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

Optimal. Leaf size=284 \[ \frac {(5 A+i B) \sqrt {\cot (c+d x)}}{8 a^2 d (\cot (c+d x)+i)}+\frac {\left (\frac {1}{32}+\frac {i}{32}\right ) ((2+i) B-(7-2 i) A) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a^2 d}+\frac {((9+5 i) A-(1+3 i) B) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{32 \sqrt {2} a^2 d}+\frac {((9-5 i) A+(1-3 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{16 \sqrt {2} a^2 d}+\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) ((1+2 i) B-(2-7 i) A) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a^2 d}+\frac {(A+i B) \cot ^{\frac {3}{2}}(c+d x)}{4 d (a \cot (c+d x)+i a)^2} \]

[Out]

1/4*(A+I*B)*cot(d*x+c)^(3/2)/d/(I*a+a*cot(d*x+c))^2-1/32*((9-5*I)*A+(1-3*I)*B)*arctan(-1+2^(1/2)*cot(d*x+c)^(1
/2))/a^2/d*2^(1/2)+(1/32+1/32*I)*((-2+7*I)*A+(1+2*I)*B)*arctan(1+2^(1/2)*cot(d*x+c)^(1/2))/a^2/d*2^(1/2)+(1/64
+1/64*I)*((-7+2*I)*A+(2+I)*B)*ln(1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(1/2))/a^2/d*2^(1/2)+1/64*((9+5*I)*A-(1+3*I)*
B)*ln(1+cot(d*x+c)+2^(1/2)*cot(d*x+c)^(1/2))/a^2/d*2^(1/2)+1/8*(5*A+I*B)*cot(d*x+c)^(1/2)/a^2/d/(I+cot(d*x+c))

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Rubi [A]  time = 0.61, antiderivative size = 284, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3581, 3595, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac {(5 A+i B) \sqrt {\cot (c+d x)}}{8 a^2 d (\cot (c+d x)+i)}+\frac {\left (\frac {1}{32}+\frac {i}{32}\right ) ((2+i) B-(7-2 i) A) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a^2 d}+\frac {((9+5 i) A-(1+3 i) B) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{32 \sqrt {2} a^2 d}+\frac {((9-5 i) A+(1-3 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{16 \sqrt {2} a^2 d}+\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) ((1+2 i) B-(2-7 i) A) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a^2 d}+\frac {(A+i B) \cot ^{\frac {3}{2}}(c+d x)}{4 d (a \cot (c+d x)+i a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((9 - 5*I)*A + (1 - 3*I)*B)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(16*Sqrt[2]*a^2*d) + ((1/16 + I/16)*((-2
+ 7*I)*A + (1 + 2*I)*B)*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*a^2*d) + ((5*A + I*B)*Sqrt[Cot[c + d*
x]])/(8*a^2*d*(I + Cot[c + d*x])) + ((A + I*B)*Cot[c + d*x]^(3/2))/(4*d*(I*a + a*Cot[c + d*x])^2) + ((1/32 + I
/32)*((-7 + 2*I)*A + (2 + I)*B)*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(Sqrt[2]*a^2*d) + (((9 + 5
*I)*A - (1 + 3*I)*B)*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(32*Sqrt[2]*a^2*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]

Rubi steps

\begin {align*} \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx &=\int \frac {\cot ^{\frac {3}{2}}(c+d x) (B+A \cot (c+d x))}{(i a+a \cot (c+d x))^2} \, dx\\ &=\frac {(A+i B) \cot ^{\frac {3}{2}}(c+d x)}{4 d (i a+a \cot (c+d x))^2}+\frac {\int \frac {\sqrt {\cot (c+d x)} \left (-\frac {3}{2} a (i A-B)+\frac {1}{2} a (7 A-i B) \cot (c+d x)\right )}{i a+a \cot (c+d x)} \, dx}{4 a^2}\\ &=\frac {(5 A+i B) \sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {(A+i B) \cot ^{\frac {3}{2}}(c+d x)}{4 d (i a+a \cot (c+d x))^2}+\frac {\int \frac {-\frac {1}{2} a^2 (5 i A-B)+\frac {3}{2} a^2 (3 A-i B) \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx}{8 a^4}\\ &=\frac {(5 A+i B) \sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {(A+i B) \cot ^{\frac {3}{2}}(c+d x)}{4 d (i a+a \cot (c+d x))^2}+\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{2} a^2 (5 i A-B)-\frac {3}{2} a^2 (3 A-i B) x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{4 a^4 d}\\ &=\frac {(5 A+i B) \sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {(A+i B) \cot ^{\frac {3}{2}}(c+d x)}{4 d (i a+a \cot (c+d x))^2}+\frac {((9+5 i) A-(1+3 i) B) \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{16 a^2 d}+\frac {\left (\left (\frac {1}{16}+\frac {i}{16}\right ) ((-2+7 i) A+(1+2 i) B)\right ) \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{a^2 d}\\ &=\frac {(5 A+i B) \sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {(A+i B) \cot ^{\frac {3}{2}}(c+d x)}{4 d (i a+a \cot (c+d x))^2}-\frac {((9+5 i) A-(1+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 )}{32 \sqrt {2} a^2 d}-\frac {((9+5 i) A-(1+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 )}{32 \sqrt {2} a^2 d}+\frac {\left (\left (\frac {1}{32}+\frac {i}{32}\right ) ((-2+7 i) A+(1+2 i) B)\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{a^2 d}+\frac {\left (\left (\frac {1}{32}+\frac {i}{32}\right ) ((-2+7 i) A+(1+2 i) B)\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{a^2 d}\\ &=\frac {(5 A+i B) \sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {(A+i B) \cot ^{\frac {3}{2}}(c+d x)}{4 d (i a+a \cot (c+d x))^2}-\frac {((9+5 i) A-(1+3 i) B) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{32 \sqrt {2} a^2 d}+\frac {((9+5 i) A-(1+3 i) B) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{32 \sqrt {2} a^2 d}+\frac {\left (\left (\frac {1}{16}+\frac {i}{16}\right ) ((-2+7 i) A+(1+2 i) 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^2 d}+\frac {((9-5 i) A+(1-3 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^2 d}\\ &=\frac {((9-5 i) A+(1-3 i) B) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{16 \sqrt {2} a^2 d}-\frac {((9-5 i) A+(1-3 i) B) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{16 \sqrt {2} a^2 d}+\frac {(5 A+i B) \sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {(A+i B) \cot ^{\frac {3}{2}}(c+d x)}{4 d (i a+a \cot (c+d x))^2}-\frac {((9+5 i) A-(1+3 i) B) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{32 \sqrt {2} a^2 d}+\frac {((9+5 i) A-(1+3 i) B) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{32 \sqrt {2} a^2 d}\\ \end {align*}

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Mathematica [A]  time = 2.36, size = 243, normalized size = 0.86 \[ \frac {\sec (c+d x) (\cos (d x)+i \sin (d x))^2 (A+B \tan (c+d x)) \left (4 \cos (c+d x) (\cos (2 d x)-i \sin (2 d x)) ((-B+5 i A) \sin (c+d x)+(7 A+3 i B) \cos (c+d x))+(-\sin (2 c)+i \cos (2 c)) \sqrt {\sin (2 (c+d x))} \csc (c+d x) \left (((5+9 i) A+(3+i) B) \sin ^{-1}(\cos (c+d x)-\sin (c+d x))-(1+i) ((2+7 i) A+(1-2 i) B) \log \left (\sin (c+d x)+\sqrt {\sin (2 (c+d x))}+\cos (c+d x)\right )\right )\right )}{32 d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^2 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]*(Cos[d*x] + I*Sin[d*x])^2*(4*Cos[c + d*x]*(Cos[2*d*x] - I*Sin[2*d*x])*((7*A + (3*I)*B)*Cos[c + d
*x] + ((5*I)*A - B)*Sin[c + d*x]) + Csc[c + d*x]*(((5 + 9*I)*A + (3 + I)*B)*ArcSin[Cos[c + d*x] - Sin[c + d*x]
] - (1 + I)*((2 + 7*I)*A + (1 - 2*I)*B)*Log[Cos[c + d*x] + Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]])*(I*Cos[2*c]
- Sin[2*c])*Sqrt[Sin[2*(c + d*x)]])*(A + B*Tan[c + d*x]))/(32*d*Sqrt[Cot[c + d*x]]*(A*Cos[c + d*x] + B*Sin[c
+ d*x])*(a + I*a*Tan[c + d*x])^2)

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fricas [B]  time = 1.08, size = 666, normalized size = 2.35 \[ \frac {{\left (2 \, a^{2} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {2 \, {\left ({\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2} 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^{4} 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 ) - 2 \, a^{2} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\frac {2 \, {\left ({\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2} 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^{4} 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 ) - a^{2} d \sqrt {\frac {49 i \, A^{2} + 14 \, A B - i \, B^{2}}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {{\left ({\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2} 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 {49 i \, A^{2} + 14 \, A B - i \, B^{2}}{a^{4} d^{2}}} + 7 \, A - i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{2} d}\right ) + a^{2} d \sqrt {\frac {49 i \, A^{2} + 14 \, A B - i \, B^{2}}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\frac {{\left ({\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2} 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 {49 i \, A^{2} + 14 \, A B - i \, B^{2}}{a^{4} d^{2}}} - 7 \, A + i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{2} d}\right ) + 2 \, {\left ({\left (-6 i \, A + 2 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (5 i \, A - B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A - 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 (-4 i \, d x - 4 i \, c\right )}}{32 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/32*(2*a^2*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^4*d^2))*e^(4*I*d*x + 4*I*c)*log(-2*((a^2*d*e^(2*I*d*x + 2*I*c)
- a^2*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^4*d^2))
+ (A - I*B)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(I*A + B)) - 2*a^2*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^4*
d^2))*e^(4*I*d*x + 4*I*c)*log(2*((a^2*d*e^(2*I*d*x + 2*I*c) - a^2*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^4*d^2)) - (A - I*B)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I
*c)/(I*A + B)) - a^2*d*sqrt((49*I*A^2 + 14*A*B - I*B^2)/(a^4*d^2))*e^(4*I*d*x + 4*I*c)*log(-1/8*((a^2*d*e^(2*I
*d*x + 2*I*c) - a^2*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt((49*I*A^2 + 14*A*B - I
*B^2)/(a^4*d^2)) + 7*A - I*B)*e^(-2*I*d*x - 2*I*c)/(a^2*d)) + a^2*d*sqrt((49*I*A^2 + 14*A*B - I*B^2)/(a^4*d^2)
)*e^(4*I*d*x + 4*I*c)*log(1/8*((a^2*d*e^(2*I*d*x + 2*I*c) - a^2*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*
x + 2*I*c) - 1))*sqrt((49*I*A^2 + 14*A*B - I*B^2)/(a^4*d^2)) - 7*A + I*B)*e^(-2*I*d*x - 2*I*c)/(a^2*d)) + 2*((
-6*I*A + 2*B)*e^(4*I*d*x + 4*I*c) + (5*I*A - B)*e^(2*I*d*x + 2*I*c) + I*A - B)*sqrt((I*e^(2*I*d*x + 2*I*c) + I
)/(e^(2*I*d*x + 2*I*c) - 1)))*e^(-4*I*d*x - 4*I*c)/(a^2*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

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

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maple [C]  time = 4.47, size = 1517, normalized size = 5.34 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16/a^2/d*(cos(d*x+c)/sin(d*x+c))^(1/2)*(1+cos(d*x+c))^2*(-1+cos(d*x+c))*(2*A*(-(-sin(d*x+c)-1+cos(d*x+c))/s
in(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)-4*A*2^(1/2)*cos(d*x+c)^5+4*A*2
^(1/2)*cos(d*x+c)^4-3*A*2^(1/2)*cos(d*x+c)^3+3*A*2^(1/2)*cos(d*x+c)^2+4*I*B*cos(d*x+c)^4*2^(1/2)-2*I*A*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))*((-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)-I*B*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))*((-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)-2*I
*B*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))*((-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*(-(-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)+7*
I*A*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))*((-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)-2*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)+4*B*cos(d*x+c)^3*sin(d*x+c)*2^(1/2)-4*B*2^(1/2)*cos(d*x+c)^4*sin(d*x+c)-4*I*B*2^(1/2)*cos(d*x+c)^5+B*2
^(1/2)*cos(d*x+c)*sin(d*x+c)+5*I*A*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)-I*B*2^(1/2)*cos(d*x+c)^2-B*2^(1/2)*cos(d*x+
c)^2*sin(d*x+c)-9*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))*si
n(d*x+c)+I*B*cos(d*x+c)^3*2^(1/2)+4*I*A*2^(1/2)*cos(d*x+c)^4*sin(d*x+c)-5*I*A*2^(1/2)*cos(d*x+c)*sin(d*x+c)-4*
I*A*cos(d*x+c)^3*sin(d*x+c)*2^(1/2)+7*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))/s
in(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))+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)))/sin(d*x+c)^3/cos(d*x+c)*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)^(1/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^2,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 \[ \int \frac {\sqrt {\mathrm {cot}\left (c+d\,x\right )}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {A \sqrt {\cot {\left (c + d x \right )}}}{\tan ^{2}{\left (c + d x \right )} - 2 i \tan {\left (c + d x \right )} - 1}\, dx + \int \frac {B \tan {\left (c + d x \right )} \sqrt {\cot {\left (c + d x \right )}}}{\tan ^{2}{\left (c + d x \right )} - 2 i \tan {\left (c + d x \right )} - 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(Integral(A*sqrt(cot(c + d*x))/(tan(c + d*x)**2 - 2*I*tan(c + d*x) - 1), x) + Integral(B*tan(c + d*x)*sqrt(co
t(c + d*x))/(tan(c + d*x)**2 - 2*I*tan(c + d*x) - 1), x))/a**2

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