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

Optimal. Leaf size=111 \[ -\frac {i \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {i \sqrt {2} \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {\cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 111, 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 = {3561, 21, 3554, 3480, 206, 3599, 63, 208} \[ -\frac {i \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {i \sqrt {2} \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {\cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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 3554

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(3/2)/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(2
*a^2)/(a*c - b*d), Int[Sqrt[a + b*Tan[e + f*x]], x], x] - Dist[(2*b*c*d + a*(c^2 - d^2))/(a*(c^2 + d^2)), Int[
((a - b*Tan[e + f*x])*Sqrt[a + b*Tan[e + f*x]])/(c + d*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] &
& NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]

Rule 3561

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

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]

Rubi steps

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

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Mathematica [A]  time = 4.10, size = 197, normalized size = 1.77 \[ \frac {\sqrt {a+i a \tan (c+d x)} \left (-4 \cot (c+d x)+i e^{-i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \left (\sqrt {2} \left (\log \left (1-e^{i (c+d x)}\right )-\log \left (1+e^{i (c+d x)}\right )+\log \left (\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}-e^{i (c+d x)}+1\right )-\log \left (\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}+e^{i (c+d x)}+1\right )\right )+4 \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*Cot[c + d*x] + (I*Sqrt[1 + E^((2*I)*(c + d*x))]*(4*ArcSinh[E^(I*(c + d*x))] + Sqrt[2]*(Log[1 - E^(I*(c +
d*x))] - Log[1 + E^(I*(c + d*x))] + Log[1 - E^(I*(c + d*x)) + Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))]] - Log[1 +
E^(I*(c + d*x)) + Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))]])))/E^(I*(c + d*x)))*Sqrt[a + I*a*Tan[c + d*x]])/(4*d
)

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fricas [B]  time = 0.45, size = 476, normalized size = 4.29 \[ -\frac {2 \, \sqrt {2} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {-\frac {a}{d^{2}}} \log \left (4 \, {\left ({\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {a}{d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 2 \, \sqrt {2} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {-\frac {a}{d^{2}}} \log \left (4 \, {\left ({\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {a}{d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {-\frac {a}{d^{2}}} \log \left ({\left (48 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {2} {\left (32 i \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + 32 i \, a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {a}{d^{2}}} + 16 \, a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {-\frac {a}{d^{2}}} \log \left ({\left (48 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {2} {\left (-32 i \, a d e^{\left (3 i \, d x + 3 i \, c\right )} - 32 i \, a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {a}{d^{2}}} + 16 \, a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-4 i \, e^{\left (3 i \, d x + 3 i \, c\right )} - 4 i \, e^{\left (i \, d x + i \, c\right )}\right )}}{4 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*sqrt(2)*(d*e^(2*I*d*x + 2*I*c) - d)*sqrt(-a/d^2)*log(4*((I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt(a/(e^(2*I
*d*x + 2*I*c) + 1))*sqrt(-a/d^2) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 2*sqrt(2)*(d*e^(2*I*d*x + 2*I*c) - d
)*sqrt(-a/d^2)*log(4*((-I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(-a/d^2) + a*e^(I
*d*x + I*c))*e^(-I*d*x - I*c)) - (d*e^(2*I*d*x + 2*I*c) - d)*sqrt(-a/d^2)*log((48*a^2*e^(2*I*d*x + 2*I*c) + sq
rt(2)*(32*I*a*d*e^(3*I*d*x + 3*I*c) + 32*I*a*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(-a/d^2)
+ 16*a^2)*e^(-2*I*d*x - 2*I*c)) + (d*e^(2*I*d*x + 2*I*c) - d)*sqrt(-a/d^2)*log((48*a^2*e^(2*I*d*x + 2*I*c) +
sqrt(2)*(-32*I*a*d*e^(3*I*d*x + 3*I*c) - 32*I*a*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(-a/d
^2) + 16*a^2)*e^(-2*I*d*x - 2*I*c)) - sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-4*I*e^(3*I*d*x + 3*I*c) - 4*
I*e^(I*d*x + I*c)))/(d*e^(2*I*d*x + 2*I*c) - d)

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

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

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maple [B]  time = 1.42, size = 586, normalized size = 5.28 \[ -\frac {\sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (2 i \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right )+i \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \ln \left (-\frac {-\sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right )+2 \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right )-2 i \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right )+\left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+2 i \cos \left (d x +c \right ) \sin \left (d x +c \right )-i \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \ln \left (-\frac {-\sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right )-2 \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right )+2 \left (\cos ^{2}\left (d x +c \right )\right )-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )-2 \cos \left (d x +c \right )\right )}{2 d \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )-1\right ) \sin \left (d x +c \right )} \]

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]

-1/2/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(2*I*cos(d*x+c)^2*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))
^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))+I*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))+2
*cos(d*x+c)^2*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(
1/2))-2*I*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*
x+c)/cos(d*x+c)*2^(1/2))+cos(d*x+c)^2*(-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))+2*I*cos(d*x+c)*sin(d*x+c)-I*(-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))-2*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-2*c
os(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))+2*cos(d*x+c)^2-(-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))-2*cos(d*x+c))/(I*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c)

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maxima [A]  time = 0.60, size = 134, normalized size = 1.21 \[ -\frac {i \, a {\left (\frac {\sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{\sqrt {a}} - \frac {\log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{\sqrt {a}} - \frac {2 i \, \sqrt {i \, a \tan \left (d x + c\right ) + a}}{a \tan \left (d x + c\right )}\right )}}{2 \, d} \]

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]

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

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mupad [B]  time = 4.02, size = 97, normalized size = 0.87 \[ -\frac {\sqrt {-a}\,\mathrm {atan}\left (\frac {\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{\sqrt {-a}}\right )\,1{}\mathrm {i}}{d}-\frac {\mathrm {cot}\left (c+d\,x\right )\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{d}+\frac {\sqrt {2}\,\sqrt {-a}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{d} \]

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]

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

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

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

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