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

Optimal. Leaf size=100 \[ \frac{(10 A-4 B+C) \sin (c+d x)}{3 a^2 d}-\frac{(2 A-B) \sin (c+d x)}{a^2 d (\sec (c+d x)+1)}-\frac{x (2 A-B)}{a^2}-\frac{(A-B+C) \sin (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

-(((2*A - B)*x)/a^2) + ((10*A - 4*B + C)*Sin[c + d*x])/(3*a^2*d) - ((2*A - B)*Sin[c + d*x])/(a^2*d*(1 + Sec[c
+ d*x])) - ((A - B + C)*Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2)

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Rubi [A]  time = 0.25863, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.128, Rules used = {4084, 4020, 3787, 2637, 8} \[ \frac{(10 A-4 B+C) \sin (c+d x)}{3 a^2 d}-\frac{(2 A-B) \sin (c+d x)}{a^2 d (\sec (c+d x)+1)}-\frac{x (2 A-B)}{a^2}-\frac{(A-B+C) \sin (c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((2*A - B)*x)/a^2) + ((10*A - 4*B + C)*Sin[c + d*x])/(3*a^2*d) - ((2*A - B)*Sin[c + d*x])/(a^2*d*(1 + Sec[c
+ d*x])) - ((A - B + C)*Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2)

Rule 4084

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[((a*A - b*B + a*C)*Cot[e + f*x]*(a + b*Cs
c[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[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(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, B, 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+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx &=-\frac{(A-B+C) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{\int \frac{\cos (c+d x) (a (4 A-B+C)-a (2 A-2 B-C) \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=-\frac{(2 A-B) \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac{(A-B+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-4 B+C)-3 a^2 (2 A-B) \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac{(2 A-B) \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac{(A-B+C) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{(2 A-B) \int 1 \, dx}{a^2}+\frac{(10 A-4 B+C) \int \cos (c+d x) \, dx}{3 a^2}\\ &=-\frac{(2 A-B) x}{a^2}+\frac{(10 A-4 B+C) \sin (c+d x)}{3 a^2 d}-\frac{(2 A-B) \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac{(A-B+C) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}\\ \end{align*}

Mathematica [B]  time = 0.831113, size = 279, normalized size = 2.79 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (-18 d x (2 A-B) \cos \left (c+\frac{d x}{2}\right )-18 d x (2 A-B) \cos \left (\frac{d x}{2}\right )-30 A \sin \left (c+\frac{d x}{2}\right )+41 A \sin \left (c+\frac{3 d x}{2}\right )+9 A \sin \left (2 c+\frac{3 d x}{2}\right )+3 A \sin \left (2 c+\frac{5 d x}{2}\right )+3 A \sin \left (3 c+\frac{5 d x}{2}\right )-12 A d x \cos \left (c+\frac{3 d x}{2}\right )-12 A d x \cos \left (2 c+\frac{3 d x}{2}\right )+66 A \sin \left (\frac{d x}{2}\right )+24 B \sin \left (c+\frac{d x}{2}\right )-20 B \sin \left (c+\frac{3 d x}{2}\right )+6 B d x \cos \left (c+\frac{3 d x}{2}\right )+6 B d x \cos \left (2 c+\frac{3 d x}{2}\right )-36 B \sin \left (\frac{d x}{2}\right )-12 C \sin \left (c+\frac{d x}{2}\right )+8 C \sin \left (c+\frac{3 d x}{2}\right )+12 C \sin \left (\frac{d x}{2}\right )\right )}{12 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(-18*(2*A - B)*d*x*Cos[(d*x)/2] - 18*(2*A - B)*d*x*Cos[c + (d*x)/2] - 12*A*d*x*Cos[
c + (3*d*x)/2] + 6*B*d*x*Cos[c + (3*d*x)/2] - 12*A*d*x*Cos[2*c + (3*d*x)/2] + 6*B*d*x*Cos[2*c + (3*d*x)/2] + 6
6*A*Sin[(d*x)/2] - 36*B*Sin[(d*x)/2] + 12*C*Sin[(d*x)/2] - 30*A*Sin[c + (d*x)/2] + 24*B*Sin[c + (d*x)/2] - 12*
C*Sin[c + (d*x)/2] + 41*A*Sin[c + (3*d*x)/2] - 20*B*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]))/(12*a^2*d*(1 + Cos[c + d*x])^2)

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Maple [A]  time = 0.098, size = 187, normalized size = 1.9 \begin{align*} -{\frac{A}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{B}{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{3\,B}{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}}}+2\,{\frac{B\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+B*sec(d*x+c)+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*tan(1/2*d*x+1/2*c)^3*B-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)-3/2/d/a^2*B*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))+2/d/a^2*B*arctan(tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.44759, size = 317, normalized size = 3.17 \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 )} - B{\left (\frac{\frac{9 \, \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{12 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\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+B*sec(d*x+c)+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
))) - B*((9*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 12*arctan(sin(d*x + c
)/(cos(d*x + c) + 1))/a^2) + 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

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Fricas [A]  time = 0.487391, size = 309, normalized size = 3.09 \begin{align*} -\frac{3 \,{\left (2 \, A - B\right )} d x \cos \left (d x + c\right )^{2} + 6 \,{\left (2 \, A - B\right )} d x \cos \left (d x + c\right ) + 3 \,{\left (2 \, A - B\right )} d x -{\left (3 \, A \cos \left (d x + c\right )^{2} +{\left (14 \, A - 5 \, B + 2 \, C\right )} \cos \left (d x + c\right ) + 10 \, A - 4 \, B + 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+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/3*(3*(2*A - B)*d*x*cos(d*x + c)^2 + 6*(2*A - B)*d*x*cos(d*x + c) + 3*(2*A - B)*d*x - (3*A*cos(d*x + c)^2 +
(14*A - 5*B + 2*C)*cos(d*x + c) + 10*A - 4*B + C)*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{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{B \cos{\left (c + d x \right )} \sec{\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+B*sec(d*x+c)+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(B*cos(c + d*x)*sec(c + d*x)/(se
c(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

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Giac [A]  time = 1.26524, size = 205, normalized size = 2.05 \begin{align*} -\frac{\frac{6 \,{\left (d x + c\right )}{\left (2 \, A - B\right )}}{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} - B 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 ) + 9 \, B 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+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-1/6*(6*(d*x + c)*(2*A - B)/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 - B*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)
+ 9*B*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