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

Optimal. Leaf size=65 \[ \frac{2 i a}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{2 a}{d \sqrt{\cot (c+d x)}}-\frac{2 (-1)^{3/4} a \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.0997634, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3673, 3529, 3533, 208} \[ \frac{2 i a}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{2 a}{d \sqrt{\cot (c+d x)}}-\frac{2 (-1)^{3/4} a \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(-1)^(3/4)*a*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d + (((2*I)/3)*a)/(d*Cot[c + d*x]^(3/2)) + (2*a)/(d*S
qrt[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)}{\cot ^{\frac{3}{2}}(c+d x)} \, dx &=\int \frac{i a+a \cot (c+d x)}{\cot ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 i a}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\int \frac{a-i a \cot (c+d x)}{\cot ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 i a}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{2 a}{d \sqrt{\cot (c+d x)}}+\int \frac{-i a-a \cot (c+d x)}{\sqrt{\cot (c+d x)}} \, dx\\ &=\frac{2 i a}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{2 a}{d \sqrt{\cot (c+d x)}}-\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{i a-a x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{d}\\ &=-\frac{2 (-1)^{3/4} a \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d}+\frac{2 i a}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{2 a}{d \sqrt{\cot (c+d x)}}\\ \end{align*}

Mathematica [B]  time = 1.22124, size = 136, normalized size = 2.09 \[ \frac{a e^{-i c} (\cot (c+d x)+i) \sec (c+d x) (\sin (d x)+i \cos (d x)) \left (-3 i \sin (2 (c+d x))-\cos (2 (c+d x))+6 \cos ^2(c+d x) \sqrt{i \tan (c+d x)} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )+1\right )}{3 d \sqrt{\cot (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(I + Cot[c + d*x])*Sec[c + d*x]*(I*Cos[d*x] + Sin[d*x])*(1 - Cos[2*(c + d*x)] - (3*I)*Sin[2*(c + d*x)] + 6*
ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]]*Cos[c + d*x]^2*Sqrt[I*Tan[c + d*x]]))/(3*d
*E^(I*c)*Sqrt[Cot[c + d*x]])

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Maple [C]  time = 0.217, size = 469, normalized size = 7.2 \begin{align*}{\frac{a\sqrt{2} \left ( \cos \left ( dx+c \right ) -1 \right ) \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2}}{3\,d \left ( \sin \left ( dx+c \right ) \right ) ^{5}} \left ( -3\,i\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \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 ) }}}{\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 ) -3\,{\it EllipticF} \left ( \sqrt{{\frac{1-\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}},1/2\,\sqrt{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 ) \cos \left ( dx+c \right ) +3\,{\it EllipticPi} \left ( \sqrt{{\frac{1-\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}},1/2+i/2,1/2\,\sqrt{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 ) \cos \left ( dx+c \right ) +i\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) \sqrt{2}-i\sin \left ( dx+c \right ) \sqrt{2}+3\,\sqrt{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}-3\,\sqrt{2}\cos \left ( dx+c \right ) \right ) \left ({\frac{\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) ^{-{\frac{3}{2}}}} \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)^(3/2),x)

[Out]

1/3*a/d*2^(1/2)*(cos(d*x+c)-1)*(-3*I*sin(d*x+c)*cos(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+si
n(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/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))-3*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),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)/si
n(d*x+c))^(1/2)*sin(d*x+c)*cos(d*x+c)+3*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)*cos(d*x+c)+I*sin(d*x+c)*cos(d*x+c)*2^(1/2)-I*sin(d*x+c)*2^(1/2)+3*2^(1/2)*co
s(d*x+c)^2-3*2^(1/2)*cos(d*x+c))*(cos(d*x+c)+1)^2/(cos(d*x+c)/sin(d*x+c))^(3/2)/sin(d*x+c)^5

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Maxima [B]  time = 1.52196, size = 192, normalized size = 2.95 \begin{align*} \frac{8 \,{\left (i \, a + \frac{3 \, a}{\tan \left (d x + c\right )}\right )} \tan \left (d x + c\right )^{\frac{3}{2}} - 3 \,{\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}{12 \, 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)^(3/2),x, algorithm="maxima")

[Out]

1/12*(8*(I*a + 3*a/tan(d*x + c))*tan(d*x + c)^(3/2) - 3*(-(2*I + 2)*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sq
rt(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/ta
n(d*x + c) + 1))*a)/d

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Fricas [B]  time = 1.49569, size = 923, normalized size = 14.2 \begin{align*} -\frac{3 \,{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, 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 (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, 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 ) - 3 \,{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, 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 (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, 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 (-32 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 16 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 16 i \, 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}}}{12 \,{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, 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)^(3/2),x, algorithm="fricas")

[Out]

-1/12*(3*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-4*I*a^2/d^2)*log(((I*d*e^(2*I*d*x + 2*I*c
) - I*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) - 3*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-4*I*a^2/d^2)*lo
g(((-I*d*e^(2*I*d*x + 2*I*c) + I*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) - (-32*I*a*e^(4*I*d*x + 4*I*c) + 16*I*a*e^(2*I*d*x +
2*I*c) + 16*I*a)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))/(d*e^(4*I*d*x + 4*I*c) + 2*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 )}}{\cot ^{\frac{3}{2}}{\left (c + d x \right )}}\, dx + \int \frac{1}{\cot ^{\frac{3}{2}}{\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)**(3/2),x)

[Out]

a*(Integral(I*tan(c + d*x)/cot(c + d*x)**(3/2), x) + Integral(cot(c + d*x)**(-3/2), 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}{\cot \left (d x + c\right )^{\frac{3}{2}}}\,{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)^(3/2),x, algorithm="giac")

[Out]

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