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

Optimal. Leaf size=65 \[ -\frac{2 a \cot ^{\frac{3}{2}}(c+d x)}{3 d}-\frac{2 i a \sqrt{\cot (c+d x)}}{d}-\frac{2 (-1)^{3/4} a \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.102828, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3673, 3528, 3533, 208} \[ -\frac{2 a \cot ^{\frac{3}{2}}(c+d x)}{3 d}-\frac{2 i a \sqrt{\cot (c+d x)}}{d}-\frac{2 (-1)^{3/4} a \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 3528

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

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

Mathematica [A]  time = 1.15187, size = 116, normalized size = 1.78 \[ -\frac{2 a e^{-i c} \sin (c+d x) \sqrt{\cot (c+d x)} (\cot (c+d x)+i) (\cos (d x)-i \sin (d x)) \left (\cot (c+d x)-3 i \sqrt{i \tan (c+d x)} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )+3 i\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.268, size = 777, normalized size = 12. \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)),x)

[Out]

1/3*a/d*2^(1/2)*(3*I*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))*((1-cos(d*
x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1
/2)*sin(d*x+c)*cos(d*x+c)+3*I*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))*(
(1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d
*x+c))^(1/2)*sin(d*x+c)-3*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))*((1-c
os(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c
))^(1/2)*sin(d*x+c)*cos(d*x+c)+3*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*((1-cos(d
*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(
1/2)*sin(d*x+c)*cos(d*x+c)-3*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))*((
1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*
x+c))^(1/2)*sin(d*x+c)+3*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*((1-cos(d*x+c)+si
n(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*sin
(d*x+c)-3*I*2^(1/2)*sin(d*x+c)*cos(d*x+c)-2^(1/2)*cos(d*x+c)^2)*(cos(d*x+c)/sin(d*x+c))^(5/2)*sin(d*x+c)/cos(d
*x+c)^3

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Maxima [B]  time = 1.49962, size = 188, normalized size = 2.89 \begin{align*} -\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 + \frac{24 i \, a}{\sqrt{\tan \left (d x + c\right )}} + \frac{8 \, a}{\tan \left (d x + c\right )^{\frac{3}{2}}}}{12 \, d} \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)),x, algorithm="maxima")

[Out]

-1/12*(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 + 24*I*a/sqrt(tan(d*x + c))
+ 8*a/tan(d*x + c)^(3/2))/d

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Fricas [B]  time = 1.45333, size = 779, normalized size = 11.98 \begin{align*} -\frac{3 \,{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt{-\frac{4 i \, a^{2}}{d^{2}}} \log \left (\frac{{\left ({\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt{-\frac{4 i \, a^{2}}{d^{2}}} \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}} + 2 i \, a e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a}\right ) - 3 \,{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt{-\frac{4 i \, a^{2}}{d^{2}}} \log \left (\frac{{\left ({\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt{-\frac{4 i \, a^{2}}{d^{2}}} \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}} + 2 i \, a e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a}\right ) -{\left (-32 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 16 i \, a\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 (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)),x, algorithm="fricas")

[Out]

-1/12*(3*(d*e^(2*I*d*x + 2*I*c) - d)*sqrt(-4*I*a^2/d^2)*log(((I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt(-4*I*a^2/d^2
)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)) + 2*I*a*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c
)/a) - 3*(d*e^(2*I*d*x + 2*I*c) - d)*sqrt(-4*I*a^2/d^2)*log(((-I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt(-4*I*a^2/d^
2)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)) + 2*I*a*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*
c)/a) - (-32*I*a*e^(2*I*d*x + 2*I*c) + 16*I*a)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))/(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)),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 )} \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)),x, algorithm="giac")

[Out]

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