Integrand size = 24, antiderivative size = 55 \[ \int (a+a \sec (e+f x))^2 (c-c \sec (e+f x)) \, dx=a^2 c x+\frac {a^2 c \text {arctanh}(\sin (e+f x))}{2 f}-\frac {c \left (2 a^2+a^2 \sec (e+f x)\right ) \tan (e+f x)}{2 f} \]
Time = 0.26 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.76 \[ \int (a+a \sec (e+f x))^2 (c-c \sec (e+f x)) \, dx=\frac {a^2 c \sec ^2(e+f x) \left (e+f x+6 \arctan (\tan (e+f x)) \cos ^2(e+f x)+4 \text {arctanh}(\sin (e+f x)) \cos ^2(e+f x)+e \cos (2 (e+f x))+f x \cos (2 (e+f x))-4 \sin (e+f x)-4 \sin (2 (e+f x))\right )}{8 f} \]
(a^2*c*Sec[e + f*x]^2*(e + f*x + 6*ArcTan[Tan[e + f*x]]*Cos[e + f*x]^2 + 4 *ArcTanh[Sin[e + f*x]]*Cos[e + f*x]^2 + e*Cos[2*(e + f*x)] + f*x*Cos[2*(e + f*x)] - 4*Sin[e + f*x] - 4*Sin[2*(e + f*x)]))/(8*f)
Time = 0.31 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3042, 4392, 3042, 4369, 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.
\(\displaystyle \int (a \sec (e+f x)+a)^2 (c-c \sec (e+f x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^2 \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )dx\) |
\(\Big \downarrow \) 4392 |
\(\displaystyle -a c \int (\sec (e+f x) a+a) \tan ^2(e+f x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -a c \int \cot \left (e+f x+\frac {\pi }{2}\right )^2 \left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right )dx\) |
\(\Big \downarrow \) 4369 |
\(\displaystyle -a c \left (\frac {\tan (e+f x) (a \sec (e+f x)+2 a)}{2 f}-\frac {1}{2} \int (\sec (e+f x) a+2 a)dx\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -a c \left (\frac {1}{2} \left (-\frac {a \text {arctanh}(\sin (e+f x))}{f}-2 a x\right )+\frac {\tan (e+f x) (a \sec (e+f x)+2 a)}{2 f}\right )\) |
-(a*c*((-2*a*x - (a*ArcTanh[Sin[e + f*x]])/f)/2 + ((2*a + a*Sec[e + f*x])* Tan[e + f*x])/(2*f)))
3.1.5.3.1 Defintions of rubi rules used
Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( a_)), x_Symbol] :> Simp[(-e)*(e*Cot[c + d*x])^(m - 1)*((a*m + b*(m - 1)*Csc [c + d*x])/(d*m*(m - 1))), x] - Simp[e^2/m Int[(e*Cot[c + d*x])^(m - 2)*( a*m + b*(m - 1)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[m , 1]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*( d_.) + (c_))^(n_.), x_Symbol] :> Simp[((-a)*c)^m Int[Cot[e + f*x]^(2*m)*( c + d*Csc[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && E qQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && RationalQ[n] && !( IntegerQ[n] && GtQ[m - n, 0])
Time = 1.65 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.53
method | result | size |
derivativedivides | \(\frac {-a^{2} c \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )-a^{2} c \tan \left (f x +e \right )+a^{2} c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )+a^{2} c \left (f x +e \right )}{f}\) | \(84\) |
default | \(\frac {-a^{2} c \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )-a^{2} c \tan \left (f x +e \right )+a^{2} c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )+a^{2} c \left (f x +e \right )}{f}\) | \(84\) |
parts | \(a^{2} c x +\frac {a^{2} c \ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f}-\frac {a^{2} c \tan \left (f x +e \right )}{f}-\frac {a^{2} c \left (\frac {\sec \left (f x +e \right ) \tan \left (f x +e \right )}{2}+\frac {\ln \left (\sec \left (f x +e \right )+\tan \left (f x +e \right )\right )}{2}\right )}{f}\) | \(85\) |
risch | \(a^{2} c x +\frac {i a^{2} c \left ({\mathrm e}^{3 i \left (f x +e \right )}-2 \,{\mathrm e}^{2 i \left (f x +e \right )}-{\mathrm e}^{i \left (f x +e \right )}-2\right )}{f \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )^{2}}+\frac {a^{2} c \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{2 f}-\frac {a^{2} c \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{2 f}\) | \(108\) |
parallelrisch | \(-\frac {\left (\left (1+\cos \left (2 f x +2 e \right )\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )+\left (-1-\cos \left (2 f x +2 e \right )\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )-2 f x \cos \left (2 f x +2 e \right )-2 f x +2 \sin \left (f x +e \right )+2 \sin \left (2 f x +2 e \right )\right ) a^{2} c}{2 f \left (1+\cos \left (2 f x +2 e \right )\right )}\) | \(110\) |
norman | \(\frac {a^{2} c x +a^{2} c x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+\frac {a^{2} c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{f}-2 a^{2} c x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-\frac {3 a^{2} c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{2}}-\frac {a^{2} c \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2 f}+\frac {a^{2} c \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2 f}\) | \(139\) |
1/f*(-a^2*c*(1/2*sec(f*x+e)*tan(f*x+e)+1/2*ln(sec(f*x+e)+tan(f*x+e)))-a^2* c*tan(f*x+e)+a^2*c*ln(sec(f*x+e)+tan(f*x+e))+a^2*c*(f*x+e))
Leaf count of result is larger than twice the leaf count of optimal. 103 vs. \(2 (51) = 102\).
Time = 0.27 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.87 \[ \int (a+a \sec (e+f x))^2 (c-c \sec (e+f x)) \, dx=\frac {4 \, a^{2} c f x \cos \left (f x + e\right )^{2} + a^{2} c \cos \left (f x + e\right )^{2} \log \left (\sin \left (f x + e\right ) + 1\right ) - a^{2} c \cos \left (f x + e\right )^{2} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (2 \, a^{2} c \cos \left (f x + e\right ) + a^{2} c\right )} \sin \left (f x + e\right )}{4 \, f \cos \left (f x + e\right )^{2}} \]
1/4*(4*a^2*c*f*x*cos(f*x + e)^2 + a^2*c*cos(f*x + e)^2*log(sin(f*x + e) + 1) - a^2*c*cos(f*x + e)^2*log(-sin(f*x + e) + 1) - 2*(2*a^2*c*cos(f*x + e) + a^2*c)*sin(f*x + e))/(f*cos(f*x + e)^2)
\[ \int (a+a \sec (e+f x))^2 (c-c \sec (e+f x)) \, dx=- a^{2} c \left (\int \left (-1\right )\, dx + \int \left (- \sec {\left (e + f x \right )}\right )\, dx + \int \sec ^{2}{\left (e + f x \right )}\, dx + \int \sec ^{3}{\left (e + f x \right )}\, dx\right ) \]
-a**2*c*(Integral(-1, x) + Integral(-sec(e + f*x), x) + Integral(sec(e + f *x)**2, x) + Integral(sec(e + f*x)**3, x))
Time = 0.19 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.73 \[ \int (a+a \sec (e+f x))^2 (c-c \sec (e+f x)) \, dx=\frac {4 \, {\left (f x + e\right )} a^{2} c + a^{2} c {\left (\frac {2 \, \sin \left (f x + e\right )}{\sin \left (f x + e\right )^{2} - 1} - \log \left (\sin \left (f x + e\right ) + 1\right ) + \log \left (\sin \left (f x + e\right ) - 1\right )\right )} + 4 \, a^{2} c \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) - 4 \, a^{2} c \tan \left (f x + e\right )}{4 \, f} \]
1/4*(4*(f*x + e)*a^2*c + a^2*c*(2*sin(f*x + e)/(sin(f*x + e)^2 - 1) - log( sin(f*x + e) + 1) + log(sin(f*x + e) - 1)) + 4*a^2*c*log(sec(f*x + e) + ta n(f*x + e)) - 4*a^2*c*tan(f*x + e))/f
Leaf count of result is larger than twice the leaf count of optimal. 103 vs. \(2 (51) = 102\).
Time = 0.30 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.87 \[ \int (a+a \sec (e+f x))^2 (c-c \sec (e+f x)) \, dx=\frac {2 \, {\left (f x + e\right )} a^{2} c + a^{2} c \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right ) - a^{2} c \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right ) + \frac {2 \, {\left (a^{2} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 3 \, a^{2} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{2}}}{2 \, f} \]
1/2*(2*(f*x + e)*a^2*c + a^2*c*log(abs(tan(1/2*f*x + 1/2*e) + 1)) - a^2*c* log(abs(tan(1/2*f*x + 1/2*e) - 1)) + 2*(a^2*c*tan(1/2*f*x + 1/2*e)^3 - 3*a ^2*c*tan(1/2*f*x + 1/2*e))/(tan(1/2*f*x + 1/2*e)^2 - 1)^2)/f
Time = 14.43 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.65 \[ \int (a+a \sec (e+f x))^2 (c-c \sec (e+f x)) \, dx=a^2\,c\,x-\frac {3\,a^2\,c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-a^2\,c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}+\frac {a^2\,c\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{f} \]