3.187 \(\int \frac{\cot ^3(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=176 \[ -\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{a}}\right )}{a^{3/2} d}+\frac{11 \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{16 \sqrt{2} a^{3/2} d}-\frac{3 a}{20 d (a \sec (c+d x)+a)^{5/2}}+\frac{a}{2 d (1-\sec (c+d x)) (a \sec (c+d x)+a)^{5/2}}+\frac{5}{24 d (a \sec (c+d x)+a)^{3/2}}+\frac{21}{16 a d \sqrt{a \sec (c+d x)+a}} \]

[Out]

(-2*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/(a^(3/2)*d) + (11*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/(Sqrt[2]*Sqr
t[a])])/(16*Sqrt[2]*a^(3/2)*d) - (3*a)/(20*d*(a + a*Sec[c + d*x])^(5/2)) + a/(2*d*(1 - Sec[c + d*x])*(a + a*Se
c[c + d*x])^(5/2)) + 5/(24*d*(a + a*Sec[c + d*x])^(3/2)) + 21/(16*a*d*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.162197, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3880, 103, 152, 156, 63, 207} \[ -\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{a}}\right )}{a^{3/2} d}+\frac{11 \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{16 \sqrt{2} a^{3/2} d}-\frac{3 a}{20 d (a \sec (c+d x)+a)^{5/2}}+\frac{a}{2 d (1-\sec (c+d x)) (a \sec (c+d x)+a)^{5/2}}+\frac{5}{24 d (a \sec (c+d x)+a)^{3/2}}+\frac{21}{16 a d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^3/(a + a*Sec[c + d*x])^(3/2),x]

[Out]

(-2*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/(a^(3/2)*d) + (11*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/(Sqrt[2]*Sqr
t[a])])/(16*Sqrt[2]*a^(3/2)*d) - (3*a)/(20*d*(a + a*Sec[c + d*x])^(5/2)) + a/(2*d*(1 - Sec[c + d*x])*(a + a*Se
c[c + d*x])^(5/2)) + 5/(24*d*(a + a*Sec[c + d*x])^(3/2)) + 21/(16*a*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 103

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

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 ^3(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx &=\frac{a^4 \operatorname{Subst}\left (\int \frac{1}{x (-a+a x)^2 (a+a x)^{7/2}} \, dx,x,\sec (c+d x)\right )}{d}\\ &=\frac{a}{2 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{5/2}}-\frac{a \operatorname{Subst}\left (\int \frac{2 a^2+\frac{7 a^2 x}{2}}{x (-a+a x) (a+a x)^{7/2}} \, dx,x,\sec (c+d x)\right )}{2 d}\\ &=-\frac{3 a}{20 d (a+a \sec (c+d x))^{5/2}}+\frac{a}{2 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{5/2}}+\frac{\operatorname{Subst}\left (\int \frac{-10 a^4-\frac{15 a^4 x}{4}}{x (-a+a x) (a+a x)^{5/2}} \, dx,x,\sec (c+d x)\right )}{10 a^2 d}\\ &=-\frac{3 a}{20 d (a+a \sec (c+d x))^{5/2}}+\frac{a}{2 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{5/2}}+\frac{5}{24 d (a+a \sec (c+d x))^{3/2}}-\frac{\operatorname{Subst}\left (\int \frac{30 a^6-\frac{75 a^6 x}{8}}{x (-a+a x) (a+a x)^{3/2}} \, dx,x,\sec (c+d x)\right )}{30 a^5 d}\\ &=-\frac{3 a}{20 d (a+a \sec (c+d x))^{5/2}}+\frac{a}{2 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{5/2}}+\frac{5}{24 d (a+a \sec (c+d x))^{3/2}}+\frac{21}{16 a d \sqrt{a+a \sec (c+d x)}}+\frac{\operatorname{Subst}\left (\int \frac{-30 a^8+\frac{315 a^8 x}{16}}{x (-a+a x) \sqrt{a+a x}} \, dx,x,\sec (c+d x)\right )}{30 a^8 d}\\ &=-\frac{3 a}{20 d (a+a \sec (c+d x))^{5/2}}+\frac{a}{2 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{5/2}}+\frac{5}{24 d (a+a \sec (c+d x))^{3/2}}+\frac{21}{16 a d \sqrt{a+a \sec (c+d x)}}-\frac{11 \operatorname{Subst}\left (\int \frac{1}{(-a+a x) \sqrt{a+a x}} \, dx,x,\sec (c+d x)\right )}{32 d}+\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+a x}} \, dx,x,\sec (c+d x)\right )}{a d}\\ &=-\frac{3 a}{20 d (a+a \sec (c+d x))^{5/2}}+\frac{a}{2 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{5/2}}+\frac{5}{24 d (a+a \sec (c+d x))^{3/2}}+\frac{21}{16 a 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^2 d}-\frac{11 \operatorname{Subst}\left (\int \frac{1}{-2 a+x^2} \, dx,x,\sqrt{a+a \sec (c+d x)}\right )}{16 a d}\\ &=-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a+a \sec (c+d x)}}{\sqrt{a}}\right )}{a^{3/2} d}+\frac{11 \tanh ^{-1}\left (\frac{\sqrt{a+a \sec (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{16 \sqrt{2} a^{3/2} d}-\frac{3 a}{20 d (a+a \sec (c+d x))^{5/2}}+\frac{a}{2 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{5/2}}+\frac{5}{24 d (a+a \sec (c+d x))^{3/2}}+\frac{21}{16 a d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 0.139785, size = 90, normalized size = 0.51 \[ \frac{a \left (-11 (\sec (c+d x)-1) \text{Hypergeometric2F1}\left (-\frac{5}{2},1,-\frac{3}{2},\frac{1}{2} (\sec (c+d x)+1)\right )+8 (\sec (c+d x)-1) \text{Hypergeometric2F1}\left (-\frac{5}{2},1,-\frac{3}{2},\sec (c+d x)+1\right )-10\right )}{20 d (\sec (c+d x)-1) (a (\sec (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^3/(a + a*Sec[c + d*x])^(3/2),x]

[Out]

(a*(-10 - 11*Hypergeometric2F1[-5/2, 1, -3/2, (1 + Sec[c + d*x])/2]*(-1 + Sec[c + d*x]) + 8*Hypergeometric2F1[
-5/2, 1, -3/2, 1 + Sec[c + d*x]]*(-1 + Sec[c + d*x])))/(20*d*(-1 + Sec[c + d*x])*(a*(1 + Sec[c + d*x]))^(5/2))

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Maple [B]  time = 0.265, size = 514, normalized size = 2.9 \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)^3/(a+a*sec(d*x+c))^(3/2),x)

[Out]

1/480/d/a^2*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(cos(d*x+c)+1)*(-1+cos(d*x+c))^3*(480*2^(1/2)*cos(d*x+c)^4*(-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))+960*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))+165*cos(d*
x+c)^4*(-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))+330*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))-960*2^(1/2)*cos(d*x+c)*(-2*c
os(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))+898*cos(d*x+c)^4-480*
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))-330*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))+702*cos(d*x+c)^3-16
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))-730*cos(d*x+c)^2-630*cos
(d*x+c))/sin(d*x+c)^8

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot \left (d x + c\right )^{3}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3/(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(cot(d*x + c)^3/(a*sec(d*x + c) + a)^(3/2), 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)^3/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot ^{3}{\left (c + d x \right )}}{\left (a \left (\sec{\left (c + d x \right )} + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**3/(a+a*sec(d*x+c))**(3/2),x)

[Out]

Integral(cot(c + d*x)**3/(a*(sec(c + d*x) + 1))**(3/2), x)

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Giac [B]  time = 9.82777, size = 387, normalized size = 2.2 \begin{align*} \frac{\frac{165 \, \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 \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} - \frac{960 \, \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 \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} + \frac{15 \, \sqrt{2} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}{a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}} - \frac{2 \,{\left (3 \, \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^{16} + 20 \, \sqrt{2}{\left (-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{\frac{3}{2}} a^{17} + 165 \, \sqrt{2} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} a^{18}\right )}}{a^{20} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3/(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

1/480*(165*sqrt(2)*arctan(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/(sqrt(-a)*a*sgn(tan(1/2*d*x + 1/2*c)^2
- 1)) - 960*arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/(sqrt(-a)*a*sgn(tan(1/2*d*x + 1/
2*c)^2 - 1)) + 15*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/(a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)*tan(1/2*d*x
+ 1/2*c)^2) - 2*(3*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^16 + 20*sqr
t(2)*(-a*tan(1/2*d*x + 1/2*c)^2 + a)^(3/2)*a^17 + 165*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*a^18)/(a^20*
sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))/d