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

Optimal. Leaf size=285 \[ \frac{\left (\frac{1}{16}+\frac{i}{16}\right ) ((1+2 i) B-(2-7 i) A) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} a^2 d}+\frac{((9-5 i) A+(1-3 i) B) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{16 \sqrt{2} a^2 d}+\frac{(5 A+i B) \sqrt{\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac{\left (\frac{1}{32}+\frac{i}{32}\right ) ((2+i) B-(7-2 i) A) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} a^2 d}+\frac{((9+5 i) A-(1+3 i) B) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{32 \sqrt{2} a^2 d}+\frac{(A+i B) \sqrt{\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

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

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Rubi [A]  time = 0.499608, antiderivative size = 285, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {3596, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac{\left (\frac{1}{16}+\frac{i}{16}\right ) ((1+2 i) B-(2-7 i) A) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} a^2 d}+\frac{((9-5 i) A+(1-3 i) B) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{16 \sqrt{2} a^2 d}+\frac{(5 A+i B) \sqrt{\tan (c+d x)}}{8 a^2 d (1+i \tan (c+d x))}+\frac{\left (\frac{1}{32}+\frac{i}{32}\right ) ((2+i) B-(7-2 i) A) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} a^2 d}+\frac{((9+5 i) A-(1+3 i) B) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{32 \sqrt{2} a^2 d}+\frac{(A+i B) \sqrt{\tan (c+d x)}}{4 d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 2.17374, size = 243, normalized size = 0.85 \[ \frac{\sec (c+d x) (\cos (d x)+i \sin (d x))^2 (A+B \tan (c+d x)) \left (4 \sin (c+d x) (\sin (2 d x)+i \cos (2 d x)) ((5 A+i B) \sin (c+d x)+(3 B-7 i A) \cos (c+d x))+(-\sin (2 c)+i \cos (2 c)) \sqrt{\sin (2 (c+d x))} \sec (c+d x) \left (((5+9 i) A+(3+i) B) \sin ^{-1}(\cos (c+d x)-\sin (c+d x))-(1+i) ((2+7 i) A+(1-2 i) B) \log \left (\sin (c+d x)+\sqrt{\sin (2 (c+d x))}+\cos (c+d x)\right )\right )\right )}{32 d \sqrt{\tan (c+d x)} (a+i a \tan (c+d x))^2 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tan[c + d*x])/(Sqrt[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*(I*Cos[2*d*x] + Sin[2*d*x])*Sin[c + d*x]*(((-7*I)*A + 3*B)*Cos[c +
d*x] + (5*A + I*B)*Sin[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)]]])*Sec[c + d*x]*(I*Cos[2*c]
- Sin[2*c])*Sqrt[Sin[2*(c + d*x)]])*(A + B*Tan[c + d*x]))/(32*d*(A*Cos[c + d*x] + B*Sin[c + d*x])*Sqrt[Tan[c +
d*x]]*(a + I*a*Tan[c + d*x])^2)

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Maple [A]  time = 0.065, size = 294, normalized size = 1. \begin{align*}{\frac{-{\frac{5\,i}{8}}A}{{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}} \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{B}{8\,{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}} \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{7\,A}{8\,{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}\sqrt{\tan \left ( dx+c \right ) }}-{\frac{{\frac{3\,i}{8}}B}{{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}\sqrt{\tan \left ( dx+c \right ) }}-{\frac{{\frac{7\,i}{4}}A}{{a}^{2}d \left ( \sqrt{2}-i\sqrt{2} \right ) }\arctan \left ( 2\,{\frac{\sqrt{\tan \left ( dx+c \right ) }}{\sqrt{2}-i\sqrt{2}}} \right ) }-{\frac{B}{4\,{a}^{2}d \left ( \sqrt{2}-i\sqrt{2} \right ) }\arctan \left ( 2\,{\frac{\sqrt{\tan \left ( dx+c \right ) }}{\sqrt{2}-i\sqrt{2}}} \right ) }+{\frac{B}{2\,{a}^{2}d \left ( \sqrt{2}+i\sqrt{2} \right ) }\arctan \left ( 2\,{\frac{\sqrt{\tan \left ( dx+c \right ) }}{\sqrt{2}+i\sqrt{2}}} \right ) }+{\frac{{\frac{i}{2}}A}{{a}^{2}d \left ( \sqrt{2}+i\sqrt{2} \right ) }\arctan \left ( 2\,{\frac{\sqrt{\tan \left ( dx+c \right ) }}{\sqrt{2}+i\sqrt{2}}} \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)^(1/2)/(a+I*a*tan(d*x+c))^2,x)

[Out]

-5/8*I/d/a^2/(tan(d*x+c)-I)^2*tan(d*x+c)^(3/2)*A+1/8/d/a^2/(tan(d*x+c)-I)^2*tan(d*x+c)^(3/2)*B-7/8/d/a^2/(tan(
d*x+c)-I)^2*tan(d*x+c)^(1/2)*A-3/8*I/d/a^2/(tan(d*x+c)-I)^2*tan(d*x+c)^(1/2)*B-7/4*I/d/a^2/(2^(1/2)-I*2^(1/2))
*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)-I*2^(1/2)))*A-1/4/d/a^2*B/(2^(1/2)-I*2^(1/2))*arctan(2*tan(d*x+c)^(1/2)/(2
^(1/2)-I*2^(1/2)))+1/2/d/a^2/(2^(1/2)+I*2^(1/2))*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)+I*2^(1/2)))*B+1/2*I/d/a^2/
(2^(1/2)+I*2^(1/2))*arctan(2*tan(d*x+c)^(1/2)/(2^(1/2)+I*2^(1/2)))*A

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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)^(1/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 1.86404, size = 1755, normalized size = 6.16 \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)^(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((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))*s
qrt((49*I*A^2 + 14*A*B - I*B^2)/(a^4*d^2)) + 7*I*A + 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*I*A - B)*e^(-2*I*
d*x - 2*I*c)/(a^2*d)) + 2*(2*(3*A + I*B)*e^(4*I*d*x + 4*I*c) + (7*A + 3*I*B)*e^(2*I*d*x + 2*I*c) + A + I*B)*sq
rt((-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: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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Giac [A]  time = 1.24192, size = 170, normalized size = 0.6 \begin{align*} \frac{\left (i + 1\right ) \, \sqrt{2}{\left (A - i \, B\right )} \arctan \left (-\left (\frac{1}{2} i - \frac{1}{2}\right ) \, \sqrt{2} \sqrt{\tan \left (d x + c\right )}\right )}{8 \, a^{2} d} + \frac{\left (i - 1\right ) \, \sqrt{2}{\left (7 \, A - i \, B\right )} \arctan \left (-\left (\frac{1}{2} i + \frac{1}{2}\right ) \, \sqrt{2} \sqrt{\tan \left (d x + c\right )}\right )}{16 \, a^{2} d} - \frac{5 i \, A \tan \left (d x + c\right )^{\frac{3}{2}} - B \tan \left (d x + c\right )^{\frac{3}{2}} + 7 \, A \sqrt{\tan \left (d x + c\right )} + 3 i \, B \sqrt{\tan \left (d x + c\right )}}{8 \, a^{2} d{\left (\tan \left (d x + c\right ) - i\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1/8*I + 1/8)*sqrt(2)*(A - I*B)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/(a^2*d) + (1/16*I - 1/16)*sq
rt(2)*(7*A - I*B)*arctan(-(1/2*I + 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/(a^2*d) - 1/8*(5*I*A*tan(d*x + c)^(3/2) -
B*tan(d*x + c)^(3/2) + 7*A*sqrt(tan(d*x + c)) + 3*I*B*sqrt(tan(d*x + c)))/(a^2*d*(tan(d*x + c) - I)^2)