\(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec (c+d x)}{a+a \cos (c+d x)} \, dx\) [343]

Optimal result

Integrand size = 39, antiderivative size = 51 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{a+a \cos (c+d x)} \, dx=\frac {C x}{a}+\frac {A \text {arctanh}(\sin (c+d x))}{a d}-\frac {(A-B+C) \sin (c+d x)}{d (a+a \cos (c+d x))} \]

[Out]

C*x/a+A*arctanh(sin(d*x+c))/a/d-(A-B+C)*sin(d*x+c)/d/(a+a*cos(d*x+c))

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3120, 2814, 3855} \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{a+a \cos (c+d x)} \, dx=\frac {A \text {arctanh}(\sin (c+d x))}{a d}-\frac {(A-B+C) \sin (c+d x)}{d (a \cos (c+d x)+a)}+\frac {C x}{a} \]

[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 2814

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 3120

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 3855

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

Rubi steps \begin{align*} \text {integral}& = -\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 \text {arctanh}(\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] (verified)

Leaf count is larger than twice the leaf count of optimal. \(163\) vs. \(2(51)=102\).

Time = 1.47 (sec) , antiderivative size = 163, normalized size of antiderivative = 3.20 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{a+a \cos (c+d x)} \, dx=\frac {4 \cos \left (\frac {1}{2} (c+d x)\right ) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right ) \left (C d x-A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-(A-B+C) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )\right )}{a d (1+\cos (c+d x)) (2 A+C+2 B \cos (c+d x)+C \cos (2 (c+d x)))} \]

[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)]))

Maple [A] (verified)

Time = 2.79 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.18

method	result	size
parallelrisch	\(\frac {-A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-A +B -C \right )+d x C}{a d}\)	\(60\)
derivativedivides	\(\frac {-A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 C \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\)	\(86\)
default	\(\frac {-A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 C \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\)	\(86\)
risch	\(\frac {C x}{a}-\frac {2 i A}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {2 i B}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}-\frac {2 i C}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a d}-\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a d}\)	\(120\)
norman	\(\frac {\frac {C x}{a}+\frac {C x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {2 C x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {\left (A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {2 \left (A -B +C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {\left (A -B +C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a d}-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a d}\)	\(174\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.78 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{a+a \cos (c+d x)} \, dx=\frac {2 \, C d x \cos \left (d x + c\right ) + 2 \, C d x + {\left (A \cos \left (d x + c\right ) + A\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A \cos \left (d x + c\right ) + A\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (A - B + C\right )} \sin \left (d x + c\right )}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]

[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)

Sympy [F]

\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\int \frac {A \sec {\left (c + d x \right )}}{\cos {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}}{\cos {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\cos {\left (c + d x \right )} + 1}\, dx}{a} \]

[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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 146 vs. \(2 (51) = 102\).

Time = 0.29 (sec) , antiderivative size = 146, normalized size of antiderivative = 2.86 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{a+a \cos (c+d x)} \, dx=\frac {C {\left (\frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + A {\left (\frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + \frac {B \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}}{d} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.80 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\frac {{\left (d x + c\right )} C}{a} + \frac {A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac {A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a}}{d} \]

[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

Mupad [B] (verification not implemented)

Time = 1.44 (sec) , antiderivative size = 113, normalized size of antiderivative = 2.22 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x)}{a+a \cos (c+d x)} \, dx=\frac {2\,A\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+2\,C\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a\,d}-\frac {A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-B\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a\,d\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )} \]

[In]

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

[Out]

(2*A*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 2*C*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(a*d) - (
A*sin(c/2 + (d*x)/2) - B*sin(c/2 + (d*x)/2) + C*sin(c/2 + (d*x)/2))/(a*d*cos(c/2 + (d*x)/2))