3.764 \(\int \cot ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\)
Optimal. Leaf size=142 \[ -\frac{4 i a^2 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{d}-\frac{(4-4 i) a^{5/2} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 a \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d} \]
[Out]
((-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 - ((4*I)*a^2*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/d - (2*a*Cot[c + d*x]^(3/2
)*(a + I*a*Tan[c + d*x])^(3/2))/(3*d)
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Rubi [A] time = 0.276094, antiderivative size = 142, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used =
{4241, 3545, 3544, 205} \[ -\frac{4 i a^2 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{d}-\frac{(4-4 i) a^{5/2} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 a \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d} \]
Antiderivative was successfully verified.
[In]
Int[Cot[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^(5/2),x]
[Out]
((-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 - ((4*I)*a^2*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/d - (2*a*Cot[c + d*x]^(3/2
)*(a + I*a*Tan[c + d*x])^(3/2))/(3*d)
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 3545
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(a*b*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m - 1)*(a*c - b*d)), x] + Dist[(2*a^2)/(
a*c - b*d), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1), 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] && EqQ[m + n, 0] && GtQ[m, 1/2]
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]
Rubi steps
\begin{align*} \int \cot ^{\frac{5}{2}}(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}}{\tan ^{\frac{5}{2}}(c+d x)} \, dx\\ &=-\frac{2 a \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}+\left (2 i a \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac{3}{2}}(c+d x)} \, dx\\ &=-\frac{4 i a^2 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{d}-\frac{2 a \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}-\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{4 i a^2 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{d}-\frac{2 a \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac{\left (8 i a^4 \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{(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{4 i a^2 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{d}-\frac{2 a \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}\\ \end{align*}
Mathematica [A] time = 1.43406, size = 142, normalized size = 1. \[ -\frac{4 i a^2 e^{-i (c+d x)} \sqrt{\cot (c+d x)} \left (e^{i (c+d x)} \left (-3+4 e^{2 i (c+d x)}\right )-3 \left (-1+e^{2 i (c+d x)}\right )^{3/2} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right ) \sqrt{a+i a \tan (c+d x)}}{3 d \left (-1+e^{2 i (c+d x)}\right )} \]
Antiderivative was successfully verified.
[In]
Integrate[Cot[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^(5/2),x]
[Out]
(((-4*I)/3)*a^2*(E^(I*(c + d*x))*(-3 + 4*E^((2*I)*(c + d*x))) - 3*(-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[a + I*a*Tan[c + d*x]])/(d*E^(I*(c + d*x)
)*(-1 + E^((2*I)*(c + d*x))))
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Maple [B] time = 0.393, size = 1044, normalized size = 7.4 \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)^(5/2)*(a+I*a*tan(d*x+c))^(5/2),x)
[Out]
-1/3/d*2^(1/2)*a^2*(8*I*sin(d*x+c)*cos(d*x+c)*2^(1/2)-6*I*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*ln(-(((cos(d*x+c)-
1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)/(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*si
n(d*x+c)-cos(d*x+c)-sin(d*x+c)+1))+12*I*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(
1/2)*2^(1/2)-1)*cos(d*x+c)^2+12*cos(d*x+c)^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*arctan(((cos(d*x+c)-1)/sin(d*x+
c))^(1/2)*2^(1/2)-1)+12*cos(d*x+c)^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2
)*2^(1/2)+1)+6*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)*2^(1/2)*s
in(d*x+c)-cos(d*x+c)-sin(d*x+c)+1)/(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)
-1))+12*I*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)+1)*cos(d*x+c)^2-1
2*I*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)-1)-12*I*((cos(d*x+c)-1)
/sin(d*x+c))^(1/2)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)+1)-7*I*2^(1/2)*sin(d*x+c)+8*2^(1/2)*cos(d*
x+c)^2+6*I*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*ln(-(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+cos(d*x
+c)+sin(d*x+c)-1)/(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1))*cos(d*x+c)^2
-12*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)-1)-12*((cos(d*x+c)-1)/s
in(d*x+c))^(1/2)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)+1)-6*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*ln(-(
((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1)/(((cos(d*x+c)-1)/sin(d*x+c))^(1/
2)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1))-2^(1/2)*cos(d*x+c)-7*2^(1/2))*(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)-1)/cos(d*x+c)^2
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Maxima [B] time = 2.26926, size = 1422, normalized size = 10.01 \begin{align*} \text{result too large to display} \end{align*}
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))^(5/2),x, algorithm="maxima")
[Out]
-1/9*(sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*(((36*I - 36)*a^2*cos(3*d*x + 3*c
) - (24*I - 24)*a^2*cos(d*x + c) - (36*I + 36)*a^2*sin(3*d*x + 3*c) + (24*I + 24)*a^2*sin(d*x + c))*cos(3/2*ar
ctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + ((36*I + 36)*a^2*cos(3*d*x + 3*c) - (24*I + 24)*a^2*cos(d*x +
c) + (36*I - 36)*a^2*sin(3*d*x + 3*c) - (24*I - 24)*a^2*sin(d*x + c))*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2
*d*x + 2*c) - 1)))*sqrt(a) + (((36*I + 36)*a^2*cos(2*d*x + 2*c)^2 + (36*I + 36)*a^2*sin(2*d*x + 2*c)^2 - (72*I
+ 72)*a^2*cos(2*d*x + 2*c) + (36*I + 36)*a^2)*arctan2(2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*
x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + 2*sin(d*x + c), 2*(cos(2*d*x +
2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)
- 1)) + 2*cos(d*x + c)) + (-(18*I - 18)*a^2*cos(2*d*x + 2*c)^2 - (18*I - 18)*a^2*sin(2*d*x + 2*c)^2 + (36*I -
36)*a^2*cos(2*d*x + 2*c) - (18*I - 18)*a^2)*log(4*cos(d*x + c)^2 + 4*sin(d*x + c)^2 + 4*sqrt(cos(2*d*x + 2*c)
^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1))^2
+ sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1))^2) + 8*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2
*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(d*x + c)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + sin(d*x
+ c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)))))*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*
cos(2*d*x + 2*c) + 1)^(1/4)*sqrt(a) + ((((12*I - 12)*a^2*cos(d*x + c) - (12*I + 12)*a^2*sin(d*x + c))*cos(2*d*
x + 2*c)^2 + (12*I - 12)*a^2*cos(d*x + c) + ((12*I - 12)*a^2*cos(d*x + c) - (12*I + 12)*a^2*sin(d*x + c))*sin(
2*d*x + 2*c)^2 - (12*I + 12)*a^2*sin(d*x + c) + (-(24*I - 24)*a^2*cos(d*x + c) + (24*I + 24)*a^2*sin(d*x + c))
*cos(2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + (((12*I + 12)*a^2*cos(d*x + c) +
(12*I - 12)*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^2 + (12*I + 12)*a^2*cos(d*x + c) + ((12*I + 12)*a^2*cos(d*x +
c) + (12*I - 12)*a^2*sin(d*x + c))*sin(2*d*x + 2*c)^2 + (12*I - 12)*a^2*sin(d*x + c) + (-(24*I + 24)*a^2*cos(d
*x + c) - (24*I - 24)*a^2*sin(d*x + c))*cos(2*d*x + 2*c))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) -
1)))*sqrt(a))/((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(5/4)*d)
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Fricas [B] time = 1.3795, size = 1037, normalized size = 7.3 \begin{align*} \frac{\sqrt{2}{\left (-32 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 24 i \, a^{2}\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}} e^{\left (i \, d x + i \, c\right )} - 3 \, \sqrt{-\frac{32 i \, a^{5}}{d^{2}}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac{{\left (4 \, \sqrt{2}{\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2}\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}} e^{\left (i \, d x + i \, c\right )} + i \, \sqrt{-\frac{32 i \, a^{5}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a^{2}}\right ) + 3 \, \sqrt{-\frac{32 i \, a^{5}}{d^{2}}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac{{\left (4 \, \sqrt{2}{\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2}\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}} e^{\left (i \, d x + i \, c\right )} - i \, \sqrt{-\frac{32 i \, a^{5}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a^{2}}\right )}{6 \,{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \end{align*}
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))^(5/2),x, algorithm="fricas")
[Out]
1/6*(sqrt(2)*(-32*I*a^2*e^(2*I*d*x + 2*I*c) + 24*I*a^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^(I*d*x + I*c) - 3*sqrt(-32*I*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*lo
g(1/4*(4*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) - a^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^(I*d*x + I*c) + I*sqrt(-32*I*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x -
2*I*c)/a^2) + 3*sqrt(-32*I*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log(1/4*(4*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) -
a^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^(I*d*x +
I*c) - I*sqrt(-32*I*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/a^2))/(d*e^(2*I*d*x + 2*I*c) - d)
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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(cot(d*x+c)**(5/2)*(a+I*a*tan(d*x+c))**(5/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \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(cot(d*x+c)^(5/2)*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")
[Out]
integrate((I*a*tan(d*x + c) + a)^(5/2)*cot(d*x + c)^(5/2), x)