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

Optimal. Leaf size=52 \[ \frac{(2 A+C) \sin (c+d x)}{a d}-\frac{(A+C) \sin (c+d x)}{d (a \sec (c+d x)+a)}-\frac{A x}{a} \]

[Out]

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

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Rubi [A]  time = 0.108917, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {4085, 3787, 2637, 8} \[ \frac{(2 A+C) \sin (c+d x)}{a d}-\frac{(A+C) \sin (c+d x)}{d (a \sec (c+d x)+a)}-\frac{A x}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [B]  time = 0.279259, size = 108, normalized size = 2.08 \[ \frac{\sec \left (\frac{c}{2}\right ) \sec \left (\frac{1}{2} (c+d x)\right ) \left (A \sin \left (c+\frac{d x}{2}\right )+A \sin \left (c+\frac{3 d x}{2}\right )+A \sin \left (2 c+\frac{3 d x}{2}\right )-2 A d x \cos \left (c+\frac{d x}{2}\right )+5 A \sin \left (\frac{d x}{2}\right )-2 A d x \cos \left (\frac{d x}{2}\right )+4 C \sin \left (\frac{d x}{2}\right )\right )}{4 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.08, size = 88, normalized size = 1.7 \begin{align*}{\frac{A}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{C}{ad}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{A\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{A\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{ad}} \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)),x)

[Out]

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

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Maxima [B]  time = 1.42461, size = 158, normalized size = 3.04 \begin{align*} -\frac{A{\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 )} - \frac{C \sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}}{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)),x, algorithm="maxima")

[Out]

-(A*(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*sin(d*x + c)/(a*(cos(d*x + c) + 1)))/d

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Fricas [A]  time = 0.47804, size = 132, normalized size = 2.54 \begin{align*} -\frac{A d x \cos \left (d x + c\right ) + A d x -{\left (A \cos \left (d x + c\right ) + 2 \, A + 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)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

-(A*d*x*cos(d*x + c) + A*d*x - (A*cos(d*x + c) + 2*A + 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{A \cos{\left (c + d x \right )}}{\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{\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)*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c)),x)

[Out]

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

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Giac [A]  time = 1.17157, size = 100, normalized size = 1.92 \begin{align*} -\frac{\frac{{\left (d x + c\right )} A}{a} - \frac{A \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 \, 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}}{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)),x, algorithm="giac")

[Out]

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