3.779 \(\int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=221 \[ \frac {2 (-1)^{3/4} \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}-\frac {1}{3 d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {3 i}{2 a d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.61, antiderivative size = 221, 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 = {4241, 3558, 3595, 3601, 3544, 205, 3599, 63, 217, 203} \[ \frac {2 (-1)^{3/4} \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}-\frac {1}{3 d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {3 i}{2 a d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 63

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 203

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

Rule 205

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

Rule 217

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 3544

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 3558

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[((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 3595

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 3599

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 3601

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(A*b + a*B)/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x]
, x] - Dist[B/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[e + f*x]), 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] && NeQ[A*b + a*B, 0]

Rule 4241

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 {1}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx &=\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\\ &=-\frac {1}{3 d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}-\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {\tan (c+d x)} \left (-\frac {3 a}{2}+3 i a \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx}{3 a^2}\\ &=-\frac {1}{3 d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {3 i}{2 a d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {9 i a^2}{4}-3 a^2 \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{3 a^4}\\ &=-\frac {1}{3 d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {3 i}{2 a d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {\left (i \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {(a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{a^3}+\frac {\left (i \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}{4 a^2}\\ &=-\frac {1}{3 d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {3 i}{2 a d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \operatorname {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 )}{2 d}-\frac {\left (i \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{a d}\\ &=\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) \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)}}{a^{3/2} d}-\frac {1}{3 d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {3 i}{2 a d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {\left (2 i \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+i a x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{a d}\\ &=\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) \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)}}{a^{3/2} d}-\frac {1}{3 d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {3 i}{2 a d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {\left (2 i \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \operatorname {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 )}{a d}\\ &=\frac {2 (-1)^{3/4} \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)}}{a^{3/2} d}+\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) \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)}}{a^{3/2} d}-\frac {1}{3 d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {3 i}{2 a d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 2.50, size = 226, normalized size = 1.02 \[ \frac {e^{-4 i (c+d x)} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {\cot (c+d x)} \left (-11 e^{2 i (c+d x)}+10 e^{4 i (c+d x)}+3 e^{3 i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )-12 \sqrt {2} e^{3 i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )+1\right )}{6 \sqrt {2} a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(1 - 11*E^((2*I)*(c + d*x)) + 10*E^((4*I)*(c + d*x))
+ 3*E^((3*I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]]
- 12*Sqrt[2]*E^((3*I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]*ArcTanh[(Sqrt[2]*E^(I*(c + d*x)))/Sqrt[-1 + E
^((2*I)*(c + d*x))]])*Sqrt[Cot[c + d*x]])/(6*Sqrt[2]*a^2*d*E^((4*I)*(c + d*x)))

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fricas [B]  time = 0.62, size = 648, normalized size = 2.93 \[ \frac {{\left (3 \, a^{2} d \sqrt {\frac {i}{2 \, a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2} 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 {i}{2 \, a^{3} d^{2}}} + i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 3 \, a^{2} d \sqrt {\frac {i}{2 \, a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2} 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 {i}{2 \, a^{3} d^{2}}} - i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 3 \, a^{2} d \sqrt {\frac {4 i}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left ({\left (\sqrt {2} {\left (16 i \, a^{3} d e^{\left (3 i \, d x + 3 i \, c\right )} - 16 i \, a^{3} 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 {4 i}{a^{3} d^{2}}} - 48 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 16 \, a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + 3 \, a^{2} d \sqrt {\frac {4 i}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left ({\left (\sqrt {2} {\left (-16 i \, a^{3} d e^{\left (3 i \, d x + 3 i \, c\right )} + 16 i \, a^{3} 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 {4 i}{a^{3} d^{2}}} - 48 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 16 \, a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + \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 (10 \, e^{\left (4 i \, d x + 4 i \, c\right )} - 11 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{12 \, a^{2} d} \]

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))^(3/2),x, algorithm="fricas")

[Out]

1/12*(3*a^2*d*sqrt(1/2*I/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*log(4*(sqrt(2)*(a^2*d*e^(2*I*d*x + 2*I*c) - a^2*d)*sqr
t(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(1/2*I/(a^3*d^2
)) + I*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 3*a^2*d*sqrt(1/2*I/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*log(-4*(sqrt(2
)*(a^2*d*e^(2*I*d*x + 2*I*c) - a^2*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(1/2*I/(a^3*d^2)) - I*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 3*a^2*d*sqrt(4*I/(a^3*d
^2))*e^(3*I*d*x + 3*I*c)*log((sqrt(2)*(16*I*a^3*d*e^(3*I*d*x + 3*I*c) - 16*I*a^3*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(4*I/(a^3*d^2)) - 48*a
^2*e^(2*I*d*x + 2*I*c) + 16*a^2)*e^(-2*I*d*x - 2*I*c)) + 3*a^2*d*sqrt(4*I/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*log((
sqrt(2)*(-16*I*a^3*d*e^(3*I*d*x + 3*I*c) + 16*I*a^3*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(4*I/(a^3*d^2)) - 48*a^2*e^(2*I*d*x + 2*I*c) + 16*a
^2)*e^(-2*I*d*x - 2*I*c)) + 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))*(10*e^(4*I*d*x + 4*I*c) - 11*e^(2*I*d*x + 2*I*c) + 1))*e^(-3*I*d*x - 3*I*c)/(a^2*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cot \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

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))^(3/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 2.26, size = 1111, normalized size = 5.03 \[ \text {result too large to display} \]

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))^(3/2),x)

[Out]

(-1/12-1/12*I)/d*(3*sin(d*x+c)*2^(1/2)*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))-3*I*2^(1
/2)*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))-11*I*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-6*c
os(d*x+c)*sin(d*x+c)*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))*2^(1/2)+12*I*ln(((-1+cos(d
*x+c))/sin(d*x+c))^(1/2)+I)*cos(d*x+c)*sin(d*x+c)-12*I*cos(d*x+c)*sin(d*x+c)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(
1/2)-1)+12*I*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*cos(d*x+c)*sin(d*x+c)-12*I*cos(d*x+c)*sin(d*x+c)*ln(((-1
+cos(d*x+c))/sin(d*x+c))^(1/2)-I)-9*cos(d*x+c)*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+6*I*2^(1/2)*arcta
n((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))*cos(d*x+c)^2-3*I*2^(1/2)*arctan((1/2+1/2*I)*((-1+cos
(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))*cos(d*x+c)+6*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-I)-6*ln(((-1+cos(d*x+c)
)/sin(d*x+c))^(1/2)+I)-11*cos(d*x+c)^2*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+12*ln(((-1+cos(d*x+c))/sin(d*x+c))^(
1/2)+I)*cos(d*x+c)^2+12*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*cos(d*x+c)^2-12*ln(((-1+cos(d*x+c))/sin(d*x+c
))^(1/2)-1)*cos(d*x+c)^2-12*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-I)*cos(d*x+c)^2-6*cos(d*x+c)*ln(((-1+cos(d*x
+c))/sin(d*x+c))^(1/2)+I)-6*cos(d*x+c)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)+6*cos(d*x+c)*ln(((-1+cos(d*x+c
))/sin(d*x+c))^(1/2)-1)+6*cos(d*x+c)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-I)+11*((-1+cos(d*x+c))/sin(d*x+c))^
(1/2)-9*I*cos(d*x+c)*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+6*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)-
6*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)+11*I*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2-6*I*ln(((-1+co
s(d*x+c))/sin(d*x+c))^(1/2)+I)*sin(d*x+c)+6*I*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*sin(d*x+c)-6*I*ln(((-1+
cos(d*x+c))/sin(d*x+c))^(1/2)+1)*sin(d*x+c)+6*I*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-I)*sin(d*x+c))*cos(d*x+c
)^3*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/(2*I*cos(d*x+c)*sin(d*x+c)+2*cos(d*x+c)^2-1)/((-1+cos(d*x+c
))/sin(d*x+c))^(1/2)/(cos(d*x+c)/sin(d*x+c))^(5/2)/sin(d*x+c)^3/a^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)^(5/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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

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)^(3/2)),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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))**(3/2),x)

[Out]

Timed out

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