3.743 \(\int \frac {1}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=234 \[ \frac {\sqrt {\cot (c+d x)}}{8 a^2 d (\cot (c+d x)+i)}+\frac {\left (\frac {1}{32}-\frac {3 i}{32}\right ) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a^2 d}-\frac {\left (\frac {1}{32}-\frac {3 i}{32}\right ) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a^2 d}-\frac {\left (\frac {1}{16}+\frac {3 i}{16}\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^2 d}+\frac {\left (\frac {1}{16}+\frac {3 i}{16}\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a^2 d}+\frac {i \sqrt {\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2} \]

[Out]

(1/32+3/32*I)*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))/a^2/d*2^(1/2)+(1/32+3/32*I)*arctan(1+2^(1/2)*cot(d*x+c)^(1/2
))/a^2/d*2^(1/2)+(1/64-3/64*I)*ln(1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(1/2))/a^2/d*2^(1/2)+(-1/64+3/64*I)*ln(1+cot
(d*x+c)+2^(1/2)*cot(d*x+c)^(1/2))/a^2/d*2^(1/2)+1/8*cot(d*x+c)^(1/2)/a^2/d/(I+cot(d*x+c))+1/4*I*cot(d*x+c)^(1/
2)/d/(I*a+a*cot(d*x+c))^2

________________________________________________________________________________________

Rubi [A]  time = 0.31, antiderivative size = 234, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3673, 3557, 3596, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac {\sqrt {\cot (c+d x)}}{8 a^2 d (\cot (c+d x)+i)}+\frac {\left (\frac {1}{32}-\frac {3 i}{32}\right ) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a^2 d}-\frac {\left (\frac {1}{32}-\frac {3 i}{32}\right ) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a^2 d}-\frac {\left (\frac {1}{16}+\frac {3 i}{16}\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^2 d}+\frac {\left (\frac {1}{16}+\frac {3 i}{16}\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a^2 d}+\frac {i \sqrt {\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(Cot[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^2),x]

[Out]

((-1/16 - (3*I)/16)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*a^2*d) + ((1/16 + (3*I)/16)*ArcTan[1 + Sq
rt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*a^2*d) + Sqrt[Cot[c + d*x]]/(8*a^2*d*(I + Cot[c + d*x])) + ((I/4)*Sqrt[Cot
[c + d*x]])/(d*(I*a + a*Cot[c + d*x])^2) + ((1/32 - (3*I)/32)*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x
]])/(Sqrt[2]*a^2*d) - ((1/32 - (3*I)/32)*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(Sqrt[2]*a^2*d)

Rule 204

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

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1168

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[-(a*c)]

Rule 3534

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 3557

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> -Sim
p[(b*(a + b*Tan[e + f*x])^m*Sqrt[c + d*Tan[e + f*x]])/(2*a*f*m), x] + Dist[1/(4*a^2*m), Int[((a + b*Tan[e + f*
x])^(m + 1)*Simp[2*a*c*m + b*d + a*d*(2*m + 1)*Tan[e + f*x], 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] && LtQ[m, 0] && IntegersQ[2
*m]

Rule 3596

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*A + b*B)*(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*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x
] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] &&  !GtQ[n,
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 \frac {1}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx &=\int \frac {\sqrt {\cot (c+d x)}}{(i a+a \cot (c+d x))^2} \, dx\\ &=\frac {i \sqrt {\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2}+\frac {\int \frac {a-3 i a \cot (c+d x)}{\sqrt {\cot (c+d x)} (i a+a \cot (c+d x))} \, dx}{8 a^2}\\ &=\frac {\sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {i \sqrt {\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2}+\frac {\int \frac {-3 i a^2-a^2 \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx}{16 a^4}\\ &=\frac {\sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {i \sqrt {\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2}+\frac {\operatorname {Subst}\left (\int \frac {3 i a^2+a^2 x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{8 a^4 d}\\ &=\frac {\sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {i \sqrt {\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2}+-\frac {\left (\frac {1}{16}-\frac {3 i}{16}\right ) \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{a^2 d}+\frac {\left (\frac {1}{16}+\frac {3 i}{16}\right ) \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{a^2 d}\\ &=\frac {\sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {i \sqrt {\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2}+\frac {\left (\frac {1}{32}+\frac {3 i}{32}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{a^2 d}+\frac {\left (\frac {1}{32}+\frac {3 i}{32}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{a^2 d}+\frac {\left (\frac {1}{32}-\frac {3 i}{32}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^2 d}+\frac {\left (\frac {1}{32}-\frac {3 i}{32}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^2 d}\\ &=\frac {\sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {i \sqrt {\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2}+\frac {\left (\frac {1}{32}-\frac {3 i}{32}\right ) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{\sqrt {2} a^2 d}-\frac {\left (\frac {1}{32}-\frac {3 i}{32}\right ) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{\sqrt {2} a^2 d}+-\frac {\left (\frac {1}{16}+\frac {3 i}{16}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^2 d}+\frac {\left (\frac {1}{16}+\frac {3 i}{16}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^2 d}\\ &=-\frac {\left (\frac {1}{16}+\frac {3 i}{16}\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^2 d}+\frac {\left (\frac {1}{16}+\frac {3 i}{16}\right ) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^2 d}+\frac {\sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {i \sqrt {\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2}+\frac {\left (\frac {1}{32}-\frac {3 i}{32}\right ) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{\sqrt {2} a^2 d}-\frac {\left (\frac {1}{32}-\frac {3 i}{32}\right ) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{\sqrt {2} a^2 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.83, size = 222, normalized size = 0.95 \[ \frac {\csc ^3(c+d x) \left (\sin (c+d x)+\sin (3 (c+d x))+3 i \cos (c+d x)-3 i \cos (3 (c+d x))+(1+3 i) \sqrt {\sin (2 (c+d x))} \sin ^{-1}(\cos (c+d x)-\sin (c+d x)) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+(-3-i) \sin ^{\frac {3}{2}}(2 (c+d x)) \log \left (\sin (c+d x)+\sqrt {\sin (2 (c+d x))}+\cos (c+d x)\right )-(1-3 i) \sqrt {\sin (2 (c+d x))} \cos (2 (c+d x)) \log \left (\sin (c+d x)+\sqrt {\sin (2 (c+d x))}+\cos (c+d x)\right )\right )}{32 a^2 d \sqrt {\cot (c+d x)} (\cot (c+d x)+i)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Cot[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^2),x]

[Out]

(Csc[c + d*x]^3*((3*I)*Cos[c + d*x] - (3*I)*Cos[3*(c + d*x)] + Sin[c + d*x] - (1 - 3*I)*Cos[2*(c + d*x)]*Log[C
os[c + d*x] + Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]]*Sqrt[Sin[2*(c + d*x)]] + (1 + 3*I)*ArcSin[Cos[c + d*x] -
Sin[c + d*x]]*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)])*Sqrt[Sin[2*(c + d*x)]] - (3 + I)*Log[Cos[c + d*x] + Sin[
c + d*x] + Sqrt[Sin[2*(c + d*x)]]]*Sin[2*(c + d*x)]^(3/2) + Sin[3*(c + d*x)]))/(32*a^2*d*Sqrt[Cot[c + d*x]]*(I
+ Cot[c + d*x])^2)

________________________________________________________________________________________

fricas [B]  time = 1.76, size = 507, normalized size = 2.17 \[ -\frac {{\left (4 \, a^{2} d \sqrt {-\frac {i}{16 \, a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left ({\left ({\left (8 i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - 8 i \, a^{2} 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}} \sqrt {-\frac {i}{16 \, a^{4} d^{2}}} + 2 i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - 4 \, a^{2} d \sqrt {-\frac {i}{16 \, a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left ({\left ({\left (-8 i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + 8 i \, a^{2} 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}} \sqrt {-\frac {i}{16 \, a^{4} d^{2}}} + 2 i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + 4 \, a^{2} d \sqrt {\frac {i}{64 \, a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {{\left (8 \, {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2} 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}} \sqrt {\frac {i}{64 \, a^{4} d^{2}}} + 1\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{2} d}\right ) - 4 \, a^{2} d \sqrt {\frac {i}{64 \, a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\frac {{\left (8 \, {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2} 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}} \sqrt {\frac {i}{64 \, a^{4} d^{2}}} - 1\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{2} 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}} {\left (-2 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 3 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{16 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(4*a^2*d*sqrt(-1/16*I/(a^4*d^2))*e^(4*I*d*x + 4*I*c)*log(((8*I*a^2*d*e^(2*I*d*x + 2*I*c) - 8*I*a^2*d)*sq
rt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(-1/16*I/(a^4*d^2)) + 2*I*e^(2*I*d*x + 2*I*c))*e
^(-2*I*d*x - 2*I*c)) - 4*a^2*d*sqrt(-1/16*I/(a^4*d^2))*e^(4*I*d*x + 4*I*c)*log(((-8*I*a^2*d*e^(2*I*d*x + 2*I*c
) + 8*I*a^2*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(-1/16*I/(a^4*d^2)) + 2*I*e^(2*
I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)) + 4*a^2*d*sqrt(1/64*I/(a^4*d^2))*e^(4*I*d*x + 4*I*c)*log(-1/8*(8*(a^2*d*
e^(2*I*d*x + 2*I*c) - a^2*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(1/64*I/(a^4*d^2)
) + 1)*e^(-2*I*d*x - 2*I*c)/(a^2*d)) - 4*a^2*d*sqrt(1/64*I/(a^4*d^2))*e^(4*I*d*x + 4*I*c)*log(1/8*(8*(a^2*d*e^
(2*I*d*x + 2*I*c) - a^2*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(1/64*I/(a^4*d^2))
- 1)*e^(-2*I*d*x - 2*I*c)/(a^2*d)) - sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*(-2*I*e^(4*I*
d*x + 4*I*c) + 3*I*e^(2*I*d*x + 2*I*c) - I))*e^(-4*I*d*x - 4*I*c)/(a^2*d)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} \cot \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 1.46, size = 2502, normalized size = 10.69 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16/a^2/d*(-1+cos(d*x+c))*(-2*I*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
))*((-1+cos(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))
/sin(d*x+c))^(1/2)*cos(d*x+c)^3+2*I*((-1+cos(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))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*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))+4*I*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))*((-1+cos(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))/sin(d*x+c))^(1/2)*cos(d*x+c)^3+2*I*cos(d*x+c)*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))
^(1/2),1/2*2^(1/2))*((-1+cos(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))/sin(d*x+c))^(1/2)-4*I*cos(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)*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))+4*cos(d*x+c)^3*((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))+2*((-1+cos(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))/sin(d*x+c))^(1/2)*cos(d*x+c)^3*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))-4*sin(d*x+c)*((-1+cos(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))/sin(d*x+c))^(1/2)*cos(d*x+c)^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))+2*((-1+cos(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))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*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))+2*sin(d*x+c)*((-1+cos(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))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticF(
((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))+2*I*cos(d*x+c)*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))*((-1+cos(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))/sin(d*x+c))^(1/2)-I*((-1+cos(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))/sin(d*x+c))^(1/2)*sin(d*x+c)*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))-2*I*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(
1/2),1/2*2^(1/2))*((-1+cos(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))/sin(d*x+c))^(1/2)*cos(d*x+c)^3-2*I*((-1+cos(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))/sin(d*x+c))^(1/2)*sin(d*x+c)*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))+4*I*((-1+cos(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))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*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))+3*I*2^(1/2)*sin(d*x+c)*cos(d*x+c)-4*cos(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)*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))-2*((1-cos(d*x+c)+s
in(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))*cos(d*x+c)+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)*El
lipticPi(((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)-((-1+cos(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))/sin(d*x+c))^(1/2)*Ellipt
icPi(((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)-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)
*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))-3*I*sin(d*x+c)*2^(1/2)*cos(d*x+c)^2-2^(1/
2)*cos(d*x+c)^3+cos(d*x+c)^2*2^(1/2))*cos(d*x+c)*(1+cos(d*x+c))^2/(2*I*cos(d*x+c)*sin(d*x+c)+2*cos(d*x+c)^2-1)
/(cos(d*x+c)/sin(d*x+c))^(3/2)/sin(d*x+c)^5*2^(1/2)

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {1}{\tan ^{2}{\left (c + d x \right )} \cot ^{\frac {3}{2}}{\left (c + d x \right )} - 2 i \tan {\left (c + d x \right )} \cot ^{\frac {3}{2}}{\left (c + d x \right )} - \cot ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________