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

Optimal. Leaf size=297 \[ \frac {(A+i B) \cot ^{\frac {5}{2}}(c+d x)}{2 d (a \cot (c+d x)+i a)}-\frac {(7 A+3 i B) \cot ^{\frac {3}{2}}(c+d x)}{6 a d}+\frac {5 (-B+i A) \sqrt {\cot (c+d x)}}{2 a d}+\frac {((7+5 i) A-(5-3 i) B) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{8 \sqrt {2} a d}+\frac {((5-3 i) B-(7+5 i) A) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{8 \sqrt {2} a d}-\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) ((6+i) A+(1+4 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a d}+\frac {((7-5 i) A+(5+3 i) B) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{4 \sqrt {2} a d} \]

[Out]

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

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Rubi [A]  time = 0.52, antiderivative size = 297, 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, 3595, 3528, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac {(A+i B) \cot ^{\frac {5}{2}}(c+d x)}{2 d (a \cot (c+d x)+i a)}-\frac {(7 A+3 i B) \cot ^{\frac {3}{2}}(c+d x)}{6 a d}+\frac {5 (-B+i A) \sqrt {\cot (c+d x)}}{2 a d}+\frac {((7+5 i) A-(5-3 i) B) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{8 \sqrt {2} a d}+\frac {((5-3 i) B-(7+5 i) A) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{8 \sqrt {2} a d}-\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) ((6+i) A+(1+4 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a d}+\frac {((7-5 i) A+(5+3 i) B) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{4 \sqrt {2} a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/4 + I/4)*((6 + I)*A + (1 + 4*I)*B)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*a*d) + (((7 - 5*I)*A
+ (5 + 3*I)*B)*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]])/(4*Sqrt[2]*a*d) + (5*(I*A - B)*Sqrt[Cot[c + d*x]])/(2*a
*d) - ((7*A + (3*I)*B)*Cot[c + d*x]^(3/2))/(6*a*d) + ((A + I*B)*Cot[c + d*x]^(5/2))/(2*d*(I*a + a*Cot[c + d*x]
)) + (((7 + 5*I)*A - (5 - 3*I)*B)*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(8*Sqrt[2]*a*d) + (((-7
- 5*I)*A + (5 - 3*I)*B)*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(8*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 3528

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

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 3595

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*b - a*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(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 - 1)*Simp[A*(a*c*m + b*d*n
) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a*A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A,
B}, 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 {\cot ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{a+i a \tan (c+d x)} \, dx &=\int \frac {\cot ^{\frac {5}{2}}(c+d x) (B+A \cot (c+d x))}{i a+a \cot (c+d x)} \, dx\\ &=\frac {(A+i B) \cot ^{\frac {5}{2}}(c+d x)}{2 d (i a+a \cot (c+d x))}+\frac {\int \cot ^{\frac {3}{2}}(c+d x) \left (-\frac {5}{2} a (i A-B)+\frac {1}{2} a (7 A+3 i B) \cot (c+d x)\right ) \, dx}{2 a^2}\\ &=-\frac {(7 A+3 i B) \cot ^{\frac {3}{2}}(c+d x)}{6 a d}+\frac {(A+i B) \cot ^{\frac {5}{2}}(c+d x)}{2 d (i a+a \cot (c+d x))}+\frac {\int \sqrt {\cot (c+d x)} \left (-\frac {1}{2} a (7 A+3 i B)-\frac {5}{2} a (i A-B) \cot (c+d x)\right ) \, dx}{2 a^2}\\ &=\frac {5 (i A-B) \sqrt {\cot (c+d x)}}{2 a d}-\frac {(7 A+3 i B) \cot ^{\frac {3}{2}}(c+d x)}{6 a d}+\frac {(A+i B) \cot ^{\frac {5}{2}}(c+d x)}{2 d (i a+a \cot (c+d x))}+\frac {\int \frac {\frac {5}{2} a (i A-B)-\frac {1}{2} a (7 A+3 i B) \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx}{2 a^2}\\ &=\frac {5 (i A-B) \sqrt {\cot (c+d x)}}{2 a d}-\frac {(7 A+3 i B) \cot ^{\frac {3}{2}}(c+d x)}{6 a d}+\frac {(A+i B) \cot ^{\frac {5}{2}}(c+d x)}{2 d (i a+a \cot (c+d x))}+\frac {\operatorname {Subst}\left (\int \frac {-\frac {5}{2} a (i A-B)+\frac {1}{2} a (7 A+3 i B) x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{a^2 d}\\ &=\frac {5 (i A-B) \sqrt {\cot (c+d x)}}{2 a d}-\frac {(7 A+3 i B) \cot ^{\frac {3}{2}}(c+d x)}{6 a d}+\frac {(A+i B) \cot ^{\frac {5}{2}}(c+d x)}{2 d (i a+a \cot (c+d x))}-\frac {((7+5 i) A-(5-3 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 {((7-5 i) A+(5+3 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 {5 (i A-B) \sqrt {\cot (c+d x)}}{2 a d}-\frac {(7 A+3 i B) \cot ^{\frac {3}{2}}(c+d x)}{6 a d}+\frac {(A+i B) \cot ^{\frac {5}{2}}(c+d x)}{2 d (i a+a \cot (c+d x))}+\frac {((7+5 i) A-(5-3 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 {((7+5 i) A-(5-3 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 {((7-5 i) A+(5+3 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 {((7-5 i) A+(5+3 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 {5 (i A-B) \sqrt {\cot (c+d x)}}{2 a d}-\frac {(7 A+3 i B) \cot ^{\frac {3}{2}}(c+d x)}{6 a d}+\frac {(A+i B) \cot ^{\frac {5}{2}}(c+d x)}{2 d (i a+a \cot (c+d x))}+\frac {((7+5 i) A-(5-3 i) B) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{8 \sqrt {2} a d}-\frac {((7+5 i) A-(5-3 i) B) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{8 \sqrt {2} a d}+\frac {((7-5 i) A+(5+3 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 {((7-5 i) A+(5+3 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 {((7-5 i) A+(5+3 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{4 \sqrt {2} a d}+\frac {((7-5 i) A+(5+3 i) B) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{4 \sqrt {2} a d}+\frac {5 (i A-B) \sqrt {\cot (c+d x)}}{2 a d}-\frac {(7 A+3 i B) \cot ^{\frac {3}{2}}(c+d x)}{6 a d}+\frac {(A+i B) \cot ^{\frac {5}{2}}(c+d x)}{2 d (i a+a \cot (c+d x))}+\frac {((7+5 i) A-(5-3 i) B) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{8 \sqrt {2} a d}-\frac {((7+5 i) A-(5-3 i) B) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{8 \sqrt {2} a d}\\ \end {align*}

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Mathematica [A]  time = 3.10, size = 247, normalized size = 0.83 \[ \frac {(\cos (d x)+i \sin (d x)) (A+B \tan (c+d x)) \left (\frac {2}{3} \cot (c+d x) \csc (c+d x) (\cos (d x)-i \sin (d x)) ((-12 B+8 i A) \sin (2 (c+d x))+(11 A+15 i B) \cos (2 (c+d x))-19 A-15 i B)+(1-i) (\cos (c)+i \sin (c)) \sqrt {\sin (2 (c+d x))} \csc (c+d x) \left (((6+i) A+(1+4 i) B) \sin ^{-1}(\cos (c+d x)-\sin (c+d x))+((4+i) B-(1+6 i) A) \log \left (\sin (c+d x)+\sqrt {\sin (2 (c+d x))}+\cos (c+d x)\right )\right )\right )}{8 d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x)) (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[d*x] + I*Sin[d*x])*((1 - I)*Csc[c + d*x]*(((6 + I)*A + (1 + 4*I)*B)*ArcSin[Cos[c + d*x] - Sin[c + d*x]]
+ ((-1 - 6*I)*A + (4 + I)*B)*Log[Cos[c + d*x] + Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]])*(Cos[c] + I*Sin[c])*Sq
rt[Sin[2*(c + d*x)]] + (2*Cot[c + d*x]*Csc[c + d*x]*(Cos[d*x] - I*Sin[d*x])*(-19*A - (15*I)*B + (11*A + (15*I)
*B)*Cos[2*(c + d*x)] + ((8*I)*A - 12*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]))

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fricas [B]  time = 0.57, size = 715, normalized size = 2.41 \[ -\frac {3 \, {\left (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 {2 \, {\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 {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{2} d^{2}}} + {\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 )}}{i \, A + B}\right ) - 3 \, {\left (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 {2 \, {\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 {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{2} d^{2}}} - {\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 )}}{i \, A + B}\right ) - 6 \, {\left (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 {9 i \, A^{2} - 12 \, A B - 4 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 {9 i \, A^{2} - 12 \, A B - 4 i \, B^{2}}{a^{2} d^{2}}} + 3 \, A + 2 i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a d}\right ) + 6 \, {\left (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 {9 i \, A^{2} - 12 \, A B - 4 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 {9 i \, A^{2} - 12 \, A B - 4 i \, B^{2}}{a^{2} d^{2}}} - 3 \, A - 2 i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a d}\right ) - 2 \, {\left ({\left (19 i \, A - 27 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-38 i \, A + 30 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, A - 3 \, 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 (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(cot(d*x+c)^(5/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

-1/24*(3*(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(-2*(
(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((-I*A^2 - 2*A
*B + I*B^2)/(a^2*d^2)) + (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^(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(2*((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((-I*A^2 - 2*A*B + I*B^2)/(a^2*d^2)) - (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^(4*I*d*x + 4*I*c) - a*d*e^(2*I*d*x + 2*
I*c))*sqrt((9*I*A^2 - 12*A*B - 4*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((9*I*A^2 - 12*A*B - 4*I*B^2)/(a^2*d^2)) + 3*A + 2*I*B)*e^(-2*I*d*x
- 2*I*c)/(a*d)) + 6*(a*d*e^(4*I*d*x + 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))*sqrt((9*I*A^2 - 12*A*B - 4*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(
(9*I*A^2 - 12*A*B - 4*I*B^2)/(a^2*d^2)) - 3*A - 2*I*B)*e^(-2*I*d*x - 2*I*c)/(a*d)) - 2*((19*I*A - 27*B)*e^(4*I
*d*x + 4*I*c) + (-38*I*A + 30*B)*e^(2*I*d*x + 2*I*c) + 3*I*A - 3*B)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d
*x + 2*I*c) - 1)))/(a*d*e^(4*I*d*x + 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 4.45, size = 2598, normalized size = 8.75 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12/a/d*(cos(d*x+c)/sin(d*x+c))^(5/2)*sin(d*x+c)*(-3*A*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+co
s(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))*sin(d*x+c)+3*A*2^(1/2)*cos(d*x+c)^4-7*A*2^(1/2)*cos(d*x+c)^2+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))*cos(d*x+c)*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)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d
*x+c))^(1/2)-3*I*B*2^(1/2)*cos(d*x+c)^2+3*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+cos(d*x+c))/sin(d*x+
c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*sin(d*x+c)+3*I*B*2^(1/2)*cos(d*x+c)^4+3*B*cos(d*x+c)^3*sin(d*x+c)*2^(1/2)-15*
B*2^(1/2)*cos(d*x+c)*sin(d*x+c)+21*A*(-(-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)*EllipticF((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2
),1/2*2^(1/2))*sin(d*x+c)-3*A*(-(-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+cos(d*x+c))/sin(d*x+c))^(1/2),1/2+
1/2*I,1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)+21*A*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+s
in(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticF((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x
+c))^(1/2),1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)+3*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+cos(d*x+c))/si
n(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)-18*A*cos(d*x+c)*sin(d*x+c)*(-(-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)*Elli
pticPi((-(-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))+12*B*cos(d*x+c)*sin(d*x+c)*(-(-si
n(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+cos(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))+3*I*A*sin(d*x+c)*
(-(-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+cos(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))-18*I*A*sin(
d*x+c)*(-(-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+cos(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))+3*I*
B*sin(d*x+c)*(-(-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+cos(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2)
)-12*I*B*sin(d*x+c)*(-(-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+cos(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*
2^(1/2))+9*I*B*sin(d*x+c)*(-(-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)*EllipticF((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/
2))-3*I*A*2^(1/2)*cos(d*x+c)^3*sin(d*x+c)+15*I*A*2^(1/2)*cos(d*x+c)*sin(d*x+c)-18*A*sin(d*x+c)*(-(-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+cos(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))+12*B*sin(d*x+c)*(-(-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+cos(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))-18*I*A*cos(d*x+c)*sin
(d*x+c)*(-(-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+cos(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))+3*I
*B*cos(d*x+c)*sin(d*x+c)*(-(-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+cos(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I
,1/2*2^(1/2))-12*I*B*cos(d*x+c)*sin(d*x+c)*(-(-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+cos(d*x+c))/sin(d*x+c
))^(1/2),1/2-1/2*I,1/2*2^(1/2))+9*I*B*cos(d*x+c)*sin(d*x+c)*(-(-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)*EllipticF((-(-sin(d*x+c)-1+cos(d
*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2)))/cos(d*x+c)^3*2^(1/2)

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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(cot(d*x+c)^(5/2)*(A+B*tan(d*x+c))/(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.

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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