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

Optimal. Leaf size=110 \[ -\frac{16 \sin (c+d x)}{3 a^2 d}+\frac{7 \sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac{8 \sin (c+d x) \cos (c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac{7 x}{2 a^2}-\frac{\sin (c+d x) \cos (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

(7*x)/(2*a^2) - (16*Sin[c + d*x])/(3*a^2*d) + (7*Cos[c + d*x]*Sin[c + d*x])/(2*a^2*d) - (8*Cos[c + d*x]*Sin[c
+ d*x])/(3*a^2*d*(1 + Sec[c + d*x])) - (Cos[c + d*x]*Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2)

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Rubi [A]  time = 0.175024, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3817, 4020, 3787, 2635, 8, 2637} \[ -\frac{16 \sin (c+d x)}{3 a^2 d}+\frac{7 \sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac{8 \sin (c+d x) \cos (c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac{7 x}{2 a^2}-\frac{\sin (c+d x) \cos (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*x)/(2*a^2) - (16*Sin[c + d*x])/(3*a^2*d) + (7*Cos[c + d*x]*Sin[c + d*x])/(2*a^2*d) - (8*Cos[c + d*x]*Sin[c
+ d*x])/(3*a^2*d*(1 + Sec[c + d*x])) - (Cos[c + d*x]*Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2)

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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

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

Rule 2637

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

Rubi steps

\begin{align*} \int \frac{\cos ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx &=-\frac{\cos (c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{\int \frac{\cos ^2(c+d x) (-5 a+3 a \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=-\frac{8 \cos (c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac{\cos (c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{\int \cos ^2(c+d x) \left (-21 a^2+16 a^2 \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac{8 \cos (c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac{\cos (c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{16 \int \cos (c+d x) \, dx}{3 a^2}+\frac{7 \int \cos ^2(c+d x) \, dx}{a^2}\\ &=-\frac{16 \sin (c+d x)}{3 a^2 d}+\frac{7 \cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac{8 \cos (c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac{\cos (c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{7 \int 1 \, dx}{2 a^2}\\ &=\frac{7 x}{2 a^2}-\frac{16 \sin (c+d x)}{3 a^2 d}+\frac{7 \cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac{8 \cos (c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac{\cos (c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}\\ \end{align*}

Mathematica [A]  time = 0.371987, size = 177, normalized size = 1.61 \[ \frac{\sec \left (\frac{c}{2}\right ) \sec ^3\left (\frac{1}{2} (c+d x)\right ) \left (147 \sin \left (c+\frac{d x}{2}\right )-239 \sin \left (c+\frac{3 d x}{2}\right )-63 \sin \left (2 c+\frac{3 d x}{2}\right )-15 \sin \left (2 c+\frac{5 d x}{2}\right )-15 \sin \left (3 c+\frac{5 d x}{2}\right )+3 \sin \left (3 c+\frac{7 d x}{2}\right )+3 \sin \left (4 c+\frac{7 d x}{2}\right )+252 d x \cos \left (c+\frac{d x}{2}\right )+84 d x \cos \left (c+\frac{3 d x}{2}\right )+84 d x \cos \left (2 c+\frac{3 d x}{2}\right )-381 \sin \left (\frac{d x}{2}\right )+252 d x \cos \left (\frac{d x}{2}\right )\right )}{192 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^3*(252*d*x*Cos[(d*x)/2] + 252*d*x*Cos[c + (d*x)/2] + 84*d*x*Cos[c + (3*d*x)/2] + 84
*d*x*Cos[2*c + (3*d*x)/2] - 381*Sin[(d*x)/2] + 147*Sin[c + (d*x)/2] - 239*Sin[c + (3*d*x)/2] - 63*Sin[2*c + (3
*d*x)/2] - 15*Sin[2*c + (5*d*x)/2] - 15*Sin[3*c + (5*d*x)/2] + 3*Sin[3*c + (7*d*x)/2] + 3*Sin[4*c + (7*d*x)/2]
))/(192*a^2*d)

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Maple [A]  time = 0.061, size = 122, normalized size = 1.1 \begin{align*}{\frac{1}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{7}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-5\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-3\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+7\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6/a^2/d*tan(1/2*d*x+1/2*c)^3-7/2/a^2/d*tan(1/2*d*x+1/2*c)-5/a^2/d/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2
*c)^3-3/a^2/d/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)+7/a^2/d*arctan(tan(1/2*d*x+1/2*c))

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Maxima [A]  time = 1.73415, size = 221, normalized size = 2.01 \begin{align*} -\frac{\frac{6 \,{\left (\frac{3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} + \frac{2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac{\frac{21 \, \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{42 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) + 5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 + 2*a^2*sin(d*x + c)^
2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) - sin(
d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 42*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

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Fricas [A]  time = 1.64118, size = 257, normalized size = 2.34 \begin{align*} \frac{21 \, d x \cos \left (d x + c\right )^{2} + 42 \, d x \cos \left (d x + c\right ) + 21 \, d x +{\left (3 \, \cos \left (d x + c\right )^{3} - 6 \, \cos \left (d x + c\right )^{2} - 43 \, \cos \left (d x + c\right ) - 32\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(21*d*x*cos(d*x + c)^2 + 42*d*x*cos(d*x + c) + 21*d*x + (3*cos(d*x + c)^3 - 6*cos(d*x + c)^2 - 43*cos(d*x
+ c) - 32)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.361, size = 128, normalized size = 1.16 \begin{align*} \frac{\frac{21 \,{\left (d x + c\right )}}{a^{2}} - \frac{6 \,{\left (5 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, \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^{2}} + \frac{a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 21 \, 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(cos(d*x+c)^2/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/6*(21*(d*x + c)/a^2 - 6*(5*tan(1/2*d*x + 1/2*c)^3 + 3*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*
a^2) + (a^4*tan(1/2*d*x + 1/2*c)^3 - 21*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d