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

Optimal. Leaf size=133 \[ \frac{4 \tan ^3(c+d x)}{a^2 d}+\frac{12 \tan (c+d x)}{a^2 d}-\frac{5 \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac{5 \tan (c+d x) \sec (c+d x)}{a^2 d}-\frac{10 \tan (c+d x) \sec ^2(c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac{\tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[Out]

(-5*ArcTanh[Sin[c + d*x]])/(a^2*d) + (12*Tan[c + d*x])/(a^2*d) - (5*Sec[c + d*x]*Tan[c + d*x])/(a^2*d) - (10*S
ec[c + d*x]^2*Tan[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])) - (Sec[c + d*x]^2*Tan[c + d*x])/(3*d*(a + a*Cos[c + d
*x])^2) + (4*Tan[c + d*x]^3)/(a^2*d)

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Rubi [A]  time = 0.199197, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2766, 2978, 2748, 3767, 3768, 3770} \[ \frac{4 \tan ^3(c+d x)}{a^2 d}+\frac{12 \tan (c+d x)}{a^2 d}-\frac{5 \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac{5 \tan (c+d x) \sec (c+d x)}{a^2 d}-\frac{10 \tan (c+d x) \sec ^2(c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac{\tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*ArcTanh[Sin[c + d*x]])/(a^2*d) + (12*Tan[c + d*x])/(a^2*d) - (5*Sec[c + d*x]*Tan[c + d*x])/(a^2*d) - (10*S
ec[c + d*x]^2*Tan[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])) - (Sec[c + d*x]^2*Tan[c + d*x])/(3*d*(a + a*Cos[c + d
*x])^2) + (4*Tan[c + d*x]^3)/(a^2*d)

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 3768

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 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 ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx &=-\frac{\sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \frac{(6 a-4 a \cos (c+d x)) \sec ^4(c+d x)}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=-\frac{10 \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{\sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \left (36 a^2-30 a^2 \cos (c+d x)\right ) \sec ^4(c+d x) \, dx}{3 a^4}\\ &=-\frac{10 \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{\sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac{10 \int \sec ^3(c+d x) \, dx}{a^2}+\frac{12 \int \sec ^4(c+d x) \, dx}{a^2}\\ &=-\frac{5 \sec (c+d x) \tan (c+d x)}{a^2 d}-\frac{10 \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{\sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac{5 \int \sec (c+d x) \, dx}{a^2}-\frac{12 \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{a^2 d}\\ &=-\frac{5 \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac{12 \tan (c+d x)}{a^2 d}-\frac{5 \sec (c+d x) \tan (c+d x)}{a^2 d}-\frac{10 \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{\sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{4 \tan ^3(c+d x)}{a^2 d}\\ \end{align*}

Mathematica [B]  time = 3.95732, size = 343, normalized size = 2.58 \[ \frac{960 \cos ^4\left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+\sec \left (\frac{c}{2}\right ) \sec (c) \left (-153 \sin \left (c-\frac{d x}{2}\right )+21 \sin \left (c+\frac{d x}{2}\right )-135 \sin \left (2 c+\frac{d x}{2}\right )+25 \sin \left (c+\frac{3 d x}{2}\right )+45 \sin \left (2 c+\frac{3 d x}{2}\right )-85 \sin \left (3 c+\frac{3 d x}{2}\right )+99 \sin \left (c+\frac{5 d x}{2}\right )+21 \sin \left (2 c+\frac{5 d x}{2}\right )+33 \sin \left (3 c+\frac{5 d x}{2}\right )-45 \sin \left (4 c+\frac{5 d x}{2}\right )+57 \sin \left (2 c+\frac{7 d x}{2}\right )+18 \sin \left (3 c+\frac{7 d x}{2}\right )+24 \sin \left (4 c+\frac{7 d x}{2}\right )-15 \sin \left (5 c+\frac{7 d x}{2}\right )+24 \sin \left (3 c+\frac{9 d x}{2}\right )+11 \sin \left (4 c+\frac{9 d x}{2}\right )+13 \sin \left (5 c+\frac{9 d x}{2}\right )-3 \sin \left (\frac{d x}{2}\right )+155 \sin \left (\frac{3 d x}{2}\right )\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \sec ^3(c+d x)}{48 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(960*Cos[(c + d*x)/2]^4*(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]^3*(-3*Sin[(d*x)/2] + 155*Sin[(3*d*x)/2] - 153*Sin[c - (d*x)/2]
+ 21*Sin[c + (d*x)/2] - 135*Sin[2*c + (d*x)/2] + 25*Sin[c + (3*d*x)/2] + 45*Sin[2*c + (3*d*x)/2] - 85*Sin[3*c
+ (3*d*x)/2] + 99*Sin[c + (5*d*x)/2] + 21*Sin[2*c + (5*d*x)/2] + 33*Sin[3*c + (5*d*x)/2] - 45*Sin[4*c + (5*d*
x)/2] + 57*Sin[2*c + (7*d*x)/2] + 18*Sin[3*c + (7*d*x)/2] + 24*Sin[4*c + (7*d*x)/2] - 15*Sin[5*c + (7*d*x)/2]
+ 24*Sin[3*c + (9*d*x)/2] + 11*Sin[4*c + (9*d*x)/2] + 13*Sin[5*c + (9*d*x)/2]))/(48*a^2*d*(1 + Cos[c + d*x])^2
)

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Maple [A]  time = 0.072, size = 204, normalized size = 1.5 \begin{align*}{\frac{1}{6\,{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{9}{2\,{a}^{2}d}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{1}{3\,{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-{\frac{3}{2\,{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-5\,{\frac{1}{{a}^{2}d \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }}+5\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }{{a}^{2}d}}-{\frac{1}{3\,{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+{\frac{3}{2\,{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-5\,{\frac{1}{{a}^{2}d \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }}-5\,{\frac{\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }{{a}^{2}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/d/a^2*tan(1/2*d*x+1/2*c)^3+9/2/d/a^2*tan(1/2*d*x+1/2*c)-1/3/d/a^2/(tan(1/2*d*x+1/2*c)-1)^3-3/2/d/a^2/(tan(
1/2*d*x+1/2*c)-1)^2-5/d/a^2/(tan(1/2*d*x+1/2*c)-1)+5/d/a^2*ln(tan(1/2*d*x+1/2*c)-1)-1/3/d/a^2/(tan(1/2*d*x+1/2
*c)+1)^3+3/2/d/a^2/(tan(1/2*d*x+1/2*c)+1)^2-5/d/a^2/(tan(1/2*d*x+1/2*c)+1)-5/d/a^2*ln(tan(1/2*d*x+1/2*c)+1)

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Maxima [A]  time = 1.1031, size = 316, normalized size = 2.38 \begin{align*} \frac{\frac{4 \,{\left (\frac{9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{2} - \frac{3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac{\frac{27 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{30 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac{30 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(4*(9*sin(d*x + c)/(cos(d*x + c) + 1) - 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos(d*
x + c) + 1)^5)/(a^2 - 3*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 -
a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + (27*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c)
+ 1)^3)/a^2 - 30*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 30*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^
2)/d

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Fricas [A]  time = 1.67939, size = 450, normalized size = 3.38 \begin{align*} -\frac{15 \,{\left (\cos \left (d x + c\right )^{5} + 2 \, \cos \left (d x + c\right )^{4} + \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (\cos \left (d x + c\right )^{5} + 2 \, \cos \left (d x + c\right )^{4} + \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (24 \, \cos \left (d x + c\right )^{4} + 33 \, \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{2} d \cos \left (d x + c\right )^{5} + 2 \, a^{2} d \cos \left (d x + c\right )^{4} + a^{2} d \cos \left (d x + c\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(15*(cos(d*x + c)^5 + 2*cos(d*x + c)^4 + cos(d*x + c)^3)*log(sin(d*x + c) + 1) - 15*(cos(d*x + c)^5 + 2*c
os(d*x + c)^4 + cos(d*x + c)^3)*log(-sin(d*x + c) + 1) - 2*(24*cos(d*x + c)^4 + 33*cos(d*x + c)^3 + 6*cos(d*x
+ c)^2 - cos(d*x + c) + 1)*sin(d*x + c))/(a^2*d*cos(d*x + c)^5 + 2*a^2*d*cos(d*x + c)^4 + a^2*d*cos(d*x + c)^3
)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\sec ^{4}{\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos{\left (c + d x \right )} + 1}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.42684, size = 182, normalized size = 1.37 \begin{align*} -\frac{\frac{30 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac{30 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac{4 \,{\left (15 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 20 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, \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 )}^{3} a^{2}} - \frac{a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 27 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{6}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(30*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 30*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2 + 4*(15*tan(1/2*d*
x + 1/2*c)^5 - 20*tan(1/2*d*x + 1/2*c)^3 + 9*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*a^2) - (a^4
*tan(1/2*d*x + 1/2*c)^3 + 27*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d