Integrand size = 31, antiderivative size = 160 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {1}{2} a^4 (13 A+8 B) x+\frac {a^4 (8 A+13 B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {5 a^4 (A-B) \sin (c+d x)}{2 d}+\frac {a A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac {(A-B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac {(A+6 B) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{2 d} \] Output:
1/2*a^4*(13*A+8*B)*x+1/2*a^4*(8*A+13*B)*arctanh(sin(d*x+c))/d+5/2*a^4*(A-B )*sin(d*x+c)/d+1/2*a*A*cos(d*x+c)*(a+a*sec(d*x+c))^3*sin(d*x+c)/d-1/2*(A-B )*(a^2+a^2*sec(d*x+c))^2*sin(d*x+c)/d+1/2*(A+6*B)*(a^4+a^4*sec(d*x+c))*sin (d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(373\) vs. \(2(160)=320\).
Time = 8.57 (sec) , antiderivative size = 373, normalized size of antiderivative = 2.33 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {a^4 \cos ^5(c+d x) \sec ^8\left (\frac {1}{2} (c+d x)\right ) (1+\sec (c+d x))^4 (A+B \sec (c+d x)) \left (2 (13 A+8 B) x-\frac {2 (8 A+13 B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {2 (8 A+13 B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {4 (4 A+B) \cos (d x) \sin (c)}{d}+\frac {A \cos (2 d x) \sin (2 c)}{d}+\frac {4 (4 A+B) \cos (c) \sin (d x)}{d}+\frac {A \cos (2 c) \sin (2 d x)}{d}+\frac {B}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 (A+4 B) \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {B}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 (A+4 B) \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )}{64 (B+A \cos (c+d x))} \] Input:
Integrate[Cos[c + d*x]^2*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x]),x]
Output:
(a^4*Cos[c + d*x]^5*Sec[(c + d*x)/2]^8*(1 + Sec[c + d*x])^4*(A + B*Sec[c + d*x])*(2*(13*A + 8*B)*x - (2*(8*A + 13*B)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (2*(8*A + 13*B)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/d + (4*(4*A + B)*Cos[d*x]*Sin[c])/d + (A*Cos[2*d*x]*Sin[2*c])/d + (4*(4*A + B)*Cos[c]*Sin[d*x])/d + (A*Cos[2*c]*Sin[2*d*x])/d + B/(d*(Cos[(c + d*x)/2 ] - Sin[(c + d*x)/2])^2) + (4*(A + 4*B)*Sin[(d*x)/2])/(d*(Cos[c/2] - Sin[c /2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - B/(d*(Cos[(c + d*x)/2] + Sin [(c + d*x)/2])^2) + (4*(A + 4*B)*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(C os[(c + d*x)/2] + Sin[(c + d*x)/2]))))/(64*(B + A*Cos[c + d*x]))
Time = 0.93 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.95, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {3042, 4505, 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.
| \(\displaystyle \int \cos ^2(c+d x) (a \sec (c+d x)+a)^4 (A+B \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^4 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {1}{2} \int \cos (c+d x) (\sec (c+d x) a+a)^3 (a (5 A+2 B)-2 a (A-B) \sec (c+d x))dx+\frac {a A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (a (5 A+2 B)-2 a (A-B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int 2 \cos (c+d x) (\sec (c+d x) a+a)^2 \left ((6 A+B) a^2+(A+6 B) \sec (c+d x) a^2\right )dx-\frac {(A-B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d}\right )+\frac {a A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (\int \cos (c+d x) (\sec (c+d x) a+a)^2 \left ((6 A+B) a^2+(A+6 B) \sec (c+d x) a^2\right )dx-\frac {(A-B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d}\right )+\frac {a A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left ((6 A+B) a^2+(A+6 B) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {(A-B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d}\right )+\frac {a A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {1}{2} \left (\int \cos (c+d x) (\sec (c+d x) a+a) \left (5 (A-B) a^3+(8 A+13 B) \sec (c+d x) a^3\right )dx+\frac {(A+6 B) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{d}-\frac {(A-B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d}\right )+\frac {a A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (5 (A-B) a^3+(8 A+13 B) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {(A+6 B) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{d}-\frac {(A-B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d}\right )+\frac {a A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 4484 |
| \(\displaystyle \frac {1}{2} \left (-\int \left (-\left ((13 A+8 B) a^4\right )-(8 A+13 B) \sec (c+d x) a^4\right )dx+\frac {5 a^4 (A-B) \sin (c+d x)}{d}+\frac {(A+6 B) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{d}-\frac {(A-B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d}\right )+\frac {a A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \left (\frac {a^4 (8 A+13 B) \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^4 (A-B) \sin (c+d x)}{d}+\frac {(A+6 B) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{d}+a^4 x (13 A+8 B)-\frac {(A-B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d}\right )+\frac {a A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d}\) |
Input:
Int[Cos[c + d*x]^2*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x]),x]
Output:
(a*A*Cos[c + d*x]*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(2*d) + (a^4*(13*A + 8*B)*x + (a^4*(8*A + 13*B)*ArcTanh[Sin[c + d*x]])/d + (5*a^4*(A - B)*Sin [c + d*x])/d - ((A - B)*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/d + ((A + 6*B)*(a^4 + a^4*Sec[c + d*x])*Sin[c + d*x])/d)/2
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]
Time = 0.83 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.04
| method | result | size |
| parallelrisch | \(\frac {2 a^{4} \left (-2 \left (A +\frac {13 B}{8}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 \left (A +\frac {13 B}{8}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {13 \left (A +\frac {8 B}{13}\right ) d x \cos \left (2 d x +2 c \right )}{4}+\left (\frac {5 A}{8}+2 B \right ) \sin \left (2 d x +2 c \right )+\left (A +\frac {B}{4}\right ) \sin \left (3 d x +3 c \right )+\frac {A \sin \left (4 d x +4 c \right )}{16}+\left (A +\frac {3 B}{4}\right ) \sin \left (d x +c \right )+\frac {13 \left (A +\frac {8 B}{13}\right ) d x}{4}\right )}{d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(166\) |
| derivativedivides | \(\frac {a^{4} A \tan \left (d x +c \right )+B \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 B \,a^{4} \tan \left (d x +c \right )+6 a^{4} A \left (d x +c \right )+6 B \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{4} A \sin \left (d x +c \right )+4 B \,a^{4} \left (d x +c \right )+a^{4} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{4} \sin \left (d x +c \right )}{d}\) | \(177\) |
| default | \(\frac {a^{4} A \tan \left (d x +c \right )+B \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 B \,a^{4} \tan \left (d x +c \right )+6 a^{4} A \left (d x +c \right )+6 B \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{4} A \sin \left (d x +c \right )+4 B \,a^{4} \left (d x +c \right )+a^{4} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{4} \sin \left (d x +c \right )}{d}\) | \(177\) |
| risch | \(\frac {13 a^{4} A x}{2}+4 a^{4} x B -\frac {i a^{4} A \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {2 i a^{4} A \,{\mathrm e}^{i \left (d x +c \right )}}{d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B \,a^{4}}{2 d}+\frac {2 i a^{4} A \,{\mathrm e}^{-i \left (d x +c \right )}}{d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{4}}{2 d}+\frac {i a^{4} A \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {i a^{4} \left (B \,{\mathrm e}^{3 i \left (d x +c \right )}-2 A \,{\mathrm e}^{2 i \left (d x +c \right )}-8 B \,{\mathrm e}^{2 i \left (d x +c \right )}-B \,{\mathrm e}^{i \left (d x +c \right )}-2 A -8 B \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {4 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {13 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}+\frac {4 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {13 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}\) | \(294\) |
| norman | \(\frac {\left (\frac {13}{2} a^{4} A +4 B \,a^{4}\right ) x +\left (-\frac {13}{2} a^{4} A -4 B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-\frac {13}{2} a^{4} A -4 B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (\frac {13}{2} a^{4} A +4 B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (-13 a^{4} A -8 B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-13 a^{4} A -8 B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (26 a^{4} A +16 B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {5 a^{4} \left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}+\frac {11 a^{4} \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a^{4} \left (5 A +9 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {2 a^{4} \left (11 A -7 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {a^{4} \left (17 A -3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}-\frac {a^{4} \left (31 A +13 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {a^{4} \left (8 A +13 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{4} \left (8 A +13 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(407\) |
Input:
int(cos(d*x+c)^2*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x,method=_RETURNVERBO SE)
Output:
2*a^4*(-2*(A+13/8*B)*(1+cos(2*d*x+2*c))*ln(tan(1/2*d*x+1/2*c)-1)+2*(A+13/8 *B)*(1+cos(2*d*x+2*c))*ln(tan(1/2*d*x+1/2*c)+1)+13/4*(A+8/13*B)*d*x*cos(2* d*x+2*c)+(5/8*A+2*B)*sin(2*d*x+2*c)+(A+1/4*B)*sin(3*d*x+3*c)+1/16*A*sin(4* d*x+4*c)+(A+3/4*B)*sin(d*x+c)+13/4*(A+8/13*B)*d*x)/d/(1+cos(2*d*x+2*c))
Time = 0.11 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.98 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {2 \, {\left (13 \, A + 8 \, B\right )} a^{4} d x \cos \left (d x + c\right )^{2} + {\left (8 \, A + 13 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (8 \, A + 13 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (A a^{4} \cos \left (d x + c\right )^{3} + 2 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right )^{2} + 2 \, {\left (A + 4 \, B\right )} a^{4} \cos \left (d x + c\right ) + B a^{4}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="f ricas")
Output:
1/4*(2*(13*A + 8*B)*a^4*d*x*cos(d*x + c)^2 + (8*A + 13*B)*a^4*cos(d*x + c) ^2*log(sin(d*x + c) + 1) - (8*A + 13*B)*a^4*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(A*a^4*cos(d*x + c)^3 + 2*(4*A + B)*a^4*cos(d*x + c)^2 + 2*(A + 4*B)*a^4*cos(d*x + c) + B*a^4)*sin(d*x + c))/(d*cos(d*x + c)^2)
Timed out. \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**2*(a+a*sec(d*x+c))**4*(A+B*sec(d*x+c)),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.24 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} + 24 \, {\left (d x + c\right )} A a^{4} + 16 \, {\left (d x + c\right )} B a^{4} - B a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 8 \, A a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, B a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 16 \, A a^{4} \sin \left (d x + c\right ) + 4 \, B a^{4} \sin \left (d x + c\right ) + 4 \, A a^{4} \tan \left (d x + c\right ) + 16 \, B a^{4} \tan \left (d x + c\right )}{4 \, d} \] Input:
integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="m axima")
Output:
1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^4 + 24*(d*x + c)*A*a^4 + 16*(d*x + c)*B*a^4 - B*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c ) + 1) + log(sin(d*x + c) - 1)) + 8*A*a^4*(log(sin(d*x + c) + 1) - log(sin (d*x + c) - 1)) + 12*B*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 16*A*a^4*sin(d*x + c) + 4*B*a^4*sin(d*x + c) + 4*A*a^4*tan(d*x + c) + 1 6*B*a^4*tan(d*x + c))/d
Time = 0.18 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.44 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {{\left (13 \, A a^{4} + 8 \, B a^{4}\right )} {\left (d x + c\right )} + {\left (8 \, A a^{4} + 13 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (8 \, A a^{4} + 13 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (5 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 5 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 7 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 7 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 11 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 11 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1\right )}^{2}}}{2 \, d} \] Input:
integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="g iac")
Output:
1/2*((13*A*a^4 + 8*B*a^4)*(d*x + c) + (8*A*a^4 + 13*B*a^4)*log(abs(tan(1/2 *d*x + 1/2*c) + 1)) - (8*A*a^4 + 13*B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(5*A*a^4*tan(1/2*d*x + 1/2*c)^7 - 5*B*a^4*tan(1/2*d*x + 1/2*c)^7 - 7*A*a^4*tan(1/2*d*x + 1/2*c)^5 - 7*B*a^4*tan(1/2*d*x + 1/2*c)^5 - 9*A*a^4 *tan(1/2*d*x + 1/2*c)^3 + 9*B*a^4*tan(1/2*d*x + 1/2*c)^3 + 11*A*a^4*tan(1/ 2*d*x + 1/2*c) + 11*B*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^4 - 1)^2)/d
Time = 11.25 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.52 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {4\,A\,a^4\,\sin \left (c+d\,x\right )}{d}+\frac {B\,a^4\,\sin \left (c+d\,x\right )}{d}+\frac {13\,A\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {8\,A\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {8\,B\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {13\,B\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a^4\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {4\,B\,a^4\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {B\,a^4\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {A\,a^4\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d} \] Input:
int(cos(c + d*x)^2*(A + B/cos(c + d*x))*(a + a/cos(c + d*x))^4,x)
Output:
(4*A*a^4*sin(c + d*x))/d + (B*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 + (8*B*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) )/d + (13*B*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (A*a^4*s in(c + d*x))/(d*cos(c + d*x)) + (4*B*a^4*sin(c + d*x))/(d*cos(c + d*x)) + (B*a^4*sin(c + d*x))/(2*d*cos(c + d*x)^2) + (A*a^4*cos(c + d*x)*sin(c + d* x))/(2*d)
Time = 0.18 (sec) , antiderivative size = 327, normalized size of antiderivative = 2.04 \[ \int \cos ^2(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {a^{4} \left (\cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a -3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a -8 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b -8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a -13 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b +8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +13 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a +13 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b -8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -13 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b +8 \sin \left (d x +c \right )^{3} a +2 \sin \left (d x +c \right )^{3} b +13 \sin \left (d x +c \right )^{2} a c +13 \sin \left (d x +c \right )^{2} a d x +8 \sin \left (d x +c \right )^{2} b c +8 \sin \left (d x +c \right )^{2} b d x -8 \sin \left (d x +c \right ) a -3 \sin \left (d x +c \right ) b -13 a c -13 a d x -8 b c -8 b d x \right )}{2 d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(cos(d*x+c)^2*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x)
Output:
(a**4*(cos(c + d*x)*sin(c + d*x)**3*a - 3*cos(c + d*x)*sin(c + d*x)*a - 8* cos(c + d*x)*sin(c + d*x)*b - 8*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2* a - 13*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b + 8*log(tan((c + d*x)/2 ) - 1)*a + 13*log(tan((c + d*x)/2) - 1)*b + 8*log(tan((c + d*x)/2) + 1)*si n(c + d*x)**2*a + 13*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b - 8*log(t an((c + d*x)/2) + 1)*a - 13*log(tan((c + d*x)/2) + 1)*b + 8*sin(c + d*x)** 3*a + 2*sin(c + d*x)**3*b + 13*sin(c + d*x)**2*a*c + 13*sin(c + d*x)**2*a* d*x + 8*sin(c + d*x)**2*b*c + 8*sin(c + d*x)**2*b*d*x - 8*sin(c + d*x)*a - 3*sin(c + d*x)*b - 13*a*c - 13*a*d*x - 8*b*c - 8*b*d*x))/(2*d*(sin(c + d* x)**2 - 1))