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

Optimal. Leaf size=217 \[ -\frac{1}{d \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{(-1)^{3/4} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d}-\frac{2 i \sqrt{a+i a \tan (c+d x)}}{a d \sqrt{\cot (c+d x)}}+\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d} \]

[Out]

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

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Rubi [A]  time = 0.583552, antiderivative size = 217, normalized size of antiderivative = 1., 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, 3597, 3601, 3544, 205, 3599, 63, 217, 203} \[ -\frac{1}{d \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}-\frac{(-1)^{3/4} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d}-\frac{2 i \sqrt{a+i a \tan (c+d x)}}{a d \sqrt{\cot (c+d x)}}+\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{\sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 3597

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(B*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(f*(m + n)), x] +
Dist[1/(a*(m + n)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Simp[a*A*c*(m + n) - B*(b*c*m + a*
d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*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] && GtQ[n, 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 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 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 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 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 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 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])

Rubi steps

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

Mathematica [A]  time = 2.06605, size = 248, normalized size = 1.14 \[ \frac{\sqrt{\cot (c+d x)} \left (2 e^{2 i (c+d x)}-3 e^{4 i (c+d x)}+e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \left (1+e^{2 i (c+d x)}\right ) \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )+\sqrt{2} e^{i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \left (1+e^{2 i (c+d x)}\right ) \tanh ^{-1}\left (\frac{\sqrt{2} e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )+1\right )}{\sqrt{2} d \left (1+e^{2 i (c+d x)}\right )^2 \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + 2*E^((2*I)*(c + d*x)) - 3*E^((4*I)*(c + d*x)) + E^(I*(c + d*x))*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))]] + Sqrt[2]*E^(I*(c + d*x))*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))]
])*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^((2*I)*(c + d
*x)))^2)

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Maple [B]  time = 0.47, size = 801, normalized size = 3.7 \begin{align*} \text{result too large to display} \end{align*}

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))^(1/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{i \, a \tan \left (d x + c\right ) + a} \cot \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

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))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

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))^(1/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{i \, a \tan \left (d x + c\right ) + a} \cot \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

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))^(1/2),x, algorithm="giac")

[Out]

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