3.8.56 \(\int \frac {\sqrt {a+i a \tan (c+d x)}}{\cot ^{\frac {3}{2}}(c+d x)} \, dx\) [756]

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

[Out]

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

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Rubi [A]
time = 0.27, antiderivative size = 175, normalized size of antiderivative = 1.00, 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 = {4326, 3641, 21, 3636, 3625, 211, 3680, 65, 223, 209} \begin {gather*} -\frac {\sqrt [4]{-1} \sqrt {a} \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \text {ArcTan}\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} \end {gather*}

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 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 65

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 209

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

Rule 211

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

Rule 223

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 3625

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 3636

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 3641

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 3680

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 4326

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]

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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 2.91, size = 223, normalized size = 1.27 \begin {gather*} \frac {\left (8-\frac {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right ) \left (8 \log \left (e^{i (c+d x)}+\sqrt {-1+e^{2 i (c+d x)}}\right )+\sqrt {2} \left (-\log \left (1-3 e^{2 i (c+d x)}-2 \sqrt {2} e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}}\right )+\log \left (1-3 e^{2 i (c+d x)}+2 \sqrt {2} e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}}\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)}} \end {gather*}

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] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2092 vs. \(2 (140 ) = 280\).
time = 41.99, size = 2093, normalized size = 11.96

method	result	size
default	\(\text {Expression too large to display}\)	\(2093\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 610 vs. \(2 (133) = 266\).
time = 0.99, size = 610, normalized size = 3.49 \begin {gather*} -\frac {4 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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}} {\left (i \, e^{\left (3 i \, d x + 3 i \, c\right )} - i \, 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 {8 i \, a}{d^{2}}} \log \left ({\left (\sqrt {2} {\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 {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {8 i \, a}{d^{2}}} + 4 i \, 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 {8 i \, a}{d^{2}}} \log \left ({\left (\sqrt {2} {\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 {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {8 i \, a}{d^{2}}} + 4 i \, 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 {i \, a}{d^{2}}} \log \left (-16 \, {\left (2 \, \sqrt {2} {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} - 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 {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {i \, a}{d^{2}}} + 3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 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 {i \, a}{d^{2}}} \log \left (16 \, {\left (2 \, \sqrt {2} {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} - 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 {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {i \, a}{d^{2}}} - 3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right )}{4 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end {gather*}

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]

-1/4*(4*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))*
(I*e^(3*I*d*x + 3*I*c) - I*e^(I*d*x + I*c)) + (d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-8*I*a/d^2)*log((sqrt(2)*(I*d*e
^(2*I*d*x + 2*I*c) - I*d)*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))*sqrt(-8*I*a/d^2) + 4*I*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - (d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-8*I*
a/d^2)*log((sqrt(2)*(-I*d*e^(2*I*d*x + 2*I*c) + I*d)*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))*sqrt(-8*I*a/d^2) + 4*I*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - (d*e^(2*I*d
*x + 2*I*c) + d)*sqrt(-I*a/d^2)*log(-16*(2*sqrt(2)*(a*d*e^(3*I*d*x + 3*I*c) - a*d*e^(I*d*x + I*c))*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))*sqrt(-I*a/d^2) + 3*a^2*e^(2
*I*d*x + 2*I*c) - a^2)*e^(-2*I*d*x - 2*I*c)) + (d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-I*a/d^2)*log(16*(2*sqrt(2)*(a
*d*e^(3*I*d*x + 3*I*c) - a*d*e^(I*d*x + I*c))*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))*sqrt(-I*a/d^2) - 3*a^2*e^(2*I*d*x + 2*I*c) + a^2)*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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}{\cot ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \end {gather*}

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{{\mathrm {cot}\left (c+d\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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