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

Optimal. Leaf size=307 \[ \frac {-B+i A}{2 d \cot ^{\frac {3}{2}}(c+d x) (a \cot (c+d x)+i a)}-\frac {3 A+7 i B}{6 a d \cot ^{\frac {3}{2}}(c+d x)}-\frac {5 (-B+i A)}{2 a d \sqrt {\cot (c+d x)}}+\frac {((3-5 i) A+(5+7 i) B) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{8 \sqrt {2} a d}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) ((1+4 i) A-(6+i) B) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a d}+\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) ((4+i) A+(1+6 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a d}-\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) ((4+i) A+(1+6 i) B) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a d} \]

[Out]

1/6*(-3*A-7*I*B)/a/d/cot(d*x+c)^(3/2)+1/2*(I*A-B)/d/cot(d*x+c)^(3/2)/(I*a+a*cot(d*x+c))-(1/8+1/8*I)*((4+I)*A+(
1+6*I)*B)*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))/a/d*2^(1/2)-(1/8+1/8*I)*((4+I)*A+(1+6*I)*B)*arctan(1+2^(1/2)*cot
(d*x+c)^(1/2))/a/d*2^(1/2)+1/16*((3-5*I)*A+(5+7*I)*B)*ln(1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(1/2))/a/d*2^(1/2)+(1
/16+1/16*I)*((1+4*I)*A-(6+I)*B)*ln(1+cot(d*x+c)+2^(1/2)*cot(d*x+c)^(1/2))/a/d*2^(1/2)-5/2*(I*A-B)/a/d/cot(d*x+
c)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.51, antiderivative size = 307, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {3581, 3596, 3529, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac {-B+i A}{2 d \cot ^{\frac {3}{2}}(c+d x) (a \cot (c+d x)+i a)}-\frac {3 A+7 i B}{6 a d \cot ^{\frac {3}{2}}(c+d x)}-\frac {5 (-B+i A)}{2 a d \sqrt {\cot (c+d x)}}+\frac {((3-5 i) A+(5+7 i) B) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{8 \sqrt {2} a d}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) ((1+4 i) A-(6+i) B) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a d}+\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) ((4+i) A+(1+6 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a d}-\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) ((4+i) A+(1+6 i) B) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1/4 + I/4)*((4 + I)*A + (1 + 6*I)*B)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*a*d) - ((1/4 + I/4)*((
4 + I)*A + (1 + 6*I)*B)*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*a*d) - (3*A + (7*I)*B)/(6*a*d*Cot[c +
d*x]^(3/2)) - (5*(I*A - B))/(2*a*d*Sqrt[Cot[c + d*x]]) + (I*A - B)/(2*d*Cot[c + d*x]^(3/2)*(I*a + a*Cot[c + d
*x])) + (((3 - 5*I)*A + (5 + 7*I)*B)*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(8*Sqrt[2]*a*d) + ((1
/8 + I/8)*((1 + 4*I)*A - (6 + I)*B)*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(Sqrt[2]*a*d)

Rule 204

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

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1168

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[-(a*c)]

Rule 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3534

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 3581

Int[(cot[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.)
+ (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Cot[e + f*x])^(p - m - n)*(b + a*Cot[e + f*x])^m*(d
+ c*Cot[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&  !IntegerQ[p] && IntegerQ[m] && IntegerQ
[n]

Rule 3596

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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 3.02, size = 242, normalized size = 0.79 \[ \frac {(\cos (d x)+i \sin (d x)) (A+B \tan (c+d x)) \left (\frac {2}{3} \sec (c+d x) (\cos (d x)-i \sin (d x)) (4 (3 A+2 i B) \sin (2 (c+d x))+(11 B-15 i A) \cos (2 (c+d x))-15 i A+19 B)-(1+i) (\cos (c)+i \sin (c)) \sqrt {\sin (2 (c+d x))} \csc (c+d x) \left (((4+i) A+(1+6 i) B) \sin ^{-1}(\cos (c+d x)-\sin (c+d x))+((6+i) B-(1+4 i) A) \log \left (\sin (c+d x)+\sqrt {\sin (2 (c+d x))}+\cos (c+d x)\right )\right )\right )}{8 d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x)) (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[d*x] + I*Sin[d*x])*((-1 - I)*Csc[c + d*x]*(((4 + I)*A + (1 + 6*I)*B)*ArcSin[Cos[c + d*x] - Sin[c + d*x]]
+ ((-1 - 4*I)*A + (6 + I)*B)*Log[Cos[c + d*x] + Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]])*(Cos[c] + I*Sin[c])*S
qrt[Sin[2*(c + d*x)]] + (2*Sec[c + d*x]*(Cos[d*x] - I*Sin[d*x])*((-15*I)*A + 19*B + ((-15*I)*A + 11*B)*Cos[2*(
c + d*x)] + 4*(3*A + (2*I)*B)*Sin[2*(c + d*x)]))/3)*(A + B*Tan[c + d*x]))/(8*d*Sqrt[Cot[c + d*x]]*(A*Cos[c + d
*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x]))

________________________________________________________________________________________

fricas [B]  time = 0.66, size = 796, normalized size = 2.59 \[ \frac {3 \, {\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{2} d^{2}}} \log \left (\frac {{\left ({\left (4 i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - 4 i \, a 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 \, A^{2} + 2 \, A B - i \, B^{2}}{a^{2} d^{2}}} - 4 \, {\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{2 \, {\left (i \, A + B\right )}}\right ) - 3 \, {\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{2} d^{2}}} \log \left (\frac {{\left ({\left (-4 i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + 4 i \, a 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 \, A^{2} + 2 \, A B - i \, B^{2}}{a^{2} d^{2}}} - 4 \, {\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{2 \, {\left (i \, A + B\right )}}\right ) - 6 \, {\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {-4 i \, A^{2} + 12 \, A B + 9 i \, B^{2}}{a^{2} d^{2}}} \log \left (-\frac {{\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} - a 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 {-4 i \, A^{2} + 12 \, A B + 9 i \, B^{2}}{a^{2} d^{2}}} + 2 i \, A - 3 \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a d}\right ) + 6 \, {\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {-4 i \, A^{2} + 12 \, A B + 9 i \, B^{2}}{a^{2} d^{2}}} \log \left (\frac {{\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} - a 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 {-4 i \, A^{2} + 12 \, A B + 9 i \, B^{2}}{a^{2} d^{2}}} - 2 i \, A + 3 \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a d}\right ) - 2 \, {\left ({\left (27 \, A + 19 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (3 \, A + 19 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (27 \, A + 35 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 3 \, A - 3 i \, B\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}}}{24 \, {\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(a*d*e^(6*I*d*x + 6*I*c) + 2*a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))*sqrt((I*A^2 + 2*A*B -
I*B^2)/(a^2*d^2))*log(1/2*((4*I*a*d*e^(2*I*d*x + 2*I*c) - 4*I*a*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*
x + 2*I*c) - 1))*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^2*d^2)) - 4*(A - I*B)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*
c)/(I*A + B)) - 3*(a*d*e^(6*I*d*x + 6*I*c) + 2*a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))*sqrt((I*A^2
+ 2*A*B - I*B^2)/(a^2*d^2))*log(1/2*((-4*I*a*d*e^(2*I*d*x + 2*I*c) + 4*I*a*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)
/(e^(2*I*d*x + 2*I*c) - 1))*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^2*d^2)) - 4*(A - I*B)*e^(2*I*d*x + 2*I*c))*e^(-2*I
*d*x - 2*I*c)/(I*A + B)) - 6*(a*d*e^(6*I*d*x + 6*I*c) + 2*a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))*s
qrt((-4*I*A^2 + 12*A*B + 9*I*B^2)/(a^2*d^2))*log(-((a*d*e^(2*I*d*x + 2*I*c) - a*d)*sqrt((I*e^(2*I*d*x + 2*I*c)
+ I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt((-4*I*A^2 + 12*A*B + 9*I*B^2)/(a^2*d^2)) + 2*I*A - 3*B)*e^(-2*I*d*x - 2*
I*c)/(a*d)) + 6*(a*d*e^(6*I*d*x + 6*I*c) + 2*a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))*sqrt((-4*I*A^2
+ 12*A*B + 9*I*B^2)/(a^2*d^2))*log(((a*d*e^(2*I*d*x + 2*I*c) - a*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*
d*x + 2*I*c) - 1))*sqrt((-4*I*A^2 + 12*A*B + 9*I*B^2)/(a^2*d^2)) - 2*I*A + 3*B)*e^(-2*I*d*x - 2*I*c)/(a*d)) -
2*((27*A + 19*I*B)*e^(6*I*d*x + 6*I*c) + (3*A + 19*I*B)*e^(4*I*d*x + 4*I*c) - (27*A + 35*I*B)*e^(2*I*d*x + 2*I
*c) - 3*A - 3*I*B)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))/(a*d*e^(6*I*d*x + 6*I*c) + 2*a
*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \tan \left (d x + c\right ) + A}{{\left (i \, a \tan \left (d x + c\right ) + a\right )} \cot \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 1.95, size = 3871, normalized size = 12.61 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12/a/d*(-1+cos(d*x+c))*(3*I*A*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/
2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(-sin(d*x+c)-1+cos(d*x+
c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)+15*A*EllipticF((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/
2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(-sin(d*x+c)-1+c
os(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)-11*B*2^(1/2)*cos(d*x+c)^3+11*B*2^(1/2)*cos(d*x+c)^2-4*B*2^(1/2)*cos(d*
x+c)+12*A*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*((-1+cos(d*x+c))/si
n(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*c
os(d*x+c)^3+4*B*2^(1/2)-8*I*B*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)+8*I*B*cos(d*x+c)*sin(d*x+c)*2^(1/2)-15*I*A*2^(1/
2)*cos(d*x+c)^2+15*I*A*2^(1/2)*cos(d*x+c)^3-18*B*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2
-1/2*I,1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(-sin(d
*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^3-15*A*EllipticF((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/
2),1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(-sin(d*x+c
)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^3+3*A*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1
/2+1/2*I,1/2*2^(1/2))*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(
1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^3-3*B*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(
1/2),1/2+1/2*I,1/2*2^(1/2))*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x
+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^3-3*A*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x
+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/s
in(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)+3*B*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin
(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c
))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)+12*I*A*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+
c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/si
n(d*x+c))^(1/2)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)-15*I*A*EllipticF((-(-si
n(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d
*x+c))/sin(d*x+c))^(1/2)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)-3*I*B*Elliptic
Pi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((
-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*
x+c)-18*I*B*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*((-1+cos(d*x+c))/
sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)
*cos(d*x+c)^2*sin(d*x+c)-12*A*2^(1/2)*sin(d*x+c)*cos(d*x+c)^2+12*A*2^(1/2)*cos(d*x+c)*sin(d*x+c)-18*I*B*Ellipt
icPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*
((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)+3*I*B*
EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(-(-sin(d*x+c)-1+cos(d*x+c))/
sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)-
12*I*A*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d
*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(
d*x+c)+3*A*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(-sin(d*x+c)-1+c
os(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)
,1/2+1/2*I,1/2*2^(1/2))+15*B*sin(d*x+c)*EllipticF((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*
((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(-sin(d*x+c)-1+cos(d*x+c))/
sin(d*x+c))^(1/2)*cos(d*x+c)^2+3*B*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^
(1/2)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*EllipticPi((-(-sin(d*x+c)-1+cos(d
*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))-15*I*B*EllipticF((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2
),1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(-sin(d*x+c)
-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^3+3*I*A*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),
1/2+1/2*I,1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(-si
n(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)-12*A*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+c
os(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x
+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(d*x+c)+18*B*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*(
(-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(-sin(d*x+c)-1+co
s(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(d*x+c)+12*I*A*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/s
in(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x
+c))^(1/2)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^3-18*B*EllipticPi((-(-sin(d*x+c)-1+cos(d*
x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/
sin(d*x+c))^(1/2)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)-12*A*EllipticPi((-(-s
in(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d
*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)+15*
I*B*EllipticF((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*(
(-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)-3*I*B*E
llipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(
1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^3
-3*I*A*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d
*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(
d*x+c)^3+18*I*B*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*((-1+cos(d*x+
c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(
1/2)*cos(d*x+c)^3)*cos(d*x+c)*(1+cos(d*x+c))^2/(I*sin(d*x+c)+cos(d*x+c))/(cos(d*x+c)/sin(d*x+c))^(5/2)/sin(d*x
+c)^6*2^(1/2)

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+B\,\mathrm {tan}\left (c+d\,x\right )}{{\mathrm {cot}\left (c+d\,x\right )}^{5/2}\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________