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

Optimal. Leaf size=267 \[ \frac{((5+3 i) A-(3-i) B) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{4 \sqrt{2} a d}+\frac{((3-i) B-(5+3 i) A) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{4 \sqrt{2} a d}+\frac{A+i B}{2 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))}-\frac{5 A+i B}{2 a d \sqrt{\tan (c+d x)}}-\frac{\left (\frac{1}{8}-\frac{i}{8}\right ) ((4+i) A+(1+2 i) B) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} a d}+\frac{((5-3 i) A+(3+i) B) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{8 \sqrt{2} a d} \]

[Out]

(((5 + 3*I)*A - (3 - I)*B)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(4*Sqrt[2]*a*d) + (((-5 - 3*I)*A + (3 - I)*
B)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(4*Sqrt[2]*a*d) - ((1/8 - I/8)*((4 + I)*A + (1 + 2*I)*B)*Log[1 - Sq
rt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(Sqrt[2]*a*d) + (((5 - 3*I)*A + (3 + I)*B)*Log[1 + Sqrt[2]*Sqrt[Tan[
c + d*x]] + Tan[c + d*x]])/(8*Sqrt[2]*a*d) - (5*A + I*B)/(2*a*d*Sqrt[Tan[c + d*x]]) + (A + I*B)/(2*d*Sqrt[Tan[
c + d*x]]*(a + I*a*Tan[c + d*x]))

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Rubi [A]  time = 0.365688, antiderivative size = 267, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3596, 3529, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac{((5+3 i) A-(3-i) B) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{4 \sqrt{2} a d}+\frac{((3-i) B-(5+3 i) A) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{4 \sqrt{2} a d}+\frac{A+i B}{2 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))}-\frac{5 A+i B}{2 a d \sqrt{\tan (c+d x)}}-\frac{\left (\frac{1}{8}-\frac{i}{8}\right ) ((4+i) A+(1+2 i) B) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} a d}+\frac{((5-3 i) A+(3+i) B) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{8 \sqrt{2} a d} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])),x]

[Out]

(((5 + 3*I)*A - (3 - I)*B)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(4*Sqrt[2]*a*d) + (((-5 - 3*I)*A + (3 - I)*
B)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(4*Sqrt[2]*a*d) - ((1/8 - I/8)*((4 + I)*A + (1 + 2*I)*B)*Log[1 - Sq
rt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(Sqrt[2]*a*d) + (((5 - 3*I)*A + (3 + I)*B)*Log[1 + Sqrt[2]*Sqrt[Tan[
c + d*x]] + Tan[c + d*x]])/(8*Sqrt[2]*a*d) - (5*A + I*B)/(2*a*d*Sqrt[Tan[c + d*x]]) + (A + I*B)/(2*d*Sqrt[Tan[
c + d*x]]*(a + I*a*Tan[c + d*x]))

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 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

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

Mathematica [A]  time = 2.10408, size = 217, normalized size = 0.81 \[ \frac{(\cos (d x)+i \sin (d x)) (A+B \tan (c+d x)) \left ((-4 \cos (d x)+4 i \sin (d x)) (4 A \cos (c+d x)+(-B+5 i A) \sin (c+d x))+(-\sin (c)+i \cos (c)) \sqrt{\sin (2 (c+d x))} \sec (c+d x) \left (((3-5 i) A+(1+3 i) B) \sin ^{-1}(\cos (c+d x)-\sin (c+d x))-(1+i) ((4+i) A+(1+2 i) B) \log \left (\sin (c+d x)+\sqrt{\sin (2 (c+d x))}+\cos (c+d x)\right )\right )\right )}{8 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x)) (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])),x]

[Out]

((Cos[d*x] + I*Sin[d*x])*((-4*Cos[d*x] + (4*I)*Sin[d*x])*(4*A*Cos[c + d*x] + ((5*I)*A - B)*Sin[c + d*x]) + (((
3 - 5*I)*A + (1 + 3*I)*B)*ArcSin[Cos[c + d*x] - Sin[c + d*x]] - (1 + I)*((4 + I)*A + (1 + 2*I)*B)*Log[Cos[c +
d*x] + Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]])*Sec[c + d*x]*(I*Cos[c] - Sin[c])*Sqrt[Sin[2*(c + d*x)]])*(A + B
*Tan[c + d*x]))/(8*d*(A*Cos[c + d*x] + B*Sin[c + d*x])*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x]))

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Maple [A]  time = 0.049, size = 254, normalized size = 1. \begin{align*} -{\frac{A}{2\,ad \left ( \tan \left ( dx+c \right ) -i \right ) }\sqrt{\tan \left ( dx+c \right ) }}-{\frac{{\frac{i}{2}}B}{ad \left ( \tan \left ( dx+c \right ) -i \right ) }\sqrt{\tan \left ( dx+c \right ) }}-{\frac{2\,iB}{ad \left ( \sqrt{2}-i\sqrt{2} \right ) }\arctan \left ( 2\,{\frac{\sqrt{\tan \left ( dx+c \right ) }}{\sqrt{2}-i\sqrt{2}}} \right ) }-4\,{\frac{A}{ad \left ( \sqrt{2}-i\sqrt{2} \right ) }\arctan \left ( 2\,{\frac{\sqrt{\tan \left ( dx+c \right ) }}{\sqrt{2}-i\sqrt{2}}} \right ) }-{\frac{A}{ad \left ( \sqrt{2}+i\sqrt{2} \right ) }\arctan \left ( 2\,{\frac{\sqrt{\tan \left ( dx+c \right ) }}{\sqrt{2}+i\sqrt{2}}} \right ) }+{\frac{iB}{ad \left ( \sqrt{2}+i\sqrt{2} \right ) }\arctan \left ( 2\,{\frac{\sqrt{\tan \left ( dx+c \right ) }}{\sqrt{2}+i\sqrt{2}}} \right ) }-2\,{\frac{A}{ad\sqrt{\tan \left ( dx+c \right ) }}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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))/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 1.97593, size = 1832, normalized size = 6.86 \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))/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/8*((a*d*e^(4*I*d*x + 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^2*d^2))*log(2*((a*d*e
^(2*I*d*x + 2*I*c) + a*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^2*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*d*e^(4*I*d*x + 4*I*c) -
a*d*e^(2*I*d*x + 2*I*c))*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^2*d^2))*log(-2*((a*d*e^(2*I*d*x + 2*I*c) + a*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^2*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*d*e^(4*I*d*x + 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))*sq
rt((-4*I*A^2 + 4*A*B + I*B^2)/(a^2*d^2))*log(((a*d*e^(2*I*d*x + 2*I*c) + a*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I
)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-4*I*A^2 + 4*A*B + I*B^2)/(a^2*d^2)) + 2*A + I*B)*e^(-2*I*d*x - 2*I*c)/(a*d
)) - 2*(a*d*e^(4*I*d*x + 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))*sqrt((-4*I*A^2 + 4*A*B + I*B^2)/(a^2*d^2))*log(-((a
*d*e^(2*I*d*x + 2*I*c) + a*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-4*I*A^2 + 4*
A*B + I*B^2)/(a^2*d^2)) - 2*A - I*B)*e^(-2*I*d*x - 2*I*c)/(a*d)) + 2*((-9*I*A + B)*e^(4*I*d*x + 4*I*c) - 8*I*A
*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)))/(a*d*e^(4*I*d*x
+ 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))

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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))/tan(d*x+c)**(3/2)/(a+I*a*tan(d*x+c)),x)

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.29897, size = 153, normalized size = 0.57 \begin{align*} \frac{\left (i + 1\right ) \, \sqrt{2}{\left (i \, A + B\right )} \arctan \left (-\left (\frac{1}{2} i - \frac{1}{2}\right ) \, \sqrt{2} \sqrt{\tan \left (d x + c\right )}\right )}{4 \, a d} - \frac{\left (i - 1\right ) \, \sqrt{2}{\left (4 i \, A - 2 \, B\right )} \arctan \left (-\left (\frac{1}{2} i + \frac{1}{2}\right ) \, \sqrt{2} \sqrt{\tan \left (d x + c\right )}\right )}{4 \, a d} + \frac{-5 i \, A \tan \left (d x + c\right ) + B \tan \left (d x + c\right ) - 4 \, A}{2 \,{\left (i \, \tan \left (d x + c\right )^{\frac{3}{2}} + \sqrt{\tan \left (d x + c\right )}\right )} a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

(1/4*I + 1/4)*sqrt(2)*(I*A + B)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/(a*d) - (1/4*I - 1/4)*sqrt(2
)*(4*I*A - 2*B)*arctan(-(1/2*I + 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/(a*d) + 1/2*(-5*I*A*tan(d*x + c) + B*tan(d*x
+ c) - 4*A)/((I*tan(d*x + c)^(3/2) + sqrt(tan(d*x + c)))*a*d)