Integrand size = 29, antiderivative size = 151 \[ \int \cos (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=a^4 (4 A+B) x+\frac {a^4 (13 A+12 B) \text {arctanh}(\sin (c+d x))}{2 d}-\frac {5 a^4 (A+2 B) \sin (c+d x)}{2 d}+\frac {a B (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {(A+2 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac {(9 A+11 B) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{3 d} \] Output:
a^4*(4*A+B)*x+1/2*a^4*(13*A+12*B)*arctanh(sin(d*x+c))/d-5/2*a^4*(A+2*B)*si n(d*x+c)/d+1/3*a*B*(a+a*sec(d*x+c))^3*sin(d*x+c)/d+1/2*(A+2*B)*(a^2+a^2*se c(d*x+c))^2*sin(d*x+c)/d+1/3*(9*A+11*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. \(1202\) vs. \(2(151)=302\).
Time = 11.98 (sec) , antiderivative size = 1202, normalized size of antiderivative = 7.96 \[ \int \cos (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx =\text {Too large to display} \] Input:
Integrate[Cos[c + d*x]*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x]),x]
Output:
((4*A + B)*x*Cos[c + d*x]^5*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x]))/(16*(B + A*Cos[c + d*x])) + ((-13*A - 12*B)*Cos[c + d* x]^5*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 + B*Sec[c + d*x]))/(32*d*(B + A*Cos[c + d*x])) + (( 13*A + 12*B)*Cos[c + d*x]^5*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*S ec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x]))/(32*d*(B + A*Cos[c + d*x])) + (A*Cos[d*x]*Cos[c + d*x]^5*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x])*Sin[c])/(16*d*(B + A*Cos[c + d*x])) + (A*Cos[c]*Cos[c + d*x]^5*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x])*Sin[d*x])/(16*d*(B + A*Cos[c + d*x])) + (B*Cos[c + d*x] ^5*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x])*Sin[(d *x)/2])/(96*d*(B + A*Cos[c + 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]^5*Sec[c/2 + (d*x)/2]^8*(a + a* Sec[c + d*x])^4*(A + B*Sec[c + d*x])*(3*A*Cos[c/2] + 13*B*Cos[c/2] - 3*A*S in[c/2] - 11*B*Sin[c/2]))/(192*d*(B + A*Cos[c + d*x])*(Cos[c/2] - Sin[c/2] )*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^2) + (Cos[c + d*x]^5*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x])*(3*A*Sin[(d*x)/2] + 5*B*Sin[(d*x)/2]))/(12*d*(B + A*Cos[c + d*x])*(Cos[c/2] - Sin[c/2])*(Cos [c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) + (B*Cos[c + d*x]^5*Sec[c/2 + (d*x) /2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x])*Sin[(d*x)/2])/(96*d*(...
Time = 0.91 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {3042, 4506, 3042, 4506, 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.
| \(\displaystyle \int \cos (c+d x) (a \sec (c+d x)+a)^4 (A+B \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^4 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {1}{3} \int \cos (c+d x) (\sec (c+d x) a+a)^3 (a (3 A-B)+3 a (A+2 B) \sec (c+d x))dx+\frac {a B \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (a (3 A-B)+3 a (A+2 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a B \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \cos (c+d x) (\sec (c+d x) a+a)^2 \left ((3 A-8 B) a^2+2 (9 A+11 B) \sec (c+d x) a^2\right )dx+\frac {3 (A+2 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}\right )+\frac {a B \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left ((3 A-8 B) a^2+2 (9 A+11 B) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {3 (A+2 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}\right )+\frac {a B \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\int -3 \cos (c+d x) (\sec (c+d x) a+a) \left (5 a^3 (A+2 B)-a^3 (13 A+12 B) \sec (c+d x)\right )dx+\frac {2 (9 A+11 B) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{d}\right )+\frac {3 (A+2 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}\right )+\frac {a B \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\frac {2 (9 A+11 B) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{d}-3 \int \cos (c+d x) (\sec (c+d x) a+a) \left (5 a^3 (A+2 B)-a^3 (13 A+12 B) \sec (c+d x)\right )dx\right )+\frac {3 (A+2 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}\right )+\frac {a B \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\frac {2 (9 A+11 B) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{d}-3 \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (5 a^3 (A+2 B)-a^3 (13 A+12 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx\right )+\frac {3 (A+2 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}\right )+\frac {a B \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d}\) |
| \(\Big \downarrow \) 4484 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\frac {2 (9 A+11 B) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{d}-3 \left (\frac {5 a^4 (A+2 B) \sin (c+d x)}{d}-\int \left (2 (4 A+B) a^4+(13 A+12 B) \sec (c+d x) a^4\right )dx\right )\right )+\frac {3 (A+2 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}\right )+\frac {a B \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\frac {2 (9 A+11 B) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{d}-3 \left (-\frac {a^4 (13 A+12 B) \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^4 (A+2 B) \sin (c+d x)}{d}-2 a^4 x (4 A+B)\right )\right )+\frac {3 (A+2 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}\right )+\frac {a B \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d}\) |
Input:
Int[Cos[c + d*x]*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x]),x]
Output:
(a*B*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(3*d) + ((3*(A + 2*B)*(a^2 + a^2 *Sec[c + d*x])^2*Sin[c + d*x])/(2*d) + ((2*(9*A + 11*B)*(a^4 + a^4*Sec[c + d*x])*Sin[c + d*x])/d - 3*(-2*a^4*(4*A + B)*x - (a^4*(13*A + 12*B)*ArcTan h[Sin[c + d*x]])/d + (5*a^4*(A + 2*B)*Sin[c + d*x])/d))/2)/3
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]
Time = 0.94 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.28
| method | result | size |
| parallelrisch | \(\frac {2 a^{4} \left (-\frac {39 \left (A +\frac {12 B}{13}\right ) \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4}+\frac {39 \left (A +\frac {12 B}{13}\right ) \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4}+2 d \left (A +\frac {B}{4}\right ) x \cos \left (3 d x +3 c \right )+\left (A +2 B \right ) \sin \left (2 d x +2 c \right )+2 \left (A +\frac {5 B}{3}\right ) \sin \left (3 d x +3 c \right )+\frac {A \sin \left (4 d x +4 c \right )}{4}+6 d \left (A +\frac {B}{4}\right ) x \cos \left (d x +c \right )+2 \sin \left (d x +c \right ) \left (A +2 B \right )\right )}{d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\) | \(193\) |
| derivativedivides | \(\frac {a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B \,a^{4} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+4 a^{4} A \tan \left (d x +c \right )+4 B \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+6 a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 B \,a^{4} \tan \left (d x +c \right )+4 a^{4} A \left (d x +c \right )+4 B \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{4} A \sin \left (d x +c \right )+B \,a^{4} \left (d x +c \right )}{d}\) | \(199\) |
| default | \(\frac {a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B \,a^{4} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+4 a^{4} A \tan \left (d x +c \right )+4 B \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+6 a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 B \,a^{4} \tan \left (d x +c \right )+4 a^{4} A \left (d x +c \right )+4 B \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{4} A \sin \left (d x +c \right )+B \,a^{4} \left (d x +c \right )}{d}\) | \(199\) |
| risch | \(4 a^{4} A x +a^{4} x B -\frac {i a^{4} A \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i a^{4} A \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {i a^{4} \left (3 A \,{\mathrm e}^{5 i \left (d x +c \right )}+12 B \,{\mathrm e}^{5 i \left (d x +c \right )}-24 A \,{\mathrm e}^{4 i \left (d x +c \right )}-36 B \,{\mathrm e}^{4 i \left (d x +c \right )}-48 A \,{\mathrm e}^{2 i \left (d x +c \right )}-84 B \,{\mathrm e}^{2 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )} A -12 B \,{\mathrm e}^{i \left (d x +c \right )}-24 A -40 B \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {13 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 d}+\frac {6 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}-\frac {13 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 d}-\frac {6 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}\) | \(266\) |
| norman | \(\frac {\left (4 a^{4} A +B \,a^{4}\right ) x +\left (-12 a^{4} A -3 B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-12 a^{4} A -3 B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (4 a^{4} A +B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (8 a^{4} A +2 B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (8 a^{4} A +2 B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {a^{4} \left (11 A +18 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {5 a^{4} \left (A +2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}+\frac {2 a^{4} \left (5 A -4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {4 a^{4} \left (6 A +19 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}-\frac {4 a^{4} \left (18 A +19 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {a^{4} \left (13 A +12 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{4} \left (13 A +12 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(356\) |
Input:
int(cos(d*x+c)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x,method=_RETURNVERBOSE )
Output:
2*a^4*(-39/4*(A+12/13*B)*(1/3*cos(3*d*x+3*c)+cos(d*x+c))*ln(tan(1/2*d*x+1/ 2*c)-1)+39/4*(A+12/13*B)*(1/3*cos(3*d*x+3*c)+cos(d*x+c))*ln(tan(1/2*d*x+1/ 2*c)+1)+2*d*(A+1/4*B)*x*cos(3*d*x+3*c)+(A+2*B)*sin(2*d*x+2*c)+2*(A+5/3*B)* sin(3*d*x+3*c)+1/4*A*sin(4*d*x+4*c)+6*d*(A+1/4*B)*x*cos(d*x+c)+2*sin(d*x+c )*(A+2*B))/d/(cos(3*d*x+3*c)+3*cos(d*x+c))
Time = 0.11 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.05 \[ \int \cos (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {12 \, {\left (4 \, A + B\right )} a^{4} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (13 \, A + 12 \, B\right )} a^{4} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (13 \, A + 12 \, B\right )} a^{4} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (6 \, A a^{4} \cos \left (d x + c\right )^{3} + 8 \, {\left (3 \, A + 5 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} + 3 \, {\left (A + 4 \, B\right )} a^{4} \cos \left (d x + c\right ) + 2 \, B a^{4}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \] Input:
integrate(cos(d*x+c)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="fri cas")
Output:
1/12*(12*(4*A + B)*a^4*d*x*cos(d*x + c)^3 + 3*(13*A + 12*B)*a^4*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(13*A + 12*B)*a^4*cos(d*x + c)^3*log(-sin(d *x + c) + 1) + 2*(6*A*a^4*cos(d*x + c)^3 + 8*(3*A + 5*B)*a^4*cos(d*x + c)^ 2 + 3*(A + 4*B)*a^4*cos(d*x + c) + 2*B*a^4)*sin(d*x + c))/(d*cos(d*x + c)^ 3)
\[ \int \cos (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=a^{4} \left (\int A \cos {\left (c + d x \right )}\, dx + \int 4 A \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 6 A \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 4 A \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int A \cos {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int B \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 4 B \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 6 B \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int 4 B \cos {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int B \cos {\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(cos(d*x+c)*(a+a*sec(d*x+c))**4*(A+B*sec(d*x+c)),x)
Output:
a**4*(Integral(A*cos(c + d*x), x) + Integral(4*A*cos(c + d*x)*sec(c + d*x) , x) + Integral(6*A*cos(c + d*x)*sec(c + d*x)**2, x) + Integral(4*A*cos(c + d*x)*sec(c + d*x)**3, x) + Integral(A*cos(c + d*x)*sec(c + d*x)**4, x) + Integral(B*cos(c + d*x)*sec(c + d*x), x) + Integral(4*B*cos(c + d*x)*sec( c + d*x)**2, x) + Integral(6*B*cos(c + d*x)*sec(c + d*x)**3, x) + Integral (4*B*cos(c + d*x)*sec(c + d*x)**4, x) + Integral(B*cos(c + d*x)*sec(c + d* x)**5, x))
Time = 0.04 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.56 \[ \int \cos (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {48 \, {\left (d x + c\right )} A a^{4} + 4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{4} + 12 \, {\left (d x + c\right )} B a^{4} - 3 \, A a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, B a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, A a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, B a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a^{4} \sin \left (d x + c\right ) + 48 \, A a^{4} \tan \left (d x + c\right ) + 72 \, B a^{4} \tan \left (d x + c\right )}{12 \, d} \] Input:
integrate(cos(d*x+c)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="max ima")
Output:
1/12*(48*(d*x + c)*A*a^4 + 4*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^4 + 12* (d*x + c)*B*a^4 - 3*A*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)) - 12*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)) + 36*A*a^4*(l og(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 24*B*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*A*a^4*sin(d*x + c) + 48*A*a^4*tan(d*x + c) + 72*B*a^4*tan(d*x + c))/d
Time = 0.24 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.50 \[ \int \cos (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {\frac {12 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + 6 \, {\left (4 \, A a^{4} + B a^{4}\right )} {\left (d x + c\right )} + 3 \, {\left (13 \, A a^{4} + 12 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (13 \, A a^{4} + 12 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (21 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 30 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 48 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 76 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 27 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 54 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \] Input:
integrate(cos(d*x+c)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="gia c")
Output:
1/6*(12*A*a^4*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1) + 6*(4*A*a ^4 + B*a^4)*(d*x + c) + 3*(13*A*a^4 + 12*B*a^4)*log(abs(tan(1/2*d*x + 1/2* c) + 1)) - 3*(13*A*a^4 + 12*B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2* (21*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 30*B*a^4*tan(1/2*d*x + 1/2*c)^5 - 48*A* a^4*tan(1/2*d*x + 1/2*c)^3 - 76*B*a^4*tan(1/2*d*x + 1/2*c)^3 + 27*A*a^4*ta n(1/2*d*x + 1/2*c) + 54*B*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^ 2 - 1)^3)/d
Time = 10.98 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.68 \[ \int \cos (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {A\,a^4\,\sin \left (c+d\,x\right )}{d}+\frac {8\,A\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {13\,A\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,B\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {12\,B\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,A\,a^4\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {A\,a^4\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {20\,B\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {2\,B\,a^4\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^2}+\frac {B\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3} \] Input:
int(cos(c + d*x)*(A + B/cos(c + d*x))*(a + a/cos(c + d*x))^4,x)
Output:
(A*a^4*sin(c + d*x))/d + (8*A*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/ 2)))/d + (13*A*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*B* a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (12*B*a^4*atanh(sin(c /2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (4*A*a^4*sin(c + d*x))/(d*cos(c + d *x)) + (A*a^4*sin(c + d*x))/(2*d*cos(c + d*x)^2) + (20*B*a^4*sin(c + d*x)) /(3*d*cos(c + d*x)) + (2*B*a^4*sin(c + d*x))/(d*cos(c + d*x)^2) + (B*a^4*s in(c + d*x))/(3*d*cos(c + d*x)^3)
Time = 0.18 (sec) , antiderivative size = 432, normalized size of antiderivative = 2.86 \[ \int \cos (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {a^{4} \left (-39 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a -36 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b +39 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +36 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +39 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a +36 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b -39 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -36 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b +6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a +24 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a c +24 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a d x +6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b c +6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b d x -9 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a -12 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b -24 \cos \left (d x +c \right ) a c -24 \cos \left (d x +c \right ) a d x -6 \cos \left (d x +c \right ) b c -6 \cos \left (d x +c \right ) b d x +24 \sin \left (d x +c \right )^{3} a +40 \sin \left (d x +c \right )^{3} b -24 \sin \left (d x +c \right ) a -42 \sin \left (d x +c \right ) b \right )}{6 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(cos(d*x+c)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x)
Output:
(a**4*( - 39*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*b + 39*cos(c + d*x )*log(tan((c + d*x)/2) - 1)*a + 36*cos(c + d*x)*log(tan((c + d*x)/2) - 1)* b + 39*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*b - 39*cos(c + d*x)*log( tan((c + d*x)/2) + 1)*a - 36*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*b + 6* cos(c + d*x)*sin(c + d*x)**3*a + 24*cos(c + d*x)*sin(c + d*x)**2*a*c + 24* cos(c + d*x)*sin(c + d*x)**2*a*d*x + 6*cos(c + d*x)*sin(c + d*x)**2*b*c + 6*cos(c + d*x)*sin(c + d*x)**2*b*d*x - 9*cos(c + d*x)*sin(c + d*x)*a - 12* cos(c + d*x)*sin(c + d*x)*b - 24*cos(c + d*x)*a*c - 24*cos(c + d*x)*a*d*x - 6*cos(c + d*x)*b*c - 6*cos(c + d*x)*b*d*x + 24*sin(c + d*x)**3*a + 40*si n(c + d*x)**3*b - 24*sin(c + d*x)*a - 42*sin(c + d*x)*b))/(6*cos(c + d*x)* d*(sin(c + d*x)**2 - 1))