3.733 \(\int \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=86 \[ -\frac {2 \left (a^3 \cot (c+d x)+i a^3\right )}{3 d \cot ^{\frac {3}{2}}(c+d x)}-\frac {16 a^3}{3 d \sqrt {\cot (c+d x)}}+\frac {8 (-1)^{3/4} a^3 \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d} \]

[Out]

8*(-1)^(3/4)*a^3*arctanh((-1)^(3/4)*cot(d*x+c)^(1/2))/d-2/3*(I*a^3+a^3*cot(d*x+c))/d/cot(d*x+c)^(3/2)-16/3*a^3
/d/cot(d*x+c)^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3673, 3553, 3591, 3533, 208} \[ -\frac {2 \left (a^3 \cot (c+d x)+i a^3\right )}{3 d \cot ^{\frac {3}{2}}(c+d x)}-\frac {16 a^3}{3 d \sqrt {\cot (c+d x)}}+\frac {8 (-1)^{3/4} a^3 \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])^3,x]

[Out]

(8*(-1)^(3/4)*a^3*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d - (16*a^3)/(3*d*Sqrt[Cot[c + d*x]]) - (2*(I*a^3 +
a^3*Cot[c + d*x]))/(3*d*Cot[c + d*x]^(3/2))

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]

Rule 3533

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(2*c^2)/f, S
ubst[Int[1/(b*c - d*x^2), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 + d^2, 0]

Rule 3553

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(a^2*(b*c - a*d)*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(b*c + a*d)*(n + 1)), x] +
Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[b*(b*c*(m
- 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) + b^2*d*(n + 1) - a^2*d*(m + n - 1))*Tan[e + f*x], x], x], x] /; Fr
eeQ[{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] && GtQ[m, 1] && LtQ[
n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3591

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((b*c - a*d)*(A*b - a*B)*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)*(a^2
+ b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[a*A*c + b*B*c + A*b*d - a*B*d - (A*b*
c - a*B*c - a*A*d - b*B*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& LtQ[m, -1] && NeQ[a^2 + b^2, 0]

Rule 3673

Int[(cot[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Cot[e + f*x])^(m - n*p)*(b + a*Cot[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] &&  !IntegerQ[m] && IntegersQ[n, p]

Rubi steps

\begin {align*} \int \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^3 \, dx &=\int \frac {(i a+a \cot (c+d x))^3}{\cot ^{\frac {5}{2}}(c+d x)} \, dx\\ &=-\frac {2 \left (i a^3+a^3 \cot (c+d x)\right )}{3 d \cot ^{\frac {3}{2}}(c+d x)}-\frac {2}{3} \int \frac {(i a+a \cot (c+d x)) \left (-4 i a^2-2 a^2 \cot (c+d x)\right )}{\cot ^{\frac {3}{2}}(c+d x)} \, dx\\ &=-\frac {16 a^3}{3 d \sqrt {\cot (c+d x)}}-\frac {2 \left (i a^3+a^3 \cot (c+d x)\right )}{3 d \cot ^{\frac {3}{2}}(c+d x)}-\frac {2}{3} \int \frac {-6 i a^3-6 a^3 \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx\\ &=-\frac {16 a^3}{3 d \sqrt {\cot (c+d x)}}-\frac {2 \left (i a^3+a^3 \cot (c+d x)\right )}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {\left (48 a^6\right ) \operatorname {Subst}\left (\int \frac {1}{6 i a^3-6 a^3 x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d}\\ &=\frac {8 (-1)^{3/4} a^3 \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}-\frac {16 a^3}{3 d \sqrt {\cot (c+d x)}}-\frac {2 \left (i a^3+a^3 \cot (c+d x)\right )}{3 d \cot ^{\frac {3}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]  time = 2.83, size = 147, normalized size = 1.71 \[ \frac {i a^3 e^{-3 i c} \sqrt {\cot (c+d x)} (\cos (3 (c+d x))+i \sin (3 (c+d x))) \left (\sec ^2(c+d x) (9 i \sin (2 (c+d x))+\cos (2 (c+d x))-1)-24 \sqrt {i \tan (c+d x)} \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )\right )}{3 d (\cos (d x)+i \sin (d x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])^3,x]

[Out]

((I/3)*a^3*Sqrt[Cot[c + d*x]]*(Cos[3*(c + d*x)] + I*Sin[3*(c + d*x)])*(Sec[c + d*x]^2*(-1 + Cos[2*(c + d*x)] +
(9*I)*Sin[2*(c + d*x)]) - 24*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]]*Sqrt[I*Tan[c
+ d*x]]))/(d*E^((3*I)*c)*(Cos[d*x] + I*Sin[d*x])^3)

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fricas [B]  time = 2.70, size = 340, normalized size = 3.95 \[ \frac {3 \, \sqrt {-\frac {64 i \, a^{6}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (8 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {-\frac {64 i \, a^{6}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a^{3}}\right ) - 3 \, \sqrt {-\frac {64 i \, a^{6}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (8 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {-\frac {64 i \, a^{6}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a^{3}}\right ) + {\left (80 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 16 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 64 i \, a^{3}\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}}{12 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

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))^3,x, algorithm="fricas")

[Out]

1/12*(3*sqrt(-64*I*a^6/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log(1/4*(8*I*a^3*e^(2*I*d*x
+ 2*I*c) + sqrt(-64*I*a^6/d^2)*(I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x +
2*I*c) - 1)))*e^(-2*I*d*x - 2*I*c)/a^3) - 3*sqrt(-64*I*a^6/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*
c) + d)*log(1/4*(8*I*a^3*e^(2*I*d*x + 2*I*c) + sqrt(-64*I*a^6/d^2)*(-I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt((I*e^
(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))*e^(-2*I*d*x - 2*I*c)/a^3) + (80*I*a^3*e^(4*I*d*x + 4*I*c) -
16*I*a^3*e^(2*I*d*x + 2*I*c) - 64*I*a^3)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))/(d*e^(4
*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} \sqrt {\cot \left (d x + c\right )}\,{d x} \]

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))^3,x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^3*sqrt(cot(d*x + c)), x)

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maple [C]  time = 1.48, size = 479, normalized size = 5.57 \[ \frac {a^{3} \left (-1+\cos \left (d x +c \right )\right ) \left (12 i \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-12 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \EllipticPi \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+12 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \EllipticF \left (\sqrt {\frac {1-\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right )-i \sqrt {2}\, \sin \left (d x +c \right ) \cos \left (d x +c \right )+i \sin \left (d x +c \right ) \sqrt {2}-9 \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}+9 \cos \left (d x +c \right ) \sqrt {2}\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (1+\cos \left (d x +c \right )\right )^{2} \sqrt {2}}{3 d \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{3}} \]

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))^3,x)

[Out]

1/3*a^3/d*(-1+cos(d*x+c))*(12*I*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d
*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/
2*I,1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)-12*sin(d*x+c)*cos(d*x+c)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(
(-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(
d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))+12*sin(d*x+c)*cos(d*x+c)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c
))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticF(((1-cos(d*
x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))-I*2^(1/2)*cos(d*x+c)*sin(d*x+c)+I*2^(1/2)*sin(d*x+c)-9*cos(d*x
+c)^2*2^(1/2)+9*cos(d*x+c)*2^(1/2))*(cos(d*x+c)/sin(d*x+c))^(1/2)*(1+cos(d*x+c))^2/cos(d*x+c)^2/sin(d*x+c)^3*2
^(1/2)

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maxima [B]  time = 0.96, size = 148, normalized size = 1.72 \[ -\frac {3 \, {\left (\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + \left (2 i + 2\right ) \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + \left (i - 1\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - \left (i - 1\right ) \, \sqrt {2} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )\right )} a^{3} - 2 \, {\left (-i \, a^{3} - \frac {9 \, a^{3}}{\tan \left (d x + c\right )}\right )} \tan \left (d x + c\right )^{\frac {3}{2}}}{3 \, d} \]

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))^3,x, algorithm="maxima")

[Out]

-1/3*(3*((2*I + 2)*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + (2*I + 2)*sqrt(2)*arctan(-1/
2*sqrt(2)*(sqrt(2) - 2/sqrt(tan(d*x + c)))) + (I - 1)*sqrt(2)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c)
+ 1) - (I - 1)*sqrt(2)*log(-sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1))*a^3 - 2*(-I*a^3 - 9*a^3/tan(d*x
+ c))*tan(d*x + c)^(3/2))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {\mathrm {cot}\left (c+d\,x\right )}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)^3,x)

[Out]

int(cot(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - i a^{3} \left (\int i \sqrt {\cot {\left (c + d x \right )}}\, dx + \int \left (- 3 \tan {\left (c + d x \right )} \sqrt {\cot {\left (c + d x \right )}}\right )\, dx + \int \tan ^{3}{\left (c + d x \right )} \sqrt {\cot {\left (c + d x \right )}}\, dx + \int \left (- 3 i \tan ^{2}{\left (c + d x \right )} \sqrt {\cot {\left (c + d x \right )}}\right )\, dx\right ) \]

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))**3,x)

[Out]

-I*a**3*(Integral(I*sqrt(cot(c + d*x)), x) + Integral(-3*tan(c + d*x)*sqrt(cot(c + d*x)), x) + Integral(tan(c
+ d*x)**3*sqrt(cot(c + d*x)), x) + Integral(-3*I*tan(c + d*x)**2*sqrt(cot(c + d*x)), x))

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