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

Optimal. Leaf size=274 \[ \frac{(B+3 i A) \sqrt{\cot (c+d x)}}{8 a^2 d (\cot (c+d x)+i)}+\frac{((1+3 i) A+(1-3 i) B) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{32 \sqrt{2} a^2 d}-\frac{((1+3 i) A+(1-3 i) B) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{32 \sqrt{2} a^2 d}-\frac{((1+3 i) B-(1-3 i) A) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{16 \sqrt{2} a^2 d}+\frac{((1+3 i) B-(1-3 i) A) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{16 \sqrt{2} a^2 d}+\frac{(A+i B) \sqrt{\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2} \]

[Out]

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

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Rubi [A]  time = 0.575973, antiderivative size = 274, normalized size of antiderivative = 1., number of steps used = 13, 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} \[ \frac{(B+3 i A) \sqrt{\cot (c+d x)}}{8 a^2 d (\cot (c+d x)+i)}+\frac{((1+3 i) A+(1-3 i) B) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{32 \sqrt{2} a^2 d}-\frac{((1+3 i) A+(1-3 i) B) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{32 \sqrt{2} a^2 d}-\frac{((1+3 i) B-(1-3 i) A) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{16 \sqrt{2} a^2 d}+\frac{((1+3 i) B-(1-3 i) A) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{16 \sqrt{2} a^2 d}+\frac{(A+i B) \sqrt{\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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 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 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 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]

Rubi steps

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

Mathematica [A]  time = 1.91462, size = 243, normalized size = 0.89 \[ \frac{\csc (c+d x) (\cos (d x)+i \sin (d x))^2 (A \cot (c+d x)+B) \left (4 \cos (c+d x) (\sin (2 d x)+i \cos (2 d x)) ((3 B+i A) \sin (c+d x)+(3 A-i B) \cos (c+d x))+(1-i) (-\sin (2 c)+i \cos (2 c)) \sqrt{\sin (2 (c+d x))} \csc (c+d x) \left (((1+2 i) A+(2+i) B) \sin ^{-1}(\cos (c+d x)-\sin (c+d x))+((1+2 i) B-(2+i) A) \log \left (\sin (c+d x)+\sqrt{\sin (2 (c+d x))}+\cos (c+d x)\right )\right )\right )}{32 a^2 d \sqrt{\cot (c+d x)} (\cot (c+d x)+i)^2 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.575, size = 5032, normalized size = 18.4 \begin{align*} \text{output too large to display} \end{align*}

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))^2,x)

[Out]

result too large to display

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

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))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 1.599, size = 1717, normalized size = 6.27 \begin{align*} \text{result too large to display} \end{align*}

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))^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(1/4*((8*I*a^2*d*e^(2*I*d*x + 2*
I*c) - 8*I*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)) - 8*(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(1/4*((-8*I*a^2*d*e^(2*I*d*x + 2*I*c) + 8*I*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)) - 8*(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((-I*A^2 + 2*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((-I*A
^2 + 2*A*B + I*B^2)/(a^4*d^2)) + I*A - B)*e^(-2*I*d*x - 2*I*c)/(a^2*d)) + 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(-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((-I*A^2 + 2*A*B + I*B^2)/(a^4*d^2)) - I*A + B)*e^(-2*I*d*x - 2*I*c)/(a^2*d)) -
2*(2*(A - I*B)*e^(4*I*d*x + 4*I*c) - (A - 3*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^(-4*I*d*x - 4*I*c)/(a^2*d)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

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))**2,x)

[Out]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \tan \left (d x + c\right ) + A}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} \sqrt{\cot \left (d x + c\right )}}\,{d x} \end{align*}

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))^2,x, algorithm="giac")

[Out]

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