\(\int \frac {1}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx\) [748]

Optimal result

Integrand size = 26, antiderivative size = 141 \[ \int \frac {1}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=-\frac {(-1)^{3/4} \text {arctanh}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{8 a^3 d}+\frac {\sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {i \sqrt {\cot (c+d x)}}{6 a d (i a+a \cot (c+d x))^2}+\frac {\sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )} \]

[Out]

-1/8*(-1)^(3/4)*arctanh((-1)^(3/4)*cot(d*x+c)^(1/2))/a^3/d+1/6*cot(d*x+c)^(1/2)/d/(I*a+a*cot(d*x+c))^3+1/6*I*c
ot(d*x+c)^(1/2)/a/d/(I*a+a*cot(d*x+c))^2+1/8*cot(d*x+c)^(1/2)/d/(I*a^3+a^3*cot(d*x+c))

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {3754, 3639, 3677, 12, 3630, 3614, 214} \[ \int \frac {1}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=-\frac {(-1)^{3/4} \text {arctanh}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{8 a^3 d}+\frac {\sqrt {\cot (c+d x)}}{8 d \left (a^3 \cot (c+d x)+i a^3\right )}+\frac {i \sqrt {\cot (c+d x)}}{6 a d (a \cot (c+d x)+i a)^2}+\frac {\sqrt {\cot (c+d x)}}{6 d (a \cot (c+d x)+i a)^3} \]

[In]

Int[1/(Cot[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^3),x]

[Out]

-1/8*((-1)^(3/4)*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/(a^3*d) + Sqrt[Cot[c + d*x]]/(6*d*(I*a + a*Cot[c + d*
x])^3) + ((I/6)*Sqrt[Cot[c + d*x]])/(a*d*(I*a + a*Cot[c + d*x])^2) + Sqrt[Cot[c + d*x]]/(8*d*(I*a^3 + a^3*Cot[
c + d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 3614

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 3630

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(
a*c + b*d))*((c + d*Tan[e + f*x])^n/(2*(b*c - a*d)*f*(a + b*Tan[e + f*x]))), x] + Dist[1/(2*a*(b*c - a*d)), In
t[(c + d*Tan[e + f*x])^(n - 1)*Simp[a*c*d*(n - 1) + b*c^2 + b*d^2*n - d*(b*c - a*d)*(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] && LtQ[0
, n, 1]

Rule 3639

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-(b*c - a*d))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n - 1)/(2*a*f*m)), x] + Dist[1/(2*a^2*m), Int[(
a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1)
) - d*(b*d*(m - n + 1) - a*c*(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] && LtQ[m, 0] && GtQ[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 3754

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]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\cot ^{\frac {3}{2}}(c+d x)}{(i a+a \cot (c+d x))^3} \, dx \\ & = \frac {\sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {\int \frac {-\frac {i a}{2}+\frac {7}{2} a \cot (c+d x)}{\sqrt {\cot (c+d x)} (i a+a \cot (c+d x))^2} \, dx}{6 a^2} \\ & = \frac {\sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {i \sqrt {\cot (c+d x)}}{6 a d (i a+a \cot (c+d x))^2}+\frac {\int -\frac {6 i a^2 \sqrt {\cot (c+d x)}}{i a+a \cot (c+d x)} \, dx}{24 a^4} \\ & = \frac {\sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {i \sqrt {\cot (c+d x)}}{6 a d (i a+a \cot (c+d x))^2}-\frac {i \int \frac {\sqrt {\cot (c+d x)}}{i a+a \cot (c+d x)} \, dx}{4 a^2} \\ & = \frac {\sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {i \sqrt {\cot (c+d x)}}{6 a d (i a+a \cot (c+d x))^2}+\frac {\sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}+\frac {i \int \frac {-\frac {a}{2}+\frac {1}{2} i a \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx}{8 a^4} \\ & = \frac {\sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {i \sqrt {\cot (c+d x)}}{6 a d (i a+a \cot (c+d x))^2}+\frac {\sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}+\frac {i \text {Subst}\left (\int \frac {1}{\frac {a}{2}+\frac {1}{2} i a x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{16 a^2 d} \\ & = -\frac {(-1)^{3/4} \text {arctanh}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{8 a^3 d}+\frac {\sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {i \sqrt {\cot (c+d x)}}{6 a d (i a+a \cot (c+d x))^2}+\frac {\sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.64 \[ \int \frac {1}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=\frac {\sqrt {\cot (c+d x)} \left (\frac {-3+10 i \cot (c+d x)+3 \cot ^2(c+d x)}{(i+\cot (c+d x))^3}+3 \sqrt [4]{-1} \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) \sqrt {\tan (c+d x)}\right )}{24 a^3 d} \]

[In]

Integrate[1/(Cot[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^3),x]

[Out]

(Sqrt[Cot[c + d*x]]*((-3 + (10*I)*Cot[c + d*x] + 3*Cot[c + d*x]^2)/(I + Cot[c + d*x])^3 + 3*(-1)^(1/4)*ArcTan[
(-1)^(3/4)*Sqrt[Tan[c + d*x]]]*Sqrt[Tan[c + d*x]]))/(24*a^3*d)

Maple [A] (verified)

Time = 2.03 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.63

method	result	size
derivativedivides	\(\frac {\frac {\arctan \left (\frac {2 \left (\sqrt {\cot }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{4 \sqrt {2}-4 i \sqrt {2}}-\frac {-\left (\cot ^{\frac {5}{2}}\left (d x +c \right )\right )-\frac {10 i \left (\cot ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+\sqrt {\cot }\left (d x +c \right )}{8 \left (i+\cot \left (d x +c \right )\right )^{3}}}{a^{3} d}\)	\(89\)
default	\(\frac {\frac {\arctan \left (\frac {2 \left (\sqrt {\cot }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{4 \sqrt {2}-4 i \sqrt {2}}-\frac {-\left (\cot ^{\frac {5}{2}}\left (d x +c \right )\right )-\frac {10 i \left (\cot ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}+\sqrt {\cot }\left (d x +c \right )}{8 \left (i+\cot \left (d x +c \right )\right )^{3}}}{a^{3} d}\)	\(89\)

[In]

int(1/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/a^3/d*(1/4/(2^(1/2)-I*2^(1/2))*arctan(2*cot(d*x+c)^(1/2)/(2^(1/2)-I*2^(1/2)))-1/8*(-cot(d*x+c)^(5/2)-10/3*I*
cot(d*x+c)^(3/2)+cot(d*x+c)^(1/2))/(I+cot(d*x+c))^3)

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. 307 vs. \(2 (113) = 226\).

Time = 0.26 (sec) , antiderivative size = 307, normalized size of antiderivative = 2.18 \[ \int \frac {1}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=\frac {{\left (12 \, a^{3} d \sqrt {-\frac {i}{64 \, a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-2 \, {\left (8 \, {\left (i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{3} d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {i}{64 \, a^{6} d^{2}}} - i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - 12 \, a^{3} d \sqrt {-\frac {i}{64 \, a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-2 \, {\left (8 \, {\left (-i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{3} d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {i}{64 \, a^{6} d^{2}}} - i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} {\left (-4 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 4 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{48 \, a^{3} d} \]

[In]

integrate(1/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/48*(12*a^3*d*sqrt(-1/64*I/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(-2*(8*(I*a^3*d*e^(2*I*d*x + 2*I*c) - I*a^3*d)*s
qrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(-1/64*I/(a^6*d^2)) - I*e^(2*I*d*x + 2*I*c))*e^
(-2*I*d*x - 2*I*c)) - 12*a^3*d*sqrt(-1/64*I/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(-2*(8*(-I*a^3*d*e^(2*I*d*x + 2*
I*c) + I*a^3*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(-1/64*I/(a^6*d^2)) - I*e^(2*I
*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)) + sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*(-4*I*e^(6*
I*d*x + 6*I*c) + 4*I*e^(4*I*d*x + 4*I*c) + I*e^(2*I*d*x + 2*I*c) - I))*e^(-6*I*d*x - 6*I*c)/(a^3*d)

Sympy [F]

\[ \int \frac {1}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=\frac {i \int \frac {1}{\tan ^{3}{\left (c + d x \right )} \cot ^{\frac {3}{2}}{\left (c + d x \right )} - 3 i \tan ^{2}{\left (c + d x \right )} \cot ^{\frac {3}{2}}{\left (c + d x \right )} - 3 \tan {\left (c + d x \right )} \cot ^{\frac {3}{2}}{\left (c + d x \right )} + i \cot ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx}{a^{3}} \]

[In]

integrate(1/cot(d*x+c)**(3/2)/(a+I*a*tan(d*x+c))**3,x)

[Out]

I*Integral(1/(tan(c + d*x)**3*cot(c + d*x)**(3/2) - 3*I*tan(c + d*x)**2*cot(c + d*x)**(3/2) - 3*tan(c + d*x)*c
ot(c + d*x)**(3/2) + I*cot(c + d*x)**(3/2)), x)/a**3

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(1/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [F]

\[ \int \frac {1}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=\int { \frac {1}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} \cot \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate(1/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx=\int \frac {1}{{\mathrm {cot}\left (c+d\,x\right )}^{3/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3} \,d x \]

[In]

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

[Out]

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