Integrand size = 33, antiderivative size = 156 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{2} a^3 (5 A+6 C) x+\frac {3 a^3 C \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^3 A \sin (c+d x)}{2 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {A \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 a d}-\frac {(5 A-6 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{6 d} \] Output:
1/2*a^3*(5*A+6*C)*x+3*a^3*C*arctanh(sin(d*x+c))/d+5/2*a^3*A*sin(d*x+c)/d+1 /3*A*cos(d*x+c)^2*(a+a*sec(d*x+c))^3*sin(d*x+c)/d+1/2*A*cos(d*x+c)*(a^2+a^ 2*sec(d*x+c))^2*sin(d*x+c)/a/d-1/6*(5*A-6*C)*(a^3+a^3*sec(d*x+c))*sin(d*x+ c)/d
Leaf count is larger than twice the leaf count of optimal. \(1014\) vs. \(2(156)=312\).
Time = 7.49 (sec) , antiderivative size = 1014, normalized size of antiderivative = 6.50 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:
Integrate[Cos[c + d*x]^3*(a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2),x]
Output:
a^3*(((5*A + 6*C)*x*Cos[c + d*x]^2*(1 + Cos[c + d*x])^3*Sec[c/2 + (d*x)/2] ^6*(A + C*Sec[c + d*x]^2))/(8*(A + 2*C + A*Cos[2*c + 2*d*x])) - (3*C*Cos[c + d*x]^2*(1 + Cos[c + d*x])^3*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2] ]*Sec[c/2 + (d*x)/2]^6*(A + C*Sec[c + d*x]^2))/(4*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + (3*C*Cos[c + d*x]^2*(1 + Cos[c + d*x])^3*Log[Cos[c/2 + (d*x)/2 ] + Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^6*(A + C*Sec[c + d*x]^2))/(4*d* (A + 2*C + A*Cos[2*c + 2*d*x])) + ((15*A + 4*C)*Cos[d*x]*Cos[c + d*x]^2*(1 + Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(A + C*Sec[c + d*x]^2)*Sin[c])/(16 *d*(A + 2*C + A*Cos[2*c + 2*d*x])) + (3*A*Cos[2*d*x]*Cos[c + d*x]^2*(1 + C os[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(A + C*Sec[c + d*x]^2)*Sin[2*c])/(16*d *(A + 2*C + A*Cos[2*c + 2*d*x])) + (A*Cos[3*d*x]*Cos[c + d*x]^2*(1 + Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(A + C*Sec[c + d*x]^2)*Sin[3*c])/(48*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + ((15*A + 4*C)*Cos[c]*Cos[c + d*x]^2*(1 + Co s[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(A + C*Sec[c + d*x]^2)*Sin[d*x])/(16*d* (A + 2*C + A*Cos[2*c + 2*d*x])) + (3*A*Cos[2*c]*Cos[c + d*x]^2*(1 + Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(A + C*Sec[c + d*x]^2)*Sin[2*d*x])/(16*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + (A*Cos[3*c]*Cos[c + d*x]^2*(1 + Cos[c + d* x])^3*Sec[c/2 + (d*x)/2]^6*(A + C*Sec[c + d*x]^2)*Sin[3*d*x])/(48*d*(A + 2 *C + A*Cos[2*c + 2*d*x])) + (C*Cos[c + d*x]^2*(1 + Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(A + C*Sec[c + d*x]^2)*Sin[(d*x)/2])/(4*d*(A + 2*C + A*Co...
Time = 0.97 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {3042, 4575, 3042, 4505, 3042, 4506, 27, 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.
| \(\displaystyle \int \cos ^3(c+d x) (a \sec (c+d x)+a)^3 \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
| \(\Big \downarrow \) 4575 |
| \(\displaystyle \frac {\int \cos ^2(c+d x) (\sec (c+d x) a+a)^3 (3 a A-a (A-3 C) \sec (c+d x))dx}{3 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^3}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (3 a A-a (A-3 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx}{3 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^3}{3 d}\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {\frac {1}{2} \int \cos (c+d x) (\sec (c+d x) a+a)^2 \left (2 a^2 (5 A+3 C)-a^2 (5 A-6 C) \sec (c+d x)\right )dx+\frac {3 A \sin (c+d x) \cos (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}}{3 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^3}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (2 a^2 (5 A+3 C)-a^2 (5 A-6 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {3 A \sin (c+d x) \cos (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}}{3 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^3}{3 d}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {\frac {1}{2} \left (\int 3 \cos (c+d x) (\sec (c+d x) a+a) \left (5 A a^3+6 C \sec (c+d x) a^3\right )dx-\frac {(5 A-6 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{d}\right )+\frac {3 A \sin (c+d x) \cos (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}}{3 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^3}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{2} \left (3 \int \cos (c+d x) (\sec (c+d x) a+a) \left (5 A a^3+6 C \sec (c+d x) a^3\right )dx-\frac {(5 A-6 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{d}\right )+\frac {3 A \sin (c+d x) \cos (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}}{3 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^3}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \left (3 \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (5 A a^3+6 C \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {(5 A-6 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{d}\right )+\frac {3 A \sin (c+d x) \cos (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}}{3 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^3}{3 d}\) |
| \(\Big \downarrow \) 4484 |
| \(\displaystyle \frac {\frac {1}{2} \left (3 \left (\frac {5 a^4 A \sin (c+d x)}{d}-\int \left (-\left ((5 A+6 C) a^4\right )-6 C \sec (c+d x) a^4\right )dx\right )-\frac {(5 A-6 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{d}\right )+\frac {3 A \sin (c+d x) \cos (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}}{3 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^3}{3 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{2} \left (3 \left (\frac {5 a^4 A \sin (c+d x)}{d}+a^4 x (5 A+6 C)+\frac {6 a^4 C \text {arctanh}(\sin (c+d x))}{d}\right )-\frac {(5 A-6 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{d}\right )+\frac {3 A \sin (c+d x) \cos (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}}{3 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^3}{3 d}\) |
Input:
Int[Cos[c + d*x]^3*(a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2),x]
Output:
(A*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(3*d) + ((3*A*Cos[c + d*x]*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) + (-(((5*A - 6*C)*( a^4 + a^4*Sec[c + d*x])*Sin[c + d*x])/d) + 3*(a^4*(5*A + 6*C)*x + (6*a^4*C *ArcTanh[Sin[c + d*x]])/d + (5*a^4*A*Sin[c + d*x])/d))/2)/(3*a)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[a*A*Cot [e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x] - Sim p[b/(a*d*n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Sim p[a*A*(m - n - 1) - b*B*n - (a*B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0 ] && GtQ[m, 1/2] && LtQ[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 = 0.47 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.81
| method | result | size |
| parallelrisch | \(\frac {3 a^{3} \left (-8 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )+8 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )+\left (\frac {46 A}{9}+\frac {4 C}{3}\right ) \sin \left (2 d x +2 c \right )+A \sin \left (3 d x +3 c \right )+\frac {A \sin \left (4 d x +4 c \right )}{9}+\frac {20 d \left (A +\frac {6 C}{5}\right ) x \cos \left (d x +c \right )}{3}+\sin \left (d x +c \right ) \left (A +\frac {8 C}{3}\right )\right )}{8 d \cos \left (d x +c \right )}\) | \(127\) |
| derivativedivides | \(\frac {\frac {a^{3} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a^{3} C \sin \left (d x +c \right )+3 a^{3} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{3} C \left (d x +c \right )+3 a^{3} A \sin \left (d x +c \right )+3 a^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{3} A \left (d x +c \right )+a^{3} C \tan \left (d x +c \right )}{d}\) | \(131\) |
| default | \(\frac {\frac {a^{3} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a^{3} C \sin \left (d x +c \right )+3 a^{3} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{3} C \left (d x +c \right )+3 a^{3} A \sin \left (d x +c \right )+3 a^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{3} A \left (d x +c \right )+a^{3} C \tan \left (d x +c \right )}{d}\) | \(131\) |
| risch | \(\frac {5 a^{3} A x}{2}+3 a^{3} x C -\frac {3 i a^{3} A \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {15 i a^{3} A \,{\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{3} C}{2 d}+\frac {15 i a^{3} A \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{3} C}{2 d}+\frac {3 i a^{3} A \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i a^{3} C}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}+\frac {a^{3} A \sin \left (3 d x +3 c \right )}{12 d}\) | \(215\) |
| norman | \(\frac {\left (\frac {5}{2} a^{3} A +3 a^{3} C \right ) x +\left (-\frac {15}{2} a^{3} A -9 a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-\frac {15}{2} a^{3} A -9 a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (-\frac {5}{2} a^{3} A -3 a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-\frac {5}{2} a^{3} A -3 a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (\frac {5}{2} a^{3} A +3 a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}+\left (\frac {15}{2} a^{3} A +9 a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (\frac {15}{2} a^{3} A +9 a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\frac {a^{3} \left (11 A +4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {5 a^{3} A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{d}+\frac {8 a^{3} \left (2 A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {4 a^{3} \left (5 A +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}-\frac {4 a^{3} \left (23 A +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {a^{3} \left (37 A -12 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}+\frac {a^{3} \left (53 A -24 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {3 a^{3} C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {3 a^{3} C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\) | \(444\) |
Input:
int(cos(d*x+c)^3*(a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x,method=_RETURNVER BOSE)
Output:
3/8*a^3*(-8*C*ln(tan(1/2*d*x+1/2*c)-1)*cos(d*x+c)+8*C*ln(tan(1/2*d*x+1/2*c )+1)*cos(d*x+c)+(46/9*A+4/3*C)*sin(2*d*x+2*c)+A*sin(3*d*x+3*c)+1/9*A*sin(4 *d*x+4*c)+20/3*d*(A+6/5*C)*x*cos(d*x+c)+sin(d*x+c)*(A+8/3*C))/d/cos(d*x+c)
Time = 0.10 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.88 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (5 \, A + 6 \, C\right )} a^{3} d x \cos \left (d x + c\right ) + 9 \, C a^{3} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, C a^{3} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, A a^{3} \cos \left (d x + c\right )^{3} + 9 \, A a^{3} \cos \left (d x + c\right )^{2} + 2 \, {\left (11 \, A + 3 \, C\right )} a^{3} \cos \left (d x + c\right ) + 6 \, C a^{3}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )} \] Input:
integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm= "fricas")
Output:
1/6*(3*(5*A + 6*C)*a^3*d*x*cos(d*x + c) + 9*C*a^3*cos(d*x + c)*log(sin(d*x + c) + 1) - 9*C*a^3*cos(d*x + c)*log(-sin(d*x + c) + 1) + (2*A*a^3*cos(d* x + c)^3 + 9*A*a^3*cos(d*x + c)^2 + 2*(11*A + 3*C)*a^3*cos(d*x + c) + 6*C* a^3)*sin(d*x + c))/(d*cos(d*x + c))
Timed out. \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**3*(a+a*sec(d*x+c))**3*(A+C*sec(d*x+c)**2),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.88 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{3} - 9 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 12 \, {\left (d x + c\right )} A a^{3} - 36 \, {\left (d x + c\right )} C a^{3} - 18 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, A a^{3} \sin \left (d x + c\right ) - 12 \, C a^{3} \sin \left (d x + c\right ) - 12 \, C a^{3} \tan \left (d x + c\right )}{12 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm= "maxima")
Output:
-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^3 - 9*(2*d*x + 2*c + sin(2* d*x + 2*c))*A*a^3 - 12*(d*x + c)*A*a^3 - 36*(d*x + c)*C*a^3 - 18*C*a^3*(lo g(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 36*A*a^3*sin(d*x + c) - 12* C*a^3*sin(d*x + c) - 12*C*a^3*tan(d*x + c))/d
Time = 0.39 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.35 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {18 \, C a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 18 \, C a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {12 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + 3 \, {\left (5 \, A a^{3} + 6 \, C a^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (15 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 33 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm= "giac")
Output:
1/6*(18*C*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 18*C*a^3*log(abs(tan(1/ 2*d*x + 1/2*c) - 1)) - 12*C*a^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c) ^2 - 1) + 3*(5*A*a^3 + 6*C*a^3)*(d*x + c) + 2*(15*A*a^3*tan(1/2*d*x + 1/2* c)^5 + 6*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 40*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 12*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 33*A*a^3*tan(1/2*d*x + 1/2*c) + 6*C*a^3* tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d
Time = 12.90 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.21 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {11\,A\,a^3\,\sin \left (c+d\,x\right )}{3\,d}+\frac {C\,a^3\,\sin \left (c+d\,x\right )}{d}+\frac {5\,A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {6\,C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {6\,C\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a^3\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d}+\frac {C\,a^3\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {3\,A\,a^3\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d} \] Input:
int(cos(c + d*x)^3*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^3,x)
Output:
(11*A*a^3*sin(c + d*x))/(3*d) + (C*a^3*sin(c + d*x))/d + (5*A*a^3*atan(sin (c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (6*C*a^3*atan(sin(c/2 + (d*x)/2)/ cos(c/2 + (d*x)/2)))/d + (6*C*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x) /2)))/d + (A*a^3*cos(c + d*x)^2*sin(c + d*x))/(3*d) + (C*a^3*sin(c + d*x)) /(d*cos(c + d*x)) + (3*A*a^3*cos(c + d*x)*sin(c + d*x))/(2*d)
Time = 0.15 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.14 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{3} \left (-18 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) c +18 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) c -2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a +24 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a +6 \cos \left (d x +c \right ) \sin \left (d x +c \right ) c +15 \cos \left (d x +c \right ) a c +15 \cos \left (d x +c \right ) a d x +18 \cos \left (d x +c \right ) c^{2}+18 \cos \left (d x +c \right ) c d x -9 \sin \left (d x +c \right )^{3} a +9 \sin \left (d x +c \right ) a +6 \sin \left (d x +c \right ) c \right )}{6 \cos \left (d x +c \right ) d} \] Input:
int(cos(d*x+c)^3*(a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x)
Output:
(a**3*( - 18*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*c + 18*cos(c + d*x)*lo g(tan((c + d*x)/2) + 1)*c - 2*cos(c + d*x)*sin(c + d*x)**3*a + 24*cos(c + d*x)*sin(c + d*x)*a + 6*cos(c + d*x)*sin(c + d*x)*c + 15*cos(c + d*x)*a*c + 15*cos(c + d*x)*a*d*x + 18*cos(c + d*x)*c**2 + 18*cos(c + d*x)*c*d*x - 9 *sin(c + d*x)**3*a + 9*sin(c + d*x)*a + 6*sin(c + d*x)*c))/(6*cos(c + d*x) *d)