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

Optimal. Leaf size=48 \[ -\frac{(A+C) \sin (c+d x)}{d (a \cos (c+d x)+a)}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a d}+\frac{C x}{a} \]

[Out]

(C*x)/a + (A*ArcTanh[Sin[c + d*x]])/(a*d) - ((A + C)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x]))

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Rubi [A]  time = 0.103687, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {3042, 2735, 3770} \[ -\frac{(A+C) \sin (c+d x)}{d (a \cos (c+d x)+a)}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a d}+\frac{C x}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*x)/a + (A*ArcTanh[Sin[c + d*x]])/(a*d) - ((A + C)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x]))

Rule 3042

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*(A + C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x
])^(n + 1))/(f*(b*c - a*d)*(2*m + 1)), x] + Dist[1/(b*(b*c - a*d)*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)
*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(m + n + 2
) + C*(b*c*(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] &&
NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [B]  time = 0.317462, size = 114, normalized size = 2.38 \[ \frac{2 \cos \left (\frac{1}{2} (c+d x)\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right ) \left (-A \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+A \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+C d x\right )-(A+C) \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right )\right )}{a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.05, size = 98, normalized size = 2. \begin{align*} -{\frac{A}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{A}{da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{A}{da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{C}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) C}{da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a/d*A*tan(1/2*d*x+1/2*c)-1/a/d*A*ln(tan(1/2*d*x+1/2*c)-1)+1/a/d*A*ln(tan(1/2*d*x+1/2*c)+1)-1/a/d*C*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 = 1.56837, size = 169, normalized size = 3.52 \begin{align*} \frac{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 )} + A{\left (\frac{\log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac{\log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 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((A+C*cos(d*x+c)^2)*sec(d*x+c)/(a+a*cos(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.37114, size = 242, normalized size = 5.04 \begin{align*} \frac{2 \, C d x \cos \left (d x + c\right ) + 2 \, C d x +{\left (A \cos \left (d x + c\right ) + A\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (A \cos \left (d x + c\right ) + A\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (A + C\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((A+C*cos(d*x+c)^2)*sec(d*x+c)/(a+a*cos(d*x+c)),x, algorithm="fricas")

[Out]

1/2*(2*C*d*x*cos(d*x + c) + 2*C*d*x + (A*cos(d*x + c) + A)*log(sin(d*x + c) + 1) - (A*cos(d*x + c) + A)*log(-s
in(d*x + c) + 1) - 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 \sec{\left (c + d x \right )}}{\cos{\left (c + d x \right )} + 1}\, dx + \int \frac{C \cos ^{2}{\left (c + d x \right )} \sec{\left (c + d x \right )}}{\cos{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.23023, size = 108, normalized size = 2.25 \begin{align*} \frac{\frac{{\left (d x + c\right )} C}{a} + \frac{A \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac{A \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{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}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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