\(\int \frac {\sec ^3(c+d x)}{(a+a \cos (c+d x))^4} \, dx\) [81]

Optimal result

Integrand size = 21, antiderivative size = 185 \[ \int \frac {\sec ^3(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {21 \text {arctanh}(\sin (c+d x))}{2 a^4 d}-\frac {576 \tan (c+d x)}{35 a^4 d}+\frac {21 \sec (c+d x) \tan (c+d x)}{2 a^4 d}-\frac {43 \sec (c+d x) \tan (c+d x)}{35 a^4 d (1+\cos (c+d x))^2}-\frac {288 \sec (c+d x) \tan (c+d x)}{35 a^4 d (1+\cos (c+d x))}-\frac {\sec (c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 \sec (c+d x) \tan (c+d x)}{5 a d (a+a \cos (c+d x))^3} \]

[Out]

21/2*arctanh(sin(d*x+c))/a^4/d-576/35*tan(d*x+c)/a^4/d+21/2*sec(d*x+c)*tan(d*x+c)/a^4/d-43/35*sec(d*x+c)*tan(d
*x+c)/a^4/d/(1+cos(d*x+c))^2-288/35*sec(d*x+c)*tan(d*x+c)/a^4/d/(1+cos(d*x+c))-1/7*sec(d*x+c)*tan(d*x+c)/d/(a+
a*cos(d*x+c))^4-2/5*sec(d*x+c)*tan(d*x+c)/a/d/(a+a*cos(d*x+c))^3

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2845, 3057, 2827, 3853, 3855, 3852, 8} \[ \int \frac {\sec ^3(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {21 \text {arctanh}(\sin (c+d x))}{2 a^4 d}-\frac {576 \tan (c+d x)}{35 a^4 d}+\frac {21 \tan (c+d x) \sec (c+d x)}{2 a^4 d}-\frac {288 \tan (c+d x) \sec (c+d x)}{35 a^4 d (\cos (c+d x)+1)}-\frac {43 \tan (c+d x) \sec (c+d x)}{35 a^4 d (\cos (c+d x)+1)^2}-\frac {2 \tan (c+d x) \sec (c+d x)}{5 a d (a \cos (c+d x)+a)^3}-\frac {\tan (c+d x) \sec (c+d x)}{7 d (a \cos (c+d x)+a)^4} \]

[In]

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

[Out]

(21*ArcTanh[Sin[c + d*x]])/(2*a^4*d) - (576*Tan[c + d*x])/(35*a^4*d) + (21*Sec[c + d*x]*Tan[c + d*x])/(2*a^4*d
) - (43*Sec[c + d*x]*Tan[c + d*x])/(35*a^4*d*(1 + Cos[c + d*x])^2) - (288*Sec[c + d*x]*Tan[c + d*x])/(35*a^4*d
*(1 + Cos[c + d*x])) - (Sec[c + d*x]*Tan[c + d*x])/(7*d*(a + a*Cos[c + d*x])^4) - (2*Sec[c + d*x]*Tan[c + d*x]
)/(5*a*d*(a + a*Cos[c + d*x])^3)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2845

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[b^2*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] + Dis
t[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*c*(m + 1) - a*d*
(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d,
0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] &&  !GtQ[n, 0] && (IntegersQ[2*m, 2*n] || (IntegerQ
[m] && EqQ[c, 0]))

Rule 3057

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[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] + Dist[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])

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\sec (c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {\int \frac {(9 a-5 a \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx}{7 a^2} \\ & = -\frac {\sec (c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 \sec (c+d x) \tan (c+d x)}{5 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {\left (73 a^2-56 a^2 \cos (c+d x)\right ) \sec ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{35 a^4} \\ & = -\frac {43 \sec (c+d x) \tan (c+d x)}{35 a^4 d (1+\cos (c+d x))^2}-\frac {\sec (c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 \sec (c+d x) \tan (c+d x)}{5 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {\left (477 a^3-387 a^3 \cos (c+d x)\right ) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx}{105 a^6} \\ & = -\frac {43 \sec (c+d x) \tan (c+d x)}{35 a^4 d (1+\cos (c+d x))^2}-\frac {\sec (c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 \sec (c+d x) \tan (c+d x)}{5 a d (a+a \cos (c+d x))^3}-\frac {288 \sec (c+d x) \tan (c+d x)}{35 d \left (a^4+a^4 \cos (c+d x)\right )}+\frac {\int \left (2205 a^4-1728 a^4 \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{105 a^8} \\ & = -\frac {43 \sec (c+d x) \tan (c+d x)}{35 a^4 d (1+\cos (c+d x))^2}-\frac {\sec (c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 \sec (c+d x) \tan (c+d x)}{5 a d (a+a \cos (c+d x))^3}-\frac {288 \sec (c+d x) \tan (c+d x)}{35 d \left (a^4+a^4 \cos (c+d x)\right )}-\frac {576 \int \sec ^2(c+d x) \, dx}{35 a^4}+\frac {21 \int \sec ^3(c+d x) \, dx}{a^4} \\ & = \frac {21 \sec (c+d x) \tan (c+d x)}{2 a^4 d}-\frac {43 \sec (c+d x) \tan (c+d x)}{35 a^4 d (1+\cos (c+d x))^2}-\frac {\sec (c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 \sec (c+d x) \tan (c+d x)}{5 a d (a+a \cos (c+d x))^3}-\frac {288 \sec (c+d x) \tan (c+d x)}{35 d \left (a^4+a^4 \cos (c+d x)\right )}+\frac {21 \int \sec (c+d x) \, dx}{2 a^4}+\frac {576 \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{35 a^4 d} \\ & = \frac {21 \text {arctanh}(\sin (c+d x))}{2 a^4 d}-\frac {576 \tan (c+d x)}{35 a^4 d}+\frac {21 \sec (c+d x) \tan (c+d x)}{2 a^4 d}-\frac {43 \sec (c+d x) \tan (c+d x)}{35 a^4 d (1+\cos (c+d x))^2}-\frac {\sec (c+d x) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 \sec (c+d x) \tan (c+d x)}{5 a d (a+a \cos (c+d x))^3}-\frac {288 \sec (c+d x) \tan (c+d x)}{35 d \left (a^4+a^4 \cos (c+d x)\right )} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(455\) vs. \(2(185)=370\).

Time = 6.78 (sec) , antiderivative size = 455, normalized size of antiderivative = 2.46 \[ \int \frac {\sec ^3(c+d x)}{(a+a \cos (c+d x))^4} \, dx=-\frac {168 \cos ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}{d (a+a \cos (c+d x))^4}+\frac {168 \cos ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}{d (a+a \cos (c+d x))^4}+\frac {\cos \left (\frac {c}{2}+\frac {d x}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec (c) \sec ^2(c+d x) \left (24402 \sin \left (\frac {d x}{2}\right )-55556 \sin \left (\frac {3 d x}{2}\right )+61054 \sin \left (c-\frac {d x}{2}\right )-33614 \sin \left (c+\frac {d x}{2}\right )+51842 \sin \left (2 c+\frac {d x}{2}\right )+12460 \sin \left (c+\frac {3 d x}{2}\right )-33716 \sin \left (2 c+\frac {3 d x}{2}\right )+34300 \sin \left (3 c+\frac {3 d x}{2}\right )-39788 \sin \left (c+\frac {5 d x}{2}\right )+2940 \sin \left (2 c+\frac {5 d x}{2}\right )-26068 \sin \left (3 c+\frac {5 d x}{2}\right )+16660 \sin \left (4 c+\frac {5 d x}{2}\right )-21351 \sin \left (2 c+\frac {7 d x}{2}\right )-1295 \sin \left (3 c+\frac {7 d x}{2}\right )-14911 \sin \left (4 c+\frac {7 d x}{2}\right )+5145 \sin \left (5 c+\frac {7 d x}{2}\right )-7329 \sin \left (3 c+\frac {9 d x}{2}\right )-1225 \sin \left (4 c+\frac {9 d x}{2}\right )-5369 \sin \left (5 c+\frac {9 d x}{2}\right )+735 \sin \left (6 c+\frac {9 d x}{2}\right )-1152 \sin \left (4 c+\frac {11 d x}{2}\right )-280 \sin \left (5 c+\frac {11 d x}{2}\right )-872 \sin \left (6 c+\frac {11 d x}{2}\right )\right )}{2240 d (a+a \cos (c+d x))^4} \]

[In]

Integrate[Sec[c + d*x]^3/(a + a*Cos[c + d*x])^4,x]

[Out]

(-168*Cos[c/2 + (d*x)/2]^8*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]])/(d*(a + a*Cos[c + d*x])^4) + (168*Cos
[c/2 + (d*x)/2]^8*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]])/(d*(a + a*Cos[c + d*x])^4) + (Cos[c/2 + (d*x)/
2]*Sec[c/2]*Sec[c]*Sec[c + d*x]^2*(24402*Sin[(d*x)/2] - 55556*Sin[(3*d*x)/2] + 61054*Sin[c - (d*x)/2] - 33614*
Sin[c + (d*x)/2] + 51842*Sin[2*c + (d*x)/2] + 12460*Sin[c + (3*d*x)/2] - 33716*Sin[2*c + (3*d*x)/2] + 34300*Si
n[3*c + (3*d*x)/2] - 39788*Sin[c + (5*d*x)/2] + 2940*Sin[2*c + (5*d*x)/2] - 26068*Sin[3*c + (5*d*x)/2] + 16660
*Sin[4*c + (5*d*x)/2] - 21351*Sin[2*c + (7*d*x)/2] - 1295*Sin[3*c + (7*d*x)/2] - 14911*Sin[4*c + (7*d*x)/2] +
5145*Sin[5*c + (7*d*x)/2] - 7329*Sin[3*c + (9*d*x)/2] - 1225*Sin[4*c + (9*d*x)/2] - 5369*Sin[5*c + (9*d*x)/2]
+ 735*Sin[6*c + (9*d*x)/2] - 1152*Sin[4*c + (11*d*x)/2] - 280*Sin[5*c + (11*d*x)/2] - 872*Sin[6*c + (11*d*x)/2
]))/(2240*d*(a + a*Cos[c + d*x])^4)

Maple [A] (verified)

Time = 1.27 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.80

method	result	size
derivativedivides	\(\frac {-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}-\frac {9 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-13 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-111 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {36}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+84 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {36}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}-84 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d \,a^{4}}\)	\(148\)
default	\(\frac {-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}-\frac {9 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-13 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-111 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {36}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+84 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {36}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}-84 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d \,a^{4}}\)	\(148\)
parallelrisch	\(\frac {\left (-23520 \cos \left (2 d x +2 c \right )-23520\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (23520 \cos \left (2 d x +2 c \right )+23520\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-34168 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\cos \left (d x +c \right )+\frac {5885 \cos \left (2 d x +2 c \right )}{8542}+\frac {1497 \cos \left (3 d x +3 c \right )}{4271}+\frac {3873 \cos \left (4 d x +4 c \right )}{34168}+\frac {72 \cos \left (5 d x +5 c \right )}{4271}+\frac {19387}{34168}\right ) \left (\sec ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2240 a^{4} d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(149\)
norman	\(\frac {-\frac {167 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}+\frac {281 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}-\frac {217 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d a}-\frac {167 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{140 d a}-\frac {53 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{280 d a}-\frac {\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )}{56 d a}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2} a^{3}}-\frac {21 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{4} d}+\frac {21 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{4}}\)	\(174\)
risch	\(-\frac {i \left (735 \,{\mathrm e}^{10 i \left (d x +c \right )}+5145 \,{\mathrm e}^{9 i \left (d x +c \right )}+16660 \,{\mathrm e}^{8 i \left (d x +c \right )}+34300 \,{\mathrm e}^{7 i \left (d x +c \right )}+51842 \,{\mathrm e}^{6 i \left (d x +c \right )}+61054 \,{\mathrm e}^{5 i \left (d x +c \right )}+55556 \,{\mathrm e}^{4 i \left (d x +c \right )}+39788 \,{\mathrm e}^{3 i \left (d x +c \right )}+21351 \,{\mathrm e}^{2 i \left (d x +c \right )}+7329 \,{\mathrm e}^{i \left (d x +c \right )}+1152\right )}{35 d \,a^{4} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}-\frac {21 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a^{4} d}+\frac {21 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d \,a^{4}}\)	\(191\)

[In]

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

[Out]

1/8/d/a^4*(-1/7*tan(1/2*d*x+1/2*c)^7-9/5*tan(1/2*d*x+1/2*c)^5-13*tan(1/2*d*x+1/2*c)^3-111*tan(1/2*d*x+1/2*c)-4
/(tan(1/2*d*x+1/2*c)+1)^2+36/(tan(1/2*d*x+1/2*c)+1)+84*ln(tan(1/2*d*x+1/2*c)+1)+4/(tan(1/2*d*x+1/2*c)-1)^2+36/
(tan(1/2*d*x+1/2*c)-1)-84*ln(tan(1/2*d*x+1/2*c)-1))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.35 \[ \int \frac {\sec ^3(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {735 \, {\left (\cos \left (d x + c\right )^{6} + 4 \, \cos \left (d x + c\right )^{5} + 6 \, \cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 735 \, {\left (\cos \left (d x + c\right )^{6} + 4 \, \cos \left (d x + c\right )^{5} + 6 \, \cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (1152 \, \cos \left (d x + c\right )^{5} + 3873 \, \cos \left (d x + c\right )^{4} + 4548 \, \cos \left (d x + c\right )^{3} + 2012 \, \cos \left (d x + c\right )^{2} + 140 \, \cos \left (d x + c\right ) - 35\right )} \sin \left (d x + c\right )}{140 \, {\left (a^{4} d \cos \left (d x + c\right )^{6} + 4 \, a^{4} d \cos \left (d x + c\right )^{5} + 6 \, a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + a^{4} d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

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

[Out]

1/140*(735*(cos(d*x + c)^6 + 4*cos(d*x + c)^5 + 6*cos(d*x + c)^4 + 4*cos(d*x + c)^3 + cos(d*x + c)^2)*log(sin(
d*x + c) + 1) - 735*(cos(d*x + c)^6 + 4*cos(d*x + c)^5 + 6*cos(d*x + c)^4 + 4*cos(d*x + c)^3 + cos(d*x + c)^2)
*log(-sin(d*x + c) + 1) - 2*(1152*cos(d*x + c)^5 + 3873*cos(d*x + c)^4 + 4548*cos(d*x + c)^3 + 2012*cos(d*x +
c)^2 + 140*cos(d*x + c) - 35)*sin(d*x + c))/(a^4*d*cos(d*x + c)^6 + 4*a^4*d*cos(d*x + c)^5 + 6*a^4*d*cos(d*x +
c)^4 + 4*a^4*d*cos(d*x + c)^3 + a^4*d*cos(d*x + c)^2)

Sympy [F]

\[ \int \frac {\sec ^3(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {\int \frac {\sec ^{3}{\left (c + d x \right )}}{\cos ^{4}{\left (c + d x \right )} + 4 \cos ^{3}{\left (c + d x \right )} + 6 \cos ^{2}{\left (c + d x \right )} + 4 \cos {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]

[In]

integrate(sec(d*x+c)**3/(a+a*cos(d*x+c))**4,x)

[Out]

Integral(sec(c + d*x)**3/(cos(c + d*x)**4 + 4*cos(c + d*x)**3 + 6*cos(c + d*x)**2 + 4*cos(c + d*x) + 1), x)/a*
*4

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.25 \[ \int \frac {\sec ^3(c+d x)}{(a+a \cos (c+d x))^4} \, dx=-\frac {\frac {280 \, {\left (\frac {7 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {9 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{4} - \frac {2 \, a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{4} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {3885 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {455 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {63 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {2940 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {2940 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}}{280 \, d} \]

[In]

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

[Out]

-1/280*(280*(7*sin(d*x + c)/(cos(d*x + c) + 1) - 9*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^4 - 2*a^4*sin(d*x +
c)^2/(cos(d*x + c) + 1)^2 + a^4*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (3885*sin(d*x + c)/(cos(d*x + c) + 1)
+ 455*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 63*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 5*sin(d*x + c)^7/(cos(d*x
+ c) + 1)^7)/a^4 - 2940*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^4 + 2940*log(sin(d*x + c)/(cos(d*x + c) +
1) - 1)/a^4)/d

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.84 \[ \int \frac {\sec ^3(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {\frac {2940 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {2940 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac {280 \, {\left (9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{4}} - \frac {5 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 63 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 455 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3885 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{280 \, d} \]

[In]

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

[Out]

1/280*(2940*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 2940*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^4 + 280*(9*tan(
1/2*d*x + 1/2*c)^3 - 7*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^4) - (5*a^24*tan(1/2*d*x + 1/2*
c)^7 + 63*a^24*tan(1/2*d*x + 1/2*c)^5 + 455*a^24*tan(1/2*d*x + 1/2*c)^3 + 3885*a^24*tan(1/2*d*x + 1/2*c))/a^28
)/d

Mupad [B] (verification not implemented)

Time = 15.50 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.86 \[ \int \frac {\sec ^3(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {21\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4\,d}-\frac {9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{40\,a^4\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{56\,a^4\,d}-\frac {13\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{8\,a^4\,d}-\frac {7\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^4\right )}-\frac {111\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a^4\,d} \]

[In]

int(1/(cos(c + d*x)^3*(a + a*cos(c + d*x))^4),x)

[Out]

(21*atanh(tan(c/2 + (d*x)/2)))/(a^4*d) - (9*tan(c/2 + (d*x)/2)^5)/(40*a^4*d) - tan(c/2 + (d*x)/2)^7/(56*a^4*d)
- (13*tan(c/2 + (d*x)/2)^3)/(8*a^4*d) - (7*tan(c/2 + (d*x)/2) - 9*tan(c/2 + (d*x)/2)^3)/(d*(a^4*tan(c/2 + (d*
x)/2)^4 - 2*a^4*tan(c/2 + (d*x)/2)^2 + a^4)) - (111*tan(c/2 + (d*x)/2))/(8*a^4*d)