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

Optimal. Leaf size=147 \[ \frac {2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{d}-\frac {43 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{16 \sqrt {2} d}-\frac {11 a^2 \sqrt {a \sec (c+d x)+a}}{16 d (1-\sec (c+d x))}-\frac {a^2 \sqrt {a \sec (c+d x)+a}}{4 d (1-\sec (c+d x))^2} \]

[Out]

2*a^(5/2)*arctanh((a+a*sec(d*x+c))^(1/2)/a^(1/2))/d-43/32*a^(5/2)*arctanh(1/2*(a+a*sec(d*x+c))^(1/2)*2^(1/2)/a
^(1/2))*2^(1/2)/d-1/4*a^2*(a+a*sec(d*x+c))^(1/2)/d/(1-sec(d*x+c))^2-11/16*a^2*(a+a*sec(d*x+c))^(1/2)/d/(1-sec(
d*x+c))

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Rubi [A]  time = 0.13, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3880, 103, 151, 156, 63, 207} \[ -\frac {11 a^2 \sqrt {a \sec (c+d x)+a}}{16 d (1-\sec (c+d x))}-\frac {a^2 \sqrt {a \sec (c+d x)+a}}{4 d (1-\sec (c+d x))^2}+\frac {2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{d}-\frac {43 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{16 \sqrt {2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^(5/2)*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/d - (43*a^(5/2)*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/(Sqrt[2
]*Sqrt[a])])/(16*Sqrt[2]*d) - (a^2*Sqrt[a + a*Sec[c + d*x]])/(4*d*(1 - Sec[c + d*x])^2) - (11*a^2*Sqrt[a + a*S
ec[c + d*x]])/(16*d*(1 - Sec[c + d*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 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 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 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])

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]

Rubi steps

\begin {align*} \int \cot ^5(c+d x) (a+a \sec (c+d x))^{5/2} \, dx &=\frac {a^6 \operatorname {Subst}\left (\int \frac {1}{x (-a+a x)^3 \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{d}\\ &=-\frac {a^2 \sqrt {a+a \sec (c+d x)}}{4 d (1-\sec (c+d x))^2}-\frac {a^3 \operatorname {Subst}\left (\int \frac {4 a^2+\frac {3 a^2 x}{2}}{x (-a+a x)^2 \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{4 d}\\ &=-\frac {a^2 \sqrt {a+a \sec (c+d x)}}{4 d (1-\sec (c+d x))^2}-\frac {11 a^2 \sqrt {a+a \sec (c+d x)}}{16 d (1-\sec (c+d x))}+\frac {\operatorname {Subst}\left (\int \frac {8 a^4+\frac {11 a^4 x}{4}}{x (-a+a x) \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{8 d}\\ &=-\frac {a^2 \sqrt {a+a \sec (c+d x)}}{4 d (1-\sec (c+d x))^2}-\frac {11 a^2 \sqrt {a+a \sec (c+d x)}}{16 d (1-\sec (c+d x))}-\frac {a^3 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{d}+\frac {\left (43 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{(-a+a x) \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{32 d}\\ &=-\frac {a^2 \sqrt {a+a \sec (c+d x)}}{4 d (1-\sec (c+d x))^2}-\frac {11 a^2 \sqrt {a+a \sec (c+d x)}}{16 d (1-\sec (c+d x))}-\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{-1+\frac {x^2}{a}} \, dx,x,\sqrt {a+a \sec (c+d x)}\right )}{d}+\frac {\left (43 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{-2 a+x^2} \, dx,x,\sqrt {a+a \sec (c+d x)}\right )}{16 d}\\ &=\frac {2 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}-\frac {43 a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{16 \sqrt {2} d}-\frac {a^2 \sqrt {a+a \sec (c+d x)}}{4 d (1-\sec (c+d x))^2}-\frac {11 a^2 \sqrt {a+a \sec (c+d x)}}{16 d (1-\sec (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 1.39, size = 138, normalized size = 0.94 \[ \frac {(a (\sec (c+d x)+1))^{5/2} \left (\sqrt {\sec (c+d x)+1} (11 \sec (c+d x)-15)+32 (\sec (c+d x)-1)^2 \tanh ^{-1}\left (\sqrt {\sec (c+d x)+1}\right )-86 \sqrt {2} \sin ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \tanh ^{-1}\left (\frac {\sqrt {\sec (c+d x)+1}}{\sqrt {2}}\right )\right )}{16 d (\sec (c+d x)-1)^2 (\sec (c+d x)+1)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(1 + Sec[c + d*x]))^(5/2)*(32*ArcTanh[Sqrt[1 + Sec[c + d*x]]]*(-1 + Sec[c + d*x])^2 + Sqrt[1 + Sec[c + d*x
]]*(-15 + 11*Sec[c + d*x]) - 86*Sqrt[2]*ArcTanh[Sqrt[1 + Sec[c + d*x]]/Sqrt[2]]*Sec[c + d*x]^2*Sin[(c + d*x)/2
]^4))/(16*d*(-1 + Sec[c + d*x])^2*(1 + Sec[c + d*x])^(5/2))

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fricas [B]  time = 0.59, size = 503, normalized size = 3.42 \[ \left [\frac {64 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {a} \log \left (-2 \, a \cos \left (d x + c\right ) - 2 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) - a\right ) + 43 \, {\left (\sqrt {2} a^{2} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} a^{2} \cos \left (d x + c\right ) + \sqrt {2} a^{2}\right )} \sqrt {a} \log \left (-\frac {2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) - 3 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) - 1}\right ) - 4 \, {\left (15 \, a^{2} \cos \left (d x + c\right )^{2} - 11 \, a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{64 \, {\left (d \cos \left (d x + c\right )^{2} - 2 \, d \cos \left (d x + c\right ) + d\right )}}, \frac {43 \, {\left (\sqrt {2} a^{2} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} a^{2} \cos \left (d x + c\right ) + \sqrt {2} a^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{a \cos \left (d x + c\right ) + a}\right ) - 64 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{a \cos \left (d x + c\right ) + a}\right ) - 2 \, {\left (15 \, a^{2} \cos \left (d x + c\right )^{2} - 11 \, a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{32 \, {\left (d \cos \left (d x + c\right )^{2} - 2 \, d \cos \left (d x + c\right ) + d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/64*(64*(a^2*cos(d*x + c)^2 - 2*a^2*cos(d*x + c) + a^2)*sqrt(a)*log(-2*a*cos(d*x + c) - 2*sqrt(a)*sqrt((a*co
s(d*x + c) + a)/cos(d*x + c))*cos(d*x + c) - a) + 43*(sqrt(2)*a^2*cos(d*x + c)^2 - 2*sqrt(2)*a^2*cos(d*x + c)
+ sqrt(2)*a^2)*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*(15*a^2*cos(d*x + c)^2 - 11*a^2*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/
cos(d*x + c)))/(d*cos(d*x + c)^2 - 2*d*cos(d*x + c) + d), 1/32*(43*(sqrt(2)*a^2*cos(d*x + c)^2 - 2*sqrt(2)*a^2
*cos(d*x + c) + sqrt(2)*a^2)*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^2*cos(d*x + c)^2 - 2*a^2*cos(d*x + c) + a^2)*sqrt(-a)*arctan(sqrt(-a)*sqrt(
(a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(a*cos(d*x + c) + a)) - 2*(15*a^2*cos(d*x + c)^2 - 11*a^2*cos(
d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/(d*cos(d*x + c)^2 - 2*d*cos(d*x + c) + d)]

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giac [A]  time = 1.76, size = 177, normalized size = 1.20 \[ \frac {\frac {43 \, \sqrt {2} a^{3} \arctan \left (\frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{\sqrt {-a}}\right ) \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{\sqrt {-a}} - \frac {64 \, a^{3} \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right ) \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{\sqrt {-a}} - \frac {\sqrt {2} {\left (13 \, {\left (-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {3}{2}} a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 11 \, \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(43*sqrt(2)*a^3*arctan(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))*sgn(cos(d*x + c))/sqrt(-a) - 64*a^3*
arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))*sgn(cos(d*x + c))/sqrt(-a) - sqrt(2)*(13*(-a*
tan(1/2*d*x + 1/2*c)^2 + a)^(3/2)*a^3*sgn(cos(d*x + c)) - 11*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*a^4*sgn(cos(d
*x + c)))/(a^2*tan(1/2*d*x + 1/2*c)^4))/d

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maple [B]  time = 1.33, size = 376, normalized size = 2.56 \[ -\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (1+\cos \left (d x +c \right )\right )^{2} \left (32 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right )+43 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \left (\cos ^{2}\left (d x +c \right )\right )-64 \cos \left (d x +c \right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right )-86 \cos \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+32 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \sqrt {2}+30 \left (\cos ^{2}\left (d x +c \right )\right )+43 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )-22 \cos \left (d x +c \right )\right ) a^{2}}{32 d \sin \left (d x +c \right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(1+cos(d*x+c))^2*(32*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1
/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*2^(1/2)*cos(d*x+c)^2+43*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*
arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)^2-64*cos(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)
))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))-86*cos(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))
^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+32*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-2*c
os(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*2^(1/2)+30*cos(d*x+c)^2+43*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arcta
n(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))-22*cos(d*x+c))/sin(d*x+c)^4*a^2

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {cot}\left (c+d\,x\right )}^5\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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