3.769 \(\int \frac{\cot ^{\frac{3}{2}}(c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=140 \[ -\frac{3 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{a d}+\frac{\sqrt{\cot (c+d x)}}{d \sqrt{a+i a \tan (c+d x)}}+\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d} \]

[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) + Sqrt[Cot[c + d*x]]/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (3*Sqrt[Cot[c + d*x]]*Sqrt[a
+ I*a*Tan[c + d*x]])/(a*d)

________________________________________________________________________________________

Rubi [A]  time = 0.329196, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ -\frac{3 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{a d}+\frac{\sqrt{\cot (c+d x)}}{d \sqrt{a+i a \tan (c+d x)}}+\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^(3/2)/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) + Sqrt[Cot[c + d*x]]/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (3*Sqrt[Cot[c + d*x]]*Sqrt[a
+ I*a*Tan[c + d*x]])/(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{3}{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{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}} \, dx\\ &=\frac{\sqrt{\cot (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{\left (\frac{3 a}{2}-i a \tan (c+d x)\right ) \sqrt{a+i a \tan (c+d x)}}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{a^2}\\ &=\frac{\sqrt{\cot (c+d x)}}{d \sqrt{a+i a \tan (c+d x)}}-\frac{3 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{a d}+\frac{\left (2 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{i a^2 \sqrt{a+i a \tan (c+d x)}}{4 \sqrt{\tan (c+d x)}} \, dx}{a^3}\\ &=\frac{\sqrt{\cot (c+d x)}}{d \sqrt{a+i a \tan (c+d x)}}-\frac{3 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{a d}+\frac{\left (i \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{\sqrt{\cot (c+d x)}}{d \sqrt{a+i a \tan (c+d x)}}-\frac{3 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{a d}+\frac{\left (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{\sqrt{\cot (c+d x)}}{d \sqrt{a+i a \tan (c+d x)}}-\frac{3 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{a d}\\ \end{align*}

Mathematica [A]  time = 1.18751, size = 139, normalized size = 0.99 \[ \frac{e^{-2 i (c+d x)} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{\cot (c+d x)} \left (-5 e^{2 i (c+d x)}+e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )+1\right )}{\sqrt{2} a d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^(3/2)/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.415, size = 287, normalized size = 2.1 \begin{align*}{\frac{ \left ( -{\frac{1}{2}}-{\frac{i}{2}} \right ) \sin \left ( dx+c \right ) }{ad \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) \cos \left ( dx+c \right ) } \left ( i\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 ) \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 ) \sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \right ) }}}\sqrt{2}\cos \left ( dx+c \right ) +\sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \right ) }}}\sqrt{2}\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 ) -2\,i\cos \left ( dx+c \right ) +3\,i\sin \left ( dx+c \right ) +2\,\cos \left ( dx+c \right ) +3\,\sin \left ( dx+c \right ) \right ) \left ({\frac{\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) ^{{\frac{3}{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)^(3/2)/(a+I*a*tan(d*x+c))^(1/2),x)

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot \left (d x + c\right )^{\frac{3}{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)^(3/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(cot(d*x + c)^(3/2)/sqrt(I*a*tan(d*x + c) + a), x)

________________________________________________________________________________________

Fricas [B]  time = 1.58559, size = 949, normalized size = 6.78 \begin{align*} \frac{{\left (a d \sqrt{\frac{2 i}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac{1}{4} \,{\left (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 ) - a d \sqrt{\frac{2 i}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-\frac{1}{4} \,{\left (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 ) - 2 \, \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 (5 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/4*(a*d*sqrt(2*I/(a*d^2))*e^(2*I*d*x + 2*I*c)*log(1/4*(a*d*sqrt(2*I/(a*d^2))*e^(2*I*d*x + 2*I*c) + sqrt(2)*sq
rt(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*sqrt(2*I/(a*d^2))*e^(2*I*d*x + 2*I*c)*log(-1/4*(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)) - 2*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))*(5*e^(2*I*d*x + 2*I*c) - 1)*e^
(I*d*x + I*c))*e^(-2*I*d*x - 2*I*c)/(a*d)

________________________________________________________________________________________

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)**(3/2)/(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot \left (d x + c\right )^{\frac{3}{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)^(3/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(cot(d*x + c)^(3/2)/sqrt(I*a*tan(d*x + c) + a), x)