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

Optimal. Leaf size=60 \[ \frac{(2 B-C) \sin (c+d x)}{a d}-\frac{(B-C) \sin (c+d x)}{d (a \sec (c+d x)+a)}-\frac{x (B-C)}{a} \]

[Out]

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

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Rubi [A]  time = 0.195895, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {4072, 4020, 3787, 2637, 8} \[ \frac{(2 B-C) \sin (c+d x)}{a d}-\frac{(B-C) \sin (c+d x)}{d (a \sec (c+d x)+a)}-\frac{x (B-C)}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4072

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

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

Mathematica [A]  time = 0.342519, size = 76, normalized size = 1.27 \[ \frac{2 \cos \left (\frac{1}{2} (c+d x)\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right ) (B \sin (c+d x)+d x (C-B))+(B-C) \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right )\right )}{a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.083, size = 108, normalized size = 1.8 \begin{align*}{\frac{B}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{C}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+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\,{\frac{B\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 ) C}{ad}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.4173, size = 193, normalized size = 3.22 \begin{align*} -\frac{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 )} - C{\left (\frac{2 \, \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 )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(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))) - C*(2*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a -
sin(d*x + c)/(a*(cos(d*x + c) + 1))))/d

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Fricas [A]  time = 0.475969, size = 149, normalized size = 2.48 \begin{align*} -\frac{{\left (B - C\right )} d x \cos \left (d x + c\right ) +{\left (B - C\right )} d x -{\left (B \cos \left (d x + c\right ) + 2 \, B - C\right )} \sin \left (d x + c\right )}{a d \cos \left (d x + c\right ) + a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((B - C)*d*x*cos(d*x + c) + (B - C)*d*x - (B*cos(d*x + c) + 2*B - C)*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{B \cos ^{2}{\left (c + d x \right )} \sec{\left (c + d x \right )}}{\sec{\left (c + d x \right )} + 1}\, dx + \int \frac{C \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\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*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.12124, size = 107, normalized size = 1.78 \begin{align*} -\frac{\frac{{\left (d x + c\right )}{\left (B - C\right )}}{a} - \frac{B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a} - \frac{2 \, B \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}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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