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

Optimal. Leaf size=31 \[ \frac{A \sin (c+d x) \cos (c+d x)}{2 d}+\frac{1}{2} x (A+2 C) \]

[Out]

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

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Rubi [A]  time = 0.0302677, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {4045, 8} \[ \frac{A \sin (c+d x) \cos (c+d x)}{2 d}+\frac{1}{2} x (A+2 C) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0338963, size = 33, normalized size = 1.06 \[ \frac{A (c+d x)}{2 d}+\frac{A \sin (2 (c+d x))}{4 d}+C x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.044, size = 37, normalized size = 1.2 \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 ) +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+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)+C*(d*x+c))

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Maxima [A]  time = 1.41121, size = 50, normalized size = 1.61 \begin{align*} \frac{{\left (d x + c\right )}{\left (A + 2 \, C\right )} + \frac{A \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.471293, size = 72, normalized size = 2.32 \begin{align*} \frac{{\left (A + 2 \, C\right )} d x + A \cos \left (d x + c\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+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 21.1504, size = 51, normalized size = 1.65 \begin{align*} A \left (\begin{cases} \frac{x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{x \cos ^{2}{\left (c + d x \right )}}{2} + \frac{\sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} & \text{for}\: d \neq 0 \\x \cos ^{2}{\left (c \right )} & \text{otherwise} \end{cases}\right ) + C x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*Piecewise((x*sin(c + d*x)**2/2 + x*cos(c + d*x)**2/2 + sin(c + d*x)*cos(c + d*x)/(2*d), Ne(d, 0)), (x*cos(c)
**2, True)) + C*x

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Giac [A]  time = 1.16847, size = 50, normalized size = 1.61 \begin{align*} \frac{{\left (d x + c\right )}{\left (A + 2 \, C\right )} + \frac{A \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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