3.71 \(\int \frac{\sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=156 \[ -\frac{152 \tan (c+d x)}{15 a^3 d}+\frac{13 \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}+\frac{13 \tan (c+d x) \sec (c+d x)}{2 a^3 d}-\frac{76 \tan (c+d x) \sec (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac{11 \tan (c+d x) \sec (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac{\tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

[Out]

(13*ArcTanh[Sin[c + d*x]])/(2*a^3*d) - (152*Tan[c + d*x])/(15*a^3*d) + (13*Sec[c + d*x]*Tan[c + d*x])/(2*a^3*d
) - (Sec[c + d*x]*Tan[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) - (11*Sec[c + d*x]*Tan[c + d*x])/(15*a*d*(a + a*C
os[c + d*x])^2) - (76*Sec[c + d*x]*Tan[c + d*x])/(15*d*(a^3 + a^3*Cos[c + d*x]))

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Rubi [A]  time = 0.304536, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2766, 2978, 2748, 3768, 3770, 3767, 8} \[ -\frac{152 \tan (c+d x)}{15 a^3 d}+\frac{13 \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}+\frac{13 \tan (c+d x) \sec (c+d x)}{2 a^3 d}-\frac{76 \tan (c+d x) \sec (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac{11 \tan (c+d x) \sec (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac{\tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3/(a + a*Cos[c + d*x])^3,x]

[Out]

(13*ArcTanh[Sin[c + d*x]])/(2*a^3*d) - (152*Tan[c + d*x])/(15*a^3*d) + (13*Sec[c + d*x]*Tan[c + d*x])/(2*a^3*d
) - (Sec[c + d*x]*Tan[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) - (11*Sec[c + d*x]*Tan[c + d*x])/(15*a*d*(a + a*C
os[c + d*x])^2) - (76*Sec[c + d*x]*Tan[c + d*x])/(15*d*(a^3 + a^3*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 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 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]

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]

Rubi steps

\begin{align*} \int \frac{\sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx &=-\frac{\sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{\int \frac{(7 a-4 a \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{\sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{11 \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{\int \frac{\left (43 a^2-33 a^2 \cos (c+d x)\right ) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=-\frac{\sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{11 \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{76 \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac{\int \left (195 a^3-152 a^3 \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{15 a^6}\\ &=-\frac{\sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{11 \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{76 \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}-\frac{152 \int \sec ^2(c+d x) \, dx}{15 a^3}+\frac{13 \int \sec ^3(c+d x) \, dx}{a^3}\\ &=\frac{13 \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac{\sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{11 \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{76 \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac{13 \int \sec (c+d x) \, dx}{2 a^3}+\frac{152 \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d}\\ &=\frac{13 \tanh ^{-1}(\sin (c+d x))}{2 a^3 d}-\frac{152 \tan (c+d x)}{15 a^3 d}+\frac{13 \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac{\sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{11 \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{76 \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 3.83344, size = 343, normalized size = 2.2 \[ -\frac{24960 \cos ^6\left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+\sec \left (\frac{c}{2}\right ) \sec (c) \left (-4329 \sin \left (c-\frac{d x}{2}\right )+1989 \sin \left (c+\frac{d x}{2}\right )-3575 \sin \left (2 c+\frac{d x}{2}\right )-475 \sin \left (c+\frac{3 d x}{2}\right )+2005 \sin \left (2 c+\frac{3 d x}{2}\right )-2275 \sin \left (3 c+\frac{3 d x}{2}\right )+2673 \sin \left (c+\frac{5 d x}{2}\right )+105 \sin \left (2 c+\frac{5 d x}{2}\right )+1593 \sin \left (3 c+\frac{5 d x}{2}\right )-975 \sin \left (4 c+\frac{5 d x}{2}\right )+1325 \sin \left (2 c+\frac{7 d x}{2}\right )+255 \sin \left (3 c+\frac{7 d x}{2}\right )+875 \sin \left (4 c+\frac{7 d x}{2}\right )-195 \sin \left (5 c+\frac{7 d x}{2}\right )+304 \sin \left (3 c+\frac{9 d x}{2}\right )+90 \sin \left (4 c+\frac{9 d x}{2}\right )+214 \sin \left (5 c+\frac{9 d x}{2}\right )-1235 \sin \left (\frac{d x}{2}\right )+3805 \sin \left (\frac{3 d x}{2}\right )\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x)}{480 a^3 d (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3/(a + a*Cos[c + d*x])^3,x]

[Out]

-(24960*Cos[(c + d*x)/2]^6*(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]^2*(-1235*Sin[(d*x)/2] + 3805*Sin[(3*d*x)/2] - 4329*Sin[c -
(d*x)/2] + 1989*Sin[c + (d*x)/2] - 3575*Sin[2*c + (d*x)/2] - 475*Sin[c + (3*d*x)/2] + 2005*Sin[2*c + (3*d*x)/2
] - 2275*Sin[3*c + (3*d*x)/2] + 2673*Sin[c + (5*d*x)/2] + 105*Sin[2*c + (5*d*x)/2] + 1593*Sin[3*c + (5*d*x)/2]
- 975*Sin[4*c + (5*d*x)/2] + 1325*Sin[2*c + (7*d*x)/2] + 255*Sin[3*c + (7*d*x)/2] + 875*Sin[4*c + (7*d*x)/2]
- 195*Sin[5*c + (7*d*x)/2] + 304*Sin[3*c + (9*d*x)/2] + 90*Sin[4*c + (9*d*x)/2] + 214*Sin[5*c + (9*d*x)/2]))/(
480*a^3*d*(1 + Cos[c + d*x])^3)

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Maple [A]  time = 0.077, size = 181, normalized size = 1.2 \begin{align*} -{\frac{1}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{2}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{31}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{1}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{7}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{13}{2\,d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{1}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{7}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{13}{2\,d{a}^{3}}\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)^3/(a+cos(d*x+c)*a)^3,x)

[Out]

-1/20/d/a^3*tan(1/2*d*x+1/2*c)^5-2/3/d/a^3*tan(1/2*d*x+1/2*c)^3-31/4/d/a^3*tan(1/2*d*x+1/2*c)+1/2/d/a^3/(tan(1
/2*d*x+1/2*c)-1)^2+7/2/d/a^3/(tan(1/2*d*x+1/2*c)-1)-13/2/d/a^3*ln(tan(1/2*d*x+1/2*c)-1)-1/2/d/a^3/(tan(1/2*d*x
+1/2*c)+1)^2+7/2/d/a^3/(tan(1/2*d*x+1/2*c)+1)+13/2/d/a^3*ln(tan(1/2*d*x+1/2*c)+1)

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Maxima [A]  time = 1.3512, size = 285, normalized size = 1.83 \begin{align*} -\frac{\frac{60 \,{\left (\frac{5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} - \frac{2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac{\frac{465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac{390 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac{390 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/60*(60*(5*sin(d*x + c)/(cos(d*x + c) + 1) - 7*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^3 - 2*a^3*sin(d*x + c
)^2/(cos(d*x + c) + 1)^2 + a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (465*sin(d*x + c)/(cos(d*x + c) + 1) + 4
0*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 390*log(sin(d*x + c)/(cos
(d*x + c) + 1) + 1)/a^3 + 390*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3)/d

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Fricas [A]  time = 1.72531, size = 548, normalized size = 3.51 \begin{align*} \frac{195 \,{\left (\cos \left (d x + c\right )^{5} + 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 195 \,{\left (\cos \left (d x + c\right )^{5} + 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (304 \, \cos \left (d x + c\right )^{4} + 717 \, \cos \left (d x + c\right )^{3} + 479 \, \cos \left (d x + c\right )^{2} + 45 \, \cos \left (d x + c\right ) - 15\right )} \sin \left (d x + c\right )}{60 \,{\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

1/60*(195*(cos(d*x + c)^5 + 3*cos(d*x + c)^4 + 3*cos(d*x + c)^3 + cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 195*
(cos(d*x + c)^5 + 3*cos(d*x + c)^4 + 3*cos(d*x + c)^3 + cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(304*cos(d*
x + c)^4 + 717*cos(d*x + c)^3 + 479*cos(d*x + c)^2 + 45*cos(d*x + c) - 15)*sin(d*x + c))/(a^3*d*cos(d*x + c)^5
+ 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\sec ^{3}{\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos{\left (c + d x \right )} + 1}\, dx}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3/(a+a*cos(d*x+c))**3,x)

[Out]

Integral(sec(c + d*x)**3/(cos(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1), x)/a**3

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Giac [A]  time = 1.3276, size = 188, normalized size = 1.21 \begin{align*} \frac{\frac{390 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac{390 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac{60 \,{\left (7 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 5 \, \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 )}^{2} a^{3}} - \frac{3 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 40 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 465 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{15}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+a*cos(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(390*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 390*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 + 60*(7*tan(1/2*
d*x + 1/2*c)^3 - 5*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^3) - (3*a^12*tan(1/2*d*x + 1/2*c)^5
+ 40*a^12*tan(1/2*d*x + 1/2*c)^3 + 465*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d