3.1123 \(\int \frac{1}{(a+i a \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}} \, dx\)

Optimal. Leaf size=221 \[ \frac{\left (2 i c^2-6 c d-7 i d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{8 a^2 f (c+i d)^{5/2}}+\frac{(-5 d+2 i c) \sqrt{c+d \tan (e+f x)}}{8 a^2 f (c+i d)^2 (1+i \tan (e+f x))}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{4 a^2 f \sqrt{c-i d}}-\frac{\sqrt{c+d \tan (e+f x)}}{4 f (-d+i c) (a+i a \tan (e+f x))^2} \]

[Out]

((-I/4)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(a^2*Sqrt[c - I*d]*f) + (((2*I)*c^2 - 6*c*d - (7*I)*d
^2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/(8*a^2*(c + I*d)^(5/2)*f) + (((2*I)*c - 5*d)*Sqrt[c + d*T
an[e + f*x]])/(8*a^2*(c + I*d)^2*f*(1 + I*Tan[e + f*x])) - Sqrt[c + d*Tan[e + f*x]]/(4*(I*c - d)*f*(a + I*a*Ta
n[e + f*x])^2)

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Rubi [A]  time = 0.606299, antiderivative size = 221, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3559, 3596, 3539, 3537, 63, 208} \[ \frac{\left (2 i c^2-6 c d-7 i d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{8 a^2 f (c+i d)^{5/2}}+\frac{(-5 d+2 i c) \sqrt{c+d \tan (e+f x)}}{8 a^2 f (c+i d)^2 (1+i \tan (e+f x))}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{4 a^2 f \sqrt{c-i d}}-\frac{\sqrt{c+d \tan (e+f x)}}{4 f (-d+i c) (a+i a \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])^2*Sqrt[c + d*Tan[e + f*x]]),x]

[Out]

((-I/4)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(a^2*Sqrt[c - I*d]*f) + (((2*I)*c^2 - 6*c*d - (7*I)*d
^2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/(8*a^2*(c + I*d)^(5/2)*f) + (((2*I)*c - 5*d)*Sqrt[c + d*T
an[e + f*x]])/(8*a^2*(c + I*d)^2*f*(1 + I*Tan[e + f*x])) - Sqrt[c + d*Tan[e + f*x]]/(4*(I*c - d)*f*(a + I*a*Ta
n[e + f*x])^2)

Rule 3559

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(a*(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*Simp[b*c*m - a*d*(2*m + n + 1) + b*d*(m + n + 1)*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] && LtQ[m, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

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 3539

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

Rule 3537

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{1}{(a+i a \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}} \, dx &=-\frac{\sqrt{c+d \tan (e+f x)}}{4 (i c-d) f (a+i a \tan (e+f x))^2}-\frac{\int \frac{-\frac{1}{2} a (4 i c-7 d)-\frac{3}{2} i a d \tan (e+f x)}{(a+i a \tan (e+f x)) \sqrt{c+d \tan (e+f x)}} \, dx}{4 a^2 (i c-d)}\\ &=\frac{(2 i c-5 d) \sqrt{c+d \tan (e+f x)}}{8 a^2 (c+i d)^2 f (1+i \tan (e+f x))}-\frac{\sqrt{c+d \tan (e+f x)}}{4 (i c-d) f (a+i a \tan (e+f x))^2}-\frac{\int \frac{-\frac{1}{2} a^2 \left (4 c^2+10 i c d-9 d^2\right )-\frac{1}{2} a^2 (2 c+5 i d) d \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{8 a^4 (c+i d)^2}\\ &=\frac{(2 i c-5 d) \sqrt{c+d \tan (e+f x)}}{8 a^2 (c+i d)^2 f (1+i \tan (e+f x))}-\frac{\sqrt{c+d \tan (e+f x)}}{4 (i c-d) f (a+i a \tan (e+f x))^2}+\frac{\int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{8 a^2}+\frac{\left (2 c^2+6 i c d-7 d^2\right ) \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{16 a^2 (c+i d)^2}\\ &=\frac{(2 i c-5 d) \sqrt{c+d \tan (e+f x)}}{8 a^2 (c+i d)^2 f (1+i \tan (e+f x))}-\frac{\sqrt{c+d \tan (e+f x)}}{4 (i c-d) f (a+i a \tan (e+f x))^2}+\frac{i \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c-i d x}} \, dx,x,i \tan (e+f x)\right )}{8 a^2 f}-\frac{\left (i \left (2 c^2+6 i c d-7 d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{16 a^2 (c+i d)^2 f}\\ &=\frac{(2 i c-5 d) \sqrt{c+d \tan (e+f x)}}{8 a^2 (c+i d)^2 f (1+i \tan (e+f x))}-\frac{\sqrt{c+d \tan (e+f x)}}{4 (i c-d) f (a+i a \tan (e+f x))^2}-\frac{\operatorname{Subst}\left (\int \frac{1}{-1-\frac{i c}{d}+\frac{i x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{4 a^2 d f}-\frac{\left (2 c^2+6 i c d-7 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{-1+\frac{i c}{d}-\frac{i x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{8 a^2 (c+i d)^2 d f}\\ &=-\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{4 a^2 \sqrt{c-i d} f}-\frac{\left (6 c d-i \left (2 c^2-7 d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{8 a^2 (c+i d)^{5/2} f}+\frac{(2 i c-5 d) \sqrt{c+d \tan (e+f x)}}{8 a^2 (c+i d)^2 f (1+i \tan (e+f x))}-\frac{\sqrt{c+d \tan (e+f x)}}{4 (i c-d) f (a+i a \tan (e+f x))^2}\\ \end{align*}

Mathematica [A]  time = 1.97604, size = 275, normalized size = 1.24 \[ \frac{\sec ^2(e+f x) (\cos (f x)+i \sin (f x))^2 \left (\frac{2 (\cos (2 e)+i \sin (2 e)) \left (\sqrt{-c+i d} \left (-2 i c^2+6 c d+7 i d^2\right ) \tan ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{-c-i d}}\right )+2 i (-c-i d)^{5/2} \tan ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{-c+i d}}\right )\right )}{(-c-i d)^{5/2} \sqrt{-c+i d}}+\frac{2 \cos (e+f x) (\sin (2 f x)+i \cos (2 f x)) \sqrt{c+d \tan (e+f x)} ((-5 d+2 i c) \sin (e+f x)+(4 c+7 i d) \cos (e+f x))}{(c+i d)^2}\right )}{16 f (a+i a \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])^2*Sqrt[c + d*Tan[e + f*x]]),x]

[Out]

(Sec[e + f*x]^2*(Cos[f*x] + I*Sin[f*x])^2*((2*(Sqrt[-c + I*d]*((-2*I)*c^2 + 6*c*d + (7*I)*d^2)*ArcTan[Sqrt[c +
d*Tan[e + f*x]]/Sqrt[-c - I*d]] + (2*I)*(-c - I*d)^(5/2)*ArcTan[Sqrt[c + d*Tan[e + f*x]]/Sqrt[-c + I*d]])*(Co
s[2*e] + I*Sin[2*e]))/((-c - I*d)^(5/2)*Sqrt[-c + I*d]) + (2*Cos[e + f*x]*(I*Cos[2*f*x] + Sin[2*f*x])*((4*c +
(7*I)*d)*Cos[e + f*x] + ((2*I)*c - 5*d)*Sin[e + f*x])*Sqrt[c + d*Tan[e + f*x]])/(c + I*d)^2))/(16*f*(a + I*a*T
an[e + f*x])^2)

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Maple [B]  time = 0.08, size = 502, normalized size = 2.3 \begin{align*}{\frac{{\frac{i}{4}}}{f{a}^{2}}\arctan \left ({\sqrt{c+d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{id-c}}}} \right ){\frac{1}{\sqrt{id-c}}}}+{\frac{{\frac{5\,i}{8}}{d}^{2}}{f{a}^{2} \left ( -id+d\tan \left ( fx+e \right ) \right ) ^{2} \left ( -{d}^{2}+2\,icd+{c}^{2} \right ) } \left ( c+d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{cd}{4\,f{a}^{2} \left ( -id+d\tan \left ( fx+e \right ) \right ) ^{2} \left ( -{d}^{2}+2\,icd+{c}^{2} \right ) } \left ( c+d\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{{\frac{9\,i}{8}}{d}^{2}c}{f{a}^{2} \left ( -id+d\tan \left ( fx+e \right ) \right ) ^{2} \left ( -{d}^{2}+2\,icd+{c}^{2} \right ) }\sqrt{c+d\tan \left ( fx+e \right ) }}-{\frac{{c}^{2}d}{4\,f{a}^{2} \left ( -id+d\tan \left ( fx+e \right ) \right ) ^{2} \left ( -{d}^{2}+2\,icd+{c}^{2} \right ) }\sqrt{c+d\tan \left ( fx+e \right ) }}+{\frac{7\,{d}^{3}}{8\,f{a}^{2} \left ( -id+d\tan \left ( fx+e \right ) \right ) ^{2} \left ( -{d}^{2}+2\,icd+{c}^{2} \right ) }\sqrt{c+d\tan \left ( fx+e \right ) }}-{\frac{{\frac{i}{4}}{c}^{2}}{f{a}^{2} \left ( -{d}^{2}+2\,icd+{c}^{2} \right ) }\arctan \left ({\sqrt{c+d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{-id-c}}}} \right ){\frac{1}{\sqrt{-id-c}}}}+{\frac{{\frac{7\,i}{8}}{d}^{2}}{f{a}^{2} \left ( -{d}^{2}+2\,icd+{c}^{2} \right ) }\arctan \left ({\sqrt{c+d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{-id-c}}}} \right ){\frac{1}{\sqrt{-id-c}}}}+{\frac{3\,cd}{4\,f{a}^{2} \left ( -{d}^{2}+2\,icd+{c}^{2} \right ) }\arctan \left ({\sqrt{c+d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{-id-c}}}} \right ){\frac{1}{\sqrt{-id-c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*I/f/a^2/(I*d-c)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)/(I*d-c)^(1/2))+5/8*I/f/a^2*d^2/(-I*d+d*tan(f*x+e))^2/(
-d^2+2*I*c*d+c^2)*(c+d*tan(f*x+e))^(3/2)+1/4/f/a^2*d/(-I*d+d*tan(f*x+e))^2/(-d^2+2*I*c*d+c^2)*(c+d*tan(f*x+e))
^(3/2)*c-9/8*I/f/a^2*d^2/(-I*d+d*tan(f*x+e))^2/(-d^2+2*I*c*d+c^2)*(c+d*tan(f*x+e))^(1/2)*c-1/4/f/a^2*d/(-I*d+d
*tan(f*x+e))^2/(-d^2+2*I*c*d+c^2)*(c+d*tan(f*x+e))^(1/2)*c^2+7/8/f/a^2*d^3/(-I*d+d*tan(f*x+e))^2/(-d^2+2*I*c*d
+c^2)*(c+d*tan(f*x+e))^(1/2)-1/4*I/f/a^2/(-d^2+2*I*c*d+c^2)/(-I*d-c)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)/(-I*d
-c)^(1/2))*c^2+7/8*I/f/a^2*d^2/(-d^2+2*I*c*d+c^2)/(-I*d-c)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)/(-I*d-c)^(1/2))
+3/4/f/a^2*d/(-d^2+2*I*c*d+c^2)/(-I*d-c)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)/(-I*d-c)^(1/2))*c

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

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 3.5673, size = 3421, normalized size = 15.48 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((4*a^2*c^2 + 8*I*a^2*c*d - 4*a^2*d^2)*f*sqrt(1/16*I/((-I*a^4*c - a^4*d)*f^2))*e^(4*I*f*x + 4*I*e)*log(-2*(4*
((I*a^2*c + a^2*d)*f*e^(2*I*f*x + 2*I*e) + (I*a^2*c + a^2*d)*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)
/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/16*I/((-I*a^4*c - a^4*d)*f^2)) - (c - I*d)*e^(2*I*f*x + 2*I*e) - c)*e^(-2*I
*f*x - 2*I*e)) - (4*a^2*c^2 + 8*I*a^2*c*d - 4*a^2*d^2)*f*sqrt(1/16*I/((-I*a^4*c - a^4*d)*f^2))*e^(4*I*f*x + 4*
I*e)*log(-2*(4*((-I*a^2*c - a^2*d)*f*e^(2*I*f*x + 2*I*e) + (-I*a^2*c - a^2*d)*f)*sqrt(((c - I*d)*e^(2*I*f*x +
2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/16*I/((-I*a^4*c - a^4*d)*f^2)) - (c - I*d)*e^(2*I*f*x + 2*
I*e) - c)*e^(-2*I*f*x - 2*I*e)) - (4*a^2*c^2 + 8*I*a^2*c*d - 4*a^2*d^2)*f*sqrt((4*I*c^4 - 24*c^3*d - 64*I*c^2*
d^2 + 84*c*d^3 + 49*I*d^4)/((-64*I*a^4*c^5 + 320*a^4*c^4*d + 640*I*a^4*c^3*d^2 - 640*a^4*c^2*d^3 - 320*I*a^4*c
*d^4 + 64*a^4*d^5)*f^2))*e^(4*I*f*x + 4*I*e)*log(-(2*c^3 + 8*I*c^2*d - 13*c*d^2 - 7*I*d^3 - ((8*I*a^2*c^3 - 24
*a^2*c^2*d - 24*I*a^2*c*d^2 + 8*a^2*d^3)*f*e^(2*I*f*x + 2*I*e) + (8*I*a^2*c^3 - 24*a^2*c^2*d - 24*I*a^2*c*d^2
+ 8*a^2*d^3)*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt((4*I*c^4 - 24*c
^3*d - 64*I*c^2*d^2 + 84*c*d^3 + 49*I*d^4)/((-64*I*a^4*c^5 + 320*a^4*c^4*d + 640*I*a^4*c^3*d^2 - 640*a^4*c^2*d
^3 - 320*I*a^4*c*d^4 + 64*a^4*d^5)*f^2)) + (2*c^3 + 6*I*c^2*d - 7*c*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*
I*e)/((8*I*a^2*c^3 - 24*a^2*c^2*d - 24*I*a^2*c*d^2 + 8*a^2*d^3)*f)) + (4*a^2*c^2 + 8*I*a^2*c*d - 4*a^2*d^2)*f*
sqrt((4*I*c^4 - 24*c^3*d - 64*I*c^2*d^2 + 84*c*d^3 + 49*I*d^4)/((-64*I*a^4*c^5 + 320*a^4*c^4*d + 640*I*a^4*c^3
*d^2 - 640*a^4*c^2*d^3 - 320*I*a^4*c*d^4 + 64*a^4*d^5)*f^2))*e^(4*I*f*x + 4*I*e)*log(-(2*c^3 + 8*I*c^2*d - 13*
c*d^2 - 7*I*d^3 - ((-8*I*a^2*c^3 + 24*a^2*c^2*d + 24*I*a^2*c*d^2 - 8*a^2*d^3)*f*e^(2*I*f*x + 2*I*e) + (-8*I*a^
2*c^3 + 24*a^2*c^2*d + 24*I*a^2*c*d^2 - 8*a^2*d^3)*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f
*x + 2*I*e) + 1))*sqrt((4*I*c^4 - 24*c^3*d - 64*I*c^2*d^2 + 84*c*d^3 + 49*I*d^4)/((-64*I*a^4*c^5 + 320*a^4*c^4
*d + 640*I*a^4*c^3*d^2 - 640*a^4*c^2*d^3 - 320*I*a^4*c*d^4 + 64*a^4*d^5)*f^2)) + (2*c^3 + 6*I*c^2*d - 7*c*d^2)
*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/((8*I*a^2*c^3 - 24*a^2*c^2*d - 24*I*a^2*c*d^2 + 8*a^2*d^3)*f)) + ((
-3*I*c + 6*d)*e^(4*I*f*x + 4*I*e) + (-4*I*c + 7*d)*e^(2*I*f*x + 2*I*e) - I*c + d)*sqrt(((c - I*d)*e^(2*I*f*x +
2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-4*I*f*x - 4*I*e)/((16*a^2*c^2 + 32*I*a^2*c*d - 16*a^2*d^2)*
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/(c+d*tan(f*x+e))**(1/2)/(a+I*a*tan(f*x+e))**2,x)

[Out]

Exception raised: AttributeError

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Giac [B]  time = 1.59023, size = 680, normalized size = 3.08 \begin{align*} \frac{1}{4} \, d^{3}{\left (\frac{16 \,{\left (2 \, c^{2} + 6 i \, c d - 7 \, d^{2}\right )} \arctan \left (\frac{4 \,{\left (\sqrt{d \tan \left (f x + e\right ) + c} c - \sqrt{c^{2} + d^{2}} \sqrt{d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}} + i \, \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}} d - \sqrt{c^{2} + d^{2}} \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}}}\right )}{{\left (8 i \, a^{2} c^{2} d^{3} f - 16 \, a^{2} c d^{4} f - 8 i \, a^{2} d^{5} f\right )} \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}}{\left (\frac{i \, d}{c - \sqrt{c^{2} + d^{2}}} + 1\right )}} + \frac{8 \,{\left (2 \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} c - 2 \, \sqrt{d \tan \left (f x + e\right ) + c} c^{2} + 5 i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} d - 9 i \, \sqrt{d \tan \left (f x + e\right ) + c} c d + 7 \, \sqrt{d \tan \left (f x + e\right ) + c} d^{2}\right )}}{{\left (16 \, a^{2} c^{2} d^{2} f + 32 i \, a^{2} c d^{3} f - 16 \, a^{2} d^{4} f\right )}{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{2}} - \frac{\sqrt{2} \arctan \left (\frac{16 i \, \sqrt{d \tan \left (f x + e\right ) + c} c + 16 i \, \sqrt{c^{2} + d^{2}} \sqrt{d \tan \left (f x + e\right ) + c}}{8 \, \sqrt{2} \sqrt{c + \sqrt{c^{2} + d^{2}}} c - 8 i \, \sqrt{2} \sqrt{c + \sqrt{c^{2} + d^{2}}} d + 8 \, \sqrt{2} \sqrt{c^{2} + d^{2}} \sqrt{c + \sqrt{c^{2} + d^{2}}}}\right )}{a^{2} \sqrt{c + \sqrt{c^{2} + d^{2}}} d^{3} f{\left (-\frac{i \, d}{c + \sqrt{c^{2} + d^{2}}} + 1\right )}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*d^3*(16*(2*c^2 + 6*I*c*d - 7*d^2)*arctan(4*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x +
e) + c))/(c*sqrt(-8*c + 8*sqrt(c^2 + d^2)) + I*sqrt(-8*c + 8*sqrt(c^2 + d^2))*d - sqrt(c^2 + d^2)*sqrt(-8*c +
8*sqrt(c^2 + d^2))))/((8*I*a^2*c^2*d^3*f - 16*a^2*c*d^4*f - 8*I*a^2*d^5*f)*sqrt(-8*c + 8*sqrt(c^2 + d^2))*(I*d
/(c - sqrt(c^2 + d^2)) + 1)) + 8*(2*(d*tan(f*x + e) + c)^(3/2)*c - 2*sqrt(d*tan(f*x + e) + c)*c^2 + 5*I*(d*tan
(f*x + e) + c)^(3/2)*d - 9*I*sqrt(d*tan(f*x + e) + c)*c*d + 7*sqrt(d*tan(f*x + e) + c)*d^2)/((16*a^2*c^2*d^2*f
+ 32*I*a^2*c*d^3*f - 16*a^2*d^4*f)*(d*tan(f*x + e) - I*d)^2) - sqrt(2)*arctan((16*I*sqrt(d*tan(f*x + e) + c)*
c + 16*I*sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c))/(8*sqrt(2)*sqrt(c + sqrt(c^2 + d^2))*c - 8*I*sqrt(2)*sqrt(c
+ sqrt(c^2 + d^2))*d + 8*sqrt(2)*sqrt(c^2 + d^2)*sqrt(c + sqrt(c^2 + d^2))))/(a^2*sqrt(c + sqrt(c^2 + d^2))*d
^3*f*(-I*d/(c + sqrt(c^2 + d^2)) + 1)))