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

Optimal. Leaf size=47 \[ \frac{2 \sqrt [4]{-1} a \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d}+\frac{2 i a}{d \sqrt{\cot (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0775335, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3673, 3529, 3533, 208} \[ \frac{2 \sqrt [4]{-1} a \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d}+\frac{2 i a}{d \sqrt{\cot (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3673

Int[(cot[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Cot[e + f*x])^(m - n*p)*(b + a*Cot[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] &&  !IntegerQ[m] && IntegersQ[n, p]

Rule 3529

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

Rule 3533

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

Rule 208

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

Rubi steps

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

Mathematica [A]  time = 0.835434, size = 71, normalized size = 1.51 \[ \frac{a \left (2 i-\frac{2 i \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )}{\sqrt{i \tan (c+d x)}}\right )}{d \sqrt{\cot (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.25, size = 416, normalized size = 8.9 \begin{align*} -{\frac{a\sqrt{2} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2} \left ( \cos \left ( dx+c \right ) -1 \right ) }{d \left ( \sin \left ( dx+c \right ) \right ) ^{4}} \left ( i\sqrt{{\frac{1-\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}}\sqrt{{\frac{\cos \left ( dx+c \right ) -1+\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}}\sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \right ) }}}{\it EllipticF} \left ( \sqrt{{\frac{1-\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}},{\frac{\sqrt{2}}{2}} \right ) \sin \left ( dx+c \right ) -i\sqrt{{\frac{1-\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}}\sqrt{{\frac{\cos \left ( dx+c \right ) -1+\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}}\sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \right ) }}}{\it EllipticPi} \left ( \sqrt{{\frac{1-\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}},{\frac{1}{2}}+{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) \sin \left ( dx+c \right ) -{\it EllipticPi} \left ( \sqrt{{\frac{1-\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}},{\frac{1}{2}}+{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) \sqrt{{\frac{1-\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}}\sqrt{{\frac{\cos \left ( dx+c \right ) -1+\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}}\sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \right ) }}}\sin \left ( dx+c \right ) -i\cos \left ( dx+c \right ) \sqrt{2}+i\sqrt{2} \right ){\frac{1}{\sqrt{{\frac{\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a/d*2^(1/2)*(cos(d*x+c)+1)^2*(cos(d*x+c)-1)*(I*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+si
n(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))
^(1/2),1/2*2^(1/2))*sin(d*x+c)-I*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d
*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2
*I,1/2*2^(1/2))*sin(d*x+c)-EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*((1-
cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+
c))^(1/2)*sin(d*x+c)-I*cos(d*x+c)*2^(1/2)+I*2^(1/2))/sin(d*x+c)^4/(cos(d*x+c)/sin(d*x+c))^(1/2)

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Maxima [B]  time = 1.51499, size = 171, normalized size = 3.64 \begin{align*} -\frac{{\left (-\left (2 i - 2\right ) \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + \frac{2}{\sqrt{\tan \left (d x + c\right )}}\right )}\right ) - \left (2 i - 2\right ) \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - \frac{2}{\sqrt{\tan \left (d x + c\right )}}\right )}\right ) + \left (i + 1\right ) \, \sqrt{2} \log \left (\frac{\sqrt{2}}{\sqrt{\tan \left (d x + c\right )}} + \frac{1}{\tan \left (d x + c\right )} + 1\right ) - \left (i + 1\right ) \, \sqrt{2} \log \left (-\frac{\sqrt{2}}{\sqrt{\tan \left (d x + c\right )}} + \frac{1}{\tan \left (d x + c\right )} + 1\right )\right )} a - 8 i \, a \sqrt{\tan \left (d x + c\right )}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*((-(2*I - 2)*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) - (2*I - 2)*sqrt(2)*arctan(-1/2
*sqrt(2)*(sqrt(2) - 2/sqrt(tan(d*x + c)))) + (I + 1)*sqrt(2)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) +
1) - (I + 1)*sqrt(2)*log(-sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1))*a - 8*I*a*sqrt(tan(d*x + c)))/d

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Fricas [B]  time = 1.33072, size = 744, normalized size = 15.83 \begin{align*} -\frac{{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt{\frac{4 i \, a^{2}}{d^{2}}} \log \left (\frac{{\left ({\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt{\frac{4 i \, a^{2}}{d^{2}}} \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}} + 2 i \, a e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a}\right ) -{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt{\frac{4 i \, a^{2}}{d^{2}}} \log \left (-\frac{{\left ({\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt{\frac{4 i \, a^{2}}{d^{2}}} \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}} - 2 i \, a e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a}\right ) - 8 \,{\left (a e^{\left (2 i \, d x + 2 i \, c\right )} - a\right )} \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}}}{4 \,{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*((d*e^(2*I*d*x + 2*I*c) + d)*sqrt(4*I*a^2/d^2)*log(((d*e^(2*I*d*x + 2*I*c) - d)*sqrt(4*I*a^2/d^2)*sqrt((I
*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)) + 2*I*a*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/a) - (d
*e^(2*I*d*x + 2*I*c) + d)*sqrt(4*I*a^2/d^2)*log(-((d*e^(2*I*d*x + 2*I*c) - d)*sqrt(4*I*a^2/d^2)*sqrt((I*e^(2*I
*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)) - 2*I*a*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/a) - 8*(a*e^(2
*I*d*x + 2*I*c) - a)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))/(d*e^(2*I*d*x + 2*I*c) + d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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