Integrand size = 33, antiderivative size = 192 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{2} a^4 (13 A+2 C) x+\frac {2 a^4 (2 A+3 C) \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^4 (A-2 C) \sin (c+d x)}{2 d}-\frac {a (3 A-2 C) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac {(A-2 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac {(3 A+22 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d} \] Output:
1/2*a^4*(13*A+2*C)*x+2*a^4*(2*A+3*C)*arctanh(sin(d*x+c))/d+5/2*a^4*(A-2*C) *sin(d*x+c)/d-1/6*a*(3*A-2*C)*(a+a*sec(d*x+c))^3*sin(d*x+c)/d+1/2*A*cos(d* x+c)*(a+a*sec(d*x+c))^4*sin(d*x+c)/d-1/2*(A-2*C)*(a^2+a^2*sec(d*x+c))^2*si n(d*x+c)/d+1/6*(3*A+22*C)*(a^4+a^4*sec(d*x+c))*sin(d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(1420\) vs. \(2(192)=384\).
Time = 12.30 (sec) , antiderivative size = 1420, normalized size of antiderivative = 7.40 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:
Integrate[Cos[c + d*x]^2*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2),x]
Output:
((13*A + 2*C)*x*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4 *(A + C*Sec[c + d*x]^2))/(16*(A + 2*C + A*Cos[2*c + 2*d*x])) + ((-2*A - 3* C)*Cos[c + d*x]^6*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c/2 + ( d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2))/(4*d*(A + 2*C + A *Cos[2*c + 2*d*x])) + ((2*A + 3*C)*Cos[c + d*x]^6*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + C*Se c[c + d*x]^2))/(4*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + (A*Cos[d*x]*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2)* Sin[c])/(2*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + (A*Cos[2*d*x]*Cos[c + d*x]^ 6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2)*Sin[2 *c])/(32*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + (A*Cos[c]*Cos[c + d*x]^6*Sec[ c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2)*Sin[d*x])/( 2*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + (A*Cos[2*c]*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2)*Sin[2*d*x])/(32* d*(A + 2*C + A*Cos[2*c + 2*d*x])) + (C*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8 *(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2)*Sin[(d*x)/2])/(48*d*(A + 2* C + A*Cos[2*c + 2*d*x])*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/ 2 + (d*x)/2])^3) + (Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x ])^4*(A + C*Sec[c + d*x]^2)*(13*C*Cos[c/2] - 11*C*Sin[c/2]))/(96*d*(A + 2* C + A*Cos[2*c + 2*d*x])*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin...
Time = 1.26 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.04, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.394, Rules used = {3042, 4575, 3042, 4506, 3042, 4506, 27, 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.
| \(\displaystyle \int \cos ^2(c+d x) (a \sec (c+d x)+a)^4 \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^4 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 4575 |
| \(\displaystyle \frac {\int \cos (c+d x) (\sec (c+d x) a+a)^4 (4 a A-a (3 A-2 C) \sec (c+d x))dx}{2 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^4}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^4 \left (4 a A-a (3 A-2 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^4}{2 d}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {\frac {1}{3} \int \cos (c+d x) (\sec (c+d x) a+a)^3 \left (a^2 (15 A-2 C)-6 a^2 (A-2 C) \sec (c+d x)\right )dx-\frac {a^2 (3 A-2 C) \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d}}{2 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^4}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (a^2 (15 A-2 C)-6 a^2 (A-2 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {a^2 (3 A-2 C) \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d}}{2 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^4}{2 d}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \int 2 \cos (c+d x) (\sec (c+d x) a+a)^2 \left (2 (9 A-4 C) a^3+(3 A+22 C) \sec (c+d x) a^3\right )dx-\frac {3 a^3 (A-2 C) \sin (c+d x) (a \sec (c+d x)+a)^2}{d}\right )-\frac {a^2 (3 A-2 C) \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d}}{2 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^4}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{3} \left (\int \cos (c+d x) (\sec (c+d x) a+a)^2 \left (2 (9 A-4 C) a^3+(3 A+22 C) \sec (c+d x) a^3\right )dx-\frac {3 a^3 (A-2 C) \sin (c+d x) (a \sec (c+d x)+a)^2}{d}\right )-\frac {a^2 (3 A-2 C) \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d}}{2 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^4}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (2 (9 A-4 C) a^3+(3 A+22 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {3 a^3 (A-2 C) \sin (c+d x) (a \sec (c+d x)+a)^2}{d}\right )-\frac {a^2 (3 A-2 C) \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d}}{2 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^4}{2 d}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {\frac {1}{3} \left (\int 3 \cos (c+d x) (\sec (c+d x) a+a) \left (5 (A-2 C) a^4+4 (2 A+3 C) \sec (c+d x) a^4\right )dx+\frac {(3 A+22 C) \sin (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{d}-\frac {3 a^3 (A-2 C) \sin (c+d x) (a \sec (c+d x)+a)^2}{d}\right )-\frac {a^2 (3 A-2 C) \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d}}{2 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^4}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{3} \left (3 \int \cos (c+d x) (\sec (c+d x) a+a) \left (5 (A-2 C) a^4+4 (2 A+3 C) \sec (c+d x) a^4\right )dx+\frac {(3 A+22 C) \sin (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{d}-\frac {3 a^3 (A-2 C) \sin (c+d x) (a \sec (c+d x)+a)^2}{d}\right )-\frac {a^2 (3 A-2 C) \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d}}{2 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^4}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (3 \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (5 (A-2 C) a^4+4 (2 A+3 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^4\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {(3 A+22 C) \sin (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{d}-\frac {3 a^3 (A-2 C) \sin (c+d x) (a \sec (c+d x)+a)^2}{d}\right )-\frac {a^2 (3 A-2 C) \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d}}{2 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^4}{2 d}\) |
| \(\Big \downarrow \) 4484 |
| \(\displaystyle \frac {\frac {1}{3} \left (3 \left (\frac {5 a^5 (A-2 C) \sin (c+d x)}{d}-\int \left (-\left ((13 A+2 C) a^5\right )-4 (2 A+3 C) \sec (c+d x) a^5\right )dx\right )+\frac {(3 A+22 C) \sin (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{d}-\frac {3 a^3 (A-2 C) \sin (c+d x) (a \sec (c+d x)+a)^2}{d}\right )-\frac {a^2 (3 A-2 C) \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d}}{2 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^4}{2 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{3} \left (3 \left (\frac {4 a^5 (2 A+3 C) \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^5 (A-2 C) \sin (c+d x)}{d}+a^5 x (13 A+2 C)\right )+\frac {(3 A+22 C) \sin (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{d}-\frac {3 a^3 (A-2 C) \sin (c+d x) (a \sec (c+d x)+a)^2}{d}\right )-\frac {a^2 (3 A-2 C) \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d}}{2 a}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^4}{2 d}\) |
Input:
Int[Cos[c + d*x]^2*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2),x]
Output:
(A*Cos[c + d*x]*(a + a*Sec[c + d*x])^4*Sin[c + d*x])/(2*d) + (-1/3*(a^2*(3 *A - 2*C)*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/d + ((-3*a^3*(A - 2*C)*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/d + ((3*A + 22*C)*(a^5 + a^5*Sec[c + d*x] )*Sin[c + d*x])/d + 3*(a^5*(13*A + 2*C)*x + (4*a^5*(2*A + 3*C)*ArcTanh[Sin [c + d*x]])/d + (5*a^5*(A - 2*C)*Sin[c + d*x])/d))/3)/(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_)]^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 = 0.78 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.99
| method | result | size |
| derivativedivides | \(\frac {a^{4} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{4} C \left (d x +c \right )+4 a^{4} A \sin \left (d x +c \right )+4 a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 a^{4} A \left (d x +c \right )+6 a^{4} C \tan \left (d x +c \right )+4 a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{4} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{4} A \tan \left (d x +c \right )-a^{4} C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(190\) |
| default | \(\frac {a^{4} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{4} C \left (d x +c \right )+4 a^{4} A \sin \left (d x +c \right )+4 a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 a^{4} A \left (d x +c \right )+6 a^{4} C \tan \left (d x +c \right )+4 a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{4} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{4} A \tan \left (d x +c \right )-a^{4} C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(190\) |
| parallelrisch | \(\frac {4 \left (-3 \left (A +\frac {3 C}{2}\right ) \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+3 \left (A +\frac {3 C}{2}\right ) \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {13 d \left (A +\frac {2 C}{13}\right ) x \cos \left (3 d x +3 c \right )}{8}+\left (A +C \right ) \sin \left (2 d x +2 c \right )+\left (\frac {11 A}{32}+\frac {5 C}{3}\right ) \sin \left (3 d x +3 c \right )+\frac {A \sin \left (4 d x +4 c \right )}{2}+\frac {A \sin \left (5 d x +5 c \right )}{32}+\frac {39 d \left (A +\frac {2 C}{13}\right ) x \cos \left (d x +c \right )}{8}+\frac {5 \left (A +\frac {32 C}{5}\right ) \sin \left (d x +c \right )}{16}\right ) a^{4}}{d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\) | \(204\) |
| risch | \(\frac {13 a^{4} A x}{2}+a^{4} x C -\frac {i a^{4} A \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {2 i a^{4} A \,{\mathrm e}^{i \left (d x +c \right )}}{d}+\frac {2 i a^{4} A \,{\mathrm e}^{-i \left (d x +c \right )}}{d}+\frac {i a^{4} A \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {2 i a^{4} \left (6 C \,{\mathrm e}^{5 i \left (d x +c \right )}-3 A \,{\mathrm e}^{4 i \left (d x +c \right )}-18 C \,{\mathrm e}^{4 i \left (d x +c \right )}-6 A \,{\mathrm e}^{2 i \left (d x +c \right )}-42 C \,{\mathrm e}^{2 i \left (d x +c \right )}-6 C \,{\mathrm e}^{i \left (d x +c \right )}-3 A -20 C \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {4 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {6 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {4 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {6 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}\) | \(280\) |
| norman | \(\frac {\left (-\frac {13}{2} a^{4} A -a^{4} C \right ) x +\left (-\frac {65}{2} a^{4} A -5 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (-\frac {39}{2} a^{4} A -3 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (-\frac {13}{2} a^{4} A -a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (\frac {13}{2} a^{4} A +a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (\frac {13}{2} a^{4} A +a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}+\left (\frac {39}{2} a^{4} A +3 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {65}{2} a^{4} A +5 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\frac {a^{4} \left (27 A +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}+\frac {5 a^{4} \left (A -2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{d}+\frac {4 a^{4} \left (9 A -38 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}-\frac {a^{4} \left (11 A +18 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 a^{4} \left (33 A -38 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}-\frac {a^{4} \left (53 A -26 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {2 a^{4} \left (63 A +38 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}-\frac {2 a^{4} \left (2 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {2 a^{4} \left (2 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\) | \(460\) |
Input:
int(cos(d*x+c)^2*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x,method=_RETURNVER BOSE)
Output:
1/d*(a^4*A*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+a^4*C*(d*x+c)+4*a^4*A *sin(d*x+c)+4*a^4*C*ln(sec(d*x+c)+tan(d*x+c))+6*a^4*A*(d*x+c)+6*a^4*C*tan( d*x+c)+4*a^4*A*ln(sec(d*x+c)+tan(d*x+c))+4*a^4*C*(1/2*sec(d*x+c)*tan(d*x+c )+1/2*ln(sec(d*x+c)+tan(d*x+c)))+a^4*A*tan(d*x+c)-a^4*C*(-2/3-1/3*sec(d*x+ c)^2)*tan(d*x+c))
Time = 0.10 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.89 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (13 \, A + 2 \, C\right )} a^{4} d x \cos \left (d x + c\right )^{3} + 6 \, {\left (2 \, A + 3 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 6 \, {\left (2 \, A + 3 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (3 \, A a^{4} \cos \left (d x + c\right )^{4} + 24 \, A a^{4} \cos \left (d x + c\right )^{3} + 2 \, {\left (3 \, A + 20 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 12 \, C a^{4} \cos \left (d x + c\right ) + 2 \, C a^{4}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{3}} \] Input:
integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm= "fricas")
Output:
1/6*(3*(13*A + 2*C)*a^4*d*x*cos(d*x + c)^3 + 6*(2*A + 3*C)*a^4*cos(d*x + c )^3*log(sin(d*x + c) + 1) - 6*(2*A + 3*C)*a^4*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + (3*A*a^4*cos(d*x + c)^4 + 24*A*a^4*cos(d*x + c)^3 + 2*(3*A + 2 0*C)*a^4*cos(d*x + c)^2 + 12*C*a^4*cos(d*x + c) + 2*C*a^4)*sin(d*x + c))/( d*cos(d*x + c)^3)
Timed out. \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**2*(a+a*sec(d*x+c))**4*(A+C*sec(d*x+c)**2),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.10 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} + 72 \, {\left (d x + c\right )} A a^{4} + 4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{4} + 12 \, {\left (d x + c\right )} C a^{4} - 12 \, C a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, A a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, C a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, A a^{4} \sin \left (d x + c\right ) + 12 \, A a^{4} \tan \left (d x + c\right ) + 72 \, C a^{4} \tan \left (d x + c\right )}{12 \, d} \] Input:
integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm= "maxima")
Output:
1/12*(3*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^4 + 72*(d*x + c)*A*a^4 + 4*(t an(d*x + c)^3 + 3*tan(d*x + c))*C*a^4 + 12*(d*x + c)*C*a^4 - 12*C*a^4*(2*s in(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c ) - 1)) + 24*A*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 24*C* a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 48*A*a^4*sin(d*x + c ) + 12*A*a^4*tan(d*x + c) + 72*C*a^4*tan(d*x + c))/d
Time = 0.35 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.29 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (13 \, A a^{4} + 2 \, C a^{4}\right )} {\left (d x + c\right )} + 12 \, {\left (2 \, A a^{4} + 3 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 12 \, {\left (2 \, A a^{4} + 3 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {6 \, {\left (7 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}} - \frac {4 \, {\left (3 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 38 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \] Input:
integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm= "giac")
Output:
1/6*(3*(13*A*a^4 + 2*C*a^4)*(d*x + c) + 12*(2*A*a^4 + 3*C*a^4)*log(abs(tan (1/2*d*x + 1/2*c) + 1)) - 12*(2*A*a^4 + 3*C*a^4)*log(abs(tan(1/2*d*x + 1/2 *c) - 1)) + 6*(7*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 9*A*a^4*tan(1/2*d*x + 1/2* c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^2 - 4*(3*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 1 5*C*a^4*tan(1/2*d*x + 1/2*c)^5 - 6*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 38*C*a^4 *tan(1/2*d*x + 1/2*c)^3 + 3*A*a^4*tan(1/2*d*x + 1/2*c) + 27*C*a^4*tan(1/2* d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d
Time = 11.84 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.31 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {4\,A\,a^4\,\sin \left (c+d\,x\right )}{d}+\frac {13\,A\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {8\,A\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {12\,C\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a^4\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {20\,C\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {2\,C\,a^4\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^2}+\frac {C\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3}+\frac {A\,a^4\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d} \] Input:
int(cos(c + d*x)^2*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^4,x)
Output:
(4*A*a^4*sin(c + d*x))/d + (13*A*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d* x)/2)))/d + (8*A*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2* C*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (12*C*a^4*atanh(sin (c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (A*a^4*sin(c + d*x))/(d*cos(c + d *x)) + (20*C*a^4*sin(c + d*x))/(3*d*cos(c + d*x)) + (2*C*a^4*sin(c + d*x)) /(d*cos(c + d*x)^2) + (C*a^4*sin(c + d*x))/(3*d*cos(c + d*x)^3) + (A*a^4*c os(c + d*x)*sin(c + d*x))/(2*d)
Time = 0.17 (sec) , antiderivative size = 445, normalized size of antiderivative = 2.32 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{4} \left (-24 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a -36 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} c +24 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +36 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) c +24 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a +36 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} c -24 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -36 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) c +24 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a +39 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a c +39 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a d x +6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} c^{2}+6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} c d x -24 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a -12 \cos \left (d x +c \right ) \sin \left (d x +c \right ) c -39 \cos \left (d x +c \right ) a c -39 \cos \left (d x +c \right ) a d x -6 \cos \left (d x +c \right ) c^{2}-6 \cos \left (d x +c \right ) c d x -3 \sin \left (d x +c \right )^{5} a +12 \sin \left (d x +c \right )^{3} a +40 \sin \left (d x +c \right )^{3} c -9 \sin \left (d x +c \right ) a -42 \sin \left (d x +c \right ) c \right )}{6 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(cos(d*x+c)^2*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x)
Output:
(a**4*( - 24*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a - 36 *cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*c + 24*cos(c + d*x )*log(tan((c + d*x)/2) - 1)*a + 36*cos(c + d*x)*log(tan((c + d*x)/2) - 1)* c + 24*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a + 36*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*c - 24*cos(c + d*x)*log( tan((c + d*x)/2) + 1)*a - 36*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*c + 24 *cos(c + d*x)*sin(c + d*x)**3*a + 39*cos(c + d*x)*sin(c + d*x)**2*a*c + 39 *cos(c + d*x)*sin(c + d*x)**2*a*d*x + 6*cos(c + d*x)*sin(c + d*x)**2*c**2 + 6*cos(c + d*x)*sin(c + d*x)**2*c*d*x - 24*cos(c + d*x)*sin(c + d*x)*a - 12*cos(c + d*x)*sin(c + d*x)*c - 39*cos(c + d*x)*a*c - 39*cos(c + d*x)*a*d *x - 6*cos(c + d*x)*c**2 - 6*cos(c + d*x)*c*d*x - 3*sin(c + d*x)**5*a + 12 *sin(c + d*x)**3*a + 40*sin(c + d*x)**3*c - 9*sin(c + d*x)*a - 42*sin(c + d*x)*c))/(6*cos(c + d*x)*d*(sin(c + d*x)**2 - 1))