\(\int \cos ^3(c+d x) (a+a \sec (c+d x))^4 (A+C \sec ^2(c+d x)) \, dx\) [116]

Optimal result

Integrand size = 33, antiderivative size = 198 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=2 a^4 (3 A+2 C) x+\frac {a^4 (2 A+13 C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {5 a^4 (2 A-C) \sin (c+d x)}{2 d}+\frac {2 a A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac {(2 A-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}-\frac {(4 A-9 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {4172, 4102, 4103, 4081, 3855} \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^4 (2 A+13 C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {5 a^4 (2 A-C) \sin (c+d x)}{2 d}-\frac {(4 A-9 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{3 d}+2 a^4 x (3 A+2 C)-\frac {(2 A-C) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^4}{3 d}+\frac {2 a A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{3 d} \]

[In]

Int[Cos[c + d*x]^3*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2),x]

[Out]

2*a^4*(3*A + 2*C)*x + (a^4*(2*A + 13*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (5*a^4*(2*A - C)*Sin[c + d*x])/(2*d) +
(2*a*A*Cos[c + d*x]*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(3*d) + (A*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^4*Sin[
c + d*x])/(3*d) - ((2*A - C)*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) - ((4*A - 9*C)*(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 4102

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]
- Dist[b/(a*d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[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]

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]

Rule 4172

Int[((A_.) + 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] - Dis
t[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2
^(-1)] || EqQ[m + n + 1, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{3 d}+\frac {\int \cos ^2(c+d x) (a+a \sec (c+d x))^4 (4 a A-a (2 A-3 C) \sec (c+d x)) \, dx}{3 a} \\ & = \frac {2 a A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{3 d}+\frac {\int \cos (c+d x) (a+a \sec (c+d x))^3 \left (2 a^2 (8 A+3 C)-6 a^2 (2 A-C) \sec (c+d x)\right ) \, dx}{6 a} \\ & = \frac {2 a A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac {(2 A-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac {\int \cos (c+d x) (a+a \sec (c+d x))^2 \left (2 a^3 (22 A+3 C)-4 a^3 (4 A-9 C) \sec (c+d x)\right ) \, dx}{12 a} \\ & = \frac {2 a A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac {(2 A-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}-\frac {(4 A-9 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{3 d}+\frac {\int \cos (c+d x) (a+a \sec (c+d x)) \left (30 a^4 (2 A-C)+6 a^4 (2 A+13 C) \sec (c+d x)\right ) \, dx}{12 a} \\ & = \frac {5 a^4 (2 A-C) \sin (c+d x)}{2 d}+\frac {2 a A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac {(2 A-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}-\frac {(4 A-9 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{3 d}-\frac {\int \left (-24 a^5 (3 A+2 C)-6 a^5 (2 A+13 C) \sec (c+d x)\right ) \, dx}{12 a} \\ & = 2 a^4 (3 A+2 C) x+\frac {5 a^4 (2 A-C) \sin (c+d x)}{2 d}+\frac {2 a A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac {(2 A-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}-\frac {(4 A-9 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{3 d}+\frac {1}{2} \left (a^4 (2 A+13 C)\right ) \int \sec (c+d x) \, dx \\ & = 2 a^4 (3 A+2 C) x+\frac {a^4 (2 A+13 C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {5 a^4 (2 A-C) \sin (c+d x)}{2 d}+\frac {2 a A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac {(2 A-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}-\frac {(4 A-9 C) \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. \(1250\) vs. \(2(198)=396\).

Time = 12.31 (sec) , antiderivative size = 1250, normalized size of antiderivative = 6.31 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {(3 A+2 C) x \cos ^6(c+d x) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right )}{4 (A+2 C+A \cos (2 c+2 d x))}+\frac {(-2 A-13 C) \cos ^6(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 \left (A+C \sec ^2(c+d x)\right )}{16 d (A+2 C+A \cos (2 c+2 d x))}+\frac {(2 A+13 C) \cos ^6(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 \left (A+C \sec ^2(c+d x)\right )}{16 d (A+2 C+A \cos (2 c+2 d x))}+\frac {(27 A+4 C) \cos (d x) \cos ^6(c+d x) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \sin (c)}{32 d (A+2 C+A \cos (2 c+2 d x))}+\frac {A \cos (2 d x) \cos ^6(c+d x) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \sin (2 c)}{8 d (A+2 C+A \cos (2 c+2 d x))}+\frac {A \cos (3 d x) \cos ^6(c+d x) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \sin (3 c)}{96 d (A+2 C+A \cos (2 c+2 d x))}+\frac {(27 A+4 C) \cos (c) \cos ^6(c+d x) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \sin (d x)}{32 d (A+2 C+A \cos (2 c+2 d x))}+\frac {A \cos (2 c) \cos ^6(c+d x) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \sin (2 d x)}{8 d (A+2 C+A \cos (2 c+2 d x))}+\frac {A \cos (3 c) \cos ^6(c+d x) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \sin (3 d x)}{96 d (A+2 C+A \cos (2 c+2 d x))}+\frac {C \cos ^6(c+d x) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right )}{32 d (A+2 C+A \cos (2 c+2 d x)) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}+\frac {C \cos ^6(c+d x) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \sin \left (\frac {d x}{2}\right )}{2 d (A+2 C+A \cos (2 c+2 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 {C \cos ^6(c+d x) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right )}{32 d (A+2 C+A \cos (2 c+2 d x)) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}+\frac {C \cos ^6(c+d x) \sec ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \sin \left (\frac {d x}{2}\right )}{2 d (A+2 C+A \cos (2 c+2 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]^3*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2),x]

[Out]

((3*A + 2*C)*x*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2))/(4*(A + 2*C
+ A*Cos[2*c + 2*d*x])) + ((-2*A - 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 + C*Sec[c + d*x]^2))/(16*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + ((2*A + 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 +
C*Sec[c + d*x]^2))/(16*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + ((27*A + 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 + C*Sec[c + d*x]^2)*Sin[c])/(32*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + (A*Cos[
2*d*x]*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2)*Sin[2*c])/(8*d*(A + 2
*C + 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 + C*Se
c[c + d*x]^2)*Sin[3*c])/(96*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + ((27*A + 4*C)*Cos[c]*Cos[c + d*x]^6*Sec[c/2 +
(d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2)*Sin[d*x])/(32*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + (A*
Cos[2*c]*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2)*Sin[2*d*x])/(8*d*(A
+ 2*C + A*Cos[2*c + 2*d*x])) + (A*Cos[3*c]*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + C*
Sec[c + d*x]^2)*Sin[3*d*x])/(96*d*(A + 2*C + A*Cos[2*c + 2*d*x])) + (C*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a
+ a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2))/(32*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(Cos[c/2 + (d*x)/2] - Sin[c/2
+ (d*x)/2])^2) + (C*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2)*Sin[(d*
x)/2])/(2*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) -
(C*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2))/(32*d*(A + 2*C + A*Cos[2
*c + 2*d*x])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^2) + (C*Cos[c + d*x]^6*Sec[c/2 + (d*x)/2]^8*(a + a*Sec[
c + d*x])^4*(A + C*Sec[c + d*x]^2)*Sin[(d*x)/2])/(2*d*(A + 2*C + A*Cos[2*c + 2*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 = 0.62 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.91

method	result	size
parallelrisch	\(\frac {83 \left (-\frac {24 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (\frac {13 C}{2}+A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{83}+\frac {24 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (\frac {13 C}{2}+A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{83}+\frac {144 x d \left (A +\frac {2 C}{3}\right ) \cos \left (2 d x +2 c \right )}{83}+\frac {24 \left (A +4 C \right ) \sin \left (2 d x +2 c \right )}{83}+\left (A +\frac {12 C}{83}\right ) \sin \left (3 d x +3 c \right )+\frac {12 A \sin \left (4 d x +4 c \right )}{83}+\frac {A \sin \left (5 d x +5 c \right )}{83}+\frac {2 \left (41 A +18 C \right ) \sin \left (d x +c \right )}{83}+\frac {144 x d \left (A +\frac {2 C}{3}\right )}{83}\right ) a^{4}}{24 d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(180\)
derivativedivides	\(\frac {\frac {a^{4} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a^{4} C \sin \left (d x +c \right )+4 a^{4} A \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{4} C \left (d x +c \right )+6 a^{4} A \sin \left (d x +c \right )+6 a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{4} A \left (d x +c \right )+4 a^{4} C \tan \left (d x +c \right )+a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{4} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(188\)
default	\(\frac {\frac {a^{4} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a^{4} C \sin \left (d x +c \right )+4 a^{4} A \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{4} C \left (d x +c \right )+6 a^{4} A \sin \left (d x +c \right )+6 a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{4} A \left (d x +c \right )+4 a^{4} C \tan \left (d x +c \right )+a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{4} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(188\)
risch	\(6 a^{4} A x +4 a^{4} x C -\frac {i a^{4} A \,{\mathrm e}^{3 i \left (d x +c \right )}}{24 d}-\frac {i a^{4} A \,{\mathrm e}^{2 i \left (d x +c \right )}}{2 d}-\frac {27 i a^{4} A \,{\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{4} C}{2 d}+\frac {27 i a^{4} A \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{4} C}{2 d}+\frac {i a^{4} A \,{\mathrm e}^{-2 i \left (d x +c \right )}}{2 d}+\frac {i a^{4} A \,{\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}-\frac {i a^{4} C \left ({\mathrm e}^{3 i \left (d x +c \right )}-8 \,{\mathrm e}^{2 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}-8\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {13 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}-\frac {a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {13 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}\)	\(311\)
norman	\(\frac {\left (-6 a^{4} A -4 a^{4} C \right ) x +\left (-36 a^{4} A -24 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (-12 a^{4} A -8 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (-12 a^{4} A -8 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}+\left (6 a^{4} A +4 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{16}+\left (12 a^{4} A +8 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (12 a^{4} A +8 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (36 a^{4} A +24 a^{4} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\frac {5 a^{4} \left (2 A -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{d}-\frac {a^{4} \left (18 A +11 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {a^{4} \left (26 A -69 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}-\frac {5 a^{4} \left (38 A -15 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}-\frac {a^{4} \left (70 A +93 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}-\frac {a^{4} \left (74 A -9 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{3 d}+\frac {a^{4} \left (190 A -51 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}+\frac {a^{4} \left (194 A +39 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}-\frac {a^{4} \left (2 A +13 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{4} \left (2 A +13 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(491\)

[In]

int(cos(d*x+c)^3*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

83/24*(-24/83*(1+cos(2*d*x+2*c))*(13/2*C+A)*ln(tan(1/2*d*x+1/2*c)-1)+24/83*(1+cos(2*d*x+2*c))*(13/2*C+A)*ln(ta
n(1/2*d*x+1/2*c)+1)+144/83*x*d*(A+2/3*C)*cos(2*d*x+2*c)+24/83*(A+4*C)*sin(2*d*x+2*c)+(A+12/83*C)*sin(3*d*x+3*c
)+12/83*A*sin(4*d*x+4*c)+1/83*A*sin(5*d*x+5*c)+2/83*(41*A+18*C)*sin(d*x+c)+144/83*x*d*(A+2/3*C))*a^4/d/(1+cos(
2*d*x+2*c))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.86 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {24 \, {\left (3 \, A + 2 \, C\right )} a^{4} d x \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, A + 13 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, A + 13 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, A a^{4} \cos \left (d x + c\right )^{4} + 12 \, A a^{4} \cos \left (d x + c\right )^{3} + 2 \, {\left (20 \, A + 3 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 24 \, C a^{4} \cos \left (d x + c\right ) + 3 \, C a^{4}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{2}} \]

[In]

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/12*(24*(3*A + 2*C)*a^4*d*x*cos(d*x + c)^2 + 3*(2*A + 13*C)*a^4*cos(d*x + c)^2*log(sin(d*x + c) + 1) - 3*(2*A
+ 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 + 12*A*a^4*cos(d*x + c)^3 + 2*(
20*A + 3*C)*a^4*cos(d*x + c)^2 + 24*C*a^4*cos(d*x + c) + 3*C*a^4)*sin(d*x + c))/(d*cos(d*x + c)^2)

Sympy [F(-1)]

Timed out. \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**3*(a+a*sec(d*x+c))**4*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.07 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} - 12 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 48 \, {\left (d x + c\right )} A a^{4} - 48 \, {\left (d x + c\right )} C a^{4} + 3 \, C a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 6 \, A a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, C a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 72 \, A a^{4} \sin \left (d x + c\right ) - 12 \, C a^{4} \sin \left (d x + c\right ) - 48 \, C a^{4} \tan \left (d x + c\right )}{12 \, d} \]

[In]

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

-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 - 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)) - 36*C*a^4*(log(sin(d*x + c) + 1) - log(s
in(d*x + c) - 1)) - 72*A*a^4*sin(d*x + c) - 12*C*a^4*sin(d*x + c) - 48*C*a^4*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.25 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {12 \, {\left (3 \, A a^{4} + 2 \, C a^{4}\right )} {\left (d x + c\right )} + 3 \, {\left (2 \, A a^{4} + 13 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (2 \, A a^{4} + 13 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {6 \, {\left (7 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}} + \frac {4 \, {\left (15 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 38 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 27 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

[In]

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/6*(12*(3*A*a^4 + 2*C*a^4)*(d*x + c) + 3*(2*A*a^4 + 13*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(2*A*a^4
+ 13*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 6*(7*C*a^4*tan(1/2*d*x + 1/2*c)^3 - 9*C*a^4*tan(1/2*d*x + 1/
2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2 + 4*(15*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^4*tan(1/2*d*x + 1/2*c)^5 + 3
8*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 6*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 27*A*a^4*tan(1/2*d*x + 1/2*c) + 3*C*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 = 15.66 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.23 \[ \int \cos ^3(c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {20\,A\,a^4\,\sin \left (c+d\,x\right )}{3\,d}+\frac {C\,a^4\,\sin \left (c+d\,x\right )}{d}+\frac {12\,A\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,A\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {8\,C\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {13\,C\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a^4\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d}+\frac {4\,C\,a^4\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {C\,a^4\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {2\,A\,a^4\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{d} \]

[In]

int(cos(c + d*x)^3*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^4,x)

[Out]

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