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

Optimal. Leaf size=35 \[ \frac{(a B+A b) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a A \tan (c+d x)}{d}+b B x \]

[Out]

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

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Rubi [A]  time = 0.113709, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {2968, 3021, 2735, 3770} \[ \frac{(a B+A b) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a A \tan (c+d x)}{d}+b B x \]

Antiderivative was successfully verified.

[In]

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

[Out]

b*B*x + ((A*b + a*B)*ArcTanh[Sin[c + d*x]])/d + (a*A*Tan[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 3021

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

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

Mathematica [A]  time = 0.0139163, size = 43, normalized size = 1.23 \[ \frac{a A \tan (c+d x)}{d}+\frac{a B \tanh ^{-1}(\sin (c+d x))}{d}+\frac{A b \tanh ^{-1}(\sin (c+d x))}{d}+b B x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.07, size = 65, normalized size = 1.9 \begin{align*} bBx+{\frac{A\tan \left ( dx+c \right ) a}{d}}+{\frac{Ab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{aB\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{Bbc}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.10044, size = 99, normalized size = 2.83 \begin{align*} \frac{2 \,{\left (d x + c\right )} B b + B a{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + A b{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A a \tan \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.50891, size = 225, normalized size = 6.43 \begin{align*} \frac{2 \, B b d x \cos \left (d x + c\right ) +{\left (B a + A b\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (B a + A b\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, A a \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.28301, size = 113, normalized size = 3.23 \begin{align*} \frac{{\left (d x + c\right )} B b +{\left (B a + A b\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (B a + A b\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \, A 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+b*cos(d*x+c))*(A+B*cos(d*x+c))*sec(d*x+c)^2,x, algorithm="giac")

[Out]

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