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

Optimal. Leaf size=83 \[ -\frac{(4 A-B-2 C) \sin (c+d x)}{3 a^2 d (\cos (c+d x)+1)}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac{(A-B+C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[Out]

(A*ArcTanh[Sin[c + d*x]])/(a^2*d) - ((4*A - B - 2*C)*Sin[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])) - ((A - B + C)
*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2)

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Rubi [A]  time = 0.214822, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {3041, 2978, 12, 3770} \[ -\frac{(4 A-B-2 C) \sin (c+d x)}{3 a^2 d (\cos (c+d x)+1)}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac{(A-B+C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

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])^2,x]

[Out]

(A*ArcTanh[Sin[c + d*x]])/(a^2*d) - ((4*A - B - 2*C)*Sin[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])) - ((A - B + C)
*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2)

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 2978

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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))^2} \, dx &=-\frac{(A-B+C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \frac{(3 a A-a (A-B-2 C) \cos (c+d x)) \sec (c+d x)}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=-\frac{(4 A-B-2 C) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{(A-B+C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int 3 a^2 A \sec (c+d x) \, dx}{3 a^4}\\ &=-\frac{(4 A-B-2 C) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{(A-B+C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{A \int \sec (c+d x) \, dx}{a^2}\\ &=\frac{A \tanh ^{-1}(\sin (c+d x))}{a^2 d}-\frac{(4 A-B-2 C) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{(A-B+C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}\\ \end{align*}

Mathematica [B]  time = 0.849884, size = 221, normalized size = 2.66 \[ -\frac{4 \cos \left (\frac{1}{2} (c+d x)\right ) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \left (\tan \left (\frac{c}{2}\right ) (A-B+C) \cos \left (\frac{1}{2} (c+d x)\right )+\sec \left (\frac{c}{2}\right ) (A-B+C) \sin \left (\frac{d x}{2}\right )+2 \sec \left (\frac{c}{2}\right ) (4 A-B-2 C) \sin \left (\frac{d x}{2}\right ) \cos ^2\left (\frac{1}{2} (c+d x)\right )+6 A \cos ^3\left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{3 a^2 d (\cos (c+d x)+1)^2 (2 A+2 B \cos (c+d x)+C \cos (2 (c+d x))+C)} \]

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])^2,x]

[Out]

(-4*Cos[(c + d*x)/2]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*(6*A*Cos[(c + d*x)/2]^3*(Log[Cos[(c + d*x)/2] - S
in[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + (A - B + C)*Sec[c/2]*Sin[(d*x)/2] + 2*(4*A - B
- 2*C)*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[(d*x)/2] + (A - B + C)*Cos[(c + d*x)/2]*Tan[c/2]))/(3*a^2*d*(1 + Cos[c
+ d*x])^2*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*(c + d*x)]))

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Maple [A]  time = 0.057, size = 157, normalized size = 1.9 \begin{align*}{\frac{B}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{3\,A}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{A}{d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{A}{d{a}^{2}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{A}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{B}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{C}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{C}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) } \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))^2,x)

[Out]

1/2/d/a^2*B*tan(1/2*d*x+1/2*c)-3/2/d/a^2*A*tan(1/2*d*x+1/2*c)+1/d/a^2*A*ln(tan(1/2*d*x+1/2*c)+1)-1/d/a^2*A*ln(
tan(1/2*d*x+1/2*c)-1)-1/6/d/a^2*tan(1/2*d*x+1/2*c)^3*A+1/6/d/a^2*tan(1/2*d*x+1/2*c)^3*B-1/6/d/a^2*C*tan(1/2*d*
x+1/2*c)^3+1/2/d/a^2*C*tan(1/2*d*x+1/2*c)

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Maxima [B]  time = 0.996903, size = 257, normalized size = 3.1 \begin{align*} -\frac{A{\left (\frac{\frac{9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{6 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac{6 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )} - \frac{B{\left (\frac{3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}} - \frac{C{\left (\frac{3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, 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))^2,x, algorithm="maxima")

[Out]

-1/6*(A*((9*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 6*log(sin(d*x + c)/(c
os(d*x + c) + 1) + 1)/a^2 + 6*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2) - B*(3*sin(d*x + c)/(cos(d*x + c)
+ 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - C*(3*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d
*x + c) + 1)^3)/a^2)/d

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Fricas [A]  time = 2.0472, size = 351, normalized size = 4.23 \begin{align*} \frac{3 \,{\left (A \cos \left (d x + c\right )^{2} + 2 \, A \cos \left (d x + c\right ) + A\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (A \cos \left (d x + c\right )^{2} + 2 \, A \cos \left (d x + c\right ) + A\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left ({\left (4 \, A - B - 2 \, C\right )} \cos \left (d x + c\right ) + 5 \, A - 2 \, B - C\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} 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))^2,x, algorithm="fricas")

[Out]

1/6*(3*(A*cos(d*x + c)^2 + 2*A*cos(d*x + c) + A)*log(sin(d*x + c) + 1) - 3*(A*cos(d*x + c)^2 + 2*A*cos(d*x + c
) + A)*log(-sin(d*x + c) + 1) - 2*((4*A - B - 2*C)*cos(d*x + c) + 5*A - 2*B - C)*sin(d*x + c))/(a^2*d*cos(d*x
+ c)^2 + 2*a^2*d*cos(d*x + c) + a^2*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 ^{2}{\left (c + d x \right )} + 2 \cos{\left (c + d x \right )} + 1}\, dx + \int \frac{B \cos{\left (c + d x \right )} \sec{\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \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 ^{2}{\left (c + d x \right )} + 2 \cos{\left (c + d x \right )} + 1}\, dx}{a^{2}} \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))**2,x)

[Out]

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

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Giac [A]  time = 1.1866, size = 194, normalized size = 2.34 \begin{align*} \frac{\frac{6 \, A \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac{6 \, A \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} - \frac{A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{6}}}{6 \, 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))^2,x, algorithm="giac")

[Out]

1/6*(6*A*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 6*A*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2 - (A*a^4*tan(1/2*
d*x + 1/2*c)^3 - B*a^4*tan(1/2*d*x + 1/2*c)^3 + C*a^4*tan(1/2*d*x + 1/2*c)^3 + 9*A*a^4*tan(1/2*d*x + 1/2*c) -
3*B*a^4*tan(1/2*d*x + 1/2*c) - 3*C*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d