3.366 \(\int \cos (5 x) \sec ^5(x) \, dx\)

Optimal. Leaf size=16 \[ 16 x+\frac{5 \tan ^3(x)}{3}-15 \tan (x) \]

[Out]

16*x - 15*Tan[x] + (5*Tan[x]^3)/3

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Rubi [A]  time = 0.0428634, antiderivative size = 16, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {1153, 203} \[ 16 x+\frac{5 \tan ^3(x)}{3}-15 \tan (x) \]

Antiderivative was successfully verified.

[In]

Int[Cos[5*x]*Sec[x]^5,x]

[Out]

16*x - 15*Tan[x] + (5*Tan[x]^3)/3

Rule 1153

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \cos (5 x) \sec ^5(x) \, dx &=\operatorname{Subst}\left (\int \frac{1-10 x^2+5 x^4}{1+x^2} \, dx,x,\tan (x)\right )\\ &=\operatorname{Subst}\left (\int \left (-15+5 x^2+\frac{16}{1+x^2}\right ) \, dx,x,\tan (x)\right )\\ &=-15 \tan (x)+\frac{5 \tan ^3(x)}{3}+16 \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (x)\right )\\ &=16 x-15 \tan (x)+\frac{5 \tan ^3(x)}{3}\\ \end{align*}

Mathematica [A]  time = 0.0144063, size = 20, normalized size = 1.25 \[ 16 x-\frac{50 \tan (x)}{3}+\frac{5}{3} \tan (x) \sec ^2(x) \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[5*x]*Sec[x]^5,x]

[Out]

16*x - (50*Tan[x])/3 + (5*Sec[x]^2*Tan[x])/3

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Maple [A]  time = 0.059, size = 21, normalized size = 1.3 \begin{align*} 16\,x-5\, \left ( -2/3-1/3\, \left ( \sec \left ( x \right ) \right ) ^{2} \right ) \tan \left ( x \right ) -20\,\tan \left ( x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(5*x)/cos(x)^5,x)

[Out]

16*x-5*(-2/3-1/3*sec(x)^2)*tan(x)-20*tan(x)

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Maxima [A]  time = 1.43112, size = 19, normalized size = 1.19 \begin{align*} \frac{5}{3} \, \tan \left (x\right )^{3} + 16 \, x - 15 \, \tan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(5*x)/cos(x)^5,x, algorithm="maxima")

[Out]

5/3*tan(x)^3 + 16*x - 15*tan(x)

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Fricas [A]  time = 2.18106, size = 80, normalized size = 5. \begin{align*} \frac{48 \, x \cos \left (x\right )^{3} - 5 \,{\left (10 \, \cos \left (x\right )^{2} - 1\right )} \sin \left (x\right )}{3 \, \cos \left (x\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(5*x)/cos(x)^5,x, algorithm="fricas")

[Out]

1/3*(48*x*cos(x)^3 - 5*(10*cos(x)^2 - 1)*sin(x))/cos(x)^3

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Sympy [A]  time = 110.271, size = 24, normalized size = 1.5 \begin{align*} 16 x - \frac{20 \sin{\left (x \right )}}{\cos{\left (x \right )}} + \frac{5 \tan ^{3}{\left (x \right )}}{3} + 5 \tan{\left (x \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(5*x)/cos(x)**5,x)

[Out]

16*x - 20*sin(x)/cos(x) + 5*tan(x)**3/3 + 5*tan(x)

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Giac [A]  time = 1.06622, size = 19, normalized size = 1.19 \begin{align*} \frac{5}{3} \, \tan \left (x\right )^{3} + 16 \, x - 15 \, \tan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(5*x)/cos(x)^5,x, algorithm="giac")

[Out]

5/3*tan(x)^3 + 16*x - 15*tan(x)