3.777 \(\int \frac{1}{\sqrt{\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=147 \[ -\frac{\left (\frac{1}{4}+\frac{i}{4}\right ) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac{1}{3 d \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac{i}{2 a d \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}} \]

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

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Rubi [A]  time = 0.278175, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {4241, 3547, 3546, 3544, 205} \[ -\frac{\left (\frac{1}{4}+\frac{i}{4}\right ) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac{1}{3 d \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac{i}{2 a d \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[Cot[c + d*x]]*(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) + 1/(3*d*Cot[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(3/2)) + (I/2)/(a*d*Sqrt[Cot[c
+ d*x]]*Sqrt[a + I*a*Tan[c + d*x]])

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 3547

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), Int[(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && Eq
Q[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && EqQ[m + n + 1, 0] && LtQ[m, -1]

Rule 3546

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)/(2*b*f*m), x] - Dist[(a*c - b*d)/(2*b^2), Int[(a + b*Tan[e
+ f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1), 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] && EqQ[m + n, 0] && LeQ[m, -2^(-1)]

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

Mathematica [A]  time = 1.31143, size = 156, normalized size = 1.06 \[ \frac{e^{-4 i (c+d x)} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{\cot (c+d x)} \left (-e^{2 i (c+d x)}+2 e^{4 i (c+d x)}-3 e^{3 i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )-1\right )}{6 \sqrt{2} a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[Cot[c + d*x]]*(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 - E^((2*I)*(c + d*x)) + 2*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))]])*S
qrt[Cot[c + d*x]])/(6*Sqrt[2]*a^2*d*E^((4*I)*(c + d*x)))

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Maple [B]  time = 0.393, size = 482, normalized size = 3.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1/12+1/12*I)/d/a^2*(6*I*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))*2^(1/2)-I*
cos(d*x+c)^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)+3*I*cos(d*x+c)*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-3*I
*cos(d*x+c)*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))*2^(1/2)-6*cos(d*x+c)*sin(d*x+c)*arct
an((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))*2^(1/2)-3*I*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d
*x+c))^(1/2)*2^(1/2))*2^(1/2)+cos(d*x+c)^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)+3*cos(d*x+c)*sin(d*x+c)*((cos(d*x
+c)-1)/sin(d*x+c))^(1/2)+3*sin(d*x+c)*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))*2^(1/2)+I*
((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-((cos(d*x+c)-1)/sin(d*x+c))^(1/2))*cos(d*x+c)*(a*(I*sin(d*x+c)+cos(d*x+c))/c
os(d*x+c))^(1/2)/(2*I*cos(d*x+c)*sin(d*x+c)+2*cos(d*x+c)^2-1)/sin(d*x+c)/(cos(d*x+c)/sin(d*x+c))^(1/2)/((cos(d
*x+c)-1)/sin(d*x+c))^(1/2)

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

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 1.59281, size = 1025, normalized size = 6.97 \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 (2 \, e^{\left (4 i \, d x + 4 i \, c\right )} - 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(1/cot(d*x+c)^(1/2)/(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))*sqr
t((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)
)*(2*e^(4*I*d*x + 4*I*c) - 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(1/cot(d*x+c)**(1/2)/(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{1}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \sqrt{\cot \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((I*a*tan(d*x + c) + a)^(3/2)*sqrt(cot(d*x + c))), x)