3.768 \(\int \frac {\cot ^{\frac {5}{2}}(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=181 \[ -\frac {5 \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 a d}+\frac {\cot ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {7 i \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{3 a d}-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d} \]

[Out]

(-1/2+1/2*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
)/d/a^(1/2)+cot(d*x+c)^(3/2)/d/(a+I*a*tan(d*x+c))^(1/2)-5/3*cot(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(1/2)/a/d+7/3*
I*cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/a/d

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Rubi [A]  time = 0.47, antiderivative size = 181, 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 = {4241, 3559, 3598, 12, 3544, 205} \[ -\frac {5 \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 a d}+\frac {\cot ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {7 i \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{3 a d}-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/2 + I/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]])/(Sqrt[a]*d) + Cot[c + d*x]^(3/2)/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (((7*I)/3)*Sqrt[Cot[c + d*x]
]*Sqrt[a + I*a*Tan[c + d*x]])/(a*d) - (5*Cot[c + d*x]^(3/2)*Sqrt[a + I*a*Tan[c + d*x]])/(3*a*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 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 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 3559

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(a*(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*Simp[b*c*m - a*d*(2*m + n + 1) + b*d*(m + n + 1)*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] && LtQ[m, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3598

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

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 {\cot ^{\frac {5}{2}}(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx &=\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {1}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx\\ &=\frac {\cot ^{\frac {3}{2}}(c+d x)}{d \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 {5 a}{2}-2 i a \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx}{a^2}\\ &=\frac {\cot ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {5 \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 a d}+\frac {\left (2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {7 i a^2}{4}-\frac {5}{2} a^2 \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{3 a^3}\\ &=\frac {\cot ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {7 i \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{3 a d}-\frac {5 \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 a d}+\frac {\left (4 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int -\frac {3 a^3 \sqrt {a+i a \tan (c+d x)}}{8 \sqrt {\tan (c+d x)}} \, dx}{3 a^4}\\ &=\frac {\cot ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {7 i \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{3 a d}-\frac {5 \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 a d}-\frac {\left (\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}{2 a}\\ &=\frac {\cot ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {7 i \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{3 a d}-\frac {5 \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 a d}+\frac {\left (i a \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 )}{d}\\ &=-\frac {\left (\frac {1}{2}-\frac {i}{2}\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)}}{\sqrt {a} d}+\frac {\cot ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {7 i \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{3 a d}-\frac {5 \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 a d}\\ \end {align*}

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Mathematica [A]  time = 1.40, size = 159, normalized size = 0.88 \[ \frac {i \sqrt {\cot (c+d x)} \left (-18 e^{2 i (c+d x)}+7 e^{4 i (c+d x)}+3 e^{i (c+d x)} \left (-1+e^{2 i (c+d x)}\right )^{3/2} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )+3\right )}{3 \sqrt {2} d \left (-1+e^{4 i (c+d x)}\right ) \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/3)*(3 - 18*E^((2*I)*(c + d*x)) + 7*E^((4*I)*(c + d*x)) + 3*E^(I*(c + d*x))*(-1 + E^((2*I)*(c + d*x)))^(3/2
)*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])*Sqrt[Cot[c + d*x]])/(Sqrt[2]*d*Sqrt[(a*E^((2*I)*(c
+ d*x)))/(1 + E^((2*I)*(c + d*x)))]*(-1 + E^((4*I)*(c + d*x))))

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fricas [B]  time = 1.26, size = 385, normalized size = 2.13 \[ \frac {\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} {\left (14 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 36 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 6 i\right )} - 3 \, {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} - a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {-\frac {2 i}{a d^{2}}} \log \left ({\left (\sqrt {2} {\left (2 i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {2 i}{a d^{2}}} + 4 i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 3 \, {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} - a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {-\frac {2 i}{a d^{2}}} \log \left ({\left (\sqrt {2} {\left (-2 i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {2 i}{a d^{2}}} + 4 i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right )}{12 \, {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} - a d e^{\left (i \, d x + i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(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))*(1
4*I*e^(4*I*d*x + 4*I*c) - 36*I*e^(2*I*d*x + 2*I*c) + 6*I) - 3*(a*d*e^(3*I*d*x + 3*I*c) - a*d*e^(I*d*x + I*c))*
sqrt(-2*I/(a*d^2))*log((sqrt(2)*(2*I*a*d*e^(2*I*d*x + 2*I*c) - 2*I*a*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(-2*I/(a*d^2)) + 4*I*a*e^(I*d*x + I*c))*e^(-I*d*x
- I*c)) + 3*(a*d*e^(3*I*d*x + 3*I*c) - a*d*e^(I*d*x + I*c))*sqrt(-2*I/(a*d^2))*log((sqrt(2)*(-2*I*a*d*e^(2*I*d
*x + 2*I*c) + 2*I*a*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(-2*I/(a*d^2)) + 4*I*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)))/(a*d*e^(3*I*d*x + 3*I*c) - a*d*e^(I*d*x
+ I*c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 1.93, size = 374, normalized size = 2.07 \[ \frac {\left (-\frac {1}{6}-\frac {i}{6}\right ) \left (3 i \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {2}+3 i \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}+3 \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}+5 i \left (\cos ^{2}\left (d x +c \right )\right )-2 i \cos \left (d x +c \right ) \sin \left (d x +c \right )-3 \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}-5 \left (\cos ^{2}\left (d x +c \right )\right )-2 \cos \left (d x +c \right ) \sin \left (d x +c \right )+7-7 i\right ) \left (\frac {\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )^{\frac {5}{2}} \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )}{d \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right ) \cos \left (d x +c \right )^{2} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-1/6-1/6*I)/d*(3*I*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c)
)^(1/2)*cos(d*x+c)*sin(d*x+c)*2^(1/2)+3*I*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))*((-1+
cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)*2^(1/2)+3*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/
2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*2^(1/2)+5*I*cos(d*x+c)^2-2*I*cos(d*x+c)*sin(d*x+c)-3*arcta
n((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)-5*cos(d*x
+c)^2-2*cos(d*x+c)*sin(d*x+c)+7-7*I)*(cos(d*x+c)/sin(d*x+c))^(5/2)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1
/2)*sin(d*x+c)/(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c)^2/a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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