3.79 \(\int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^4} \, dx\)

Optimal. Leaf size=120 \[ \frac {\tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac {32 \sin (c+d x)}{21 a^4 d (\cos (c+d x)+1)}-\frac {11 \sin (c+d x)}{21 a^4 d (\cos (c+d x)+1)^2}-\frac {2 \sin (c+d x)}{7 a d (a \cos (c+d x)+a)^3}-\frac {\sin (c+d x)}{7 d (a \cos (c+d x)+a)^4} \]

[Out]

arctanh(sin(d*x+c))/a^4/d-11/21*sin(d*x+c)/a^4/d/(1+cos(d*x+c))^2-32/21*sin(d*x+c)/a^4/d/(1+cos(d*x+c))-1/7*si
n(d*x+c)/d/(a+a*cos(d*x+c))^4-2/7*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^3

________________________________________________________________________________________

Rubi [A]  time = 0.29, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2766, 2978, 12, 3770} \[ \frac {\tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac {32 \sin (c+d x)}{21 a^4 d (\cos (c+d x)+1)}-\frac {11 \sin (c+d x)}{21 a^4 d (\cos (c+d x)+1)^2}-\frac {2 \sin (c+d x)}{7 a d (a \cos (c+d x)+a)^3}-\frac {\sin (c+d x)}{7 d (a \cos (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[Sin[c + d*x]]/(a^4*d) - (11*Sin[c + d*x])/(21*a^4*d*(1 + Cos[c + d*x])^2) - (32*Sin[c + d*x])/(21*a^4*
d*(1 + Cos[c + d*x])) - Sin[c + d*x]/(7*d*(a + a*Cos[c + d*x])^4) - (2*Sin[c + d*x])/(7*a*d*(a + a*Cos[c + d*x
])^3)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2766

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 2978

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 3770

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

Rubi steps

\begin {align*} \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^4} \, dx &=-\frac {\sin (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {\int \frac {(7 a-3 a \cos (c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {\sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 \sin (c+d x)}{7 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {\left (35 a^2-20 a^2 \cos (c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac {11 \sin (c+d x)}{21 a^4 d (1+\cos (c+d x))^2}-\frac {\sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 \sin (c+d x)}{7 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {\left (105 a^3-55 a^3 \cos (c+d x)\right ) \sec (c+d x)}{a+a \cos (c+d x)} \, dx}{105 a^6}\\ &=-\frac {11 \sin (c+d x)}{21 a^4 d (1+\cos (c+d x))^2}-\frac {\sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 \sin (c+d x)}{7 a d (a+a \cos (c+d x))^3}-\frac {32 \sin (c+d x)}{21 d \left (a^4+a^4 \cos (c+d x)\right )}+\frac {\int 105 a^4 \sec (c+d x) \, dx}{105 a^8}\\ &=-\frac {11 \sin (c+d x)}{21 a^4 d (1+\cos (c+d x))^2}-\frac {\sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 \sin (c+d x)}{7 a d (a+a \cos (c+d x))^3}-\frac {32 \sin (c+d x)}{21 d \left (a^4+a^4 \cos (c+d x)\right )}+\frac {\int \sec (c+d x) \, dx}{a^4}\\ &=\frac {\tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac {11 \sin (c+d x)}{21 a^4 d (1+\cos (c+d x))^2}-\frac {\sin (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {2 \sin (c+d x)}{7 a d (a+a \cos (c+d x))^3}-\frac {32 \sin (c+d x)}{21 d \left (a^4+a^4 \cos (c+d x)\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.86, size = 185, normalized size = 1.54 \[ \frac {\sec \left (\frac {c}{2}\right ) \left (434 \sin \left (c+\frac {d x}{2}\right )-525 \sin \left (c+\frac {3 d x}{2}\right )+147 \sin \left (2 c+\frac {3 d x}{2}\right )-203 \sin \left (2 c+\frac {5 d x}{2}\right )+21 \sin \left (3 c+\frac {5 d x}{2}\right )-32 \sin \left (3 c+\frac {7 d x}{2}\right )-686 \sin \left (\frac {d x}{2}\right )\right ) \cos \left (\frac {1}{2} (c+d x)\right )-1344 \cos ^8\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{84 a^4 d (\cos (c+d x)+1)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-1344*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]*(-686*Sin[(d*x)/2] + 434*Sin[c + (d*x)/2] - 525*Sin[c + (3*d*x)/2] + 147*Sin[2*c
+ (3*d*x)/2] - 203*Sin[2*c + (5*d*x)/2] + 21*Sin[3*c + (5*d*x)/2] - 32*Sin[3*c + (7*d*x)/2]))/(84*a^4*d*(1 +
Cos[c + d*x])^4)

________________________________________________________________________________________

fricas [A]  time = 1.04, size = 202, normalized size = 1.68 \[ \frac {21 \, {\left (\cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 21 \, {\left (\cos \left (d x + c\right )^{4} + 4 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right ) + 1\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (32 \, \cos \left (d x + c\right )^{3} + 107 \, \cos \left (d x + c\right )^{2} + 124 \, \cos \left (d x + c\right ) + 52\right )} \sin \left (d x + c\right )}{42 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/42*(21*(cos(d*x + c)^4 + 4*cos(d*x + c)^3 + 6*cos(d*x + c)^2 + 4*cos(d*x + c) + 1)*log(sin(d*x + c) + 1) - 2
1*(cos(d*x + c)^4 + 4*cos(d*x + c)^3 + 6*cos(d*x + c)^2 + 4*cos(d*x + c) + 1)*log(-sin(d*x + c) + 1) - 2*(32*c
os(d*x + c)^3 + 107*cos(d*x + c)^2 + 124*cos(d*x + c) + 52)*sin(d*x + c))/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(
d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) + a^4*d)

________________________________________________________________________________________

giac [A]  time = 0.69, size = 110, normalized size = 0.92 \[ \frac {\frac {168 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {168 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} - \frac {3 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 21 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 77 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 315 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{168 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/168*(168*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 168*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^4 - (3*a^24*tan(1
/2*d*x + 1/2*c)^7 + 21*a^24*tan(1/2*d*x + 1/2*c)^5 + 77*a^24*tan(1/2*d*x + 1/2*c)^3 + 315*a^24*tan(1/2*d*x + 1
/2*c))/a^28)/d

________________________________________________________________________________________

maple [A]  time = 0.09, size = 115, normalized size = 0.96 \[ -\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{56 d \,a^{4}}-\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}-\frac {11 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d \,a^{4}}-\frac {15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{4}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/56/d/a^4*tan(1/2*d*x+1/2*c)^7-1/8/d/a^4*tan(1/2*d*x+1/2*c)^5-11/24/d/a^4*tan(1/2*d*x+1/2*c)^3-15/8/d/a^4*ta
n(1/2*d*x+1/2*c)-1/d/a^4*ln(tan(1/2*d*x+1/2*c)-1)+1/d/a^4*ln(tan(1/2*d*x+1/2*c)+1)

________________________________________________________________________________________

maxima [A]  time = 1.48, size = 139, normalized size = 1.16 \[ -\frac {\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}}{168 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/168*((315*sin(d*x + c)/(cos(d*x + c) + 1) + 77*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5/(cos
(d*x + c) + 1)^5 + 3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 168*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a
^4 + 168*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^4)/d

________________________________________________________________________________________

mupad [B]  time = 0.37, size = 83, normalized size = 0.69 \[ -\frac {\frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{8\,a^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{56\,a^4}-\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^4}+\frac {15\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a^4}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((11*tan(c/2 + (d*x)/2)^3)/(24*a^4) + tan(c/2 + (d*x)/2)^5/(8*a^4) + tan(c/2 + (d*x)/2)^7/(56*a^4) - (2*atanh
(tan(c/2 + (d*x)/2)))/a^4 + (15*tan(c/2 + (d*x)/2))/(8*a^4))/d

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sec {\left (c + d x \right )}}{\cos ^{4}{\left (c + d x \right )} + 4 \cos ^{3}{\left (c + d x \right )} + 6 \cos ^{2}{\left (c + d x \right )} + 4 \cos {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sec(c + d*x)/(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

________________________________________________________________________________________