Integrand size = 31, antiderivative size = 107 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {(3 A-2 B) \text {arctanh}(\sin (c+d x))}{2 a d}-\frac {2 (A-B) \tan (c+d x)}{a d}+\frac {(3 A-2 B) \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{d (a+a \cos (c+d x))} \] Output:
1/2*(3*A-2*B)*arctanh(sin(d*x+c))/a/d-2*(A-B)*tan(d*x+c)/a/d+1/2*(3*A-2*B) *sec(d*x+c)*tan(d*x+c)/a/d-(A-B)*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(289\) vs. \(2(107)=214\).
Time = 4.80 (sec) , antiderivative size = 289, normalized size of antiderivative = 2.70 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \left (4 (-A+B) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+\cos \left (\frac {1}{2} (c+d x)\right ) \left ((-6 A+4 B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-4 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {4 (A-B) \sin (d x)}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )\right )}{2 a d (1+\cos (c+d x))} \] Input:
Integrate[((A + B*Cos[c + d*x])*Sec[c + d*x]^3)/(a + a*Cos[c + d*x]),x]
Output:
(Cos[(c + d*x)/2]*(4*(-A + B)*Sec[c/2]*Sin[(d*x)/2] + Cos[(c + d*x)/2]*((- 6*A + 4*B)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 6*A*Log[Cos[(c + d*x )/2] + Sin[(c + d*x)/2]] - 4*B*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + A/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 - A/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - (4*(A - B)*Sin[d*x])/((Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[ c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] + Sin[(c + d *x)/2])))))/(2*a*d*(1 + Cos[c + d*x]))
Time = 0.64 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.93, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {3042, 3457, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x) (A+B \cos (c+d x))}{a \cos (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )}dx\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\int (a (3 A-2 B)-2 a (A-B) \cos (c+d x)) \sec ^3(c+d x)dx}{a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (3 A-2 B)-2 a (A-B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx}{a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {a (3 A-2 B) \int \sec ^3(c+d x)dx-2 a (A-B) \int \sec ^2(c+d x)dx}{a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a (3 A-2 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-2 a (A-B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx}{a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\frac {2 a (A-B) \int 1d(-\tan (c+d x))}{d}+a (3 A-2 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {a (3 A-2 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {2 a (A-B) \tan (c+d x)}{d}}{a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {a (3 A-2 B) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {2 a (A-B) \tan (c+d x)}{d}}{a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a (3 A-2 B) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {2 a (A-B) \tan (c+d x)}{d}}{a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {a (3 A-2 B) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {2 a (A-B) \tan (c+d x)}{d}}{a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}\) |
Input:
Int[((A + B*Cos[c + d*x])*Sec[c + d*x]^3)/(a + a*Cos[c + d*x]),x]
Output:
-(((A - B)*Sec[c + d*x]*Tan[c + d*x])/(d*(a + a*Cos[c + d*x]))) + ((-2*a*( A - B)*Tan[c + d*x])/d + a*(3*A - 2*B)*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec [c + d*x]*Tan[c + d*x])/(2*d)))/a^2
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_)])^(n_), x_Symbol] :> Sim p[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b *d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ [b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[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 = 1.62 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.18
| method | result | size |
| parallelrisch | \(\frac {-3 \left (A -\frac {2 B}{3}\right ) \left (\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+3 \left (A -\frac {2 B}{3}\right ) \left (\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-2 \left (\left (2 A -2 B \right ) \cos \left (2 d x +2 c \right )+\left (\cos \left (d x +c \right )+1\right ) \left (A -2 B \right )\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a \left (\cos \left (2 d x +2 c \right )+1\right )}\) | \(126\) |
| derivativedivides | \(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\frac {A}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-\frac {3 A}{2}+B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (\frac {3 A}{2}-B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {A}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-\frac {3 A}{2}+B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-\frac {3 A}{2}+B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d a}\) | \(142\) |
| default | \(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\frac {A}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-\frac {3 A}{2}+B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (\frac {3 A}{2}-B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {A}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-\frac {3 A}{2}+B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-\frac {3 A}{2}+B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d a}\) | \(142\) |
| norman | \(\frac {\frac {\left (3 A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{a d}+\frac {\left (4 A -3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a d}-\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a d}-\frac {\left (2 A -3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}-\frac {\left (3 A -2 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a d}+\frac {\left (3 A -2 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a d}\) | \(186\) |
| risch | \(-\frac {i \left (3 A \,{\mathrm e}^{4 i \left (d x +c \right )}-2 B \,{\mathrm e}^{4 i \left (d x +c \right )}+3 A \,{\mathrm e}^{3 i \left (d x +c \right )}-2 B \,{\mathrm e}^{3 i \left (d x +c \right )}+5 A \,{\mathrm e}^{2 i \left (d x +c \right )}-6 B \,{\mathrm e}^{2 i \left (d x +c \right )}+A \,{\mathrm e}^{i \left (d x +c \right )}-2 B \,{\mathrm e}^{i \left (d x +c \right )}+4 A -4 B \right )}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {3 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a d}-\frac {3 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{a d}\) | \(226\) |
Input:
int((A+B*cos(d*x+c))*sec(d*x+c)^3/(a+a*cos(d*x+c)),x,method=_RETURNVERBOSE )
Output:
1/2*(-3*(A-2/3*B)*(cos(2*d*x+2*c)+1)*ln(tan(1/2*d*x+1/2*c)-1)+3*(A-2/3*B)* (cos(2*d*x+2*c)+1)*ln(tan(1/2*d*x+1/2*c)+1)-2*((2*A-2*B)*cos(2*d*x+2*c)+(c os(d*x+c)+1)*(A-2*B))*tan(1/2*d*x+1/2*c))/d/a/(cos(2*d*x+2*c)+1)
Time = 0.08 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.46 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {{\left ({\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, {\left (A - B\right )} \cos \left (d x + c\right )^{2} + {\left (A - 2 \, B\right )} \cos \left (d x + c\right ) - A\right )} \sin \left (d x + c\right )}{4 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}} \] Input:
integrate((A+B*cos(d*x+c))*sec(d*x+c)^3/(a+a*cos(d*x+c)),x, algorithm="fri cas")
Output:
1/4*(((3*A - 2*B)*cos(d*x + c)^3 + (3*A - 2*B)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - ((3*A - 2*B)*cos(d*x + c)^3 + (3*A - 2*B)*cos(d*x + c)^2)*log (-sin(d*x + c) + 1) - 2*(4*(A - B)*cos(d*x + c)^2 + (A - 2*B)*cos(d*x + c) - A)*sin(d*x + c))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2)
\[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\int \frac {A \sec ^{3}{\left (c + d x \right )}}{\cos {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}}{\cos {\left (c + d x \right )} + 1}\, dx}{a} \] Input:
integrate((A+B*cos(d*x+c))*sec(d*x+c)**3/(a+a*cos(d*x+c)),x)
Output:
(Integral(A*sec(c + d*x)**3/(cos(c + d*x) + 1), x) + Integral(B*cos(c + d* x)*sec(c + d*x)**3/(cos(c + d*x) + 1), x))/a
Leaf count of result is larger than twice the leaf count of optimal. 282 vs. \(2 (103) = 206\).
Time = 0.06 (sec) , antiderivative size = 282, normalized size of antiderivative = 2.64 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {A {\left (\frac {2 \, {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a - \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac {3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac {2 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + 2 \, B {\left (\frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} - \frac {2 \, \sin \left (d x + c\right )}{{\left (a - \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{2 \, d} \] Input:
integrate((A+B*cos(d*x+c))*sec(d*x+c)^3/(a+a*cos(d*x+c)),x, algorithm="max ima")
Output:
-1/2*(A*(2*(sin(d*x + c)/(cos(d*x + c) + 1) - 3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a - 2*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a*sin(d*x + c)^4 /(cos(d*x + c) + 1)^4) - 3*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a + 3* log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a + 2*sin(d*x + c)/(a*(cos(d*x + c) + 1))) + 2*B*(log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a - log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a - 2*sin(d*x + c)/((a - a*sin(d*x + c)^2/(co s(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) - sin(d*x + c)/(a*(cos(d*x + c) + 1 ))))/d
Time = 0.32 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.47 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\frac {{\left (3 \, A - 2 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {{\left (3 \, A - 2 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac {2 \, {\left (A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a} + \frac {2 \, {\left (3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a}}{2 \, d} \] Input:
integrate((A+B*cos(d*x+c))*sec(d*x+c)^3/(a+a*cos(d*x+c)),x, algorithm="gia c")
Output:
1/2*((3*A - 2*B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a - (3*A - 2*B)*log(ab s(tan(1/2*d*x + 1/2*c) - 1))/a - 2*(A*tan(1/2*d*x + 1/2*c) - B*tan(1/2*d*x + 1/2*c))/a + 2*(3*A*tan(1/2*d*x + 1/2*c)^3 - 2*B*tan(1/2*d*x + 1/2*c)^3 - A*tan(1/2*d*x + 1/2*c) + 2*B*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c )^2 - 1)^2*a))/d
Time = 41.31 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.11 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,A-2\,B\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A-2\,B\right )}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}+\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {3\,A}{2}-B\right )}{a\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A-B\right )}{a\,d} \] Input:
int((A + B*cos(c + d*x))/(cos(c + d*x)^3*(a + a*cos(c + d*x))),x)
Output:
(tan(c/2 + (d*x)/2)^3*(3*A - 2*B) - tan(c/2 + (d*x)/2)*(A - 2*B))/(d*(a - 2*a*tan(c/2 + (d*x)/2)^2 + a*tan(c/2 + (d*x)/2)^4)) + (2*atanh(tan(c/2 + ( d*x)/2))*((3*A)/2 - B))/(a*d) - (tan(c/2 + (d*x)/2)*(A - B))/(a*d)
Time = 0.18 (sec) , antiderivative size = 285, normalized size of antiderivative = 2.66 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {4 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a -4 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b -2 \cos \left (d x +c \right ) a +2 \cos \left (d x +c \right ) b -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{3} a +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{3} b +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) a -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) b +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3} a -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3} b -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) a +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) b -3 \sin \left (d x +c \right )^{2} a +2 \sin \left (d x +c \right )^{2} b +2 a -2 b}{2 \sin \left (d x +c \right ) a d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int((A+B*cos(d*x+c))*sec(d*x+c)^3/(a+a*cos(d*x+c)),x)
Output:
(4*cos(c + d*x)*sin(c + d*x)**2*a - 4*cos(c + d*x)*sin(c + d*x)**2*b - 2*c os(c + d*x)*a + 2*cos(c + d*x)*b - 3*log(tan((c + d*x)/2) - 1)*sin(c + d*x )**3*a + 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*b + 3*log(tan((c + d* x)/2) - 1)*sin(c + d*x)*a - 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*b + 3 *log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3*a - 2*log(tan((c + d*x)/2) + 1) *sin(c + d*x)**3*b - 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)*a + 2*log(ta n((c + d*x)/2) + 1)*sin(c + d*x)*b - 3*sin(c + d*x)**2*a + 2*sin(c + d*x)* *2*b + 2*a - 2*b)/(2*sin(c + d*x)*a*d*(sin(c + d*x)**2 - 1))