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

Optimal. Leaf size=135 \[ \frac{664 \tan (c+d x)}{105 a^4 d}-\frac{4 \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac{4 \tan (c+d x)}{a^4 d (\cos (c+d x)+1)}-\frac{88 \tan (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}-\frac{12 \tan (c+d x)}{35 a d (a \cos (c+d x)+a)^3}-\frac{\tan (c+d x)}{7 d (a \cos (c+d x)+a)^4} \]

[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)

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Rubi [A]  time = 0.391206, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2766, 2978, 2748, 3767, 8, 3770} \[ \frac{664 \tan (c+d x)}{105 a^4 d}-\frac{4 \tanh ^{-1}(\sin (c+d x))}{a^4 d}-\frac{4 \tan (c+d x)}{a^4 d (\cos (c+d x)+1)}-\frac{88 \tan (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}-\frac{12 \tan (c+d x)}{35 a d (a \cos (c+d x)+a)^3}-\frac{\tan (c+d x)}{7 d (a \cos (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[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 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 2748

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 3767

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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx &=-\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 \tanh ^{-1}(\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 \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{105 a^4 d}\\ &=-\frac{4 \tanh ^{-1}(\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]  time = 4.08718, size = 341, normalized size = 2.53 \[ \frac{107520 \cos ^8\left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+\sec \left (\frac{c}{2}\right ) \sec (c) \left (-20524 \sin \left (c-\frac{d x}{2}\right )+14644 \sin \left (c+\frac{d x}{2}\right )-16660 \sin \left (2 c+\frac{d x}{2}\right )-4690 \sin \left (c+\frac{3 d x}{2}\right )+14378 \sin \left (2 c+\frac{3 d x}{2}\right )-9100 \sin \left (3 c+\frac{3 d x}{2}\right )+11668 \sin \left (c+\frac{5 d x}{2}\right )-630 \sin \left (2 c+\frac{5 d x}{2}\right )+9358 \sin \left (3 c+\frac{5 d x}{2}\right )-2940 \sin \left (4 c+\frac{5 d x}{2}\right )+4228 \sin \left (2 c+\frac{7 d x}{2}\right )+315 \sin \left (3 c+\frac{7 d x}{2}\right )+3493 \sin \left (4 c+\frac{7 d x}{2}\right )-420 \sin \left (5 c+\frac{7 d x}{2}\right )+664 \sin \left (3 c+\frac{9 d x}{2}\right )+105 \sin \left (4 c+\frac{9 d x}{2}\right )+559 \sin \left (5 c+\frac{9 d x}{2}\right )-10780 \sin \left (\frac{d x}{2}\right )+18788 \sin \left (\frac{3 d x}{2}\right )\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)}{1680 a^4 d (\cos (c+d x)+1)^4} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.072, size = 158, normalized size = 1.2 \begin{align*}{\frac{1}{56\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}+{\frac{7}{40\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{23}{24\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{49}{8\,d{a}^{4}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{1}{d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+4\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }{d{a}^{4}}}-{\frac{1}{d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-4\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }{d{a}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/56/d/a^4*tan(1/2*d*x+1/2*c)^7+7/40/d/a^4*tan(1/2*d*x+1/2*c)^5+23/24/d/a^4*tan(1/2*d*x+1/2*c)^3+49/8/d/a^4*ta
n(1/2*d*x+1/2*c)-1/d/a^4/(tan(1/2*d*x+1/2*c)-1)+4/d/a^4*ln(tan(1/2*d*x+1/2*c)-1)-1/d/a^4/(tan(1/2*d*x+1/2*c)+1
)-4/d/a^4*ln(tan(1/2*d*x+1/2*c)+1)

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Maxima [A]  time = 1.17412, size = 251, normalized size = 1.86 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.72812, size = 632, normalized size = 4.68 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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

[Out]

Timed out

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Giac [A]  time = 1.53379, size = 188, normalized size = 1.39 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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