3.340 \(\int \frac{1}{(-\cos (x)+\sec (x))^{7/2}} \, dx\)

Optimal. Leaf size=110 \[ -\frac{5 \sin (x) \tan ^{-1}\left (\sqrt{\cos (x)}\right )}{128 \sqrt{\cos (x)} \sqrt{\sin (x) \tan (x)}}+\frac{5 \csc ^3(x)}{48 \sqrt{\sin (x) \tan (x)}}-\frac{5 \csc (x)}{192 \sqrt{\sin (x) \tan (x)}}-\frac{5 \sin (x) \tanh ^{-1}\left (\sqrt{\cos (x)}\right )}{128 \sqrt{\cos (x)} \sqrt{\sin (x) \tan (x)}}-\frac{\cot ^2(x) \csc ^3(x)}{6 \sqrt{\sin (x) \tan (x)}} \]

[Out]

(-5*Csc[x])/(192*Sqrt[Sin[x]*Tan[x]]) + (5*Csc[x]^3)/(48*Sqrt[Sin[x]*Tan[x]]) - (Cot[x]^2*Csc[x]^3)/(6*Sqrt[Si
n[x]*Tan[x]]) - (5*ArcTan[Sqrt[Cos[x]]]*Sin[x])/(128*Sqrt[Cos[x]]*Sqrt[Sin[x]*Tan[x]]) - (5*ArcTanh[Sqrt[Cos[x
]]]*Sin[x])/(128*Sqrt[Cos[x]]*Sqrt[Sin[x]*Tan[x]])

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Rubi [A]  time = 0.140367, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 11, 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, 212, 206, 203} \[ -\frac{5 \sin (x) \tan ^{-1}\left (\sqrt{\cos (x)}\right )}{128 \sqrt{\cos (x)} \sqrt{\sin (x) \tan (x)}}+\frac{5 \csc ^3(x)}{48 \sqrt{\sin (x) \tan (x)}}-\frac{5 \csc (x)}{192 \sqrt{\sin (x) \tan (x)}}-\frac{5 \sin (x) \tanh ^{-1}\left (\sqrt{\cos (x)}\right )}{128 \sqrt{\cos (x)} \sqrt{\sin (x) \tan (x)}}-\frac{\cot ^2(x) \csc ^3(x)}{6 \sqrt{\sin (x) \tan (x)}} \]

Antiderivative was successfully verified.

[In]

Int[(-Cos[x] + Sec[x])^(-7/2),x]

[Out]

(-5*Csc[x])/(192*Sqrt[Sin[x]*Tan[x]]) + (5*Csc[x]^3)/(48*Sqrt[Sin[x]*Tan[x]]) - (Cot[x]^2*Csc[x]^3)/(6*Sqrt[Si
n[x]*Tan[x]]) - (5*ArcTan[Sqrt[Cos[x]]]*Sin[x])/(128*Sqrt[Cos[x]]*Sqrt[Sin[x]*Tan[x]]) - (5*ArcTanh[Sqrt[Cos[x
]]]*Sin[x])/(128*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 212

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

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 \frac{1}{(-\cos (x)+\sec (x))^{7/2}} \, dx &=\int \frac{1}{(\sin (x) \tan (x))^{7/2}} \, dx\\ &=\frac{\left (\sqrt{\sin (x)} \sqrt{\tan (x)}\right ) \int \frac{1}{\sin ^{\frac{7}{2}}(x) \tan ^{\frac{7}{2}}(x)} \, dx}{\sqrt{\sin (x) \tan (x)}}\\ &=-\frac{\cot ^2(x) \csc ^3(x)}{6 \sqrt{\sin (x) \tan (x)}}-\frac{\left (5 \sqrt{\sin (x)} \sqrt{\tan (x)}\right ) \int \frac{1}{\sin ^{\frac{7}{2}}(x) \tan ^{\frac{3}{2}}(x)} \, dx}{12 \sqrt{\sin (x) \tan (x)}}\\ &=\frac{5 \csc ^3(x)}{48 \sqrt{\sin (x) \tan (x)}}-\frac{\cot ^2(x) \csc ^3(x)}{6 \sqrt{\sin (x) \tan (x)}}+\frac{\left (5 \sqrt{\sin (x)} \sqrt{\tan (x)}\right ) \int \frac{\sqrt{\tan (x)}}{\sin ^{\frac{7}{2}}(x)} \, dx}{96 \sqrt{\sin (x) \tan (x)}}\\ &=-\frac{5 \csc (x)}{192 \sqrt{\sin (x) \tan (x)}}+\frac{5 \csc ^3(x)}{48 \sqrt{\sin (x) \tan (x)}}-\frac{\cot ^2(x) \csc ^3(x)}{6 \sqrt{\sin (x) \tan (x)}}+\frac{\left (5 \sqrt{\sin (x)} \sqrt{\tan (x)}\right ) \int \frac{\sqrt{\tan (x)}}{\sin ^{\frac{3}{2}}(x)} \, dx}{128 \sqrt{\sin (x) \tan (x)}}\\ &=-\frac{5 \csc (x)}{192 \sqrt{\sin (x) \tan (x)}}+\frac{5 \csc ^3(x)}{48 \sqrt{\sin (x) \tan (x)}}-\frac{\cot ^2(x) \csc ^3(x)}{6 \sqrt{\sin (x) \tan (x)}}+\frac{(5 \sin (x)) \int \frac{\csc (x)}{\sqrt{\cos (x)}} \, dx}{128 \sqrt{\cos (x)} \sqrt{\sin (x) \tan (x)}}\\ &=-\frac{5 \csc (x)}{192 \sqrt{\sin (x) \tan (x)}}+\frac{5 \csc ^3(x)}{48 \sqrt{\sin (x) \tan (x)}}-\frac{\cot ^2(x) \csc ^3(x)}{6 \sqrt{\sin (x) \tan (x)}}-\frac{(5 \sin (x)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (1-x^2\right )} \, dx,x,\cos (x)\right )}{128 \sqrt{\cos (x)} \sqrt{\sin (x) \tan (x)}}\\ &=-\frac{5 \csc (x)}{192 \sqrt{\sin (x) \tan (x)}}+\frac{5 \csc ^3(x)}{48 \sqrt{\sin (x) \tan (x)}}-\frac{\cot ^2(x) \csc ^3(x)}{6 \sqrt{\sin (x) \tan (x)}}-\frac{(5 \sin (x)) \operatorname{Subst}\left (\int \frac{1}{1-x^4} \, dx,x,\sqrt{\cos (x)}\right )}{64 \sqrt{\cos (x)} \sqrt{\sin (x) \tan (x)}}\\ &=-\frac{5 \csc (x)}{192 \sqrt{\sin (x) \tan (x)}}+\frac{5 \csc ^3(x)}{48 \sqrt{\sin (x) \tan (x)}}-\frac{\cot ^2(x) \csc ^3(x)}{6 \sqrt{\sin (x) \tan (x)}}-\frac{(5 \sin (x)) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt{\cos (x)}\right )}{128 \sqrt{\cos (x)} \sqrt{\sin (x) \tan (x)}}-\frac{(5 \sin (x)) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sqrt{\cos (x)}\right )}{128 \sqrt{\cos (x)} \sqrt{\sin (x) \tan (x)}}\\ &=-\frac{5 \csc (x)}{192 \sqrt{\sin (x) \tan (x)}}+\frac{5 \csc ^3(x)}{48 \sqrt{\sin (x) \tan (x)}}-\frac{\cot ^2(x) \csc ^3(x)}{6 \sqrt{\sin (x) \tan (x)}}-\frac{5 \tan ^{-1}\left (\sqrt{\cos (x)}\right ) \sin (x)}{128 \sqrt{\cos (x)} \sqrt{\sin (x) \tan (x)}}-\frac{5 \tanh ^{-1}\left (\sqrt{\cos (x)}\right ) \sin (x)}{128 \sqrt{\cos (x)} \sqrt{\sin (x) \tan (x)}}\\ \end{align*}

Mathematica [A]  time = 0.369063, size = 74, normalized size = 0.67 \[ -\frac{\cot (x) \sqrt{\sin (x) \tan (x)} \left (15 \tan ^{-1}\left (\sqrt [4]{\cos ^2(x)}\right )+2 \sqrt [4]{\cos ^2(x)} \left (32 \csc ^4(x)-52 \csc ^2(x)+5\right ) \csc ^2(x)+15 \tanh ^{-1}\left (\sqrt [4]{\cos ^2(x)}\right )\right )}{384 \sqrt [4]{\cos ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(-Cos[x] + Sec[x])^(-7/2),x]

[Out]

-(Cot[x]*(15*ArcTan[(Cos[x]^2)^(1/4)] + 15*ArcTanh[(Cos[x]^2)^(1/4)] + 2*(Cos[x]^2)^(1/4)*Csc[x]^2*(5 - 52*Csc
[x]^2 + 32*Csc[x]^4))*Sqrt[Sin[x]*Tan[x]])/(384*(Cos[x]^2)^(1/4))

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Maple [B]  time = 0.167, size = 494, normalized size = 4.5 \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))^(7/2),x)

[Out]

1/768*(56*cos(x)^4*(-cos(x)/(1+cos(x))^2)^(3/2)-16*cos(x)^3*(-cos(x)/(1+cos(x))^2)^(3/2)-15*cos(x)^4*ln(-(2*co
s(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)+15*cos(x)^4*
arctan(1/2/(-cos(x)/(1+cos(x))^2)^(1/2))-192*cos(x)^2*(-cos(x)/(1+cos(x))^2)^(3/2)+76*cos(x)^3*(-cos(x)/(1+cos
(x))^2)^(1/2)+30*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)-30*cos(x)^3*arctan(1/2/(-cos(x)/(1+cos(x))^2)^(1/2))+16*cos(x)*(-cos(x)/(1+cos(x))^2)^(
3/2)-148*cos(x)^2*(-cos(x)/(1+cos(x))^2)^(1/2)+136*(-cos(x)/(1+cos(x))^2)^(3/2)+196*cos(x)*(-cos(x)/(1+cos(x))
^2)^(1/2)-30*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)+30*cos(x)*arctan(1/2/(-cos(x)/(1+cos(x))^2)^(1/2))-60*(-cos(x)/(1+cos(x))^2)^(1/2)+15*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)-15*arctan(
1/2/(-cos(x)/(1+cos(x))^2)^(1/2)))*sin(x)^3/(-1+cos(x))/cos(x)^3/(-(-1+cos(x)^2)/cos(x))^(7/2)/(-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{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-cos(x)+sec(x))^(7/2),x, algorithm="maxima")

[Out]

integrate((-cos(x) + sec(x))^(-7/2), x)

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Fricas [B]  time = 2.57107, size = 529, normalized size = 4.81 \begin{align*} \frac{15 \,{\left (\cos \left (x\right )^{6} - 3 \, \cos \left (x\right )^{4} + 3 \, \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 ) + 15 \,{\left (\cos \left (x\right )^{6} - 3 \, \cos \left (x\right )^{4} + 3 \, \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 (5 \, \cos \left (x\right )^{5} + 42 \, \cos \left (x\right )^{3} - 15 \, \cos \left (x\right )\right )} \sqrt{-\frac{\cos \left (x\right )^{2} - 1}{\cos \left (x\right )}}}{768 \,{\left (\cos \left (x\right )^{6} - 3 \, \cos \left (x\right )^{4} + 3 \, \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))^(7/2),x, algorithm="fricas")

[Out]

1/768*(15*(cos(x)^6 - 3*cos(x)^4 + 3*cos(x)^2 - 1)*arctan(2*sqrt(-(cos(x)^2 - 1)/cos(x))*cos(x)/((cos(x) - 1)*
sin(x)))*sin(x) + 15*(cos(x)^6 - 3*cos(x)^4 + 3*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*(5*cos(x)^5 + 42*cos(x)^3 - 15*cos(x))*sqrt(-(cos(x)^2 - 1
)/cos(x)))/((cos(x)^6 - 3*cos(x)^4 + 3*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))**(7/2),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{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-cos(x)+sec(x))^(7/2),x, algorithm="giac")

[Out]

integrate((-cos(x) + sec(x))^(-7/2), x)