\(\int \cos (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx\) [75]

Optimal result

Integrand size = 29, antiderivative size = 151 \[ \int \cos (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=a^4 (4 A+B) x+\frac {a^4 (13 A+12 B) \text {arctanh}(\sin (c+d x))}{2 d}-\frac {5 a^4 (A+2 B) \sin (c+d x)}{2 d}+\frac {a B (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {(A+2 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac {(9 A+11 B) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{3 d} \]

[Out]

a^4*(4*A+B)*x+1/2*a^4*(13*A+12*B)*arctanh(sin(d*x+c))/d-5/2*a^4*(A+2*B)*sin(d*x+c)/d+1/3*a*B*(a+a*sec(d*x+c))^
3*sin(d*x+c)/d+1/2*(A+2*B)*(a^2+a^2*sec(d*x+c))^2*sin(d*x+c)/d+1/3*(9*A+11*B)*(a^4+a^4*sec(d*x+c))*sin(d*x+c)/
d

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {4103, 4081, 3855} \[ \int \cos (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {a^4 (13 A+12 B) \text {arctanh}(\sin (c+d x))}{2 d}-\frac {5 a^4 (A+2 B) \sin (c+d x)}{2 d}+\frac {(9 A+11 B) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{3 d}+a^4 x (4 A+B)+\frac {(A+2 B) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}+\frac {a B \sin (c+d x) (a \sec (c+d x)+a)^3}{3 d} \]

[In]

Int[Cos[c + d*x]*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x]),x]

[Out]

a^4*(4*A + B)*x + (a^4*(13*A + 12*B)*ArcTanh[Sin[c + d*x]])/(2*d) - (5*a^4*(A + 2*B)*Sin[c + d*x])/(2*d) + (a*
B*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(3*d) + ((A + 2*B)*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) + ((9
*A + 11*B)*(a^4 + a^4*Sec[c + d*x])*Sin[c + d*x])/(3*d)

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4081

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] + Dist[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]

Rule 4103

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] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {a B (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {1}{3} \int \cos (c+d x) (a+a \sec (c+d x))^3 (a (3 A-B)+3 a (A+2 B) \sec (c+d x)) \, dx \\ & = \frac {a B (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {(A+2 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac {1}{6} \int \cos (c+d x) (a+a \sec (c+d x))^2 \left (a^2 (3 A-8 B)+2 a^2 (9 A+11 B) \sec (c+d x)\right ) \, dx \\ & = \frac {a B (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {(A+2 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac {(9 A+11 B) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{3 d}+\frac {1}{6} \int \cos (c+d x) (a+a \sec (c+d x)) \left (-15 a^3 (A+2 B)+3 a^3 (13 A+12 B) \sec (c+d x)\right ) \, dx \\ & = -\frac {5 a^4 (A+2 B) \sin (c+d x)}{2 d}+\frac {a B (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {(A+2 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac {(9 A+11 B) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{3 d}-\frac {1}{6} \int \left (-6 a^4 (4 A+B)-3 a^4 (13 A+12 B) \sec (c+d x)\right ) \, dx \\ & = a^4 (4 A+B) x-\frac {5 a^4 (A+2 B) \sin (c+d x)}{2 d}+\frac {a B (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {(A+2 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac {(9 A+11 B) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{3 d}+\frac {1}{2} \left (a^4 (13 A+12 B)\right ) \int \sec (c+d x) \, dx \\ & = a^4 (4 A+B) x+\frac {a^4 (13 A+12 B) \text {arctanh}(\sin (c+d x))}{2 d}-\frac {5 a^4 (A+2 B) \sin (c+d x)}{2 d}+\frac {a B (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {(A+2 B) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac {(9 A+11 B) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{3 d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(1202\) vs. \(2(151)=302\).

Time = 12.29 (sec) , antiderivative size = 1202, normalized size of antiderivative = 7.96 \[ \int \cos (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {(4 A+B) x \cos ^5(c+d x) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 (A+B \sec (c+d x))}{16 (B+A \cos (c+d x))}+\frac {(-13 A-12 B) \cos ^5(c+d x) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 (A+B \sec (c+d x))}{32 d (B+A \cos (c+d x))}+\frac {(13 A+12 B) \cos ^5(c+d x) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 (A+B \sec (c+d x))}{32 d (B+A \cos (c+d x))}+\frac {A \cos (d x) \cos ^5(c+d x) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \sin (c)}{16 d (B+A \cos (c+d x))}+\frac {A \cos (c) \cos ^5(c+d x) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \sin (d x)}{16 d (B+A \cos (c+d x))}+\frac {B \cos ^5(c+d x) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \sin \left (\frac {d x}{2}\right )}{96 d (B+A \cos (c+d x)) \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^3}+\frac {\cos ^5(c+d x) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \left (3 A \cos \left (\frac {c}{2}\right )+13 B \cos \left (\frac {c}{2}\right )-3 A \sin \left (\frac {c}{2}\right )-11 B \sin \left (\frac {c}{2}\right )\right )}{192 d (B+A \cos (c+d x)) \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}+\frac {\cos ^5(c+d x) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \left (3 A \sin \left (\frac {d x}{2}\right )+5 B \sin \left (\frac {d x}{2}\right )\right )}{12 d (B+A \cos (c+d x)) \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}+\frac {B \cos ^5(c+d x) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \sin \left (\frac {d x}{2}\right )}{96 d (B+A \cos (c+d x)) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^3}+\frac {\cos ^5(c+d x) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \left (-3 A \cos \left (\frac {c}{2}\right )-13 B \cos \left (\frac {c}{2}\right )-3 A \sin \left (\frac {c}{2}\right )-11 B \sin \left (\frac {c}{2}\right )\right )}{192 d (B+A \cos (c+d x)) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}+\frac {\cos ^5(c+d x) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \left (3 A \sin \left (\frac {d x}{2}\right )+5 B \sin \left (\frac {d x}{2}\right )\right )}{12 d (B+A \cos (c+d x)) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )} \]

[In]

Integrate[Cos[c + d*x]*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x]),x]

[Out]

((4*A + B)*x*Cos[c + d*x]^5*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x]))/(16*(B + A*Cos[c
+ d*x])) + ((-13*A - 12*B)*Cos[c + d*x]^5*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^8*(
a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x]))/(32*d*(B + A*Cos[c + d*x])) + ((13*A + 12*B)*Cos[c + d*x]^5*Log[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]))/(32*d
*(B + A*Cos[c + d*x])) + (A*Cos[d*x]*Cos[c + d*x]^5*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c +
d*x])*Sin[c])/(16*d*(B + A*Cos[c + d*x])) + (A*Cos[c]*Cos[c + d*x]^5*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x]
)^4*(A + B*Sec[c + d*x])*Sin[d*x])/(16*d*(B + A*Cos[c + d*x])) + (B*Cos[c + d*x]^5*Sec[c/2 + (d*x)/2]^8*(a + a
*Sec[c + d*x])^4*(A + B*Sec[c + d*x])*Sin[(d*x)/2])/(96*d*(B + A*Cos[c + d*x])*(Cos[c/2] - Sin[c/2])*(Cos[c/2
+ (d*x)/2] - Sin[c/2 + (d*x)/2])^3) + (Cos[c + d*x]^5*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c
+ d*x])*(3*A*Cos[c/2] + 13*B*Cos[c/2] - 3*A*Sin[c/2] - 11*B*Sin[c/2]))/(192*d*(B + A*Cos[c + d*x])*(Cos[c/2]
- Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^2) + (Cos[c + d*x]^5*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c +
d*x])^4*(A + B*Sec[c + d*x])*(3*A*Sin[(d*x)/2] + 5*B*Sin[(d*x)/2]))/(12*d*(B + A*Cos[c + d*x])*(Cos[c/2] - Si
n[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) + (B*Cos[c + d*x]^5*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x
])^4*(A + B*Sec[c + d*x])*Sin[(d*x)/2])/(96*d*(B + A*Cos[c + d*x])*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] +
Sin[c/2 + (d*x)/2])^3) + (Cos[c + d*x]^5*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x])*(-3
*A*Cos[c/2] - 13*B*Cos[c/2] - 3*A*Sin[c/2] - 11*B*Sin[c/2]))/(192*d*(B + A*Cos[c + d*x])*(Cos[c/2] + Sin[c/2])
*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^2) + (Cos[c + d*x]^5*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A
+ B*Sec[c + d*x])*(3*A*Sin[(d*x)/2] + 5*B*Sin[(d*x)/2]))/(12*d*(B + A*Cos[c + d*x])*(Cos[c/2] + Sin[c/2])*(Co
s[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]))

Maple [A] (verified)

Time = 3.15 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.28

method	result	size
parallelrisch	\(\frac {2 \left (-\frac {39 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (A +\frac {12 B}{13}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4}+\frac {39 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (A +\frac {12 B}{13}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4}+2 d \left (A +\frac {B}{4}\right ) x \cos \left (3 d x +3 c \right )+\left (A +2 B \right ) \sin \left (2 d x +2 c \right )+2 \left (A +\frac {5 B}{3}\right ) \sin \left (3 d x +3 c \right )+\frac {A \sin \left (4 d x +4 c \right )}{4}+6 d \left (A +\frac {B}{4}\right ) x \cos \left (d x +c \right )+2 \sin \left (d x +c \right ) \left (A +2 B \right )\right ) a^{4}}{d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\)	\(193\)
derivativedivides	\(\frac {a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B \,a^{4} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+4 a^{4} A \tan \left (d x +c \right )+4 B \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+6 a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 B \,a^{4} \tan \left (d x +c \right )+4 a^{4} A \left (d x +c \right )+4 B \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{4} A \sin \left (d x +c \right )+B \,a^{4} \left (d x +c \right )}{d}\)	\(199\)
default	\(\frac {a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B \,a^{4} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+4 a^{4} A \tan \left (d x +c \right )+4 B \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+6 a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 B \,a^{4} \tan \left (d x +c \right )+4 a^{4} A \left (d x +c \right )+4 B \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{4} A \sin \left (d x +c \right )+B \,a^{4} \left (d x +c \right )}{d}\)	\(199\)
risch	\(4 a^{4} A x +a^{4} x B -\frac {i a^{4} A \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i a^{4} A \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {i a^{4} \left (3 A \,{\mathrm e}^{5 i \left (d x +c \right )}+12 B \,{\mathrm e}^{5 i \left (d x +c \right )}-24 A \,{\mathrm e}^{4 i \left (d x +c \right )}-36 B \,{\mathrm e}^{4 i \left (d x +c \right )}-48 A \,{\mathrm e}^{2 i \left (d x +c \right )}-84 B \,{\mathrm e}^{2 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )} A -12 B \,{\mathrm e}^{i \left (d x +c \right )}-24 A -40 B \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {13 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 d}+\frac {6 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}-\frac {13 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 d}-\frac {6 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}\)	\(266\)
norman	\(\frac {\left (4 a^{4} A +B \,a^{4}\right ) x +\left (-12 a^{4} A -3 B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-12 a^{4} A -3 B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (4 a^{4} A +B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (8 a^{4} A +2 B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (8 a^{4} A +2 B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {a^{4} \left (11 A +18 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {5 a^{4} \left (A +2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}+\frac {2 a^{4} \left (5 A -4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {4 a^{4} \left (6 A +19 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}-\frac {4 a^{4} \left (18 A +19 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}-\frac {a^{4} \left (13 A +12 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{4} \left (13 A +12 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(356\)

[In]

int(cos(d*x+c)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

2*(-39/4*(1/3*cos(3*d*x+3*c)+cos(d*x+c))*(A+12/13*B)*ln(tan(1/2*d*x+1/2*c)-1)+39/4*(1/3*cos(3*d*x+3*c)+cos(d*x
+c))*(A+12/13*B)*ln(tan(1/2*d*x+1/2*c)+1)+2*d*(A+1/4*B)*x*cos(3*d*x+3*c)+(A+2*B)*sin(2*d*x+2*c)+2*(A+5/3*B)*si
n(3*d*x+3*c)+1/4*A*sin(4*d*x+4*c)+6*d*(A+1/4*B)*x*cos(d*x+c)+2*sin(d*x+c)*(A+2*B))*a^4/d/(cos(3*d*x+3*c)+3*cos
(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.05 \[ \int \cos (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {12 \, {\left (4 \, A + B\right )} a^{4} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (13 \, A + 12 \, B\right )} a^{4} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (13 \, A + 12 \, B\right )} a^{4} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (6 \, A a^{4} \cos \left (d x + c\right )^{3} + 8 \, {\left (3 \, A + 5 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} + 3 \, {\left (A + 4 \, B\right )} a^{4} \cos \left (d x + c\right ) + 2 \, B a^{4}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]

[In]

integrate(cos(d*x+c)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/12*(12*(4*A + B)*a^4*d*x*cos(d*x + c)^3 + 3*(13*A + 12*B)*a^4*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(13*A
+ 12*B)*a^4*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(6*A*a^4*cos(d*x + c)^3 + 8*(3*A + 5*B)*a^4*cos(d*x + c
)^2 + 3*(A + 4*B)*a^4*cos(d*x + c) + 2*B*a^4)*sin(d*x + c))/(d*cos(d*x + c)^3)

Sympy [F]

\[ \int \cos (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=a^{4} \left (\int A \cos {\left (c + d x \right )}\, dx + \int 4 A \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 6 A \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 4 A \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int A \cos {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int B \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 4 B \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 6 B \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int 4 B \cos {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int B \cos {\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}\, dx\right ) \]

[In]

integrate(cos(d*x+c)*(a+a*sec(d*x+c))**4*(A+B*sec(d*x+c)),x)

[Out]

a**4*(Integral(A*cos(c + d*x), x) + Integral(4*A*cos(c + d*x)*sec(c + d*x), x) + Integral(6*A*cos(c + d*x)*sec
(c + d*x)**2, x) + Integral(4*A*cos(c + d*x)*sec(c + d*x)**3, x) + Integral(A*cos(c + d*x)*sec(c + d*x)**4, x)
+ Integral(B*cos(c + d*x)*sec(c + d*x), x) + Integral(4*B*cos(c + d*x)*sec(c + d*x)**2, x) + Integral(6*B*cos
(c + d*x)*sec(c + d*x)**3, x) + Integral(4*B*cos(c + d*x)*sec(c + d*x)**4, x) + Integral(B*cos(c + d*x)*sec(c
+ d*x)**5, x))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.56 \[ \int \cos (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {48 \, {\left (d x + c\right )} A a^{4} + 4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{4} + 12 \, {\left (d x + c\right )} B a^{4} - 3 \, A a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, B a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, A a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, B a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a^{4} \sin \left (d x + c\right ) + 48 \, A a^{4} \tan \left (d x + c\right ) + 72 \, B a^{4} \tan \left (d x + c\right )}{12 \, d} \]

[In]

integrate(cos(d*x+c)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/12*(48*(d*x + c)*A*a^4 + 4*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^4 + 12*(d*x + c)*B*a^4 - 3*A*a^4*(2*sin(d*x
+ c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 12*B*a^4*(2*sin(d*x + c)/(sin(d*
x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 36*A*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x
+ c) - 1)) + 24*B*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*A*a^4*sin(d*x + c) + 48*A*a^4*tan(
d*x + c) + 72*B*a^4*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.50 \[ \int \cos (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {\frac {12 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + 6 \, {\left (4 \, A a^{4} + B a^{4}\right )} {\left (d x + c\right )} + 3 \, {\left (13 \, A a^{4} + 12 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (13 \, A a^{4} + 12 \, B a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (21 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 30 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 48 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 76 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 27 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 54 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

[In]

integrate(cos(d*x+c)*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/6*(12*A*a^4*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1) + 6*(4*A*a^4 + B*a^4)*(d*x + c) + 3*(13*A*a^4
+ 12*B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(13*A*a^4 + 12*B*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) -
2*(21*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 30*B*a^4*tan(1/2*d*x + 1/2*c)^5 - 48*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 76*B*
a^4*tan(1/2*d*x + 1/2*c)^3 + 27*A*a^4*tan(1/2*d*x + 1/2*c) + 54*B*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2
*c)^2 - 1)^3)/d

Mupad [B] (verification not implemented)

Time = 13.54 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.68 \[ \int \cos (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx=\frac {A\,a^4\,\sin \left (c+d\,x\right )}{d}+\frac {8\,A\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {13\,A\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,B\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {12\,B\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,A\,a^4\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {A\,a^4\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {20\,B\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {2\,B\,a^4\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^2}+\frac {B\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3} \]

[In]

int(cos(c + d*x)*(A + B/cos(c + d*x))*(a + a/cos(c + d*x))^4,x)

[Out]

(A*a^4*sin(c + d*x))/d + (8*A*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (13*A*a^4*atanh(sin(c/2 + (
d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*B*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (12*B*a^4*atanh(sin
(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (4*A*a^4*sin(c + d*x))/(d*cos(c + d*x)) + (A*a^4*sin(c + d*x))/(2*d*c
os(c + d*x)^2) + (20*B*a^4*sin(c + d*x))/(3*d*cos(c + d*x)) + (2*B*a^4*sin(c + d*x))/(d*cos(c + d*x)^2) + (B*a
^4*sin(c + d*x))/(3*d*cos(c + d*x)^3)