Integrand size = 32, antiderivative size = 105 \[ \int \sec ^2(e+f x) (a+a \sec (e+f x))^3 (c-c \sec (e+f x)) \, dx=\frac {a^3 c \text {arctanh}(\sin (e+f x))}{4 f}+\frac {a^3 c \sec (e+f x) \tan (e+f x)}{4 f}-\frac {a^3 c \sec ^3(e+f x) \tan (e+f x)}{2 f}-\frac {2 a^3 c \tan ^3(e+f x)}{3 f}-\frac {a^3 c \tan ^5(e+f x)}{5 f} \]
1/4*a^3*c*arctanh(sin(f*x+e))/f+1/4*a^3*c*sec(f*x+e)*tan(f*x+e)/f-1/2*a^3* c*sec(f*x+e)^3*tan(f*x+e)/f-2/3*a^3*c*tan(f*x+e)^3/f-1/5*a^3*c*tan(f*x+e)^ 5/f
Time = 0.23 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.65 \[ \int \sec ^2(e+f x) (a+a \sec (e+f x))^3 (c-c \sec (e+f x)) \, dx=\frac {a^3 c \left (15 \text {arctanh}(\sin (e+f x))-\tan (e+f x) \left (-15 \sec (e+f x)+30 \sec ^3(e+f x)+40 \tan ^2(e+f x)+12 \tan ^4(e+f x)\right )\right )}{60 f} \]
(a^3*c*(15*ArcTanh[Sin[e + f*x]] - Tan[e + f*x]*(-15*Sec[e + f*x] + 30*Sec [e + f*x]^3 + 40*Tan[e + f*x]^2 + 12*Tan[e + f*x]^4)))/(60*f)
Time = 0.40 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {3042, 4450, 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.
| \(\displaystyle \int \sec ^2(e+f x) (a \sec (e+f x)+a)^3 (c-c \sec (e+f x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (e+f x+\frac {\pi }{2}\right )^2 \left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^3 \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4450 |
| \(\displaystyle -a c \int \left (a^2 \tan ^2(e+f x) \sec ^4(e+f x)+2 a^2 \tan ^2(e+f x) \sec ^3(e+f x)+a^2 \tan ^2(e+f x) \sec ^2(e+f x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -a c \left (-\frac {a^2 \text {arctanh}(\sin (e+f x))}{4 f}+\frac {a^2 \tan ^5(e+f x)}{5 f}+\frac {2 a^2 \tan ^3(e+f x)}{3 f}+\frac {a^2 \tan (e+f x) \sec ^3(e+f x)}{2 f}-\frac {a^2 \tan (e+f x) \sec (e+f x)}{4 f}\right )\) |
-(a*c*(-1/4*(a^2*ArcTanh[Sin[e + f*x]])/f - (a^2*Sec[e + f*x]*Tan[e + f*x] )/(4*f) + (a^2*Sec[e + f*x]^3*Tan[e + f*x])/(2*f) + (2*a^2*Tan[e + f*x]^3) /(3*f) + (a^2*Tan[e + f*x]^5)/(5*f)))
3.2.68.3.1 Defintions of rubi rules used
Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Simp [((-a)*c)^m Int[ExpandTrig[(g*csc[e + f*x])^p*cot[e + f*x]^(2*m), (c + d* csc[e + f*x])^(n - m), x], x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegersQ[m, n] && GeQ[n - m, 0] && GtQ[m*n, 0]
Time = 4.05 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.30
| method | result | size |
| derivativedivides | \(\frac {a^{3} c \left (-\frac {8}{15}-\frac {\sec \left (f x +e \right )^{4}}{5}-\frac {4 \sec \left (f x +e \right )^{2}}{15}\right ) \tan \left (f x +e \right )-2 a^{3} c \left (-\left (-\frac {\sec \left (f x +e \right )^{3}}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )+2 a^{3} c \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )+a^{3} c \tan \left (f x +e \right )}{f}\) | \(137\) |
| default | \(\frac {a^{3} c \left (-\frac {8}{15}-\frac {\sec \left (f x +e \right )^{4}}{5}-\frac {4 \sec \left (f x +e \right )^{2}}{15}\right ) \tan \left (f x +e \right )-2 a^{3} c \left (-\left (-\frac {\sec \left (f x +e \right )^{3}}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )+2 a^{3} c \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )+a^{3} c \tan \left (f x +e \right )}{f}\) | \(137\) |
| parts | \(\frac {a^{3} c \tan \left (f x +e \right )}{f}+\frac {a^{3} c \sec \left (f x +e \right ) \tan \left (f x +e \right )}{f}+\frac {a^{3} c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}-\frac {2 a^{3} c \left (-\left (-\frac {\sec \left (f x +e \right )^{3}}{4}-\frac {3 \sec \left (f x +e \right )}{8}\right ) \tan \left (f x +e \right )+\frac {3 \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{8}\right )}{f}+\frac {a^{3} c \left (-\frac {8}{15}-\frac {\sec \left (f x +e \right )^{4}}{5}-\frac {4 \sec \left (f x +e \right )^{2}}{15}\right ) \tan \left (f x +e \right )}{f}\) | \(147\) |
| norman | \(\frac {\frac {a^{3} c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 f}+\frac {25 a^{3} c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3 f}-\frac {64 a^{3} c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{15 f}+\frac {7 a^{3} c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{3 f}-\frac {a^{3} c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{2 f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{5}}-\frac {a^{3} c \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{4 f}+\frac {a^{3} c \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{4 f}\) | \(159\) |
| risch | \(-\frac {i a^{3} c \left (15 \,{\mathrm e}^{9 i \left (f x +e \right )}-60 \,{\mathrm e}^{8 i \left (f x +e \right )}-90 \,{\mathrm e}^{7 i \left (f x +e \right )}-240 \,{\mathrm e}^{6 i \left (f x +e \right )}-40 \,{\mathrm e}^{4 i \left (f x +e \right )}+90 \,{\mathrm e}^{3 i \left (f x +e \right )}-80 \,{\mathrm e}^{2 i \left (f x +e \right )}-15 \,{\mathrm e}^{i \left (f x +e \right )}-28\right )}{30 f \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )^{5}}+\frac {a^{3} c \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{4 f}-\frac {a^{3} c \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{4 f}\) | \(159\) |
| parallelrisch | \(-\frac {3 \left (\left (\frac {5 \cos \left (f x +e \right )}{6}+\frac {5 \cos \left (3 f x +3 e \right )}{12}+\frac {\cos \left (5 f x +5 e \right )}{12}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )+\left (-\frac {5 \cos \left (f x +e \right )}{6}-\frac {5 \cos \left (3 f x +3 e \right )}{12}-\frac {\cos \left (5 f x +5 e \right )}{12}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )+\sin \left (2 f x +2 e \right )-\frac {\sin \left (3 f x +3 e \right )}{9}-\frac {\sin \left (4 f x +4 e \right )}{6}-\frac {7 \sin \left (5 f x +5 e \right )}{45}+\frac {10 \sin \left (f x +e \right )}{9}\right ) a^{3} c}{f \left (\cos \left (5 f x +5 e \right )+5 \cos \left (3 f x +3 e \right )+10 \cos \left (f x +e \right )\right )}\) | \(180\) |
1/f*(a^3*c*(-8/15-1/5*sec(f*x+e)^4-4/15*sec(f*x+e)^2)*tan(f*x+e)-2*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*(1/2*sec(f*x+e)*tan(f*x+e)+1/2*ln(sec(f*x+e)+tan(f*x+e)))+a^3* c*tan(f*x+e))
Time = 0.27 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.25 \[ \int \sec ^2(e+f x) (a+a \sec (e+f x))^3 (c-c \sec (e+f x)) \, dx=\frac {15 \, a^{3} c \cos \left (f x + e\right )^{5} \log \left (\sin \left (f x + e\right ) + 1\right ) - 15 \, a^{3} c \cos \left (f x + e\right )^{5} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (28 \, a^{3} c \cos \left (f x + e\right )^{4} + 15 \, a^{3} c \cos \left (f x + e\right )^{3} - 16 \, a^{3} c \cos \left (f x + e\right )^{2} - 30 \, a^{3} c \cos \left (f x + e\right ) - 12 \, a^{3} c\right )} \sin \left (f x + e\right )}{120 \, f \cos \left (f x + e\right )^{5}} \]
1/120*(15*a^3*c*cos(f*x + e)^5*log(sin(f*x + e) + 1) - 15*a^3*c*cos(f*x + e)^5*log(-sin(f*x + e) + 1) + 2*(28*a^3*c*cos(f*x + e)^4 + 15*a^3*c*cos(f* x + e)^3 - 16*a^3*c*cos(f*x + e)^2 - 30*a^3*c*cos(f*x + e) - 12*a^3*c)*sin (f*x + e))/(f*cos(f*x + e)^5)
\[ \int \sec ^2(e+f x) (a+a \sec (e+f x))^3 (c-c \sec (e+f x)) \, dx=- a^{3} c \left (\int \left (- \sec ^{2}{\left (e + f x \right )}\right )\, dx + \int \left (- 2 \sec ^{3}{\left (e + f x \right )}\right )\, dx + \int 2 \sec ^{5}{\left (e + f x \right )}\, dx + \int \sec ^{6}{\left (e + f x \right )}\, dx\right ) \]
-a**3*c*(Integral(-sec(e + f*x)**2, x) + Integral(-2*sec(e + f*x)**3, x) + Integral(2*sec(e + f*x)**5, x) + Integral(sec(e + f*x)**6, x))
Time = 0.23 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.64 \[ \int \sec ^2(e+f x) (a+a \sec (e+f x))^3 (c-c \sec (e+f x)) \, dx=-\frac {8 \, {\left (3 \, \tan \left (f x + e\right )^{5} + 10 \, \tan \left (f x + e\right )^{3} + 15 \, \tan \left (f x + e\right )\right )} a^{3} c - 15 \, a^{3} c {\left (\frac {2 \, {\left (3 \, \sin \left (f x + e\right )^{3} - 5 \, \sin \left (f x + e\right )\right )}}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1} - 3 \, \log \left (\sin \left (f x + e\right ) + 1\right ) + 3 \, \log \left (\sin \left (f x + e\right ) - 1\right )\right )} + 60 \, a^{3} c {\left (\frac {2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} - 120 \, a^{3} c \tan \left (f x + e\right )}{120 \, f} \]
-1/120*(8*(3*tan(f*x + e)^5 + 10*tan(f*x + e)^3 + 15*tan(f*x + e))*a^3*c - 15*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(sin(f*x + e) + 1) + 3*log(sin(f*x + e) - 1)) + 60*a ^3*c*(2*sin(f*x + e)/(sin(f*x + e)^2 - 1) - log(sin(f*x + e) + 1) + log(si n(f*x + e) - 1)) - 120*a^3*c*tan(f*x + e))/f
Time = 0.35 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.38 \[ \int \sec ^2(e+f x) (a+a \sec (e+f x))^3 (c-c \sec (e+f x)) \, dx=\frac {15 \, a^{3} c \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right ) - 15 \, a^{3} c \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right ) - \frac {2 \, {\left (15 \, a^{3} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9} - 70 \, a^{3} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} + 128 \, a^{3} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 250 \, a^{3} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 15 \, a^{3} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{5}}}{60 \, f} \]
1/60*(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)^9 - 70*a^3*c*tan( 1/2*f*x + 1/2*e)^7 + 128*a^3*c*tan(1/2*f*x + 1/2*e)^5 - 250*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)^5)/f
Time = 17.84 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.67 \[ \int \sec ^2(e+f x) (a+a \sec (e+f x))^3 (c-c \sec (e+f x)) \, dx=\frac {-\frac {c\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9}{2}+\frac {7\,c\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7}{3}-\frac {64\,c\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{15}+\frac {25\,c\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{3}+\frac {c\,a^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{2}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )}+\frac {a^3\,c\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{2\,f} \]
((a^3*c*tan(e/2 + (f*x)/2))/2 + (25*a^3*c*tan(e/2 + (f*x)/2)^3)/3 - (64*a^ 3*c*tan(e/2 + (f*x)/2)^5)/15 + (7*a^3*c*tan(e/2 + (f*x)/2)^7)/3 - (a^3*c*t an(e/2 + (f*x)/2)^9)/2)/(f*(5*tan(e/2 + (f*x)/2)^2 - 10*tan(e/2 + (f*x)/2) ^4 + 10*tan(e/2 + (f*x)/2)^6 - 5*tan(e/2 + (f*x)/2)^8 + tan(e/2 + (f*x)/2) ^10 - 1)) + (a^3*c*atanh(tan(e/2 + (f*x)/2)))/(2*f)