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

Optimal. Leaf size=103 \[ \frac{24 \sin (c+d x)}{5 a^3 d}-\frac{3 \sin (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}-\frac{3 x}{a^3}-\frac{3 \sin (c+d x)}{5 a d (a \sec (c+d x)+a)^2}-\frac{\sin (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

[Out]

(-3*x)/a^3 + (24*Sin[c + d*x])/(5*a^3*d) - Sin[c + d*x]/(5*d*(a + a*Sec[c + d*x])^3) - (3*Sin[c + d*x])/(5*a*d
*(a + a*Sec[c + d*x])^2) - (3*Sin[c + d*x])/(d*(a^3 + a^3*Sec[c + d*x]))

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Rubi [A]  time = 0.220573, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {3817, 4020, 3787, 2637, 8} \[ \frac{24 \sin (c+d x)}{5 a^3 d}-\frac{3 \sin (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}-\frac{3 x}{a^3}-\frac{3 \sin (c+d x)}{5 a d (a \sec (c+d x)+a)^2}-\frac{\sin (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*x)/a^3 + (24*Sin[c + d*x])/(5*a^3*d) - Sin[c + d*x]/(5*d*(a + a*Sec[c + d*x])^3) - (3*Sin[c + d*x])/(5*a*d
*(a + a*Sec[c + d*x])^2) - (3*Sin[c + d*x])/(d*(a^3 + a^3*Sec[c + d*x]))

Rule 3817

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(Cot[
e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*(2*m + 1)), x] + Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc
[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*(a*(2*m + n + 1) - b*(m + n + 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d
, e, f, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m])

Rule 4020

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(b*f*(2
*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*(2
*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*
b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{\cos (c+d x)}{(a+a \sec (c+d x))^3} \, dx &=-\frac{\sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{\int \frac{\cos (c+d x) (-6 a+3 a \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{\sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{3 \sin (c+d x)}{5 a d (a+a \sec (c+d x))^2}-\frac{\int \frac{\cos (c+d x) \left (-27 a^2+18 a^2 \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac{\sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{3 \sin (c+d x)}{5 a d (a+a \sec (c+d x))^2}-\frac{3 \sin (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}-\frac{\int \cos (c+d x) \left (-72 a^3+45 a^3 \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac{\sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{3 \sin (c+d x)}{5 a d (a+a \sec (c+d x))^2}-\frac{3 \sin (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}-\frac{3 \int 1 \, dx}{a^3}+\frac{24 \int \cos (c+d x) \, dx}{5 a^3}\\ &=-\frac{3 x}{a^3}+\frac{24 \sin (c+d x)}{5 a^3 d}-\frac{\sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{3 \sin (c+d x)}{5 a d (a+a \sec (c+d x))^2}-\frac{3 \sin (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.561546, size = 169, normalized size = 1.64 \[ \frac{2 \cos \left (\frac{1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (20 (\sin (c+d x)-3 d x) \cos ^5\left (\frac{1}{2} (c+d x)\right )-12 \tan \left (\frac{c}{2}\right ) \cos ^3\left (\frac{1}{2} (c+d x)\right )+\tan \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right )+\sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right )+96 \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \cos ^4\left (\frac{1}{2} (c+d x)\right )-12 \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \cos ^2\left (\frac{1}{2} (c+d x)\right )\right )}{5 a^3 d (\sec (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[(c + d*x)/2]*Sec[c + d*x]^3*(Sec[c/2]*Sin[(d*x)/2] - 12*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[(d*x)/2] + 96*C
os[(c + d*x)/2]^4*Sec[c/2]*Sin[(d*x)/2] + 20*Cos[(c + d*x)/2]^5*(-3*d*x + Sin[c + d*x]) + Cos[(c + d*x)/2]*Tan
[c/2] - 12*Cos[(c + d*x)/2]^3*Tan[c/2]))/(5*a^3*d*(1 + Sec[c + d*x])^3)

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Maple [A]  time = 0.06, size = 107, normalized size = 1. \begin{align*}{\frac{1}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{1}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{17}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}-6\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20/d/a^3*tan(1/2*d*x+1/2*c)^5-1/2/d/a^3*tan(1/2*d*x+1/2*c)^3+17/4/d/a^3*tan(1/2*d*x+1/2*c)+2/d/a^3*tan(1/2*d
*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2)-6/d/a^3*arctan(tan(1/2*d*x+1/2*c))

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Maxima [A]  time = 1.71797, size = 185, normalized size = 1.8 \begin{align*} \frac{\frac{40 \, \sin \left (d x + c\right )}{{\left (a^{3} + \frac{a^{3} \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{85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac{120 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{20 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*(40*sin(d*x + c)/((a^3 + a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (85*sin(d*x + c)/
(cos(d*x + c) + 1) - 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 120*a
rctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

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Fricas [A]  time = 1.90152, size = 325, normalized size = 3.16 \begin{align*} -\frac{15 \, d x \cos \left (d x + c\right )^{3} + 45 \, d x \cos \left (d x + c\right )^{2} + 45 \, d x \cos \left (d x + c\right ) + 15 \, d x -{\left (5 \, \cos \left (d x + c\right )^{3} + 39 \, \cos \left (d x + c\right )^{2} + 57 \, \cos \left (d x + c\right ) + 24\right )} \sin \left (d x + c\right )}{5 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*(15*d*x*cos(d*x + c)^3 + 45*d*x*cos(d*x + c)^2 + 45*d*x*cos(d*x + c) + 15*d*x - (5*cos(d*x + c)^3 + 39*co
s(d*x + c)^2 + 57*cos(d*x + c) + 24)*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*co
s(d*x + c) + a^3*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.37433, size = 130, normalized size = 1.26 \begin{align*} -\frac{\frac{60 \,{\left (d x + c\right )}}{a^{3}} - \frac{40 \, \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^{3}} - \frac{a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 10 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 85 \, a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{15}}}{20 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20*(60*(d*x + c)/a^3 - 40*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^3) - (a^12*tan(1/2*d*x + 1/2
*c)^5 - 10*a^12*tan(1/2*d*x + 1/2*c)^3 + 85*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d