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

Optimal. Leaf size=51 \[ -\frac{(A-B+C) \sin (c+d x)}{d (a \cos (c+d x)+a)}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a d}+\frac{C x}{a} \]

[Out]

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

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Rubi [A]  time = 0.119369, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {3041, 2735, 3770} \[ -\frac{(A-B+C) \sin (c+d x)}{d (a \cos (c+d x)+a)}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a d}+\frac{C x}{a} \]

Antiderivative was successfully verified.

[In]

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

Rule 3041

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

Mathematica [B]  time = 0.540545, size = 163, normalized size = 3.2 \[ \frac{4 \cos \left (\frac{1}{2} (c+d x)\right ) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right ) \left (-A \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+A \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+C d x\right )-\sec \left (\frac{c}{2}\right ) (A-B+C) \sin \left (\frac{d x}{2}\right )\right )}{a d (\cos (c+d x)+1) (2 A+2 B \cos (c+d x)+C \cos (2 (c+d x))+C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.055, size = 115, normalized size = 2.3 \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{B}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \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+B*cos(d*x+c)+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*B*tan(1/2*
d*x+1/2*c)-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 = 1.48401, size = 197, normalized size = 3.86 \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 )} + \frac{B \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((A+B*cos(d*x+c)+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))) +
B*sin(d*x + c)/(a*(cos(d*x + c) + 1)))/d

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Fricas [A]  time = 1.97946, size = 247, normalized size = 4.84 \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 - B + 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+B*cos(d*x+c)+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 - 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{A \sec{\left (c + d x \right )}}{\cos{\left (c + d x \right )} + 1}\, dx + \int \frac{B \cos{\left (c + d x \right )} \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+B*cos(d*x+c)+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(B*cos(c + d*x)*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 = 1.19232, size = 124, normalized size = 2.43 \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 ) - 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}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+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) - B*tan(1/2*d*x + 1/2*c) + C*tan(1/2*d*x + 1/2*c))/a)/d