3.163 \(\int \cos ^4(e+f x) (a+b \sec ^2(e+f x)) \, dx\)

Optimal. Leaf size=61 \[ \frac{(3 a+4 b) \sin (e+f x) \cos (e+f x)}{8 f}+\frac{1}{8} x (3 a+4 b)+\frac{a \sin (e+f x) \cos ^3(e+f x)}{4 f} \]

[Out]

((3*a + 4*b)*x)/8 + ((3*a + 4*b)*Cos[e + f*x]*Sin[e + f*x])/(8*f) + (a*Cos[e + f*x]^3*Sin[e + f*x])/(4*f)

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Rubi [A]  time = 0.041084, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4045, 2635, 8} \[ \frac{(3 a+4 b) \sin (e+f x) \cos (e+f x)}{8 f}+\frac{1}{8} x (3 a+4 b)+\frac{a \sin (e+f x) \cos ^3(e+f x)}{4 f} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^4*(a + b*Sec[e + f*x]^2),x]

[Out]

((3*a + 4*b)*x)/8 + ((3*a + 4*b)*Cos[e + f*x]*Sin[e + f*x])/(8*f) + (a*Cos[e + f*x]^3*Sin[e + f*x])/(4*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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

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

Rubi steps

\begin{align*} \int \cos ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx &=\frac{a \cos ^3(e+f x) \sin (e+f x)}{4 f}+\frac{1}{4} (3 a+4 b) \int \cos ^2(e+f x) \, dx\\ &=\frac{(3 a+4 b) \cos (e+f x) \sin (e+f x)}{8 f}+\frac{a \cos ^3(e+f x) \sin (e+f x)}{4 f}+\frac{1}{8} (3 a+4 b) \int 1 \, dx\\ &=\frac{1}{8} (3 a+4 b) x+\frac{(3 a+4 b) \cos (e+f x) \sin (e+f x)}{8 f}+\frac{a \cos ^3(e+f x) \sin (e+f x)}{4 f}\\ \end{align*}

Mathematica [A]  time = 0.0884978, size = 45, normalized size = 0.74 \[ \frac{4 (3 a+4 b) (e+f x)+8 (a+b) \sin (2 (e+f x))+a \sin (4 (e+f x))}{32 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^4*(a + b*Sec[e + f*x]^2),x]

[Out]

(4*(3*a + 4*b)*(e + f*x) + 8*(a + b)*Sin[2*(e + f*x)] + a*Sin[4*(e + f*x)])/(32*f)

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Maple [A]  time = 0.058, size = 65, normalized size = 1.1 \begin{align*}{\frac{1}{f} \left ( a \left ({\frac{\sin \left ( fx+e \right ) }{4} \left ( \left ( \cos \left ( fx+e \right ) \right ) ^{3}+{\frac{3\,\cos \left ( fx+e \right ) }{2}} \right ) }+{\frac{3\,fx}{8}}+{\frac{3\,e}{8}} \right ) +b \left ({\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^4*(a+b*sec(f*x+e)^2),x)

[Out]

1/f*(a*(1/4*(cos(f*x+e)^3+3/2*cos(f*x+e))*sin(f*x+e)+3/8*f*x+3/8*e)+b*(1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*e
))

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Maxima [A]  time = 1.48162, size = 99, normalized size = 1.62 \begin{align*} \frac{{\left (f x + e\right )}{\left (3 \, a + 4 \, b\right )} + \frac{{\left (3 \, a + 4 \, b\right )} \tan \left (f x + e\right )^{3} +{\left (5 \, a + 4 \, b\right )} \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}}{8 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^4*(a+b*sec(f*x+e)^2),x, algorithm="maxima")

[Out]

1/8*((f*x + e)*(3*a + 4*b) + ((3*a + 4*b)*tan(f*x + e)^3 + (5*a + 4*b)*tan(f*x + e))/(tan(f*x + e)^4 + 2*tan(f
*x + e)^2 + 1))/f

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Fricas [A]  time = 0.483855, size = 119, normalized size = 1.95 \begin{align*} \frac{{\left (3 \, a + 4 \, b\right )} f x +{\left (2 \, a \cos \left (f x + e\right )^{3} +{\left (3 \, a + 4 \, b\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^4*(a+b*sec(f*x+e)^2),x, algorithm="fricas")

[Out]

1/8*((3*a + 4*b)*f*x + (2*a*cos(f*x + e)^3 + (3*a + 4*b)*cos(f*x + e))*sin(f*x + e))/f

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)**4*(a+b*sec(f*x+e)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.34478, size = 107, normalized size = 1.75 \begin{align*} \frac{{\left (f x + e\right )}{\left (3 \, a + 4 \, b\right )} + \frac{3 \, a \tan \left (f x + e\right )^{3} + 4 \, b \tan \left (f x + e\right )^{3} + 5 \, a \tan \left (f x + e\right ) + 4 \, b \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{2}}}{8 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^4*(a+b*sec(f*x+e)^2),x, algorithm="giac")

[Out]

1/8*((f*x + e)*(3*a + 4*b) + (3*a*tan(f*x + e)^3 + 4*b*tan(f*x + e)^3 + 5*a*tan(f*x + e) + 4*b*tan(f*x + e))/(
tan(f*x + e)^2 + 1)^2)/f