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

Optimal. Leaf size=34 \[ \frac {(B-C) \sin (c+d x)}{d (a \cos (c+d x)+a)}+\frac {C x}{a} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {3029, 2735, 2648} \[ \frac {(B-C) \sin (c+d x)}{d (a \cos (c+d x)+a)}+\frac {C x}{a} \]

Antiderivative was successfully verified.

[In]

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

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

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 3029

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

Rubi steps

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

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Mathematica [B]  time = 0.12, size = 72, normalized size = 2.12 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (2 (B-C) \sin \left (\frac {d x}{2}\right )+C d x \cos \left (c+\frac {d x}{2}\right )+C d x \cos \left (\frac {d x}{2}\right )\right )}{a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.47, size = 43, normalized size = 1.26 \[ \frac {C d x \cos \left (d x + c\right ) + C d x + {\left (B - C\right )} \sin \left (d x + c\right )}{a d \cos \left (d x + c\right ) + a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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]

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

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giac [A]  time = 0.37, size = 43, normalized size = 1.26 \[ \frac {\frac {{\left (d x + c\right )} C}{a} + \frac {B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [A]  time = 0.17, size = 56, normalized size = 1.65 \[ \frac {B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{a d}-\frac {C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [B]  time = 0.96, size = 73, normalized size = 2.15 \[ \frac {C {\left (\frac {2 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + \frac {B \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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))) + B*sin(d*x + c)/(a*(co
s(d*x + c) + 1)))/d

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mupad [B]  time = 1.06, size = 30, normalized size = 0.88 \[ \frac {\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (B-C\right )}{a}+\frac {C\,d\,x}{a}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {B \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}}{\cos {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\cos {\left (c + d x \right )} + 1}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(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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