3.534 \(\int \frac {A+B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=310 \[ -\frac {(2 A-(5+7 i) B) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}+\frac {(2 A-(5+7 i) B) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}-\frac {(2 A+(5-7 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{16 \sqrt {2} a^3 d}+\frac {(2 A+(5-7 i) B) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{16 \sqrt {2} a^3 d}+\frac {5 B \sqrt {\cot (c+d x)}}{8 d \left (a^3 \cot (c+d x)+i a^3\right )}+\frac {(-B+i A) \sqrt {\cot (c+d x)}}{6 d (a \cot (c+d x)+i a)^3}+\frac {(A+4 i B) \sqrt {\cot (c+d x)}}{12 a d (a \cot (c+d x)+i a)^2} \]

[Out]

1/32*(2*A+(5-7*I)*B)*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))/a^3/d*2^(1/2)+1/32*(2*A+(5-7*I)*B)*arctan(1+2^(1/2)*c
ot(d*x+c)^(1/2))/a^3/d*2^(1/2)-1/64*(2*A-(5+7*I)*B)*ln(1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(1/2))/a^3/d*2^(1/2)+1/
64*(2*A-(5+7*I)*B)*ln(1+cot(d*x+c)+2^(1/2)*cot(d*x+c)^(1/2))/a^3/d*2^(1/2)+1/6*(I*A-B)*cot(d*x+c)^(1/2)/d/(I*a
+a*cot(d*x+c))^3+1/12*(A+4*I*B)*cot(d*x+c)^(1/2)/a/d/(I*a+a*cot(d*x+c))^2+5/8*B*cot(d*x+c)^(1/2)/d/(I*a^3+a^3*
cot(d*x+c))

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Rubi [A]  time = 0.76, antiderivative size = 310, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3581, 3596, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac {(2 A-(5+7 i) B) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}+\frac {(2 A-(5+7 i) B) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}-\frac {(2 A+(5-7 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{16 \sqrt {2} a^3 d}+\frac {(2 A+(5-7 i) B) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{16 \sqrt {2} a^3 d}+\frac {5 B \sqrt {\cot (c+d x)}}{8 d \left (a^3 \cot (c+d x)+i a^3\right )}+\frac {(-B+i A) \sqrt {\cot (c+d x)}}{6 d (a \cot (c+d x)+i a)^3}+\frac {(A+4 i B) \sqrt {\cot (c+d x)}}{12 a d (a \cot (c+d x)+i a)^2} \]

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])^3),x]

[Out]

-((2*A + (5 - 7*I)*B)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(16*Sqrt[2]*a^3*d) + ((2*A + (5 - 7*I)*B)*ArcTan
[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]])/(16*Sqrt[2]*a^3*d) + ((I*A - B)*Sqrt[Cot[c + d*x]])/(6*d*(I*a + a*Cot[c + d*
x])^3) + ((A + (4*I)*B)*Sqrt[Cot[c + d*x]])/(12*a*d*(I*a + a*Cot[c + d*x])^2) + (5*B*Sqrt[Cot[c + d*x]])/(8*d*
(I*a^3 + a^3*Cot[c + d*x])) - ((2*A - (5 + 7*I)*B)*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(32*Sqr
t[2]*a^3*d) + ((2*A - (5 + 7*I)*B)*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(32*Sqrt[2]*a^3*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 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))^3} \, dx &=\int \frac {B+A \cot (c+d x)}{\sqrt {\cot (c+d x)} (i a+a \cot (c+d x))^3} \, dx\\ &=\frac {(i A-B) \sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {\int \frac {\frac {1}{2} a (A-11 i B)-\frac {5}{2} a (i A-B) \cot (c+d x)}{\sqrt {\cot (c+d x)} (i a+a \cot (c+d x))^2} \, dx}{6 a^2}\\ &=\frac {(i A-B) \sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {(A+4 i B) \sqrt {\cot (c+d x)}}{12 a d (i a+a \cot (c+d x))^2}+\frac {\int \frac {-3 a^2 (i A+6 B)-3 a^2 (A+4 i B) \cot (c+d x)}{\sqrt {\cot (c+d x)} (i a+a \cot (c+d x))} \, dx}{24 a^4}\\ &=\frac {(i A-B) \sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {(A+4 i B) \sqrt {\cot (c+d x)}}{12 a d (i a+a \cot (c+d x))^2}+\frac {5 B \sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}+\frac {\int \frac {-3 a^3 (2 A-7 i B)-15 a^3 B \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx}{48 a^6}\\ &=\frac {(i A-B) \sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {(A+4 i B) \sqrt {\cot (c+d x)}}{12 a d (i a+a \cot (c+d x))^2}+\frac {5 B \sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {3 a^3 (2 A-7 i B)+15 a^3 B x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{24 a^6 d}\\ &=\frac {(i A-B) \sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {(A+4 i B) \sqrt {\cot (c+d x)}}{12 a d (i a+a \cot (c+d x))^2}+\frac {5 B \sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}+\frac {(2 A-(5+7 i) B) \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{16 a^3 d}+\frac {(2 A+(5-7 i) B) \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{16 a^3 d}\\ &=\frac {(i A-B) \sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {(A+4 i B) \sqrt {\cot (c+d x)}}{12 a d (i a+a \cot (c+d x))^2}+\frac {5 B \sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}-\frac {(2 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 )}{32 \sqrt {2} a^3 d}-\frac {(2 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 )}{32 \sqrt {2} a^3 d}+\frac {(2 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 )}{32 a^3 d}+\frac {(2 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 )}{32 a^3 d}\\ &=\frac {(i A-B) \sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {(A+4 i B) \sqrt {\cot (c+d x)}}{12 a d (i a+a \cot (c+d x))^2}+\frac {5 B \sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}-\frac {(2 A-(5+7 i) B) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {(2 A-(5+7 i) B) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {(2 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 )}{16 \sqrt {2} a^3 d}-\frac {(2 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 )}{16 \sqrt {2} a^3 d}\\ &=-\frac {(2 A+(5-7 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{16 \sqrt {2} a^3 d}+\frac {(2 A+(5-7 i) B) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{16 \sqrt {2} a^3 d}+\frac {(i A-B) \sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {(A+4 i B) \sqrt {\cot (c+d x)}}{12 a d (i a+a \cot (c+d x))^2}+\frac {5 B \sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}-\frac {(2 A-(5+7 i) B) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {(2 A-(5+7 i) B) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{32 \sqrt {2} a^3 d}\\ \end {align*}

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Mathematica [A]  time = 4.78, size = 415, normalized size = 1.34 \[ \frac {\cot ^{\frac {3}{2}}(c+d x) \csc ^2(c+d x) \sec ^3(c+d x) (A \cos (c+d x)+B \sin (c+d x)) \left (-(A+19 i B) \cos (4 (c+d x))+(3+3 i) ((1+i) A+(6-i) B) \sqrt {\sin (2 (c+d x))} \sin ^{-1}(\cos (c+d x)-\sin (c+d x)) (\sin (3 (c+d x))-i \cos (3 (c+d x)))+6 i A \sin (2 (c+d x))-3 i A \sin (4 (c+d x))+6 i A \sqrt {\sin (2 (c+d x))} \sin (3 (c+d x)) \log \left (\sin (c+d x)+\sqrt {\sin (2 (c+d x))}+\cos (c+d x)\right )+6 A \sqrt {\sin (2 (c+d x))} \cos (3 (c+d x)) \log \left (\sin (c+d x)+\sqrt {\sin (2 (c+d x))}+\cos (c+d x)\right )+A-12 B \sin (2 (c+d x))+21 B \sin (4 (c+d x))+(21-15 i) B \sqrt {\sin (2 (c+d x))} \sin (3 (c+d x)) \log \left (\sin (c+d x)+\sqrt {\sin (2 (c+d x))}+\cos (c+d x)\right )-(15+21 i) B \sqrt {\sin (2 (c+d x))} \cos (3 (c+d x)) \log \left (\sin (c+d x)+\sqrt {\sin (2 (c+d x))}+\cos (c+d x)\right )+19 i B\right )}{96 a^3 d (\cot (c+d x)+i)^3 (A+B \tan (c+d x))} \]

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])^3),x]

[Out]

(Cot[c + d*x]^(3/2)*Csc[c + d*x]^2*Sec[c + d*x]^3*(A*Cos[c + d*x] + B*Sin[c + d*x])*(A + (19*I)*B - (A + (19*I
)*B)*Cos[4*(c + d*x)] + 6*A*Cos[3*(c + d*x)]*Log[Cos[c + d*x] + Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]]*Sqrt[Si
n[2*(c + d*x)]] - (15 + 21*I)*B*Cos[3*(c + d*x)]*Log[Cos[c + d*x] + Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]]*Sqr
t[Sin[2*(c + d*x)]] + (6*I)*A*Sin[2*(c + d*x)] - 12*B*Sin[2*(c + d*x)] + (6*I)*A*Log[Cos[c + d*x] + Sin[c + d*
x] + Sqrt[Sin[2*(c + d*x)]]]*Sqrt[Sin[2*(c + d*x)]]*Sin[3*(c + d*x)] + (21 - 15*I)*B*Log[Cos[c + d*x] + Sin[c
+ d*x] + Sqrt[Sin[2*(c + d*x)]]]*Sqrt[Sin[2*(c + d*x)]]*Sin[3*(c + d*x)] + (3 + 3*I)*((1 + I)*A + (6 - I)*B)*A
rcSin[Cos[c + d*x] - Sin[c + d*x]]*Sqrt[Sin[2*(c + d*x)]]*((-I)*Cos[3*(c + d*x)] + Sin[3*(c + d*x)]) - (3*I)*A
*Sin[4*(c + d*x)] + 21*B*Sin[4*(c + d*x)]))/(96*a^3*d*(I + Cot[c + d*x])^3*(A + B*Tan[c + d*x]))

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fricas [B]  time = 0.55, size = 687, normalized size = 2.22 \[ \frac {{\left (3 \, a^{3} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac {{\left ({\left (16 i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - 16 i \, a^{3} 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^{6} d^{2}}} - 16 \, {\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 )}}{8 \, {\left (i \, A + B\right )}}\right ) - 3 \, a^{3} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac {{\left ({\left (-16 i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + 16 i \, a^{3} 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^{6} d^{2}}} - 16 \, {\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 )}}{8 \, {\left (i \, A + B\right )}}\right ) + 3 \, a^{3} d \sqrt {\frac {-i \, A^{2} - 12 \, A B + 36 i \, B^{2}}{a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac {{\left ({\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{3} 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} - 12 \, A B + 36 i \, B^{2}}{a^{6} d^{2}}} + i \, A + 6 \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{3} d}\right ) - 3 \, a^{3} d \sqrt {\frac {-i \, A^{2} - 12 \, A B + 36 i \, B^{2}}{a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-\frac {{\left ({\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{3} 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} - 12 \, A B + 36 i \, B^{2}}{a^{6} d^{2}}} - i \, A - 6 \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{3} d}\right ) - 2 \, {\left (2 \, {\left (A + 10 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - {\left (5 \, A + 26 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (4 \, A + 7 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - A - 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}}\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \]

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))^3,x, algorithm="fricas")

[Out]

1/96*(3*a^3*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(1/8*((16*I*a^3*d*e^(2*I*d*x + 2*
I*c) - 16*I*a^3*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
^6*d^2)) - 16*(A - I*B)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(I*A + B)) - 3*a^3*d*sqrt((I*A^2 + 2*A*B - I
*B^2)/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(1/8*((-16*I*a^3*d*e^(2*I*d*x + 2*I*c) + 16*I*a^3*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^6*d^2)) - 16*(A - I*B)*e^(2*I*d*x +
2*I*c))*e^(-2*I*d*x - 2*I*c)/(I*A + B)) + 3*a^3*d*sqrt((-I*A^2 - 12*A*B + 36*I*B^2)/(a^6*d^2))*e^(6*I*d*x + 6
*I*c)*log(1/8*((a^3*d*e^(2*I*d*x + 2*I*c) - a^3*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 - 12*A*B + 36*I*B^2)/(a^6*d^2)) + I*A + 6*B)*e^(-2*I*d*x - 2*I*c)/(a^3*d)) - 3*a^3*d*sqrt((-I*A^
2 - 12*A*B + 36*I*B^2)/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(-1/8*((a^3*d*e^(2*I*d*x + 2*I*c) - a^3*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 - 12*A*B + 36*I*B^2)/(a^6*d^2)) - I*A - 6*B)*e^
(-2*I*d*x - 2*I*c)/(a^3*d)) - 2*(2*(A + 10*I*B)*e^(6*I*d*x + 6*I*c) - (5*A + 26*I*B)*e^(4*I*d*x + 4*I*c) + (4*
A + 7*I*B)*e^(2*I*d*x + 2*I*c) - A - I*B)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))*e^(-6*I
*d*x - 6*I*c)/(a^3*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \tan \left (d x + c\right ) + A}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} \cot \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

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))^3,x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)/((I*a*tan(d*x + c) + a)^3*cot(d*x + c)^(5/2)), x)

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maple [C]  time = 4.02, size = 5731, normalized size = 18.49 \[ \text {output too large to display} \]

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))^3,x)

[Out]

result too large to display

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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((A+B*tan(d*x+c))/cot(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^3,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 \[ \int \frac {A+B\,\mathrm {tan}\left (c+d\,x\right )}{{\mathrm {cot}\left (c+d\,x\right )}^{5/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3} \,d x \]

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)^3),x)

[Out]

int((A + B*tan(c + d*x))/(cot(c + d*x)^(5/2)*(a + a*tan(c + d*x)*1i)^3), 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((A+B*tan(d*x+c))/cot(d*x+c)**(5/2)/(a+I*a*tan(d*x+c))**3,x)

[Out]

Timed out

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