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

Optimal. Leaf size=316 \[ \frac {(B+2 i A) \sqrt {\cot (c+d x)}}{8 d \left (a^3 \cot (c+d x)+i a^3\right )}+\frac {(2 i A+(1-i) B) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}-\frac {(2 i A+(1-i) B) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}-\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) (B+(1+i) A) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^3 d}+\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) (B+(1+i) A) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a^3 d}+\frac {(A+i B) \cot ^{\frac {3}{2}}(c+d x)}{6 d (a \cot (c+d x)+i a)^3}+\frac {A \sqrt {\cot (c+d x)}}{4 a d (a \cot (c+d x)+i a)^2} \]

[Out]

1/6*(A+I*B)*cot(d*x+c)^(3/2)/d/(I*a+a*cot(d*x+c))^3+(1/32+1/32*I)*((1+I)*A+B)*arctan(-1+2^(1/2)*cot(d*x+c)^(1/
2))/a^3/d*2^(1/2)+(1/32+1/32*I)*((1+I)*A+B)*arctan(1+2^(1/2)*cot(d*x+c)^(1/2))/a^3/d*2^(1/2)+1/64*(2*I*A+(1-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*I*A+(1-I)*B)*ln(1+cot(d*x+c)+2^(1/2)*cot(d
*x+c)^(1/2))/a^3/d*2^(1/2)+1/4*A*cot(d*x+c)^(1/2)/a/d/(I*a+a*cot(d*x+c))^2+1/8*(2*I*A+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 = 316, 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, 3596, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac {(B+2 i A) \sqrt {\cot (c+d x)}}{8 d \left (a^3 \cot (c+d x)+i a^3\right )}+\frac {(2 i A+(1-i) B) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}-\frac {(2 i A+(1-i) B) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}-\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) (B+(1+i) A) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^3 d}+\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) (B+(1+i) A) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a^3 d}+\frac {(A+i B) \cot ^{\frac {3}{2}}(c+d x)}{6 d (a \cot (c+d x)+i a)^3}+\frac {A \sqrt {\cot (c+d x)}}{4 a d (a \cot (c+d x)+i a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[c + d*x])/(Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])^3),x]

[Out]

((-1/16 - I/16)*((1 + I)*A + B)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*a^3*d) + ((1/16 + I/16)*((1 +
I)*A + B)*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*a^3*d) + ((A + I*B)*Cot[c + d*x]^(3/2))/(6*d*(I*a
+ a*Cot[c + d*x])^3) + (A*Sqrt[Cot[c + d*x]])/(4*a*d*(I*a + a*Cot[c + d*x])^2) + (((2*I)*A + B)*Sqrt[Cot[c + d
*x]])/(8*d*(I*a^3 + a^3*Cot[c + d*x])) + (((2*I)*A + (1 - I)*B)*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d
*x]])/(32*Sqrt[2]*a^3*d) - (((2*I)*A + (1 - 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 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]

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)}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^3} \, dx &=\int \frac {\cot ^{\frac {3}{2}}(c+d x) (B+A \cot (c+d x))}{(i a+a \cot (c+d x))^3} \, dx\\ &=\frac {(A+i B) \cot ^{\frac {3}{2}}(c+d x)}{6 d (i a+a \cot (c+d x))^3}+\frac {\int \frac {\sqrt {\cot (c+d x)} \left (-\frac {3}{2} a (i A-B)+\frac {3}{2} a (3 A-i B) \cot (c+d x)\right )}{(i a+a \cot (c+d x))^2} \, dx}{6 a^2}\\ &=\frac {(A+i B) \cot ^{\frac {3}{2}}(c+d x)}{6 d (i a+a \cot (c+d x))^3}+\frac {A \sqrt {\cot (c+d x)}}{4 a d (i a+a \cot (c+d x))^2}+\frac {\int \frac {-3 i a^2 A+3 a^2 (3 A-2 i B) \cot (c+d x)}{\sqrt {\cot (c+d x)} (i a+a \cot (c+d x))} \, dx}{24 a^4}\\ &=\frac {(A+i B) \cot ^{\frac {3}{2}}(c+d x)}{6 d (i a+a \cot (c+d x))^3}+\frac {A \sqrt {\cot (c+d x)}}{4 a d (i a+a \cot (c+d x))^2}+\frac {(2 i A+B) \sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}+\frac {\int \frac {-3 i a^3 B-3 a^3 (2 i A+B) \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx}{48 a^6}\\ &=\frac {(A+i B) \cot ^{\frac {3}{2}}(c+d x)}{6 d (i a+a \cot (c+d x))^3}+\frac {A \sqrt {\cot (c+d x)}}{4 a d (i a+a \cot (c+d x))^2}+\frac {(2 i A+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 i a^3 B+3 a^3 (2 i A+B) x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{24 a^6 d}\\ &=\frac {(A+i B) \cot ^{\frac {3}{2}}(c+d x)}{6 d (i a+a \cot (c+d x))^3}+\frac {A \sqrt {\cot (c+d x)}}{4 a d (i a+a \cot (c+d x))^2}+\frac {(2 i A+B) \sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}+\frac {\left (\left (\frac {1}{16}+\frac {i}{16}\right ) ((1+i) A+B)\right ) \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{a^3 d}-\frac {(2 i A+(1-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 {(A+i B) \cot ^{\frac {3}{2}}(c+d x)}{6 d (i a+a \cot (c+d x))^3}+\frac {A \sqrt {\cot (c+d x)}}{4 a d (i a+a \cot (c+d x))^2}+\frac {(2 i A+B) \sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}+\frac {\left (\left (\frac {1}{32}+\frac {i}{32}\right ) ((1+i) A+B)\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{a^3 d}+\frac {\left (\left (\frac {1}{32}+\frac {i}{32}\right ) ((1+i) A+B)\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{a^3 d}+\frac {(2 i A+(1-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 i A+(1-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 {(A+i B) \cot ^{\frac {3}{2}}(c+d x)}{6 d (i a+a \cot (c+d x))^3}+\frac {A \sqrt {\cot (c+d x)}}{4 a d (i a+a \cot (c+d x))^2}+\frac {(2 i A+B) \sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}+\frac {(2 i A+(1-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 i A+(1-i) B) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{32 \sqrt {2} a^3 d}+-\frac {\left (\left (\frac {1}{16}+\frac {i}{16}\right ) ((1+i) A+B)\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^3 d}+\frac {\left (\left (\frac {1}{16}+\frac {i}{16}\right ) ((1+i) A+B)\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^3 d}\\ &=-\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) ((1+i) A+B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^3 d}+\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) ((1+i) A+B) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^3 d}+\frac {(A+i B) \cot ^{\frac {3}{2}}(c+d x)}{6 d (i a+a \cot (c+d x))^3}+\frac {A \sqrt {\cot (c+d x)}}{4 a d (i a+a \cot (c+d x))^2}+\frac {(2 i A+B) \sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}+\frac {(2 i A+(1-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 i A+(1-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 = 4.26, size = 272, normalized size = 0.86 \[ \frac {e^{-4 i (c+d x)} \sqrt {\cot (c+d x)} \sec (c+d x) (\cos (3 (c+d x))-i \sin (3 (c+d x))) \left (\left (-2 e^{2 i (c+d x)}+e^{4 i (c+d x)}+2 e^{6 i (c+d x)}-1\right ) \left (A e^{2 i (c+d x)}+A-2 i B e^{2 i (c+d x)}+i B\right )-6 (A-i B) e^{6 i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \sqrt {1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )-3 A e^{6 i (c+d x)} \sqrt {-1+e^{4 i (c+d x)}} \tan ^{-1}\left (\sqrt {-1+e^{4 i (c+d x)}}\right )\right )}{96 a^3 d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(A + B*Tan[c + d*x])/(Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])^3),x]

[Out]

(((A + I*B + A*E^((2*I)*(c + d*x)) - (2*I)*B*E^((2*I)*(c + d*x)))*(-1 - 2*E^((2*I)*(c + d*x)) + E^((4*I)*(c +
d*x)) + 2*E^((6*I)*(c + d*x))) - 3*A*E^((6*I)*(c + d*x))*Sqrt[-1 + E^((4*I)*(c + d*x))]*ArcTan[Sqrt[-1 + E^((4
*I)*(c + d*x))]] - 6*(A - I*B)*E^((6*I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[1 + E^((2*I)*(c + d*x))
]*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]])*Sqrt[Cot[c + d*x]]*Sec[c + d*x]*(Cos[3*
(c + d*x)] - I*Sin[3*(c + d*x)]))/(96*a^3*d*E^((4*I)*(c + d*x)))

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fricas [B]  time = 0.54, size = 637, normalized size = 2.02 \[ -\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 ) - 24 \, a^{3} d \sqrt {-\frac {i \, A^{2}}{64 \, a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac {{\left (8 \, {\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}}{64 \, a^{6} d^{2}}} + i \, A\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{3} d}\right ) + 24 \, a^{3} d \sqrt {-\frac {i \, A^{2}}{64 \, a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-\frac {{\left (8 \, {\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}}{64 \, a^{6} d^{2}}} - i \, A\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{3} d}\right ) - 2 \, {\left (2 \, {\left (A - 2 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (A + 4 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (2 \, A - 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)^(1/2)/(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)) - 24*a^3*d*sqrt(-1/64*I*A^2/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(1/8*(
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(-1/64*I
*A^2/(a^6*d^2)) + I*A)*e^(-2*I*d*x - 2*I*c)/(a^3*d)) + 24*a^3*d*sqrt(-1/64*I*A^2/(a^6*d^2))*e^(6*I*d*x + 6*I*c
)*log(-1/8*(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(-1/64*I*A^2/(a^6*d^2)) - I*A)*e^(-2*I*d*x - 2*I*c)/(a^3*d)) - 2*(2*(A - 2*I*B)*e^(6*I*d*x + 6*I*c) + (A +
4*I*B)*e^(4*I*d*x + 4*I*c) - (2*A - 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 \[ \int \frac {B \tan \left (d x + c\right ) + A}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} \sqrt {\cot \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/cot(d*x+c)^(1/2)/(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*sqrt(cot(d*x + c))), x)

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maple [C]  time = 4.10, size = 5075, normalized size = 16.06 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/cot(d*x+c)**(1/2)/(a+I*a*tan(d*x+c))**3,x)

[Out]

I*(Integral(A/(tan(c + d*x)**3*sqrt(cot(c + d*x)) - 3*I*tan(c + d*x)**2*sqrt(cot(c + d*x)) - 3*tan(c + d*x)*sq
rt(cot(c + d*x)) + I*sqrt(cot(c + d*x))), x) + Integral(B*tan(c + d*x)/(tan(c + d*x)**3*sqrt(cot(c + d*x)) - 3
*I*tan(c + d*x)**2*sqrt(cot(c + d*x)) - 3*tan(c + d*x)*sqrt(cot(c + d*x)) + I*sqrt(cot(c + d*x))), x))/a**3

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