3.768 \(\int \frac{\cot ^{\frac{5}{2}}(c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=181 \[ -\frac{5 \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 a d}+\frac{\cot ^{\frac{3}{2}}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{7 i \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{3 a d}-\frac{\left (\frac{1}{2}-\frac{i}{2}\right ) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d} \]

[Out]

((-1/2 + I/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]*Sqrt
[Tan[c + d*x]])/(Sqrt[a]*d) + Cot[c + d*x]^(3/2)/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (((7*I)/3)*Sqrt[Cot[c + d*x]
]*Sqrt[a + I*a*Tan[c + d*x]])/(a*d) - (5*Cot[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/(3*a*d)

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Rubi [A]  time = 0.474986, antiderivative size = 181, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {4241, 3559, 3598, 12, 3544, 205} \[ -\frac{5 \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 a d}+\frac{\cot ^{\frac{3}{2}}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{7 i \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{3 a d}-\frac{\left (\frac{1}{2}-\frac{i}{2}\right ) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/2 + I/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]*Sqrt
[Tan[c + d*x]])/(Sqrt[a]*d) + Cot[c + d*x]^(3/2)/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (((7*I)/3)*Sqrt[Cot[c + d*x]
]*Sqrt[a + I*a*Tan[c + d*x]])/(a*d) - (5*Cot[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/(3*a*d)

Rule 4241

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ
[u, x]

Rule 3559

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

Rule 3598

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*d - B*c)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(f
*(n + 1)*(c^2 + d^2)), x] - Dist[1/(a*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n
+ 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c*m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[n, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3544

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*a*b)/f, Subst[Int[1/(a*c - b*d - 2*a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\cot ^{\frac{5}{2}}(c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx &=\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{1}{\tan ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}} \, dx\\ &=\frac{\cot ^{\frac{3}{2}}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{5 a}{2}-2 i a \tan (c+d x)\right )}{\tan ^{\frac{5}{2}}(c+d x)} \, dx}{a^2}\\ &=\frac{\cot ^{\frac{3}{2}}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}-\frac{5 \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 a d}+\frac{\left (2 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\sqrt{a+i a \tan (c+d x)} \left (-\frac{7 i a^2}{4}-\frac{5}{2} a^2 \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{3 a^3}\\ &=\frac{\cot ^{\frac{3}{2}}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{7 i \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{3 a d}-\frac{5 \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 a d}+\frac{\left (4 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int -\frac{3 a^3 \sqrt{a+i a \tan (c+d x)}}{8 \sqrt{\tan (c+d x)}} \, dx}{3 a^4}\\ &=\frac{\cot ^{\frac{3}{2}}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{7 i \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{3 a d}-\frac{5 \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 a d}-\frac{\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{2 a}\\ &=\frac{\cot ^{\frac{3}{2}}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{7 i \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{3 a d}-\frac{5 \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 a d}+\frac{\left (i a \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{-i a-2 a^2 x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}\\ &=-\frac{\left (\frac{1}{2}-\frac{i}{2}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right ) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}}{\sqrt{a} d}+\frac{\cot ^{\frac{3}{2}}(c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{7 i \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{3 a d}-\frac{5 \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 a d}\\ \end{align*}

Mathematica [A]  time = 1.4526, size = 159, normalized size = 0.88 \[ \frac{i \sqrt{\cot (c+d x)} \left (-18 e^{2 i (c+d x)}+7 e^{4 i (c+d x)}+3 e^{i (c+d x)} \left (-1+e^{2 i (c+d x)}\right )^{3/2} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )+3\right )}{3 \sqrt{2} d \left (-1+e^{4 i (c+d x)}\right ) \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/3)*(3 - 18*E^((2*I)*(c + d*x)) + 7*E^((4*I)*(c + d*x)) + 3*E^(I*(c + d*x))*(-1 + E^((2*I)*(c + d*x)))^(3/2
)*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])*Sqrt[Cot[c + d*x]])/(Sqrt[2]*d*Sqrt[(a*E^((2*I)*(c
+ d*x)))/(1 + E^((2*I)*(c + d*x)))]*(-1 + E^((4*I)*(c + d*x))))

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Maple [B]  time = 0.441, size = 374, normalized size = 2.1 \begin{align*}{\frac{ \left ( -{\frac{1}{6}}-{\frac{i}{6}} \right ) \sin \left ( dx+c \right ) }{ad \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}} \left ( 3\,i\sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \right ) }}}\sqrt{2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \arctan \left ( \left ({\frac{1}{2}}+{\frac{i}{2}} \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \right ) }}}\sqrt{2} \right ) +3\,i\sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \right ) }}}\sqrt{2}\sin \left ( dx+c \right ) \arctan \left ( \left ({\frac{1}{2}}+{\frac{i}{2}} \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \right ) }}}\sqrt{2} \right ) +3\,\sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \right ) }}}\sqrt{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}\arctan \left ( \left ( 1/2+i/2 \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \right ) }}}\sqrt{2} \right ) +5\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}-2\,i\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) -3\,\sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \right ) }}}\sqrt{2}\arctan \left ( \left ( 1/2+i/2 \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \right ) }}}\sqrt{2} \right ) -5\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +7-7\,i \right ) \left ({\frac{\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) ^{{\frac{5}{2}}}\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^(1/2),x)

[Out]

(-1/6-1/6*I)/d/a*(3*I*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*cos(d*x+c)*sin(d*x+c)*arctan((1/2+1/2*I)*((cos
(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))+3*I*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)*arctan((1/2+1/2
*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))+3*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*cos(d*x+c)^2*arctan
((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))+5*I*cos(d*x+c)^2-2*I*cos(d*x+c)*sin(d*x+c)-3*((cos(d*x
+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))-5*cos(d*x+c)^2-
2*cos(d*x+c)*sin(d*x+c)+7-7*I)*sin(d*x+c)*(cos(d*x+c)/sin(d*x+c))^(5/2)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c
))^(1/2)/(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c)^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot \left (d x + c\right )^{\frac{5}{2}}}{\sqrt{i \, a \tan \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(cot(d*x + c)^(5/2)/sqrt(I*a*tan(d*x + c) + a), x)

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Fricas [B]  time = 1.64936, size = 1116, normalized size = 6.17 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}{\left (14 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 36 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 6 i\right )} e^{\left (i \, d x + i \, c\right )} - 3 \,{\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt{-\frac{2 i}{a d^{2}}} \log \left (\frac{1}{4} \,{\left (i \, a d \sqrt{-\frac{2 i}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 3 \,{\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt{-\frac{2 i}{a d^{2}}} \log \left (\frac{1}{4} \,{\left (-i \, a d \sqrt{-\frac{2 i}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right )}{12 \,{\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/12*(sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*(1
4*I*e^(4*I*d*x + 4*I*c) - 36*I*e^(2*I*d*x + 2*I*c) + 6*I)*e^(I*d*x + I*c) - 3*(a*d*e^(4*I*d*x + 4*I*c) - a*d*e
^(2*I*d*x + 2*I*c))*sqrt(-2*I/(a*d^2))*log(1/4*(I*a*d*sqrt(-2*I/(a*d^2))*e^(2*I*d*x + 2*I*c) + sqrt(2)*sqrt(a/
(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*(e^(2*I*d*x + 2*I*c) -
1)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + 3*(a*d*e^(4*I*d*x + 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))*sqrt(-2*I/(a*d^2
))*log(1/4*(-I*a*d*sqrt(-2*I/(a*d^2))*e^(2*I*d*x + 2*I*c) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*
e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*(e^(2*I*d*x + 2*I*c) - 1)*e^(I*d*x + I*c))*e^(-I*d*x - I*c
)))/(a*d*e^(4*I*d*x + 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))

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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)**(5/2)/(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot \left (d x + c\right )^{\frac{5}{2}}}{\sqrt{i \, a \tan \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(cot(d*x + c)^(5/2)/sqrt(I*a*tan(d*x + c) + a), x)