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

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

[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]))

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Rubi [A]  time = 0.126855, antiderivative size = 147, normalized size of antiderivative = 1., 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 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 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))^{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*}

Mathematica [A]  time = 1.29451, 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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Maple [B]  time = 0.262, size = 376, normalized size = 2.6 \begin{align*} -{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2}}{32\,d \left ( \sin \left ( dx+c \right ) \right ) ^{4}}\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( 32\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\arctan \left ( 1/2\,\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \right ) -64\,\sqrt{2}\cos \left ( dx+c \right ) \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\arctan \left ( 1/2\,\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \right ) +43\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}} \right ) +32\,\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\arctan \left ( 1/2\,\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \right ) -86\,\cos \left ( dx+c \right ) \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}} \right ) +43\,\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\arctan \left ({\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}} \right ) +30\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-22\,\cos \left ( dx+c \right ) \right ) } \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))^(5/2),x)

[Out]

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

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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))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 1.79357, size = 1319, normalized size = 8.97 \begin{align*} \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 ] \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))^(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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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))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 6.68825, size = 198, normalized size = 1.35 \begin{align*} -\frac{\sqrt{2} a^{3}{\left (\frac{32 \, \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}} - \frac{43 \, \arctan \left (\frac{\sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} + \frac{13 \,{\left (-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{\frac{3}{2}} - 11 \, \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} a}{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4}}\right )} \mathrm{sgn}\left (\cos \left (d x + c\right )\right )}{32 \, 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))^(5/2),x, algorithm="giac")

[Out]

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