Integrand size = 38, antiderivative size = 93 \[ \int (a+b \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {(b B+a C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {(2 a B+3 b C) \tan (c+d x)}{3 d}+\frac {(b B+a C) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a B \sec ^2(c+d x) \tan (c+d x)}{3 d} \] Output:
1/2*(B*b+C*a)*arctanh(sin(d*x+c))/d+1/3*(2*B*a+3*C*b)*tan(d*x+c)/d+1/2*(B* b+C*a)*sec(d*x+c)*tan(d*x+c)/d+1/3*a*B*sec(d*x+c)^2*tan(d*x+c)/d
Time = 0.28 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.72 \[ \int (a+b \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {3 (b B+a C) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (6 a B+6 b C+3 (b B+a C) \sec (c+d x)+2 a B \tan ^2(c+d x)\right )}{6 d} \] Input:
Integrate[(a + b*Cos[c + d*x])*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]
Output:
(3*(b*B + a*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(6*a*B + 6*b*C + 3*(b* B + a*C)*Sec[c + d*x] + 2*a*B*Tan[c + d*x]^2))/(6*d)
Time = 0.74 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.98, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3042, 3508, 3042, 3447, 3042, 3500, 3042, 3227, 3042, 4254, 24, 4255, 3042, 4257}
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.
| \(\displaystyle \int \sec ^5(c+d x) (a+b \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 3508 |
| \(\displaystyle \int \sec ^4(c+d x) (a+b \cos (c+d x)) (B+C \cos (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \int \sec ^4(c+d x) \left ((a C+b B) \cos (c+d x)+a B+b C \cos ^2(c+d x)\right )dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a C+b B) \sin \left (c+d x+\frac {\pi }{2}\right )+a B+b C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{3} \int (3 (b B+a C)+(2 a B+3 b C) \cos (c+d x)) \sec ^3(c+d x)dx+\frac {a B \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \frac {3 (b B+a C)+(2 a B+3 b C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a B \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{3} \left (3 (a C+b B) \int \sec ^3(c+d x)dx+(2 a B+3 b C) \int \sec ^2(c+d x)dx\right )+\frac {a B \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left ((2 a B+3 b C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+3 (a C+b B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx\right )+\frac {a B \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{3} \left (3 (a C+b B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {(2 a B+3 b C) \int 1d(-\tan (c+d x))}{d}\right )+\frac {a B \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{3} \left (3 (a C+b B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {(2 a B+3 b C) \tan (c+d x)}{d}\right )+\frac {a B \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{3} \left (3 (a C+b B) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {(2 a B+3 b C) \tan (c+d x)}{d}\right )+\frac {a B \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (3 (a C+b B) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {(2 a B+3 b C) \tan (c+d x)}{d}\right )+\frac {a B \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{3} \left (3 (a C+b B) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {(2 a B+3 b C) \tan (c+d x)}{d}\right )+\frac {a B \tan (c+d x) \sec ^2(c+d x)}{3 d}\) |
Input:
Int[(a + b*Cos[c + d*x])*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^ 5,x]
Output:
(a*B*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + (((2*a*B + 3*b*C)*Tan[c + d*x])/ d + 3*(b*B + a*C)*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x ])/(2*d)))/3
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* (a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A *b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ .) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b^2 Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.60 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.87
| method | result | size |
| parts | \(\frac {\left (B b +C a \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {B a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {C b \tan \left (d x +c \right )}{d}\) | \(81\) |
| derivativedivides | \(\frac {C a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+C b \tan \left (d x +c \right )+B b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(105\) |
| default | \(\frac {C a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+C b \tan \left (d x +c \right )+B b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(105\) |
| parallelrisch | \(\frac {-9 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (B b +C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+9 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (B b +C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+6 \left (B b +C a \right ) \sin \left (2 d x +2 c \right )+2 \left (2 B a +3 C b \right ) \sin \left (3 d x +3 c \right )+6 \sin \left (d x +c \right ) \left (2 B a +C b \right )}{6 d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\) | \(159\) |
| risch | \(-\frac {i \left (3 B b \,{\mathrm e}^{5 i \left (d x +c \right )}+3 C a \,{\mathrm e}^{5 i \left (d x +c \right )}-6 C b \,{\mathrm e}^{4 i \left (d x +c \right )}-12 B a \,{\mathrm e}^{2 i \left (d x +c \right )}-12 C b \,{\mathrm e}^{2 i \left (d x +c \right )}-3 B b \,{\mathrm e}^{i \left (d x +c \right )}-3 C a \,{\mathrm e}^{i \left (d x +c \right )}-4 B a -6 C b \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B b}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C a}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B b}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C a}{2 d}\) | \(201\) |
| norman | \(\frac {\frac {\left (2 B a +B b +C a +2 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {4 \left (B b +C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {2 \left (4 B a -3 B b -3 C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}+\frac {2 \left (4 B a +3 B b +3 C a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {\left (2 B a -3 B b -3 C a +18 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}-\frac {\left (2 B a -B b -C a +2 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{d}-\frac {\left (2 B a +3 B b +3 C a +18 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {\left (B b +C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (B b +C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(290\) |
Input:
int((a+b*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x,method=_ RETURNVERBOSE)
Output:
(B*b+C*a)/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))-B*a/ d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+C*b/d*tan(d*x+c)
Time = 0.14 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.24 \[ \int (a+b \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {3 \, {\left (C a + B b\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (C a + B b\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (2 \, B a + 3 \, C b\right )} \cos \left (d x + c\right )^{2} + 2 \, B a + 3 \, {\left (C a + B b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \] Input:
integrate((a+b*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, a lgorithm="fricas")
Output:
1/12*(3*(C*a + B*b)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(C*a + B*b)*c os(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*(2*B*a + 3*C*b)*cos(d*x + c)^2 + 2*B*a + 3*(C*a + B*b)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3)
Timed out. \[ \int (a+b \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**5,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.37 \[ \int (a+b \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a - 3 \, C a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 3 \, B b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, C b \tan \left (d x + c\right )}{12 \, d} \] Input:
integrate((a+b*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, a lgorithm="maxima")
Output:
1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a - 3*C*a*(2*sin(d*x + c)/(sin (d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 3*B*b* (2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 12*C*b*tan(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 210 vs. \(2 (85) = 170\).
Time = 0.16 (sec) , antiderivative size = 210, normalized size of antiderivative = 2.26 \[ \int (a+b \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {3 \, {\left (C a + B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (C a + B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \] Input:
integrate((a+b*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, a lgorithm="giac")
Output:
1/6*(3*(C*a + B*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(C*a + B*b)*log( abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(6*B*a*tan(1/2*d*x + 1/2*c)^5 - 3*C*a*t an(1/2*d*x + 1/2*c)^5 - 3*B*b*tan(1/2*d*x + 1/2*c)^5 + 6*C*b*tan(1/2*d*x + 1/2*c)^5 - 4*B*a*tan(1/2*d*x + 1/2*c)^3 - 12*C*b*tan(1/2*d*x + 1/2*c)^3 + 6*B*a*tan(1/2*d*x + 1/2*c) + 3*C*a*tan(1/2*d*x + 1/2*c) + 3*B*b*tan(1/2*d *x + 1/2*c) + 6*C*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/ d
Time = 1.89 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.56 \[ \int (a+b \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (B\,b+C\,a\right )}{d}-\frac {\left (2\,B\,a-B\,b-C\,a+2\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {4\,B\,a}{3}-4\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,B\,a+B\,b+C\,a+2\,C\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:
int(((B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + b*cos(c + d*x)))/cos(c + d*x )^5,x)
Output:
(atanh(tan(c/2 + (d*x)/2))*(B*b + C*a))/d - (tan(c/2 + (d*x)/2)*(2*B*a + B *b + C*a + 2*C*b) - tan(c/2 + (d*x)/2)^3*((4*B*a)/3 + 4*C*b) + tan(c/2 + ( d*x)/2)^5*(2*B*a - B*b - C*a + 2*C*b))/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan( c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 - 1))
Time = 0.17 (sec) , antiderivative size = 315, normalized size of antiderivative = 3.39 \[ \int (a+b \cos (c+d x)) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {-3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a c -3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b^{2}+3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a c +3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{2}+3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a c +3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b^{2}-3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a c -3 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{2}-3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a c -3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{2}+4 \sin \left (d x +c \right )^{3} a b +6 \sin \left (d x +c \right )^{3} b c -6 \sin \left (d x +c \right ) a b -6 \sin \left (d x +c \right ) b c}{6 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int((a+b*cos(d*x+c))*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x)
Output:
( - 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*c - 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**2 + 3*cos(c + d*x)*lo g(tan((c + d*x)/2) - 1)*a*c + 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*b** 2 + 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a*c + 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b**2 - 3*cos(c + d*x)*lo g(tan((c + d*x)/2) + 1)*a*c - 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*b** 2 - 3*cos(c + d*x)*sin(c + d*x)*a*c - 3*cos(c + d*x)*sin(c + d*x)*b**2 + 4 *sin(c + d*x)**3*a*b + 6*sin(c + d*x)**3*b*c - 6*sin(c + d*x)*a*b - 6*sin( c + d*x)*b*c)/(6*cos(c + d*x)*d*(sin(c + d*x)**2 - 1))