3.90 \(\int \frac{\sec (c+d x)}{(a+a \cos (c+d x))^5} \, dx\)

Optimal. Leaf size=153 \[ \frac{\tanh ^{-1}(\sin (c+d x))}{a^5 d}-\frac{488 \sin (c+d x)}{315 d \left (a^5 \cos (c+d x)+a^5\right )}-\frac{173 \sin (c+d x)}{315 a^3 d (a \cos (c+d x)+a)^2}-\frac{34 \sin (c+d x)}{105 a^2 d (a \cos (c+d x)+a)^3}-\frac{13 \sin (c+d x)}{63 a d (a \cos (c+d x)+a)^4}-\frac{\sin (c+d x)}{9 d (a \cos (c+d x)+a)^5} \]

[Out]

ArcTanh[Sin[c + d*x]]/(a^5*d) - Sin[c + d*x]/(9*d*(a + a*Cos[c + d*x])^5) - (13*Sin[c + d*x])/(63*a*d*(a + a*C
os[c + d*x])^4) - (34*Sin[c + d*x])/(105*a^2*d*(a + a*Cos[c + d*x])^3) - (173*Sin[c + d*x])/(315*a^3*d*(a + a*
Cos[c + d*x])^2) - (488*Sin[c + d*x])/(315*d*(a^5 + a^5*Cos[c + d*x]))

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Rubi [A]  time = 0.377481, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2766, 2978, 12, 3770} \[ \frac{\tanh ^{-1}(\sin (c+d x))}{a^5 d}-\frac{488 \sin (c+d x)}{315 d \left (a^5 \cos (c+d x)+a^5\right )}-\frac{173 \sin (c+d x)}{315 a^3 d (a \cos (c+d x)+a)^2}-\frac{34 \sin (c+d x)}{105 a^2 d (a \cos (c+d x)+a)^3}-\frac{13 \sin (c+d x)}{63 a d (a \cos (c+d x)+a)^4}-\frac{\sin (c+d x)}{9 d (a \cos (c+d x)+a)^5} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]/(a + a*Cos[c + d*x])^5,x]

[Out]

ArcTanh[Sin[c + d*x]]/(a^5*d) - Sin[c + d*x]/(9*d*(a + a*Cos[c + d*x])^5) - (13*Sin[c + d*x])/(63*a*d*(a + a*C
os[c + d*x])^4) - (34*Sin[c + d*x])/(105*a^2*d*(a + a*Cos[c + d*x])^3) - (173*Sin[c + d*x])/(315*a^3*d*(a + a*
Cos[c + d*x])^2) - (488*Sin[c + d*x])/(315*d*(a^5 + a^5*Cos[c + d*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 12

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

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))^5} \, dx &=-\frac{\sin (c+d x)}{9 d (a+a \cos (c+d x))^5}+\frac{\int \frac{(9 a-4 a \cos (c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^4} \, dx}{9 a^2}\\ &=-\frac{\sin (c+d x)}{9 d (a+a \cos (c+d x))^5}-\frac{13 \sin (c+d x)}{63 a d (a+a \cos (c+d x))^4}+\frac{\int \frac{\left (63 a^2-39 a^2 \cos (c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx}{63 a^4}\\ &=-\frac{\sin (c+d x)}{9 d (a+a \cos (c+d x))^5}-\frac{13 \sin (c+d x)}{63 a d (a+a \cos (c+d x))^4}-\frac{34 \sin (c+d x)}{105 a^2 d (a+a \cos (c+d x))^3}+\frac{\int \frac{\left (315 a^3-204 a^3 \cos (c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^2} \, dx}{315 a^6}\\ &=-\frac{\sin (c+d x)}{9 d (a+a \cos (c+d x))^5}-\frac{13 \sin (c+d x)}{63 a d (a+a \cos (c+d x))^4}-\frac{34 \sin (c+d x)}{105 a^2 d (a+a \cos (c+d x))^3}-\frac{173 \sin (c+d x)}{315 a^3 d (a+a \cos (c+d x))^2}+\frac{\int \frac{\left (945 a^4-519 a^4 \cos (c+d x)\right ) \sec (c+d x)}{a+a \cos (c+d x)} \, dx}{945 a^8}\\ &=-\frac{\sin (c+d x)}{9 d (a+a \cos (c+d x))^5}-\frac{13 \sin (c+d x)}{63 a d (a+a \cos (c+d x))^4}-\frac{34 \sin (c+d x)}{105 a^2 d (a+a \cos (c+d x))^3}-\frac{173 \sin (c+d x)}{315 a^3 d (a+a \cos (c+d x))^2}-\frac{488 \sin (c+d x)}{315 d \left (a^5+a^5 \cos (c+d x)\right )}+\frac{\int 945 a^5 \sec (c+d x) \, dx}{945 a^{10}}\\ &=-\frac{\sin (c+d x)}{9 d (a+a \cos (c+d x))^5}-\frac{13 \sin (c+d x)}{63 a d (a+a \cos (c+d x))^4}-\frac{34 \sin (c+d x)}{105 a^2 d (a+a \cos (c+d x))^3}-\frac{173 \sin (c+d x)}{315 a^3 d (a+a \cos (c+d x))^2}-\frac{488 \sin (c+d x)}{315 d \left (a^5+a^5 \cos (c+d x)\right )}+\frac{\int \sec (c+d x) \, dx}{a^5}\\ &=\frac{\tanh ^{-1}(\sin (c+d x))}{a^5 d}-\frac{\sin (c+d x)}{9 d (a+a \cos (c+d x))^5}-\frac{13 \sin (c+d x)}{63 a d (a+a \cos (c+d x))^4}-\frac{34 \sin (c+d x)}{105 a^2 d (a+a \cos (c+d x))^3}-\frac{173 \sin (c+d x)}{315 a^3 d (a+a \cos (c+d x))^2}-\frac{488 \sin (c+d x)}{315 d \left (a^5+a^5 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 1.82841, size = 211, normalized size = 1.38 \[ -\frac{\cos \left (\frac{1}{2} (c+d x)\right ) \left (\sec \left (\frac{c}{2}\right ) \left (-25515 \sin \left (c+\frac{d x}{2}\right )+29757 \sin \left (c+\frac{3 d x}{2}\right )-11235 \sin \left (2 c+\frac{3 d x}{2}\right )+14733 \sin \left (2 c+\frac{5 d x}{2}\right )-2835 \sin \left (3 c+\frac{5 d x}{2}\right )+4077 \sin \left (3 c+\frac{7 d x}{2}\right )-315 \sin \left (4 c+\frac{7 d x}{2}\right )+488 \sin \left (4 c+\frac{9 d x}{2}\right )+35973 \sin \left (\frac{d x}{2}\right )\right )+80640 \cos ^9\left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{2520 a^5 d (\cos (c+d x)+1)^5} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]/(a + a*Cos[c + d*x])^5,x]

[Out]

-(Cos[(c + d*x)/2]*(80640*Cos[(c + d*x)/2]^9*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2]]) + Sec[c/2]*(35973*Sin[(d*x)/2] - 25515*Sin[c + (d*x)/2] + 29757*Sin[c + (3*d*x)/2] - 1123
5*Sin[2*c + (3*d*x)/2] + 14733*Sin[2*c + (5*d*x)/2] - 2835*Sin[3*c + (5*d*x)/2] + 4077*Sin[3*c + (7*d*x)/2] -
315*Sin[4*c + (7*d*x)/2] + 488*Sin[4*c + (9*d*x)/2])))/(2520*a^5*d*(1 + Cos[c + d*x])^5)

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Maple [A]  time = 0.068, size = 134, normalized size = 0.9 \begin{align*} -{\frac{1}{144\,d{a}^{5}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{9}}-{\frac{3}{56\,d{a}^{5}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}-{\frac{1}{5\,d{a}^{5}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{13}{24\,d{a}^{5}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{31}{16\,d{a}^{5}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{1}{d{a}^{5}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{1}{d{a}^{5}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)/(a+cos(d*x+c)*a)^5,x)

[Out]

-1/144/d/a^5*tan(1/2*d*x+1/2*c)^9-3/56/d/a^5*tan(1/2*d*x+1/2*c)^7-1/5/d/a^5*tan(1/2*d*x+1/2*c)^5-13/24/d/a^5*t
an(1/2*d*x+1/2*c)^3-31/16/d/a^5*tan(1/2*d*x+1/2*c)-1/d/a^5*ln(tan(1/2*d*x+1/2*c)-1)+1/d/a^5*ln(tan(1/2*d*x+1/2
*c)+1)

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Maxima [A]  time = 1.15144, size = 215, normalized size = 1.41 \begin{align*} -\frac{\frac{\frac{9765 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{2730 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{1008 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{270 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac{35 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}}{a^{5}} - \frac{5040 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{5}} + \frac{5040 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{5}}}{5040 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)/(a+a*cos(d*x+c))^5,x, algorithm="maxima")

[Out]

-1/5040*((9765*sin(d*x + c)/(cos(d*x + c) + 1) + 2730*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 1008*sin(d*x + c)^
5/(cos(d*x + c) + 1)^5 + 270*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 35*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)/a^5
- 5040*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^5 + 5040*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^5)/d

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Fricas [A]  time = 1.69057, size = 671, normalized size = 4.39 \begin{align*} \frac{315 \,{\left (\cos \left (d x + c\right )^{5} + 5 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{3} + 10 \, \cos \left (d x + c\right )^{2} + 5 \, \cos \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 315 \,{\left (\cos \left (d x + c\right )^{5} + 5 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{3} + 10 \, \cos \left (d x + c\right )^{2} + 5 \, \cos \left (d x + c\right ) + 1\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (488 \, \cos \left (d x + c\right )^{4} + 2125 \, \cos \left (d x + c\right )^{3} + 3549 \, \cos \left (d x + c\right )^{2} + 2740 \, \cos \left (d x + c\right ) + 863\right )} \sin \left (d x + c\right )}{630 \,{\left (a^{5} d \cos \left (d x + c\right )^{5} + 5 \, a^{5} d \cos \left (d x + c\right )^{4} + 10 \, a^{5} d \cos \left (d x + c\right )^{3} + 10 \, a^{5} d \cos \left (d x + c\right )^{2} + 5 \, a^{5} d \cos \left (d x + c\right ) + a^{5} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)/(a+a*cos(d*x+c))^5,x, algorithm="fricas")

[Out]

1/630*(315*(cos(d*x + c)^5 + 5*cos(d*x + c)^4 + 10*cos(d*x + c)^3 + 10*cos(d*x + c)^2 + 5*cos(d*x + c) + 1)*lo
g(sin(d*x + c) + 1) - 315*(cos(d*x + c)^5 + 5*cos(d*x + c)^4 + 10*cos(d*x + c)^3 + 10*cos(d*x + c)^2 + 5*cos(d
*x + c) + 1)*log(-sin(d*x + c) + 1) - 2*(488*cos(d*x + c)^4 + 2125*cos(d*x + c)^3 + 3549*cos(d*x + c)^2 + 2740
*cos(d*x + c) + 863)*sin(d*x + c))/(a^5*d*cos(d*x + c)^5 + 5*a^5*d*cos(d*x + c)^4 + 10*a^5*d*cos(d*x + c)^3 +
10*a^5*d*cos(d*x + c)^2 + 5*a^5*d*cos(d*x + c) + a^5*d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)/(a+a*cos(d*x+c))**5,x)

[Out]

Timed out

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Giac [A]  time = 1.51151, size = 170, normalized size = 1.11 \begin{align*} \frac{\frac{5040 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{5}} - \frac{5040 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{5}} - \frac{35 \, a^{40} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 270 \, a^{40} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 1008 \, a^{40} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 2730 \, a^{40} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9765 \, a^{40} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{45}}}{5040 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)/(a+a*cos(d*x+c))^5,x, algorithm="giac")

[Out]

1/5040*(5040*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^5 - 5040*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^5 - (35*a^40*t
an(1/2*d*x + 1/2*c)^9 + 270*a^40*tan(1/2*d*x + 1/2*c)^7 + 1008*a^40*tan(1/2*d*x + 1/2*c)^5 + 2730*a^40*tan(1/2
*d*x + 1/2*c)^3 + 9765*a^40*tan(1/2*d*x + 1/2*c))/a^45)/d