\(\int (-\cos (x)+\sec (x))^{5/2} \, dx\) [334]

Optimal result

Integrand size = 11, antiderivative size = 50 \[ \int (-\cos (x)+\sec (x))^{5/2} \, dx=\frac {64}{15} \cot (x) \sqrt {\sin (x) \tan (x)}+\frac {16}{15} \tan (x) \sqrt {\sin (x) \tan (x)}-\frac {2}{5} \sin ^2(x) \tan (x) \sqrt {\sin (x) \tan (x)} \]

[Out]

64/15*cot(x)*(sin(x)*tan(x))^(1/2)+16/15*(sin(x)*tan(x))^(1/2)*tan(x)-2/5*sin(x)^2*(sin(x)*tan(x))^(1/2)*tan(x
)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {4482, 4485, 2678, 2674, 2669} \[ \int (-\cos (x)+\sec (x))^{5/2} \, dx=-\frac {2}{5} \sin ^2(x) \tan (x) \sqrt {\sin (x) \tan (x)}+\frac {16}{15} \tan (x) \sqrt {\sin (x) \tan (x)}+\frac {64}{15} \cot (x) \sqrt {\sin (x) \tan (x)} \]

[In]

Int[(-Cos[x] + Sec[x])^(5/2),x]

[Out]

(64*Cot[x]*Sqrt[Sin[x]*Tan[x]])/15 + (16*Tan[x]*Sqrt[Sin[x]*Tan[x]])/15 - (2*Sin[x]^2*Tan[x]*Sqrt[Sin[x]*Tan[x
]])/5

Rule 2669

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*(a*Sin[e
+ f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*m)), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n - 1, 0]

Rule 2674

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a*Sin[e +
f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(n - 1))), x] - Dist[b^2*((m + n - 1)/(n - 1)), Int[(a*Sin[e + f*x])^m*(
b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && IntegersQ[2*m, 2*n] &&  !(GtQ[m,
1] &&  !IntegerQ[(m - 1)/2])

Rule 2678

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[(-b)*(a*Sin
[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*m)), x] + Dist[a^2*((m + n - 1)/m), Int[(a*Sin[e + f*x])^(m - 2)*(b*
Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (GtQ[m, 1] || (EqQ[m, 1] && EqQ[n, 1/2])) && IntegersQ
[2*m, 2*n]

Rule 4482

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

Rule 4485

Int[(u_.)*((v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> With[{uu = ActivateTrig[u], vv = ActivateTrig[v], ww = Ac
tivateTrig[w]}, Dist[(vv^m*ww^n)^FracPart[p]/(vv^(m*FracPart[p])*ww^(n*FracPart[p])), Int[uu*vv^(m*p)*ww^(n*p)
, x], x]] /; FreeQ[{m, n, p}, x] &&  !IntegerQ[p] && ( !InertTrigFreeQ[v] ||  !InertTrigFreeQ[w])

Rubi steps \begin{align*} \text {integral}& = \int (\sin (x) \tan (x))^{5/2} \, dx \\ & = \frac {\sqrt {\sin (x) \tan (x)} \int \sin ^{\frac {5}{2}}(x) \tan ^{\frac {5}{2}}(x) \, dx}{\sqrt {\sin (x)} \sqrt {\tan (x)}} \\ & = -\frac {2}{5} \sin ^2(x) \tan (x) \sqrt {\sin (x) \tan (x)}+\frac {\left (8 \sqrt {\sin (x) \tan (x)}\right ) \int \sqrt {\sin (x)} \tan ^{\frac {5}{2}}(x) \, dx}{5 \sqrt {\sin (x)} \sqrt {\tan (x)}} \\ & = \frac {16}{15} \tan (x) \sqrt {\sin (x) \tan (x)}-\frac {2}{5} \sin ^2(x) \tan (x) \sqrt {\sin (x) \tan (x)}-\frac {\left (32 \sqrt {\sin (x) \tan (x)}\right ) \int \sqrt {\sin (x)} \sqrt {\tan (x)} \, dx}{15 \sqrt {\sin (x)} \sqrt {\tan (x)}} \\ & = \frac {64}{15} \cot (x) \sqrt {\sin (x) \tan (x)}+\frac {16}{15} \tan (x) \sqrt {\sin (x) \tan (x)}-\frac {2}{5} \sin ^2(x) \tan (x) \sqrt {\sin (x) \tan (x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.58 \[ \int (-\cos (x)+\sec (x))^{5/2} \, dx=\frac {2}{15} \left (5+3 \cos ^2(x)+32 \cot ^2(x)\right ) \tan (x) \sqrt {\sin (x) \tan (x)} \]

[In]

Integrate[(-Cos[x] + Sec[x])^(5/2),x]

[Out]

(2*(5 + 3*Cos[x]^2 + 32*Cot[x]^2)*Tan[x]*Sqrt[Sin[x]*Tan[x]])/15

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(279\) vs. \(2(38)=76\).

Time = 9.23 (sec) , antiderivative size = 280, normalized size of antiderivative = 5.60

method	result	size
default	\(\frac {\tan \left (x \right ) \left (6 \cos \left (x \right )^{4}-15 \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}\, \ln \left (\frac {2 \cos \left (x \right ) \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}-\cos \left (x \right )+1}{\cos \left (x \right )+1}\right ) \cos \left (x \right )^{2}+15 \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}\, \ln \left (\frac {4 \cos \left (x \right ) \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}-2 \cos \left (x \right )+4 \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}+2}{\cos \left (x \right )+1}\right ) \cos \left (x \right )^{2}-15 \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}\, \ln \left (\frac {2 \cos \left (x \right ) \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}-\cos \left (x \right )+1}{\cos \left (x \right )+1}\right ) \cos \left (x \right )+15 \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}\, \ln \left (\frac {4 \cos \left (x \right ) \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}-2 \cos \left (x \right )+4 \sqrt {-\frac {\cos \left (x \right )}{\left (\cos \left (x \right )+1\right )^{2}}}+2}{\cos \left (x \right )+1}\right ) \cos \left (x \right )-60 \cos \left (x \right )^{2}-10\right ) \sqrt {\sin \left (x \right ) \tan \left (x \right )}}{15 \cos \left (x \right )^{2}-15}\)	\(280\)

[In]

int((-cos(x)+sec(x))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/15*tan(x)*(6*cos(x)^4-15*(-cos(x)/(cos(x)+1)^2)^(1/2)*ln((2*cos(x)*(-cos(x)/(cos(x)+1)^2)^(1/2)+2*(-cos(x)/(
cos(x)+1)^2)^(1/2)-cos(x)+1)/(cos(x)+1))*cos(x)^2+15*(-cos(x)/(cos(x)+1)^2)^(1/2)*ln(2*(2*cos(x)*(-cos(x)/(cos
(x)+1)^2)^(1/2)+2*(-cos(x)/(cos(x)+1)^2)^(1/2)-cos(x)+1)/(cos(x)+1))*cos(x)^2-15*(-cos(x)/(cos(x)+1)^2)^(1/2)*
ln((2*cos(x)*(-cos(x)/(cos(x)+1)^2)^(1/2)+2*(-cos(x)/(cos(x)+1)^2)^(1/2)-cos(x)+1)/(cos(x)+1))*cos(x)+15*(-cos
(x)/(cos(x)+1)^2)^(1/2)*ln(2*(2*cos(x)*(-cos(x)/(cos(x)+1)^2)^(1/2)+2*(-cos(x)/(cos(x)+1)^2)^(1/2)-cos(x)+1)/(
cos(x)+1))*cos(x)-60*cos(x)^2-10)*(sin(x)*tan(x))^(1/2)/(cos(x)^2-1)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.76 \[ \int (-\cos (x)+\sec (x))^{5/2} \, dx=-\frac {2 \, {\left (3 \, \cos \left (x\right )^{4} - 30 \, \cos \left (x\right )^{2} - 5\right )} \sqrt {-\frac {\cos \left (x\right )^{2} - 1}{\cos \left (x\right )}}}{15 \, \cos \left (x\right ) \sin \left (x\right )} \]

[In]

integrate((-cos(x)+sec(x))^(5/2),x, algorithm="fricas")

[Out]

-2/15*(3*cos(x)^4 - 30*cos(x)^2 - 5)*sqrt(-(cos(x)^2 - 1)/cos(x))/(cos(x)*sin(x))

Sympy [F(-1)]

Timed out. \[ \int (-\cos (x)+\sec (x))^{5/2} \, dx=\text {Timed out} \]

[In]

integrate((-cos(x)+sec(x))**(5/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (38) = 76\).

Time = 0.32 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.64 \[ \int (-\cos (x)+\sec (x))^{5/2} \, dx=-\frac {32 \, {\left (\frac {5 \, \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} - \frac {5 \, \sin \left (x\right )^{6}}{{\left (\cos \left (x\right ) + 1\right )}^{6}} + \frac {2 \, \sin \left (x\right )^{10}}{{\left (\cos \left (x\right ) + 1\right )}^{10}} - 2\right )}}{15 \, {\left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1\right )}^{\frac {5}{2}} {\left (-\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1\right )}^{\frac {5}{2}} {\left (\frac {\sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + 1\right )}^{\frac {5}{2}}} \]

[In]

integrate((-cos(x)+sec(x))^(5/2),x, algorithm="maxima")

[Out]

-32/15*(5*sin(x)^4/(cos(x) + 1)^4 - 5*sin(x)^6/(cos(x) + 1)^6 + 2*sin(x)^10/(cos(x) + 1)^10 - 2)/((sin(x)/(cos
(x) + 1) + 1)^(5/2)*(-sin(x)/(cos(x) + 1) + 1)^(5/2)*(sin(x)^2/(cos(x) + 1)^2 + 1)^(5/2))

Giac [F]

\[ \int (-\cos (x)+\sec (x))^{5/2} \, dx=\int { {\left (-\cos \left (x\right ) + \sec \left (x\right )\right )}^{\frac {5}{2}} \,d x } \]

[In]

integrate((-cos(x)+sec(x))^(5/2),x, algorithm="giac")

[Out]

integrate((-cos(x) + sec(x))^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int (-\cos (x)+\sec (x))^{5/2} \, dx=\int {\left (\frac {1}{\cos \left (x\right )}-\cos \left (x\right )\right )}^{5/2} \,d x \]

[In]

int((1/cos(x) - cos(x))^(5/2),x)

[Out]

int((1/cos(x) - cos(x))^(5/2), x)