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

Optimal. Leaf size=42 \[ \frac{A \sin (c+d x) \cos (c+d x)}{2 d}+\frac{1}{2} x (A+2 C)+\frac{B \sin (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.0592603, antiderivative size = 42, 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 = {4047, 2637, 4045, 8} \[ \frac{A \sin (c+d x) \cos (c+d x)}{2 d}+\frac{1}{2} x (A+2 C)+\frac{B \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [A]  time = 0.0543058, size = 55, normalized size = 1.31 \[ \frac{A (c+d x)}{2 d}+\frac{A \sin (2 (c+d x))}{4 d}+\frac{B \sin (c) \cos (d x)}{d}+\frac{B \cos (c) \sin (d x)}{d}+C x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.047, size = 45, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ( A \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +B\sin \left ( dx+c \right ) +C \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.92903, size = 57, normalized size = 1.36 \begin{align*} \frac{{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A + 4 \,{\left (d x + c\right )} C + 4 \, B \sin \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*A + 4*(d*x + c)*C + 4*B*sin(d*x + c))/d

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Fricas [A]  time = 0.477398, size = 82, normalized size = 1.95 \begin{align*} \frac{{\left (A + 2 \, C\right )} d x +{\left (A \cos \left (d x + c\right ) + 2 \, B\right )} \sin \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.15828, size = 116, normalized size = 2.76 \begin{align*} \frac{{\left (d x + c\right )}{\left (A + 2 \, C\right )} - \frac{2 \,{\left (A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, B \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(cos(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

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