3.198 \(\int \frac{\cot (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=144 \[ -\frac{7}{4 a^2 d \sqrt{a \sec (c+d x)+a}}+\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{a}}\right )}{a^{5/2} d}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} a^{5/2} d}-\frac{1}{2 a d (a \sec (c+d x)+a)^{3/2}}-\frac{1}{5 d (a \sec (c+d x)+a)^{5/2}} \]

[Out]

(2*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/(a^(5/2)*d) - ArcTanh[Sqrt[a + a*Sec[c + d*x]]/(Sqrt[2]*Sqrt[a])
]/(4*Sqrt[2]*a^(5/2)*d) - 1/(5*d*(a + a*Sec[c + d*x])^(5/2)) - 1/(2*a*d*(a + a*Sec[c + d*x])^(3/2)) - 7/(4*a^2
*d*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.125346, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3880, 85, 152, 156, 63, 207} \[ -\frac{7}{4 a^2 d \sqrt{a \sec (c+d x)+a}}+\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{a}}\right )}{a^{5/2} d}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} a^{5/2} d}-\frac{1}{2 a d (a \sec (c+d x)+a)^{3/2}}-\frac{1}{5 d (a \sec (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

(2*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/(a^(5/2)*d) - ArcTanh[Sqrt[a + a*Sec[c + d*x]]/(Sqrt[2]*Sqrt[a])
]/(4*Sqrt[2]*a^(5/2)*d) - 1/(5*d*(a + a*Sec[c + d*x])^(5/2)) - 1/(2*a*d*(a + a*Sec[c + d*x])^(3/2)) - 7/(4*a^2
*d*Sqrt[a + a*Sec[c + d*x]])

Rule 3880

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Dist[(d*b^(m - 1)
)^(-1), Subst[Int[((-a + b*x)^((m - 1)/2)*(a + b*x)^((m - 1)/2 + n))/x, x], x, Csc[c + d*x]], x] /; FreeQ[{a,
b, c, d, n}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[n]

Rule 85

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[(f*(e + f*x)^(p +
1))/((p + 1)*(b*e - a*f)*(d*e - c*f)), x] + Dist[1/((b*e - a*f)*(d*e - c*f)), Int[((b*d*e - b*c*f - a*d*f - b
*d*f*x)*(e + f*x)^(p + 1))/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[p, -1]

Rule 152

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 207

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

Rubi steps

\begin{align*} \int \frac{\cot (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx &=\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{x (-a+a x) (a+a x)^{7/2}} \, dx,x,\sec (c+d x)\right )}{d}\\ &=-\frac{1}{5 d (a+a \sec (c+d x))^{5/2}}+\frac{\operatorname{Subst}\left (\int \frac{2 a^2-a^2 x}{x (-a+a x) (a+a x)^{5/2}} \, dx,x,\sec (c+d x)\right )}{2 a d}\\ &=-\frac{1}{5 d (a+a \sec (c+d x))^{5/2}}-\frac{1}{2 a d (a+a \sec (c+d x))^{3/2}}-\frac{\operatorname{Subst}\left (\int \frac{-6 a^4+\frac{9 a^4 x}{2}}{x (-a+a x) (a+a x)^{3/2}} \, dx,x,\sec (c+d x)\right )}{6 a^4 d}\\ &=-\frac{1}{5 d (a+a \sec (c+d x))^{5/2}}-\frac{1}{2 a d (a+a \sec (c+d x))^{3/2}}-\frac{7}{4 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{\operatorname{Subst}\left (\int \frac{6 a^6-\frac{21 a^6 x}{4}}{x (-a+a x) \sqrt{a+a x}} \, dx,x,\sec (c+d x)\right )}{6 a^7 d}\\ &=-\frac{1}{5 d (a+a \sec (c+d x))^{5/2}}-\frac{1}{2 a d (a+a \sec (c+d x))^{3/2}}-\frac{7}{4 a^2 d \sqrt{a+a \sec (c+d x)}}-\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+a x}} \, dx,x,\sec (c+d x)\right )}{a^2 d}+\frac{\operatorname{Subst}\left (\int \frac{1}{(-a+a x) \sqrt{a+a x}} \, dx,x,\sec (c+d x)\right )}{8 a d}\\ &=-\frac{1}{5 d (a+a \sec (c+d x))^{5/2}}-\frac{1}{2 a d (a+a \sec (c+d x))^{3/2}}-\frac{7}{4 a^2 d \sqrt{a+a \sec (c+d x)}}-\frac{2 \operatorname{Subst}\left (\int \frac{1}{-1+\frac{x^2}{a}} \, dx,x,\sqrt{a+a \sec (c+d x)}\right )}{a^3 d}+\frac{\operatorname{Subst}\left (\int \frac{1}{-2 a+x^2} \, dx,x,\sqrt{a+a \sec (c+d x)}\right )}{4 a^2 d}\\ &=\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a+a \sec (c+d x)}}{\sqrt{a}}\right )}{a^{5/2} d}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+a \sec (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} a^{5/2} d}-\frac{1}{5 d (a+a \sec (c+d x))^{5/2}}-\frac{1}{2 a d (a+a \sec (c+d x))^{3/2}}-\frac{7}{4 a^2 d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 0.0617843, size = 60, normalized size = 0.42 \[ \frac{\text{Hypergeometric2F1}\left (-\frac{5}{2},1,-\frac{3}{2},\frac{1}{2} (\sec (c+d x)+1)\right )-2 \text{Hypergeometric2F1}\left (-\frac{5}{2},1,-\frac{3}{2},\sec (c+d x)+1\right )}{5 d (a (\sec (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

(Hypergeometric2F1[-5/2, 1, -3/2, (1 + Sec[c + d*x])/2] - 2*Hypergeometric2F1[-5/2, 1, -3/2, 1 + Sec[c + d*x]]
)/(5*d*(a*(1 + Sec[c + d*x]))^(5/2))

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Maple [B]  time = 0.224, size = 496, normalized size = 3.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)/(a+a*sec(d*x+c))^(5/2),x)

[Out]

1/40/d/a^3*(-1+cos(d*x+c))^3*(40*2^(1/2)*cos(d*x+c)^3*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*
(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+5*cos(d*x+c)^3*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x
+c)/(cos(d*x+c)+1))^(1/2))+120*cos(d*x+c)^2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(-
2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+15*cos(d*x+c)^2*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+
c)/(cos(d*x+c)+1))^(1/2))+120*2^(1/2)*cos(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(-2*c
os(d*x+c)/(cos(d*x+c)+1))^(1/2))+98*cos(d*x+c)^3+15*cos(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(
-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+40*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(-2*co
s(d*x+c)/(cos(d*x+c)+1))^(1/2))+160*cos(d*x+c)^2+5*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+c
)/(cos(d*x+c)+1))^(1/2))+70*cos(d*x+c))*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/sin(d*x+c)^6

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(cot(d*x + c)/(a*sec(d*x + c) + a)^(5/2), x)

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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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(cot(d*x+c)/(a+a*sec(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 9.0779, size = 315, normalized size = 2.19 \begin{align*} -\frac{\frac{5 \, \sqrt{2} \arctan \left (\frac{\sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} - \frac{80 \, \arctan \left (\frac{\sqrt{2} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}{2 \, \sqrt{-a}}\right )}{\sqrt{-a} a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} - \frac{\sqrt{2}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} a^{20} + 5 \, \sqrt{2}{\left (-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{\frac{3}{2}} a^{21} + 35 \, \sqrt{2} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} a^{22}}{a^{25} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}}{40 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

-1/40*(5*sqrt(2)*arctan(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/(sqrt(-a)*a^2*sgn(tan(1/2*d*x + 1/2*c)^2
- 1)) - 80*arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/(sqrt(-a)*a^2*sgn(tan(1/2*d*x + 1
/2*c)^2 - 1)) - (sqrt(2)*(a*tan(1/2*d*x + 1/2*c)^2 - a)^2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*a^20 + 5*sqrt(2)
*(-a*tan(1/2*d*x + 1/2*c)^2 + a)^(3/2)*a^21 + 35*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*a^22)/(a^25*sgn(t
an(1/2*d*x + 1/2*c)^2 - 1)))/d