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

Optimal. Leaf size=71 \[ -\frac{2 a^2}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{4 i a^2}{d \sqrt{\cot (c+d x)}}+\frac{4 \sqrt [4]{-1} a^2 \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.143445, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {3673, 3542, 3529, 3533, 208} \[ -\frac{2 a^2}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{4 i a^2}{d \sqrt{\cot (c+d x)}}+\frac{4 \sqrt [4]{-1} a^2 \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(-1)^(1/4)*a^2*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d - (2*a^2)/(3*d*Cot[c + d*x]^(3/2)) + ((4*I)*a^2)/(
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 3542

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

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))^2}{\sqrt{\cot (c+d x)}} \, dx &=\int \frac{(i a+a \cot (c+d x))^2}{\cot ^{\frac{5}{2}}(c+d x)} \, dx\\ &=-\frac{2 a^2}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\int \frac{2 i a^2+2 a^2 \cot (c+d x)}{\cot ^{\frac{3}{2}}(c+d x)} \, dx\\ &=-\frac{2 a^2}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{4 i a^2}{d \sqrt{\cot (c+d x)}}+\int \frac{2 a^2-2 i a^2 \cot (c+d x)}{\sqrt{\cot (c+d x)}} \, dx\\ &=-\frac{2 a^2}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{4 i a^2}{d \sqrt{\cot (c+d x)}}+\frac{\left (8 a^4\right ) \operatorname{Subst}\left (\int \frac{1}{-2 a^2-2 i a^2 x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{d}\\ &=\frac{4 \sqrt [4]{-1} a^2 \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d}-\frac{2 a^2}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{4 i a^2}{d \sqrt{\cot (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.91501, size = 90, normalized size = 1.27 \[ -\frac{2 a^2 \left (-6 i \cot (c+d x)+6 \sqrt{i \tan (c+d x)} \cot ^2(c+d x) \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )+1\right )}{3 d \cot ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.296, size = 485, normalized size = 6.8 \begin{align*}{\frac{{a}^{2}\sqrt{2} \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2} \left ( \cos \left ( dx+c \right ) -1 \right ) }{3\,d\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{4}} \left ( 6\,i\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 ) }}}\sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \right ) }}}{\it EllipticPi} \left ( \sqrt{-{\frac{\cos \left ( dx+c \right ) -1-\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}},{\frac{1}{2}}+{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) \cos \left ( dx+c \right ) \sin \left ( dx+c \right ) -6\,i{\it EllipticF} \left ( \sqrt{-{\frac{\cos \left ( dx+c \right ) -1-\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}},{\frac{\sqrt{2}}{2}} \right ) \cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\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 ) }}}+6\,\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 ) }}}\sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \right ) }}}{\it EllipticPi} \left ( \sqrt{-{\frac{\cos \left ( dx+c \right ) -1-\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}},1/2+i/2,1/2\,\sqrt{2} \right ) \cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +6\,i\sqrt{2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}-6\,i\cos \left ( dx+c \right ) \sqrt{2}-\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \sqrt{2}+\sqrt{2}\sin \left ( dx+c \right ) \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))^2/cot(d*x+c)^(1/2),x)

[Out]

1/3*a^2/d*2^(1/2)*(cos(d*x+c)+1)^2*(cos(d*x+c)-1)*(6*I*(-(cos(d*x+c)-1-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((-(cos(d*x+c)-1-sin(d*x+c))/s
in(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)-6*I*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*
x+c))^(1/2),1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/si
n(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)+6*(-(cos(d*x+c)-1-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((-(cos(d*x+c)-1-sin(
d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)+6*I*2^(1/2)*cos(d*x+c)^2-6*I*cos(d*x+c)
*2^(1/2)-cos(d*x+c)*sin(d*x+c)*2^(1/2)+2^(1/2)*sin(d*x+c))/cos(d*x+c)/sin(d*x+c)^4/(cos(d*x+c)/sin(d*x+c))^(1/
2)

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Maxima [B]  time = 1.70954, size = 197, normalized size = 2.77 \begin{align*} -\frac{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^{2} + 4 \,{\left (a^{2} - \frac{6 i \, a^{2}}{\tan \left (d x + c\right )}\right )} \tan \left (d x + c\right )^{\frac{3}{2}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*(-(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^2 + 4*(a^2 - 6*I*a^2/tan(d*x
+ c))*tan(d*x + c)^(3/2))/d

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Fricas [B]  time = 1.40805, size = 930, normalized size = 13.1 \begin{align*} -\frac{3 \, \sqrt{\frac{16 i \, a^{4}}{d^{2}}}{\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 )} \log \left (\frac{{\left (4 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{\frac{16 i \, a^{4}}{d^{2}}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{2 \, a^{2}}\right ) - 3 \, \sqrt{\frac{16 i \, a^{4}}{d^{2}}}{\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 )} \log \left (\frac{{\left (4 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{\frac{16 i \, a^{4}}{d^{2}}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{2 \, a^{2}}\right ) - 8 \,{\left (7 \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 5 \, a^{2}\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))^2/cot(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

-1/12*(3*sqrt(16*I*a^4/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log(1/2*(4*I*a^2*e^(2*I*d*x
+ 2*I*c) + sqrt(16*I*a^4/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*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)/a^2) - 3*sqrt(16*I*a^4/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d
)*log(1/2*(4*I*a^2*e^(2*I*d*x + 2*I*c) - sqrt(16*I*a^4/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*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)/a^2) - 8*(7*a^2*e^(4*I*d*x + 4*I*c) - 2*a^2*e^(2*I
*d*x + 2*I*c) - 5*a^2)*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^{2} \left (\int - \frac{\tan ^{2}{\left (c + d x \right )}}{\sqrt{\cot{\left (c + d x \right )}}}\, dx + \int \frac{2 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))**2/cot(d*x+c)**(1/2),x)

[Out]

a**2*(Integral(-tan(c + d*x)**2/sqrt(cot(c + d*x)), x) + Integral(2*I*tan(c + d*x)/sqrt(cot(c + d*x)), x) + In
tegral(1/sqrt(cot(c + d*x)), x))

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

[Out]

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