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

Optimal. Leaf size=82 \[ \frac{(10 A+C) \sin (c+d x)}{3 a^2 d}-\frac{2 A \sin (c+d x)}{a^2 d (\sec (c+d x)+1)}-\frac{2 A x}{a^2}-\frac{(A+C) \sin (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

(-2*A*x)/a^2 + ((10*A + C)*Sin[c + d*x])/(3*a^2*d) - (2*A*Sin[c + d*x])/(a^2*d*(1 + Sec[c + d*x])) - ((A + C)*
Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2)

________________________________________________________________________________________

Rubi [A]  time = 0.217872, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {4085, 4020, 3787, 2637, 8} \[ \frac{(10 A+C) \sin (c+d x)}{3 a^2 d}-\frac{2 A \sin (c+d x)}{a^2 d (\sec (c+d x)+1)}-\frac{2 A x}{a^2}-\frac{(A+C) \sin (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A*x)/a^2 + ((10*A + C)*Sin[c + d*x])/(3*a^2*d) - (2*A*Sin[c + d*x])/(a^2*d*(1 + Sec[c + d*x])) - ((A + C)*
Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2)

Rule 4085

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

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) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx &=-\frac{(A+C) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{\int \frac{\cos (c+d x) (-a (4 A+C)+a (2 A-C) \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=-\frac{2 A \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac{(A+C) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{\int \cos (c+d x) \left (-a^2 (10 A+C)+6 a^2 A \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac{2 A \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac{(A+C) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{(2 A) \int 1 \, dx}{a^2}+\frac{(10 A+C) \int \cos (c+d x) \, dx}{3 a^2}\\ &=-\frac{2 A x}{a^2}+\frac{(10 A+C) \sin (c+d x)}{3 a^2 d}-\frac{2 A \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac{(A+C) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}\\ \end{align*}

Mathematica [B]  time = 0.764618, size = 195, normalized size = 2.38 \[ \frac{\sec \left (\frac{c}{2}\right ) \sec ^3\left (\frac{1}{2} (c+d x)\right ) \left (-30 A \sin \left (c+\frac{d x}{2}\right )+41 A \sin \left (c+\frac{3 d x}{2}\right )+9 A \sin \left (2 c+\frac{3 d x}{2}\right )+3 A \sin \left (2 c+\frac{5 d x}{2}\right )+3 A \sin \left (3 c+\frac{5 d x}{2}\right )-36 A d x \cos \left (c+\frac{d x}{2}\right )-12 A d x \cos \left (c+\frac{3 d x}{2}\right )-12 A d x \cos \left (2 c+\frac{3 d x}{2}\right )+66 A \sin \left (\frac{d x}{2}\right )-36 A d x \cos \left (\frac{d x}{2}\right )-12 C \sin \left (c+\frac{d x}{2}\right )+8 C \sin \left (c+\frac{3 d x}{2}\right )+12 C \sin \left (\frac{d x}{2}\right )\right )}{48 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^3*(-36*A*d*x*Cos[(d*x)/2] - 36*A*d*x*Cos[c + (d*x)/2] - 12*A*d*x*Cos[c + (3*d*x)/2]
- 12*A*d*x*Cos[2*c + (3*d*x)/2] + 66*A*Sin[(d*x)/2] + 12*C*Sin[(d*x)/2] - 30*A*Sin[c + (d*x)/2] - 12*C*Sin[c
+ (d*x)/2] + 41*A*Sin[c + (3*d*x)/2] + 8*C*Sin[c + (3*d*x)/2] + 9*A*Sin[2*c + (3*d*x)/2] + 3*A*Sin[2*c + (5*d*
x)/2] + 3*A*Sin[3*c + (5*d*x)/2]))/(48*a^2*d)

________________________________________________________________________________________

Maple [A]  time = 0.088, size = 130, normalized size = 1.6 \begin{align*} -{\frac{A}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{C}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{5\,A}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{C}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{A\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 ) }}-4\,{\frac{A\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)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x)

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 1.42876, size = 223, normalized size = 2.72 \begin{align*} \frac{A{\left (\frac{\frac{15 \, \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{24 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac{12 \, \sin \left (d x + c\right )}{{\left (a^{2} + \frac{a^{2} \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 )}}\right )} + \frac{C{\left (\frac{3 \, \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}}\right )}}{a^{2}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(A*((15*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 24*arctan(sin(d*x + c
)/(cos(d*x + c) + 1))/a^2 + 12*sin(d*x + c)/((a^2 + a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1
))) + C*(3*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2)/d

________________________________________________________________________________________

Fricas [A]  time = 0.483188, size = 261, normalized size = 3.18 \begin{align*} -\frac{6 \, A d x \cos \left (d x + c\right )^{2} + 12 \, A d x \cos \left (d x + c\right ) + 6 \, A d x -{\left (3 \, A \cos \left (d x + c\right )^{2} + 2 \,{\left (7 \, A + C\right )} \cos \left (d x + c\right ) + 10 \, A + C\right )} \sin \left (d x + c\right )}{3 \,{\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)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/3*(6*A*d*x*cos(d*x + c)^2 + 12*A*d*x*cos(d*x + c) + 6*A*d*x - (3*A*cos(d*x + c)^2 + 2*(7*A + C)*cos(d*x + c
) + 10*A + C)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.21883, size = 154, normalized size = 1.88 \begin{align*} -\frac{\frac{12 \,{\left (d x + c\right )} A}{a^{2}} - \frac{12 \, A \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^{2}} + \frac{A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 15 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, C 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)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-1/6*(12*(d*x + c)*A/a^2 - 12*A*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^2) + (A*a^4*tan(1/2*d*x +
1/2*c)^3 + C*a^4*tan(1/2*d*x + 1/2*c)^3 - 15*A*a^4*tan(1/2*d*x + 1/2*c) - 3*C*a^4*tan(1/2*d*x + 1/2*c))/a^6)/
d