3.168 \(\int \frac{1}{\sqrt{d \tan (e+f x)} (a+i a \tan (e+f x))} \, dx\)

Optimal. Leaf size=262 \[ -\frac{\left (\frac{3}{4}-\frac{i}{4}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a \sqrt{d} f}+\frac{\left (\frac{3}{4}-\frac{i}{4}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} a \sqrt{d} f}+\frac{\sqrt{d \tan (e+f x)}}{2 d f (a+i a \tan (e+f x))}-\frac{\left (\frac{3}{8}+\frac{i}{8}\right ) \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a \sqrt{d} f}+\frac{\left (\frac{3}{8}+\frac{i}{8}\right ) \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a \sqrt{d} f} \]

[Out]

((-3/4 + I/4)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a*Sqrt[d]*f) + ((3/4 - I/4)*ArcTan[
1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a*Sqrt[d]*f) - ((3/8 + I/8)*Log[Sqrt[d] + Sqrt[d]*Tan[e
+ f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a*Sqrt[d]*f) + ((3/8 + I/8)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x
] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a*Sqrt[d]*f) + Sqrt[d*Tan[e + f*x]]/(2*d*f*(a + I*a*Tan[e + f*x]))

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Rubi [A]  time = 0.2377, antiderivative size = 262, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3552, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac{\left (\frac{3}{4}-\frac{i}{4}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a \sqrt{d} f}+\frac{\left (\frac{3}{4}-\frac{i}{4}\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}+1\right )}{\sqrt{2} a \sqrt{d} f}+\frac{\sqrt{d \tan (e+f x)}}{2 d f (a+i a \tan (e+f x))}-\frac{\left (\frac{3}{8}+\frac{i}{8}\right ) \log \left (\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a \sqrt{d} f}+\frac{\left (\frac{3}{8}+\frac{i}{8}\right ) \log \left (\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}+\sqrt{d}\right )}{\sqrt{2} a \sqrt{d} f} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[d*Tan[e + f*x]]*(a + I*a*Tan[e + f*x])),x]

[Out]

((-3/4 + I/4)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a*Sqrt[d]*f) + ((3/4 - I/4)*ArcTan[
1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a*Sqrt[d]*f) - ((3/8 + I/8)*Log[Sqrt[d] + Sqrt[d]*Tan[e
+ f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a*Sqrt[d]*f) + ((3/8 + I/8)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x
] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a*Sqrt[d]*f) + Sqrt[d*Tan[e + f*x]]/(2*d*f*(a + I*a*Tan[e + f*x]))

Rule 3552

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(a
*(c + d*Tan[e + f*x])^(n + 1))/(2*f*(b*c - a*d)*(a + b*Tan[e + f*x])), x] + Dist[1/(2*a*(b*c - a*d)), Int[(c +
d*Tan[e + f*x])^n*Simp[b*c + a*d*(n - 1) - b*d*n*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x]
&& NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 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{1}{\sqrt{d \tan (e+f x)} (a+i a \tan (e+f x))} \, dx &=\frac{\sqrt{d \tan (e+f x)}}{2 d f (a+i a \tan (e+f x))}-\frac{\int \frac{-\frac{3 a d}{2}+\frac{1}{2} i a d \tan (e+f x)}{\sqrt{d \tan (e+f x)}} \, dx}{2 a^2 d}\\ &=\frac{\sqrt{d \tan (e+f x)}}{2 d f (a+i a \tan (e+f x))}-\frac{\operatorname{Subst}\left (\int \frac{-\frac{3 a d^2}{2}+\frac{1}{2} i a d x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{a^2 d f}\\ &=\frac{\sqrt{d \tan (e+f x)}}{2 d f (a+i a \tan (e+f x))}--\frac{\left (\frac{3}{4}+\frac{i}{4}\right ) \operatorname{Subst}\left (\int \frac{d-x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{a f}--\frac{\left (\frac{3}{4}-\frac{i}{4}\right ) \operatorname{Subst}\left (\int \frac{d+x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{a f}\\ &=\frac{\sqrt{d \tan (e+f x)}}{2 d f (a+i a \tan (e+f x))}--\frac{\left (\frac{3}{8}-\frac{i}{8}\right ) \operatorname{Subst}\left (\int \frac{1}{d-\sqrt{2} \sqrt{d} x+x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{a f}--\frac{\left (\frac{3}{8}-\frac{i}{8}\right ) \operatorname{Subst}\left (\int \frac{1}{d+\sqrt{2} \sqrt{d} x+x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{a f}-\frac{\left (\frac{3}{8}+\frac{i}{8}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt{d}+2 x}{-d-\sqrt{2} \sqrt{d} x-x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a \sqrt{d} f}-\frac{\left (\frac{3}{8}+\frac{i}{8}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt{d}-2 x}{-d+\sqrt{2} \sqrt{d} x-x^2} \, dx,x,\sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a \sqrt{d} f}\\ &=-\frac{\left (\frac{3}{8}+\frac{i}{8}\right ) \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a \sqrt{d} f}+\frac{\left (\frac{3}{8}+\frac{i}{8}\right ) \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a \sqrt{d} f}+\frac{\sqrt{d \tan (e+f x)}}{2 d f (a+i a \tan (e+f x))}--\frac{\left (\frac{3}{4}-\frac{i}{4}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a \sqrt{d} f}-\frac{\left (\frac{3}{4}-\frac{i}{4}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a \sqrt{d} f}\\ &=-\frac{\left (\frac{3}{4}-\frac{i}{4}\right ) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a \sqrt{d} f}+\frac{\left (\frac{3}{4}-\frac{i}{4}\right ) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{d \tan (e+f x)}}{\sqrt{d}}\right )}{\sqrt{2} a \sqrt{d} f}-\frac{\left (\frac{3}{8}+\frac{i}{8}\right ) \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)-\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a \sqrt{d} f}+\frac{\left (\frac{3}{8}+\frac{i}{8}\right ) \log \left (\sqrt{d}+\sqrt{d} \tan (e+f x)+\sqrt{2} \sqrt{d \tan (e+f x)}\right )}{\sqrt{2} a \sqrt{d} f}+\frac{\sqrt{d \tan (e+f x)}}{2 d f (a+i a \tan (e+f x))}\\ \end{align*}

Mathematica [A]  time = 0.776341, size = 147, normalized size = 0.56 \[ \frac{\sqrt{\sin (2 (e+f x))} \sec (e+f x) \left (-2 i \sqrt{\sin (2 (e+f x))} \sec (e+f x)+(1+3 i) (1+i \tan (e+f x)) \sin ^{-1}(\cos (e+f x)-\sin (e+f x))+(3+i) (\tan (e+f x)-i) \log \left (\sin (e+f x)+\sqrt{\sin (2 (e+f x))}+\cos (e+f x)\right )\right )}{8 a f (\tan (e+f x)-i) \sqrt{d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[d*Tan[e + f*x]]*(a + I*a*Tan[e + f*x])),x]

[Out]

(Sec[e + f*x]*Sqrt[Sin[2*(e + f*x)]]*((-2*I)*Sec[e + f*x]*Sqrt[Sin[2*(e + f*x)]] + (1 + 3*I)*ArcSin[Cos[e + f*
x] - Sin[e + f*x]]*(1 + I*Tan[e + f*x]) + (3 + I)*Log[Cos[e + f*x] + Sin[e + f*x] + Sqrt[Sin[2*(e + f*x)]]]*(-
I + Tan[e + f*x])))/(8*a*f*Sqrt[d*Tan[e + f*x]]*(-I + Tan[e + f*x]))

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Maple [A]  time = 0.063, size = 102, normalized size = 0.4 \begin{align*}{\frac{-{\frac{i}{2}}}{fa \left ( -id+d\tan \left ( fx+e \right ) \right ) }\sqrt{d\tan \left ( fx+e \right ) }}-{\frac{i}{fa}\arctan \left ({\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{-id}}}} \right ){\frac{1}{\sqrt{-id}}}}+{\frac{{\frac{i}{2}}}{fa}\arctan \left ({\sqrt{d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{id}}}} \right ){\frac{1}{\sqrt{id}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e)),x)

[Out]

-1/2*I/f/a*(d*tan(f*x+e))^(1/2)/(-I*d+d*tan(f*x+e))-I/f/a/(-I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(-I*d)^(1/2
))+1/2*I/f/a/(I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(I*d)^(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(1/(d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 2.12274, size = 1364, normalized size = 5.21 \begin{align*} -\frac{{\left (a d f \sqrt{-\frac{i}{4 \, a^{2} d f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-2 \,{\left (2 \,{\left (a d f e^{\left (2 i \, f x + 2 i \, e\right )} + a d f\right )} \sqrt{\frac{-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{i}{4 \, a^{2} d f^{2}}} + i \, d e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) - a d f \sqrt{-\frac{i}{4 \, a^{2} d f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (2 \,{\left (2 \,{\left (a d f e^{\left (2 i \, f x + 2 i \, e\right )} + a d f\right )} \sqrt{\frac{-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{i}{4 \, a^{2} d f^{2}}} - i \, d e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) - a d f \sqrt{\frac{i}{a^{2} d f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac{{\left ({\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt{\frac{-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{i}{a^{2} d f^{2}}} + i\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a f}\right ) + a d f \sqrt{\frac{i}{a^{2} d f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac{{\left ({\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt{\frac{-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{\frac{i}{a^{2} d f^{2}}} - i\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a f}\right ) - \sqrt{\frac{-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}{\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a d f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e)),x, algorithm="fricas")

[Out]

-1/4*(a*d*f*sqrt(-1/4*I/(a^2*d*f^2))*e^(2*I*f*x + 2*I*e)*log(-2*(2*(a*d*f*e^(2*I*f*x + 2*I*e) + a*d*f)*sqrt((-
I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-1/4*I/(a^2*d*f^2)) + I*d*e^(2*I*f*x + 2*I*e))*
e^(-2*I*f*x - 2*I*e)) - a*d*f*sqrt(-1/4*I/(a^2*d*f^2))*e^(2*I*f*x + 2*I*e)*log(2*(2*(a*d*f*e^(2*I*f*x + 2*I*e)
+ a*d*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-1/4*I/(a^2*d*f^2)) - I*d*e^(2
*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)) - a*d*f*sqrt(I/(a^2*d*f^2))*e^(2*I*f*x + 2*I*e)*log(((a*f*e^(2*I*f*x +
2*I*e) + a*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(I/(a^2*d*f^2)) + I)*e^(-2*
I*f*x - 2*I*e)/(a*f)) + a*d*f*sqrt(I/(a^2*d*f^2))*e^(2*I*f*x + 2*I*e)*log(-((a*f*e^(2*I*f*x + 2*I*e) + a*f)*sq
rt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(I/(a^2*d*f^2)) - I)*e^(-2*I*f*x - 2*I*e)/(
a*f)) - sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*(e^(2*I*f*x + 2*I*e) + 1))*e^(-2*I*f*
x - 2*I*e)/(a*d*f)

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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(1/(d*tan(f*x+e))**(1/2)/(a+I*a*tan(f*x+e)),x)

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.16944, size = 244, normalized size = 0.93 \begin{align*} -\frac{1}{2} \, d^{2}{\left (-\frac{i \, \sqrt{2} \arctan \left (\frac{16 \, \sqrt{d^{2}} \sqrt{d \tan \left (f x + e\right )}}{8 i \, \sqrt{2} d^{\frac{3}{2}} + 8 \, \sqrt{2} \sqrt{d^{2}} \sqrt{d}}\right )}{a d^{\frac{5}{2}} f{\left (\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} + \frac{2 i \, \sqrt{2} \arctan \left (\frac{16 \, \sqrt{d^{2}} \sqrt{d \tan \left (f x + e\right )}}{-8 i \, \sqrt{2} d^{\frac{3}{2}} + 8 \, \sqrt{2} \sqrt{d^{2}} \sqrt{d}}\right )}{a d^{\frac{5}{2}} f{\left (-\frac{i \, d}{\sqrt{d^{2}}} + 1\right )}} + \frac{i \, \sqrt{d \tan \left (f x + e\right )}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )} a d^{2} f}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e)),x, algorithm="giac")

[Out]

-1/2*d^2*(-I*sqrt(2)*arctan(16*sqrt(d^2)*sqrt(d*tan(f*x + e))/(8*I*sqrt(2)*d^(3/2) + 8*sqrt(2)*sqrt(d^2)*sqrt(
d)))/(a*d^(5/2)*f*(I*d/sqrt(d^2) + 1)) + 2*I*sqrt(2)*arctan(16*sqrt(d^2)*sqrt(d*tan(f*x + e))/(-8*I*sqrt(2)*d^
(3/2) + 8*sqrt(2)*sqrt(d^2)*sqrt(d)))/(a*d^(5/2)*f*(-I*d/sqrt(d^2) + 1)) + I*sqrt(d*tan(f*x + e))/((d*tan(f*x
+ e) - I*d)*a*d^2*f))