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

Optimal. Leaf size=175 \[ -\frac{\sqrt [4]{-1} \sqrt{a} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{\sqrt{a+i a \tan (c+d x)}}{d \sqrt{\cot (c+d x)}}-\frac{(1-i) \sqrt{a} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.406614, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {4241, 3560, 21, 3555, 3544, 205, 3599, 63, 217, 203} \[ -\frac{\sqrt [4]{-1} \sqrt{a} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{\sqrt{a+i a \tan (c+d x)}}{d \sqrt{\cot (c+d x)}}-\frac{(1-i) \sqrt{a} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4241

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ
[u, x]

Rule 3560

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*(m + n - 1)), x] - Dist[1/(a*(m + n - 1)), Int[(a
+ b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 2)*Simp[d*(b*c*m + a*d*(-1 + n)) - a*c^2*(m + n - 1) + d*(b*d*m
- a*c*(m + 2*n - 2))*Tan[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] && GtQ[n, 1] && NeQ[m + n - 1, 0] && (IntegerQ[n] || IntegersQ[2*m, 2*n])

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 3555

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(3/2)/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dis
t[2*a, Int[Sqrt[a + b*Tan[e + f*x]]/Sqrt[c + d*Tan[e + f*x]], x], x] + Dist[b/a, Int[((b + a*Tan[e + f*x])*Sqr
t[a + b*Tan[e + f*x]])/Sqrt[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 3544

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*a*b)/f, Subst[Int[1/(a*c - b*d - 2*a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*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 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

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 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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

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

Rubi steps

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

Mathematica [A]  time = 2.53801, size = 223, normalized size = 1.27 \[ \frac{\left (8-\frac{e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right ) \left (8 \log \left (\sqrt{-1+e^{2 i (c+d x)}}+e^{i (c+d x)}\right )+\sqrt{2} \left (\log \left (2 \sqrt{2} e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )-\log \left (-2 \sqrt{2} e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )\right )\right )}{\sqrt{-1+e^{2 i (c+d x)}}}\right ) \sqrt{a+i a \tan (c+d x)}}{8 d \sqrt{\cot (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8 - ((1 + E^((2*I)*(c + d*x)))*(8*Log[E^(I*(c + d*x)) + Sqrt[-1 + E^((2*I)*(c + d*x))]] + Sqrt[2]*(-Log[1 -
3*E^((2*I)*(c + d*x)) - 2*Sqrt[2]*E^(I*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]] + Log[1 - 3*E^((2*I)*(c + d*
x)) + 2*Sqrt[2]*E^(I*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]])))/(E^(I*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*
x))]))*Sqrt[a + I*a*Tan[c + d*x]])/(8*d*Sqrt[Cot[c + d*x]])

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Maple [B]  time = 0.464, size = 2093, normalized size = 12. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^(1/2)/cot(d*x+c)^(3/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{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))^(1/2)/cot(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.52289, size = 1661, normalized size = 9.49 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(1/2)/cot(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

1/2*(sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*(-2
*I*e^(2*I*d*x + 2*I*c) + 2*I)*e^(I*d*x + I*c) + (d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-I*a/d^2)*log((sqrt(2)*sqrt(a
/(e^(2*I*d*x + 2*I*c) + 1))*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) -
1)*e^(I*d*x + I*c) + 2*I*d*sqrt(-I*a/d^2)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)) - (d*e^(2*I*d*x + 2*I*c)
+ d)*sqrt(-I*a/d^2)*log((sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*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) - 1)*e^(I*d*x + I*c) - 2*I*d*sqrt(-I*a/d^2)*e^(2*I*d*x + 2*I*c))*e^(-2*
I*d*x - 2*I*c)) - (d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-2*I*a/d^2)*log((sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*
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) - 1)*e^(I*d*x + I*c) + I*d*sq
rt(-2*I*a/d^2)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)) + (d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-2*I*a/d^2)*log((
sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*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) - 1)*e^(I*d*x + I*c) - I*d*sqrt(-2*I*a/d^2)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)))/(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*} \int \frac{\sqrt{a \left (i \tan{\left (c + d x \right )} + 1\right )}}{\cot ^{\frac{3}{2}}{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**(1/2)/cot(d*x+c)**(3/2),x)

[Out]

Integral(sqrt(a*(I*tan(c + d*x) + 1))/cot(c + d*x)**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{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))^(1/2)/cot(d*x+c)^(3/2),x, algorithm="giac")

[Out]

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