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

Optimal. Leaf size=200 \[ -\frac {\sqrt {\cot (c+d x)}}{2 d (a \cot (c+d x)+i a)}-\frac {\left (\frac {1}{8}+\frac {3 i}{8}\right ) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a d}+\frac {\left (\frac {1}{8}+\frac {3 i}{8}\right ) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a d}+\frac {\left (\frac {1}{4}-\frac {3 i}{4}\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a d}-\frac {\left (\frac {1}{4}-\frac {3 i}{4}\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a d} \]

[Out]

(-1/8+3/8*I)*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))/a/d*2^(1/2)+(-1/8+3/8*I)*arctan(1+2^(1/2)*cot(d*x+c)^(1/2))/a
/d*2^(1/2)-(1/16+3/16*I)*ln(1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(1/2))/a/d*2^(1/2)+(1/16+3/16*I)*ln(1+cot(d*x+c)+2
^(1/2)*cot(d*x+c)^(1/2))/a/d*2^(1/2)-1/2*cot(d*x+c)^(1/2)/d/(I*a+a*cot(d*x+c))

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Rubi [A]  time = 0.20, antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3673, 3552, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac {\sqrt {\cot (c+d x)}}{2 d (a \cot (c+d x)+i a)}-\frac {\left (\frac {1}{8}+\frac {3 i}{8}\right ) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a d}+\frac {\left (\frac {1}{8}+\frac {3 i}{8}\right ) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a d}+\frac {\left (\frac {1}{4}-\frac {3 i}{4}\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a d}-\frac {\left (\frac {1}{4}-\frac {3 i}{4}\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1/4 - (3*I)/4)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*a*d) - ((1/4 - (3*I)/4)*ArcTan[1 + Sqrt[2]*S
qrt[Cot[c + d*x]]])/(Sqrt[2]*a*d) - Sqrt[Cot[c + d*x]]/(2*d*(I*a + a*Cot[c + d*x])) - ((1/8 + (3*I)/8)*Log[1 -
Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(Sqrt[2]*a*d) + ((1/8 + (3*I)/8)*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x
]] + Cot[c + d*x]])/(Sqrt[2]*a*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 3552

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(a
*(c + d*Tan[e + f*x])^(n + 1))/(2*f*(b*c - a*d)*(a + b*Tan[e + f*x])), x] + Dist[1/(2*a*(b*c - a*d)), Int[(c +
d*Tan[e + f*x])^n*Simp[b*c + a*d*(n - 1) - b*d*n*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] &&  !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))} \, dx &=\int \frac {1}{\sqrt {\cot (c+d x)} (i a+a \cot (c+d x))} \, dx\\ &=-\frac {\sqrt {\cot (c+d x)}}{2 d (i a+a \cot (c+d x))}-\frac {\int \frac {\frac {3 i a}{2}-\frac {1}{2} a \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {\sqrt {\cot (c+d x)}}{2 d (i a+a \cot (c+d x))}-\frac {\operatorname {Subst}\left (\int \frac {-\frac {3 i a}{2}+\frac {a x^2}{2}}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{a^2 d}\\ &=-\frac {\sqrt {\cot (c+d x)}}{2 d (i a+a \cot (c+d x))}--\frac {\left (\frac {1}{4}+\frac {3 i}{4}\right ) \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{a d}-\frac {\left (\frac {1}{4}-\frac {3 i}{4}\right ) \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{a d}\\ &=-\frac {\sqrt {\cot (c+d x)}}{2 d (i a+a \cot (c+d x))}-\frac {\left (\frac {1}{8}-\frac {3 i}{8}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{a d}-\frac {\left (\frac {1}{8}-\frac {3 i}{8}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{a d}-\frac {\left (\frac {1}{8}+\frac {3 i}{8}\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 d}-\frac {\left (\frac {1}{8}+\frac {3 i}{8}\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 d}\\ &=-\frac {\sqrt {\cot (c+d x)}}{2 d (i a+a \cot (c+d x))}-\frac {\left (\frac {1}{8}+\frac {3 i}{8}\right ) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{\sqrt {2} a d}+\frac {\left (\frac {1}{8}+\frac {3 i}{8}\right ) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{\sqrt {2} a d}--\frac {\left (\frac {1}{4}-\frac {3 i}{4}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a d}-\frac {\left (\frac {1}{4}-\frac {3 i}{4}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a d}\\ &=\frac {\left (\frac {1}{4}-\frac {3 i}{4}\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a d}-\frac {\left (\frac {1}{4}-\frac {3 i}{4}\right ) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a d}-\frac {\sqrt {\cot (c+d x)}}{2 d (i a+a \cot (c+d x))}-\frac {\left (\frac {1}{8}+\frac {3 i}{8}\right ) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{\sqrt {2} a d}+\frac {\left (\frac {1}{8}+\frac {3 i}{8}\right ) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{\sqrt {2} a d}\\ \end {align*}

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Mathematica [A]  time = 1.07, size = 174, normalized size = 0.87 \[ \frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \sqrt {\sin (2 (c+d x))} \sqrt {\cot (c+d x)} \left (-\frac {2-2 i}{\sqrt {\sin (2 (c+d x))}}-(1-2 i) \sec (c+d x) \log \left (\sin (c+d x)+\sqrt {\sin (2 (c+d x))}+\cos (c+d x)\right )+(2+i) \csc (c+d x) \log \left (\sin (c+d x)+\sqrt {\sin (2 (c+d x))}+\cos (c+d x)\right )+(1+2 i) \sin ^{-1}(\cos (c+d x)-\sin (c+d x)) (\csc (c+d x)+i \sec (c+d x))\right )}{a d (\cot (c+d x)+i)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1/8 + I/8)*Sqrt[Cot[c + d*x]]*((2 + I)*Csc[c + d*x]*Log[Cos[c + d*x] + Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]
] + (1 + 2*I)*ArcSin[Cos[c + d*x] - Sin[c + d*x]]*(Csc[c + d*x] + I*Sec[c + d*x]) - (1 - 2*I)*Log[Cos[c + d*x]
+ Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]]*Sec[c + d*x] - (2 - 2*I)/Sqrt[Sin[2*(c + d*x)]])*Sqrt[Sin[2*(c + d*x
)]])/(a*d*(I + Cot[c + d*x]))

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fricas [B]  time = 1.11, size = 467, normalized size = 2.34 \[ -\frac {{\left (a d \sqrt {-\frac {i}{4 \, a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left ({\left ({\left (4 i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - 4 i \, a d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {i}{4 \, a^{2} d^{2}}} + 2 i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - a d \sqrt {-\frac {i}{4 \, a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left ({\left ({\left (-4 i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + 4 i \, a d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {i}{4 \, a^{2} d^{2}}} + 2 i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + a d \sqrt {\frac {i}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-\frac {{\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} - a d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {i}{a^{2} d^{2}}} + 1\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a d}\right ) - a d \sqrt {\frac {i}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac {{\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} - a d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {i}{a^{2} d^{2}}} - 1\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a d}\right ) - \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} {\left (i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a d} \]

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)),x, algorithm="fricas")

[Out]

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

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

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)),x, algorithm="giac")

[Out]

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

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maple [C]  time = 1.27, size = 1836, normalized size = 9.18 \[ \text {result too large to display} \]

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)),x)

[Out]

-1/4/a/d*(-1+cos(d*x+c))*(-2*I*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)*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))+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)*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))+I*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)*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(
d*x+c))^(1/2),1/2*2^(1/2))+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)*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))-I*((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))+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)*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*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)*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))-sin(d*x+c)*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))-sin(d*x+c)*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)*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d
*x+c))^(1/2),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)*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))+I*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))*((-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)*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/
2*2^(1/2))+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)*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))/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))/si
n(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))/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*2^(1/2)-cos(d*x
+c)*2^(1/2))*cos(d*x+c)*(1+cos(d*x+c))^2/(I*sin(d*x+c)+cos(d*x+c))/(cos(d*x+c)/sin(d*x+c))^(3/2)/sin(d*x+c)^5*
2^(1/2)

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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)),x, algorithm="maxima")

[Out]

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

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

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)),x)

[Out]

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

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

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)),x)

[Out]

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

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