3.5 \(\int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec (c+d x) \, dx\)

Optimal. Leaf size=32 \[ a x (A+B)+\frac{a A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a B \sin (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.0913122, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {2968, 3023, 2735, 3770} \[ a x (A+B)+\frac{a A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a B \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2968

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

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

Mathematica [A]  time = 0.0241613, size = 46, normalized size = 1.44 \[ \frac{a A \tanh ^{-1}(\sin (c+d x))}{d}+a A x+\frac{a B \sin (c) \cos (d x)}{d}+\frac{a B \cos (c) \sin (d x)}{d}+a B x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.076, size = 56, normalized size = 1.8 \begin{align*} aAx+aBx+{\frac{aA\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{Aac}{d}}+{\frac{aB\sin \left ( dx+c \right ) }{d}}+{\frac{Bac}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.989942, size = 63, normalized size = 1.97 \begin{align*} \frac{{\left (d x + c\right )} A a +{\left (d x + c\right )} B a + A a \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + B a \sin \left (d x + c\right )}{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))*sec(d*x+c),x, algorithm="maxima")

[Out]

((d*x + c)*A*a + (d*x + c)*B*a + A*a*log(sec(d*x + c) + tan(d*x + c)) + B*a*sin(d*x + c))/d

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Fricas [A]  time = 1.45065, size = 139, normalized size = 4.34 \begin{align*} \frac{2 \,{\left (A + B\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 ) + 2 \, B a \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))*sec(d*x+c),x, algorithm="fricas")

[Out]

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

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Giac [B]  time = 1.23301, size = 107, normalized size = 3.34 \begin{align*} \frac{A a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - A a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) +{\left (A a + B a\right )}{\left (d x + c\right )} + \frac{2 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1}}{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))*sec(d*x+c),x, algorithm="giac")

[Out]

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