Integrand size = 31, antiderivative size = 179 \[ \int \frac {(A+B \cos (c+d x)) \sec ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {(10 A-7 B) \text {arctanh}(\sin (c+d x))}{2 a^2 d}+\frac {4 (3 A-2 B) \tan (c+d x)}{a^2 d}-\frac {(10 A-7 B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}-\frac {(10 A-7 B) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A-B) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {4 (3 A-2 B) \tan ^3(c+d x)}{3 a^2 d} \] Output:
-1/2*(10*A-7*B)*arctanh(sin(d*x+c))/a^2/d+4*(3*A-2*B)*tan(d*x+c)/a^2/d-1/2 *(10*A-7*B)*sec(d*x+c)*tan(d*x+c)/a^2/d-1/3*(10*A-7*B)*sec(d*x+c)^2*tan(d* x+c)/a^2/d/(1+cos(d*x+c))-1/3*(A-B)*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*cos(d*x +c))^2+4/3*(3*A-2*B)*tan(d*x+c)^3/a^2/d
Leaf count is larger than twice the leaf count of optimal. \(609\) vs. \(2(179)=358\).
Time = 6.43 (sec) , antiderivative size = 609, normalized size of antiderivative = 3.40 \[ \int \frac {(A+B \cos (c+d x)) \sec ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {192 (10 A-7 B) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sec (c) \sec ^3(c+d x) \left ((-6 A+45 B) \sin \left (\frac {d x}{2}\right )+(310 A-201 B) \sin \left (\frac {3 d x}{2}\right )-306 A \sin \left (c-\frac {d x}{2}\right )+195 B \sin \left (c-\frac {d x}{2}\right )+42 A \sin \left (c+\frac {d x}{2}\right )-51 B \sin \left (c+\frac {d x}{2}\right )-270 A \sin \left (2 c+\frac {d x}{2}\right )+189 B \sin \left (2 c+\frac {d x}{2}\right )+50 A \sin \left (c+\frac {3 d x}{2}\right )-B \sin \left (c+\frac {3 d x}{2}\right )+90 A \sin \left (2 c+\frac {3 d x}{2}\right )-81 B \sin \left (2 c+\frac {3 d x}{2}\right )-170 A \sin \left (3 c+\frac {3 d x}{2}\right )+119 B \sin \left (3 c+\frac {3 d x}{2}\right )+198 A \sin \left (c+\frac {5 d x}{2}\right )-129 B \sin \left (c+\frac {5 d x}{2}\right )+42 A \sin \left (2 c+\frac {5 d x}{2}\right )-9 B \sin \left (2 c+\frac {5 d x}{2}\right )+66 A \sin \left (3 c+\frac {5 d x}{2}\right )-57 B \sin \left (3 c+\frac {5 d x}{2}\right )-90 A \sin \left (4 c+\frac {5 d x}{2}\right )+63 B \sin \left (4 c+\frac {5 d x}{2}\right )+114 A \sin \left (2 c+\frac {7 d x}{2}\right )-75 B \sin \left (2 c+\frac {7 d x}{2}\right )+36 A \sin \left (3 c+\frac {7 d x}{2}\right )-15 B \sin \left (3 c+\frac {7 d x}{2}\right )+48 A \sin \left (4 c+\frac {7 d x}{2}\right )-39 B \sin \left (4 c+\frac {7 d x}{2}\right )-30 A \sin \left (5 c+\frac {7 d x}{2}\right )+21 B \sin \left (5 c+\frac {7 d x}{2}\right )+48 A \sin \left (3 c+\frac {9 d x}{2}\right )-32 B \sin \left (3 c+\frac {9 d x}{2}\right )+22 A \sin \left (4 c+\frac {9 d x}{2}\right )-12 B \sin \left (4 c+\frac {9 d x}{2}\right )+26 A \sin \left (5 c+\frac {9 d x}{2}\right )-20 B \sin \left (5 c+\frac {9 d x}{2}\right )\right )}{96 a^2 d (1+\cos (c+d x))^2} \] Input:
Integrate[((A + B*Cos[c + d*x])*Sec[c + d*x]^4)/(a + a*Cos[c + d*x])^2,x]
Output:
(192*(10*A - 7*B)*Cos[(c + d*x)/2]^4*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x) /2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Cos[(c + d*x)/2]*Sec[c/ 2]*Sec[c]*Sec[c + d*x]^3*((-6*A + 45*B)*Sin[(d*x)/2] + (310*A - 201*B)*Sin [(3*d*x)/2] - 306*A*Sin[c - (d*x)/2] + 195*B*Sin[c - (d*x)/2] + 42*A*Sin[c + (d*x)/2] - 51*B*Sin[c + (d*x)/2] - 270*A*Sin[2*c + (d*x)/2] + 189*B*Sin [2*c + (d*x)/2] + 50*A*Sin[c + (3*d*x)/2] - B*Sin[c + (3*d*x)/2] + 90*A*Si n[2*c + (3*d*x)/2] - 81*B*Sin[2*c + (3*d*x)/2] - 170*A*Sin[3*c + (3*d*x)/2 ] + 119*B*Sin[3*c + (3*d*x)/2] + 198*A*Sin[c + (5*d*x)/2] - 129*B*Sin[c + (5*d*x)/2] + 42*A*Sin[2*c + (5*d*x)/2] - 9*B*Sin[2*c + (5*d*x)/2] + 66*A*S in[3*c + (5*d*x)/2] - 57*B*Sin[3*c + (5*d*x)/2] - 90*A*Sin[4*c + (5*d*x)/2 ] + 63*B*Sin[4*c + (5*d*x)/2] + 114*A*Sin[2*c + (7*d*x)/2] - 75*B*Sin[2*c + (7*d*x)/2] + 36*A*Sin[3*c + (7*d*x)/2] - 15*B*Sin[3*c + (7*d*x)/2] + 48* A*Sin[4*c + (7*d*x)/2] - 39*B*Sin[4*c + (7*d*x)/2] - 30*A*Sin[5*c + (7*d*x )/2] + 21*B*Sin[5*c + (7*d*x)/2] + 48*A*Sin[3*c + (9*d*x)/2] - 32*B*Sin[3* c + (9*d*x)/2] + 22*A*Sin[4*c + (9*d*x)/2] - 12*B*Sin[4*c + (9*d*x)/2] + 2 6*A*Sin[5*c + (9*d*x)/2] - 20*B*Sin[5*c + (9*d*x)/2]))/(96*a^2*d*(1 + Cos[ c + d*x])^2)
Time = 1.01 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.96, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.419, Rules used = {3042, 3457, 3042, 3457, 27, 3042, 3227, 3042, 4254, 2009, 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.
| \(\displaystyle \int \frac {\sec ^4(c+d x) (A+B \cos (c+d x))}{(a \cos (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\int \frac {(3 a (2 A-B)-4 a (A-B) \cos (c+d x)) \sec ^4(c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {3 a (2 A-B)-4 a (A-B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\int 3 \left (4 a^2 (3 A-2 B)-a^2 (10 A-7 B) \cos (c+d x)\right ) \sec ^4(c+d x)dx}{a^2}-\frac {(10 A-7 B) \tan (c+d x) \sec ^2(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {3 \int \left (4 a^2 (3 A-2 B)-a^2 (10 A-7 B) \cos (c+d x)\right ) \sec ^4(c+d x)dx}{a^2}-\frac {(10 A-7 B) \tan (c+d x) \sec ^2(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \int \frac {4 a^2 (3 A-2 B)-a^2 (10 A-7 B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx}{a^2}-\frac {(10 A-7 B) \tan (c+d x) \sec ^2(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {\frac {3 \left (4 a^2 (3 A-2 B) \int \sec ^4(c+d x)dx-a^2 (10 A-7 B) \int \sec ^3(c+d x)dx\right )}{a^2}-\frac {(10 A-7 B) \tan (c+d x) \sec ^2(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \left (4 a^2 (3 A-2 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx-a^2 (10 A-7 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx\right )}{a^2}-\frac {(10 A-7 B) \tan (c+d x) \sec ^2(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\frac {3 \left (-\frac {4 a^2 (3 A-2 B) \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}-\left (a^2 (10 A-7 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx\right )\right )}{a^2}-\frac {(10 A-7 B) \tan (c+d x) \sec ^2(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {3 \left (-a^2 (10 A-7 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {4 a^2 (3 A-2 B) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )}{a^2}-\frac {(10 A-7 B) \tan (c+d x) \sec ^2(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {\frac {3 \left (-a^2 (10 A-7 B) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 a^2 (3 A-2 B) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )}{a^2}-\frac {(10 A-7 B) \tan (c+d x) \sec ^2(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \left (-a^2 (10 A-7 B) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 a^2 (3 A-2 B) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )}{a^2}-\frac {(10 A-7 B) \tan (c+d x) \sec ^2(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {3 \left (-\left (a^2 (10 A-7 B) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )\right )-\frac {4 a^2 (3 A-2 B) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )}{a^2}-\frac {(10 A-7 B) \tan (c+d x) \sec ^2(c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
Input:
Int[((A + B*Cos[c + d*x])*Sec[c + d*x]^4)/(a + a*Cos[c + d*x])^2,x]
Output:
-1/3*((A - B)*Sec[c + d*x]^2*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])^2) + (- (((10*A - 7*B)*Sec[c + d*x]^2*Tan[c + d*x])/(d*(1 + Cos[c + d*x]))) + (3*( -(a^2*(10*A - 7*B)*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d* x])/(2*d))) - (4*a^2*(3*A - 2*B)*(-Tan[c + d*x] - Tan[c + d*x]^3/3))/d))/a ^2)/(3*a^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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 = 1.88 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.09
| method | result | size |
| parallelrisch | \(\frac {180 \left (\cos \left (d x +c \right )+\frac {\cos \left (3 d x +3 c \right )}{3}\right ) \left (A -\frac {7 B}{10}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-180 \left (\cos \left (d x +c \right )+\frac {\cos \left (3 d x +3 c \right )}{3}\right ) \left (A -\frac {7 B}{10}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+24 \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (\frac {11 A}{4}-\frac {43 B}{24}\right ) \cos \left (3 d x +3 c \right )+\left (5 A -\frac {19 B}{6}\right ) \cos \left (2 d x +2 c \right )+\left (A -\frac {2 B}{3}\right ) \cos \left (4 d x +4 c \right )+\left (\frac {95 A}{12}-\frac {39 B}{8}\right ) \cos \left (d x +c \right )+\frac {13 A}{3}-\frac {5 B}{2}\right )}{12 d \,a^{2} \left (3 \cos \left (d x +c \right )+\cos \left (3 d x +3 c \right )\right )}\) | \(196\) |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\frac {6 A -2 B}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\left (10 A -7 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {10 A -5 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}-\frac {2 A}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\left (-10 A +7 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {-6 A +2 B}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {10 A -5 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-\frac {2 A}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}}{2 d \,a^{2}}\) | \(222\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\frac {6 A -2 B}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\left (10 A -7 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {10 A -5 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}-\frac {2 A}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\left (-10 A +7 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {-6 A +2 B}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {10 A -5 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-\frac {2 A}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}}{2 d \,a^{2}}\) | \(222\) |
| norman | \(\frac {\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{6 a d}+\frac {\left (11 A -10 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 a d}-\frac {\left (19 A -12 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a d}-\frac {\left (21 A -13 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}+\frac {\left (25 A -19 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{6 a d}+\frac {\left (97 A -71 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3} a}+\frac {\left (10 A -7 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{2} d}-\frac {\left (10 A -7 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{2} d}\) | \(243\) |
| risch | \(\frac {i \left (30 A \,{\mathrm e}^{8 i \left (d x +c \right )}-21 B \,{\mathrm e}^{8 i \left (d x +c \right )}+90 A \,{\mathrm e}^{7 i \left (d x +c \right )}-63 B \,{\mathrm e}^{7 i \left (d x +c \right )}+170 A \,{\mathrm e}^{6 i \left (d x +c \right )}-119 B \,{\mathrm e}^{6 i \left (d x +c \right )}+270 A \,{\mathrm e}^{5 i \left (d x +c \right )}-189 B \,{\mathrm e}^{5 i \left (d x +c \right )}+306 A \,{\mathrm e}^{4 i \left (d x +c \right )}-195 B \,{\mathrm e}^{4 i \left (d x +c \right )}+310 A \,{\mathrm e}^{3 i \left (d x +c \right )}-201 B \,{\mathrm e}^{3 i \left (d x +c \right )}+198 A \,{\mathrm e}^{2 i \left (d x +c \right )}-129 B \,{\mathrm e}^{2 i \left (d x +c \right )}+114 A \,{\mathrm e}^{i \left (d x +c \right )}-75 B \,{\mathrm e}^{i \left (d x +c \right )}+48 A -32 B \right )}{3 d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}+\frac {5 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{2} d}-\frac {7 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 a^{2} d}-\frac {5 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{2} d}+\frac {7 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 a^{2} d}\) | \(324\) |
Input:
int((A+B*cos(d*x+c))*sec(d*x+c)^4/(a+a*cos(d*x+c))^2,x,method=_RETURNVERBO SE)
Output:
1/12*(180*(cos(d*x+c)+1/3*cos(3*d*x+3*c))*(A-7/10*B)*ln(tan(1/2*d*x+1/2*c) -1)-180*(cos(d*x+c)+1/3*cos(3*d*x+3*c))*(A-7/10*B)*ln(tan(1/2*d*x+1/2*c)+1 )+24*sec(1/2*d*x+1/2*c)^2*tan(1/2*d*x+1/2*c)*((11/4*A-43/24*B)*cos(3*d*x+3 *c)+(5*A-19/6*B)*cos(2*d*x+2*c)+(A-2/3*B)*cos(4*d*x+4*c)+(95/12*A-39/8*B)* cos(d*x+c)+13/3*A-5/2*B))/d/a^2/(3*cos(d*x+c)+cos(3*d*x+3*c))
Time = 0.09 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.38 \[ \int \frac {(A+B \cos (c+d x)) \sec ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {3 \, {\left ({\left (10 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (10 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{4} + {\left (10 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (10 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (10 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{4} + {\left (10 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (16 \, {\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{4} + {\left (66 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{3} + 6 \, {\left (2 \, A - B\right )} \cos \left (d x + c\right )^{2} - {\left (2 \, A - 3 \, B\right )} \cos \left (d x + c\right ) + 2 \, A\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} + 2 \, a^{2} d \cos \left (d x + c\right )^{4} + a^{2} d \cos \left (d x + c\right )^{3}\right )}} \] Input:
integrate((A+B*cos(d*x+c))*sec(d*x+c)^4/(a+a*cos(d*x+c))^2,x, algorithm="f ricas")
Output:
-1/12*(3*((10*A - 7*B)*cos(d*x + c)^5 + 2*(10*A - 7*B)*cos(d*x + c)^4 + (1 0*A - 7*B)*cos(d*x + c)^3)*log(sin(d*x + c) + 1) - 3*((10*A - 7*B)*cos(d*x + c)^5 + 2*(10*A - 7*B)*cos(d*x + c)^4 + (10*A - 7*B)*cos(d*x + c)^3)*log (-sin(d*x + c) + 1) - 2*(16*(3*A - 2*B)*cos(d*x + c)^4 + (66*A - 43*B)*cos (d*x + c)^3 + 6*(2*A - B)*cos(d*x + c)^2 - (2*A - 3*B)*cos(d*x + c) + 2*A) *sin(d*x + c))/(a^2*d*cos(d*x + c)^5 + 2*a^2*d*cos(d*x + c)^4 + a^2*d*cos( d*x + c)^3)
\[ \int \frac {(A+B \cos (c+d x)) \sec ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\int \frac {A \sec ^{4}{\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:
integrate((A+B*cos(d*x+c))*sec(d*x+c)**4/(a+a*cos(d*x+c))**2,x)
Output:
(Integral(A*sec(c + d*x)**4/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1), x) + I ntegral(B*cos(c + d*x)*sec(c + d*x)**4/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1), x))/a**2
Leaf count of result is larger than twice the leaf count of optimal. 425 vs. \(2 (169) = 338\).
Time = 0.05 (sec) , antiderivative size = 425, normalized size of antiderivative = 2.37 \[ \int \frac {(A+B \cos (c+d x)) \sec ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {A {\left (\frac {4 \, {\left (\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{2} - \frac {3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {\frac {27 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {30 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {30 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )} - B {\left (\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} - \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {21 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {21 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )}}{6 \, d} \] Input:
integrate((A+B*cos(d*x+c))*sec(d*x+c)^4/(a+a*cos(d*x+c))^2,x, algorithm="m axima")
Output:
1/6*(A*(4*(9*sin(d*x + c)/(cos(d*x + c) + 1) - 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/(a^2 - 3*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + (27*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 30*log(sin(d*x + c)/(co s(d*x + c) + 1) + 1)/a^2 + 30*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2 ) - B*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) - 5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 - 2*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d *x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 21*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 21*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2))/d
Time = 0.34 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.26 \[ \int \frac {(A+B \cos (c+d x)) \sec ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {\frac {3 \, {\left (10 \, A - 7 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {3 \, {\left (10 \, A - 7 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {2 \, {\left (30 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a^{2}} - \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 27 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 21 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \] Input:
integrate((A+B*cos(d*x+c))*sec(d*x+c)^4/(a+a*cos(d*x+c))^2,x, algorithm="g iac")
Output:
-1/6*(3*(10*A - 7*B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 3*(10*A - 7* B)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2 + 2*(30*A*tan(1/2*d*x + 1/2*c)^5 - 15*B*tan(1/2*d*x + 1/2*c)^5 - 40*A*tan(1/2*d*x + 1/2*c)^3 + 24*B*tan(1/ 2*d*x + 1/2*c)^3 + 18*A*tan(1/2*d*x + 1/2*c) - 9*B*tan(1/2*d*x + 1/2*c))/( (tan(1/2*d*x + 1/2*c)^2 - 1)^3*a^2) - (A*a^4*tan(1/2*d*x + 1/2*c)^3 - B*a^ 4*tan(1/2*d*x + 1/2*c)^3 + 27*A*a^4*tan(1/2*d*x + 1/2*c) - 21*B*a^4*tan(1/ 2*d*x + 1/2*c))/a^6)/d
Time = 41.97 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.13 \[ \int \frac {(A+B \cos (c+d x)) \sec ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {2\,\left (A-B\right )}{a^2}+\frac {5\,A-3\,B}{2\,a^2}\right )}{d}-\frac {\left (10\,A-5\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (8\,B-\frac {40\,A}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (6\,A-3\,B\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B\right )}{6\,a^2\,d}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (10\,A-7\,B\right )}{a^2\,d} \] Input:
int((A + B*cos(c + d*x))/(cos(c + d*x)^4*(a + a*cos(c + d*x))^2),x)
Output:
(tan(c/2 + (d*x)/2)*((2*(A - B))/a^2 + (5*A - 3*B)/(2*a^2)))/d - (tan(c/2 + (d*x)/2)^5*(10*A - 5*B) - tan(c/2 + (d*x)/2)^3*((40*A)/3 - 8*B) + tan(c/ 2 + (d*x)/2)*(6*A - 3*B))/(d*(3*a^2*tan(c/2 + (d*x)/2)^2 - 3*a^2*tan(c/2 + (d*x)/2)^4 + a^2*tan(c/2 + (d*x)/2)^6 - a^2)) + (tan(c/2 + (d*x)/2)^3*(A - B))/(6*a^2*d) - (atanh(tan(c/2 + (d*x)/2))*(10*A - 7*B))/(a^2*d)
Time = 0.24 (sec) , antiderivative size = 557, normalized size of antiderivative = 3.11 \[ \int \frac {(A+B \cos (c+d x)) \sec ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx =\text {Too large to display} \] Input:
int((A+B*cos(d*x+c))*sec(d*x+c)^4/(a+a*cos(d*x+c))^2,x)
Output:
(30*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**6*a - 21*log(tan((c + d*x) /2) - 1)*tan((c + d*x)/2)**6*b - 90*log(tan((c + d*x)/2) - 1)*tan((c + d*x )/2)**4*a + 63*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**4*b + 90*log(ta n((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*a - 63*log(tan((c + d*x)/2) - 1)*t an((c + d*x)/2)**2*b - 30*log(tan((c + d*x)/2) - 1)*a + 21*log(tan((c + d* x)/2) - 1)*b - 30*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**6*a + 21*log (tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**6*b + 90*log(tan((c + d*x)/2) + 1 )*tan((c + d*x)/2)**4*a - 63*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**4 *b - 90*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*a + 63*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*b + 30*log(tan((c + d*x)/2) + 1)*a - 21*l og(tan((c + d*x)/2) + 1)*b + tan((c + d*x)/2)**9*a - tan((c + d*x)/2)**9*b + 24*tan((c + d*x)/2)**7*a - 18*tan((c + d*x)/2)**7*b - 138*tan((c + d*x) /2)**5*a + 90*tan((c + d*x)/2)**5*b + 160*tan((c + d*x)/2)**3*a - 110*tan( (c + d*x)/2)**3*b - 63*tan((c + d*x)/2)*a + 39*tan((c + d*x)/2)*b)/(6*a**2 *d*(tan((c + d*x)/2)**6 - 3*tan((c + d*x)/2)**4 + 3*tan((c + d*x)/2)**2 - 1))