3.8.49 \(\int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx\) [749]

Optimal. Leaf size=222 \[ -\frac {\text {ArcTan}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{8 \sqrt {2} a^3 d}+\frac {\text {ArcTan}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{8 \sqrt {2} a^3 d}+\frac {i \sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {\sqrt {\cot (c+d x)}}{12 a d (i a+a \cot (c+d x))^2}-\frac {\log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{16 \sqrt {2} a^3 d}+\frac {\log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{16 \sqrt {2} a^3 d} \]

[Out]

1/16*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))/a^3/d*2^(1/2)+1/16*arctan(1+2^(1/2)*cot(d*x+c)^(1/2))/a^3/d*2^(1/2)-1
/32*ln(1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(1/2))/a^3/d*2^(1/2)+1/32*ln(1+cot(d*x+c)+2^(1/2)*cot(d*x+c)^(1/2))/a^3
/d*2^(1/2)+1/6*I*cot(d*x+c)^(1/2)/d/(I*a+a*cot(d*x+c))^3+1/12*cot(d*x+c)^(1/2)/a/d/(I*a+a*cot(d*x+c))^2

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Rubi [A]
time = 0.21, antiderivative size = 222, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 12, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3754, 3638, 3677, 21, 3557, 335, 217, 1179, 642, 1176, 631, 210} \begin {gather*} -\frac {\text {ArcTan}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{8 \sqrt {2} a^3 d}+\frac {\text {ArcTan}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{8 \sqrt {2} a^3 d}-\frac {\log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{16 \sqrt {2} a^3 d}+\frac {\log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{16 \sqrt {2} a^3 d}+\frac {\sqrt {\cot (c+d x)}}{12 a d (a \cot (c+d x)+i a)^2}+\frac {i \sqrt {\cot (c+d x)}}{6 d (a \cot (c+d x)+i a)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Cot[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^3),x]

[Out]

-1/8*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]]/(Sqrt[2]*a^3*d) + ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]]/(8*Sqrt[2
]*a^3*d) + ((I/6)*Sqrt[Cot[c + d*x]])/(d*(I*a + a*Cot[c + d*x])^3) + Sqrt[Cot[c + d*x]]/(12*a*d*(I*a + a*Cot[c
+ d*x])^2) - Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]]/(16*Sqrt[2]*a^3*d) + Log[1 + Sqrt[2]*Sqrt[Cot
[c + d*x]] + Cot[c + d*x]]/(16*Sqrt[2]*a^3*d)

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 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 217

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{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 3638

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp
[(-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 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}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx &=\int \frac {\sqrt {\cot (c+d x)}}{(i a+a \cot (c+d x))^3} \, dx\\ &=\frac {i \sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {\int \frac {a-5 i a \cot (c+d x)}{\sqrt {\cot (c+d x)} (i a+a \cot (c+d x))^2} \, dx}{12 a^2}\\ &=\frac {i \sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {\sqrt {\cot (c+d x)}}{12 a d (i a+a \cot (c+d x))^2}+\frac {\int \frac {-6 i a^2-6 a^2 \cot (c+d x)}{\sqrt {\cot (c+d x)} (i a+a \cot (c+d x))} \, dx}{48 a^4}\\ &=\frac {i \sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {\sqrt {\cot (c+d x)}}{12 a d (i a+a \cot (c+d x))^2}-\frac {\int \frac {1}{\sqrt {\cot (c+d x)}} \, dx}{8 a^3}\\ &=\frac {i \sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {\sqrt {\cot (c+d x)}}{12 a d (i a+a \cot (c+d x))^2}+\frac {\text {Subst}\left (\int \frac {1}{\sqrt {x} \left (1+x^2\right )} \, dx,x,\cot (c+d x)\right )}{8 a^3 d}\\ &=\frac {i \sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {\sqrt {\cot (c+d x)}}{12 a d (i a+a \cot (c+d x))^2}+\frac {\text {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{4 a^3 d}\\ &=\frac {i \sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {\sqrt {\cot (c+d x)}}{12 a d (i a+a \cot (c+d x))^2}+\frac {\text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{8 a^3 d}+\frac {\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 \sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {\sqrt {\cot (c+d x)}}{12 a d (i a+a \cot (c+d x))^2}+\frac {\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 {\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 {\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 {\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 \sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {\sqrt {\cot (c+d x)}}{12 a d (i a+a \cot (c+d x))^2}-\frac {\log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{16 \sqrt {2} a^3 d}+\frac {\log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{16 \sqrt {2} a^3 d}+\frac {\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 {\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 {\tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{8 \sqrt {2} a^3 d}+\frac {\tan ^{-1}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{8 \sqrt {2} a^3 d}+\frac {i \sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {\sqrt {\cot (c+d x)}}{12 a d (i a+a \cot (c+d x))^2}-\frac {\log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{16 \sqrt {2} a^3 d}+\frac {\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 = 2.06, size = 224, normalized size = 1.01 \begin {gather*} \frac {(\cos (3 (c+d x))-i \sin (3 (c+d x))) \left (6 i \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right ) (\cos (3 (c+d x))+i \sin (3 (c+d x)))+6 \text {ArcTan}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right ) (-i \cos (3 (c+d x))+\sin (3 (c+d x)))+(3 i \cos (c+d x)-3 i \cos (3 (c+d x))+\sin (c+d x)+\sin (3 (c+d x))) \sqrt {i \tan (c+d x)}\right )}{48 a^3 d \sqrt {\cot (c+d x)} \sqrt {i \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Cot[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^3),x]

[Out]

((Cos[3*(c + d*x)] - I*Sin[3*(c + d*x)])*((6*I)*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x
)))]]*(Cos[3*(c + d*x)] + I*Sin[3*(c + d*x)]) + 6*ArcTan[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*
x)))]]*((-I)*Cos[3*(c + d*x)] + Sin[3*(c + d*x)]) + ((3*I)*Cos[c + d*x] - (3*I)*Cos[3*(c + d*x)] + Sin[c + d*x
] + Sin[3*(c + d*x)])*Sqrt[I*Tan[c + d*x]]))/(48*a^3*d*Sqrt[Cot[c + d*x]]*Sqrt[I*Tan[c + d*x]])

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 14.42, size = 7710, normalized size = 34.73

method	result	size
default	\(\text {Expression too large to display}\)	\(7710\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cot(d*x+c)^(5/2)/(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)^(5/2)/(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. 520 vs. \(2 (175) = 350\).
time = 1.13, size = 520, normalized size = 2.34 \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 )} - 5 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, 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)^(5/2)/(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)
- 5*e^(4*I*d*x + 4*I*c) + 4*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(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cot(d*x+c)**(5/2)/(a+I*a*tan(d*x+c))**3,x)

[Out]

Timed out

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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)^(5/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

integrate(1/((I*a*tan(d*x + c) + a)^3*cot(d*x + c)^(5/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\mathrm {cot}\left (c+d\,x\right )}^{5/2}\,{\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)^(5/2)*(a + a*tan(c + d*x)*1i)^3),x)

[Out]

int(1/(cot(c + d*x)^(5/2)*(a + a*tan(c + d*x)*1i)^3), x)

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