3.123 \(\int \frac{\cot (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\)
Optimal. Leaf size=132 \[ -\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{a^{3/2} d}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{3}{2 a d \sqrt{a+i a \tan (c+d x)}}+\frac{1}{3 d (a+i a \tan (c+d x))^{3/2}} \]
[Out]
(-2*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/(a^(3/2)*d) + ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqr
t[a])]/(2*Sqrt[2]*a^(3/2)*d) + 1/(3*d*(a + I*a*Tan[c + d*x])^(3/2)) + 3/(2*a*d*Sqrt[a + I*a*Tan[c + d*x]])
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Rubi [A] time = 0.391173, antiderivative size = 132, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 8, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used =
{3559, 3596, 3600, 3480, 206, 3599, 63, 208} \[ -\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{a^{3/2} d}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{3}{2 a d \sqrt{a+i a \tan (c+d x)}}+\frac{1}{3 d (a+i a \tan (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[Cot[c + d*x]/(a + I*a*Tan[c + d*x])^(3/2),x]
[Out]
(-2*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/(a^(3/2)*d) + ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqr
t[a])]/(2*Sqrt[2]*a^(3/2)*d) + 1/(3*d*(a + I*a*Tan[c + d*x])^(3/2)) + 3/(2*a*d*Sqrt[a + I*a*Tan[c + d*x]])
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 3596
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 3600
Int[(((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*tan[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(A*b + a*B)/(b*c + a*d), Int[(a + b*Tan[e + f*x])^m, x], x] - Dist[(B*c
- A*d)/(b*c + a*d), Int[((a + b*Tan[e + f*x])^m*(a - b*Tan[e + f*x]))/(c + d*Tan[e + f*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] && NeQ[A*b + a*B, 0]
Rule 3480
Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0]
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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 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 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
Rubi steps
\begin{align*} \int \frac{\cot (c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx &=\frac{1}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{\int \frac{\cot (c+d x) \left (3 a-\frac{3}{2} i a \tan (c+d x)\right )}{\sqrt{a+i a \tan (c+d x)}} \, dx}{3 a^2}\\ &=\frac{1}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{3}{2 a d \sqrt{a+i a \tan (c+d x)}}+\frac{\int \cot (c+d x) \sqrt{a+i a \tan (c+d x)} \left (3 a^2-\frac{9}{4} i a^2 \tan (c+d x)\right ) \, dx}{3 a^4}\\ &=\frac{1}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{3}{2 a d \sqrt{a+i a \tan (c+d x)}}+\frac{\int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)} \, dx}{a^3}+\frac{i \int \sqrt{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac{1}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{3}{2 a d \sqrt{a+i a \tan (c+d x)}}+\frac{\operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{2 a d}+\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{a d}\\ &=\frac{\tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{1}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{3}{2 a d \sqrt{a+i a \tan (c+d x)}}-\frac{(2 i) \operatorname{Subst}\left (\int \frac{1}{i-\frac{i x^2}{a}} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{a^2 d}\\ &=-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{a^{3/2} d}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{1}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac{3}{2 a d \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}
Mathematica [A] time = 1.87642, size = 196, normalized size = 1.48 \[ -\frac{i \left (11 e^{2 i (c+d x)}+10 e^{4 i (c+d x)}+3 e^{3 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )-12 \sqrt{2} e^{3 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{\sqrt{2} e^{i (c+d x)}}{\sqrt{1+e^{2 i (c+d x)}}}\right )+1\right )}{3 a d \left (1+e^{2 i (c+d x)}\right )^2 (\tan (c+d x)-i) \sqrt{a+i a \tan (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[Cot[c + d*x]/(a + I*a*Tan[c + d*x])^(3/2),x]
[Out]
((-I/3)*(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))]*ArcSinh[E^(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))]]))/(a*d*(1 + E^((2*I)*(c + d*x)))^2*(-I + Tan[c + d*x])*Sqrt[a
+ I*a*Tan[c + d*x]])
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Maple [B] time = 0.369, size = 703, normalized size = 5.3 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(d*x+c)/(a+I*a*tan(d*x+c))^(3/2),x)
[Out]
-1/24/d/a^2*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(-3*I*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*
arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*sin(d*x+c)+16*
I*cos(d*x+c)^3*sin(d*x+c)+3*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I-si
n(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)+12*I*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2
)*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*cos(d*x+c)-12*I*(-2*cos(d*x+c)
/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*sin(d*x+c)-16*cos(d*x+c)^4+3*2^(1/2)*(-2
*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(co
s(d*x+c)+1))^(1/2))+12*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d
*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+12*I*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln(((-2*cos(d*x+c)/(cos(d*x+c
)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))+12*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+
c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)+36*I*sin(d*x+c)*cos(d*x+c)+12*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan
(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-28*cos(d*x+c)^2)
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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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(cot(d*x+c)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")
[Out]
Exception raised: UnboundLocalError
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot{\left (c + d x \right )}}{\left (a \left (i \tan{\left (c + d x \right )} + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)/(a+I*a*tan(d*x+c))**(3/2),x)
[Out]
Integral(cot(c + d*x)/(a*(I*tan(c + d*x) + 1))**(3/2), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot \left (d x + c\right )}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")
[Out]
integrate(cot(d*x + c)/(I*a*tan(d*x + c) + a)^(3/2), x)