3.754 \(\int \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)} \, dx\)

Optimal. Leaf size=69 \[ \frac{(1-i) \sqrt{a} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d} \]

[Out]

((1 - I)*Sqrt[a]*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]*S
qrt[Tan[c + d*x]])/d

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Rubi [A]  time = 0.129696, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {4241, 3544, 205} \[ \frac{(1-i) \sqrt{a} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

((1 - I)*Sqrt[a]*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]*S
qrt[Tan[c + d*x]])/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 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 \sqrt{\cot (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{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{\left (2 i a^2 \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{(1-i) \sqrt{a} \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)}}{d}\\ \end{align*}

Mathematica [A]  time = 0.705258, size = 118, normalized size = 1.71 \[ -\frac{i e^{-i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \sqrt{\frac{i \left (1+e^{2 i (c+d x)}\right )}{-1+e^{2 i (c+d x)}}} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right ) \sqrt{a+i a \tan (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

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

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Maple [B]  time = 0.374, size = 408, normalized size = 5.9 \begin{align*} -{\frac{\sqrt{2} \left ( \cos \left ( dx+c \right ) -1 \right ) }{2\,d \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) -1 \right ) }\sqrt{{\frac{\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}}\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}} \left ( i\ln \left ( -{ \left ( \sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \right ) }}}\sqrt{2}\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) -1 \right ) \left ( \sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \right ) }}}\sqrt{2}\sin \left ( dx+c \right ) -\cos \left ( dx+c \right ) -\sin \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) +2\,i\arctan \left ( \sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \right ) }}}\sqrt{2}+1 \right ) +2\,i\arctan \left ( \sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \right ) }}}\sqrt{2}-1 \right ) +2\,\arctan \left ( \sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \right ) }}}\sqrt{2}+1 \right ) +2\,\arctan \left ( \sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \right ) }}}\sqrt{2}-1 \right ) +\ln \left ( -{ \left ( \sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \right ) }}}\sqrt{2}\sin \left ( dx+c \right ) -\cos \left ( dx+c \right ) -\sin \left ( dx+c \right ) +1 \right ) \left ( \sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \right ) }}}\sqrt{2}\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) -1 \right ) ^{-1}} \right ) \right ){\frac{1}{\sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \right ) }}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 2.33207, size = 505, normalized size = 7.32 \begin{align*} -\frac{\sqrt{a}{\left (-\left (2 i + 2\right ) \, \arctan \left (2 \,{\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} - 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac{1}{4}} \sin \left (\frac{1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) - 1\right )\right ) + 2 \, \sin \left (d x + c\right ), 2 \,{\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} - 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac{1}{4}} \cos \left (\frac{1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) - 1\right )\right ) + 2 \, \cos \left (d x + c\right )\right ) + \left (i - 1\right ) \, \log \left (4 \, \cos \left (d x + c\right )^{2} + 4 \, \sin \left (d x + c\right )^{2} + 4 \, \sqrt{\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} - 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1}{\left (\cos \left (\frac{1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) - 1\right )\right )^{2} + \sin \left (\frac{1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) - 1\right )\right )^{2}\right )} + 8 \,{\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} - 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac{1}{4}}{\left (\cos \left (d x + c\right ) \cos \left (\frac{1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) - 1\right )\right ) + \sin \left (d x + c\right ) \sin \left (\frac{1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) - 1\right )\right )\right )}\right )\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(a)*(-(2*I + 2)*arctan2(2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*si
n(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + 2*sin(d*x + c), 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2
*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + 2*cos(d*x + c
)) + (I - 1)*log(4*cos(d*x + c)^2 + 4*sin(d*x + c)^2 + 4*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(
2*d*x + 2*c) + 1)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1))^2 + sin(1/2*arctan2(sin(2*d*x + 2*
c), cos(2*d*x + 2*c) - 1))^2) + 8*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*(co
s(d*x + c)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + sin(d*x + c)*sin(1/2*arctan2(sin(2*d*x +
2*c), cos(2*d*x + 2*c) - 1)))))/d

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Fricas [B]  time = 1.26761, size = 641, normalized size = 9.29 \begin{align*} \frac{1}{2} \, \sqrt{-\frac{2 i \, a}{d^{2}}} \log \left ({\left (\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 )} + i \, d \sqrt{-\frac{2 i \, a}{d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - \frac{1}{2} \, \sqrt{-\frac{2 i \, a}{d^{2}}} \log \left ({\left (\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 )} - i \, d \sqrt{-\frac{2 i \, a}{d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} 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)^(1/2)*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \left (i \tan{\left (c + d x \right )} + 1\right )} \sqrt{\cot{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a*(I*tan(c + d*x) + 1))*sqrt(cot(c + d*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(I*a*tan(d*x + c) + a)*sqrt(cot(d*x + c)), x)