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

Optimal. Leaf size=27 \[ \frac{A \tan (c+d x)}{d}+\frac{B \tanh ^{-1}(\sin (c+d x))}{d}+C x \]

[Out]

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

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Rubi [A]  time = 0.0540409, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {3021, 2735, 3770} \[ \frac{A \tan (c+d x)}{d}+\frac{B \tanh ^{-1}(\sin (c+d x))}{d}+C x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0185488, size = 27, normalized size = 1. \[ \frac{A \tan (c+d x)}{d}+\frac{B \tanh ^{-1}(\sin (c+d x))}{d}+C x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.035, size = 41, normalized size = 1.5 \begin{align*} Cx+{\frac{A\tan \left ( dx+c \right ) }{d}}+{\frac{B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{Cc}{d}} \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)^2,x)

[Out]

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

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Maxima [A]  time = 0.972369, size = 62, normalized size = 2.3 \begin{align*} \frac{2 \,{\left (d x + c\right )} C + 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 \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)+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.94841, size = 193, normalized size = 7.15 \begin{align*} \frac{2 \, C d x \cos \left (d x + c\right ) + B \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - B \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, 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)+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algorithm="fricas")

[Out]

1/2*(2*C*d*x*cos(d*x + c) + B*cos(d*x + c)*log(sin(d*x + c) + 1) - B*cos(d*x + c)*log(-sin(d*x + c) + 1) + 2*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 )} + C \cos ^{2}{\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)+C*cos(d*x+c)**2)*sec(d*x+c)**2,x)

[Out]

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

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Giac [B]  time = 1.26282, size = 95, normalized size = 3.52 \begin{align*} \frac{{\left (d x + c\right )} C + B \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - B \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \, 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)+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algorithm="giac")

[Out]

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