3.162 \(\int \cos ^2(e+f x) (a+b \sec ^2(e+f x)) \, dx\)

Optimal. Leaf size=31 \[ \frac{1}{2} x (a+2 b)+\frac{a \sin (e+f x) \cos (e+f x)}{2 f} \]

[Out]

((a + 2*b)*x)/2 + (a*Cos[e + f*x]*Sin[e + f*x])/(2*f)

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Rubi [A]  time = 0.0273794, 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} \[ \frac{1}{2} x (a+2 b)+\frac{a \sin (e+f x) \cos (e+f x)}{2 f} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^2*(a + b*Sec[e + f*x]^2),x]

[Out]

((a + 2*b)*x)/2 + (a*Cos[e + f*x]*Sin[e + f*x])/(2*f)

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(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx &=\frac{a \cos (e+f x) \sin (e+f x)}{2 f}+\frac{1}{2} (a+2 b) \int 1 \, dx\\ &=\frac{1}{2} (a+2 b) x+\frac{a \cos (e+f x) \sin (e+f x)}{2 f}\\ \end{align*}

Mathematica [A]  time = 0.0305511, size = 33, normalized size = 1.06 \[ \frac{a (e+f x)}{2 f}+\frac{a \sin (2 (e+f x))}{4 f}+b x \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^2*(a + b*Sec[e + f*x]^2),x]

[Out]

b*x + (a*(e + f*x))/(2*f) + (a*Sin[2*(e + f*x)])/(4*f)

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Maple [A]  time = 0.054, size = 37, normalized size = 1.2 \begin{align*}{\frac{1}{f} \left ( a \left ({\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) + \left ( fx+e \right ) b \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^2*(a+b*sec(f*x+e)^2),x)

[Out]

1/f*(a*(1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*e)+(f*x+e)*b)

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Maxima [A]  time = 1.5004, size = 50, normalized size = 1.61 \begin{align*} \frac{{\left (f x + e\right )}{\left (a + 2 \, b\right )} + \frac{a \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{2} + 1}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(a+b*sec(f*x+e)^2),x, algorithm="maxima")

[Out]

1/2*((f*x + e)*(a + 2*b) + a*tan(f*x + e)/(tan(f*x + e)^2 + 1))/f

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Fricas [A]  time = 0.475012, size = 72, normalized size = 2.32 \begin{align*} \frac{{\left (a + 2 \, b\right )} f x + a \cos \left (f x + e\right ) \sin \left (f x + e\right )}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(a+b*sec(f*x+e)^2),x, algorithm="fricas")

[Out]

1/2*((a + 2*b)*f*x + a*cos(f*x + e)*sin(f*x + e))/f

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Sympy [A]  time = 21.2814, size = 51, normalized size = 1.65 \begin{align*} a \left (\begin{cases} \frac{x \sin ^{2}{\left (e + f x \right )}}{2} + \frac{x \cos ^{2}{\left (e + f x \right )}}{2} + \frac{\sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} & \text{for}\: f \neq 0 \\x \cos ^{2}{\left (e \right )} & \text{otherwise} \end{cases}\right ) + b x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)**2*(a+b*sec(f*x+e)**2),x)

[Out]

a*Piecewise((x*sin(e + f*x)**2/2 + x*cos(e + f*x)**2/2 + sin(e + f*x)*cos(e + f*x)/(2*f), Ne(f, 0)), (x*cos(e)
**2, True)) + b*x

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Giac [A]  time = 1.24232, size = 54, normalized size = 1.74 \begin{align*} \frac{{\left (f x + e\right )}{\left (a + 2 \, b\right )} + \frac{a \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{2} + 1}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(a+b*sec(f*x+e)^2),x, algorithm="giac")

[Out]

1/2*((f*x + e)*(a + 2*b) + a*tan(f*x + e)/(tan(f*x + e)^2 + 1))/f