{8 f}-\frac {a^3 c \sec ^3(e+f x) \tan (e+f x)}{4 f}-\frac {2 a^3 c \tan ^3(e+f x)}{3 f} \]
5/8*a^3*c*arctanh(sin(f*x+e))/f-3/8*a^3*c*sec(f*x+e)*tan(f*x+e)/f-1/4*a^3* c*sec(f*x+e)^3*tan(f*x+e)/f-2/3*a^3*c*tan(f*x+e)^3/f
Time = 0.40 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.81 \[ \int \sec (e+f x) (a+a \sec (e+f x))^3 (c-c \sec (e+f x)) \, dx=\frac {a^3 c \left (60 \text {arctanh}(\sin (e+f x))-\sec ^4(e+f x) (33 \sin (e+f x)+16 \sin (2 (e+f x))+9 \sin (3 (e+f x))-8 \sin (4 (e+f x)))\right )}{96 f} \]
(a^3*c*(60*ArcTanh[Sin[e + f*x]] - Sec[e + f*x]^4*(33*Sin[e + f*x] + 16*Si n[2*(e + f*x)] + 9*Sin[3*(e + f*x)] - 8*Sin[4*(e + f*x)])))/(96*f)
Time = 0.34 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3042, 4446, 2009}
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.
| \(\displaystyle \int \sec (e+f x) (a \sec (e+f x)+a)^3 (c-c \sec (e+f x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^3 \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4446 |
| \(\displaystyle -a c \int \left (a^2 \tan ^2(e+f x) \sec ^3(e+f x)+2 a^2 \tan ^2(e+f x) \sec ^2(e+f x)+a^2 \tan ^2(e+f x) \sec (e+f x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -a c \left (-\frac {5 a^2 \text {arctanh}(\sin (e+f x))}{8 f}+\frac {2 a^2 \tan ^3(e+f x)}{3 f}+\frac {a^2 \tan (e+f x) \sec ^3(e+f x)}{4 f}+\frac {3 a^2 \tan (e+f x) \sec (e+f x)}{8 f}\right )\) |
-(a*c*((-5*a^2*ArcTanh[Sin[e + f*x]])/(8*f) + (3*a^2*Sec[e + f*x]*Tan[e + f*x])/(8*f) + (a^2*Sec[e + f*x]^3*Tan[e + f*x])/(4*f) + (2*a^2*Tan[e + f*x ]^3)/(3*f)))
3.1.26.3.1 Defintions of rubi rules used
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 = 4.16 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.29
| method | result | size |
| derivativedivides | \(\frac {-a^{3} c \left (-\left (-\frac {\sec \left (f x +e \right )^{3}}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )+2 a^{3} c \left (-\frac {2}{3}-\frac {\sec \left (f x +e \right )^{2}}{3}\right ) \tan \left (f x +e \right )+2 a^{3} c \tan \left (f x +e \right )+a^{3} c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}\) | \(111\) |
| default | \(\frac {-a^{3} c \left (-\left (-\frac {\sec \left (f x +e \right )^{3}}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )+2 a^{3} c \left (-\frac {2}{3}-\frac {\sec \left (f x +e \right )^{2}}{3}\right ) \tan \left (f x +e \right )+2 a^{3} c \tan \left (f x +e \right )+a^{3} c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}\) | \(111\) |
| parts | \(\frac {a^{3} c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}+\frac {2 a^{3} c \tan \left (f x +e \right )}{f}+\frac {2 a^{3} c \left (-\frac {2}{3}-\frac {\sec \left (f x +e \right )^{2}}{3}\right ) \tan \left (f x +e \right )}{f}-\frac {a^{3} c \left (-\left (-\frac {\sec \left (f x +e \right )^{3}}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )}{f}\) | \(119\) |
| norman | \(\frac {-\frac {5 a^{3} c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{4 f}-\frac {73 a^{3} c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{12 f}+\frac {55 a^{3} c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{12 f}-\frac {5 a^{3} c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{4 f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{4}}-\frac {5 a^{3} c \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{8 f}+\frac {5 a^{3} c \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{8 f}\) | \(139\) |
| risch | \(\frac {i a^{3} c \left (9 \,{\mathrm e}^{7 i \left (f x +e \right )}+48 \,{\mathrm e}^{6 i \left (f x +e \right )}+33 \,{\mathrm e}^{5 i \left (f x +e \right )}+48 \,{\mathrm e}^{4 i \left (f x +e \right )}-33 \,{\mathrm e}^{3 i \left (f x +e \right )}+16 \,{\mathrm e}^{2 i \left (f x +e \right )}-9 \,{\mathrm e}^{i \left (f x +e \right )}+16\right )}{12 f \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )^{4}}-\frac {5 a^{3} c \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{8 f}+\frac {5 a^{3} c \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{8 f}\) | \(148\) |
| parallelrisch | \(-\frac {3 a^{3} c \left (\left (\frac {10 \cos \left (2 f x +2 e \right )}{3}+\frac {5 \cos \left (4 f x +4 e \right )}{6}+\frac {5}{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )+\left (-\frac {10 \cos \left (2 f x +2 e \right )}{3}-\frac {5 \cos \left (4 f x +4 e \right )}{6}-\frac {5}{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )+\sin \left (3 f x +3 e \right )-\frac {8 \sin \left (4 f x +4 e \right )}{9}+\frac {11 \sin \left (f x +e \right )}{3}+\frac {16 \sin \left (2 f x +2 e \right )}{9}\right )}{4 f \left (3+\cos \left (4 f x +4 e \right )+4 \cos \left (2 f x +2 e \right )\right )}\) | \(148\) |
1/f*(-a^3*c*(-(-1/4*sec(f*x+e)^3-3/8*sec(f*x+e))*tan(f*x+e)+3/8*ln(sec(f*x +e)+tan(f*x+e)))+2*a^3*c*(-2/3-1/3*sec(f*x+e)^2)*tan(f*x+e)+2*a^3*c*tan(f* x+e)+a^3*c*ln(sec(f*x+e)+tan(f*x+e)))
Time = 0.29 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.36 \[ \int \sec (e+f x) (a+a \sec (e+f x))^3 (c-c \sec (e+f x)) \, dx=\frac {15 \, a^{3} c \cos \left (f x + e\right )^{4} \log \left (\sin \left (f x + e\right ) + 1\right ) - 15 \, a^{3} c \cos \left (f x + e\right )^{4} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (16 \, a^{3} c \cos \left (f x + e\right )^{3} - 9 \, a^{3} c \cos \left (f x + e\right )^{2} - 16 \, a^{3} c \cos \left (f x + e\right ) - 6 \, a^{3} c\right )} \sin \left (f x + e\right )}{48 \, f \cos \left (f x + e\right )^{4}} \]
1/48*(15*a^3*c*cos(f*x + e)^4*log(sin(f*x + e) + 1) - 15*a^3*c*cos(f*x + e )^4*log(-sin(f*x + e) + 1) + 2*(16*a^3*c*cos(f*x + e)^3 - 9*a^3*c*cos(f*x + e)^2 - 16*a^3*c*cos(f*x + e) - 6*a^3*c)*sin(f*x + e))/(f*cos(f*x + e)^4)
\[ \int \sec (e+f x) (a+a \sec (e+f x))^3 (c-c \sec (e+f x)) \, dx=- a^{3} c \left (\int \left (- \sec {\left (e + f x \right )}\right )\, dx + \int \left (- 2 \sec ^{2}{\left (e + f x \right )}\right )\, dx + \int 2 \sec ^{4}{\left (e + f x \right )}\, dx + \int \sec ^{5}{\left (e + f x \right )}\, dx\right ) \]
-a**3*c*(Integral(-sec(e + f*x), x) + Integral(-2*sec(e + f*x)**2, x) + In tegral(2*sec(e + f*x)**4, x) + Integral(sec(e + f*x)**5, x))
Time = 0.21 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.55 \[ \int \sec (e+f x) (a+a \sec (e+f x))^3 (c-c \sec (e+f x)) \, dx=-\frac {32 \, {\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} a^{3} c - 3 \, a^{3} c {\left (\frac {2 \, {\left (3 \, \sin \left (f x + e\right )^{3} - 5 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1} - 3 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 3 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 48 \, a^{3} c \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) - 96 \, a^{3} c \tan \left (f x + e\right )}{48 \, f} \]
-1/48*(32*(tan(f*x + e)^3 + 3*tan(f*x + e))*a^3*c - 3*a^3*c*(2*(3*sin(f*x + e)^3 - 5*sin(f*x + e))/(sin(f*x + e)^4 - 2*sin(f*x + e)^2 + 1) - 3*log(s in(f*x + e) + 1) + 3*log(sin(f*x + e) - 1)) - 48*a^3*c*log(sec(f*x + e) + tan(f*x + e)) - 96*a^3*c*tan(f*x + e))/f
Time = 0.33 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.49 \[ \int \sec (e+f x) (a+a \sec (e+f x))^3 (c-c \sec (e+f x)) \, dx=\frac {15 \, a^{3} c \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right ) - 15 \, a^{3} c \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right ) - \frac {2 \, {\left (15 \, a^{3} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} - 55 \, a^{3} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 73 \, a^{3} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 15 \, a^{3} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{4}}}{24 \, f} \]
1/24*(15*a^3*c*log(abs(tan(1/2*f*x + 1/2*e) + 1)) - 15*a^3*c*log(abs(tan(1 /2*f*x + 1/2*e) - 1)) - 2*(15*a^3*c*tan(1/2*f*x + 1/2*e)^7 - 55*a^3*c*tan( 1/2*f*x + 1/2*e)^5 + 73*a^3*c*tan(1/2*f*x + 1/2*e)^3 + 15*a^3*c*tan(1/2*f* x + 1/2*e))/(tan(1/2*f*x + 1/2*e)^2 - 1)^4)/f
Time = 16.37 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.70 \[ \int \sec (e+f x) (a+a \sec (e+f x))^3 (c-c \sec (e+f x)) \, dx=\frac {5\,a^3\,c\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{4\,f}-\frac {\frac {5\,c\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7}{4}-\frac {55\,c\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{12}+\frac {73\,c\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{12}+\frac {5\,c\,a^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{4}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )} \]
(5*a^3*c*atanh(tan(e/2 + (f*x)/2)))/(4*f) - ((5*a^3*c*tan(e/2 + (f*x)/2))/ 4 + (73*a^3*c*tan(e/2 + (f*x)/2)^3)/12 - (55*a^3*c*tan(e/2 + (f*x)/2)^5)/1 2 + (5*a^3*c*tan(e/2 + (f*x)/2)^7)/4)/(f*(6*tan(e/2 + (f*x)/2)^4 - 4*tan(e /2 + (f*x)/2)^2 - 4*tan(e/2 + (f*x)/2)^6 + tan(e/2 + (f*x)/2)^8 + 1))