\(\int \frac {\sec ^4(c+d x) (A+C \sec ^2(c+d x))}{a+a \sec (c+d x)} \, dx\) [121]

Optimal result

Integrand size = 33, antiderivative size = 165 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {3 (4 A+5 C) \text {arctanh}(\sin (c+d x))}{8 a d}-\frac {(3 A+4 C) \tan (c+d x)}{a d}+\frac {3 (4 A+5 C) \sec (c+d x) \tan (c+d x)}{8 a d}+\frac {(4 A+5 C) \sec ^3(c+d x) \tan (c+d x)}{4 a d}-\frac {(A+C) \sec ^4(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}-\frac {(3 A+4 C) \tan ^3(c+d x)}{3 a d} \]

[Out]

3/8*(4*A+5*C)*arctanh(sin(d*x+c))/a/d-(3*A+4*C)*tan(d*x+c)/a/d+3/8*(4*A+5*C)*sec(d*x+c)*tan(d*x+c)/a/d+1/4*(4*
A+5*C)*sec(d*x+c)^3*tan(d*x+c)/a/d-(A+C)*sec(d*x+c)^4*tan(d*x+c)/d/(a+a*sec(d*x+c))-1/3*(3*A+4*C)*tan(d*x+c)^3
/a/d

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 165, 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 = {4170, 3872, 3852, 3853, 3855} \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {3 (4 A+5 C) \text {arctanh}(\sin (c+d x))}{8 a d}-\frac {(3 A+4 C) \tan ^3(c+d x)}{3 a d}-\frac {(3 A+4 C) \tan (c+d x)}{a d}-\frac {(A+C) \tan (c+d x) \sec ^4(c+d x)}{d (a \sec (c+d x)+a)}+\frac {(4 A+5 C) \tan (c+d x) \sec ^3(c+d x)}{4 a d}+\frac {3 (4 A+5 C) \tan (c+d x) \sec (c+d x)}{8 a d} \]

[In]

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

[Out]

(3*(4*A + 5*C)*ArcTanh[Sin[c + d*x]])/(8*a*d) - ((3*A + 4*C)*Tan[c + d*x])/(a*d) + (3*(4*A + 5*C)*Sec[c + d*x]
*Tan[c + d*x])/(8*a*d) + ((4*A + 5*C)*Sec[c + d*x]^3*Tan[c + d*x])/(4*a*d) - ((A + C)*Sec[c + d*x]^4*Tan[c + d
*x])/(d*(a + a*Sec[c + d*x])) - ((3*A + 4*C)*Tan[c + d*x]^3)/(3*a*d)

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3853

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] + Dist[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]

Rule 3855

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

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4170

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

Rubi steps \begin{align*} \text {integral}& = -\frac {(A+C) \sec ^4(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}-\frac {\int \sec ^4(c+d x) (a (3 A+4 C)-a (4 A+5 C) \sec (c+d x)) \, dx}{a^2} \\ & = -\frac {(A+C) \sec ^4(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}-\frac {(3 A+4 C) \int \sec ^4(c+d x) \, dx}{a}+\frac {(4 A+5 C) \int \sec ^5(c+d x) \, dx}{a} \\ & = \frac {(4 A+5 C) \sec ^3(c+d x) \tan (c+d x)}{4 a d}-\frac {(A+C) \sec ^4(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}+\frac {(3 (4 A+5 C)) \int \sec ^3(c+d x) \, dx}{4 a}+\frac {(3 A+4 C) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{a d} \\ & = -\frac {(3 A+4 C) \tan (c+d x)}{a d}+\frac {3 (4 A+5 C) \sec (c+d x) \tan (c+d x)}{8 a d}+\frac {(4 A+5 C) \sec ^3(c+d x) \tan (c+d x)}{4 a d}-\frac {(A+C) \sec ^4(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}-\frac {(3 A+4 C) \tan ^3(c+d x)}{3 a d}+\frac {(3 (4 A+5 C)) \int \sec (c+d x) \, dx}{8 a} \\ & = \frac {3 (4 A+5 C) \text {arctanh}(\sin (c+d x))}{8 a d}-\frac {(3 A+4 C) \tan (c+d x)}{a d}+\frac {3 (4 A+5 C) \sec (c+d x) \tan (c+d x)}{8 a d}+\frac {(4 A+5 C) \sec ^3(c+d x) \tan (c+d x)}{4 a d}-\frac {(A+C) \sec ^4(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}-\frac {(3 A+4 C) \tan ^3(c+d x)}{3 a d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(792\) vs. \(2(165)=330\).

Time = 7.74 (sec) , antiderivative size = 792, normalized size of antiderivative = 4.80 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=-\frac {3 (4 A+5 C) \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \cos (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 ) \left (A+C \sec ^2(c+d x)\right )}{2 d (A+2 C+A \cos (2 c+2 d x)) (a+a \sec (c+d x))}+\frac {3 (4 A+5 C) \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \cos (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 ) \left (A+C \sec ^2(c+d x)\right )}{2 d (A+2 C+A \cos (2 c+2 d x)) (a+a \sec (c+d x))}+\frac {\cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec (c) \sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right ) \left (-60 A \sin \left (\frac {d x}{2}\right )-75 C \sin \left (\frac {d x}{2}\right )-60 A \sin \left (\frac {3 d x}{2}\right )-91 C \sin \left (\frac {3 d x}{2}\right )+204 A \sin \left (c-\frac {d x}{2}\right )+219 C \sin \left (c-\frac {d x}{2}\right )-60 A \sin \left (c+\frac {d x}{2}\right )+21 C \sin \left (c+\frac {d x}{2}\right )+84 A \sin \left (2 c+\frac {d x}{2}\right )+165 C \sin \left (2 c+\frac {d x}{2}\right )+36 A \sin \left (c+\frac {3 d x}{2}\right )+5 C \sin \left (c+\frac {3 d x}{2}\right )+36 A \sin \left (2 c+\frac {3 d x}{2}\right )+69 C \sin \left (2 c+\frac {3 d x}{2}\right )+132 A \sin \left (3 c+\frac {3 d x}{2}\right )+165 C \sin \left (3 c+\frac {3 d x}{2}\right )-156 A \sin \left (c+\frac {5 d x}{2}\right )-211 C \sin \left (c+\frac {5 d x}{2}\right )-60 A \sin \left (2 c+\frac {5 d x}{2}\right )-115 C \sin \left (2 c+\frac {5 d x}{2}\right )-60 A \sin \left (3 c+\frac {5 d x}{2}\right )-51 C \sin \left (3 c+\frac {5 d x}{2}\right )+36 A \sin \left (4 c+\frac {5 d x}{2}\right )+45 C \sin \left (4 c+\frac {5 d x}{2}\right )-12 A \sin \left (2 c+\frac {7 d x}{2}\right )-19 C \sin \left (2 c+\frac {7 d x}{2}\right )+12 A \sin \left (3 c+\frac {7 d x}{2}\right )+5 C \sin \left (3 c+\frac {7 d x}{2}\right )+12 A \sin \left (4 c+\frac {7 d x}{2}\right )+21 C \sin \left (4 c+\frac {7 d x}{2}\right )+36 A \sin \left (5 c+\frac {7 d x}{2}\right )+45 C \sin \left (5 c+\frac {7 d x}{2}\right )-48 A \sin \left (3 c+\frac {9 d x}{2}\right )-64 C \sin \left (3 c+\frac {9 d x}{2}\right )-24 A \sin \left (4 c+\frac {9 d x}{2}\right )-40 C \sin \left (4 c+\frac {9 d x}{2}\right )-24 A \sin \left (5 c+\frac {9 d x}{2}\right )-24 C \sin \left (5 c+\frac {9 d x}{2}\right )\right )}{192 d (A+2 C+A \cos (2 c+2 d x)) (a+a \sec (c+d x))} \]

[In]

Integrate[(Sec[c + d*x]^4*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x]),x]

[Out]

(-3*(4*A + 5*C)*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*(A + C*Sec[c +
d*x]^2))/(2*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) + (3*(4*A + 5*C)*Cos[c/2 + (d*x)/2]^2*Cos[c
+ d*x]*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*(A + C*Sec[c + d*x]^2))/(2*d*(A + 2*C + A*Cos[2*c + 2*d*x
])*(a + a*Sec[c + d*x])) + (Cos[c/2 + (d*x)/2]*Sec[c/2]*Sec[c]*Sec[c + d*x]^3*(A + C*Sec[c + d*x]^2)*(-60*A*Si
n[(d*x)/2] - 75*C*Sin[(d*x)/2] - 60*A*Sin[(3*d*x)/2] - 91*C*Sin[(3*d*x)/2] + 204*A*Sin[c - (d*x)/2] + 219*C*Si
n[c - (d*x)/2] - 60*A*Sin[c + (d*x)/2] + 21*C*Sin[c + (d*x)/2] + 84*A*Sin[2*c + (d*x)/2] + 165*C*Sin[2*c + (d*
x)/2] + 36*A*Sin[c + (3*d*x)/2] + 5*C*Sin[c + (3*d*x)/2] + 36*A*Sin[2*c + (3*d*x)/2] + 69*C*Sin[2*c + (3*d*x)/
2] + 132*A*Sin[3*c + (3*d*x)/2] + 165*C*Sin[3*c + (3*d*x)/2] - 156*A*Sin[c + (5*d*x)/2] - 211*C*Sin[c + (5*d*x
)/2] - 60*A*Sin[2*c + (5*d*x)/2] - 115*C*Sin[2*c + (5*d*x)/2] - 60*A*Sin[3*c + (5*d*x)/2] - 51*C*Sin[3*c + (5*
d*x)/2] + 36*A*Sin[4*c + (5*d*x)/2] + 45*C*Sin[4*c + (5*d*x)/2] - 12*A*Sin[2*c + (7*d*x)/2] - 19*C*Sin[2*c + (
7*d*x)/2] + 12*A*Sin[3*c + (7*d*x)/2] + 5*C*Sin[3*c + (7*d*x)/2] + 12*A*Sin[4*c + (7*d*x)/2] + 21*C*Sin[4*c +
(7*d*x)/2] + 36*A*Sin[5*c + (7*d*x)/2] + 45*C*Sin[5*c + (7*d*x)/2] - 48*A*Sin[3*c + (9*d*x)/2] - 64*C*Sin[3*c
+ (9*d*x)/2] - 24*A*Sin[4*c + (9*d*x)/2] - 40*C*Sin[4*c + (9*d*x)/2] - 24*A*Sin[5*c + (9*d*x)/2] - 24*C*Sin[5*
c + (9*d*x)/2]))/(192*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x]))

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.19

method	result	size
parallelrisch	\(\frac {-3 \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right ) \left (A +\frac {5 C}{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+3 \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right ) \left (A +\frac {5 C}{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-4 \left (\left (3 A +\frac {17 C}{4}\right ) \cos \left (2 d x +2 c \right )+\left (A +\frac {4 C}{3}\right ) \cos \left (4 d x +4 c \right )+\frac {\left (A +\frac {19 C}{12}\right ) \cos \left (3 d x +3 c \right )}{2}+\frac {\left (3 A +\frac {65 C}{12}\right ) \cos \left (d x +c \right )}{2}+2 A +\frac {23 C}{12}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(197\)
derivativedivides	\(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\frac {C}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {5 C}{6 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\left (-\frac {15 C}{8}-\frac {3 A}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {-\frac {15 C}{4}-A}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-\frac {25 C}{8}-\frac {3 A}{2}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}-\frac {C}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {5 C}{6 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {A +\frac {15 C}{4}}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\left (\frac {15 C}{8}+\frac {3 A}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {-\frac {25 C}{8}-\frac {3 A}{2}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{d a}\)	\(223\)
default	\(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\frac {C}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {5 C}{6 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\left (-\frac {15 C}{8}-\frac {3 A}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {-\frac {15 C}{4}-A}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-\frac {25 C}{8}-\frac {3 A}{2}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}-\frac {C}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {5 C}{6 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {A +\frac {15 C}{4}}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\left (\frac {15 C}{8}+\frac {3 A}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {-\frac {25 C}{8}-\frac {3 A}{2}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{d a}\)	\(223\)
norman	\(\frac {-\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{a d}+\frac {\left (8 A +11 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {5 \left (24 A +31 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 a d}+\frac {\left (32 A +45 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 a d}+\frac {2 \left (33 A +43 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 a d}-\frac {\left (66 A +95 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}-\frac {3 \left (4 A +5 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 a d}+\frac {3 \left (4 A +5 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 a d}\)	\(223\)
risch	\(-\frac {i \left (36 A \,{\mathrm e}^{8 i \left (d x +c \right )}+45 C \,{\mathrm e}^{8 i \left (d x +c \right )}+36 A \,{\mathrm e}^{7 i \left (d x +c \right )}+45 C \,{\mathrm e}^{7 i \left (d x +c \right )}+132 A \,{\mathrm e}^{6 i \left (d x +c \right )}+165 C \,{\mathrm e}^{6 i \left (d x +c \right )}+84 A \,{\mathrm e}^{5 i \left (d x +c \right )}+165 C \,{\mathrm e}^{5 i \left (d x +c \right )}+204 A \,{\mathrm e}^{4 i \left (d x +c \right )}+219 C \,{\mathrm e}^{4 i \left (d x +c \right )}+60 A \,{\mathrm e}^{3 i \left (d x +c \right )}+91 C \,{\mathrm e}^{3 i \left (d x +c \right )}+156 A \,{\mathrm e}^{2 i \left (d x +c \right )}+211 C \,{\mathrm e}^{2 i \left (d x +c \right )}+12 A \,{\mathrm e}^{i \left (d x +c \right )}+19 C \,{\mathrm e}^{i \left (d x +c \right )}+48 A +64 C \right )}{12 d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 a d}+\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 a d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 a d}-\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 a d}\)	\(324\)

[In]

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

[Out]

1/2*(-3*(cos(4*d*x+4*c)+4*cos(2*d*x+2*c)+3)*(A+5/4*C)*ln(tan(1/2*d*x+1/2*c)-1)+3*(cos(4*d*x+4*c)+4*cos(2*d*x+2
*c)+3)*(A+5/4*C)*ln(tan(1/2*d*x+1/2*c)+1)-4*((3*A+17/4*C)*cos(2*d*x+2*c)+(A+4/3*C)*cos(4*d*x+4*c)+1/2*(A+19/12
*C)*cos(3*d*x+3*c)+1/2*(3*A+65/12*C)*cos(d*x+c)+2*A+23/12*C)*tan(1/2*d*x+1/2*c))/a/d/(cos(4*d*x+4*c)+4*cos(2*d
*x+2*c)+3)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.15 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {9 \, {\left ({\left (4 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{5} + {\left (4 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, {\left ({\left (4 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{5} + {\left (4 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (16 \, {\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (12 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{3} - {\left (12 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, C \cos \left (d x + c\right ) - 6 \, C\right )} \sin \left (d x + c\right )}{48 \, {\left (a d \cos \left (d x + c\right )^{5} + a d \cos \left (d x + c\right )^{4}\right )}} \]

[In]

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

[Out]

1/48*(9*((4*A + 5*C)*cos(d*x + c)^5 + (4*A + 5*C)*cos(d*x + c)^4)*log(sin(d*x + c) + 1) - 9*((4*A + 5*C)*cos(d
*x + c)^5 + (4*A + 5*C)*cos(d*x + c)^4)*log(-sin(d*x + c) + 1) - 2*(16*(3*A + 4*C)*cos(d*x + c)^4 + (12*A + 19
*C)*cos(d*x + c)^3 - (12*A + 13*C)*cos(d*x + c)^2 + 2*C*cos(d*x + c) - 6*C)*sin(d*x + c))/(a*d*cos(d*x + c)^5
+ a*d*cos(d*x + c)^4)

Sympy [F]

\[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {A \sec ^{4}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{6}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \]

[In]

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

[Out]

(Integral(A*sec(c + d*x)**4/(sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)**6/(sec(c + d*x) + 1), x))/a

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 408 vs. \(2 (157) = 314\).

Time = 0.21 (sec) , antiderivative size = 408, normalized size of antiderivative = 2.47 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=-\frac {C {\left (\frac {2 \, {\left (\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {109 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {115 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {75 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a - \frac {4 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {4 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} - \frac {45 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac {45 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac {24 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + 12 \, 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 )}}{24 \, d} \]

[In]

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

[Out]

-1/24*(C*(2*(21*sin(d*x + c)/(cos(d*x + c) + 1) - 109*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 115*sin(d*x + c)^5
/(cos(d*x + c) + 1)^5 - 75*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/(a - 4*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 +
6*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 4*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a*sin(d*x + c)^8/(cos(d*x
+ c) + 1)^8) - 45*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a + 45*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a
+ 24*sin(d*x + c)/(a*(cos(d*x + c) + 1))) + 12*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))))/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.29 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\frac {9 \, {\left (4 \, A + 5 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {9 \, {\left (4 \, A + 5 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac {24 \, {\left (A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a} + \frac {2 \, {\left (36 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 75 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 84 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 115 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 60 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 109 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 21 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4} a}}{24 \, d} \]

[In]

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

[Out]

1/24*(9*(4*A + 5*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a - 9*(4*A + 5*C)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a
- 24*(A*tan(1/2*d*x + 1/2*c) + C*tan(1/2*d*x + 1/2*c))/a + 2*(36*A*tan(1/2*d*x + 1/2*c)^7 + 75*C*tan(1/2*d*x +
1/2*c)^7 - 84*A*tan(1/2*d*x + 1/2*c)^5 - 115*C*tan(1/2*d*x + 1/2*c)^5 + 60*A*tan(1/2*d*x + 1/2*c)^3 + 109*C*t
an(1/2*d*x + 1/2*c)^3 - 12*A*tan(1/2*d*x + 1/2*c) - 21*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^4
*a))/d

Mupad [B] (verification not implemented)

Time = 16.28 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.12 \[ \int \frac {\sec ^4(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\left (3\,A+\frac {25\,C}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (-7\,A-\frac {115\,C}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (5\,A+\frac {109\,C}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (-A-\frac {7\,C}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A+C\right )}{a\,d}+\frac {3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (4\,A+5\,C\right )}{4\,a\,d} \]

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^7*(3*A + (25*C)/4) + tan(c/2 + (d*x)/2)^3*(5*A + (109*C)/12) - tan(c/2 + (d*x)/2)^5*(7*A +
(115*C)/12) - tan(c/2 + (d*x)/2)*(A + (7*C)/4))/(d*(a - 4*a*tan(c/2 + (d*x)/2)^2 + 6*a*tan(c/2 + (d*x)/2)^4 -
4*a*tan(c/2 + (d*x)/2)^6 + a*tan(c/2 + (d*x)/2)^8)) - (tan(c/2 + (d*x)/2)*(A + C))/(a*d) + (3*atanh(tan(c/2 +
(d*x)/2))*(4*A + 5*C))/(4*a*d)