3.157 \(\int \cos ^5(e+f x) (a+b \sec ^2(e+f x)) \, dx\)

Optimal. Leaf size=50 \[ -\frac{(2 a+b) \sin ^3(e+f x)}{3 f}+\frac{(a+b) \sin (e+f x)}{f}+\frac{a \sin ^5(e+f x)}{5 f} \]

[Out]

((a + b)*Sin[e + f*x])/f - ((2*a + b)*Sin[e + f*x]^3)/(3*f) + (a*Sin[e + f*x]^5)/(5*f)

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Rubi [A]  time = 0.0662125, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4044, 3013, 373} \[ -\frac{(2 a+b) \sin ^3(e+f x)}{3 f}+\frac{(a+b) \sin (e+f x)}{f}+\frac{a \sin ^5(e+f x)}{5 f} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^5*(a + b*Sec[e + f*x]^2),x]

[Out]

((a + b)*Sin[e + f*x])/f - ((2*a + b)*Sin[e + f*x]^3)/(3*f) + (a*Sin[e + f*x]^5)/(5*f)

Rule 4044

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Int[(C + A*Sin[e + f*
x]^2)/Sin[e + f*x]^(m + 2), x] /; FreeQ[{e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && ILtQ[(m + 1)/2, 0]

Rule 3013

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Dist[f^(-1), Subst[I
nt[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2
, 0]

Rule 373

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n
)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]

Rubi steps

\begin{align*} \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx &=\int \cos ^3(e+f x) \left (b+a \cos ^2(e+f x)\right ) \, dx\\ &=-\frac{\operatorname{Subst}\left (\int \left (1-x^2\right ) \left (a+b-a x^2\right ) \, dx,x,-\sin (e+f x)\right )}{f}\\ &=-\frac{\operatorname{Subst}\left (\int \left (a \left (1+\frac{b}{a}\right )-(2 a+b) x^2+a x^4\right ) \, dx,x,-\sin (e+f x)\right )}{f}\\ &=\frac{(a+b) \sin (e+f x)}{f}-\frac{(2 a+b) \sin ^3(e+f x)}{3 f}+\frac{a \sin ^5(e+f x)}{5 f}\\ \end{align*}

Mathematica [A]  time = 0.0238516, size = 71, normalized size = 1.42 \[ \frac{a \sin ^5(e+f x)}{5 f}-\frac{2 a \sin ^3(e+f x)}{3 f}+\frac{a \sin (e+f x)}{f}-\frac{b \sin ^3(e+f x)}{3 f}+\frac{b \sin (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^5*(a + b*Sec[e + f*x]^2),x]

[Out]

(a*Sin[e + f*x])/f + (b*Sin[e + f*x])/f - (2*a*Sin[e + f*x]^3)/(3*f) - (b*Sin[e + f*x]^3)/(3*f) + (a*Sin[e + f
*x]^5)/(5*f)

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Maple [A]  time = 0.055, size = 54, normalized size = 1.1 \begin{align*}{\frac{1}{f} \left ({\frac{\sin \left ( fx+e \right ) a}{5} \left ({\frac{8}{3}}+ \left ( \cos \left ( fx+e \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{3}} \right ) }+{\frac{b \left ( 2+ \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) \sin \left ( fx+e \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^5*(a+b*sec(f*x+e)^2),x)

[Out]

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

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Maxima [A]  time = 1.00095, size = 58, normalized size = 1.16 \begin{align*} \frac{3 \, a \sin \left (f x + e\right )^{5} - 5 \,{\left (2 \, a + b\right )} \sin \left (f x + e\right )^{3} + 15 \,{\left (a + b\right )} \sin \left (f x + e\right )}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^5*(a+b*sec(f*x+e)^2),x, algorithm="maxima")

[Out]

1/15*(3*a*sin(f*x + e)^5 - 5*(2*a + b)*sin(f*x + e)^3 + 15*(a + b)*sin(f*x + e))/f

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Fricas [A]  time = 0.477763, size = 113, normalized size = 2.26 \begin{align*} \frac{{\left (3 \, a \cos \left (f x + e\right )^{4} +{\left (4 \, a + 5 \, b\right )} \cos \left (f x + e\right )^{2} + 8 \, a + 10 \, b\right )} \sin \left (f x + e\right )}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^5*(a+b*sec(f*x+e)^2),x, algorithm="fricas")

[Out]

1/15*(3*a*cos(f*x + e)^4 + (4*a + 5*b)*cos(f*x + e)^2 + 8*a + 10*b)*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)**5*(a+b*sec(f*x+e)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.22154, size = 84, normalized size = 1.68 \begin{align*} \frac{3 \, a \sin \left (f x + e\right )^{5} - 10 \, a \sin \left (f x + e\right )^{3} - 5 \, b \sin \left (f x + e\right )^{3} + 15 \, a \sin \left (f x + e\right ) + 15 \, b \sin \left (f x + e\right )}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^5*(a+b*sec(f*x+e)^2),x, algorithm="giac")

[Out]

1/15*(3*a*sin(f*x + e)^5 - 10*a*sin(f*x + e)^3 - 5*b*sin(f*x + e)^3 + 15*a*sin(f*x + e) + 15*b*sin(f*x + e))/f