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

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

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 16, normalized size of antiderivative = 1.00, 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 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])

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]

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 = 0.02, 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

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \cos (5 x) \sec ^5(x) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

________________________________________________________________________________________

fricas [A]  time = 0.68, size = 26, normalized size = 1.62 \[ \frac {48 \, x \cos \relax (x)^{3} - 5 \, {\left (10 \, \cos \relax (x)^{2} - 1\right )} \sin \relax (x)}{3 \, \cos \relax (x)^{3}} \]

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

________________________________________________________________________________________

giac [A]  time = 0.62, size = 14, normalized size = 0.88 \[ \frac {5}{3} \, \tan \relax (x)^{3} + 16 \, x - 15 \, \tan \relax (x) \]

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)

________________________________________________________________________________________

maple [A]  time = 0.33, size = 21, normalized size = 1.31

method	result	size
default	\(16 x -5 \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\relax (x )\right )}{3}\right ) \tan \relax (x )-20 \tan \relax (x )\)	\(21\)
risch	\(16 x -\frac {20 i \left (6 \,{\mathrm e}^{4 i x}+9 \,{\mathrm e}^{2 i x}+5\right )}{3 \left (1+{\mathrm e}^{2 i x}\right )^{3}}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(5*x)/cos(x)^5,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 1.28, size = 14, normalized size = 0.88 \[ \frac {5}{3} \, \tan \relax (x)^{3} + 16 \, x - 15 \, \tan \relax (x) \]

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)

________________________________________________________________________________________

mupad [B]  time = 0.31, size = 26, normalized size = 1.62 \[ \frac {48\,x\,{\cos \relax (x)}^3-50\,\sin \relax (x)\,{\cos \relax (x)}^2+5\,\sin \relax (x)}{3\,{\cos \relax (x)}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*sin(x) + 48*x*cos(x)^3 - 50*cos(x)^2*sin(x))/(3*cos(x)^3)

________________________________________________________________________________________

sympy [A]  time = 17.46, size = 24, normalized size = 1.50 \[ 16 x - \frac {20 \sin {\relax (x )}}{\cos {\relax (x )}} + \frac {5 \tan ^{3}{\relax (x )}}{3} + 5 \tan {\relax (x )} \]

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)

________________________________________________________________________________________