Integrand size = 33, antiderivative size = 119 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{2} a^2 (3 A+2 C) x+\frac {2 a^2 C \text {arctanh}(\sin (c+d x))}{d}+\frac {a^2 (3 A-2 C) \sin (c+d x)}{2 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {(A-2 C) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{2 d} \]
1/2*a^2*(3*A+2*C)*x+2*a^2*C*arctanh(sin(d*x+c))/d+1/2*a^2*(3*A-2*C)*sin(d* x+c)/d+1/2*A*cos(d*x+c)*(a+a*sec(d*x+c))^2*sin(d*x+c)/d-1/2*(A-2*C)*(a^2+a ^2*sec(d*x+c))*sin(d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(292\) vs. \(2(119)=238\).
Time = 2.10 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.45 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {a^2 \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (4 \cos (d x) \left (3 A d x+2 C d x-4 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+4 \cos (2 c+d x) \left (3 A d x+2 C d x-4 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+A \sin (d x)+16 C \sin (d x)+A \sin (2 c+d x)+8 A \sin (c+2 d x)+8 A \sin (3 c+2 d x)+A \sin (2 c+3 d x)+A \sin (4 c+3 d x)\right )}{16 d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (-1+\tan \left (\frac {1}{2} (c+d x)\right )\right ) \left (1+\tan \left (\frac {1}{2} (c+d x)\right )\right )} \]
-1/16*(a^2*Sec[(c + d*x)/2]^2*(4*Cos[d*x]*(3*A*d*x + 2*C*d*x - 4*C*Log[Cos [(c + d*x)/2] - Sin[(c + d*x)/2]] + 4*C*Log[Cos[(c + d*x)/2] + Sin[(c + d* x)/2]]) + 4*Cos[2*c + d*x]*(3*A*d*x + 2*C*d*x - 4*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 4*C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + A*Sin [d*x] + 16*C*Sin[d*x] + A*Sin[2*c + d*x] + 8*A*Sin[c + 2*d*x] + 8*A*Sin[3* c + 2*d*x] + A*Sin[2*c + 3*d*x] + A*Sin[4*c + 3*d*x]))/(d*(Cos[c/2] - Sin[ c/2])*(Cos[c/2] + Sin[c/2])*(-1 + Tan[(c + d*x)/2])*(1 + Tan[(c + d*x)/2]) )
Time = 0.68 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3042, 4575, 3042, 4506, 3042, 4484, 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.
| \(\displaystyle \int \cos ^2(c+d x) (a \sec (c+d x)+a)^2 \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 4575 |
| \(\displaystyle \frac {\int \cos (c+d x) (\sec (c+d x) a+a)^2 (2 a A-a (A-2 C) \sec (c+d x))dx}{2 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (2 a A-a (A-2 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {\int \cos (c+d x) (\sec (c+d x) a+a) \left ((3 A-2 C) a^2+4 C \sec (c+d x) a^2\right )dx-\frac {(A-2 C) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{d}}{2 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left ((3 A-2 C) a^2+4 C \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {(A-2 C) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{d}}{2 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 4484 |
| \(\displaystyle \frac {-\int \left (-\left ((3 A+2 C) a^3\right )-4 C \sec (c+d x) a^3\right )dx+\frac {a^3 (3 A-2 C) \sin (c+d x)}{d}-\frac {(A-2 C) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{d}}{2 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {a^3 (3 A-2 C) \sin (c+d x)}{d}-\frac {(A-2 C) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{d}+a^3 x (3 A+2 C)+\frac {4 a^3 C \text {arctanh}(\sin (c+d x))}{d}}{2 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^2}{2 d}\) |
(A*Cos[c + d*x]*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) + (a^3*(3*A + 2 *C)*x + (4*a^3*C*ArcTanh[Sin[c + d*x]])/d + (a^3*(3*A - 2*C)*Sin[c + d*x]) /d - ((A - 2*C)*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/d)/(2*a)
3.1.97.3.1 Defintions of rubi rules used
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n)), x] + Simp[1/(d*n) Int[(d*Csc[e + f*x])^( n + 1)*Simp[n*(B*a + A*b) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && LeQ[n, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B* Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x] )^n*Simp[a*A*d*(m + n) + B*(b*d*n) + (A*b*d*(m + n) + a*B*d*(2*m + n - 1))* Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. ))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[A*Co t[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/( b*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b *(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])
Time = 0.34 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.81
| method | result | size |
| derivativedivides | \(\frac {a^{2} A \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C \,a^{2} \left (d x +c \right )+2 a^{2} A \sin \left (d x +c \right )+2 C \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{2} A \left (d x +c \right )+C \,a^{2} \tan \left (d x +c \right )}{d}\) | \(96\) |
| default | \(\frac {a^{2} A \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C \,a^{2} \left (d x +c \right )+2 a^{2} A \sin \left (d x +c \right )+2 C \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{2} A \left (d x +c \right )+C \,a^{2} \tan \left (d x +c \right )}{d}\) | \(96\) |
| parallelrisch | \(\frac {a^{2} \left (-16 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )+16 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )+8 A \sin \left (2 d x +2 c \right )+A \sin \left (3 d x +3 c \right )+12 x d \left (A +\frac {2 C}{3}\right ) \cos \left (d x +c \right )+\sin \left (d x +c \right ) \left (A +8 C \right )\right )}{8 d \cos \left (d x +c \right )}\) | \(110\) |
| risch | \(\frac {3 a^{2} A x}{2}+a^{2} x C -\frac {i a^{2} A \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {i a^{2} A \,{\mathrm e}^{i \left (d x +c \right )}}{d}+\frac {i a^{2} A \,{\mathrm e}^{-i \left (d x +c \right )}}{d}+\frac {i a^{2} A \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i C \,a^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}+\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}\) | \(158\) |
| norman | \(\frac {\left (-\frac {3}{2} a^{2} A -C \,a^{2}\right ) x +\left (-\frac {3}{2} a^{2} A -C \,a^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (\frac {3}{2} a^{2} A +C \,a^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {3}{2} a^{2} A +C \,a^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (-3 a^{2} A -2 C \,a^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (3 a^{2} A +2 C \,a^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {a^{2} \left (3 A -2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}+\frac {12 a^{2} A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-\frac {4 a^{2} A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {2 a^{2} \left (3 A -2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {a^{2} \left (5 A +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}-\frac {2 C \,a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {2 C \,a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\) | \(334\) |
1/d*(a^2*A*(1/2*sin(d*x+c)*cos(d*x+c)+1/2*d*x+1/2*c)+C*a^2*(d*x+c)+2*a^2*A *sin(d*x+c)+2*C*a^2*ln(sec(d*x+c)+tan(d*x+c))+a^2*A*(d*x+c)+C*a^2*tan(d*x+ c))
Time = 0.29 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.97 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (3 \, A + 2 \, C\right )} a^{2} d x \cos \left (d x + c\right ) + 2 \, C a^{2} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 2 \, C a^{2} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (A a^{2} \cos \left (d x + c\right )^{2} + 4 \, A a^{2} \cos \left (d x + c\right ) + 2 \, C a^{2}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]
1/2*((3*A + 2*C)*a^2*d*x*cos(d*x + c) + 2*C*a^2*cos(d*x + c)*log(sin(d*x + c) + 1) - 2*C*a^2*cos(d*x + c)*log(-sin(d*x + c) + 1) + (A*a^2*cos(d*x + c)^2 + 4*A*a^2*cos(d*x + c) + 2*C*a^2)*sin(d*x + c))/(d*cos(d*x + c))
\[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=a^{2} \left (\int A \cos ^{2}{\left (c + d x \right )}\, dx + \int 2 A \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int A \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 2 C \cos ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx\right ) \]
a**2*(Integral(A*cos(c + d*x)**2, x) + Integral(2*A*cos(c + d*x)**2*sec(c + d*x), x) + Integral(A*cos(c + d*x)**2*sec(c + d*x)**2, x) + Integral(C*c os(c + d*x)**2*sec(c + d*x)**2, x) + Integral(2*C*cos(c + d*x)**2*sec(c + d*x)**3, x) + Integral(C*cos(c + d*x)**2*sec(c + d*x)**4, x))
Time = 0.22 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.85 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} + 4 \, {\left (d x + c\right )} A a^{2} + 4 \, {\left (d x + c\right )} C a^{2} + 4 \, C a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 8 \, A a^{2} \sin \left (d x + c\right ) + 4 \, C a^{2} \tan \left (d x + c\right )}{4 \, d} \]
1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2 + 4*(d*x + c)*A*a^2 + 4*(d*x + c)*C*a^2 + 4*C*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 8*A* a^2*sin(d*x + c) + 4*C*a^2*tan(d*x + c))/d
Time = 0.34 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.20 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {4 \, C a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 4 \, C a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {4 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + {\left (3 \, A a^{2} + 2 \, C a^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]
1/2*(4*C*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 4*C*a^2*log(abs(tan(1/2* d*x + 1/2*c) - 1)) - 4*C*a^2*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) + (3*A*a^2 + 2*C*a^2)*(d*x + c) + 2*(3*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 5*A*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^2)/d
Time = 15.18 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.28 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2\,A\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {3\,A\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,C\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {C\,a^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {A\,a^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d} \]