3.22 \(\int \cot ^3(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=123 \[ -\frac{a^3 (4 A-3 i B) \log (\sin (c+d x))}{d}-\frac{(B+2 i A) \cot (c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{d}-4 a^3 x (B+i A)+\frac{i a^3 B \log (\cos (c+d x))}{d}-\frac{a A \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d} \]

[Out]

-4*a^3*(I*A + B)*x + (I*a^3*B*Log[Cos[c + d*x]])/d - (a^3*(4*A - (3*I)*B)*Log[Sin[c + d*x]])/d - (a*A*Cot[c +
d*x]^2*(a + I*a*Tan[c + d*x])^2)/(2*d) - (((2*I)*A + B)*Cot[c + d*x]*(a^3 + I*a^3*Tan[c + d*x]))/d

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Rubi [A]  time = 0.315215, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {3593, 3589, 3475, 3531} \[ -\frac{a^3 (4 A-3 i B) \log (\sin (c+d x))}{d}-\frac{(B+2 i A) \cot (c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{d}-4 a^3 x (B+i A)+\frac{i a^3 B \log (\cos (c+d x))}{d}-\frac{a A \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-4*a^3*(I*A + B)*x + (I*a^3*B*Log[Cos[c + d*x]])/d - (a^3*(4*A - (3*I)*B)*Log[Sin[c + d*x]])/d - (a*A*Cot[c +
d*x]^2*(a + I*a*Tan[c + d*x])^2)/(2*d) - (((2*I)*A + B)*Cot[c + d*x]*(a^3 + I*a^3*Tan[c + d*x]))/d

Rule 3593

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^2*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^
(n + 1))/(d*f*(b*c + a*d)*(n + 1)), x] - Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^(n + 1)*Simp[A*b*d*(m - n - 2) - B*(b*c*(m - 1) + a*d*(n + 1)) + (a*A*d*(m + n) - B*(a*c*(m -
1) + b*d*(n + 1)))*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] && GtQ[m, 1] && LtQ[n, -1]

Rule 3589

Int[(((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]))/((a_.) + (b_.)*tan[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Dist[(B*d)/b, Int[Tan[e + f*x], x], x] + Dist[1/b, Int[Simp[A*b*c + (A*b*d + B*(
b*c - a*d))*Tan[e + f*x], x]/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a
*d, 0]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3531

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

Rubi steps

\begin{align*} \int \cot ^3(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx &=-\frac{a A \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}+\frac{1}{2} \int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 (2 a (2 i A+B)+2 i a B \tan (c+d x)) \, dx\\ &=-\frac{a A \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}-\frac{(2 i A+B) \cot (c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{d}+\frac{1}{2} \int \cot (c+d x) (a+i a \tan (c+d x)) \left (-2 a^2 (4 A-3 i B)-2 a^2 B \tan (c+d x)\right ) \, dx\\ &=-\frac{a A \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}-\frac{(2 i A+B) \cot (c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{d}+\frac{1}{2} \int \cot (c+d x) \left (-2 a^3 (4 A-3 i B)-8 a^3 (i A+B) \tan (c+d x)\right ) \, dx-\left (i a^3 B\right ) \int \tan (c+d x) \, dx\\ &=-4 a^3 (i A+B) x+\frac{i a^3 B \log (\cos (c+d x))}{d}-\frac{a A \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}-\frac{(2 i A+B) \cot (c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{d}-\left (a^3 (4 A-3 i B)\right ) \int \cot (c+d x) \, dx\\ &=-4 a^3 (i A+B) x+\frac{i a^3 B \log (\cos (c+d x))}{d}-\frac{a^3 (4 A-3 i B) \log (\sin (c+d x))}{d}-\frac{a A \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}-\frac{(2 i A+B) \cot (c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{d}\\ \end{align*}

Mathematica [B]  time = 8.58067, size = 1010, normalized size = 8.21 \[ a^3 \left (\frac{x (\cot (c+d x)+i)^3 (B+A \cot (c+d x)) \left (-16 i A \cos ^3(c)-\frac{25}{2} B \cos ^3(c)+4 A \cot (c) \cos ^3(c)-3 i B \cot (c) \cos ^3(c)-24 A \sin (c) \cos ^2(c)+20 i B \sin (c) \cos ^2(c)+16 i A \sin ^2(c) \cos (c)+15 B \sin ^2(c) \cos (c)+\frac{1}{2} B \cos (c)+4 A \sin ^3(c)-5 i B \sin ^3(c)-i B \sin (c)+(2 \cos (2 c) A+2 A-i B-2 i B \cos (2 c)) \csc (c) \sec (c) (i \sin (3 c)-\cos (3 c))-\frac{1}{2} B \sin ^3(c) \tan (c)-\frac{1}{2} B \sin (c) \tan (c)\right ) \sin ^4(c+d x)}{(\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac{i B \cos (3 c) (\cot (c+d x)+i)^3 (B+A \cot (c+d x)) \log \left (\cos ^2(c+d x)\right ) \sin ^4(c+d x)}{2 d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac{(\cot (c+d x)+i)^3 (B+A \cot (c+d x)) \left (4 A \cos \left (\frac{3 c}{2}\right )-3 i B \cos \left (\frac{3 c}{2}\right )-4 i A \sin \left (\frac{3 c}{2}\right )-3 B \sin \left (\frac{3 c}{2}\right )\right ) \left (i \tan ^{-1}(\tan (4 c+d x)) \cos \left (\frac{3 c}{2}\right )+\tan ^{-1}(\tan (4 c+d x)) \sin \left (\frac{3 c}{2}\right )\right ) \sin ^4(c+d x)}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac{(\cot (c+d x)+i)^3 (B+A \cot (c+d x)) \left (4 A \cos \left (\frac{3 c}{2}\right )-3 i B \cos \left (\frac{3 c}{2}\right )-4 i A \sin \left (\frac{3 c}{2}\right )-3 B \sin \left (\frac{3 c}{2}\right )\right ) \left (\frac{1}{2} i \log \left (\sin ^2(c+d x)\right ) \sin \left (\frac{3 c}{2}\right )-\frac{1}{2} \cos \left (\frac{3 c}{2}\right ) \log \left (\sin ^2(c+d x)\right )\right ) \sin ^4(c+d x)}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac{B (\cot (c+d x)+i)^3 (B+A \cot (c+d x)) \log \left (\cos ^2(c+d x)\right ) \sin (3 c) \sin ^4(c+d x)}{2 d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac{(A-i B) (\cot (c+d x)+i)^3 (B+A \cot (c+d x)) (-4 i d x \cos (3 c)-4 d x \sin (3 c)) \sin ^4(c+d x)}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac{(\cot (c+d x)+i)^3 (B+A \cot (c+d x)) \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \left (\frac{1}{2} \cos (3 c)-\frac{1}{2} i \sin (3 c)\right ) (3 i A \sin (d x)+B \sin (d x)) \sin ^3(c+d x)}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}+\frac{(\cot (c+d x)+i)^3 (B+A \cot (c+d x)) \left (\frac{1}{2} i A \sin (3 c)-\frac{1}{2} A \cos (3 c)\right ) \sin ^2(c+d x)}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^3*(((I + Cot[c + d*x])^3*(B + A*Cot[c + d*x])*(-(A*Cos[3*c])/2 + (I/2)*A*Sin[3*c])*Sin[c + d*x]^2)/(d*(Cos[d
*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I + Cot[c + d*x])^3*(B + A*Cot[c + d*x])*Csc[c/2]*S
ec[c/2]*(Cos[3*c]/2 - (I/2)*Sin[3*c])*((3*I)*A*Sin[d*x] + B*Sin[d*x])*Sin[c + d*x]^3)/(d*(Cos[d*x] + I*Sin[d*x
])^3*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I/2)*B*Cos[3*c]*(I + Cot[c + d*x])^3*(B + A*Cot[c + d*x])*Log[Cos[
c + d*x]^2]*Sin[c + d*x]^4)/(d*(Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I + Cot[c + d*
x])^3*(B + A*Cot[c + d*x])*(4*A*Cos[(3*c)/2] - (3*I)*B*Cos[(3*c)/2] - (4*I)*A*Sin[(3*c)/2] - 3*B*Sin[(3*c)/2])
*(I*ArcTan[Tan[4*c + d*x]]*Cos[(3*c)/2] + ArcTan[Tan[4*c + d*x]]*Sin[(3*c)/2])*Sin[c + d*x]^4)/(d*(Cos[d*x] +
I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((I + Cot[c + d*x])^3*(B + A*Cot[c + d*x])*(4*A*Cos[(3*c)/2
] - (3*I)*B*Cos[(3*c)/2] - (4*I)*A*Sin[(3*c)/2] - 3*B*Sin[(3*c)/2])*(-(Cos[(3*c)/2]*Log[Sin[c + d*x]^2])/2 + (
I/2)*Log[Sin[c + d*x]^2]*Sin[(3*c)/2])*Sin[c + d*x]^4)/(d*(Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c
+ d*x])) + (B*(I + Cot[c + d*x])^3*(B + A*Cot[c + d*x])*Log[Cos[c + d*x]^2]*Sin[3*c]*Sin[c + d*x]^4)/(2*d*(Cos
[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c + d*x])) + ((A - I*B)*(I + Cot[c + d*x])^3*(B + A*Cot[c + d*x]
)*((-4*I)*d*x*Cos[3*c] - 4*d*x*Sin[3*c])*Sin[c + d*x]^4)/(d*(Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[
c + d*x])) + (x*(I + Cot[c + d*x])^3*(B + A*Cot[c + d*x])*Sin[c + d*x]^4*((B*Cos[c])/2 - (16*I)*A*Cos[c]^3 - (
25*B*Cos[c]^3)/2 + 4*A*Cos[c]^3*Cot[c] - (3*I)*B*Cos[c]^3*Cot[c] - I*B*Sin[c] - 24*A*Cos[c]^2*Sin[c] + (20*I)*
B*Cos[c]^2*Sin[c] + (16*I)*A*Cos[c]*Sin[c]^2 + 15*B*Cos[c]*Sin[c]^2 + 4*A*Sin[c]^3 - (5*I)*B*Sin[c]^3 + (2*A -
I*B + 2*A*Cos[2*c] - (2*I)*B*Cos[2*c])*Csc[c]*Sec[c]*(-Cos[3*c] + I*Sin[3*c]) - (B*Sin[c]*Tan[c])/2 - (B*Sin[
c]^3*Tan[c])/2))/((Cos[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c + d*x])))

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Maple [A]  time = 0.075, size = 136, normalized size = 1.1 \begin{align*} -4\,iAx{a}^{3}-{\frac{4\,iA{a}^{3}c}{d}}+{\frac{iB{a}^{3}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}-4\,{\frac{A{a}^{3}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-4\,B{a}^{3}x-4\,{\frac{B{a}^{3}c}{d}}-{\frac{3\,iA\cot \left ( dx+c \right ){a}^{3}}{d}}+{\frac{3\,iB{a}^{3}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{A{a}^{3} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{\cot \left ( dx+c \right ) B{a}^{3}}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x)

[Out]

-4*I*A*x*a^3-4*I/d*A*a^3*c+I*a^3*B*ln(cos(d*x+c))/d-4*a^3*A*ln(sin(d*x+c))/d-4*B*a^3*x-4/d*B*a^3*c-3*I/d*A*cot
(d*x+c)*a^3+3*I/d*B*a^3*ln(sin(d*x+c))-1/2/d*A*a^3*cot(d*x+c)^2-1/d*B*cot(d*x+c)*a^3

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Maxima [A]  time = 1.66136, size = 132, normalized size = 1.07 \begin{align*} -\frac{8 \,{\left (d x + c\right )}{\left (i \, A + B\right )} a^{3} - 2 \,{\left (2 \, A - 2 i \, B\right )} a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \,{\left (4 \, A - 3 i \, B\right )} a^{3} \log \left (\tan \left (d x + c\right )\right ) - \frac{2 \,{\left (-3 i \, A - B\right )} a^{3} \tan \left (d x + c\right ) - A a^{3}}{\tan \left (d x + c\right )^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.50599, size = 474, normalized size = 3.85 \begin{align*} \frac{2 \,{\left (4 \, A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 \,{\left (3 \, A - i \, B\right )} a^{3} +{\left (i \, B a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 i \, B a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, B a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) -{\left ({\left (4 \, A - 3 i \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \,{\left (4 \, A - 3 i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (4 \, A - 3 i \, B\right )} a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}{d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 8.86526, size = 207, normalized size = 1.68 \begin{align*} \frac{- \frac{\left (6 A a^{3} - 2 i B a^{3}\right ) e^{- 4 i c}}{d} + \frac{\left (8 A a^{3} - 2 i B a^{3}\right ) e^{- 2 i c} e^{2 i d x}}{d}}{e^{4 i d x} - 2 e^{- 2 i c} e^{2 i d x} + e^{- 4 i c}} + \operatorname{RootSum}{\left (z^{2} d^{2} + z \left (4 A a^{3} d - 4 i B a^{3} d\right ) - 4 i A B a^{6} - 3 B^{2} a^{6}, \left ( i \mapsto i \log{\left (\frac{i i d}{2 i A a^{3} e^{2 i c} + B a^{3} e^{2 i c}} + \frac{2 i A + 2 B}{2 i A e^{2 i c} + B e^{2 i c}} + e^{2 i d x} \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-(6*A*a**3 - 2*I*B*a**3)*exp(-4*I*c)/d + (8*A*a**3 - 2*I*B*a**3)*exp(-2*I*c)*exp(2*I*d*x)/d)/(exp(4*I*d*x) -
2*exp(-2*I*c)*exp(2*I*d*x) + exp(-4*I*c)) + RootSum(_z**2*d**2 + _z*(4*A*a**3*d - 4*I*B*a**3*d) - 4*I*A*B*a**6
- 3*B**2*a**6, Lambda(_i, _i*log(_i*I*d/(2*I*A*a**3*exp(2*I*c) + B*a**3*exp(2*I*c)) + (2*I*A + 2*B)/(2*I*A*ex
p(2*I*c) + B*exp(2*I*c)) + exp(2*I*d*x))))

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Giac [B]  time = 1.65034, size = 306, normalized size = 2.49 \begin{align*} -\frac{A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 8 i \, B a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 8 i \, B a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - 12 i \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 16 \,{\left (4 \, A a^{3} - 4 i \, B a^{3}\right )} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + i\right ) + 8 \,{\left (4 \, A a^{3} - 3 i \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - \frac{48 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 36 i \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 i \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - A a^{3}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/8*(A*a^3*tan(1/2*d*x + 1/2*c)^2 - 8*I*B*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 8*I*B*a^3*log(abs(tan(1/2*
d*x + 1/2*c) - 1)) - 12*I*A*a^3*tan(1/2*d*x + 1/2*c) - 4*B*a^3*tan(1/2*d*x + 1/2*c) - 16*(4*A*a^3 - 4*I*B*a^3)
*log(tan(1/2*d*x + 1/2*c) + I) + 8*(4*A*a^3 - 3*I*B*a^3)*log(abs(tan(1/2*d*x + 1/2*c))) - (48*A*a^3*tan(1/2*d*
x + 1/2*c)^2 - 36*I*B*a^3*tan(1/2*d*x + 1/2*c)^2 - 12*I*A*a^3*tan(1/2*d*x + 1/2*c) - 4*B*a^3*tan(1/2*d*x + 1/2
*c) - A*a^3)/tan(1/2*d*x + 1/2*c)^2)/d