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

Optimal. Leaf size=182 \[ -\frac{25 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{6 a^2 d}+\frac{\left (\frac{1}{4}+\frac{i}{4}\right ) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac{11 \sqrt{\cot (c+d x)}}{6 a d \sqrt{a+i a \tan (c+d x)}}+\frac{\sqrt{\cot (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}} \]

[Out]

((1/4 + I/4)*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]])/(a^(3/2)*d) + Sqrt[Cot[c + d*x]]/(3*d*(a + I*a*Tan[c + d*x])^(3/2)) + (11*Sqrt[Cot[c + d*x]])/(
6*a*d*Sqrt[a + I*a*Tan[c + d*x]]) - (25*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(6*a^2*d)

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Rubi [A]  time = 0.500433, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4241, 3559, 3596, 3598, 12, 3544, 205} \[ -\frac{25 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{6 a^2 d}+\frac{\left (\frac{1}{4}+\frac{i}{4}\right ) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac{11 \sqrt{\cot (c+d x)}}{6 a d \sqrt{a+i a \tan (c+d x)}}+\frac{\sqrt{\cot (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1/4 + I/4)*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]])/(a^(3/2)*d) + Sqrt[Cot[c + d*x]]/(3*d*(a + I*a*Tan[c + d*x])^(3/2)) + (11*Sqrt[Cot[c + d*x]])/(
6*a*d*Sqrt[a + I*a*Tan[c + d*x]]) - (25*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(6*a^2*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 3596

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*A + b*B)*(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*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x
] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] &&  !GtQ[n,
0]

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)}{(a+i a \tan (c+d x))^{3/2}} \, dx &=\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{1}{\tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx\\ &=\frac{\sqrt{\cot (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\frac{7 a}{2}-2 i a \tan (c+d x)}{\tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}} \, dx}{3 a^2}\\ &=\frac{\sqrt{\cot (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{11 \sqrt{\cot (c+d x)}}{6 a 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{25 a^2}{4}-\frac{11}{2} i a^2 \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{3 a^4}\\ &=\frac{\sqrt{\cot (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{11 \sqrt{\cot (c+d x)}}{6 a d \sqrt{a+i a \tan (c+d x)}}-\frac{25 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{6 a^2 d}+\frac{\left (2 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{3 i a^3 \sqrt{a+i a \tan (c+d x)}}{8 \sqrt{\tan (c+d x)}} \, dx}{3 a^5}\\ &=\frac{\sqrt{\cot (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{11 \sqrt{\cot (c+d x)}}{6 a d \sqrt{a+i a \tan (c+d x)}}-\frac{25 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{6 a^2 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}{4 a^2}\\ &=\frac{\sqrt{\cot (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{11 \sqrt{\cot (c+d x)}}{6 a d \sqrt{a+i a \tan (c+d x)}}-\frac{25 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{6 a^2 d}+\frac{\left (\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 )}{2 d}\\ &=\frac{\left (\frac{1}{4}+\frac{i}{4}\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)}}{a^{3/2} d}+\frac{\sqrt{\cot (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{11 \sqrt{\cot (c+d x)}}{6 a d \sqrt{a+i a \tan (c+d x)}}-\frac{25 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{6 a^2 d}\\ \end{align*}

Mathematica [A]  time = 1.41441, size = 156, normalized size = 0.86 \[ \frac{e^{-4 i (c+d x)} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{\cot (c+d x)} \left (13 e^{2 i (c+d x)}-38 e^{4 i (c+d x)}+3 e^{3 i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )+1\right )}{6 \sqrt{2} a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(1 + 13*E^((2*I)*(c + d*x)) - 38*E^((4*I)*(c + d*x))
+ 3*E^((3*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]])/(6*Sqrt[2]*a^2*d*E^((4*I)*(c + d*x)))

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Maple [B]  time = 0.381, size = 344, normalized size = 1.9 \begin{align*}{\frac{ \left ( -{\frac{1}{12}}-{\frac{i}{12}} \right ) \sin \left ( dx+c \right ) }{{a}^{2}d\cos \left ( dx+c \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 ) }}} \left ( 4\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}+4\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \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 ) -4\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}+4\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +3\,\arctan \left ( \left ( 1/2+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 ) +9\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}+11\,i\sin \left ( dx+c \right ) \cos \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 ) -9\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+11\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +25-25\,i \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))^(3/2),x)

[Out]

(-1/12-1/12*I)/d/a^2*(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)*sin(d*x+c)*(
4*I*cos(d*x+c)^4+4*I*cos(d*x+c)^3*sin(d*x+c)-3*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)-4*cos(d*x+c)^4+4*cos(d*x+c)^3*sin(d*x+c)+3*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)+9*I*cos(
d*x+c)^2+11*I*sin(d*x+c)*cos(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))-9*cos(d*x+c)^2+11*cos(d*x+c)*sin(d*x+c)+25-25*I)/cos(d*x+c)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \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))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 1.56012, size = 1029, normalized size = 5.65 \begin{align*} \frac{{\left (3 \, a^{2} d \sqrt{\frac{i}{2 \, a^{3} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\frac{1}{4} \,{\left (2 \, a^{2} d \sqrt{\frac{i}{2 \, a^{3} 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 \, a^{2} d \sqrt{\frac{i}{2 \, a^{3} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac{1}{4} \,{\left (2 \, a^{2} d \sqrt{\frac{i}{2 \, a^{3} 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 ) - \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 (38 \, e^{\left (4 i \, d x + 4 i \, c\right )} - 13 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{12 \, a^{2} 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))^(3/2),x, algorithm="fricas")

[Out]

1/12*(3*a^2*d*sqrt(1/2*I/(a^3*d^2))*e^(4*I*d*x + 4*I*c)*log(1/4*(2*a^2*d*sqrt(1/2*I/(a^3*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^2*d*sqrt(1/2*I/(a^3*d^2))*e^(4*I*d*x + 4*I*c
)*log(-1/4*(2*a^2*d*sqrt(1/2*I/(a^3*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)) - 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))
*(38*e^(4*I*d*x + 4*I*c) - 13*e^(2*I*d*x + 2*I*c) - 1)*e^(I*d*x + I*c))*e^(-4*I*d*x - 4*I*c)/(a^2*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)**(3/2)/(a+I*a*tan(d*x+c))**(3/2),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{3}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{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))^(3/2),x, algorithm="giac")

[Out]

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