\(\int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [109]

Optimal result

Integrand size = 34, antiderivative size = 192 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {2 A \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{5/2} d}+\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {3 A+i B}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac {7 A+i B}{4 a^2 d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

-2*A*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))/a^(5/2)/d+1/8*(A-I*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)+1/4*(7*A+I*B)/a^2/d/(a+I*a*tan(d*x+c))^(1/2)+1/5*(A+I*B)/d/(a+I*a*tan(d*x+c))^(
5/2)+1/6*(3*A+I*B)/a/d/(a+I*a*tan(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 1.11 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.206, Rules used = {3677, 3681, 3561, 212, 3680, 65, 214} \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}-\frac {2 A \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{5/2} d}+\frac {7 A+i B}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {3 A+i B}{6 a d (a+i a \tan (c+d x))^{3/2}} \]

[In]

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

[Out]

(-2*A*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/(a^(5/2)*d) + ((A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]
/(Sqrt[2]*Sqrt[a])])/(4*Sqrt[2]*a^(5/2)*d) + (A + I*B)/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + (3*A + I*B)/(6*a*d
*(a + I*a*Tan[c + d*x])^(3/2)) + (7*A + I*B)/(4*a^2*d*Sqrt[a + I*a*Tan[c + d*x]])

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 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}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {\int \frac {\cot (c+d x) \left (5 a A-\frac {5}{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}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {3 A+i B}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac {\int \frac {\cot (c+d x) \left (15 a^2 A-\frac {15}{4} a^2 (3 i A-B) \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx}{15 a^4} \\ & = \frac {A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {3 A+i B}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac {7 A+i B}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {\int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} \left (15 a^3 A-\frac {15}{8} a^3 (7 i A-B) \tan (c+d x)\right ) \, dx}{15 a^6} \\ & = \frac {A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {3 A+i B}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac {7 A+i B}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {A \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)} \, dx}{a^4}+\frac {(i A+B) \int \sqrt {a+i a \tan (c+d x)} \, dx}{8 a^3} \\ & = \frac {A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {3 A+i B}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac {7 A+i B}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {A \text {Subst}\left (\int \frac {1}{x \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{a^2 d}+\frac {(A-i 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 {(A-i 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}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {3 A+i B}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac {7 A+i B}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {(2 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 )}{a^3 d} \\ & = -\frac {2 A \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{5/2} d}+\frac {(A-i 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}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {3 A+i B}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac {7 A+i B}{4 a^2 d \sqrt {a+i a \tan (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.28 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.13 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {3 a^{13/2} (A+i B)+(a+i a \tan (c+d x)) \left (\frac {5}{2} a^{11/2} (3 A+i B)+\frac {15}{8} a^4 (a+i a \tan (c+d x)) \left (2 \sqrt {a} (7 A+i B)-16 A \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {a+i a \tan (c+d x)}+\sqrt {2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right ) \sqrt {a+i a \tan (c+d x)}\right )\right )}{15 a^{13/2} d (a+i a \tan (c+d x))^{5/2}} \]

[In]

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

[Out]

(3*a^(13/2)*(A + I*B) + (a + I*a*Tan[c + d*x])*((5*a^(11/2)*(3*A + I*B))/2 + (15*a^4*(a + I*a*Tan[c + d*x])*(2
*Sqrt[a]*(7*A + I*B) - 16*A*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]]*Sqrt[a + I*a*Tan[c + d*x]] + Sqrt[2]*(
A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])]*Sqrt[a + I*a*Tan[c + d*x]]))/8))/(15*a^(13/2)*d
*(a + I*a*Tan[c + d*x])^(5/2))

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.80

method	result	size
derivativedivides	\(\frac {2 a \left (-\frac {-i B -7 A}{8 a^{3} \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {-i B -3 A}{12 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {-i B -A}{10 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {\left (i B -A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 a^{\frac {7}{2}}}-\frac {A \,\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{a^{\frac {7}{2}}}\right )}{d}\)	\(153\)
default	\(\frac {2 a \left (-\frac {-i B -7 A}{8 a^{3} \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {-i B -3 A}{12 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {-i B -A}{10 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {\left (i B -A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 a^{\frac {7}{2}}}-\frac {A \,\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{a^{\frac {7}{2}}}\right )}{d}\)	\(153\)

[In]

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

[Out]

2/d*a*(-1/8/a^3*(-7*A-I*B)/(a+I*a*tan(d*x+c))^(1/2)-1/12*(-3*A-I*B)/a^2/(a+I*a*tan(d*x+c))^(3/2)-1/10*(-A-I*B)
/a/(a+I*a*tan(d*x+c))^(5/2)-1/16*(-A+I*B)/a^(7/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2)
)-1/a^(7/2)*A*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. 644 vs. \(2 (145) = 290\).

Time = 0.27 (sec) , antiderivative size = 644, normalized size of antiderivative = 3.35 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {{\left (15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} + {\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} + {\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) + 60 \, a^{3} d \sqrt {\frac {A^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac {16 \, {\left (3 \, A a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + A a^{2} + 2 \, \sqrt {2} {\left (a^{4} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{4} d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {A^{2}}{a^{5} d^{2}}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{A}\right ) - 60 \, a^{3} d \sqrt {\frac {A^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac {16 \, {\left (3 \, A a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + A a^{2} - 2 \, \sqrt {2} {\left (a^{4} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{4} d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {A^{2}}{a^{5} d^{2}}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{A}\right ) - \sqrt {2} {\left ({\left (123 \, A + 23 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, {\left (72 \, A + 17 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, {\left (12 \, A + 7 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 \, A + 3 i \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{120 \, a^{3} d} \]

[In]

integrate(cot(d*x+c)*(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*sqrt((A^2 - 2*I*A*B - B^2)/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(-4*(sqrt(2)*sqrt(1/2)
*(I*a^3*d*e^(2*I*d*x + 2*I*c) + I*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*sqrt((A^2 - 2*I*A*B - B^2)
/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(-4*(sqrt(2)*sqrt(1/2)*(-I*a^3*d*e^(2*I*d*x + 2*I*c) - I*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)) + 60*a^3*d*sqrt(A^2/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(16*(3*A*a^2*e^(2*I*d*x + 2*I*c) + A*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(A^2/(a^5*d^
2)))*e^(-2*I*d*x - 2*I*c)/A) - 60*a^3*d*sqrt(A^2/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(16*(3*A*a^2*e^(2*I*d*x + 2
*I*c) + A*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(A^2/(a^5*d^2)))*e^(-2*I*d*x - 2*I*c)/A) - sqrt(2)*((123*A + 23*I*B)*e^(6*I*d*x + 6*I*c) + 2*(72*A + 17*
I*B)*e^(4*I*d*x + 4*I*c) + 2*(12*A + 7*I*B)*e^(2*I*d*x + 2*I*c) + 3*A + 3*I*B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1
)))*e^(-5*I*d*x - 5*I*c)/(a^3*d)

Sympy [F]

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

[In]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.97 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\frac {15 \, \sqrt {2} {\left (A - i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {5}{2}}} - \frac {240 \, A \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{a^{\frac {5}{2}}} - \frac {4 \, {\left (15 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} {\left (7 \, A + i \, B\right )} + 10 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} {\left (3 \, A + i \, B\right )} a + 12 \, {\left (A + i \, B\right )} a^{2}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2}}}{240 \, d} \]

[In]

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

[Out]

-1/240*(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*t
an(d*x + c) + a)))/a^(5/2) - 240*A*log((sqrt(I*a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + c) + a) + sq
rt(a)))/a^(5/2) - 4*(15*(I*a*tan(d*x + c) + a)^2*(7*A + I*B) + 10*(I*a*tan(d*x + c) + a)*(3*A + I*B)*a + 12*(A
+ I*B)*a^2)/((I*a*tan(d*x + c) + a)^(5/2)*a^2))/d

Giac [F]

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

[In]

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

Mupad [B] (verification not implemented)

Time = 7.83 (sec) , antiderivative size = 528, normalized size of antiderivative = 2.75 \[ \int \frac {\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\frac {A+B\,1{}\mathrm {i}}{5\,d}+\frac {\left (3\,A+B\,1{}\mathrm {i}\right )\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{6\,a\,d}+\frac {\left (7\,A+B\,1{}\mathrm {i}\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2}{4\,a^2\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}-\frac {2\,A\,\mathrm {atanh}\left (\frac {127\,A^3\,a\,d\,\sqrt {\frac {1}{a^3}}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{127\,d\,A^3+2{}\mathrm {i}\,d\,A^2\,B+d\,A\,B^2}+\frac {A\,B^2\,a\,d\,\sqrt {\frac {1}{a^3}}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{127\,d\,A^3+2{}\mathrm {i}\,d\,A^2\,B+d\,A\,B^2}+\frac {A^2\,B\,a\,d\,\sqrt {\frac {1}{a^3}}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,2{}\mathrm {i}}{127\,d\,A^3+2{}\mathrm {i}\,d\,A^2\,B+d\,A\,B^2}\right )\,\sqrt {\frac {1}{a^3}}}{a\,d}+\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,A^3\,a\,d\,\sqrt {-\frac {1}{a^3}}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,127{}\mathrm {i}}{32\,\left (\frac {127\,d\,A^3}{16}-\frac {125{}\mathrm {i}\,d\,A^2\,B}{16}+\frac {3\,d\,A\,B^2}{16}-\frac {1{}\mathrm {i}\,d\,B^3}{16}\right )}+\frac {\sqrt {2}\,B^3\,a\,d\,\sqrt {-\frac {1}{a^3}}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{32\,\left (\frac {127\,d\,A^3}{16}-\frac {125{}\mathrm {i}\,d\,A^2\,B}{16}+\frac {3\,d\,A\,B^2}{16}-\frac {1{}\mathrm {i}\,d\,B^3}{16}\right )}+\frac {\sqrt {2}\,A\,B^2\,a\,d\,\sqrt {-\frac {1}{a^3}}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,3{}\mathrm {i}}{32\,\left (\frac {127\,d\,A^3}{16}-\frac {125{}\mathrm {i}\,d\,A^2\,B}{16}+\frac {3\,d\,A\,B^2}{16}-\frac {1{}\mathrm {i}\,d\,B^3}{16}\right )}+\frac {125\,\sqrt {2}\,A^2\,B\,a\,d\,\sqrt {-\frac {1}{a^3}}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{32\,\left (\frac {127\,d\,A^3}{16}-\frac {125{}\mathrm {i}\,d\,A^2\,B}{16}+\frac {3\,d\,A\,B^2}{16}-\frac {1{}\mathrm {i}\,d\,B^3}{16}\right )}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,\sqrt {-\frac {1}{a^3}}}{8\,a\,d} \]

[In]

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

[Out]

((A + B*1i)/(5*d) + ((3*A + B*1i)*(a + a*tan(c + d*x)*1i))/(6*a*d) + ((7*A + B*1i)*(a + a*tan(c + d*x)*1i)^2)/
(4*a^2*d))/(a + a*tan(c + d*x)*1i)^(5/2) - (2*A*atanh((127*A^3*a*d*(1/a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)
)/(127*A^3*d + A*B^2*d + A^2*B*d*2i) + (A*B^2*a*d*(1/a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(127*A^3*d + A*
B^2*d + A^2*B*d*2i) + (A^2*B*a*d*(1/a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*2i)/(127*A^3*d + A*B^2*d + A^2*B*
d*2i))*(1/a^3)^(1/2))/(a*d) + (2^(1/2)*atanh((2^(1/2)*A^3*a*d*(-1/a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*127
i)/(32*((127*A^3*d)/16 - (B^3*d*1i)/16 + (3*A*B^2*d)/16 - (A^2*B*d*125i)/16)) + (2^(1/2)*B^3*a*d*(-1/a^3)^(1/2
)*(a + a*tan(c + d*x)*1i)^(1/2))/(32*((127*A^3*d)/16 - (B^3*d*1i)/16 + (3*A*B^2*d)/16 - (A^2*B*d*125i)/16)) +
(2^(1/2)*A*B^2*a*d*(-1/a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*3i)/(32*((127*A^3*d)/16 - (B^3*d*1i)/16 + (3*A
*B^2*d)/16 - (A^2*B*d*125i)/16)) + (125*2^(1/2)*A^2*B*a*d*(-1/a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(32*((
127*A^3*d)/16 - (B^3*d*1i)/16 + (3*A*B^2*d)/16 - (A^2*B*d*125i)/16)))*(A*1i + B)*(-1/a^3)^(1/2))/(8*a*d)