Integrand size = 41, antiderivative size = 217 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{2} a^4 (12 A+13 B+8 C) x+\frac {a^4 (2 A+8 B+13 C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {5 a^4 (2 A+B-C) \sin (c+d x)}{2 d}+\frac {a (4 A+3 B) \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac {(2 A+B-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}-\frac {(8 A-3 B-18 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d} \] Output:
1/2*a^4*(12*A+13*B+8*C)*x+1/2*a^4*(2*A+8*B+13*C)*arctanh(sin(d*x+c))/d+5/2 *a^4*(2*A+B-C)*sin(d*x+c)/d+1/6*a*(4*A+3*B)*cos(d*x+c)*(a+a*sec(d*x+c))^3* sin(d*x+c)/d+1/3*A*cos(d*x+c)^2*(a+a*sec(d*x+c))^4*sin(d*x+c)/d-1/2*(2*A+B -C)*(a^2+a^2*sec(d*x+c))^2*sin(d*x+c)/d-1/6*(8*A-3*B-18*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. \(1518\) vs. \(2(217)=434\).
Time = 12.18 (sec) , antiderivative size = 1518, normalized size of antiderivative = 7.00 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:
Integrate[Cos[c + d*x]^3*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Se c[c + d*x]^2),x]
Output:
((12*A + 13*B + 8*C)*x*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(16*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + ((-2*A - 8*B - 13*C)*Cos[c + d*x]^6*Log[Co s[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] + C*Sec[c + d*x]^2))/(16*d*(A + 2*C + 2*B*Cos [c + d*x] + A*Cos[2*c + 2*d*x])) + ((2*A + 8*B + 13*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 + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(16*d*(A + 2*C + 2*B*C os[c + d*x] + A*Cos[2*c + 2*d*x])) + ((27*A + 16*B + 4*C)*Cos[d*x]*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sin[c])/(32*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + ((4*A + B)*Cos[2*d*x]*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sin[2*c])/(32*d* (A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + (A*Cos[3*d*x]*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sin[3*c])/(96*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + ((27*A + 16*B + 4*C)*Cos[c]*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2] ^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sin[d*x] )/(32*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])) + ((4*A + B)*Co s[2*c]*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + ...
Time = 1.37 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.02, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.317, Rules used = {3042, 4574, 3042, 4505, 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.
| \(\displaystyle \int \cos ^3(c+d x) (a \sec (c+d x)+a)^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^4 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
| \(\Big \downarrow \) 4574 |
| \(\displaystyle \frac {\int \cos ^2(c+d x) (\sec (c+d x) a+a)^4 (a (4 A+3 B)-a (2 A-3 C) \sec (c+d x))dx}{3 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^4}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^4 \left (a (4 A+3 B)-a (2 A-3 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx}{3 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^4}{3 d}\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {\frac {1}{2} \int \cos (c+d x) (\sec (c+d x) a+a)^3 \left (a^2 (16 A+15 B+6 C)-6 a^2 (2 A+B-C) \sec (c+d x)\right )dx+\frac {a^2 (4 A+3 B) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d}}{3 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^4}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (a^2 (16 A+15 B+6 C)-6 a^2 (2 A+B-C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a^2 (4 A+3 B) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d}}{3 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^4}{3 d}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{2} \int 2 \cos (c+d x) (\sec (c+d x) a+a)^2 \left (a^3 (22 A+18 B+3 C)-a^3 (8 A-3 B-18 C) \sec (c+d x)\right )dx-\frac {3 a^3 (2 A+B-C) \sin (c+d x) (a \sec (c+d x)+a)^2}{d}\right )+\frac {a^2 (4 A+3 B) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d}}{3 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^4}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{2} \left (\int \cos (c+d x) (\sec (c+d x) a+a)^2 \left (a^3 (22 A+18 B+3 C)-a^3 (8 A-3 B-18 C) \sec (c+d x)\right )dx-\frac {3 a^3 (2 A+B-C) \sin (c+d x) (a \sec (c+d x)+a)^2}{d}\right )+\frac {a^2 (4 A+3 B) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d}}{3 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^4}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \left (\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (a^3 (22 A+18 B+3 C)-a^3 (8 A-3 B-18 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {3 a^3 (2 A+B-C) \sin (c+d x) (a \sec (c+d x)+a)^2}{d}\right )+\frac {a^2 (4 A+3 B) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d}}{3 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^4}{3 d}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {\frac {1}{2} \left (\int 3 \cos (c+d x) (\sec (c+d x) a+a) \left (5 (2 A+B-C) a^4+(2 A+8 B+13 C) \sec (c+d x) a^4\right )dx-\frac {(8 A-3 B-18 C) \sin (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{d}-\frac {3 a^3 (2 A+B-C) \sin (c+d x) (a \sec (c+d x)+a)^2}{d}\right )+\frac {a^2 (4 A+3 B) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d}}{3 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^4}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{2} \left (3 \int \cos (c+d x) (\sec (c+d x) a+a) \left (5 (2 A+B-C) a^4+(2 A+8 B+13 C) \sec (c+d x) a^4\right )dx-\frac {(8 A-3 B-18 C) \sin (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{d}-\frac {3 a^3 (2 A+B-C) \sin (c+d x) (a \sec (c+d x)+a)^2}{d}\right )+\frac {a^2 (4 A+3 B) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d}}{3 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^4}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \left (3 \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (5 (2 A+B-C) a^4+(2 A+8 B+13 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^4\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {(8 A-3 B-18 C) \sin (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{d}-\frac {3 a^3 (2 A+B-C) \sin (c+d x) (a \sec (c+d x)+a)^2}{d}\right )+\frac {a^2 (4 A+3 B) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d}}{3 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^4}{3 d}\) |
| \(\Big \downarrow \) 4484 |
| \(\displaystyle \frac {\frac {1}{2} \left (3 \left (\frac {5 a^5 (2 A+B-C) \sin (c+d x)}{d}-\int \left (-\left ((12 A+13 B+8 C) a^5\right )-(2 A+8 B+13 C) \sec (c+d x) a^5\right )dx\right )-\frac {(8 A-3 B-18 C) \sin (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{d}-\frac {3 a^3 (2 A+B-C) \sin (c+d x) (a \sec (c+d x)+a)^2}{d}\right )+\frac {a^2 (4 A+3 B) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d}}{3 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^4}{3 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {a^2 (4 A+3 B) \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d}+\frac {1}{2} \left (3 \left (\frac {a^5 (2 A+8 B+13 C) \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^5 (2 A+B-C) \sin (c+d x)}{d}+a^5 x (12 A+13 B+8 C)\right )-\frac {(8 A-3 B-18 C) \sin (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{d}-\frac {3 a^3 (2 A+B-C) \sin (c+d x) (a \sec (c+d x)+a)^2}{d}\right )}{3 a}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^4}{3 d}\) |
Input:
Int[Cos[c + d*x]^3*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
Output:
(A*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^4*Sin[c + d*x])/(3*d) + ((a^2*(4*A + 3*B)*Cos[c + d*x]*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(2*d) + ((-3*a^3* (2*A + B - C)*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/d - ((8*A - 3*B - 18*C) *(a^5 + a^5*Sec[c + d*x])*Sin[c + d*x])/d + 3*(a^5*(12*A + 13*B + 8*C)*x + (a^5*(2*A + 8*B + 13*C)*ArcTanh[Sin[c + d*x]])/d + (5*a^5*(2*A + B - C)*S in[c + d*x])/d))/2)/(3*a)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n)), x] + Simp[1/(d*n) Int[(d*Csc[e + f*x])^( n + 1)*Simp[n*(B*a + A*b) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && LeQ[n, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[a*A*Cot [e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x] - Sim p[b/(a*d*n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Sim p[a*A*(m - n - 1) - b*B*n - (a*B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0 ] && GtQ[m, 1/2] && LtQ[n, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B* Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x] )^n*Simp[a*A*d*(m + n) + B*(b*d*n) + (A*b*d*(m + n) + a*B*d*(2*m + n - 1))* Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(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 = 1.61 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.94
| method | result | size |
| parallelrisch | \(\frac {\left (-\left (A +4 B +\frac {13 C}{2}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (A +4 B +\frac {13 C}{2}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+6 x d \left (A +\frac {13 B}{12}+\frac {2 C}{3}\right ) \cos \left (2 d x +2 c \right )+\left (4 C +A +\frac {5 B}{4}\right ) \sin \left (2 d x +2 c \right )+\left (2 B +\frac {83 A}{24}+\frac {C}{2}\right ) \sin \left (3 d x +3 c \right )+\left (\frac {B}{8}+\frac {A}{2}\right ) \sin \left (4 d x +4 c \right )+\frac {A \sin \left (5 d x +5 c \right )}{24}+\left (2 B +\frac {41 A}{12}+\frac {3 C}{2}\right ) \sin \left (d x +c \right )+6 x d \left (A +\frac {13 B}{12}+\frac {2 C}{3}\right )\right ) a^{4}}{d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(204\) |
| derivativedivides | \(\frac {a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{4} \tan \left (d x +c \right )+a^{4} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 a^{4} A \left (d x +c \right )+4 B \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{4} C \tan \left (d x +c \right )+6 a^{4} A \sin \left (d x +c \right )+6 B \,a^{4} \left (d x +c \right )+6 a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{4} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 B \,a^{4} \sin \left (d x +c \right )+4 a^{4} C \left (d x +c \right )+\frac {a^{4} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+B \,a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{4} C \sin \left (d x +c \right )}{d}\) | \(269\) |
| default | \(\frac {a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{4} \tan \left (d x +c \right )+a^{4} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 a^{4} A \left (d x +c \right )+4 B \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{4} C \tan \left (d x +c \right )+6 a^{4} A \sin \left (d x +c \right )+6 B \,a^{4} \left (d x +c \right )+6 a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{4} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 B \,a^{4} \sin \left (d x +c \right )+4 a^{4} C \left (d x +c \right )+\frac {a^{4} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+B \,a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{4} C \sin \left (d x +c \right )}{d}\) | \(269\) |
| risch | \(6 a^{4} A x +\frac {13 a^{4} x B}{2}+4 a^{4} x C -\frac {i a^{4} \left (C \,{\mathrm e}^{3 i \left (d x +c \right )}-2 B \,{\mathrm e}^{2 i \left (d x +c \right )}-8 C \,{\mathrm e}^{2 i \left (d x +c \right )}-C \,{\mathrm e}^{i \left (d x +c \right )}-2 B -8 C \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {i a^{4} A \,{\mathrm e}^{-2 i \left (d x +c \right )}}{2 d}+\frac {i a^{4} A \,{\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{4} C}{2 d}-\frac {i a^{4} A \,{\mathrm e}^{3 i \left (d x +c \right )}}{24 d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} B \,a^{4}}{8 d}-\frac {i a^{4} A \,{\mathrm e}^{2 i \left (d x +c \right )}}{2 d}-\frac {27 i a^{4} A \,{\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{4} C}{2 d}+\frac {27 i a^{4} A \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d}-\frac {2 i {\mathrm e}^{i \left (d x +c \right )} B \,a^{4}}{d}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} B \,a^{4}}{8 d}+\frac {2 i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{4}}{d}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {4 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}-\frac {13 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {4 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}+\frac {13 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}\) | \(458\) |
| norman | \(\frac {\left (-6 a^{4} A -\frac {13}{2} B \,a^{4}-4 a^{4} C \right ) x +\left (6 a^{4} A +\frac {13}{2} B \,a^{4}+4 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{16}+\left (-36 a^{4} A -39 B \,a^{4}-24 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (-12 a^{4} A -13 B \,a^{4}-8 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (-12 a^{4} A -13 B \,a^{4}-8 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}+\left (12 a^{4} A +13 B \,a^{4}+8 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (12 a^{4} A +13 B \,a^{4}+8 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (36 a^{4} A +39 B \,a^{4}+24 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\frac {5 a^{4} \left (2 A +B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{d}-\frac {a^{4} \left (18 A +11 B +11 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {a^{4} \left (26 A -15 B -69 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}-\frac {a^{4} \left (70 A +123 B +93 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}-\frac {a^{4} \left (74 A +51 B -9 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{3 d}-\frac {a^{4} \left (190 A +33 B -75 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}+\frac {a^{4} \left (190 A +117 B -51 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}+\frac {a^{4} \left (194 A +93 B +39 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 )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}-\frac {a^{4} \left (2 A +8 B +13 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{4} \left (2 A +8 B +13 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(567\) |
Input:
int(cos(d*x+c)^3*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,meth od=_RETURNVERBOSE)
Output:
(-(A+4*B+13/2*C)*(1+cos(2*d*x+2*c))*ln(tan(1/2*d*x+1/2*c)-1)+(A+4*B+13/2*C )*(1+cos(2*d*x+2*c))*ln(tan(1/2*d*x+1/2*c)+1)+6*x*d*(A+13/12*B+2/3*C)*cos( 2*d*x+2*c)+(4*C+A+5/4*B)*sin(2*d*x+2*c)+(2*B+83/24*A+1/2*C)*sin(3*d*x+3*c) +(1/8*B+1/2*A)*sin(4*d*x+4*c)+1/24*A*sin(5*d*x+5*c)+(2*B+41/12*A+3/2*C)*si n(d*x+c)+6*x*d*(A+13/12*B+2/3*C))*a^4/d/(1+cos(2*d*x+2*c))
Time = 0.10 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.88 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {6 \, {\left (12 \, A + 13 \, B + 8 \, C\right )} a^{4} d x \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, A + 8 \, B + 13 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, A + 8 \, B + 13 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, A a^{4} \cos \left (d x + c\right )^{4} + 3 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right )^{3} + 2 \, {\left (20 \, A + 12 \, B + 3 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 6 \, {\left (B + 4 \, C\right )} a^{4} \cos \left (d x + c\right ) + 3 \, C a^{4}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="fricas")
Output:
1/12*(6*(12*A + 13*B + 8*C)*a^4*d*x*cos(d*x + c)^2 + 3*(2*A + 8*B + 13*C)* a^4*cos(d*x + c)^2*log(sin(d*x + c) + 1) - 3*(2*A + 8*B + 13*C)*a^4*cos(d* x + c)^2*log(-sin(d*x + c) + 1) + 2*(2*A*a^4*cos(d*x + c)^4 + 3*(4*A + B)* a^4*cos(d*x + c)^3 + 2*(20*A + 12*B + 3*C)*a^4*cos(d*x + c)^2 + 6*(B + 4*C )*a^4*cos(d*x + c) + 3*C*a^4)*sin(d*x + c))/(d*cos(d*x + c)^2)
Timed out. \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**3*(a+a*sec(d*x+c))**4*(A+B*sec(d*x+c)+C*sec(d*x+c)** 2),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.36 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} - 12 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 48 \, {\left (d x + c\right )} A a^{4} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} - 72 \, {\left (d x + c\right )} B a^{4} - 48 \, {\left (d x + c\right )} C a^{4} + 3 \, C a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 6 \, A a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 24 \, B a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, C a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 72 \, A a^{4} \sin \left (d x + c\right ) - 48 \, B a^{4} \sin \left (d x + c\right ) - 12 \, C a^{4} \sin \left (d x + c\right ) - 12 \, B a^{4} \tan \left (d x + c\right ) - 48 \, C a^{4} \tan \left (d x + c\right )}{12 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="maxima")
Output:
-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^4 - 12*(2*d*x + 2*c + sin(2 *d*x + 2*c))*A*a^4 - 48*(d*x + c)*A*a^4 - 3*(2*d*x + 2*c + sin(2*d*x + 2*c ))*B*a^4 - 72*(d*x + c)*B*a^4 - 48*(d*x + c)*C*a^4 + 3*C*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)) - 6*A*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 24*B*a^4*(log(s in(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 36*C*a^4*(log(sin(d*x + c) + 1 ) - log(sin(d*x + c) - 1)) - 72*A*a^4*sin(d*x + c) - 48*B*a^4*sin(d*x + c) - 12*C*a^4*sin(d*x + c) - 12*B*a^4*tan(d*x + c) - 48*C*a^4*tan(d*x + c))/ d
Time = 0.37 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.60 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (12 \, A a^{4} + 13 \, B a^{4} + 8 \, C a^{4}\right )} {\left (d x + c\right )} + 3 \, {\left (2 \, A a^{4} + 8 \, B a^{4} + 13 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (2 \, A a^{4} + 8 \, B a^{4} + 13 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {6 \, {\left (2 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 7 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}} + \frac {2 \, {\left (30 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 21 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 76 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 48 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 54 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="giac")
Output:
1/6*(3*(12*A*a^4 + 13*B*a^4 + 8*C*a^4)*(d*x + c) + 3*(2*A*a^4 + 8*B*a^4 + 13*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(2*A*a^4 + 8*B*a^4 + 13*C *a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 6*(2*B*a^4*tan(1/2*d*x + 1/2*c) ^3 + 7*C*a^4*tan(1/2*d*x + 1/2*c)^3 - 2*B*a^4*tan(1/2*d*x + 1/2*c) - 9*C*a ^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2 + 2*(30*A*a^4*tan( 1/2*d*x + 1/2*c)^5 + 21*B*a^4*tan(1/2*d*x + 1/2*c)^5 + 6*C*a^4*tan(1/2*d*x + 1/2*c)^5 + 76*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 48*B*a^4*tan(1/2*d*x + 1/2 *c)^3 + 12*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 54*A*a^4*tan(1/2*d*x + 1/2*c) + 27*B*a^4*tan(1/2*d*x + 1/2*c) + 6*C*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d
Time = 13.46 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.72 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2\,\left (6\,A\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-A\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,1{}\mathrm {i}+\frac {13\,B\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}-B\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,4{}\mathrm {i}+4\,C\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-\frac {C\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,13{}\mathrm {i}}{2}\right )}{d}+\frac {\frac {A\,a^4\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {83\,A\,a^4\,\sin \left (3\,c+3\,d\,x\right )}{48}+\frac {A\,a^4\,\sin \left (4\,c+4\,d\,x\right )}{4}+\frac {A\,a^4\,\sin \left (5\,c+5\,d\,x\right )}{48}+\frac {5\,B\,a^4\,\sin \left (2\,c+2\,d\,x\right )}{8}+B\,a^4\,\sin \left (3\,c+3\,d\,x\right )+\frac {B\,a^4\,\sin \left (4\,c+4\,d\,x\right )}{16}+2\,C\,a^4\,\sin \left (2\,c+2\,d\,x\right )+\frac {C\,a^4\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {41\,A\,a^4\,\sin \left (c+d\,x\right )}{24}+B\,a^4\,\sin \left (c+d\,x\right )+\frac {3\,C\,a^4\,\sin \left (c+d\,x\right )}{4}}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )} \] Input:
int(cos(c + d*x)^3*(a + a/cos(c + d*x))^4*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)
Output:
(2*(6*A*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - A*a^4*atan((sin( c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*1i + (13*B*a^4*atan(sin(c/2 + (d*x) /2)/cos(c/2 + (d*x)/2)))/2 - B*a^4*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*4i + 4*C*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - (C*a^ 4*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*13i)/2))/d + ((A*a^4*si n(2*c + 2*d*x))/2 + (83*A*a^4*sin(3*c + 3*d*x))/48 + (A*a^4*sin(4*c + 4*d* x))/4 + (A*a^4*sin(5*c + 5*d*x))/48 + (5*B*a^4*sin(2*c + 2*d*x))/8 + B*a^4 *sin(3*c + 3*d*x) + (B*a^4*sin(4*c + 4*d*x))/16 + 2*C*a^4*sin(2*c + 2*d*x) + (C*a^4*sin(3*c + 3*d*x))/4 + (41*A*a^4*sin(c + d*x))/24 + B*a^4*sin(c + d*x) + (3*C*a^4*sin(c + d*x))/4)/(d*(cos(2*c + 2*d*x)/2 + 1/2))
Time = 0.15 (sec) , antiderivative size = 503, normalized size of antiderivative = 2.32 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:
int(cos(d*x+c)^3*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
Output:
(a**4*(12*cos(c + d*x)*sin(c + d*x)**3*a + 3*cos(c + d*x)*sin(c + d*x)**3* b - 12*cos(c + d*x)*sin(c + d*x)*a - 9*cos(c + d*x)*sin(c + d*x)*b - 24*co s(c + d*x)*sin(c + d*x)*c - 6*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a - 24*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b - 39*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*c + 6*log(tan((c + d*x)/2) - 1)*a + 24*log(tan((c + d*x)/2) - 1)*b + 39*log(tan((c + d*x)/2) - 1)*c + 6*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a + 24*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b + 3 9*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*c - 6*log(tan((c + d*x)/2) + 1 )*a - 24*log(tan((c + d*x)/2) + 1)*b - 39*log(tan((c + d*x)/2) + 1)*c - 2* sin(c + d*x)**5*a + 44*sin(c + d*x)**3*a + 24*sin(c + d*x)**3*b + 6*sin(c + d*x)**3*c + 36*sin(c + d*x)**2*a*c + 36*sin(c + d*x)**2*a*d*x + 39*sin(c + d*x)**2*b*c + 39*sin(c + d*x)**2*b*d*x + 24*sin(c + d*x)**2*c**2 + 24*s in(c + d*x)**2*c*d*x - 42*sin(c + d*x)*a - 24*sin(c + d*x)*b - 9*sin(c + d *x)*c - 36*a*c - 36*a*d*x - 39*b*c - 39*b*d*x - 24*c**2 - 24*c*d*x))/(6*d* (sin(c + d*x)**2 - 1))