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

Optimal. Leaf size=141 \[ \frac {i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}+\frac {i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}-\frac {2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{a d}+\frac {\cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

I*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))/d/a^(1/2)+1/2*I*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/
2))/d*2^(1/2)/a^(1/2)+cot(d*x+c)/d/(a+I*a*tan(d*x+c))^(1/2)-2*cot(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/a/d

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Rubi [A]  time = 0.43, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3559, 3598, 3600, 3480, 206, 3599, 63, 208} \[ \frac {i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}+\frac {i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}-\frac {2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{a d}+\frac {\cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/(Sqrt[a]*d) + (I*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*S
qrt[a])])/(Sqrt[2]*Sqrt[a]*d) + Cot[c + d*x]/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (2*Cot[c + d*x]*Sqrt[a + I*a*Tan
[c + d*x]])/(a*d)

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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]

Rule 3480

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0]

Rule 3559

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

Rule 3598

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

Rule 3599

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

Rule 3600

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

Rubi steps

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

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Mathematica [A]  time = 2.19, size = 153, normalized size = 1.09 \[ \frac {i \sec (c+d x) \left (\sqrt {1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )+\sqrt {2} \sqrt {1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {1+e^{2 i (c+d x)}}}\right )+i \csc (c+d x) (2 i \sin (2 (c+d x))+\cos (2 (c+d x))+1)\right )}{2 d \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*Sec[c + d*x]*(Sqrt[1 + E^((2*I)*(c + d*x))]*ArcSinh[E^(I*(c + d*x))] + Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d
*x))]*ArcTanh[(Sqrt[2]*E^(I*(c + d*x)))/Sqrt[1 + E^((2*I)*(c + d*x))]] + I*Csc[c + d*x]*(1 + Cos[2*(c + d*x)]
+ (2*I)*Sin[2*(c + d*x)])))/(d*Sqrt[a + I*a*Tan[c + d*x]])

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fricas [B]  time = 0.46, size = 545, normalized size = 3.87 \[ \frac {\sqrt {2} {\left (i \, a d e^{\left (3 i \, d x + 3 i \, c\right )} - i \, a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {1}{a d^{2}}} \log \left (4 \, {\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} {\left (-i \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + i \, a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {1}{a d^{2}}} \log \left (-4 \, {\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + {\left (i \, a d e^{\left (3 i \, d x + 3 i \, c\right )} - i \, a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {1}{a d^{2}}} \log \left (16 \, {\left (3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, \sqrt {2} {\left (a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{2} d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} + a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + {\left (-i \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + i \, a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {1}{a d^{2}}} \log \left (16 \, {\left (3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 \, \sqrt {2} {\left (a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{2} d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} + a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-6 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 4 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )}}{4 \, {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} - a d e^{\left (i \, d x + i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(sqrt(2)*(I*a*d*e^(3*I*d*x + 3*I*c) - I*a*d*e^(I*d*x + I*c))*sqrt(1/(a*d^2))*log(4*((a*d*e^(2*I*d*x + 2*I*
c) + a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + sqrt(2)*(
-I*a*d*e^(3*I*d*x + 3*I*c) + I*a*d*e^(I*d*x + I*c))*sqrt(1/(a*d^2))*log(-4*((a*d*e^(2*I*d*x + 2*I*c) + a*d)*sq
rt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a*d^2)) - a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + (I*a*d*e^(3*I*d*x + 3
*I*c) - I*a*d*e^(I*d*x + I*c))*sqrt(1/(a*d^2))*log(16*(3*a^2*e^(2*I*d*x + 2*I*c) + 2*sqrt(2)*(a^2*d*e^(3*I*d*x
+ 3*I*c) + a^2*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a*d^2)) + a^2)*e^(-2*I*d*x - 2*I*
c)) + (-I*a*d*e^(3*I*d*x + 3*I*c) + I*a*d*e^(I*d*x + I*c))*sqrt(1/(a*d^2))*log(16*(3*a^2*e^(2*I*d*x + 2*I*c) -
2*sqrt(2)*(a^2*d*e^(3*I*d*x + 3*I*c) + a^2*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a*d^2
)) + a^2)*e^(-2*I*d*x - 2*I*c)) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-6*I*e^(4*I*d*x + 4*I*c) - 4*I*e^
(2*I*d*x + 2*I*c) + 2*I))/(a*d*e^(3*I*d*x + 3*I*c) - a*d*e^(I*d*x + I*c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 1.54, size = 1380, normalized size = 9.79 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(4*I*cos(d*x+c)^4+2^(1/2)*cos(d*x+c)^2*sin(d*x+c)*(-2*co
s(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c))
)^(1/2)*2^(1/2))-I*cos(d*x+c)^2*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(-(-sin(d*x+c)*(-2*cos(d*x+c
)/(1+cos(d*x+c)))^(1/2)+cos(d*x+c)-1)/sin(d*x+c))+I*2^(1/2)*cos(d*x+c)^2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*
arctan(1/2*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))+I*2^(1/2)*cos(
d*x+c)^3*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)
/(1+cos(d*x+c)))^(1/2)*2^(1/2))-4*I*cos(d*x+c)^2+I*cos(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/(-
2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+cos(d*x+c)^3*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(-(-sin(d*x+c)*(-2*cos
(d*x+c)/(1+cos(d*x+c)))^(1/2)+cos(d*x+c)-1)/sin(d*x+c))-cos(d*x+c)^2*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))
^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))-I*cos(d*x+c)^2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arct
an(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))-2^(1/2)*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*
(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))+I*sin(d*x+c)*(-2*cos(d*x+
c)/(1+cos(d*x+c)))^(1/2)*ln(-(-sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+cos(d*x+c)-1)/sin(d*x+c))+4*cos
(d*x+c)^3*sin(d*x+c)+cos(d*x+c)^2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(-(-sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(
d*x+c)))^(1/2)+cos(d*x+c)-1)/sin(d*x+c))+I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos
(d*x+c)))^(1/2))-I*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x
+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))-I*2^(1/2)*cos(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ar
ctan(1/2*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))-cos(d*x+c)*(-2*c
os(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(-(-sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+cos(d*x+c)-1)/sin(d*x+c)
)+(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*sin(d*x+c)-I*cos(d*x+c)^
3*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))-8*cos(d*x+c)*sin(d*x+c)-
(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(-(-sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+cos(d*x+c)-1)/sin(d
*x+c)))/(cos(d*x+c)^2-1)/a

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maxima [A]  time = 0.91, size = 160, normalized size = 1.13 \[ \frac {i \, a {\left (\frac {4 \, {\left (-2 i \, a \tan \left (d x + c\right ) - a\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a - \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{2}} - \frac {\sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {3}{2}}} - \frac {2 \, \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{a^{\frac {3}{2}}}\right )}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.09, size = 134, normalized size = 0.95 \[ \frac {\frac {\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,2{}\mathrm {i}}{d}-\frac {a\,1{}\mathrm {i}}{d}}{a\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}-{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}-\frac {\mathrm {atan}\left (\frac {\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{\sqrt {-a}}\right )\,1{}\mathrm {i}}{\sqrt {-a}\,d}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{2\,\sqrt {-a}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((a + a*tan(c + d*x)*1i)*2i)/d - (a*1i)/d)/(a*(a + a*tan(c + d*x)*1i)^(1/2) - (a + a*tan(c + d*x)*1i)^(3/2))
- (atan((a + a*tan(c + d*x)*1i)^(1/2)/(-a)^(1/2))*1i)/((-a)^(1/2)*d) - (2^(1/2)*atan((2^(1/2)*(a + a*tan(c + d
*x)*1i)^(1/2))/(2*(-a)^(1/2)))*1i)/(2*(-a)^(1/2)*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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