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

Optimal. Leaf size=98 \[ -\frac{2 (A-B) \sin (c+d x)}{a d}+\frac{(3 A-2 B) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac{(A-B) \sin (c+d x) \cos (c+d x)}{d (a \sec (c+d x)+a)}+\frac{x (3 A-2 B)}{2 a} \]

[Out]

((3*A - 2*B)*x)/(2*a) - (2*(A - B)*Sin[c + d*x])/(a*d) + ((3*A - 2*B)*Cos[c + d*x]*Sin[c + d*x])/(2*a*d) - ((A
- B)*Cos[c + d*x]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x]))

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Rubi [A]  time = 0.149809, antiderivative size = 98, 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 = {4020, 3787, 2635, 8, 2637} \[ -\frac{2 (A-B) \sin (c+d x)}{a d}+\frac{(3 A-2 B) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac{(A-B) \sin (c+d x) \cos (c+d x)}{d (a \sec (c+d x)+a)}+\frac{x (3 A-2 B)}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*A - 2*B)*x)/(2*a) - (2*(A - B)*Sin[c + d*x])/(a*d) + ((3*A - 2*B)*Cos[c + d*x]*Sin[c + d*x])/(2*a*d) - ((A
- B)*Cos[c + d*x]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x]))

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

Mathematica [B]  time = 0.443721, size = 197, normalized size = 2.01 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (4 d x (3 A-2 B) \cos \left (c+\frac{d x}{2}\right )+4 d x (3 A-2 B) \cos \left (\frac{d x}{2}\right )-4 A \sin \left (c+\frac{d x}{2}\right )-3 A \sin \left (c+\frac{3 d x}{2}\right )-3 A \sin \left (2 c+\frac{3 d x}{2}\right )+A \sin \left (2 c+\frac{5 d x}{2}\right )+A \sin \left (3 c+\frac{5 d x}{2}\right )-20 A \sin \left (\frac{d x}{2}\right )+4 B \sin \left (c+\frac{d x}{2}\right )+4 B \sin \left (c+\frac{3 d x}{2}\right )+4 B \sin \left (2 c+\frac{3 d x}{2}\right )+20 B \sin \left (\frac{d x}{2}\right )\right )}{8 a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(4*(3*A - 2*B)*d*x*Cos[(d*x)/2] + 4*(3*A - 2*B)*d*x*Cos[c + (d*x)/2] - 20*A*Sin[(d*
x)/2] + 20*B*Sin[(d*x)/2] - 4*A*Sin[c + (d*x)/2] + 4*B*Sin[c + (d*x)/2] - 3*A*Sin[c + (3*d*x)/2] + 4*B*Sin[c +
(3*d*x)/2] - 3*A*Sin[2*c + (3*d*x)/2] + 4*B*Sin[2*c + (3*d*x)/2] + A*Sin[2*c + (5*d*x)/2] + A*Sin[3*c + (5*d*
x)/2]))/(8*a*d*(1 + Cos[c + d*x]))

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Maple [B]  time = 0.082, size = 211, normalized size = 2.2 \begin{align*} -{\frac{A}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{B}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-3\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}A}{ad \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}B}{ad \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{A}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-2}}+2\,{\frac{B\tan \left ( 1/2\,dx+c/2 \right ) }{ad \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+3\,{\frac{A\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{ad}}-2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) B}{ad}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a/d*A*tan(1/2*d*x+1/2*c)+1/a/d*B*tan(1/2*d*x+1/2*c)-3/a/d/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3*A
+2/a/d/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3*B-1/a/d/(1+tan(1/2*d*x+1/2*c)^2)^2*A*tan(1/2*d*x+1/2*c)
+2/a/d/(1+tan(1/2*d*x+1/2*c)^2)^2*B*tan(1/2*d*x+1/2*c)+3/a/d*A*arctan(tan(1/2*d*x+1/2*c))-2/a/d*arctan(tan(1/2
*d*x+1/2*c))*B

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Maxima [B]  time = 1.44804, size = 304, normalized size = 3.1 \begin{align*} -\frac{A{\left (\frac{\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a + \frac{2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac{3 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac{\sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + B{\left (\frac{2 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac{2 \, \sin \left (d x + c\right )}{{\left (a + \frac{a \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{\sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(A*((sin(d*x + c)/(cos(d*x + c) + 1) + 3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a + 2*a*sin(d*x + c)^2/(cos(d*
x + c) + 1)^2 + a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) - 3*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a + sin(d*x
+ c)/(a*(cos(d*x + c) + 1))) + B*(2*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a - 2*sin(d*x + c)/((a + a*sin(d*
x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) - sin(d*x + c)/(a*(cos(d*x + c) + 1))))/d

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Fricas [A]  time = 0.465594, size = 203, normalized size = 2.07 \begin{align*} \frac{{\left (3 \, A - 2 \, B\right )} d x \cos \left (d x + c\right ) +{\left (3 \, A - 2 \, B\right )} d x +{\left (A \cos \left (d x + c\right )^{2} -{\left (A - 2 \, B\right )} \cos \left (d x + c\right ) - 4 \, A + 4 \, B\right )} \sin \left (d x + c\right )}{2 \,{\left (a d \cos \left (d x + c\right ) + a d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((3*A - 2*B)*d*x*cos(d*x + c) + (3*A - 2*B)*d*x + (A*cos(d*x + c)^2 - (A - 2*B)*cos(d*x + c) - 4*A + 4*B)*
sin(d*x + c))/(a*d*cos(d*x + c) + a*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.29655, size = 166, normalized size = 1.69 \begin{align*} \frac{\frac{{\left (d x + c\right )}{\left (3 \, A - 2 \, B\right )}}{a} - \frac{2 \,{\left (A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{a} - \frac{2 \,{\left (3 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, B \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 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((d*x + c)*(3*A - 2*B)/a - 2*(A*tan(1/2*d*x + 1/2*c) - B*tan(1/2*d*x + 1/2*c))/a - 2*(3*A*tan(1/2*d*x + 1/
2*c)^3 - 2*B*tan(1/2*d*x + 1/2*c)^3 + A*tan(1/2*d*x + 1/2*c) - 2*B*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c
)^2 + 1)^2*a))/d