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

Optimal. Leaf size=106 \[ \frac{a^2 \sqrt{a \sec (c+d x)+a}}{d (1-\sec (c+d x))}-\frac{2 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{a}}\right )}{d}+\frac{3 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{\sqrt{2} d} \]

[Out]

(-2*a^(5/2)*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/d + (3*a^(5/2)*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/(Sqrt[2
]*Sqrt[a])])/(Sqrt[2]*d) + (a^2*Sqrt[a + a*Sec[c + d*x]])/(d*(1 - Sec[c + d*x]))

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Rubi [A]  time = 0.102885, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3880, 99, 156, 63, 207} \[ \frac{a^2 \sqrt{a \sec (c+d x)+a}}{d (1-\sec (c+d x))}-\frac{2 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{a}}\right )}{d}+\frac{3 a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a \sec (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{\sqrt{2} d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^3*(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 + (3*a^(5/2)*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/(Sqrt[2
]*Sqrt[a])])/(Sqrt[2]*d) + (a^2*Sqrt[a + a*Sec[c + d*x]])/(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 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[1/((m + 1)*(b*e - a*f)), Int[(a +
b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + p + 2)*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

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

Mathematica [A]  time = 0.265232, size = 115, normalized size = 1.08 \[ -\frac{(a (\sec (c+d x)+1))^{5/2} \left (2 \sqrt{\sec (c+d x)+1}+4 (\sec (c+d x)-1) \tanh ^{-1}\left (\sqrt{\sec (c+d x)+1}\right )-3 \sqrt{2} (\sec (c+d x)-1) \tanh ^{-1}\left (\frac{\sqrt{\sec (c+d x)+1}}{\sqrt{2}}\right )\right )}{2 d (\sec (c+d x)-1) (\sec (c+d x)+1)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*(1 + Sec[c + d*x]))^(5/2)*(4*ArcTanh[Sqrt[1 + Sec[c + d*x]]]*(-1 + Sec[c + d*x]) - 3*Sqrt[2]*ArcTanh[Sqrt
[1 + Sec[c + d*x]]/Sqrt[2]]*(-1 + Sec[c + d*x]) + 2*Sqrt[1 + Sec[c + d*x]]))/(2*d*(-1 + Sec[c + d*x])*(1 + Sec
[c + d*x])^(5/2))

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Maple [B]  time = 0.204, size = 248, normalized size = 2.3 \begin{align*}{\frac{{a}^{2}}{2\,d \left ( -1+\cos \left ( dx+c \right ) \right ) }\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( 2\,\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 ) -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 ) +3\,\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 ) -3\,\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 ) +2\,\cos \left ( dx+c \right ) \right ) } \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))^(5/2),x)

[Out]

1/2/d*a^2*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(2*2^(1/2)*cos(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arcta
n(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-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))+3*cos(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*c
os(d*x+c)/(cos(d*x+c)+1))^(1/2))-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))+2*cos(d*x+c))/(-1+cos(d*x+c))

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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)^3*(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 1.77089, size = 1029, normalized size = 9.71 \begin{align*} \left [\frac{4 \, a^{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) + 4 \,{\left (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 ) + 3 \,{\left (\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 (d \cos \left (d x + c\right ) - d\right )}}, \frac{2 \, a^{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) - 3 \,{\left (\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 ) + 4 \,{\left (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 (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)^3*(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[1/4*(4*a^2*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c) + 4*(a^2*cos(d*x + c) - a^2)*sqrt(a)*log(-2*a
*cos(d*x + c) + 2*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c) - a) + 3*(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*c
os(d*x + c) + a)/(cos(d*x + c) - 1)))/(d*cos(d*x + c) - d), 1/2*(2*a^2*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))
*cos(d*x + c) - 3*(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)) + 4*(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)))/(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)**3*(a+a*sec(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 5.70923, size = 176, normalized size = 1.66 \begin{align*} \frac{\sqrt{2} a^{4}{\left (\frac{2 \, \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} a} - \frac{3 \, \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} - \frac{\sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}\right )} \mathrm{sgn}\left (\cos \left (d x + c\right )\right )}{2 \, 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))^(5/2),x, algorithm="giac")

[Out]

1/2*sqrt(2)*a^4*(2*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/(sqrt(-a)*a) - 3*a
rctan(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/(sqrt(-a)*a) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/(a^2*ta
n(1/2*d*x + 1/2*c)^2))*sgn(cos(d*x + c))/d