\(\int \frac {\sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx\) [80]

Optimal result

Integrand size = 21, antiderivative size = 135 \[ \int \frac {\sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=-\frac {4 \text {arctanh}(\sin (c+d x))}{a^4 d}+\frac {664 \tan (c+d x)}{105 a^4 d}-\frac {88 \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {4 \tan (c+d x)}{a^4 d (1+\cos (c+d x))}-\frac {\tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {12 \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3} \]

[Out]

-4*arctanh(sin(d*x+c))/a^4/d+664/105*tan(d*x+c)/a^4/d-88/105*tan(d*x+c)/a^4/d/(1+cos(d*x+c))^2-4*tan(d*x+c)/a^
4/d/(1+cos(d*x+c))-1/7*tan(d*x+c)/d/(a+a*cos(d*x+c))^4-12/35*tan(d*x+c)/a/d/(a+a*cos(d*x+c))^3

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2845, 3057, 2827, 3852, 8, 3855} \[ \int \frac {\sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=-\frac {4 \text {arctanh}(\sin (c+d x))}{a^4 d}+\frac {664 \tan (c+d x)}{105 a^4 d}-\frac {4 \tan (c+d x)}{a^4 d (\cos (c+d x)+1)}-\frac {88 \tan (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}-\frac {12 \tan (c+d x)}{35 a d (a \cos (c+d x)+a)^3}-\frac {\tan (c+d x)}{7 d (a \cos (c+d x)+a)^4} \]

[In]

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

[Out]

(-4*ArcTanh[Sin[c + d*x]])/(a^4*d) + (664*Tan[c + d*x])/(105*a^4*d) - (88*Tan[c + d*x])/(105*a^4*d*(1 + Cos[c
+ d*x])^2) - (4*Tan[c + d*x])/(a^4*d*(1 + Cos[c + d*x])) - Tan[c + d*x]/(7*d*(a + a*Cos[c + d*x])^4) - (12*Tan
[c + d*x])/(35*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 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 {\tan (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {\int \frac {(8 a-4 a \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx}{7 a^2} \\ & = -\frac {\tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {12 \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {\left (52 a^2-36 a^2 \cos (c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{35 a^4} \\ & = -\frac {88 \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {\tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {12 \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {\left (244 a^3-176 a^3 \cos (c+d x)\right ) \sec ^2(c+d x)}{a+a \cos (c+d x)} \, dx}{105 a^6} \\ & = -\frac {88 \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {\tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {12 \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac {4 \tan (c+d x)}{d \left (a^4+a^4 \cos (c+d x)\right )}+\frac {\int \left (664 a^4-420 a^4 \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{105 a^8} \\ & = -\frac {88 \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {\tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {12 \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac {4 \tan (c+d x)}{d \left (a^4+a^4 \cos (c+d x)\right )}-\frac {4 \int \sec (c+d x) \, dx}{a^4}+\frac {664 \int \sec ^2(c+d x) \, dx}{105 a^4} \\ & = -\frac {4 \text {arctanh}(\sin (c+d x))}{a^4 d}-\frac {88 \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {\tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {12 \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac {4 \tan (c+d x)}{d \left (a^4+a^4 \cos (c+d x)\right )}-\frac {664 \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{105 a^4 d} \\ & = -\frac {4 \text {arctanh}(\sin (c+d x))}{a^4 d}+\frac {664 \tan (c+d x)}{105 a^4 d}-\frac {88 \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {\tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {12 \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac {4 \tan (c+d x)}{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. \(341\) vs. \(2(135)=270\).

Time = 3.38 (sec) , antiderivative size = 341, normalized size of antiderivative = 2.53 \[ \int \frac {\sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {107520 \cos ^8\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sec (c) \sec (c+d x) \left (-10780 \sin \left (\frac {d x}{2}\right )+18788 \sin \left (\frac {3 d x}{2}\right )-20524 \sin \left (c-\frac {d x}{2}\right )+14644 \sin \left (c+\frac {d x}{2}\right )-16660 \sin \left (2 c+\frac {d x}{2}\right )-4690 \sin \left (c+\frac {3 d x}{2}\right )+14378 \sin \left (2 c+\frac {3 d x}{2}\right )-9100 \sin \left (3 c+\frac {3 d x}{2}\right )+11668 \sin \left (c+\frac {5 d x}{2}\right )-630 \sin \left (2 c+\frac {5 d x}{2}\right )+9358 \sin \left (3 c+\frac {5 d x}{2}\right )-2940 \sin \left (4 c+\frac {5 d x}{2}\right )+4228 \sin \left (2 c+\frac {7 d x}{2}\right )+315 \sin \left (3 c+\frac {7 d x}{2}\right )+3493 \sin \left (4 c+\frac {7 d x}{2}\right )-420 \sin \left (5 c+\frac {7 d x}{2}\right )+664 \sin \left (3 c+\frac {9 d x}{2}\right )+105 \sin \left (4 c+\frac {9 d x}{2}\right )+559 \sin \left (5 c+\frac {9 d x}{2}\right )\right )}{1680 a^4 d (1+\cos (c+d x))^4} \]

[In]

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

[Out]

(107520*Cos[(c + d*x)/2]^8*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]
]) + Cos[(c + d*x)/2]*Sec[c/2]*Sec[c]*Sec[c + d*x]*(-10780*Sin[(d*x)/2] + 18788*Sin[(3*d*x)/2] - 20524*Sin[c -
(d*x)/2] + 14644*Sin[c + (d*x)/2] - 16660*Sin[2*c + (d*x)/2] - 4690*Sin[c + (3*d*x)/2] + 14378*Sin[2*c + (3*d
*x)/2] - 9100*Sin[3*c + (3*d*x)/2] + 11668*Sin[c + (5*d*x)/2] - 630*Sin[2*c + (5*d*x)/2] + 9358*Sin[3*c + (5*d
*x)/2] - 2940*Sin[4*c + (5*d*x)/2] + 4228*Sin[2*c + (7*d*x)/2] + 315*Sin[3*c + (7*d*x)/2] + 3493*Sin[4*c + (7*
d*x)/2] - 420*Sin[5*c + (7*d*x)/2] + 664*Sin[3*c + (9*d*x)/2] + 105*Sin[4*c + (9*d*x)/2] + 559*Sin[5*c + (9*d*
x)/2]))/(1680*a^4*d*(1 + Cos[c + d*x])^4)

Maple [A] (verified)

Time = 1.06 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.87

method	result	size
derivativedivides	\(\frac {\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {7 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {23 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+49 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {8}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-32 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {8}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+32 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d \,a^{4}}\)	\(118\)
default	\(\frac {\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{7}+\frac {7 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {23 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+49 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {8}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-32 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {8}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+32 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d \,a^{4}}\)	\(118\)
parallelrisch	\(\frac {3360 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )-3360 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )+2861 \left (\cos \left (d x +c \right )+\frac {1650 \cos \left (2 d x +2 c \right )}{2861}+\frac {559 \cos \left (3 d x +3 c \right )}{2861}+\frac {83 \cos \left (4 d x +4 c \right )}{2861}+\frac {1672}{2861}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sec ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{840 a^{4} d \cos \left (d x +c \right )}\)	\(121\)
norman	\(\frac {-\frac {65 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}+\frac {31 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d a}+\frac {47 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 d a}+\frac {11 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{70 d a}+\frac {\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )}{56 d a}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{3}}+\frac {4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4} d}-\frac {4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{4}}\)	\(155\)
risch	\(\frac {8 i \left (105 \,{\mathrm e}^{8 i \left (d x +c \right )}+735 \,{\mathrm e}^{7 i \left (d x +c \right )}+2275 \,{\mathrm e}^{6 i \left (d x +c \right )}+4165 \,{\mathrm e}^{5 i \left (d x +c \right )}+5131 \,{\mathrm e}^{4 i \left (d x +c \right )}+4697 \,{\mathrm e}^{3 i \left (d x +c \right )}+2917 \,{\mathrm e}^{2 i \left (d x +c \right )}+1057 \,{\mathrm e}^{i \left (d x +c \right )}+166\right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{4} d}-\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{4}}\)	\(169\)

[In]

int(sec(d*x+c)^2/(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+7/5*tan(1/2*d*x+1/2*c)^5+23/3*tan(1/2*d*x+1/2*c)^3+49*tan(1/2*d*x+1/2*c)-8
/(tan(1/2*d*x+1/2*c)+1)-32*ln(tan(1/2*d*x+1/2*c)+1)-8/(tan(1/2*d*x+1/2*c)-1)+32*ln(tan(1/2*d*x+1/2*c)-1))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.73 \[ \int \frac {\sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=-\frac {210 \, {\left (\cos \left (d x + c\right )^{5} + 4 \, \cos \left (d x + c\right )^{4} + 6 \, \cos \left (d x + c\right )^{3} + 4 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 210 \, {\left (\cos \left (d x + c\right )^{5} + 4 \, \cos \left (d x + c\right )^{4} + 6 \, \cos \left (d x + c\right )^{3} + 4 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - {\left (664 \, \cos \left (d x + c\right )^{4} + 2236 \, \cos \left (d x + c\right )^{3} + 2636 \, \cos \left (d x + c\right )^{2} + 1184 \, \cos \left (d x + c\right ) + 105\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{5} + 4 \, a^{4} d \cos \left (d x + c\right )^{4} + 6 \, a^{4} d \cos \left (d x + c\right )^{3} + 4 \, a^{4} d \cos \left (d x + c\right )^{2} + a^{4} d \cos \left (d x + c\right )\right )}} \]

[In]

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

[Out]

-1/105*(210*(cos(d*x + c)^5 + 4*cos(d*x + c)^4 + 6*cos(d*x + c)^3 + 4*cos(d*x + c)^2 + cos(d*x + c))*log(sin(d
*x + c) + 1) - 210*(cos(d*x + c)^5 + 4*cos(d*x + c)^4 + 6*cos(d*x + c)^3 + 4*cos(d*x + c)^2 + cos(d*x + c))*lo
g(-sin(d*x + c) + 1) - (664*cos(d*x + c)^4 + 2236*cos(d*x + c)^3 + 2636*cos(d*x + c)^2 + 1184*cos(d*x + c) + 1
05)*sin(d*x + c))/(a^4*d*cos(d*x + c)^5 + 4*a^4*d*cos(d*x + c)^4 + 6*a^4*d*cos(d*x + c)^3 + 4*a^4*d*cos(d*x +
c)^2 + a^4*d*cos(d*x + c))

Sympy [F]

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

[In]

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

[Out]

Integral(sec(c + d*x)**2/(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 = 0.25 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.38 \[ \int \frac {\sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {\frac {1680 \, \sin \left (d x + c\right )}{{\left (a^{4} - \frac {a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {5145 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {805 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {147 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {3360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {3360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}}{840 \, d} \]

[In]

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

[Out]

1/840*(1680*sin(d*x + c)/((a^4 - a^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (5145*sin(d*x
+ c)/(cos(d*x + c) + 1) + 805*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 147*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 +
15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 3360*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^4 + 3360*log(sin
(d*x + c)/(cos(d*x + c) + 1) - 1)/a^4)/d

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.03 \[ \int \frac {\sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=-\frac {\frac {3360 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {3360 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac {1680 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{4}} - \frac {15 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 147 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 805 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5145 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \]

[In]

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

[Out]

-1/840*(3360*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 3360*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^4 + 1680*tan(1
/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*a^4) - (15*a^24*tan(1/2*d*x + 1/2*c)^7 + 147*a^24*tan(1/2*d*x +
1/2*c)^5 + 805*a^24*tan(1/2*d*x + 1/2*c)^3 + 5145*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d

Mupad [B] (verification not implemented)

Time = 14.97 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.96 \[ \int \frac {\sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {23\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^4\,d}+\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{40\,a^4\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{56\,a^4\,d}-\frac {8\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4\,d}-\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^4\right )}+\frac {49\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a^4\,d} \]

[In]

int(1/(cos(c + d*x)^2*(a + a*cos(c + d*x))^4),x)

[Out]

(23*tan(c/2 + (d*x)/2)^3)/(24*a^4*d) + (7*tan(c/2 + (d*x)/2)^5)/(40*a^4*d) + tan(c/2 + (d*x)/2)^7/(56*a^4*d) -
(8*atanh(tan(c/2 + (d*x)/2)))/(a^4*d) - (2*tan(c/2 + (d*x)/2))/(d*(a^4*tan(c/2 + (d*x)/2)^2 - a^4)) + (49*tan
(c/2 + (d*x)/2))/(8*a^4*d)