3.8.84 \(\int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx\) [784]

Optimal. Leaf size=188 \[ -\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{a^{5/2} d}+\frac {i}{5 d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {i}{10 a d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}-\frac {i}{20 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}} \]

[Out]

(-1/8-1/8*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^(5/2)/d-1/20*I/a^2/d/cot(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2)+1/5*I/d/cot(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))
^(5/2)+1/10*I/a/d/cot(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]
time = 0.33, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {4326, 3638, 3677, 12, 3625, 211} \begin {gather*} -\frac {\left (\frac {1}{8}+\frac {i}{8}\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^{5/2} d}-\frac {i}{20 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {i}{10 a d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {i}{5 d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/8 - I/8)*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^(5/2)*d) + (I/5)/(d*Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2)) + (I/10)/(a*d*Sqrt[Cot
[c + d*x]]*(a + I*a*Tan[c + d*x])^(3/2)) - (I/20)/(a^2*d*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])

Rule 12

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

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 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 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 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 \frac {1}{\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 {\sqrt {\tan (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx\\ &=\frac {i}{5 d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}}-\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {i a-4 a \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}} \, dx}{10 a^2}\\ &=\frac {i}{5 d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {i}{10 a d \sqrt {\cot (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 {\frac {9 i a^2}{2}-3 a^2 \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx}{30 a^4}\\ &=\frac {i}{5 d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {i}{10 a d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}-\frac {i}{20 a^2 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 {15 i a^3 \sqrt {a+i a \tan (c+d x)}}{4 \sqrt {\tan (c+d x)}} \, dx}{30 a^6}\\ &=\frac {i}{5 d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {i}{10 a d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}-\frac {i}{20 a^2 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 {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{8 a^3}\\ &=\frac {i}{5 d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {i}{10 a d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}-\frac {i}{20 a^2 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 ) \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 )}{4 a d}\\ &=-\frac {\left (\frac {1}{8}+\frac {i}{8}\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^{5/2} d}+\frac {i}{5 d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {i}{10 a d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}-\frac {i}{20 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 1.77, size = 167, normalized size = 0.89 \begin {gather*} \frac {e^{-6 i (c+d x)} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (-1-2 e^{2 i (c+d x)}+2 e^{4 i (c+d x)}+e^{6 i (c+d x)}-5 e^{5 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 )\right ) \sqrt {\cot (c+d x)}}{20 \sqrt {2} a^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(-1 - 2*E^((2*I)*(c + d*x)) + 2*E^((4*I)*(c + d*x)) +
E^((6*I)*(c + d*x)) - 5*E^((5*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))]])*Sqrt[Cot[c + d*x]])/(20*Sqrt[2]*a^3*d*E^((6*I)*(c + d*x)))

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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. 542 vs. \(2 (147 ) = 294\).
time = 47.20, size = 543, normalized size = 2.89

method	result	size
default	\(\frac {\left (-\frac {1}{40}+\frac {i}{40}\right ) \left (20 i \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right ) \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-10 i \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {2}\, \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right )-4 i \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+20 \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right ) \sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right )-5 i \sin \left (d x +c \right ) \sqrt {2}\, \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right )+4 \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-10 \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}\, \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right )-i \sin \left (d x +c \right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}-15 \cos \left (d x +c \right ) \sqrt {2}\, \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right )+\sin \left (d x +c \right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}+5 \sqrt {2}\, \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right )\right ) \cos \left (d x +c \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 )}}}{d \left (4 i \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+4 \left (\cos ^{3}\left (d x +c \right )\right )-i \sin \left (d x +c \right )-3 \cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, a^{3}}\)	\(543\)

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))^(5/2),x,method=_RETURNVERBOSE)

[Out]

(-1/40+1/40*I)/d*(20*I*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)*cos(d
*x+c)^2-10*I*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)*cos(d*x+c)-4*I*
sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2+20*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^
(1/2)*2^(1/2))*2^(1/2)*cos(d*x+c)^3-5*I*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)+4*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)-10*cos(d*x+c)^2*2^(1/2)*arctan((1/2+1
/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))-I*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-15*cos(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))+sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c
))^(1/2)+5*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)*(a*(I*sin(d*x+c)
+cos(d*x+c))/cos(d*x+c))^(1/2)/(4*I*cos(d*x+c)^2*sin(d*x+c)+4*cos(d*x+c)^3-I*sin(d*x+c)-3*cos(d*x+c))/(cos(d*x
+c)/sin(d*x+c))^(1/2)/sin(d*x+c)/((-1+cos(d*x+c))/sin(d*x+c))^(1/2)/a^3

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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))^(5/2),x, algorithm="maxima")

[Out]

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

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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. 366 vs. \(2 (136) = 272\).
time = 0.69, size = 366, normalized size = 1.95 \begin {gather*} -\frac {{\left (10 \, a^{3} d \sqrt {\frac {i}{8 \, a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (4 \, {\left (2 \, \sqrt {2} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{3} 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}{8 \, a^{5} d^{2}}} + i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 10 \, a^{3} d \sqrt {\frac {i}{8 \, a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-4 \, {\left (2 \, \sqrt {2} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{3} 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}{8 \, a^{5} d^{2}}} - i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - 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 (e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, 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 (-5 i \, d x - 5 i \, c\right )}}{40 \, 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))^(5/2),x, algorithm="fricas")

[Out]

-1/40*(10*a^3*d*sqrt(1/8*I/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(4*(2*sqrt(2)*(a^3*d*e^(2*I*d*x + 2*I*c) - a^3*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(1/8*I/(a^5
*d^2)) + I*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 10*a^3*d*sqrt(1/8*I/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(-4*(2
*sqrt(2)*(a^3*d*e^(2*I*d*x + 2*I*c) - a^3*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(1/8*I/(a^5*d^2)) - I*a*e^(I*d*x + I*c))*e^(-I*d*x - 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))*(e^(6*I*d*x + 6*I*c) + 2
*e^(4*I*d*x + 4*I*c) - 2*e^(2*I*d*x + 2*I*c) - 1))*e^(-5*I*d*x - 5*I*c)/(a^3*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}} \sqrt {\cot {\left (c + d x \right )}}}\, dx \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))**(5/2),x)

[Out]

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

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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))^(5/2),x, algorithm="giac")

[Out]

integrate(1/((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 \frac {1}{\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(1/(cot(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)^(5/2)),x)

[Out]

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

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