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

Optimal. Leaf size=179 \[ \frac{5 \sqrt [4]{-1} a^{5/2} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{a^2 \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\cot (c+d x)}}+\frac{(4-4 i) a^{5/2} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.465381, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.321, Rules used = {4241, 3556, 3601, 3544, 205, 3599, 63, 217, 203} \[ \frac{5 \sqrt [4]{-1} a^{5/2} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tan ^{-1}\left (\frac{(-1)^{3/4} \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{a^2 \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\cot (c+d x)}}+\frac{(4-4 i) a^{5/2} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*(-1)^(1/4)*a^(5/2)*ArcTan[((-1)^(3/4)*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 - 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 - (a^2*Sqrt[a + I*a*Tan[c + d*x]])/(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 3556

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[a/(d*(m + n - 1
)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) +
b*d*(3*m + 2*n - 4))*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] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (IntegerQ[m] || Intege
rsQ[2*m, 2*n])

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

Mathematica [A]  time = 2.16965, size = 241, normalized size = 1.35 \[ \frac{2 i \sqrt{2} a^2 e^{i (c+d x)} \cos ^2(c+d x) \sqrt{\cot (c+d x)} \left (\sqrt{2} e^{i (c+d x)} \left (-1+e^{2 i (c+d x)}\right )-4 \sqrt{2} \left (1+e^{2 i (c+d x)}\right ) \sqrt{-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )+5 \left (1+e^{2 i (c+d x)}\right ) \sqrt{-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac{\sqrt{2} e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right ) \sqrt{a+i a \tan (c+d x)}}{d \left (1+e^{2 i (c+d x)}\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.418, size = 740, normalized size = 4.1 \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)^(1/2)*(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

1/4/d*2^(1/2)*a^2*(cos(d*x+c)-1)*(10*I*2^(1/2)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2))*cos(d*x+c)+5*I*2^(1/2
)*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)+1)*cos(d*x+c)-5*I*cos(d*x+c)*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-1)*2^
(1/2)+2*I*2^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)+2*I*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)-8
*I*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)-16*I*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/
2)*2^(1/2)+1)*cos(d*x+c)-16*I*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)-1)*cos(d*x+c)+10*cos(d*x+c)*arc
tan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2))*2^(1/2)-5*cos(d*x+c)*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)+1)*2^(1/2)+5*
cos(d*x+c)*ln(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-1)*2^(1/2)-2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x
+c)-16*cos(d*x+c)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)+1)-16*cos(d*x+c)*arctan(((cos(d*x+c)-1)/sin
(d*x+c))^(1/2)*2^(1/2)-1)-8*cos(d*x+c)*ln(-(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)-cos(d*x+c)-si
n(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)))*(a*(I*sin(d*x+c)+
cos(d*x+c))/cos(d*x+c))^(1/2)*(cos(d*x+c)/sin(d*x+c))^(1/2)/(I*sin(d*x+c)+cos(d*x+c)-1)/((cos(d*x+c)-1)/sin(d*
x+c))^(1/2)/cos(d*x+c)

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Maxima [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}} \sqrt{\cot \left (d x + c\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.51629, size = 1800, normalized size = 10.06 \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)^(1/2)*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/2*(sqrt(2)*(2*I*a^2*e^(2*I*d*x + 2*I*c) - 2*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) - sqrt(-25*I*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log(1/5
*(5*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) + 2*I*sqrt(-25*I*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I
*c)/a^2) + sqrt(-25*I*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log(1/5*(5*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)
- 2*I*sqrt(-25*I*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/a^2) + 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) - 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)**(1/2)*(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}} \sqrt{\cot \left (d x + c\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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