3.726 \(\int \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=49 \[ -\frac{2 a^2 \sqrt{\cot (c+d x)}}{d}-\frac{4 \sqrt [4]{-1} a^2 \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.112776, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3673, 3543, 3533, 208} \[ -\frac{2 a^2 \sqrt{\cot (c+d x)}}{d}-\frac{4 \sqrt [4]{-1} a^2 \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^2,x]

[Out]

(-4*(-1)^(1/4)*a^2*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d - (2*a^2*Sqrt[Cot[c + d*x]])/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 3543

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
(d^2*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e
+ f*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] &&  !LeQ[m, -1] &&  !(EqQ[m, 2] && EqQ
[a, 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{3}{2}}(c+d x) (a+i a \tan (c+d x))^2 \, dx &=\int \frac{(i a+a \cot (c+d x))^2}{\sqrt{\cot (c+d x)}} \, dx\\ &=-\frac{2 a^2 \sqrt{\cot (c+d x)}}{d}+\int \frac{-2 a^2+2 i a^2 \cot (c+d x)}{\sqrt{\cot (c+d x)}} \, dx\\ &=-\frac{2 a^2 \sqrt{\cot (c+d x)}}{d}+\frac{\left (8 a^4\right ) \operatorname{Subst}\left (\int \frac{1}{2 a^2+2 i a^2 x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{d}\\ &=-\frac{4 \sqrt [4]{-1} a^2 \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d}-\frac{2 a^2 \sqrt{\cot (c+d x)}}{d}\\ \end{align*}

Mathematica [C]  time = 1.69198, size = 70, normalized size = 1.43 \[ \frac{2 a^2 \sqrt{\cot (c+d x)} \left (-1+2 \sqrt{i \tan (c+d x)} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^2,x]

[Out]

(2*a^2*Sqrt[Cot[c + d*x]]*(-1 + 2*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]]*Sqrt[I*T
an[c + d*x]]))/d

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Maple [C]  time = 0.254, size = 736, normalized size = 15. \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)^(3/2)*(a+I*a*tan(d*x+c))^2,x)

[Out]

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

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Maxima [C]  time = 1.50321, size = 177, normalized size = 3.61 \begin{align*} \frac{{\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^{2} - \frac{4 \, a^{2}}{\sqrt{\tan \left (d x + c\right )}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

1/2*((-(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^2 - 4*a^2/sqrt(tan(d*x + c)))/d

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Fricas [C]  time = 1.37446, size = 632, normalized size = 12.9 \begin{align*} -\frac{8 \, a^{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}} - \sqrt{\frac{16 i \, a^{4}}{d^{2}}} d \log \left (\frac{{\left (4 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{\frac{16 i \, a^{4}}{d^{2}}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - 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 )}}{2 \, a^{2}}\right ) + \sqrt{\frac{16 i \, a^{4}}{d^{2}}} d \log \left (\frac{{\left (4 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{\frac{16 i \, a^{4}}{d^{2}}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - 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 )}}{2 \, a^{2}}\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/4*(8*a^2*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)) - sqrt(16*I*a^4/d^2)*d*log(1/2*(4*I*a^
2*e^(2*I*d*x + 2*I*c) + sqrt(16*I*a^4/d^2)*(d*e^(2*I*d*x + 2*I*c) - 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^2) + sqrt(16*I*a^4/d^2)*d*log(1/2*(4*I*a^2*e^(2*I*d*x + 2*I*c) -
sqrt(16*I*a^4/d^2)*(d*e^(2*I*d*x + 2*I*c) - 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^2))/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)**(3/2)*(a+I*a*tan(d*x+c))**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 )}^{2} \cot \left (d x + c\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^2*cot(d*x + c)^(3/2), x)