3.360 \(\int \frac{(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=124 \[ -\frac{(22 A-2 B-3 C) \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac{(7 A-2 B-3 C) \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac{(A-B+C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

[Out]

(A*ArcTanh[Sin[c + d*x]])/(a^3*d) - ((A - B + C)*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) - ((7*A - 2*B - 3*
C)*Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - ((22*A - 2*B - 3*C)*Sin[c + d*x])/(15*d*(a^3 + a^3*Cos[c +
d*x]))

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Rubi [A]  time = 0.352924, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {3041, 2978, 12, 3770} \[ -\frac{(22 A-2 B-3 C) \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{A \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac{(7 A-2 B-3 C) \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac{(A-B+C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

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])^3,x]

[Out]

(A*ArcTanh[Sin[c + d*x]])/(a^3*d) - ((A - B + C)*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) - ((7*A - 2*B - 3*
C)*Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - ((22*A - 2*B - 3*C)*Sin[c + d*x])/(15*d*(a^3 + a^3*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 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))^3} \, dx &=-\frac{(A-B+C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{\int \frac{(5 a A-a (2 A-2 B-3 C) \cos (c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(A-B+C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(7 A-2 B-3 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{\int \frac{\left (15 a^2 A-a^2 (7 A-2 B-3 C) \cos (c+d x)\right ) \sec (c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=-\frac{(A-B+C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(7 A-2 B-3 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(22 A-2 B-3 C) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac{\int 15 a^3 A \sec (c+d x) \, dx}{15 a^6}\\ &=-\frac{(A-B+C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(7 A-2 B-3 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(22 A-2 B-3 C) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac{A \int \sec (c+d x) \, dx}{a^3}\\ &=\frac{A \tanh ^{-1}(\sin (c+d x))}{a^3 d}-\frac{(A-B+C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac{(7 A-2 B-3 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{(22 A-2 B-3 C) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 1.63551, size = 276, normalized size = 2.23 \[ -\frac{\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \left (\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (5 (29 A-4 B-3 C) \sin \left (\frac{d x}{2}\right )+15 (C-5 A) \sin \left (c+\frac{d x}{2}\right )+95 A \sin \left (c+\frac{3 d x}{2}\right )-15 A \sin \left (2 c+\frac{3 d x}{2}\right )+22 A \sin \left (2 c+\frac{5 d x}{2}\right )-10 B \sin \left (c+\frac{3 d x}{2}\right )-2 B \sin \left (2 c+\frac{5 d x}{2}\right )-15 C \sin \left (c+\frac{3 d x}{2}\right )-3 C \sin \left (2 c+\frac{5 d x}{2}\right )\right )+240 A \cos ^6\left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{15 a^3 d (\cos (c+d x)+1)^3 (2 A+2 B \cos (c+d x)+C \cos (2 (c+d x))+C)} \]

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])^3,x]

[Out]

-((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*(240*A*Cos[(c + d*x)/2]^6*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]
- Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Cos[(c + d*x)/2]*Sec[c/2]*(5*(29*A - 4*B - 3*C)*Sin[(d*x)/2] + 1
5*(-5*A + C)*Sin[c + (d*x)/2] + 95*A*Sin[c + (3*d*x)/2] - 10*B*Sin[c + (3*d*x)/2] - 15*C*Sin[c + (3*d*x)/2] -
15*A*Sin[2*c + (3*d*x)/2] + 22*A*Sin[2*c + (5*d*x)/2] - 2*B*Sin[2*c + (5*d*x)/2] - 3*C*Sin[2*c + (5*d*x)/2])))
/(15*a^3*d*(1 + Cos[c + d*x])^3*(2*A + C + 2*B*Cos[c + d*x] + C*Cos[2*(c + d*x)]))

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Maple [A]  time = 0.059, size = 197, normalized size = 1.6 \begin{align*}{\frac{B}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{7\,A}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{A}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{A}{d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{A}{d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{A}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{B}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{C}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{B}{6\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{C}{4\,d{a}^{3}}\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))^3,x)

[Out]

1/4/d/a^3*B*tan(1/2*d*x+1/2*c)-7/4/d/a^3*A*tan(1/2*d*x+1/2*c)-1/20/d/a^3*A*tan(1/2*d*x+1/2*c)^5+1/d/a^3*A*ln(t
an(1/2*d*x+1/2*c)+1)-1/d/a^3*A*ln(tan(1/2*d*x+1/2*c)-1)-1/3/d/a^3*tan(1/2*d*x+1/2*c)^3*A+1/20/d/a^3*B*tan(1/2*
d*x+1/2*c)^5-1/20/d/a^3*C*tan(1/2*d*x+1/2*c)^5+1/6/d/a^3*tan(1/2*d*x+1/2*c)^3*B+1/4/d/a^3*C*tan(1/2*d*x+1/2*c)

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Maxima [A]  time = 1.02899, size = 313, normalized size = 2.52 \begin{align*} -\frac{A{\left (\frac{\frac{105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac{60 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac{60 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - \frac{B{\left (\frac{15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}} - \frac{3 \, C{\left (\frac{5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, 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))^3,x, algorithm="maxima")

[Out]

-1/60*(A*((105*sin(d*x + c)/(cos(d*x + c) + 1) + 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(co
s(d*x + c) + 1)^5)/a^3 - 60*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) +
1) - 1)/a^3) - B*(15*sin(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c
)^5/(cos(d*x + c) + 1)^5)/a^3 - 3*C*(5*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/
a^3)/d

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Fricas [A]  time = 1.98019, size = 505, normalized size = 4.07 \begin{align*} \frac{15 \,{\left (A \cos \left (d x + c\right )^{3} + 3 \, A \cos \left (d x + c\right )^{2} + 3 \, A \cos \left (d x + c\right ) + A\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (A \cos \left (d x + c\right )^{3} + 3 \, A \cos \left (d x + c\right )^{2} + 3 \, A \cos \left (d x + c\right ) + A\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left ({\left (22 \, A - 2 \, B - 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (17 \, A - 2 \, B - 3 \, C\right )} \cos \left (d x + c\right ) + 32 \, A - 7 \, B - 3 \, C\right )} \sin \left (d x + c\right )}{30 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} 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))^3,x, algorithm="fricas")

[Out]

1/30*(15*(A*cos(d*x + c)^3 + 3*A*cos(d*x + c)^2 + 3*A*cos(d*x + c) + A)*log(sin(d*x + c) + 1) - 15*(A*cos(d*x
+ c)^3 + 3*A*cos(d*x + c)^2 + 3*A*cos(d*x + c) + A)*log(-sin(d*x + c) + 1) - 2*((22*A - 2*B - 3*C)*cos(d*x + c
)^2 + 3*(17*A - 2*B - 3*C)*cos(d*x + c) + 32*A - 7*B - 3*C)*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(
d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.25786, size = 243, normalized size = 1.96 \begin{align*} \frac{\frac{60 \, A \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac{60 \, A \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac{3 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 20 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 10 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 105 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 15 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 15 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{15}}}{60 \, 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))^3,x, algorithm="giac")

[Out]

1/60*(60*A*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 60*A*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 - (3*A*a^12*ta
n(1/2*d*x + 1/2*c)^5 - 3*B*a^12*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 + 20*A*a^12*tan(1/2*d
*x + 1/2*c)^3 - 10*B*a^12*tan(1/2*d*x + 1/2*c)^3 + 105*A*a^12*tan(1/2*d*x + 1/2*c) - 15*B*a^12*tan(1/2*d*x + 1
/2*c) - 15*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d