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

Optimal. Leaf size=49 \[ \frac {4 (-1)^{3/4} a^2 \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}-\frac {2 a^2}{d \sqrt {\cot (c+d x)}} \]

[Out]

4*(-1)^(3/4)*a^2*arctanh((-1)^(3/4)*cot(d*x+c)^(1/2))/d-2*a^2/d/cot(d*x+c)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3673, 3542, 3533, 208} \[ \frac {4 (-1)^{3/4} a^2 \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}-\frac {2 a^2}{d \sqrt {\cot (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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 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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [C]  time = 1.84, size = 83, normalized size = 1.69 \[ \frac {2 a^2 (i \tan (c+d x))^{3/2} \cot ^{\frac {3}{2}}(c+d x) \left (\sqrt {i \tan (c+d x)}-2 \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [C]  time = 0.85, size = 289, normalized size = 5.90 \[ \frac {\sqrt {-\frac {16 i \, a^{4}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (4 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {-\frac {16 i \, a^{4}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{2 \, a^{2}}\right ) - \sqrt {-\frac {16 i \, a^{4}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (4 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {-\frac {16 i \, a^{4}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{2 \, a^{2}}\right ) + {\left (8 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 8 i \, a^{2}\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}}{4 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

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))^2,x, algorithm="fricas")

[Out]

1/4*(sqrt(-16*I*a^4/d^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
)*(I*d*e^(2*I*d*x + 2*I*c) - I*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) - sqrt(-16*I*a^4/d^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)*(-I*d*e^(2*I*d*x + 2*I*c) + I*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*I*a^2*e^(2*I*d*x + 2*I*c) - 8*I*a^2)*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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} \sqrt {\cot \left (d x + c\right )}\,{d x} \]

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))^2,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [C]  time = 1.33, size = 420, normalized size = 8.57 \[ \frac {a^{2} \sqrt {\frac {\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right ) \left (2 i \sin \left (d x +c \right ) \sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-2 \sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sin \left (d x +c \right )+2 \sin \left (d x +c \right ) \sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \EllipticF \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right )-\cos \left (d x +c \right ) \sqrt {2}+\sqrt {2}\right ) \sqrt {2}}{d \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}} \]

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))^2,x)

[Out]

a^2/d*(cos(d*x+c)/sin(d*x+c))^(1/2)*(1+cos(d*x+c))^2*(-1+cos(d*x+c))*(2*I*sin(d*x+c)*((1-cos(d*x+c)+sin(d*x+c)
)/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticP
i(((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))-2*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))
^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(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))*sin(d*x+c)+2*sin(d*x+c)*((1-cos(d*x+c)+sin(d*x+c))/si
n(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/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))-cos(d*x+c)*2^(1/2)+2^(1/2))/cos(d*x+c)/sin(d*x+c)^3*2^(
1/2)

________________________________________________________________________________________

maxima [C]  time = 1.03, size = 131, normalized size = 2.67 \[ -\frac {{\left (\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + \left (2 i + 2\right ) \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + \left (i - 1\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - \left (i - 1\right ) \, \sqrt {2} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )\right )} a^{2} + 4 \, a^{2} \sqrt {\tan \left (d x + c\right )}}{2 \, d} \]

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))^2,x, algorithm="maxima")

[Out]

-1/2*(((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*sqrt(tan(d*x + c)))/d

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \sqrt {\mathrm {cot}\left (c+d\,x\right )}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - a^{2} \left (\int \tan ^{2}{\left (c + d x \right )} \sqrt {\cot {\left (c + d x \right )}}\, dx + \int \left (- 2 i \tan {\left (c + d x \right )} \sqrt {\cot {\left (c + d x \right )}}\right )\, dx + \int \left (- \sqrt {\cot {\left (c + d x \right )}}\right )\, dx\right ) \]

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))**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) + I
ntegral(-sqrt(cot(c + d*x)), x))

________________________________________________________________________________________