\(\int \cot ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\) [98]

Optimal result

Integrand size = 26, antiderivative size = 184 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {11 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{4 d}-\frac {2 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {i a^2 \cot (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d} \]

[Out]

11/4*a^(3/2)*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))/d-2*a^(3/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2
)/a^(1/2))*2^(1/2)/d-1/2*I*a^2*cot(d*x+c)/d/(a+I*a*tan(d*x+c))^(1/2)-1/2*a^2*cot(d*x+c)^2/d/(a+I*a*tan(d*x+c))
^(1/2)-5/4*I*a*cot(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3634, 3677, 3679, 3681, 3561, 212, 3680, 65, 214} \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {11 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{4 d}-\frac {2 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {i a^2 \cot (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d} \]

[In]

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

[Out]

(11*a^(3/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/(4*d) - (2*Sqrt[2]*a^(3/2)*ArcTanh[Sqrt[a + I*a*Tan[c
+ d*x]]/(Sqrt[2]*Sqrt[a])])/d - ((I/2)*a^2*Cot[c + d*x])/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (a^2*Cot[c + d*x]^2
)/(2*d*Sqrt[a + I*a*Tan[c + d*x]]) - (((5*I)/4)*a*Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/d

Rule 65

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 212

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

Rule 214

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

Rule 3561

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 3634

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-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] /;
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] && GtQ[m, 1] && Lt
Q[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3677

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 3679

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*d - B*c)*(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*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n
+ 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c*m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], 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] && LtQ[n, -1]

Rule 3680

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 3681

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {1}{2} \int \frac {\cot ^2(c+d x) \left (-\frac {7 i a^2}{2}+\frac {9}{2} a^2 \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx \\ & = -\frac {i a^2 \cot (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {\int \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)} \left (-\frac {5 i a^3}{2}+\frac {3}{2} a^3 \tan (c+d x)\right ) \, dx}{2 a^2} \\ & = -\frac {i a^2 \cot (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {\int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} \left (\frac {11 a^4}{4}+\frac {5}{4} i a^4 \tan (c+d x)\right ) \, dx}{2 a^3} \\ & = -\frac {i a^2 \cot (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {11}{8} \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)} \, dx-(2 i a) \int \sqrt {a+i a \tan (c+d x)} \, dx \\ & = -\frac {i a^2 \cot (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {\left (11 a^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{8 d}-\frac {\left (4 a^2\right ) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{d} \\ & = -\frac {2 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {i a^2 \cot (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {(11 i a) \text {Subst}\left (\int \frac {1}{i-\frac {i x^2}{a}} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{4 d} \\ & = \frac {11 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{4 d}-\frac {2 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {i a^2 \cot (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {a^2 \cot ^2(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {5 i a \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.64 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {-11 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )+8 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )+a \cot (c+d x) (5 i+2 \cot (c+d x)) \sqrt {a+i a \tan (c+d x)}}{4 d} \]

[In]

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

[Out]

-1/4*(-11*a^(3/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]] + 8*Sqrt[2]*a^(3/2)*ArcTanh[Sqrt[a + I*a*Tan[c +
d*x]]/(Sqrt[2]*Sqrt[a])] + a*Cot[c + d*x]*(5*I + 2*Cot[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])/d

Maple [A] (verified)

Time = 1.08 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.64

method	result	size
derivativedivides	\(\frac {2 a^{4} \left (-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{a^{\frac {5}{2}}}+\frac {-\frac {\frac {5 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{8}-\frac {3 a \sqrt {a +i a \tan \left (d x +c \right )}}{8}}{a^{2} \tan \left (d x +c \right )^{2}}+\frac {11 \,\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{8 \sqrt {a}}}{a^{2}}\right )}{d}\)	\(117\)
default	\(\frac {2 a^{4} \left (-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{a^{\frac {5}{2}}}+\frac {-\frac {\frac {5 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{8}-\frac {3 a \sqrt {a +i a \tan \left (d x +c \right )}}{8}}{a^{2} \tan \left (d x +c \right )^{2}}+\frac {11 \,\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{8 \sqrt {a}}}{a^{2}}\right )}{d}\)	\(117\)

[In]

int(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/d*a^4*(-1/a^(5/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))+1/a^2*(-(5/8*(a+I*a*tan(d*x+
c))^(3/2)-3/8*a*(a+I*a*tan(d*x+c))^(1/2))/a^2/tan(d*x+c)^2+11/8/a^(1/2)*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/
2))))

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 546 vs. \(2 (141) = 282\).

Time = 0.25 (sec) , antiderivative size = 546, normalized size of antiderivative = 2.97 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {16 \, \sqrt {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 )} \sqrt {\frac {a^{3}}{d^{2}}} \log \left (\frac {4 \, {\left (a^{2} e^{\left (i \, d x + i \, c\right )} + {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a}\right ) - 16 \, \sqrt {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 )} \sqrt {\frac {a^{3}}{d^{2}}} \log \left (\frac {4 \, {\left (a^{2} e^{\left (i \, d x + i \, c\right )} - {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a}\right ) - 11 \, {\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 )} \sqrt {\frac {a^{3}}{d^{2}}} \log \left (16 \, {\left (3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, \sqrt {2} {\left (d e^{\left (3 i \, d x + 3 i \, c\right )} + d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + 11 \, {\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 )} \sqrt {\frac {a^{3}}{d^{2}}} \log \left (16 \, {\left (3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 \, \sqrt {2} {\left (d e^{\left (3 i \, d x + 3 i \, c\right )} + d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - 4 \, \sqrt {2} {\left (7 \, a e^{\left (5 i \, d x + 5 i \, c\right )} + 4 \, a e^{\left (3 i \, d x + 3 i \, c\right )} - 3 \, a e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{16 \, {\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 )}} \]

[In]

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

[Out]

-1/16*(16*sqrt(2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a^3/d^2)*log(4*(a^2*e^(I*d*x + I*
c) + (d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a^3/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a) - 16*sqr
t(2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a^3/d^2)*log(4*(a^2*e^(I*d*x + I*c) - (d*e^(2*
I*d*x + 2*I*c) + d)*sqrt(a^3/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a) - 11*(d*e^(4*I*d*x +
4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a^3/d^2)*log(16*(3*a^2*e^(2*I*d*x + 2*I*c) + 2*sqrt(2)*(d*e^(3*I*d*
x + 3*I*c) + d*e^(I*d*x + I*c))*sqrt(a^3/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)) + a^2)*e^(-2*I*d*x - 2*I*c)) +
11*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a^3/d^2)*log(16*(3*a^2*e^(2*I*d*x + 2*I*c) - 2*
sqrt(2)*(d*e^(3*I*d*x + 3*I*c) + d*e^(I*d*x + I*c))*sqrt(a^3/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)) + a^2)*e^(
-2*I*d*x - 2*I*c)) - 4*sqrt(2)*(7*a*e^(5*I*d*x + 5*I*c) + 4*a*e^(3*I*d*x + 3*I*c) - 3*a*e^(I*d*x + I*c))*sqrt(
a/(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)

Sympy [F]

\[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}} \cot ^{3}{\left (c + d x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.97 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {a^{2} {\left (\frac {8 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{\sqrt {a}} - \frac {11 \, \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{\sqrt {a}} + \frac {2 \, {\left (5 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} - 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a + a^{2}}\right )}}{8 \, d} \]

[In]

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

[Out]

1/8*a^2*(8*sqrt(2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(d*x + c
) + a)))/sqrt(a) - 11*log((sqrt(I*a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + c) + a) + sqrt(a)))/sqrt(
a) + 2*(5*(I*a*tan(d*x + c) + a)^(3/2) - 3*sqrt(I*a*tan(d*x + c) + a)*a)/((I*a*tan(d*x + c) + a)^2 - 2*(I*a*ta
n(d*x + c) + a)*a + a^2))/d

Giac [F]

\[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cot \left (d x + c\right )^{3} \,d x } \]

[In]

integrate(cot(d*x+c)^3*(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)^3, x)

Mupad [B] (verification not implemented)

Time = 5.07 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.74 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {\mathrm {atan}\left (\frac {\sqrt {a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{a^2}\right )\,\sqrt {a^3}\,11{}\mathrm {i}}{4\,d}-\frac {5\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{4\,d\,{\mathrm {tan}\left (c+d\,x\right )}^2}+\frac {3\,a\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{4\,d\,{\mathrm {tan}\left (c+d\,x\right )}^2}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,a^2}\right )\,\sqrt {a^3}\,2{}\mathrm {i}}{d} \]

[In]

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

[Out]

(3*a*(a + a*tan(c + d*x)*1i)^(1/2))/(4*d*tan(c + d*x)^2) - (5*(a + a*tan(c + d*x)*1i)^(3/2))/(4*d*tan(c + d*x)
^2) - (atan(((a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*1i)/a^2)*(a^3)^(1/2)*11i)/(4*d) + (2^(1/2)*atan((2^(1/2
)*(a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*1i)/(2*a^2))*(a^3)^(1/2)*2i)/d