3.346 \(\int \frac{(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^4(c+d x)}{a+a \cos (c+d x)} \, dx\)

Optimal. Leaf size=148 \[ \frac{(4 A-3 B+3 C) \tan ^3(c+d x)}{3 a d}+\frac{(4 A-3 B+3 C) \tan (c+d x)}{a d}-\frac{(3 A-3 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac{(3 A-3 B+2 C) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac{(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{d (a \cos (c+d x)+a)} \]

[Out]

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

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Rubi [A]  time = 0.216843, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.122, Rules used = {3041, 2748, 3767, 3768, 3770} \[ \frac{(4 A-3 B+3 C) \tan ^3(c+d x)}{3 a d}+\frac{(4 A-3 B+3 C) \tan (c+d x)}{a d}-\frac{(3 A-3 B+2 C) \tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac{(3 A-3 B+2 C) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac{(A-B+C) \tan (c+d x) \sec ^2(c+d x)}{d (a \cos (c+d x)+a)} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/(a + a*Cos[c + d*x]),x]

[Out]

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

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 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

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

Mathematica [B]  time = 3.87529, size = 351, normalized size = 2.37 \[ \frac{\cos ^2\left (\frac{1}{2} (c+d x)\right ) \left (12 (A-B+C) \tan \left (\frac{1}{2} (c+d x)\right )+\frac{4 (5 A-3 B+3 C) \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{4 (5 A-3 B+3 C) \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+6 (3 A-3 B+2 C) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-6 (3 A-3 B+2 C) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{2 A-3 B}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{3 B-2 A}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{2 A \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{2 A \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}\right )}{6 a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/(a + a*Cos[c + d*x]),x]

[Out]

(Cos[(c + d*x)/2]^2*(6*(3*A - 3*B + 2*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 6*(3*A - 3*B + 2*C)*Log[Co
s[(c + d*x)/2] + Sin[(c + d*x)/2]] + (-2*A + 3*B)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (2*A*Sin[(c + d*x)
/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3 + (4*(5*A - 3*B + 3*C)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[
(c + d*x)/2]) + (2*A*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 + (2*A - 3*B)/(Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2])^2 + (4*(5*A - 3*B + 3*C)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 12*(A
- B + C)*Tan[(c + d*x)/2]))/(6*a*d*(1 + Cos[c + d*x]))

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Maple [B]  time = 0.07, size = 442, normalized size = 3. \begin{align*}{\frac{A}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \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 ) }-{\frac{A}{3\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-{\frac{A}{da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{B}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{3\,A}{2\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{3\,B}{2\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{C}{da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{5\,A}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{3\,B}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{C}{da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{A}{3\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+{\frac{A}{da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{B}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{3\,A}{2\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{3\,B}{2\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{C}{da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{5\,A}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{3\,B}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{C}{da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}} \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)^4/(a+a*cos(d*x+c)),x)

[Out]

1/a/d*A*tan(1/2*d*x+1/2*c)-1/a/d*B*tan(1/2*d*x+1/2*c)+1/a/d*C*tan(1/2*d*x+1/2*c)-1/3/a/d*A/(tan(1/2*d*x+1/2*c)
-1)^3-1/a/d*A/(tan(1/2*d*x+1/2*c)-1)^2+1/2/a/d*B/(tan(1/2*d*x+1/2*c)-1)^2+3/2/a/d*A*ln(tan(1/2*d*x+1/2*c)-1)-3
/2/a/d*B*ln(tan(1/2*d*x+1/2*c)-1)+1/a/d*ln(tan(1/2*d*x+1/2*c)-1)*C-5/2/a/d*A/(tan(1/2*d*x+1/2*c)-1)+3/2/a/d*B/
(tan(1/2*d*x+1/2*c)-1)-1/a/d/(tan(1/2*d*x+1/2*c)-1)*C-1/3/a/d*A/(tan(1/2*d*x+1/2*c)+1)^3+1/a/d*A/(tan(1/2*d*x+
1/2*c)+1)^2-1/2/a/d*B/(tan(1/2*d*x+1/2*c)+1)^2-3/2/a/d*A*ln(tan(1/2*d*x+1/2*c)+1)+3/2/a/d*B*ln(tan(1/2*d*x+1/2
*c)+1)-1/a/d*ln(tan(1/2*d*x+1/2*c)+1)*C-5/2/a/d*A/(tan(1/2*d*x+1/2*c)+1)+3/2/a/d*B/(tan(1/2*d*x+1/2*c)+1)-1/a/
d/(tan(1/2*d*x+1/2*c)+1)*C

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Maxima [B]  time = 1.03175, size = 655, normalized size = 4.43 \begin{align*} \text{result too large to display} \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)^4/(a+a*cos(d*x+c)),x, algorithm="maxima")

[Out]

1/6*(A*(2*(9*sin(d*x + c)/(cos(d*x + c) + 1) - 16*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos
(d*x + c) + 1)^5)/(a - 3*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - a*s
in(d*x + c)^6/(cos(d*x + c) + 1)^6) - 9*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a + 9*log(sin(d*x + c)/(cos(d
*x + c) + 1) - 1)/a + 6*sin(d*x + c)/(a*(cos(d*x + c) + 1))) - 3*B*(2*(sin(d*x + c)/(cos(d*x + c) + 1) - 3*sin
(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a - 2*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a*sin(d*x + c)^4/(cos(d*x + c
) + 1)^4) - 3*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a + 3*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a + 2*si
n(d*x + c)/(a*(cos(d*x + c) + 1))) - 6*C*(log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a - log(sin(d*x + c)/(cos(d
*x + c) + 1) - 1)/a - 2*sin(d*x + c)/((a - a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) - sin(d*
x + c)/(a*(cos(d*x + c) + 1))))/d

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Fricas [A]  time = 1.99522, size = 489, normalized size = 3.3 \begin{align*} -\frac{3 \,{\left ({\left (3 \, A - 3 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{4} +{\left (3 \, A - 3 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left ({\left (3 \, A - 3 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{4} +{\left (3 \, A - 3 \, B + 2 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (4 \,{\left (4 \, A - 3 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (7 \, A - 3 \, B + 6 \, C\right )} \cos \left (d x + c\right )^{2} -{\left (A - 3 \, B\right )} \cos \left (d x + c\right ) + 2 \, A\right )} \sin \left (d x + c\right )}{12 \,{\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\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)^4/(a+a*cos(d*x+c)),x, algorithm="fricas")

[Out]

-1/12*(3*((3*A - 3*B + 2*C)*cos(d*x + c)^4 + (3*A - 3*B + 2*C)*cos(d*x + c)^3)*log(sin(d*x + c) + 1) - 3*((3*A
- 3*B + 2*C)*cos(d*x + c)^4 + (3*A - 3*B + 2*C)*cos(d*x + c)^3)*log(-sin(d*x + c) + 1) - 2*(4*(4*A - 3*B + 3*
C)*cos(d*x + c)^3 + (7*A - 3*B + 6*C)*cos(d*x + c)^2 - (A - 3*B)*cos(d*x + c) + 2*A)*sin(d*x + c))/(a*d*cos(d*
x + c)^4 + a*d*cos(d*x + c)^3)

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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)**4/(a+a*cos(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.24097, size = 328, normalized size = 2.22 \begin{align*} -\frac{\frac{3 \,{\left (3 \, A - 3 \, B + 2 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac{3 \,{\left (3 \, A - 3 \, B + 2 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac{6 \,{\left (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 )\right )}}{a} + \frac{2 \,{\left (15 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 9 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 16 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 12 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{3} a}}{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)^4/(a+a*cos(d*x+c)),x, algorithm="giac")

[Out]

-1/6*(3*(3*A - 3*B + 2*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a - 3*(3*A - 3*B + 2*C)*log(abs(tan(1/2*d*x + 1/2
*c) - 1))/a - 6*(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 + 2*(15*A*tan(1/2
*d*x + 1/2*c)^5 - 9*B*tan(1/2*d*x + 1/2*c)^5 + 6*C*tan(1/2*d*x + 1/2*c)^5 - 16*A*tan(1/2*d*x + 1/2*c)^3 + 12*B
*tan(1/2*d*x + 1/2*c)^3 - 12*C*tan(1/2*d*x + 1/2*c)^3 + 9*A*tan(1/2*d*x + 1/2*c) - 3*B*tan(1/2*d*x + 1/2*c) +
6*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*a))/d