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

Optimal. Leaf size=111 \[ -\frac{i \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{d}+\frac{i \sqrt{2} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{\cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{d} \]

[Out]

((-I)*Sqrt[a]*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/d + (I*Sqrt[2]*Sqrt[a]*ArcTanh[Sqrt[a + I*a*Tan[c +
d*x]]/(Sqrt[2]*Sqrt[a])])/d - (Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/d

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Rubi [A]  time = 0.251643, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {3561, 21, 3554, 3480, 206, 3599, 63, 208} \[ -\frac{i \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{d}+\frac{i \sqrt{2} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{\cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*Sqrt[a]*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/d + (I*Sqrt[2]*Sqrt[a]*ArcTanh[Sqrt[a + I*a*Tan[c +
d*x]]/(Sqrt[2]*Sqrt[a])])/d - (Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/d

Rule 3561

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(d*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(f*(n + 1)*(c^2 + d^2)), x] - Dist[1/(a*(c^2 + d^2)*
(n + 1)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1)*Simp[b*d*m - a*c*(n + 1) + a*d*(m + n + 1)*T
an[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^
2 + d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || IntegersQ[2*m, 2*n])

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 3554

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(3/2)/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(2
*a^2)/(a*c - b*d), Int[Sqrt[a + b*Tan[e + f*x]], x], x] - Dist[(2*b*c*d + a*(c^2 - d^2))/(a*(c^2 + d^2)), Int[
((a - b*Tan[e + f*x])*Sqrt[a + b*Tan[e + f*x]])/(c + d*Tan[e + f*x]), 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 3480

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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

Mathematica [A]  time = 4.04474, size = 197, normalized size = 1.77 \[ \frac{\sqrt{a+i a \tan (c+d x)} \left (-4 \cot (c+d x)+i e^{-i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \left (\sqrt{2} \left (\log \left (1-e^{i (c+d x)}\right )-\log \left (1+e^{i (c+d x)}\right )+\log \left (\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}-e^{i (c+d x)}+1\right )-\log \left (\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}+e^{i (c+d x)}+1\right )\right )+4 \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.375, size = 586, normalized size = 5.3 \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)^2*(a+I*a*tan(d*x+c))^(1/2),x)

[Out]

-1/2/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(2*I*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan
h(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*cos(d*x+c)^2+I*(-2*cos(d*x+c)/(cos(d
*x+c)+1))^(1/2)*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*cos(d*x+c)^2+2*2
^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c
)^2-2*I*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*
sin(d*x+c)/cos(d*x+c))+(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos
(d*x+c)^2-I*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c
)+1)/sin(d*x+c))+2*I*cos(d*x+c)*sin(d*x+c)-2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(
-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+
1))^(1/2))+2*cos(d*x+c)^2-2*cos(d*x+c))/(I*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.46523, size = 1268, normalized size = 11.42 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-2 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i\right )} e^{\left (i \, d x + i \, c\right )} - \sqrt{2}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt{-\frac{a}{d^{2}}} \log \left ({\left (i \, \sqrt{2} d \sqrt{-\frac{a}{d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + \sqrt{2}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt{-\frac{a}{d^{2}}} \log \left ({\left (-i \, \sqrt{2} d \sqrt{-\frac{a}{d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) +{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt{-\frac{a}{d^{2}}} \log \left ({\left (\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )} + 2 i \, d \sqrt{-\frac{a}{d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) -{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt{-\frac{a}{d^{2}}} \log \left ({\left (\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )} - 2 i \, d \sqrt{-\frac{a}{d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right )}{2 \,{\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)^2*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/2*(sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-2*I*e^(2*I*d*x + 2*I*c) - 2*I)*e^(I*d*x + I*c) - sqrt(2)*(d*e
^(2*I*d*x + 2*I*c) - d)*sqrt(-a/d^2)*log((I*sqrt(2)*d*sqrt(-a/d^2)*e^(2*I*d*x + 2*I*c) + sqrt(2)*sqrt(a/(e^(2*
I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(-2*I*d*x - 2*I*c)) + sqrt(2)*(d*e^(2*I*d*x
+ 2*I*c) - d)*sqrt(-a/d^2)*log((-I*sqrt(2)*d*sqrt(-a/d^2)*e^(2*I*d*x + 2*I*c) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2
*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(-2*I*d*x - 2*I*c)) + (d*e^(2*I*d*x + 2*I*c) - d)*sqr
t(-a/d^2)*log((sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c) + 2*I*d*sqr
t(-a/d^2)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)) - (d*e^(2*I*d*x + 2*I*c) - d)*sqrt(-a/d^2)*log((sqrt(2)*s
qrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c) - 2*I*d*sqrt(-a/d^2)*e^(2*I*d*x + 2
*I*c))*e^(-2*I*d*x - 2*I*c)))/(d*e^(2*I*d*x + 2*I*c) - d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \left (i \tan{\left (c + d x \right )} + 1\right )} \cot ^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**2*(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(a*(I*tan(c + d*x) + 1))*cot(c + d*x)**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{i \, a \tan \left (d x + c\right ) + a} \cot \left (d x + c\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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