\(\int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [110]

Optimal result

Integrand size = 36, antiderivative size = 259 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {(5 i A-2 B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{5/2} d}+\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {(A+i B) \cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(19 A+9 i B) \cot (c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {(41 A+15 i B) \cot (c+d x)}{12 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {7 (3 A+i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a^3 d} \]

[Out]

(5*I*A-2*B)*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))/a^(5/2)/d+1/8*(I*A+B)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/
2)*2^(1/2)/a^(1/2))/a^(5/2)/d*2^(1/2)+1/12*(41*A+15*I*B)*cot(d*x+c)/a^2/d/(a+I*a*tan(d*x+c))^(1/2)-7/4*(3*A+I*
B)*cot(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/a^3/d+1/5*(A+I*B)*cot(d*x+c)/d/(a+I*a*tan(d*x+c))^(5/2)+1/30*(19*A+9*I*
B)*cot(d*x+c)/a/d/(a+I*a*tan(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 1.24 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3677, 3679, 3681, 3561, 212, 3680, 65, 214} \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {(-2 B+5 i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{5/2} d}+\frac {(B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}-\frac {7 (3 A+i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a^3 d}+\frac {(41 A+15 i B) \cot (c+d x)}{12 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {(19 A+9 i B) \cot (c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {(A+i B) \cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}} \]

[In]

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

[Out]

(((5*I)*A - 2*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/(a^(5/2)*d) + ((I*A + B)*ArcTanh[Sqrt[a + I*a*Ta
n[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(4*Sqrt[2]*a^(5/2)*d) + ((A + I*B)*Cot[c + d*x])/(5*d*(a + I*a*Tan[c + d*x])^(
5/2)) + ((19*A + (9*I)*B)*Cot[c + d*x])/(30*a*d*(a + I*a*Tan[c + d*x])^(3/2)) + ((41*A + (15*I)*B)*Cot[c + d*x
])/(12*a^2*d*Sqrt[a + I*a*Tan[c + d*x]]) - (7*(3*A + I*B)*Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/(4*a^3*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 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+i B) \cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {\int \frac {\cot ^2(c+d x) \left (a (6 A+i B)-\frac {7}{2} a (i A-B) \tan (c+d x)\right )}{(a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2} \\ & = \frac {(A+i B) \cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(19 A+9 i B) \cot (c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {\int \frac {\cot ^2(c+d x) \left (\frac {5}{2} a^2 (11 A+3 i B)-\frac {5}{4} a^2 (19 i A-9 B) \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx}{15 a^4} \\ & = \frac {(A+i B) \cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(19 A+9 i B) \cot (c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {(41 A+15 i B) \cot (c+d x)}{12 a^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 {105}{4} a^3 (3 A+i B)-\frac {15}{8} a^3 (41 i A-15 B) \tan (c+d x)\right ) \, dx}{15 a^6} \\ & = \frac {(A+i B) \cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(19 A+9 i B) \cot (c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {(41 A+15 i B) \cot (c+d x)}{12 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {7 (3 A+i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a^3 d}+\frac {\int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} \left (-\frac {15}{2} a^4 (5 i A-2 B)-\frac {105}{8} a^4 (3 A+i B) \tan (c+d x)\right ) \, dx}{15 a^7} \\ & = \frac {(A+i B) \cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(19 A+9 i B) \cot (c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {(41 A+15 i B) \cot (c+d x)}{12 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {7 (3 A+i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a^3 d}-\frac {(5 i A-2 B) \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)} \, dx}{2 a^4}-\frac {(A-i B) \int \sqrt {a+i a \tan (c+d x)} \, dx}{8 a^3} \\ & = \frac {(A+i B) \cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(19 A+9 i B) \cot (c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {(41 A+15 i B) \cot (c+d x)}{12 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {7 (3 A+i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a^3 d}-\frac {(5 i A-2 B) \text {Subst}\left (\int \frac {1}{x \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{2 a^2 d}+\frac {(i A+B) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{4 a^2 d} \\ & = \frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {(A+i B) \cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(19 A+9 i B) \cot (c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {(41 A+15 i B) \cot (c+d x)}{12 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {7 (3 A+i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a^3 d}-\frac {(5 A+2 i B) \text {Subst}\left (\int \frac {1}{i-\frac {i x^2}{a}} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{a^3 d} \\ & = \frac {(5 i A-2 B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{5/2} d}+\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {(A+i B) \cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(19 A+9 i B) \cot (c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {(41 A+15 i B) \cot (c+d x)}{12 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {7 (3 A+i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.37 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.04 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {12 a^{15/2} (A+i B) \cot (c+d x)-\frac {1}{2} a^7 (i+\cot (c+d x)) \tan ^2(c+d x) \left (2 \sqrt {a} \left (-105 (3 A+i B)+5 i (85 A+27 i B) \cot (c+d x)+12 (6 A+i B) \cot ^2(c+d x)\right )+120 (5 A+2 i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right ) (1-i \cot (c+d x)) \sqrt {a+i a \tan (c+d x)}+15 \sqrt {2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right ) (1-i \cot (c+d x)) \sqrt {a+i a \tan (c+d x)}\right )}{60 a^{15/2} d (a+i a \tan (c+d x))^{5/2}} \]

[In]

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

[Out]

(12*a^(15/2)*(A + I*B)*Cot[c + d*x] - (a^7*(I + Cot[c + d*x])*Tan[c + d*x]^2*(2*Sqrt[a]*(-105*(3*A + I*B) + (5
*I)*(85*A + (27*I)*B)*Cot[c + d*x] + 12*(6*A + I*B)*Cot[c + d*x]^2) + 120*(5*A + (2*I)*B)*ArcTanh[Sqrt[a + I*a
*Tan[c + d*x]]/Sqrt[a]]*(1 - I*Cot[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]] + 15*Sqrt[2]*(A - I*B)*ArcTanh[Sqrt[a
+ I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])]*(1 - I*Cot[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]]))/2)/(60*a^(15/2)*d*(a
+ I*a*Tan[c + d*x])^(5/2))

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.75

method	result	size
derivativedivides	\(\frac {2 i a^{2} \left (-\frac {7 i B +17 A}{8 a^{4} \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {3 i B +5 A}{12 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {i B +A}{10 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {\left (i B -A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 a^{\frac {9}{2}}}+\frac {\frac {i A \sqrt {a +i a \tan \left (d x +c \right )}}{2 a \tan \left (d x +c \right )}+\frac {\left (2 i B +5 A \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{2 \sqrt {a}}}{a^{4}}\right )}{d}\)	\(195\)
default	\(\frac {2 i a^{2} \left (-\frac {7 i B +17 A}{8 a^{4} \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {3 i B +5 A}{12 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {i B +A}{10 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {\left (i B -A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 a^{\frac {9}{2}}}+\frac {\frac {i A \sqrt {a +i a \tan \left (d x +c \right )}}{2 a \tan \left (d x +c \right )}+\frac {\left (2 i B +5 A \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{2 \sqrt {a}}}{a^{4}}\right )}{d}\)	\(195\)

[In]

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

[Out]

2*I/d*a^2*(-1/8/a^4*(17*A+7*I*B)/(a+I*a*tan(d*x+c))^(1/2)-1/12/a^3*(5*A+3*I*B)/(a+I*a*tan(d*x+c))^(3/2)-1/10/a
^2*(A+I*B)/(a+I*a*tan(d*x+c))^(5/2)-1/16*(-A+I*B)/a^(9/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)
/a^(1/2))+1/a^4*(1/2*I*A*(a+I*a*tan(d*x+c))^(1/2)/a/tan(d*x+c)+1/2*(5*A+2*I*B)/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. 834 vs. \(2 (204) = 408\).

Time = 0.29 (sec) , antiderivative size = 834, normalized size of antiderivative = 3.22 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/120*(15*sqrt(1/2)*(a^3*d*e^(7*I*d*x + 7*I*c) - a^3*d*e^(5*I*d*x + 5*I*c))*sqrt(-(A^2 - 2*I*A*B - B^2)/(a^5*
d^2))*log(-4*(sqrt(2)*sqrt(1/2)*(a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(-(A
^2 - 2*I*A*B - B^2)/(a^5*d^2)) + (-I*A - B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(I*A + B)) - 15*sqrt(1/2)*(a^3
*d*e^(7*I*d*x + 7*I*c) - a^3*d*e^(5*I*d*x + 5*I*c))*sqrt(-(A^2 - 2*I*A*B - B^2)/(a^5*d^2))*log(4*(sqrt(2)*sqrt
(1/2)*(a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(-(A^2 - 2*I*A*B - B^2)/(a^5*d
^2)) - (-I*A - B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(I*A + B)) - 30*(a^3*d*e^(7*I*d*x + 7*I*c) - a^3*d*e^(5*
I*d*x + 5*I*c))*sqrt(-(25*A^2 + 20*I*A*B - 4*B^2)/(a^5*d^2))*log(-16*(3*(5*I*A - 2*B)*a^2*e^(2*I*d*x + 2*I*c)
+ (5*I*A - 2*B)*a^2 + 2*sqrt(2)*(a^4*d*e^(3*I*d*x + 3*I*c) + a^4*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c
) + 1))*sqrt(-(25*A^2 + 20*I*A*B - 4*B^2)/(a^5*d^2)))*e^(-2*I*d*x - 2*I*c)/(-5*I*A + 2*B)) + 30*(a^3*d*e^(7*I*
d*x + 7*I*c) - a^3*d*e^(5*I*d*x + 5*I*c))*sqrt(-(25*A^2 + 20*I*A*B - 4*B^2)/(a^5*d^2))*log(-16*(3*(5*I*A - 2*B
)*a^2*e^(2*I*d*x + 2*I*c) + (5*I*A - 2*B)*a^2 - 2*sqrt(2)*(a^4*d*e^(3*I*d*x + 3*I*c) + a^4*d*e^(I*d*x + I*c))*
sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(-(25*A^2 + 20*I*A*B - 4*B^2)/(a^5*d^2)))*e^(-2*I*d*x - 2*I*c)/(-5*I*A +
2*B)) - sqrt(2)*((-403*I*A + 123*B)*e^(8*I*d*x + 8*I*c) + (-151*I*A + 21*B)*e^(6*I*d*x + 6*I*c) - 40*(-7*I*A
+ 3*B)*e^(4*I*d*x + 4*I*c) + (31*I*A - 21*B)*e^(2*I*d*x + 2*I*c) + 3*I*A - 3*B)*sqrt(a/(e^(2*I*d*x + 2*I*c) +
1)))/(a^3*d*e^(7*I*d*x + 7*I*c) - a^3*d*e^(5*I*d*x + 5*I*c))

Sympy [F]

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

[In]

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

[Out]

Integral((A + B*tan(c + d*x))*cot(c + d*x)**2/(I*a*(tan(c + d*x) - I))**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 240, normalized size of antiderivative = 0.93 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {i \, a {\left (\frac {4 \, {\left (105 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} {\left (3 \, A + i \, B\right )} - 5 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} {\left (41 \, A + 15 i \, B\right )} a - 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} {\left (19 \, A + 9 i \, B\right )} a^{2} - 12 \, {\left (A + i \, B\right )} a^{3}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{3} - {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{4}} + \frac {15 \, \sqrt {2} {\left (A - i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {7}{2}}} + \frac {120 \, {\left (5 \, A + 2 i \, B\right )} \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{a^{\frac {7}{2}}}\right )}}{240 \, d} \]

[In]

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

[Out]

-1/240*I*a*(4*(105*(I*a*tan(d*x + c) + a)^3*(3*A + I*B) - 5*(I*a*tan(d*x + c) + a)^2*(41*A + 15*I*B)*a - 2*(I*
a*tan(d*x + c) + a)*(19*A + 9*I*B)*a^2 - 12*(A + I*B)*a^3)/((I*a*tan(d*x + c) + a)^(7/2)*a^3 - (I*a*tan(d*x +
c) + a)^(5/2)*a^4) + 15*sqrt(2)*(A - I*B)*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)))/a^(7/2) + 120*(5*A + 2*I*B)*log((sqrt(I*a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I
*a*tan(d*x + c) + a) + sqrt(a)))/a^(7/2))/d

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.09 (sec) , antiderivative size = 3002, normalized size of antiderivative = 11.59 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

2*atanh((12*a*d^4*(a + a*tan(c + d*x)*1i)^(1/2)*((129*B^2)/(256*a^5*d^2) - (801*A^2)/(256*a^5*d^2) - ((638401*
A^4*a^2)/(16*d^4) + (16129*B^4*a^2)/(16*d^4) - (307555*A^2*B^2*a^2)/(8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) + (A^
3*B*a^2*256479i)/(4*d^4))^(1/2)/(64*a^6) - (A*B*319i)/(128*a^5*d^2))^(1/2)*((638401*A^4*a^2)/(16*d^4) + (16129
*B^4*a^2)/(16*d^4) - (307555*A^2*B^2*a^2)/(8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) + (A^3*B*a^2*256479i)/(4*d^4))^
(1/2))/((A^3*d*31161i)/8 + (2159*B^3*d)/8 - (A*B^2*d*15867i)/8 - (38621*A^2*B*d)/8 + (A*d^3*((638401*A^4*a^2)/
(16*d^4) + (16129*B^4*a^2)/(16*d^4) - (307555*A^2*B^2*a^2)/(8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) + (A^3*B*a^2*2
56479i)/(4*d^4))^(1/2)*41i)/(2*a) - (15*B*d^3*((638401*A^4*a^2)/(16*d^4) + (16129*B^4*a^2)/(16*d^4) - (307555*
A^2*B^2*a^2)/(8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) + (A^3*B*a^2*256479i)/(4*d^4))^(1/2))/(2*a)) + (799*A^2*a^2*
d^2*(a + a*tan(c + d*x)*1i)^(1/2)*((129*B^2)/(256*a^5*d^2) - (801*A^2)/(256*a^5*d^2) - ((638401*A^4*a^2)/(16*d
^4) + (16129*B^4*a^2)/(16*d^4) - (307555*A^2*B^2*a^2)/(8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) + (A^3*B*a^2*256479
i)/(4*d^4))^(1/2)/(64*a^6) - (A*B*319i)/(128*a^5*d^2))^(1/2))/((A^3*d*31161i)/8 + (2159*B^3*d)/8 - (A*B^2*d*15
867i)/8 - (38621*A^2*B*d)/8 + (A*d^3*((638401*A^4*a^2)/(16*d^4) + (16129*B^4*a^2)/(16*d^4) - (307555*A^2*B^2*a
^2)/(8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) + (A^3*B*a^2*256479i)/(4*d^4))^(1/2)*41i)/(2*a) - (15*B*d^3*((638401*
A^4*a^2)/(16*d^4) + (16129*B^4*a^2)/(16*d^4) - (307555*A^2*B^2*a^2)/(8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) + (A^
3*B*a^2*256479i)/(4*d^4))^(1/2))/(2*a)) - (127*B^2*a^2*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*((129*B^2)/(256*a^5*d
^2) - (801*A^2)/(256*a^5*d^2) - ((638401*A^4*a^2)/(16*d^4) + (16129*B^4*a^2)/(16*d^4) - (307555*A^2*B^2*a^2)/(
8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) + (A^3*B*a^2*256479i)/(4*d^4))^(1/2)/(64*a^6) - (A*B*319i)/(128*a^5*d^2))^
(1/2))/((A^3*d*31161i)/8 + (2159*B^3*d)/8 - (A*B^2*d*15867i)/8 - (38621*A^2*B*d)/8 + (A*d^3*((638401*A^4*a^2)/
(16*d^4) + (16129*B^4*a^2)/(16*d^4) - (307555*A^2*B^2*a^2)/(8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) + (A^3*B*a^2*2
56479i)/(4*d^4))^(1/2)*41i)/(2*a) - (15*B*d^3*((638401*A^4*a^2)/(16*d^4) + (16129*B^4*a^2)/(16*d^4) - (307555*
A^2*B^2*a^2)/(8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) + (A^3*B*a^2*256479i)/(4*d^4))^(1/2))/(2*a)) + (A*B*a^2*d^2*
(a + a*tan(c + d*x)*1i)^(1/2)*((129*B^2)/(256*a^5*d^2) - (801*A^2)/(256*a^5*d^2) - ((638401*A^4*a^2)/(16*d^4)
+ (16129*B^4*a^2)/(16*d^4) - (307555*A^2*B^2*a^2)/(8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) + (A^3*B*a^2*256479i)/(
4*d^4))^(1/2)/(64*a^6) - (A*B*319i)/(128*a^5*d^2))^(1/2)*642i)/((A^3*d*31161i)/8 + (2159*B^3*d)/8 - (A*B^2*d*1
5867i)/8 - (38621*A^2*B*d)/8 + (A*d^3*((638401*A^4*a^2)/(16*d^4) + (16129*B^4*a^2)/(16*d^4) - (307555*A^2*B^2*
a^2)/(8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) + (A^3*B*a^2*256479i)/(4*d^4))^(1/2)*41i)/(2*a) - (15*B*d^3*((638401
*A^4*a^2)/(16*d^4) + (16129*B^4*a^2)/(16*d^4) - (307555*A^2*B^2*a^2)/(8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) + (A
^3*B*a^2*256479i)/(4*d^4))^(1/2))/(2*a)))*(-(4*d^2*((((801*A^2*a)/4 - (129*B^2*a)/4)/d^2 + (A*B*a*319i)/(2*d^2
))^2 + 128*a^6*((((3*A*B^3)/4 + (15*A^3*B)/8)*1i)/(a^4*d^4) - ((25*A^4)/16 + (11*A^2*B^2)/16 + B^4/4)/(a^4*d^4
)))^(1/2) + 801*A^2*a - 129*B^2*a + A*B*a*638i)/(256*a^6*d^2))^(1/2) - (((A*a + B*a*1i)*1i)/(5*d) + ((19*A + B
*9i)*(a + a*tan(c + d*x)*1i)*1i)/(30*d) - ((3*A + B*1i)*(a + a*tan(c + d*x)*1i)^3*7i)/(4*a^2*d) + ((41*A + B*1
5i)*(a + a*tan(c + d*x)*1i)^2*1i)/(12*a*d))/(a*(a + a*tan(c + d*x)*1i)^(5/2) - (a + a*tan(c + d*x)*1i)^(7/2))
- 2*atanh((12*a*d^4*(a + a*tan(c + d*x)*1i)^(1/2)*(((638401*A^4*a^2)/(16*d^4) + (16129*B^4*a^2)/(16*d^4) - (30
7555*A^2*B^2*a^2)/(8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) + (A^3*B*a^2*256479i)/(4*d^4))^(1/2)/(64*a^6) - (801*A^
2)/(256*a^5*d^2) + (129*B^2)/(256*a^5*d^2) - (A*B*319i)/(128*a^5*d^2))^(1/2)*((638401*A^4*a^2)/(16*d^4) + (161
29*B^4*a^2)/(16*d^4) - (307555*A^2*B^2*a^2)/(8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) + (A^3*B*a^2*256479i)/(4*d^4)
)^(1/2))/((A^3*d*31161i)/8 + (2159*B^3*d)/8 - (A*B^2*d*15867i)/8 - (38621*A^2*B*d)/8 - (A*d^3*((638401*A^4*a^2
)/(16*d^4) + (16129*B^4*a^2)/(16*d^4) - (307555*A^2*B^2*a^2)/(8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) + (A^3*B*a^2
*256479i)/(4*d^4))^(1/2)*41i)/(2*a) + (15*B*d^3*((638401*A^4*a^2)/(16*d^4) + (16129*B^4*a^2)/(16*d^4) - (30755
5*A^2*B^2*a^2)/(8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) + (A^3*B*a^2*256479i)/(4*d^4))^(1/2))/(2*a)) - (799*A^2*a^
2*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*(((638401*A^4*a^2)/(16*d^4) + (16129*B^4*a^2)/(16*d^4) - (307555*A^2*B^2*a
^2)/(8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) + (A^3*B*a^2*256479i)/(4*d^4))^(1/2)/(64*a^6) - (801*A^2)/(256*a^5*d^
2) + (129*B^2)/(256*a^5*d^2) - (A*B*319i)/(128*a^5*d^2))^(1/2))/((A^3*d*31161i)/8 + (2159*B^3*d)/8 - (A*B^2*d*
15867i)/8 - (38621*A^2*B*d)/8 - (A*d^3*((638401*A^4*a^2)/(16*d^4) + (16129*B^4*a^2)/(16*d^4) - (307555*A^2*B^2
*a^2)/(8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) + (A^3*B*a^2*256479i)/(4*d^4))^(1/2)*41i)/(2*a) + (15*B*d^3*((63840
1*A^4*a^2)/(16*d^4) + (16129*B^4*a^2)/(16*d^4) - (307555*A^2*B^2*a^2)/(8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) + (
A^3*B*a^2*256479i)/(4*d^4))^(1/2))/(2*a)) + (127*B^2*a^2*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*(((638401*A^4*a^2)/
(16*d^4) + (16129*B^4*a^2)/(16*d^4) - (307555*A^2*B^2*a^2)/(8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) + (A^3*B*a^2*2
56479i)/(4*d^4))^(1/2)/(64*a^6) - (801*A^2)/(256*a^5*d^2) + (129*B^2)/(256*a^5*d^2) - (A*B*319i)/(128*a^5*d^2)
)^(1/2))/((A^3*d*31161i)/8 + (2159*B^3*d)/8 - (A*B^2*d*15867i)/8 - (38621*A^2*B*d)/8 - (A*d^3*((638401*A^4*a^2
)/(16*d^4) + (16129*B^4*a^2)/(16*d^4) - (307555*A^2*B^2*a^2)/(8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) + (A^3*B*a^2
*256479i)/(4*d^4))^(1/2)*41i)/(2*a) + (15*B*d^3*((638401*A^4*a^2)/(16*d^4) + (16129*B^4*a^2)/(16*d^4) - (30755
5*A^2*B^2*a^2)/(8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) + (A^3*B*a^2*256479i)/(4*d^4))^(1/2))/(2*a)) - (A*B*a^2*d^
2*(a + a*tan(c + d*x)*1i)^(1/2)*(((638401*A^4*a^2)/(16*d^4) + (16129*B^4*a^2)/(16*d^4) - (307555*A^2*B^2*a^2)/
(8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) + (A^3*B*a^2*256479i)/(4*d^4))^(1/2)/(64*a^6) - (801*A^2)/(256*a^5*d^2) +
(129*B^2)/(256*a^5*d^2) - (A*B*319i)/(128*a^5*d^2))^(1/2)*642i)/((A^3*d*31161i)/8 + (2159*B^3*d)/8 - (A*B^2*d
*15867i)/8 - (38621*A^2*B*d)/8 - (A*d^3*((638401*A^4*a^2)/(16*d^4) + (16129*B^4*a^2)/(16*d^4) - (307555*A^2*B^
2*a^2)/(8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) + (A^3*B*a^2*256479i)/(4*d^4))^(1/2)*41i)/(2*a) + (15*B*d^3*((6384
01*A^4*a^2)/(16*d^4) + (16129*B^4*a^2)/(16*d^4) - (307555*A^2*B^2*a^2)/(8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) +
(A^3*B*a^2*256479i)/(4*d^4))^(1/2))/(2*a)))*((4*d^2*((((801*A^2*a)/4 - (129*B^2*a)/4)/d^2 + (A*B*a*319i)/(2*d^
2))^2 + 128*a^6*((((3*A*B^3)/4 + (15*A^3*B)/8)*1i)/(a^4*d^4) - ((25*A^4)/16 + (11*A^2*B^2)/16 + B^4/4)/(a^4*d^
4)))^(1/2) - 801*A^2*a + 129*B^2*a - A*B*a*638i)/(256*a^6*d^2))^(1/2)