3.1105 \(\int \frac{\sqrt{c+d \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=211 \[ -\frac{\left (2 c d-i \left (2 c^2+d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{8 a^2 f (c+i d)^{3/2}}+\frac{(-d+2 i c) \sqrt{c+d \tan (e+f x)}}{8 a^2 f (c+i d) (1+i \tan (e+f x))}-\frac{i \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{4 a^2 f}+\frac{i \sqrt{c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2} \]

[Out]

((-I/4)*Sqrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(a^2*f) - ((2*c*d - I*(2*c^2 + d^2))*Ar
cTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/(8*a^2*(c + I*d)^(3/2)*f) + (((2*I)*c - d)*Sqrt[c + d*Tan[e + f
*x]])/(8*a^2*(c + I*d)*f*(1 + I*Tan[e + f*x])) + ((I/4)*Sqrt[c + d*Tan[e + f*x]])/(f*(a + I*a*Tan[e + f*x])^2)

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Rubi [A]  time = 0.596016, antiderivative size = 211, 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 = {3557, 3596, 3539, 3537, 63, 208} \[ -\frac{\left (2 c d-i \left (2 c^2+d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{8 a^2 f (c+i d)^{3/2}}+\frac{(-d+2 i c) \sqrt{c+d \tan (e+f x)}}{8 a^2 f (c+i d) (1+i \tan (e+f x))}-\frac{i \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{4 a^2 f}+\frac{i \sqrt{c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/4)*Sqrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(a^2*f) - ((2*c*d - I*(2*c^2 + d^2))*Ar
cTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/(8*a^2*(c + I*d)^(3/2)*f) + (((2*I)*c - d)*Sqrt[c + d*Tan[e + f
*x]])/(8*a^2*(c + I*d)*f*(1 + I*Tan[e + f*x])) + ((I/4)*Sqrt[c + d*Tan[e + f*x]])/(f*(a + I*a*Tan[e + f*x])^2)

Rule 3557

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> -Sim
p[(b*(a + b*Tan[e + f*x])^m*Sqrt[c + d*Tan[e + f*x]])/(2*a*f*m), x] + Dist[1/(4*a^2*m), Int[((a + b*Tan[e + f*
x])^(m + 1)*Simp[2*a*c*m + b*d + a*d*(2*m + 1)*Tan[e + f*x], x])/Sqrt[c + d*Tan[e + f*x]], x], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, 0] && IntegersQ[2
*m]

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

Mathematica [A]  time = 1.69978, size = 281, normalized size = 1.33 \[ \frac{\sec ^2(e+f x) (\cos (f x)+i \sin (f x))^2 \left (\frac{2 \cos (e+f x) (\sin (2 f x)+i \cos (2 f x)) \sqrt{c+d \tan (e+f x)} ((-d+2 i c) \sin (e+f x)+(4 c+3 i d) \cos (e+f x))}{c+i d}-\frac{2 (\cos (2 e)+i \sin (2 e)) \left (2 i \sqrt{-c-i d} \left (c^2+d^2\right ) \tan ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{-c+i d}}\right )-i \sqrt{-c+i d} \left (2 c^2+2 i c d+d^2\right ) \tan ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{-c-i d}}\right )\right )}{(-c-i d)^{3/2} \sqrt{-c+i d}}\right )}{16 f (a+i a \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[e + f*x]^2*(Cos[f*x] + I*Sin[f*x])^2*((-2*((-I)*Sqrt[-c + I*d]*(2*c^2 + (2*I)*c*d + d^2)*ArcTan[Sqrt[c +
d*Tan[e + f*x]]/Sqrt[-c - I*d]] + (2*I)*Sqrt[-c - I*d]*(c^2 + d^2)*ArcTan[Sqrt[c + d*Tan[e + f*x]]/Sqrt[-c + I
*d]])*(Cos[2*e] + I*Sin[2*e]))/((-c - I*d)^(3/2)*Sqrt[-c + I*d]) + (2*Cos[e + f*x]*(I*Cos[2*f*x] + Sin[2*f*x])
*((4*c + (3*I)*d)*Cos[e + f*x] + ((2*I)*c - d)*Sin[e + f*x])*Sqrt[c + d*Tan[e + f*x]])/(c + I*d)))/(16*f*(a +
I*a*Tan[e + f*x])^2)

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Maple [B]  time = 0.083, size = 715, normalized size = 3.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((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))+1/4*I/f/a^2*d/(-I*d+d*tan(f*x+e))^2/(I
*c^2-I*d^2-2*c*d)*(c+d*tan(f*x+e))^(3/2)*c^2-1/8*I/f/a^2*d^3/(-I*d+d*tan(f*x+e))^2/(I*c^2-I*d^2-2*c*d)*(c+d*ta
n(f*x+e))^(3/2)-3/8/f/a^2*d^2/(-I*d+d*tan(f*x+e))^2/(I*c^2-I*d^2-2*c*d)*(c+d*tan(f*x+e))^(3/2)*c+7/8/f/a^2*d^2
/(-I*d+d*tan(f*x+e))^2/(I*c^2-I*d^2-2*c*d)*(c+d*tan(f*x+e))^(1/2)*c^2-3/8/f/a^2*d^4/(-I*d+d*tan(f*x+e))^2/(I*c
^2-I*d^2-2*c*d)*(c+d*tan(f*x+e))^(1/2)-1/4*I/f/a^2*d/(-I*d+d*tan(f*x+e))^2/(I*c^2-I*d^2-2*c*d)*(c+d*tan(f*x+e)
)^(1/2)*c^3+I/f/a^2*d^3/(-I*d+d*tan(f*x+e))^2/(I*c^2-I*d^2-2*c*d)*(c+d*tan(f*x+e))^(1/2)*c+1/4/f/a^2/(I*c^2-I*
d^2-2*c*d)/(-I*d-c)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)/(-I*d-c)^(1/2))*c^3-1/8/f/a^2*d^2/(I*c^2-I*d^2-2*c*d)/
(-I*d-c)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)/(-I*d-c)^(1/2))*c+1/2*I/f/a^2*d/(I*c^2-I*d^2-2*c*d)/(-I*d-c)^(1/2
)*arctan((c+d*tan(f*x+e))^(1/2)/(-I*d-c)^(1/2))*c^2+1/8*I/f/a^2*d^3/(I*c^2-I*d^2-2*c*d)/(-I*d-c)^(1/2)*arctan(
(c+d*tan(f*x+e))^(1/2)/(-I*d-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((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 = 4.10874, size = 2708, normalized size = 12.83 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((I*a^2*c - a^2*d)*f*sqrt(-(c - I*d)/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(1/4*((8*I*a^2*f*e^(2*I*f*x + 2*I*
e) + 8*I*a^2*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(-(c - I*d)/(a^4
*f^2)) + 8*(c - I*d)*e^(2*I*f*x + 2*I*e) + 8*c)*e^(-2*I*f*x - 2*I*e)) + (-I*a^2*c + a^2*d)*f*sqrt(-(c - I*d)/(
a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(1/4*((-8*I*a^2*f*e^(2*I*f*x + 2*I*e) - 8*I*a^2*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(-(c - I*d)/(a^4*f^2)) + 8*(c - I*d)*e^(2*I*f*x + 2*I*e) +
8*c)*e^(-2*I*f*x - 2*I*e)) + 4*(-I*a^2*c + a^2*d)*f*sqrt((-4*I*c^4 + 8*c^3*d + 4*c*d^3 - I*d^4)/((64*I*a^4*c^
3 - 192*a^4*c^2*d - 192*I*a^4*c*d^2 + 64*a^4*d^3)*f^2))*e^(4*I*f*x + 4*I*e)*log(-(-2*I*c^3 + 4*c^2*d + I*c*d^2
+ d^3 + ((8*a^2*c^2 + 16*I*a^2*c*d - 8*a^2*d^2)*f*e^(2*I*f*x + 2*I*e) + (8*a^2*c^2 + 16*I*a^2*c*d - 8*a^2*d^2
)*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 + 8*c^3*d + 4*c*
d^3 - I*d^4)/((64*I*a^4*c^3 - 192*a^4*c^2*d - 192*I*a^4*c*d^2 + 64*a^4*d^3)*f^2)) + (-2*I*c^3 + 2*c^2*d - I*c*
d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/((8*a^2*c^2 + 16*I*a^2*c*d - 8*a^2*d^2)*f)) + 4*(I*a^2*c - a^2*
d)*f*sqrt((-4*I*c^4 + 8*c^3*d + 4*c*d^3 - I*d^4)/((64*I*a^4*c^3 - 192*a^4*c^2*d - 192*I*a^4*c*d^2 + 64*a^4*d^3
)*f^2))*e^(4*I*f*x + 4*I*e)*log(-(-2*I*c^3 + 4*c^2*d + I*c*d^2 + d^3 - ((8*a^2*c^2 + 16*I*a^2*c*d - 8*a^2*d^2)
*f*e^(2*I*f*x + 2*I*e) + (8*a^2*c^2 + 16*I*a^2*c*d - 8*a^2*d^2)*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 + 8*c^3*d + 4*c*d^3 - I*d^4)/((64*I*a^4*c^3 - 192*a^4*c^2*d - 19
2*I*a^4*c*d^2 + 64*a^4*d^3)*f^2)) + (-2*I*c^3 + 2*c^2*d - I*c*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(
(8*a^2*c^2 + 16*I*a^2*c*d - 8*a^2*d^2)*f)) - ((3*c + 2*I*d)*e^(4*I*f*x + 4*I*e) + (4*c + 3*I*d)*e^(2*I*f*x + 2
*I*e) + c + I*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)/((I*a^2*c - a^2*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((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 = 1.59577, size = 649, normalized size = 3.08 \begin{align*} \frac{1}{8} \, d^{3}{\left (\frac{32 \,{\left (2 \, c^{2} + 2 i \, c d + 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 d^{3} f - 8 \, a^{2} d^{4} f\right )} \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}}{\left (\frac{i \, d}{c - \sqrt{c^{2} + d^{2}}} + 1\right )}} + \frac{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} + i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} d - 5 i \, \sqrt{d \tan \left (f x + e\right ) + c} c d + 3 \, \sqrt{d \tan \left (f x + e\right ) + c} d^{2}}{{\left (a^{2} c d^{2} f + i \, a^{2} d^{3} f\right )}{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{2}} + \frac{2 \, \sqrt{2}{\left (c - i \, d\right )} \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((c+d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^2,x, algorithm="giac")

[Out]

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