3.176 \(\int \frac{\cot ^5(c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx\)

Optimal. Leaf size=214 \[ \frac{87 a^2}{160 d (a \sec (c+d x)+a)^{5/2}}-\frac{17 a^2}{16 d (1-\sec (c+d x)) (a \sec (c+d x)+a)^{5/2}}-\frac{a^2}{4 d (1-\sec (c+d x))^2 (a \sec (c+d x)+a)^{5/2}}+\frac{23 a}{192 d (a \sec (c+d x)+a)^{3/2}}-\frac{105}{128 d \sqrt{a \sec (c+d x)+a}}+\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{a}}\right )}{\sqrt{a} d}-\frac{151 \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{128 \sqrt{2} \sqrt{a} d} \]

[Out]

(2*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/(Sqrt[a]*d) - (151*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/(Sqrt[2]*Sqr
t[a])])/(128*Sqrt[2]*Sqrt[a]*d) + (87*a^2)/(160*d*(a + a*Sec[c + d*x])^(5/2)) - a^2/(4*d*(1 - Sec[c + d*x])^2*
(a + a*Sec[c + d*x])^(5/2)) - (17*a^2)/(16*d*(1 - Sec[c + d*x])*(a + a*Sec[c + d*x])^(5/2)) + (23*a)/(192*d*(a
+ a*Sec[c + d*x])^(3/2)) - 105/(128*d*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.177853, antiderivative size = 214, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {3880, 103, 151, 152, 156, 63, 207} \[ \frac{87 a^2}{160 d (a \sec (c+d x)+a)^{5/2}}-\frac{17 a^2}{16 d (1-\sec (c+d x)) (a \sec (c+d x)+a)^{5/2}}-\frac{a^2}{4 d (1-\sec (c+d x))^2 (a \sec (c+d x)+a)^{5/2}}+\frac{23 a}{192 d (a \sec (c+d x)+a)^{3/2}}-\frac{105}{128 d \sqrt{a \sec (c+d x)+a}}+\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{a}}\right )}{\sqrt{a} d}-\frac{151 \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{128 \sqrt{2} \sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5/Sqrt[a + a*Sec[c + d*x]],x]

[Out]

(2*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/(Sqrt[a]*d) - (151*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/(Sqrt[2]*Sqr
t[a])])/(128*Sqrt[2]*Sqrt[a]*d) + (87*a^2)/(160*d*(a + a*Sec[c + d*x])^(5/2)) - a^2/(4*d*(1 - Sec[c + d*x])^2*
(a + a*Sec[c + d*x])^(5/2)) - (17*a^2)/(16*d*(1 - Sec[c + d*x])*(a + a*Sec[c + d*x])^(5/2)) + (23*a)/(192*d*(a
+ a*Sec[c + d*x])^(3/2)) - 105/(128*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 151

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] && IntegerQ[m]

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 ^5(c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx &=\frac{a^6 \operatorname{Subst}\left (\int \frac{1}{x (-a+a x)^3 (a+a x)^{7/2}} \, dx,x,\sec (c+d x)\right )}{d}\\ &=-\frac{a^2}{4 d (1-\sec (c+d x))^2 (a+a \sec (c+d x))^{5/2}}-\frac{a^3 \operatorname{Subst}\left (\int \frac{4 a^2+\frac{9 a^2 x}{2}}{x (-a+a x)^2 (a+a x)^{7/2}} \, dx,x,\sec (c+d x)\right )}{4 d}\\ &=-\frac{a^2}{4 d (1-\sec (c+d x))^2 (a+a \sec (c+d x))^{5/2}}-\frac{17 a^2}{16 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{5/2}}+\frac{\operatorname{Subst}\left (\int \frac{8 a^4+\frac{119 a^4 x}{4}}{x (-a+a x) (a+a x)^{7/2}} \, dx,x,\sec (c+d x)\right )}{8 d}\\ &=\frac{87 a^2}{160 d (a+a \sec (c+d x))^{5/2}}-\frac{a^2}{4 d (1-\sec (c+d x))^2 (a+a \sec (c+d x))^{5/2}}-\frac{17 a^2}{16 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{5/2}}-\frac{\operatorname{Subst}\left (\int \frac{-40 a^6-\frac{435 a^6 x}{8}}{x (-a+a x) (a+a x)^{5/2}} \, dx,x,\sec (c+d x)\right )}{40 a^3 d}\\ &=\frac{87 a^2}{160 d (a+a \sec (c+d x))^{5/2}}-\frac{a^2}{4 d (1-\sec (c+d x))^2 (a+a \sec (c+d x))^{5/2}}-\frac{17 a^2}{16 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{5/2}}+\frac{23 a}{192 d (a+a \sec (c+d x))^{3/2}}+\frac{\operatorname{Subst}\left (\int \frac{120 a^8+\frac{345 a^8 x}{16}}{x (-a+a x) (a+a x)^{3/2}} \, dx,x,\sec (c+d x)\right )}{120 a^6 d}\\ &=\frac{87 a^2}{160 d (a+a \sec (c+d x))^{5/2}}-\frac{a^2}{4 d (1-\sec (c+d x))^2 (a+a \sec (c+d x))^{5/2}}-\frac{17 a^2}{16 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{5/2}}+\frac{23 a}{192 d (a+a \sec (c+d x))^{3/2}}-\frac{105}{128 d \sqrt{a+a \sec (c+d x)}}-\frac{\operatorname{Subst}\left (\int \frac{-120 a^{10}+\frac{1575 a^{10} x}{32}}{x (-a+a x) \sqrt{a+a x}} \, dx,x,\sec (c+d x)\right )}{120 a^9 d}\\ &=\frac{87 a^2}{160 d (a+a \sec (c+d x))^{5/2}}-\frac{a^2}{4 d (1-\sec (c+d x))^2 (a+a \sec (c+d x))^{5/2}}-\frac{17 a^2}{16 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{5/2}}+\frac{23 a}{192 d (a+a \sec (c+d x))^{3/2}}-\frac{105}{128 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 )}{d}+\frac{(151 a) \operatorname{Subst}\left (\int \frac{1}{(-a+a x) \sqrt{a+a x}} \, dx,x,\sec (c+d x)\right )}{256 d}\\ &=\frac{87 a^2}{160 d (a+a \sec (c+d x))^{5/2}}-\frac{a^2}{4 d (1-\sec (c+d x))^2 (a+a \sec (c+d x))^{5/2}}-\frac{17 a^2}{16 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{5/2}}+\frac{23 a}{192 d (a+a \sec (c+d x))^{3/2}}-\frac{105}{128 d \sqrt{a+a \sec (c+d x)}}+\frac{151 \operatorname{Subst}\left (\int \frac{1}{-2 a+x^2} \, dx,x,\sqrt{a+a \sec (c+d x)}\right )}{128 d}-\frac{2 \operatorname{Subst}\left (\int \frac{1}{-1+\frac{x^2}{a}} \, dx,x,\sqrt{a+a \sec (c+d x)}\right )}{a d}\\ &=\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a+a \sec (c+d x)}}{\sqrt{a}}\right )}{\sqrt{a} d}-\frac{151 \tanh ^{-1}\left (\frac{\sqrt{a+a \sec (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{128 \sqrt{2} \sqrt{a} d}+\frac{87 a^2}{160 d (a+a \sec (c+d x))^{5/2}}-\frac{a^2}{4 d (1-\sec (c+d x))^2 (a+a \sec (c+d x))^{5/2}}-\frac{17 a^2}{16 d (1-\sec (c+d x)) (a+a \sec (c+d x))^{5/2}}+\frac{23 a}{192 d (a+a \sec (c+d x))^{3/2}}-\frac{105}{128 d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 0.255118, size = 102, normalized size = 0.48 \[ \frac{\cot ^4(c+d x) \left (151 (\sec (c+d x)-1)^2 \text{Hypergeometric2F1}\left (-\frac{5}{2},1,-\frac{3}{2},\frac{1}{2} (\sec (c+d x)+1)\right )-2 \left (32 (\sec (c+d x)-1)^2 \text{Hypergeometric2F1}\left (-\frac{5}{2},1,-\frac{3}{2},\sec (c+d x)+1\right )-85 \sec (c+d x)+105\right )\right )}{160 d \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^5/Sqrt[a + a*Sec[c + d*x]],x]

[Out]

(Cot[c + d*x]^4*(-2*(105 + 32*Hypergeometric2F1[-5/2, 1, -3/2, 1 + Sec[c + d*x]]*(-1 + Sec[c + d*x])^2 - 85*Se
c[c + d*x]) + 151*Hypergeometric2F1[-5/2, 1, -3/2, (1 + Sec[c + d*x])/2]*(-1 + Sec[c + d*x])^2))/(160*d*Sqrt[a
*(1 + Sec[c + d*x])])

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Maple [B]  time = 0.382, size = 746, normalized size = 3.5 \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)^5/(a+a*sec(d*x+c))^(1/2),x)

[Out]

1/3840/d/a*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(cos(d*x+c)+1)^2*(-1+cos(d*x+c))^3*(3840*cos(d*x+c)^5*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))+2265*cos(d*x+c)^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))+3840*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))+2265*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))-7680*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))+5642*cos(d*x+
c)^5-4530*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))-768
0*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))+556*cos(d*x+c)^4-4530*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))+3840*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*cos(d*x+c)
/(cos(d*x+c)+1))^(1/2))-7928*cos(d*x+c)^3+2265*cos(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*co
s(d*x+c)/(cos(d*x+c)+1))^(1/2))+3840*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))-460*cos(d*x+c)^2+2265*(-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))+3150*cos(d*x+c))/sin(d*x+c)^10

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Maxima [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)^5/(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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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)^5/(a+a*sec(d*x+c))^(1/2),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)**5/(a+a*sec(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 10.1823, size = 400, normalized size = 1.87 \begin{align*} \frac{\sqrt{2}{\left (\frac{3840 \, \sqrt{2} \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} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} - \frac{2265 \, \arctan \left (\frac{\sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} + \frac{15 \,{\left (25 \,{\left (-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{\frac{3}{2}} - 23 \, \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} a\right )}}{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 )^{4}} + \frac{8 \,{\left (3 \,{\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^{12} + 25 \,{\left (-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{\frac{3}{2}} a^{13} + 240 \, \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} a^{14}\right )}}{a^{15} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}\right )}}{3840 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

1/3840*sqrt(2)*(3840*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/(sqrt(-a)*sgn(ta
n(1/2*d*x + 1/2*c)^2 - 1)) - 2265*arctan(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/(sqrt(-a)*sgn(tan(1/2*d
*x + 1/2*c)^2 - 1)) + 15*(25*(-a*tan(1/2*d*x + 1/2*c)^2 + a)^(3/2) - 23*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*a)
/(a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)*tan(1/2*d*x + 1/2*c)^4) + 8*(3*(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^12 + 25*(-a*tan(1/2*d*x + 1/2*c)^2 + a)^(3/2)*a^13 + 240*sqrt(-a*tan(1/2*d*x + 1
/2*c)^2 + a)*a^14)/(a^15*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))/d