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

Optimal. Leaf size=141 \[ -\frac{i a^2}{d \sqrt{a+i a \tan (c+d x)}}-\frac{3 i a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{d}+\frac{2 i \sqrt{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{a^2 \cot (c+d x)}{d \sqrt{a+i a \tan (c+d x)}} \]

[Out]

((-3*I)*a^(3/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/d + ((2*I)*Sqrt[2]*a^(3/2)*ArcTanh[Sqrt[a + I*a*T
an[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (I*a^2)/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (a^2*Cot[c + d*x])/(d*Sqrt[a + I
*a*Tan[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.408563, antiderivative size = 141, 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 = {3553, 3596, 3600, 3480, 206, 3599, 63, 208} \[ -\frac{i a^2}{d \sqrt{a+i a \tan (c+d x)}}-\frac{3 i a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{d}+\frac{2 i \sqrt{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{a^2 \cot (c+d x)}{d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*I)*a^(3/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/d + ((2*I)*Sqrt[2]*a^(3/2)*ArcTanh[Sqrt[a + I*a*T
an[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (I*a^2)/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (a^2*Cot[c + d*x])/(d*Sqrt[a + I
*a*Tan[c + d*x]])

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 3596

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

Rule 3600

Int[(((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*tan[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(A*b + a*B)/(b*c + a*d), Int[(a + b*Tan[e + f*x])^m, x], x] - Dist[(B*c
- A*d)/(b*c + a*d), Int[((a + b*Tan[e + f*x])^m*(a - b*Tan[e + f*x]))/(c + d*Tan[e + f*x]), x], x] /; FreeQ[{
a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 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) (a+i a \tan (c+d x))^{3/2} \, dx &=-\frac{a^2 \cot (c+d x)}{d \sqrt{a+i a \tan (c+d x)}}-\int \frac{\cot (c+d x) \left (-\frac{3 i a^2}{2}+\frac{5}{2} a^2 \tan (c+d x)\right )}{\sqrt{a+i a \tan (c+d x)}} \, dx\\ &=-\frac{i a^2}{d \sqrt{a+i a \tan (c+d x)}}-\frac{a^2 \cot (c+d x)}{d \sqrt{a+i a \tan (c+d x)}}-\frac{\int \cot (c+d x) \sqrt{a+i a \tan (c+d x)} \left (-\frac{3 i a^3}{2}+\frac{1}{2} a^3 \tan (c+d x)\right ) \, dx}{a^2}\\ &=-\frac{i a^2}{d \sqrt{a+i a \tan (c+d x)}}-\frac{a^2 \cot (c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{3}{2} 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{i a^2}{d \sqrt{a+i a \tan (c+d x)}}-\frac{a^2 \cot (c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{\left (3 i a^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{\left (4 i a^2\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{d}\\ &=\frac{2 i \sqrt{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{i a^2}{d \sqrt{a+i a \tan (c+d x)}}-\frac{a^2 \cot (c+d x)}{d \sqrt{a+i a \tan (c+d x)}}+\frac{(3 a) \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{3 i a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{d}+\frac{2 i \sqrt{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{i a^2}{d \sqrt{a+i a \tan (c+d x)}}-\frac{a^2 \cot (c+d x)}{d \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.27515, size = 178, normalized size = 1.26 \[ \frac{a e^{-\frac{1}{2} i (2 c+3 d x)} \sqrt{1+e^{2 i (c+d x)}} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (\sin \left (\frac{d x}{2}\right )-i \cos \left (\frac{d x}{2}\right )\right ) \left (-i \sqrt{1+e^{2 i (c+d x)}} \csc (c+d x)-4 \sinh ^{-1}\left (e^{i (c+d x)}\right )+3 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{2} e^{i (c+d x)}}{\sqrt{1+e^{2 i (c+d x)}}}\right )\right )}{\sqrt{2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*(-4*ArcSinh[E^(I*(c +
d*x))] + 3*Sqrt[2]*ArcTanh[(Sqrt[2]*E^(I*(c + d*x)))/Sqrt[1 + E^((2*I)*(c + d*x))]] - I*Sqrt[1 + E^((2*I)*(c
+ d*x))]*Csc[c + d*x])*((-I)*Cos[(d*x)/2] + Sin[(d*x)/2]))/(Sqrt[2]*d*E^((I/2)*(2*c + 3*d*x)))

________________________________________________________________________________________

Maple [B]  time = 0.311, size = 631, normalized size = 4.5 \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))^(3/2),x)

[Out]

1/2/d*a*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(4*I*(-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^(1/2)*cos(d*x+c)*sin(d*x+c)+4*I*(-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^
(1/2)*sin(d*x+c)+3*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)*sin(d*x+c)+4*(-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^(1/2)*cos(d*x+c)*sin(d*x+c)+3*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))*sin(d*x+c)+4*2^(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)*sin(d*x+c)+3*(-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)*sin(d*x+c)+3*(-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))*sin(d*x+c)-2*I*cos(d*x+c)^2-2*I*cos(d*x+c)+2*co
s(d*x+c)*sin(d*x+c))/(I*sin(d*x+c)+cos(d*x+c)-1)/(cos(d*x+c)+1)

________________________________________________________________________________________

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))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.36901, size = 1365, normalized size = 9.68 \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)^2*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/2*(sqrt(2)*(-2*I*a*e^(2*I*d*x + 2*I*c) - 2*I*a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - 2*sqrt(2
)*(d*e^(2*I*d*x + 2*I*c) - d)*sqrt(-a^3/d^2)*log(1/2*(2*I*sqrt(2)*sqrt(-a^3/d^2)*d*e^(2*I*d*x + 2*I*c) + 2*sqr
t(2)*(a*e^(2*I*d*x + 2*I*c) + a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c))*e^(-2*I*d*x - 2*I*c)/a) +
2*sqrt(2)*(d*e^(2*I*d*x + 2*I*c) - d)*sqrt(-a^3/d^2)*log(1/2*(-2*I*sqrt(2)*sqrt(-a^3/d^2)*d*e^(2*I*d*x + 2*I*c
) + 2*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c))*e^(-2*I*d*x - 2*I
*c)/a) + 3*(d*e^(2*I*d*x + 2*I*c) - d)*sqrt(-a^3/d^2)*log(1/3*(3*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*sqrt(a/(e
^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + 6*I*sqrt(-a^3/d^2)*d*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(-a^3/d^2)*log(1/3*(3*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*sqrt(a/(e^(2*I*d*
x + 2*I*c) + 1))*e^(I*d*x + I*c) - 6*I*sqrt(-a^3/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/a))/(d*e^(2*
I*d*x + 2*I*c) - d)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \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))^(3/2),x, algorithm="giac")

[Out]

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