\(\int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\) [104]

Optimal result

Integrand size = 26, antiderivative size = 114 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {5 i a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {4 i \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 114, 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 = {3634, 3681, 3561, 212, 3680, 65, 214} \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {5 i a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {4 i \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d} \]

[In]

Int[Cot[c + d*x]^2*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

((-5*I)*a^(5/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/d + ((4*I)*Sqrt[2]*a^(5/2)*ArcTanh[Sqrt[a + I*a*T
an[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (a^2*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 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 (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}-\int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} \left (-\frac {5 i a^2}{2}+\frac {3}{2} a^2 \tan (c+d x)\right ) \, dx \\ & = -\frac {a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {1}{2} (5 i a) \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)} \, dx-\left (4 a^2\right ) \int \sqrt {a+i a \tan (c+d x)} \, dx \\ & = -\frac {a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {\left (5 i a^3\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {\left (8 i a^3\right ) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{d} \\ & = \frac {4 i \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {\left (5 a^2\right ) \text {Subst}\left (\int \frac {1}{i-\frac {i x^2}{a}} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{d} \\ & = -\frac {5 i a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {4 i \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.96 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {-5 i a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )+4 i \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )-a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d} \]

[In]

Integrate[Cot[c + d*x]^2*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

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

Maple [A] (verified)

Time = 1.07 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.82

method	result	size
derivativedivides	\(\frac {2 i a^{3} \left (\frac {2 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{\sqrt {a}}+\frac {i \sqrt {a +i a \tan \left (d x +c \right )}}{2 a \tan \left (d x +c \right )}-\frac {5 \,\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )}{d}\)	\(94\)
default	\(\frac {2 i a^{3} \left (\frac {2 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{\sqrt {a}}+\frac {i \sqrt {a +i a \tan \left (d x +c \right )}}{2 a \tan \left (d x +c \right )}-\frac {5 \,\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{2 \sqrt {a}}\right )}{d}\)	\(94\)

[In]

int(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2*I/d*a^3*(2*2^(1/2)/a^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))+1/2*I*(a+I*a*tan(d*x+c))^(1
/2)/a/tan(d*x+c)-5/2/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. 511 vs. \(2 (87) = 174\).

Time = 0.25 (sec) , antiderivative size = 511, normalized size of antiderivative = 4.48 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {8 \, \sqrt {2} \sqrt {-\frac {a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {4 \, {\left (a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {-\frac {a^{5}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2}}\right ) - 8 \, \sqrt {2} \sqrt {-\frac {a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {4 \, {\left (a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {-\frac {a^{5}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2}}\right ) + 5 \, \sqrt {-\frac {a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {16 \, {\left (3 \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} - 2 \, \sqrt {2} \sqrt {-\frac {a^{5}}{d^{2}}} {\left (i \, d e^{\left (3 i \, d x + 3 i \, c\right )} + i \, d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a}\right ) - 5 \, \sqrt {-\frac {a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {16 \, {\left (3 \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} - 2 \, \sqrt {2} \sqrt {-\frac {a^{5}}{d^{2}}} {\left (-i \, d e^{\left (3 i \, d x + 3 i \, c\right )} - i \, d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a}\right ) + 4 \, \sqrt {2} {\left (i \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + i \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \]

[In]

integrate(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-1/4*(8*sqrt(2)*sqrt(-a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log(4*(a^3*e^(I*d*x + I*c) + sqrt(-a^5/d^2)*(I*d*e^
(2*I*d*x + 2*I*c) + I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^2) - 8*sqrt(2)*sqrt(-a^5/d^2)*(
d*e^(2*I*d*x + 2*I*c) - d)*log(4*(a^3*e^(I*d*x + I*c) + sqrt(-a^5/d^2)*(-I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt(a
/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^2) + 5*sqrt(-a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log(16*(3*a^
3*e^(2*I*d*x + 2*I*c) + a^3 - 2*sqrt(2)*sqrt(-a^5/d^2)*(I*d*e^(3*I*d*x + 3*I*c) + I*d*e^(I*d*x + I*c))*sqrt(a/
(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/a) - 5*sqrt(-a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log(16*(3*a
^3*e^(2*I*d*x + 2*I*c) + a^3 - 2*sqrt(2)*sqrt(-a^5/d^2)*(-I*d*e^(3*I*d*x + 3*I*c) - I*d*e^(I*d*x + I*c))*sqrt(
a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/a) + 4*sqrt(2)*(I*a^2*e^(3*I*d*x + 3*I*c) + I*a^2*e^(I*d*x
+ I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/(d*e^(2*I*d*x + 2*I*c) - d)

Sympy [F(-1)]

Timed out. \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\text {Timed out} \]

[In]

integrate(cot(d*x+c)**2*(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.17 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {i \, {\left (4 \, \sqrt {2} a^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - 5 \, a^{\frac {3}{2}} \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right ) - \frac {2 i \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a}{\tan \left (d x + c\right )}\right )} a}{2 \, d} \]

[In]

integrate(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

-1/2*I*(4*sqrt(2)*a^(3/2)*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))) - 5*a^(3/2)*log((sqrt(I*a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + c) + a) + sqrt(a)))
- 2*I*sqrt(I*a*tan(d*x + c) + a)*a/tan(d*x + c))*a/d

Giac [F]

\[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cot \left (d x + c\right )^{2} \,d x } \]

[In]

integrate(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(5/2)*cot(d*x + c)^2, x)

Mupad [B] (verification not implemented)

Time = 4.74 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00 \[ \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {-a^5}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a^3}\right )\,\sqrt {-a^5}\,5{}\mathrm {i}}{d}-\frac {a^2\,\mathrm {cot}\left (c+d\,x\right )\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{d}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {-a^5}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,a^3}\right )\,\sqrt {-a^5}\,4{}\mathrm {i}}{d} \]

[In]

int(cot(c + d*x)^2*(a + a*tan(c + d*x)*1i)^(5/2),x)

[Out]

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