3.8.47 \(\int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^3} \, dx\) [747]

Optimal. Leaf size=267 \[ -\frac {i \text {ArcTan}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{8 \sqrt {2} a^3 d}+\frac {i \text {ArcTan}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{8 \sqrt {2} a^3 d}+\frac {\cot ^{\frac {3}{2}}(c+d x)}{6 d (i a+a \cot (c+d x))^3}+\frac {\sqrt {\cot (c+d x)}}{4 a d (i a+a \cot (c+d x))^2}+\frac {i \sqrt {\cot (c+d x)}}{4 d \left (i a^3+a^3 \cot (c+d x)\right )}+\frac {i \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{16 \sqrt {2} a^3 d}-\frac {i \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{16 \sqrt {2} a^3 d} \]

[Out]

1/6*cot(d*x+c)^(3/2)/d/(I*a+a*cot(d*x+c))^3+1/16*I*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))/a^3/d*2^(1/2)+1/16*I*ar
ctan(1+2^(1/2)*cot(d*x+c)^(1/2))/a^3/d*2^(1/2)+1/32*I*ln(1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(1/2))/a^3/d*2^(1/2)-
1/32*I*ln(1+cot(d*x+c)+2^(1/2)*cot(d*x+c)^(1/2))/a^3/d*2^(1/2)+1/4*cot(d*x+c)^(1/2)/a/d/(I*a+a*cot(d*x+c))^2+1
/4*I*cot(d*x+c)^(1/2)/d/(I*a^3+a^3*cot(d*x+c))

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Rubi [A]
time = 0.31, antiderivative size = 267, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3754, 3639, 3676, 3677, 12, 3557, 335, 303, 1176, 631, 210, 1179, 642} \begin {gather*} -\frac {i \text {ArcTan}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{8 \sqrt {2} a^3 d}+\frac {i \text {ArcTan}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{8 \sqrt {2} a^3 d}+\frac {i \sqrt {\cot (c+d x)}}{4 d \left (a^3 \cot (c+d x)+i a^3\right )}+\frac {i \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{16 \sqrt {2} a^3 d}-\frac {i \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{16 \sqrt {2} a^3 d}+\frac {\cot ^{\frac {3}{2}}(c+d x)}{6 d (a \cot (c+d x)+i a)^3}+\frac {\sqrt {\cot (c+d x)}}{4 a d (a \cot (c+d x)+i a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/8*I)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*a^3*d) + ((I/8)*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x
]]])/(Sqrt[2]*a^3*d) + Cot[c + d*x]^(3/2)/(6*d*(I*a + a*Cot[c + d*x])^3) + Sqrt[Cot[c + d*x]]/(4*a*d*(I*a + a*
Cot[c + d*x])^2) + ((I/4)*Sqrt[Cot[c + d*x]])/(d*(I*a^3 + a^3*Cot[c + d*x])) + ((I/16)*Log[1 - Sqrt[2]*Sqrt[Co
t[c + d*x]] + Cot[c + d*x]])/(Sqrt[2]*a^3*d) - ((I/16)*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(Sq
rt[2]*a^3*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 210

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

Rule 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 631

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 642

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 1176

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 1179

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 3557

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 3639

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

Rule 3676

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*b - a*B))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(2*a*f
*m)), x] + Dist[1/(2*a^2*m), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1)*Simp[A*(a*c*m + b*d
*n) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a*A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A
, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && GtQ[n, 0]

Rule 3677

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 3754

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}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^3} \, dx &=\int \frac {\cot ^{\frac {5}{2}}(c+d x)}{(i a+a \cot (c+d x))^3} \, dx\\ &=\frac {\cot ^{\frac {3}{2}}(c+d x)}{6 d (i a+a \cot (c+d x))^3}+\frac {\int \frac {\sqrt {\cot (c+d x)} \left (-\frac {3 i a}{2}+\frac {9}{2} a \cot (c+d x)\right )}{(i a+a \cot (c+d x))^2} \, dx}{6 a^2}\\ &=\frac {\cot ^{\frac {3}{2}}(c+d x)}{6 d (i a+a \cot (c+d x))^3}+\frac {\sqrt {\cot (c+d x)}}{4 a d (i a+a \cot (c+d x))^2}+\frac {\int \frac {-3 i a^2+9 a^2 \cot (c+d x)}{\sqrt {\cot (c+d x)} (i a+a \cot (c+d x))} \, dx}{24 a^4}\\ &=\frac {\cot ^{\frac {3}{2}}(c+d x)}{6 d (i a+a \cot (c+d x))^3}+\frac {\sqrt {\cot (c+d x)}}{4 a d (i a+a \cot (c+d x))^2}+\frac {i \sqrt {\cot (c+d x)}}{4 d \left (i a^3+a^3 \cot (c+d x)\right )}+\frac {\int -6 i a^3 \sqrt {\cot (c+d x)} \, dx}{48 a^6}\\ &=\frac {\cot ^{\frac {3}{2}}(c+d x)}{6 d (i a+a \cot (c+d x))^3}+\frac {\sqrt {\cot (c+d x)}}{4 a d (i a+a \cot (c+d x))^2}+\frac {i \sqrt {\cot (c+d x)}}{4 d \left (i a^3+a^3 \cot (c+d x)\right )}-\frac {i \int \sqrt {\cot (c+d x)} \, dx}{8 a^3}\\ &=\frac {\cot ^{\frac {3}{2}}(c+d x)}{6 d (i a+a \cot (c+d x))^3}+\frac {\sqrt {\cot (c+d x)}}{4 a d (i a+a \cot (c+d x))^2}+\frac {i \sqrt {\cot (c+d x)}}{4 d \left (i a^3+a^3 \cot (c+d x)\right )}+\frac {i \text {Subst}\left (\int \frac {\sqrt {x}}{1+x^2} \, dx,x,\cot (c+d x)\right )}{8 a^3 d}\\ &=\frac {\cot ^{\frac {3}{2}}(c+d x)}{6 d (i a+a \cot (c+d x))^3}+\frac {\sqrt {\cot (c+d x)}}{4 a d (i a+a \cot (c+d x))^2}+\frac {i \sqrt {\cot (c+d x)}}{4 d \left (i a^3+a^3 \cot (c+d x)\right )}+\frac {i \text {Subst}\left (\int \frac {x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{4 a^3 d}\\ &=\frac {\cot ^{\frac {3}{2}}(c+d x)}{6 d (i a+a \cot (c+d x))^3}+\frac {\sqrt {\cot (c+d x)}}{4 a d (i a+a \cot (c+d x))^2}+\frac {i \sqrt {\cot (c+d x)}}{4 d \left (i a^3+a^3 \cot (c+d x)\right )}-\frac {i \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{8 a^3 d}+\frac {i \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{8 a^3 d}\\ &=\frac {\cot ^{\frac {3}{2}}(c+d x)}{6 d (i a+a \cot (c+d x))^3}+\frac {\sqrt {\cot (c+d x)}}{4 a d (i a+a \cot (c+d x))^2}+\frac {i \sqrt {\cot (c+d x)}}{4 d \left (i a^3+a^3 \cot (c+d x)\right )}+\frac {i \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{16 a^3 d}+\frac {i \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{16 a^3 d}+\frac {i \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{16 \sqrt {2} a^3 d}+\frac {i \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{16 \sqrt {2} a^3 d}\\ &=\frac {\cot ^{\frac {3}{2}}(c+d x)}{6 d (i a+a \cot (c+d x))^3}+\frac {\sqrt {\cot (c+d x)}}{4 a d (i a+a \cot (c+d x))^2}+\frac {i \sqrt {\cot (c+d x)}}{4 d \left (i a^3+a^3 \cot (c+d x)\right )}+\frac {i \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{16 \sqrt {2} a^3 d}-\frac {i \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{16 \sqrt {2} a^3 d}+\frac {i \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{8 \sqrt {2} a^3 d}-\frac {i \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{8 \sqrt {2} a^3 d}\\ &=-\frac {i \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{8 \sqrt {2} a^3 d}+\frac {i \tan ^{-1}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{8 \sqrt {2} a^3 d}+\frac {\cot ^{\frac {3}{2}}(c+d x)}{6 d (i a+a \cot (c+d x))^3}+\frac {\sqrt {\cot (c+d x)}}{4 a d (i a+a \cot (c+d x))^2}+\frac {i \sqrt {\cot (c+d x)}}{4 d \left (i a^3+a^3 \cot (c+d x)\right )}+\frac {i \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{16 \sqrt {2} a^3 d}-\frac {i \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{16 \sqrt {2} a^3 d}\\ \end {align*}

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Mathematica [A]
time = 1.03, size = 231, normalized size = 0.87 \begin {gather*} \frac {\sqrt {\cot (c+d x)} \csc ^3(c+d x) \sec (c+d x) \left (-1+\cos (4 (c+d x))-6 i \cos (3 (c+d x)) \log \left (\cos (c+d x)+\sin (c+d x)+\sqrt {\sin (2 (c+d x))}\right ) \sqrt {\sin (2 (c+d x))}+6 i \sin (2 (c+d x))+6 i \text {ArcSin}(\cos (c+d x)-\sin (c+d x)) \sqrt {\sin (2 (c+d x))} (\cos (3 (c+d x))+i \sin (3 (c+d x)))+6 \log \left (\cos (c+d x)+\sin (c+d x)+\sqrt {\sin (2 (c+d x))}\right ) \sqrt {\sin (2 (c+d x))} \sin (3 (c+d x))+3 i \sin (4 (c+d x))\right )}{96 a^3 d (i+\cot (c+d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Cot[c + d*x]]*Csc[c + d*x]^3*Sec[c + d*x]*(-1 + Cos[4*(c + d*x)] - (6*I)*Cos[3*(c + d*x)]*Log[Cos[c + d*
x] + Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]]*Sqrt[Sin[2*(c + d*x)]] + (6*I)*Sin[2*(c + d*x)] + (6*I)*ArcSin[Cos
[c + d*x] - Sin[c + d*x]]*Sqrt[Sin[2*(c + d*x)]]*(Cos[3*(c + d*x)] + I*Sin[3*(c + d*x)]) + 6*Log[Cos[c + d*x]
+ Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]]*Sqrt[Sin[2*(c + d*x)]]*Sin[3*(c + d*x)] + (3*I)*Sin[4*(c + d*x)]))/(9
6*a^3*d*(I + Cot[c + d*x])^3)

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 14.03, size = 9445, normalized size = 35.37

method	result	size
default	\(\text {Expression too large to display}\)	\(9445\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cot(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^3,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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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 518 vs. \(2 (206) = 412\).
time = 1.26, size = 518, normalized size = 1.94 \begin {gather*} -\frac {{\left (12 \, a^{3} d \sqrt {\frac {i}{64 \, a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (2 \, {\left (8 \, {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{3} 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^{6} d^{2}}} + i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - 12 \, a^{3} d \sqrt {\frac {i}{64 \, a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-2 \, {\left (8 \, {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{3} 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^{6} d^{2}}} - i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - 12 \, a^{3} d \sqrt {-\frac {i}{64 \, a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac {{\left (8 \, {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{3} 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^{6} d^{2}}} + i\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{3} d}\right ) + 12 \, a^{3} d \sqrt {-\frac {i}{64 \, a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-\frac {{\left (8 \, {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{3} 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^{6} d^{2}}} - i\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{3} 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 \, e^{\left (6 i \, d x + 6 i \, c\right )} + e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{48 \, a^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(12*a^3*d*sqrt(1/64*I/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(2*(8*(a^3*d*e^(2*I*d*x + 2*I*c) - a^3*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^6*d^2)) + I*e^(2*I*d*x + 2*I*c))*e^(-2*I*
d*x - 2*I*c)) - 12*a^3*d*sqrt(1/64*I/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(-2*(8*(a^3*d*e^(2*I*d*x + 2*I*c) - a^3
*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^6*d^2)) - I*e^(2*I*d*x + 2*I*c)
)*e^(-2*I*d*x - 2*I*c)) - 12*a^3*d*sqrt(-1/64*I/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(1/8*(8*(a^3*d*e^(2*I*d*x +
2*I*c) - a^3*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^6*d^2)) + I)*e^(-2
*I*d*x - 2*I*c)/(a^3*d)) + 12*a^3*d*sqrt(-1/64*I/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(-1/8*(8*(a^3*d*e^(2*I*d*x
+ 2*I*c) - a^3*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^6*d^2)) - I)*e^(
-2*I*d*x - 2*I*c)/(a^3*d)) - sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*(2*e^(6*I*d*x + 6*I*c
) + e^(4*I*d*x + 4*I*c) - 2*e^(2*I*d*x + 2*I*c) - 1))*e^(-6*I*d*x - 6*I*c)/(a^3*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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