3.304 \(\int (a+a \cos (c+d x)) (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec (c+d x) \, dx\)

Optimal. Leaf size=63 \[ \frac{1}{2} a x (2 A+2 B+C)+\frac{a A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a (B+C) \sin (c+d x)}{d}+\frac{a C \sin (c+d x) \cos (c+d x)}{2 d} \]

[Out]

(a*(2*A + 2*B + C)*x)/2 + (a*A*ArcTanh[Sin[c + d*x]])/d + (a*(B + C)*Sin[c + d*x])/d + (a*C*Cos[c + d*x]*Sin[c
+ d*x])/(2*d)

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Rubi [A]  time = 0.135698, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.108, Rules used = {3033, 3023, 2735, 3770} \[ \frac{1}{2} a x (2 A+2 B+C)+\frac{a A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a (B+C) \sin (c+d x)}{d}+\frac{a C \sin (c+d x) \cos (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(2*A + 2*B + C)*x)/2 + (a*A*ArcTanh[Sin[c + d*x]])/d + (a*(B + C)*Sin[c + d*x])/d + (a*C*Cos[c + d*x]*Sin[c
+ d*x])/(2*d)

Rule 3033

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*
(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!LtQ[m, -1]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

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

Mathematica [A]  time = 0.134322, size = 59, normalized size = 0.94 \[ \frac{a \left (4 A \tanh ^{-1}(\sin (c+d x))+4 A d x+4 (B+C) \sin (c+d x)+4 B d x+C \sin (2 (c+d x))+2 c C+2 C d x\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(2*c*C + 4*A*d*x + 4*B*d*x + 2*C*d*x + 4*A*ArcTanh[Sin[c + d*x]] + 4*(B + C)*Sin[c + d*x] + C*Sin[2*(c + d*
x)]))/(4*d)

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Maple [A]  time = 0.051, size = 100, normalized size = 1.6 \begin{align*}{\frac{aA\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+Bax+{\frac{Bac}{d}}+{\frac{aC\sin \left ( dx+c \right ) }{d}}+aAx+{\frac{Aac}{d}}+{\frac{Ba\sin \left ( dx+c \right ) }{d}}+{\frac{aC\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{aCx}{2}}+{\frac{aCc}{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.979101, size = 111, normalized size = 1.76 \begin{align*} \frac{4 \,{\left (d x + c\right )} A a + 4 \,{\left (d x + c\right )} B a +{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a + 4 \, A a \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 4 \, B a \sin \left (d x + c\right ) + 4 \, C a \sin \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c),x, algorithm="maxima")

[Out]

1/4*(4*(d*x + c)*A*a + 4*(d*x + c)*B*a + (2*d*x + 2*c + sin(2*d*x + 2*c))*C*a + 4*A*a*log(sec(d*x + c) + tan(d
*x + c)) + 4*B*a*sin(d*x + c) + 4*C*a*sin(d*x + c))/d

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Fricas [A]  time = 2.34583, size = 184, normalized size = 2.92 \begin{align*} \frac{{\left (2 \, A + 2 \, B + C\right )} a d x + A a \log \left (\sin \left (d x + c\right ) + 1\right ) - A a \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (C a \cos \left (d x + c\right ) + 2 \,{\left (B + C\right )} a\right )} \sin \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c),x, algorithm="fricas")

[Out]

1/2*((2*A + 2*B + C)*a*d*x + A*a*log(sin(d*x + c) + 1) - A*a*log(-sin(d*x + c) + 1) + (C*a*cos(d*x + c) + 2*(B
+ C)*a)*sin(d*x + c))/d

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a \left (\int A \sec{\left (c + d x \right )}\, dx + \int A \cos{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx + \int B \cos{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx + \int B \cos ^{2}{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx + \int C \cos ^{3}{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(Integral(A*sec(c + d*x), x) + Integral(A*cos(c + d*x)*sec(c + d*x), x) + Integral(B*cos(c + d*x)*sec(c + d*
x), x) + Integral(B*cos(c + d*x)**2*sec(c + d*x), x) + Integral(C*cos(c + d*x)**2*sec(c + d*x), x) + Integral(
C*cos(c + d*x)**3*sec(c + d*x), x))

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Giac [B]  time = 1.25072, size = 177, normalized size = 2.81 \begin{align*} \frac{2 \, A a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 2 \, A a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) +{\left (2 \, A a + 2 \, B a + C a\right )}{\left (d x + c\right )} + \frac{2 \,{\left (2 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, C a \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 )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c),x, algorithm="giac")

[Out]

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