Integrand size = 41, antiderivative size = 171 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{2} a^3 (7 A+6 B+2 C) x+\frac {a^3 (2 A+6 B+7 C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {5 a^3 (A-C) \sin (c+d x)}{2 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac {(A-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 a d}-\frac {(A-2 B-4 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{2 d} \] Output:
1/2*a^3*(7*A+6*B+2*C)*x+1/2*a^3*(2*A+6*B+7*C)*arctanh(sin(d*x+c))/d+5/2*a^ 3*(A-C)*sin(d*x+c)/d+1/2*A*cos(d*x+c)*(a+a*sec(d*x+c))^3*sin(d*x+c)/d-1/2* (A-C)*(a^2+a^2*sec(d*x+c))^2*sin(d*x+c)/a/d-1/2*(A-2*B-4*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. \(406\) vs. \(2(171)=342\).
Time = 7.25 (sec) , antiderivative size = 406, normalized size of antiderivative = 2.37 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^3 \cos ^5(c+d x) \sec ^6\left (\frac {1}{2} (c+d x)\right ) (1+\sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (2 (7 A+6 B+2 C) x-\frac {2 (2 A+6 B+7 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {2 (2 A+6 B+7 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {4 (3 A+B) \cos (d x) \sin (c)}{d}+\frac {A \cos (2 d x) \sin (2 c)}{d}+\frac {4 (3 A+B) \cos (c) \sin (d x)}{d}+\frac {A \cos (2 c) \sin (2 d x)}{d}+\frac {C}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 (B+3 C) \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {C}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 (B+3 C) \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )}{16 (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x)))} \] Input:
Integrate[Cos[c + d*x]^2*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Se c[c + d*x]^2),x]
Output:
(a^3*Cos[c + d*x]^5*Sec[(c + d*x)/2]^6*(1 + Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(2*(7*A + 6*B + 2*C)*x - (2*(2*A + 6*B + 7*C)*Lo g[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (2*(2*A + 6*B + 7*C)*Log[Cos[( c + d*x)/2] + Sin[(c + d*x)/2]])/d + (4*(3*A + B)*Cos[d*x]*Sin[c])/d + (A* Cos[2*d*x]*Sin[2*c])/d + (4*(3*A + B)*Cos[c]*Sin[d*x])/d + (A*Cos[2*c]*Sin [2*d*x])/d + C/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (4*(B + 3*C)* Sin[(d*x)/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2 ])) - C/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (4*(B + 3*C)*Sin[(d* x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/( 16*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)]))
Time = 1.01 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.96, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.244, Rules used = {3042, 4574, 3042, 4506, 27, 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.
| \(\displaystyle \int \cos ^2(c+d x) (a \sec (c+d x)+a)^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 4574 |
| \(\displaystyle \frac {\int \cos (c+d x) (\sec (c+d x) a+a)^3 (a (3 A+2 B)-2 a (A-C) \sec (c+d x))dx}{2 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (a (3 A+2 B)-2 a (A-C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {\frac {1}{2} \int 2 \cos (c+d x) (\sec (c+d x) a+a)^2 \left (a^2 (4 A+2 B-C)-a^2 (A-2 B-4 C) \sec (c+d x)\right )dx-\frac {(A-C) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d}}{2 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \cos (c+d x) (\sec (c+d x) a+a)^2 \left (a^2 (4 A+2 B-C)-a^2 (A-2 B-4 C) \sec (c+d x)\right )dx-\frac {(A-C) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d}}{2 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (a^2 (4 A+2 B-C)-a^2 (A-2 B-4 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {(A-C) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d}}{2 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {\int \cos (c+d x) (\sec (c+d x) a+a) \left (5 (A-C) a^3+(2 A+6 B+7 C) \sec (c+d x) a^3\right )dx-\frac {(A-2 B-4 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{d}-\frac {(A-C) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d}}{2 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (5 (A-C) a^3+(2 A+6 B+7 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {(A-2 B-4 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{d}-\frac {(A-C) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d}}{2 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 4484 |
| \(\displaystyle \frac {-\int \left (-\left ((7 A+6 B+2 C) a^4\right )-(2 A+6 B+7 C) \sec (c+d x) a^4\right )dx-\frac {(A-2 B-4 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{d}+\frac {5 a^4 (A-C) \sin (c+d x)}{d}-\frac {(A-C) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d}}{2 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {a^4 (2 A+6 B+7 C) \text {arctanh}(\sin (c+d x))}{d}-\frac {(A-2 B-4 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{d}+a^4 x (7 A+6 B+2 C)+\frac {5 a^4 (A-C) \sin (c+d x)}{d}-\frac {(A-C) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d}}{2 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d}\) |
Input:
Int[Cos[c + d*x]^2*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
Output:
(A*Cos[c + d*x]*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(2*d) + (a^4*(7*A + 6 *B + 2*C)*x + (a^4*(2*A + 6*B + 7*C)*ArcTanh[Sin[c + d*x]])/d + (5*a^4*(A - C)*Sin[c + d*x])/d - ((A - C)*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/d - ((A - 2*B - 4*C)*(a^4 + a^4*Sec[c + d*x])*Sin[c + d*x])/d)/(2*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[(-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_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[A*Cot[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*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x] , x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])
Time = 0.97 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.09
| method | result | size |
| parallelrisch | \(\frac {a^{3} \left (-4 \left (A +3 B +\frac {7 C}{2}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+4 \left (A +3 B +\frac {7 C}{2}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+14 x \left (A +\frac {6 B}{7}+\frac {2 C}{7}\right ) d \cos \left (2 d x +2 c \right )+\left (12 C +A +4 B \right ) \sin \left (2 d x +2 c \right )+\left (2 B +6 A \right ) \sin \left (3 d x +3 c \right )+\frac {A \sin \left (4 d x +4 c \right )}{2}+\left (6 A +2 B +4 C \right ) \sin \left (d x +c \right )+14 x \left (A +\frac {6 B}{7}+\frac {2 C}{7}\right ) d \right )}{4 d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(186\) |
| derivativedivides | \(\frac {a^{3} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{3} \tan \left (d x +c \right )+a^{3} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a^{3} A \left (d x +c \right )+3 B \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{3} C \tan \left (d x +c \right )+3 a^{3} A \sin \left (d x +c \right )+3 B \,a^{3} \left (d x +c \right )+3 a^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{3} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{3} \sin \left (d x +c \right )+a^{3} C \left (d x +c \right )}{d}\) | \(206\) |
| default | \(\frac {a^{3} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{3} \tan \left (d x +c \right )+a^{3} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a^{3} A \left (d x +c \right )+3 B \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{3} C \tan \left (d x +c \right )+3 a^{3} A \sin \left (d x +c \right )+3 B \,a^{3} \left (d x +c \right )+3 a^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{3} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{3} \sin \left (d x +c \right )+a^{3} C \left (d x +c \right )}{d}\) | \(206\) |
| risch | \(\frac {7 a^{3} A x}{2}+3 a^{3} B x +a^{3} x C -\frac {i a^{3} \left (C \,{\mathrm e}^{3 i \left (d x +c \right )}-2 B \,{\mathrm e}^{2 i \left (d x +c \right )}-6 C \,{\mathrm e}^{2 i \left (d x +c \right )}-C \,{\mathrm e}^{i \left (d x +c \right )}-2 B -6 C \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {3 i a^{3} A \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i a^{3} A \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {i a^{3} A \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {i B \,a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{3}}{2 d}+\frac {3 i a^{3} A \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}-\frac {7 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}+\frac {7 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}\) | \(343\) |
| norman | \(\frac {\left (\frac {7}{2} a^{3} A +a^{3} C +3 B \,a^{3}\right ) x +\left (-\frac {7}{2} a^{3} A -a^{3} C -3 B \,a^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-\frac {7}{2} a^{3} A -a^{3} C -3 B \,a^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (\frac {7}{2} a^{3} A +a^{3} C +3 B \,a^{3}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (-7 a^{3} A -6 B \,a^{3}-2 a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-7 a^{3} A -6 B \,a^{3}-2 a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (14 a^{3} A +12 B \,a^{3}+4 a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {a^{3} \left (7 A +4 B +7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {5 a^{3} \left (A -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}+\frac {2 a^{3} \left (11 A -7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {2 a^{3} \left (A +4 B +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {a^{3} \left (13 A -7 C +4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}-\frac {a^{3} \left (23 A +8 B +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {a^{3} \left (2 A +6 B +7 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{3} \left (2 A +6 B +7 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(466\) |
Input:
int(cos(d*x+c)^2*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,meth od=_RETURNVERBOSE)
Output:
1/4*a^3*(-4*(A+3*B+7/2*C)*(1+cos(2*d*x+2*c))*ln(tan(1/2*d*x+1/2*c)-1)+4*(A +3*B+7/2*C)*(1+cos(2*d*x+2*c))*ln(tan(1/2*d*x+1/2*c)+1)+14*x*(A+6/7*B+2/7* C)*d*cos(2*d*x+2*c)+(12*C+A+4*B)*sin(2*d*x+2*c)+(2*B+6*A)*sin(3*d*x+3*c)+1 /2*A*sin(4*d*x+4*c)+(6*A+2*B+4*C)*sin(d*x+c)+14*x*(A+6/7*B+2/7*C)*d)/d/(1+ cos(2*d*x+2*c))
Time = 0.10 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.96 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (7 \, A + 6 \, B + 2 \, C\right )} a^{3} d x \cos \left (d x + c\right )^{2} + {\left (2 \, A + 6 \, B + 7 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, A + 6 \, B + 7 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (A a^{3} \cos \left (d x + c\right )^{3} + 2 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right )^{2} + 2 \, {\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right ) + C a^{3}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="fricas")
Output:
1/4*(2*(7*A + 6*B + 2*C)*a^3*d*x*cos(d*x + c)^2 + (2*A + 6*B + 7*C)*a^3*co s(d*x + c)^2*log(sin(d*x + c) + 1) - (2*A + 6*B + 7*C)*a^3*cos(d*x + c)^2* log(-sin(d*x + c) + 1) + 2*(A*a^3*cos(d*x + c)^3 + 2*(3*A + B)*a^3*cos(d*x + c)^2 + 2*(B + 3*C)*a^3*cos(d*x + c) + C*a^3)*sin(d*x + c))/(d*cos(d*x + c)^2)
Timed out. \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**2*(a+a*sec(d*x+c))**3*(A+B*sec(d*x+c)+C*sec(d*x+c)** 2),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.39 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 12 \, {\left (d x + c\right )} A a^{3} + 12 \, {\left (d x + c\right )} B a^{3} + 4 \, {\left (d x + c\right )} C a^{3} - C a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, B a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a^{3} \sin \left (d x + c\right ) + 4 \, B a^{3} \sin \left (d x + c\right ) + 4 \, B a^{3} \tan \left (d x + c\right ) + 12 \, C a^{3} \tan \left (d x + c\right )}{4 \, d} \] Input:
integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="maxima")
Output:
1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^3 + 12*(d*x + c)*A*a^3 + 12*(d*x + c)*B*a^3 + 4*(d*x + c)*C*a^3 - C*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 2*A*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 6*B*a^3*(log(sin(d*x + c) + 1) - log (sin(d*x + c) - 1)) + 6*C*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*A*a^3*sin(d*x + c) + 4*B*a^3*sin(d*x + c) + 4*B*a^3*tan(d*x + c) + 12*C*a^3*tan(d*x + c))/d
Time = 0.30 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.64 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (7 \, A a^{3} + 6 \, B a^{3} + 2 \, C a^{3}\right )} {\left (d x + c\right )} + {\left (2 \, A a^{3} + 6 \, B a^{3} + 7 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (2 \, A a^{3} + 6 \, B a^{3} + 7 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (5 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 5 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 3 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 7 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1\right )}^{2}}}{2 \, d} \] Input:
integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="giac")
Output:
1/2*((7*A*a^3 + 6*B*a^3 + 2*C*a^3)*(d*x + c) + (2*A*a^3 + 6*B*a^3 + 7*C*a^ 3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (2*A*a^3 + 6*B*a^3 + 7*C*a^3)*log( abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(5*A*a^3*tan(1/2*d*x + 1/2*c)^7 - 5*C*a ^3*tan(1/2*d*x + 1/2*c)^7 - 3*A*a^3*tan(1/2*d*x + 1/2*c)^5 - 4*B*a^3*tan(1 /2*d*x + 1/2*c)^5 - 3*C*a^3*tan(1/2*d*x + 1/2*c)^5 - 9*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 9*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 7*A*a^3*tan(1/2*d*x + 1/2*c) + 4*B*a^3*tan(1/2*d*x + 1/2*c) + 7*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d* x + 1/2*c)^4 - 1)^2)/d
Time = 13.52 (sec) , antiderivative size = 329, normalized size of antiderivative = 1.92 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2\,\left (\frac {7\,A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}-A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,1{}\mathrm {i}+3\,B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,3{}\mathrm {i}+C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-\frac {C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,7{}\mathrm {i}}{2}\right )}{d}+\frac {\frac {A\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{8}+\frac {3\,A\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {A\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{16}+\frac {B\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {B\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {3\,C\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {3\,A\,a^3\,\sin \left (c+d\,x\right )}{4}+\frac {B\,a^3\,\sin \left (c+d\,x\right )}{4}+\frac {C\,a^3\,\sin \left (c+d\,x\right )}{2}}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )} \] Input:
int(cos(c + d*x)^2*(a + a/cos(c + d*x))^3*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)
Output:
(2*((7*A*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 - A*a^3*atan(( sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*1i + 3*B*a^3*atan(sin(c/2 + (d* x)/2)/cos(c/2 + (d*x)/2)) - B*a^3*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + ( d*x)/2))*3i + C*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - (C*a^3*a tan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*7i)/2))/d + ((A*a^3*sin(2* c + 2*d*x))/8 + (3*A*a^3*sin(3*c + 3*d*x))/4 + (A*a^3*sin(4*c + 4*d*x))/16 + (B*a^3*sin(2*c + 2*d*x))/2 + (B*a^3*sin(3*c + 3*d*x))/4 + (3*C*a^3*sin( 2*c + 2*d*x))/2 + (3*A*a^3*sin(c + d*x))/4 + (B*a^3*sin(c + d*x))/4 + (C*a ^3*sin(c + d*x))/2)/(d*(cos(2*c + 2*d*x)/2 + 1/2))
Time = 0.16 (sec) , antiderivative size = 463, normalized size of antiderivative = 2.71 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{3} \left (-2 c d x -7 a c +\cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a -6 \cos \left (d x +c \right ) \sin \left (d x +c \right ) c -6 b c -\cos \left (d x +c \right ) \sin \left (d x +c \right ) a -7 a d x -6 b d x +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +7 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) c -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -7 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) c +2 \sin \left (d x +c \right )^{2} c d x -6 \sin \left (d x +c \right ) a -2 c^{2}-2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b -6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b +6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b +2 \sin \left (d x +c \right )^{2} c^{2}+6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b -6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b +7 \sin \left (d x +c \right )^{2} a d x +6 \sin \left (d x +c \right )^{2} b d x -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a -7 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} c +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a +7 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} c +2 \sin \left (d x +c \right )^{3} b -\sin \left (d x +c \right ) c -2 \sin \left (d x +c \right ) b +7 \sin \left (d x +c \right )^{2} a c +6 \sin \left (d x +c \right )^{2} b c +6 \sin \left (d x +c \right )^{3} a \right )}{2 d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(cos(d*x+c)^2*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
Output:
(a**3*(cos(c + d*x)*sin(c + d*x)**3*a - cos(c + d*x)*sin(c + d*x)*a - 2*co s(c + d*x)*sin(c + d*x)*b - 6*cos(c + d*x)*sin(c + d*x)*c - 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a - 6*log(tan((c + d*x)/2) - 1)*sin(c + d*x) **2*b - 7*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*c + 2*log(tan((c + d*x )/2) - 1)*a + 6*log(tan((c + d*x)/2) - 1)*b + 7*log(tan((c + d*x)/2) - 1)* c + 2*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a + 6*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b + 7*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*c - 2*log(tan((c + d*x)/2) + 1)*a - 6*log(tan((c + d*x)/2) + 1)*b - 7*log(tan( (c + d*x)/2) + 1)*c + 6*sin(c + d*x)**3*a + 2*sin(c + d*x)**3*b + 7*sin(c + d*x)**2*a*c + 7*sin(c + d*x)**2*a*d*x + 6*sin(c + d*x)**2*b*c + 6*sin(c + d*x)**2*b*d*x + 2*sin(c + d*x)**2*c**2 + 2*sin(c + d*x)**2*c*d*x - 6*sin (c + d*x)*a - 2*sin(c + d*x)*b - sin(c + d*x)*c - 7*a*c - 7*a*d*x - 6*b*c - 6*b*d*x - 2*c**2 - 2*c*d*x))/(2*d*(sin(c + d*x)**2 - 1))