3.152 \(\int \cot ^5(c+d x) (a+a \sec (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=171 \[ \frac{7 a^2}{32 d \sqrt{a \sec (c+d x)+a}}-\frac{13 a^2}{16 d (1-\sec (c+d x)) \sqrt{a \sec (c+d x)+a}}-\frac{a^2}{4 d (1-\sec (c+d x))^2 \sqrt{a \sec (c+d x)+a}}+\frac{2 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{a}}\right )}{d}-\frac{71 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{32 \sqrt{2} d} \]

[Out]

(2*a^(3/2)*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/d - (71*a^(3/2)*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/(Sqrt[2
]*Sqrt[a])])/(32*Sqrt[2]*d) + (7*a^2)/(32*d*Sqrt[a + a*Sec[c + d*x]]) - a^2/(4*d*(1 - Sec[c + d*x])^2*Sqrt[a +
a*Sec[c + d*x]]) - (13*a^2)/(16*d*(1 - Sec[c + d*x])*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.143066, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 9, 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} \[ \frac{7 a^2}{32 d \sqrt{a \sec (c+d x)+a}}-\frac{13 a^2}{16 d (1-\sec (c+d x)) \sqrt{a \sec (c+d x)+a}}-\frac{a^2}{4 d (1-\sec (c+d x))^2 \sqrt{a \sec (c+d x)+a}}+\frac{2 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{a}}\right )}{d}-\frac{71 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{32 \sqrt{2} d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5*(a + a*Sec[c + d*x])^(3/2),x]

[Out]

(2*a^(3/2)*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/d - (71*a^(3/2)*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/(Sqrt[2
]*Sqrt[a])])/(32*Sqrt[2]*d) + (7*a^2)/(32*d*Sqrt[a + a*Sec[c + d*x]]) - a^2/(4*d*(1 - Sec[c + d*x])^2*Sqrt[a +
a*Sec[c + d*x]]) - (13*a^2)/(16*d*(1 - Sec[c + d*x])*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 \cot ^5(c+d x) (a+a \sec (c+d x))^{3/2} \, dx &=\frac{a^6 \operatorname{Subst}\left (\int \frac{1}{x (-a+a x)^3 (a+a x)^{3/2}} \, dx,x,\sec (c+d x)\right )}{d}\\ &=-\frac{a^2}{4 d (1-\sec (c+d x))^2 \sqrt{a+a \sec (c+d x)}}-\frac{a^3 \operatorname{Subst}\left (\int \frac{4 a^2+\frac{5 a^2 x}{2}}{x (-a+a x)^2 (a+a x)^{3/2}} \, dx,x,\sec (c+d x)\right )}{4 d}\\ &=-\frac{a^2}{4 d (1-\sec (c+d x))^2 \sqrt{a+a \sec (c+d x)}}-\frac{13 a^2}{16 d (1-\sec (c+d x)) \sqrt{a+a \sec (c+d x)}}+\frac{\operatorname{Subst}\left (\int \frac{8 a^4+\frac{39 a^4 x}{4}}{x (-a+a x) (a+a x)^{3/2}} \, dx,x,\sec (c+d x)\right )}{8 d}\\ &=\frac{7 a^2}{32 d \sqrt{a+a \sec (c+d x)}}-\frac{a^2}{4 d (1-\sec (c+d x))^2 \sqrt{a+a \sec (c+d x)}}-\frac{13 a^2}{16 d (1-\sec (c+d x)) \sqrt{a+a \sec (c+d x)}}-\frac{\operatorname{Subst}\left (\int \frac{-8 a^6-\frac{7 a^6 x}{8}}{x (-a+a x) \sqrt{a+a x}} \, dx,x,\sec (c+d x)\right )}{8 a^3 d}\\ &=\frac{7 a^2}{32 d \sqrt{a+a \sec (c+d x)}}-\frac{a^2}{4 d (1-\sec (c+d x))^2 \sqrt{a+a \sec (c+d x)}}-\frac{13 a^2}{16 d (1-\sec (c+d x)) \sqrt{a+a \sec (c+d x)}}-\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+a x}} \, dx,x,\sec (c+d x)\right )}{d}+\frac{\left (71 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{(-a+a x) \sqrt{a+a x}} \, dx,x,\sec (c+d x)\right )}{64 d}\\ &=\frac{7 a^2}{32 d \sqrt{a+a \sec (c+d x)}}-\frac{a^2}{4 d (1-\sec (c+d x))^2 \sqrt{a+a \sec (c+d x)}}-\frac{13 a^2}{16 d (1-\sec (c+d x)) \sqrt{a+a \sec (c+d x)}}-\frac{(2 a) \operatorname{Subst}\left (\int \frac{1}{-1+\frac{x^2}{a}} \, dx,x,\sqrt{a+a \sec (c+d x)}\right )}{d}+\frac{\left (71 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{-2 a+x^2} \, dx,x,\sqrt{a+a \sec (c+d x)}\right )}{32 d}\\ &=\frac{2 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+a \sec (c+d x)}}{\sqrt{a}}\right )}{d}-\frac{71 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+a \sec (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{32 \sqrt{2} d}+\frac{7 a^2}{32 d \sqrt{a+a \sec (c+d x)}}-\frac{a^2}{4 d (1-\sec (c+d x))^2 \sqrt{a+a \sec (c+d x)}}-\frac{13 a^2}{16 d (1-\sec (c+d x)) \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 0.272561, size = 104, normalized size = 0.61 \[ \frac{a^2 \left (71 (\sec (c+d x)-1)^2 \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\frac{1}{2} (\sec (c+d x)+1)\right )-64 (\sec (c+d x)-1)^2 \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\sec (c+d x)+1\right )+26 \sec (c+d x)-34\right )}{32 d (\sec (c+d x)-1)^2 \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^5*(a + a*Sec[c + d*x])^(3/2),x]

[Out]

(a^2*(-34 + 71*Hypergeometric2F1[-1/2, 1, 1/2, (1 + Sec[c + d*x])/2]*(-1 + Sec[c + d*x])^2 - 64*Hypergeometric
2F1[-1/2, 1, 1/2, 1 + Sec[c + d*x]]*(-1 + Sec[c + d*x])^2 + 26*Sec[c + d*x]))/(32*d*(-1 + Sec[c + d*x])^2*Sqrt
[a*(1 + Sec[c + d*x])])

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Maple [B]  time = 0.283, size = 502, 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)^5*(a+a*sec(d*x+c))^(3/2),x)

[Out]

1/64/d*a*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))*(cos(d*x+c)+1)^2*(64*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))-64*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))+71*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))-64*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))-71*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))+64*2^(1/2)*(-2*cos(d*x+c)/(co
s(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+54*cos(d*x+c)^3-71*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))-24*cos(d*x+c)^2+71*(-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))-14*cos(d*x+c))/sin(d*x+c)^6

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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))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 2.58474, size = 1555, normalized size = 9.09 \begin{align*} \text{result too large to display} \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))^(3/2),x, algorithm="fricas")

[Out]

[1/128*(64*(a*cos(d*x + c)^3 - a*cos(d*x + c)^2 - a*cos(d*x + c) + a)*sqrt(a)*log(-8*a*cos(d*x + c)^2 - 4*(2*c
os(d*x + c)^2 + cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)) - 8*a*cos(d*x + c) - a) + 71*(sq
rt(2)*a*cos(d*x + c)^3 - sqrt(2)*a*cos(d*x + c)^2 - sqrt(2)*a*cos(d*x + c) + sqrt(2)*a)*sqrt(a)*log(-(2*sqrt(2
)*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c) - 3*a*cos(d*x + c) - a)/(cos(d*x + c) - 1)) - 4
*(27*a*cos(d*x + c)^3 - 12*a*cos(d*x + c)^2 - 7*a*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/(d*co
s(d*x + c)^3 - d*cos(d*x + c)^2 - d*cos(d*x + c) + d), 1/64*(71*(sqrt(2)*a*cos(d*x + c)^3 - sqrt(2)*a*cos(d*x
+ c)^2 - sqrt(2)*a*cos(d*x + c) + sqrt(2)*a)*sqrt(-a)*arctan(sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*
x + c))*cos(d*x + c)/(a*cos(d*x + c) + a)) - 64*(a*cos(d*x + c)^3 - a*cos(d*x + c)^2 - a*cos(d*x + c) + a)*sqr
t(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(2*a*cos(d*x + c) + a)) - 2*(27*a
*cos(d*x + c)^3 - 12*a*cos(d*x + c)^2 - 7*a*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/(d*cos(d*x
+ c)^3 - d*cos(d*x + c)^2 - d*cos(d*x + c) + d)]

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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))**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 5.04413, size = 230, normalized size = 1.35 \begin{align*} -\frac{\sqrt{2}{\left (\frac{64 \, \sqrt{2} a \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}} - \frac{71 \, a \arctan \left (\frac{\sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} + 8 \, \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} + \frac{17 \,{\left (-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{\frac{3}{2}} a - 15 \, \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} a^{2}}{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4}}\right )} a \mathrm{sgn}\left (\cos \left (d x + c\right )\right )}{64 \, 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))^(3/2),x, algorithm="giac")

[Out]

-1/64*sqrt(2)*(64*sqrt(2)*a*arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/sqrt(-a) - 71*a*a
rctan(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/sqrt(-a) + 8*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a) + (17*(-a
*tan(1/2*d*x + 1/2*c)^2 + a)^(3/2)*a - 15*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*a^2)/(a^2*tan(1/2*d*x + 1/2*c)^4
))*a*sgn(cos(d*x + c))/d