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

Optimal. Leaf size=176 \[ -\frac{576 \sin (c+d x)}{35 a^4 d}+\frac{21 \sin (c+d x) \cos (c+d x)}{2 a^4 d}-\frac{288 \sin (c+d x) \cos (c+d x)}{35 a^4 d (\sec (c+d x)+1)}-\frac{43 \sin (c+d x) \cos (c+d x)}{35 a^4 d (\sec (c+d x)+1)^2}+\frac{21 x}{2 a^4}-\frac{2 \sin (c+d x) \cos (c+d x)}{5 a d (a \sec (c+d x)+a)^3}-\frac{\sin (c+d x) \cos (c+d x)}{7 d (a \sec (c+d x)+a)^4} \]

[Out]

(21*x)/(2*a^4) - (576*Sin[c + d*x])/(35*a^4*d) + (21*Cos[c + d*x]*Sin[c + d*x])/(2*a^4*d) - (43*Cos[c + d*x]*S
in[c + d*x])/(35*a^4*d*(1 + Sec[c + d*x])^2) - (288*Cos[c + d*x]*Sin[c + d*x])/(35*a^4*d*(1 + Sec[c + d*x])) -
(Cos[c + d*x]*Sin[c + d*x])/(7*d*(a + a*Sec[c + d*x])^4) - (2*Cos[c + d*x]*Sin[c + d*x])/(5*a*d*(a + a*Sec[c
+ d*x])^3)

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Rubi [A]  time = 0.392953, antiderivative size = 176, 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 = {3817, 4020, 3787, 2635, 8, 2637} \[ -\frac{576 \sin (c+d x)}{35 a^4 d}+\frac{21 \sin (c+d x) \cos (c+d x)}{2 a^4 d}-\frac{288 \sin (c+d x) \cos (c+d x)}{35 a^4 d (\sec (c+d x)+1)}-\frac{43 \sin (c+d x) \cos (c+d x)}{35 a^4 d (\sec (c+d x)+1)^2}+\frac{21 x}{2 a^4}-\frac{2 \sin (c+d x) \cos (c+d x)}{5 a d (a \sec (c+d x)+a)^3}-\frac{\sin (c+d x) \cos (c+d x)}{7 d (a \sec (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(21*x)/(2*a^4) - (576*Sin[c + d*x])/(35*a^4*d) + (21*Cos[c + d*x]*Sin[c + d*x])/(2*a^4*d) - (43*Cos[c + d*x]*S
in[c + d*x])/(35*a^4*d*(1 + Sec[c + d*x])^2) - (288*Cos[c + d*x]*Sin[c + d*x])/(35*a^4*d*(1 + Sec[c + d*x])) -
(Cos[c + d*x]*Sin[c + d*x])/(7*d*(a + a*Sec[c + d*x])^4) - (2*Cos[c + d*x]*Sin[c + d*x])/(5*a*d*(a + a*Sec[c
+ d*x])^3)

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))^4} \, dx &=-\frac{\cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{\int \frac{\cos ^2(c+d x) (-9 a+5 a \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac{\cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{2 \cos (c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac{\int \frac{\cos ^2(c+d x) \left (-73 a^2+56 a^2 \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac{43 \cos (c+d x) \sin (c+d x)}{35 a^4 d (1+\sec (c+d x))^2}-\frac{\cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{2 \cos (c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac{\int \frac{\cos ^2(c+d x) \left (-477 a^3+387 a^3 \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{105 a^6}\\ &=-\frac{43 \cos (c+d x) \sin (c+d x)}{35 a^4 d (1+\sec (c+d x))^2}-\frac{\cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{2 \cos (c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac{288 \cos (c+d x) \sin (c+d x)}{35 d \left (a^4+a^4 \sec (c+d x)\right )}-\frac{\int \cos ^2(c+d x) \left (-2205 a^4+1728 a^4 \sec (c+d x)\right ) \, dx}{105 a^8}\\ &=-\frac{43 \cos (c+d x) \sin (c+d x)}{35 a^4 d (1+\sec (c+d x))^2}-\frac{\cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{2 \cos (c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac{288 \cos (c+d x) \sin (c+d x)}{35 d \left (a^4+a^4 \sec (c+d x)\right )}-\frac{576 \int \cos (c+d x) \, dx}{35 a^4}+\frac{21 \int \cos ^2(c+d x) \, dx}{a^4}\\ &=-\frac{576 \sin (c+d x)}{35 a^4 d}+\frac{21 \cos (c+d x) \sin (c+d x)}{2 a^4 d}-\frac{43 \cos (c+d x) \sin (c+d x)}{35 a^4 d (1+\sec (c+d x))^2}-\frac{\cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{2 \cos (c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac{288 \cos (c+d x) \sin (c+d x)}{35 d \left (a^4+a^4 \sec (c+d x)\right )}+\frac{21 \int 1 \, dx}{2 a^4}\\ &=\frac{21 x}{2 a^4}-\frac{576 \sin (c+d x)}{35 a^4 d}+\frac{21 \cos (c+d x) \sin (c+d x)}{2 a^4 d}-\frac{43 \cos (c+d x) \sin (c+d x)}{35 a^4 d (1+\sec (c+d x))^2}-\frac{\cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac{2 \cos (c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac{288 \cos (c+d x) \sin (c+d x)}{35 d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.551801, size = 289, normalized size = 1.64 \[ \frac{\sec \left (\frac{c}{2}\right ) \sec ^7\left (\frac{1}{2} (c+d x)\right ) \left (128730 \sin \left (c+\frac{d x}{2}\right )-140826 \sin \left (c+\frac{3 d x}{2}\right )+44310 \sin \left (2 c+\frac{3 d x}{2}\right )-60487 \sin \left (2 c+\frac{5 d x}{2}\right )+1225 \sin \left (3 c+\frac{5 d x}{2}\right )-12001 \sin \left (3 c+\frac{7 d x}{2}\right )-3185 \sin \left (4 c+\frac{7 d x}{2}\right )-315 \sin \left (4 c+\frac{9 d x}{2}\right )-315 \sin \left (5 c+\frac{9 d x}{2}\right )+35 \sin \left (5 c+\frac{11 d x}{2}\right )+35 \sin \left (6 c+\frac{11 d x}{2}\right )+102900 d x \cos \left (c+\frac{d x}{2}\right )+61740 d x \cos \left (c+\frac{3 d x}{2}\right )+61740 d x \cos \left (2 c+\frac{3 d x}{2}\right )+20580 d x \cos \left (2 c+\frac{5 d x}{2}\right )+20580 d x \cos \left (3 c+\frac{5 d x}{2}\right )+2940 d x \cos \left (3 c+\frac{7 d x}{2}\right )+2940 d x \cos \left (4 c+\frac{7 d x}{2}\right )-179830 \sin \left (\frac{d x}{2}\right )+102900 d x \cos \left (\frac{d x}{2}\right )\right )}{35840 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^7*(102900*d*x*Cos[(d*x)/2] + 102900*d*x*Cos[c + (d*x)/2] + 61740*d*x*Cos[c + (3*d*x
)/2] + 61740*d*x*Cos[2*c + (3*d*x)/2] + 20580*d*x*Cos[2*c + (5*d*x)/2] + 20580*d*x*Cos[3*c + (5*d*x)/2] + 2940
*d*x*Cos[3*c + (7*d*x)/2] + 2940*d*x*Cos[4*c + (7*d*x)/2] - 179830*Sin[(d*x)/2] + 128730*Sin[c + (d*x)/2] - 14
0826*Sin[c + (3*d*x)/2] + 44310*Sin[2*c + (3*d*x)/2] - 60487*Sin[2*c + (5*d*x)/2] + 1225*Sin[3*c + (5*d*x)/2]
- 12001*Sin[3*c + (7*d*x)/2] - 3185*Sin[4*c + (7*d*x)/2] - 315*Sin[4*c + (9*d*x)/2] - 315*Sin[5*c + (9*d*x)/2]
+ 35*Sin[5*c + (11*d*x)/2] + 35*Sin[6*c + (11*d*x)/2]))/(35840*a^4*d)

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Maple [A]  time = 0.063, size = 160, normalized size = 0.9 \begin{align*}{\frac{1}{56\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}-{\frac{9}{40\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{13}{8\,d{a}^{4}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{111}{8\,d{a}^{4}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-9\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{d{a}^{4} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-7\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{4} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+21\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{4}}} \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))^4,x)

[Out]

1/56/d/a^4*tan(1/2*d*x+1/2*c)^7-9/40/d/a^4*tan(1/2*d*x+1/2*c)^5+13/8/d/a^4*tan(1/2*d*x+1/2*c)^3-111/8/d/a^4*ta
n(1/2*d*x+1/2*c)-9/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3-7/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^2*ta
n(1/2*d*x+1/2*c)+21/d/a^4*arctan(tan(1/2*d*x+1/2*c))

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Maxima [A]  time = 1.6581, size = 275, normalized size = 1.56 \begin{align*} -\frac{\frac{280 \,{\left (\frac{7 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{9 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{4} + \frac{2 \, a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a^{4} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac{\frac{3885 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{455 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{63 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac{5880 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}}{280 \, 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))^4,x, algorithm="maxima")

[Out]

-1/280*(280*(7*sin(d*x + c)/(cos(d*x + c) + 1) + 9*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^4 + 2*a^4*sin(d*x +
c)^2/(cos(d*x + c) + 1)^2 + a^4*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (3885*sin(d*x + c)/(cos(d*x + c) + 1)
- 455*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 63*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 5*sin(d*x + c)^7/(cos(d*x
+ c) + 1)^7)/a^4 - 5880*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^4)/d

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Fricas [A]  time = 1.68707, size = 470, normalized size = 2.67 \begin{align*} \frac{735 \, d x \cos \left (d x + c\right )^{4} + 2940 \, d x \cos \left (d x + c\right )^{3} + 4410 \, d x \cos \left (d x + c\right )^{2} + 2940 \, d x \cos \left (d x + c\right ) + 735 \, d x +{\left (35 \, \cos \left (d x + c\right )^{5} - 140 \, \cos \left (d x + c\right )^{4} - 2012 \, \cos \left (d x + c\right )^{3} - 4548 \, \cos \left (d x + c\right )^{2} - 3873 \, \cos \left (d x + c\right ) - 1152\right )} \sin \left (d x + c\right )}{70 \,{\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} 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))^4,x, algorithm="fricas")

[Out]

1/70*(735*d*x*cos(d*x + c)^4 + 2940*d*x*cos(d*x + c)^3 + 4410*d*x*cos(d*x + c)^2 + 2940*d*x*cos(d*x + c) + 735
*d*x + (35*cos(d*x + c)^5 - 140*cos(d*x + c)^4 - 2012*cos(d*x + c)^3 - 4548*cos(d*x + c)^2 - 3873*cos(d*x + c)
- 1152)*sin(d*x + c))/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d
*x + c) + a^4*d)

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

[Out]

Timed out

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Giac [A]  time = 1.36911, size = 173, normalized size = 0.98 \begin{align*} \frac{\frac{2940 \,{\left (d x + c\right )}}{a^{4}} - \frac{280 \,{\left (9 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 7 \, \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^{4}} + \frac{5 \, a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 63 \, a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 455 \, a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3885 \, a^{24} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{28}}}{280 \, 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))^4,x, algorithm="giac")

[Out]

1/280*(2940*(d*x + c)/a^4 - 280*(9*tan(1/2*d*x + 1/2*c)^3 + 7*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 +
1)^2*a^4) + (5*a^24*tan(1/2*d*x + 1/2*c)^7 - 63*a^24*tan(1/2*d*x + 1/2*c)^5 + 455*a^24*tan(1/2*d*x + 1/2*c)^3
- 3885*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d