3.8.66 \(\int \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2} \, dx\) [766]

Optimal. Leaf size=179 \[ \frac {5 \sqrt [4]{-1} a^{5/2} \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {(4-4 i) a^{5/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {a^2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\cot (c+d x)}} \]

[Out]

5*(-1)^(1/4)*a^(5/2)*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)/d+(4-4*I)*a^(5/2)*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)/d-a^2*(a+I*a*tan(d*x+c))^(1/2)/d/cot(d*x+c)^(1/2)

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Rubi [A]
time = 0.31, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {4326, 3637, 3682, 3625, 211, 3680, 65, 223, 209} \begin {gather*} \frac {5 \sqrt [4]{-1} a^{5/2} \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {(4-4 i) a^{5/2} \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 )}{d}-\frac {a^2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\cot (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*(-1)^(1/4)*a^(5/2)*ArcTan[((-1)^(3/4)*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]])/d + ((4 - 4*I)*a^(5/2)*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]])/d - (a^2*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[Cot[c + d*x]])

Rule 65

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 209

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

Rule 211

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

Rule 223

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 3625

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 3637

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Dist[a/(d*(m + n - 1
)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) +
b*d*(3*m + 2*n - 4))*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] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (IntegerQ[m] || Intege
rsQ[2*m, 2*n])

Rule 3680

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 3682

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 4326

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 \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2} \, dx &=\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {(a+i a \tan (c+d x))^{5/2}}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {a^2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\cot (c+d x)}}+\left (a \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {3 a}{2}+\frac {5}{2} i a \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {a^2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\cot (c+d x)}}-\frac {1}{2} \left (5 a \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+\left (4 a^2 \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\\ &=-\frac {a^2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\cot (c+d x)}}-\frac {\left (5 a^3 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{2 d}-\frac {\left (8 i a^4 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {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 )}{d}\\ &=\frac {(4-4 i) a^{5/2} \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)}}{d}-\frac {a^2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\cot (c+d x)}}-\frac {\left (5 a^3 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x^2}} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=\frac {(4-4 i) a^{5/2} \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)}}{d}-\frac {a^2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\cot (c+d x)}}-\frac {\left (5 a^3 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {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 )}{d}\\ &=\frac {5 \sqrt [4]{-1} a^{5/2} \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)}}{d}+\frac {(4-4 i) a^{5/2} \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)}}{d}-\frac {a^2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\cot (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 2.41, size = 241, normalized size = 1.35 \begin {gather*} \frac {2 i \sqrt {2} a^2 e^{i (c+d x)} \left (\sqrt {2} e^{i (c+d x)} \left (-1+e^{2 i (c+d x)}\right )-4 \sqrt {2} \sqrt {-1+e^{2 i (c+d x)}} \left (1+e^{2 i (c+d x)}\right ) \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )+5 \sqrt {-1+e^{2 i (c+d x)}} \left (1+e^{2 i (c+d x)}\right ) \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right ) \cos ^2(c+d x) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d \left (1+e^{2 i (c+d x)}\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*I)*Sqrt[2]*a^2*E^(I*(c + d*x))*(Sqrt[2]*E^(I*(c + d*x))*(-1 + E^((2*I)*(c + d*x))) - 4*Sqrt[2]*Sqrt[-1 + E
^((2*I)*(c + d*x))]*(1 + E^((2*I)*(c + d*x)))*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]] + 5*Sqrt
[-1 + E^((2*I)*(c + d*x))]*(1 + E^((2*I)*(c + d*x)))*ArcTanh[(Sqrt[2]*E^(I*(c + d*x)))/Sqrt[-1 + E^((2*I)*(c +
d*x))]])*Cos[c + d*x]^2*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(d*(1 + E^((2*I)*(c + d*x)))^3)

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 739 vs. \(2 (144 ) = 288\).
time = 44.74, size = 740, normalized size = 4.13

method	result	size
default	\(-\frac {\left (-1+\cos \left (d x +c \right )\right ) \left (5 i \sqrt {2}\, \ln \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}-1\right ) \cos \left (d x +c \right )-10 i \sqrt {2}\, \arctan \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\right ) \cos \left (d x +c \right )-5 i \ln \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}+1\right ) \cos \left (d x +c \right ) \sqrt {2}-2 i \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \cos \left (d x +c \right ) \sqrt {2}+2 \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\, \sin \left (d x +c \right )+16 i \arctan \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}+1\right ) \cos \left (d x +c \right )+16 i \arctan \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}-1\right ) \cos \left (d x +c \right )-5 \sqrt {2}\, \ln \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}-1\right ) \cos \left (d x +c \right )-10 \sqrt {2}\, \arctan \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\right ) \cos \left (d x +c \right )+5 \sqrt {2}\, \ln \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}+1\right ) \cos \left (d x +c \right )+8 i \ln \left (-\frac {\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\, \sin \left (d x +c \right )+\sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\, \sin \left (d x +c \right )-\sin \left (d x +c \right )-\cos \left (d x +c \right )+1}\right ) \cos \left (d x +c \right )-2 i \sqrt {2}\, \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}+16 \arctan \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}+1\right ) \cos \left (d x +c \right )+16 \arctan \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}-1\right ) \cos \left (d x +c \right )+8 \ln \left (-\frac {\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\, \sin \left (d x +c \right )-\sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\, \sin \left (d x +c \right )+\sin \left (d x +c \right )+\cos \left (d x +c \right )-1}\right ) \cos \left (d x +c \right )\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\, a^{2}}{4 d \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )-1\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \cos \left (d x +c \right )}\)	\(740\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/4/d*(-1+cos(d*x+c))*(5*I*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*cos(d*x+c)*2^(1/2)-10*I*arctan(((-1+cos(d
*x+c))/sin(d*x+c))^(1/2))*cos(d*x+c)*2^(1/2)-5*I*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*cos(d*x+c)*2^(1/2)-2
*I*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)*2^(1/2)+2*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+
c)+16*I*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)+1)*cos(d*x+c)+16*I*arctan(((-1+cos(d*x+c))/sin(d*x+c
))^(1/2)*2^(1/2)-1)*cos(d*x+c)-5*2^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*cos(d*x+c)-10*2^(1/2)*arctan
(((-1+cos(d*x+c))/sin(d*x+c))^(1/2))*cos(d*x+c)+5*2^(1/2)*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*cos(d*x+c)+
8*I*ln(-(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+sin(d*x+c)+cos(d*x+c)-1)/(((-1+cos(d*x+c))/sin(
d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)-sin(d*x+c)-cos(d*x+c)+1))*cos(d*x+c)-2*I*2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))
^(1/2)+16*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)+1)*cos(d*x+c)+16*arctan(((-1+cos(d*x+c))/sin(d*x+c
))^(1/2)*2^(1/2)-1)*cos(d*x+c)+8*ln(-(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)-sin(d*x+c)-cos(d*x
+c)+1)/(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+sin(d*x+c)+cos(d*x+c)-1))*cos(d*x+c))*(a*(I*sin(
d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(cos(d*x+c)/sin(d*x+c))^(1/2)/(I*sin(d*x+c)+cos(d*x+c)-1)/((-1+cos(d*x+c)
)/sin(d*x+c))^(1/2)/cos(d*x+c)*2^(1/2)*a^2

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

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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. 648 vs. \(2 (137) = 274\).
time = 0.77, size = 648, normalized size = 3.62 \begin {gather*} -\frac {4 \, \sqrt {2} {\left (-i \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + i \, a^{2} 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 {25 i \, a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac {16 \, {\left (15 \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 5 \, a^{3} + 2 \, \sqrt {2} \sqrt {-\frac {25 i \, a^{5}}{d^{2}}} {\left (d e^{\left (3 i \, d x + 3 i \, c\right )} - 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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{5 \, a}\right ) - \sqrt {-\frac {25 i \, a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac {16 \, {\left (15 \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 5 \, a^{3} - 2 \, \sqrt {2} \sqrt {-\frac {25 i \, a^{5}}{d^{2}}} {\left (d e^{\left (3 i \, d x + 3 i \, c\right )} - 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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{5 \, a}\right ) - \sqrt {-\frac {128 i \, a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (16 i \, a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} \sqrt {-\frac {128 i \, a^{5}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, 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}}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a^{2}}\right ) + \sqrt {-\frac {128 i \, a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (16 i \, a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} \sqrt {-\frac {128 i \, a^{5}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, 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}}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a^{2}}\right )}{4 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-1/4*(4*sqrt(2)*(-I*a^2*e^(3*I*d*x + 3*I*c) + I*a^2*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(-25*I*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log(-16
/5*(15*a^3*e^(2*I*d*x + 2*I*c) - 5*a^3 + 2*sqrt(2)*sqrt(-25*I*a^5/d^2)*(d*e^(3*I*d*x + 3*I*c) - 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)))*e^(-2*I*d*
x - 2*I*c)/a) - sqrt(-25*I*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log(-16/5*(15*a^3*e^(2*I*d*x + 2*I*c) - 5*a^3
- 2*sqrt(2)*sqrt(-25*I*a^5/d^2)*(d*e^(3*I*d*x + 3*I*c) - 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)))*e^(-2*I*d*x - 2*I*c)/a) - sqrt(-128*I*a^5/d^2)*(d
*e^(2*I*d*x + 2*I*c) + d)*log(1/4*(16*I*a^3*e^(I*d*x + I*c) + sqrt(2)*sqrt(-128*I*a^5/d^2)*(I*d*e^(2*I*d*x + 2
*I*c) - I*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)))*e^
(-I*d*x - I*c)/a^2) + sqrt(-128*I*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log(1/4*(16*I*a^3*e^(I*d*x + I*c) + sqr
t(2)*sqrt(-128*I*a^5/d^2)*(-I*d*e^(2*I*d*x + 2*I*c) + I*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)))*e^(-I*d*x - I*c)/a^2))/(d*e^(2*I*d*x + 2*I*c) + d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4369 deep

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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(cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sqrt {\mathrm {cot}\left (c+d\,x\right )}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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