3.339 \(\int \frac{1}{(-\cos (x)+\sec (x))^{5/2}} \, dx\)

Optimal. Leaf size=91 \[ -\frac{3 \sin (x) \tan ^{-1}\left (\sqrt{\cos (x)}\right )}{32 \sqrt{\cos (x)} \sqrt{\sin (x) \tan (x)}}+\frac{3 \cot (x)}{16 \sqrt{\sin (x) \tan (x)}}+\frac{3 \sin (x) \tanh ^{-1}\left (\sqrt{\cos (x)}\right )}{32 \sqrt{\cos (x)} \sqrt{\sin (x) \tan (x)}}-\frac{\cot (x) \csc ^2(x)}{4 \sqrt{\sin (x) \tan (x)}} \]

[Out]

(3*Cot[x])/(16*Sqrt[Sin[x]*Tan[x]]) - (Cot[x]*Csc[x]^2)/(4*Sqrt[Sin[x]*Tan[x]]) - (3*ArcTan[Sqrt[Cos[x]]]*Sin[
x])/(32*Sqrt[Cos[x]]*Sqrt[Sin[x]*Tan[x]]) + (3*ArcTanh[Sqrt[Cos[x]]]*Sin[x])/(32*Sqrt[Cos[x]]*Sqrt[Sin[x]*Tan[
x]])

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Rubi [A]  time = 0.120849, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.909, Rules used = {4397, 4400, 2597, 2599, 2601, 2565, 329, 298, 203, 206} \[ -\frac{3 \sin (x) \tan ^{-1}\left (\sqrt{\cos (x)}\right )}{32 \sqrt{\cos (x)} \sqrt{\sin (x) \tan (x)}}+\frac{3 \cot (x)}{16 \sqrt{\sin (x) \tan (x)}}+\frac{3 \sin (x) \tanh ^{-1}\left (\sqrt{\cos (x)}\right )}{32 \sqrt{\cos (x)} \sqrt{\sin (x) \tan (x)}}-\frac{\cot (x) \csc ^2(x)}{4 \sqrt{\sin (x) \tan (x)}} \]

Antiderivative was successfully verified.

[In]

Int[(-Cos[x] + Sec[x])^(-5/2),x]

[Out]

(3*Cot[x])/(16*Sqrt[Sin[x]*Tan[x]]) - (Cot[x]*Csc[x]^2)/(4*Sqrt[Sin[x]*Tan[x]]) - (3*ArcTan[Sqrt[Cos[x]]]*Sin[
x])/(32*Sqrt[Cos[x]]*Sqrt[Sin[x]*Tan[x]]) + (3*ArcTanh[Sqrt[Cos[x]]]*Sin[x])/(32*Sqrt[Cos[x]]*Sqrt[Sin[x]*Tan[
x]])

Rule 4397

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

Rule 4400

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])

Rule 2597

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*Sin[e +
f*x])^m*(b*Tan[e + f*x])^(n + 1))/(b*f*(m + n + 1)), x] - Dist[(n + 1)/(b^2*(m + 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] && LtQ[n, -1] && NeQ[m + n + 1, 0] && Integer
sQ[2*m, 2*n] &&  !(EqQ[n, -3/2] && EqQ[m, 1])

Rule 2599

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[(b*(a*Sin[e
+ f*x])^(m + 2)*(b*Tan[e + f*x])^(n - 1))/(a^2*f*(m + n + 1)), x] + Dist[(m + 2)/(a^2*(m + n + 1)), Int[(a*Sin
[e + f*x])^(m + 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] && NeQ[m + n + 1, 0]
&& IntegersQ[2*m, 2*n]

Rule 2601

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(Cos[e + f*x
]^n*(b*Tan[e + f*x])^n)/(a*Sin[e + f*x])^n, Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] &&  !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(-1)]) || IntegersQ[m - 1/2, n -
1/2])

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

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 206

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

Rubi steps

\begin{align*} \int \frac{1}{(-\cos (x)+\sec (x))^{5/2}} \, dx &=\int \frac{1}{(\sin (x) \tan (x))^{5/2}} \, dx\\ &=\frac{\left (\sqrt{\sin (x)} \sqrt{\tan (x)}\right ) \int \frac{1}{\sin ^{\frac{5}{2}}(x) \tan ^{\frac{5}{2}}(x)} \, dx}{\sqrt{\sin (x) \tan (x)}}\\ &=-\frac{\cot (x) \csc ^2(x)}{4 \sqrt{\sin (x) \tan (x)}}-\frac{\left (3 \sqrt{\sin (x)} \sqrt{\tan (x)}\right ) \int \frac{1}{\sin ^{\frac{5}{2}}(x) \sqrt{\tan (x)}} \, dx}{8 \sqrt{\sin (x) \tan (x)}}\\ &=\frac{3 \cot (x)}{16 \sqrt{\sin (x) \tan (x)}}-\frac{\cot (x) \csc ^2(x)}{4 \sqrt{\sin (x) \tan (x)}}-\frac{\left (3 \sqrt{\sin (x)} \sqrt{\tan (x)}\right ) \int \frac{1}{\sqrt{\sin (x)} \sqrt{\tan (x)}} \, dx}{32 \sqrt{\sin (x) \tan (x)}}\\ &=\frac{3 \cot (x)}{16 \sqrt{\sin (x) \tan (x)}}-\frac{\cot (x) \csc ^2(x)}{4 \sqrt{\sin (x) \tan (x)}}-\frac{(3 \sin (x)) \int \sqrt{\cos (x)} \csc (x) \, dx}{32 \sqrt{\cos (x)} \sqrt{\sin (x) \tan (x)}}\\ &=\frac{3 \cot (x)}{16 \sqrt{\sin (x) \tan (x)}}-\frac{\cot (x) \csc ^2(x)}{4 \sqrt{\sin (x) \tan (x)}}+\frac{(3 \sin (x)) \operatorname{Subst}\left (\int \frac{\sqrt{x}}{1-x^2} \, dx,x,\cos (x)\right )}{32 \sqrt{\cos (x)} \sqrt{\sin (x) \tan (x)}}\\ &=\frac{3 \cot (x)}{16 \sqrt{\sin (x) \tan (x)}}-\frac{\cot (x) \csc ^2(x)}{4 \sqrt{\sin (x) \tan (x)}}+\frac{(3 \sin (x)) \operatorname{Subst}\left (\int \frac{x^2}{1-x^4} \, dx,x,\sqrt{\cos (x)}\right )}{16 \sqrt{\cos (x)} \sqrt{\sin (x) \tan (x)}}\\ &=\frac{3 \cot (x)}{16 \sqrt{\sin (x) \tan (x)}}-\frac{\cot (x) \csc ^2(x)}{4 \sqrt{\sin (x) \tan (x)}}+\frac{(3 \sin (x)) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt{\cos (x)}\right )}{32 \sqrt{\cos (x)} \sqrt{\sin (x) \tan (x)}}-\frac{(3 \sin (x)) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt{\cos (x)}\right )}{32 \sqrt{\cos (x)} \sqrt{\sin (x) \tan (x)}}\\ &=\frac{3 \cot (x)}{16 \sqrt{\sin (x) \tan (x)}}-\frac{\cot (x) \csc ^2(x)}{4 \sqrt{\sin (x) \tan (x)}}-\frac{3 \tan ^{-1}\left (\sqrt{\cos (x)}\right ) \sin (x)}{32 \sqrt{\cos (x)} \sqrt{\sin (x) \tan (x)}}+\frac{3 \tanh ^{-1}\left (\sqrt{\cos (x)}\right ) \sin (x)}{32 \sqrt{\cos (x)} \sqrt{\sin (x) \tan (x)}}\\ \end{align*}

Mathematica [A]  time = 0.647308, size = 73, normalized size = 0.8 \[ -\frac{\cot (x) \sqrt{\sin (x) \tan (x)} \left (3 \cos (x) \tan ^{-1}\left (\sqrt [4]{\cos ^2(x)}\right )-3 \cos (x) \tanh ^{-1}\left (\sqrt [4]{\cos ^2(x)}\right )+\cos ^2(x)^{3/4} (3 \cos (2 x)+5) \cot (x) \csc ^3(x)\right )}{32 \cos ^2(x)^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(-Cos[x] + Sec[x])^(-5/2),x]

[Out]

-(Cot[x]*(3*ArcTan[(Cos[x]^2)^(1/4)]*Cos[x] - 3*ArcTanh[(Cos[x]^2)^(1/4)]*Cos[x] + (Cos[x]^2)^(3/4)*(5 + 3*Cos
[2*x])*Cot[x]*Csc[x]^3)*Sqrt[Sin[x]*Tan[x]])/(32*(Cos[x]^2)^(3/4))

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Maple [B]  time = 0.15, size = 454, normalized size = 5. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-cos(x)+sec(x))^(5/2),x)

[Out]

1/64*(24*cos(x)^3*(-cos(x)/(1+cos(x))^2)^(3/2)+40*cos(x)^2*(-cos(x)/(1+cos(x))^2)^(3/2)-12*cos(x)^3*(-cos(x)/(
1+cos(x))^2)^(1/2)-3*cos(x)^3*ln(-(2*cos(x)^2*(-cos(x)/(1+cos(x))^2)^(1/2)-cos(x)^2+2*cos(x)-2*(-cos(x)/(1+cos
(x))^2)^(1/2)-1)/sin(x)^2)-3*cos(x)^3*arctan(1/2/(-cos(x)/(1+cos(x))^2)^(1/2))+8*cos(x)*(-cos(x)/(1+cos(x))^2)
^(3/2)+24*cos(x)^2*(-cos(x)/(1+cos(x))^2)^(1/2)+3*cos(x)^2*ln(-(2*cos(x)^2*(-cos(x)/(1+cos(x))^2)^(1/2)-cos(x)
^2+2*cos(x)-2*(-cos(x)/(1+cos(x))^2)^(1/2)-1)/sin(x)^2)+3*cos(x)^2*arctan(1/2/(-cos(x)/(1+cos(x))^2)^(1/2))-8*
(-cos(x)/(1+cos(x))^2)^(3/2)-12*cos(x)*(-cos(x)/(1+cos(x))^2)^(1/2)+3*cos(x)*ln(-(2*cos(x)^2*(-cos(x)/(1+cos(x
))^2)^(1/2)-cos(x)^2+2*cos(x)-2*(-cos(x)/(1+cos(x))^2)^(1/2)-1)/sin(x)^2)+3*cos(x)*arctan(1/2/(-cos(x)/(1+cos(
x))^2)^(1/2))-3*ln(-(2*cos(x)^2*(-cos(x)/(1+cos(x))^2)^(1/2)-cos(x)^2+2*cos(x)-2*(-cos(x)/(1+cos(x))^2)^(1/2)-
1)/sin(x)^2)-3*arctan(1/2/(-cos(x)/(1+cos(x))^2)^(1/2)))*sin(x)/cos(x)^2/(-(-1+cos(x)^2)/cos(x))^(5/2)/(-cos(x
)/(1+cos(x))^2)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (-\cos \left (x\right ) + \sec \left (x\right )\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-cos(x)+sec(x))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.47547, size = 452, normalized size = 4.97 \begin{align*} \frac{3 \,{\left (\cos \left (x\right )^{4} - 2 \, \cos \left (x\right )^{2} + 1\right )} \arctan \left (\frac{2 \, \sqrt{-\frac{\cos \left (x\right )^{2} - 1}{\cos \left (x\right )}} \cos \left (x\right )}{{\left (\cos \left (x\right ) - 1\right )} \sin \left (x\right )}\right ) \sin \left (x\right ) + 3 \,{\left (\cos \left (x\right )^{4} - 2 \, \cos \left (x\right )^{2} + 1\right )} \log \left (\frac{{\left (\cos \left (x\right ) + 1\right )} \sin \left (x\right ) + 2 \, \sqrt{-\frac{\cos \left (x\right )^{2} - 1}{\cos \left (x\right )}} \cos \left (x\right )}{{\left (\cos \left (x\right ) - 1\right )} \sin \left (x\right )}\right ) \sin \left (x\right ) - 4 \,{\left (3 \, \cos \left (x\right )^{4} + \cos \left (x\right )^{2}\right )} \sqrt{-\frac{\cos \left (x\right )^{2} - 1}{\cos \left (x\right )}}}{64 \,{\left (\cos \left (x\right )^{4} - 2 \, \cos \left (x\right )^{2} + 1\right )} \sin \left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-cos(x)+sec(x))^(5/2),x, algorithm="fricas")

[Out]

1/64*(3*(cos(x)^4 - 2*cos(x)^2 + 1)*arctan(2*sqrt(-(cos(x)^2 - 1)/cos(x))*cos(x)/((cos(x) - 1)*sin(x)))*sin(x)
+ 3*(cos(x)^4 - 2*cos(x)^2 + 1)*log(((cos(x) + 1)*sin(x) + 2*sqrt(-(cos(x)^2 - 1)/cos(x))*cos(x))/((cos(x) -
1)*sin(x)))*sin(x) - 4*(3*cos(x)^4 + cos(x)^2)*sqrt(-(cos(x)^2 - 1)/cos(x)))/((cos(x)^4 - 2*cos(x)^2 + 1)*sin(
x))

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (-\cos \left (x\right ) + \sec \left (x\right )\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-cos(x)+sec(x))^(5/2),x, algorithm="giac")

[Out]

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