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

Optimal. Leaf size=53 \[ \frac{a^2 \sin ^5(e+f x)}{5 f}-\frac{2 a (a+b) \sin ^3(e+f x)}{3 f}+\frac{(a+b)^2 \sin (e+f x)}{f} \]

[Out]

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

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Rubi [A]  time = 0.0662634, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {4147, 194} \[ \frac{a^2 \sin ^5(e+f x)}{5 f}-\frac{2 a (a+b) \sin ^3(e+f x)}{3 f}+\frac{(a+b)^2 \sin (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4147

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fr
eeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[ExpandToSum[b + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 - ff^2*x^2)^
((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n
/2] && IntegerQ[p]

Rule 194

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

Rubi steps

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

Mathematica [A]  time = 0.0261719, size = 106, normalized size = 2. \[ \frac{a^2 \sin ^5(e+f x)}{5 f}-\frac{2 a^2 \sin ^3(e+f x)}{3 f}+\frac{a^2 \sin (e+f x)}{f}-\frac{2 a b \sin ^3(e+f x)}{3 f}+\frac{2 a b \sin (e+f x)}{f}+\frac{b^2 \sin (e) \cos (f x)}{f}+\frac{b^2 \cos (e) \sin (f x)}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.065, size = 67, normalized size = 1.3 \begin{align*}{\frac{1}{f} \left ({\frac{{a}^{2}\sin \left ( fx+e \right ) }{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{2\,ab \left ( 2+ \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) \sin \left ( fx+e \right ) }{3}}+{b}^{2}\sin \left ( fx+e \right ) \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)^2,x)

[Out]

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

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Maxima [A]  time = 0.973232, size = 74, normalized size = 1.4 \begin{align*} \frac{3 \, a^{2} \sin \left (f x + e\right )^{5} - 10 \,{\left (a^{2} + a b\right )} \sin \left (f x + e\right )^{3} + 15 \,{\left (a^{2} + 2 \, a b + b^{2}\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)^2,x, algorithm="maxima")

[Out]

1/15*(3*a^2*sin(f*x + e)^5 - 10*(a^2 + a*b)*sin(f*x + e)^3 + 15*(a^2 + 2*a*b + b^2)*sin(f*x + e))/f

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Fricas [A]  time = 0.486925, size = 142, normalized size = 2.68 \begin{align*} \frac{{\left (3 \, a^{2} \cos \left (f x + e\right )^{4} + 2 \,{\left (2 \, a^{2} + 5 \, a b\right )} \cos \left (f x + e\right )^{2} + 8 \, a^{2} + 20 \, a b + 15 \, b^{2}\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)^2,x, algorithm="fricas")

[Out]

1/15*(3*a^2*cos(f*x + e)^4 + 2*(2*a^2 + 5*a*b)*cos(f*x + e)^2 + 8*a^2 + 20*a*b + 15*b^2)*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)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.29164, size = 111, normalized size = 2.09 \begin{align*} \frac{3 \, a^{2} \sin \left (f x + e\right )^{5} - 10 \, a^{2} \sin \left (f x + e\right )^{3} - 10 \, a b \sin \left (f x + e\right )^{3} + 15 \, a^{2} \sin \left (f x + e\right ) + 30 \, a b \sin \left (f x + e\right ) + 15 \, b^{2} \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)^2,x, algorithm="giac")

[Out]

1/15*(3*a^2*sin(f*x + e)^5 - 10*a^2*sin(f*x + e)^3 - 10*a*b*sin(f*x + e)^3 + 15*a^2*sin(f*x + e) + 30*a*b*sin(
f*x + e) + 15*b^2*sin(f*x + e))/f