3.1129 \(\int \frac{1}{(a+i a \tan (e+f x))^2 (c+d \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=281 \[ \frac{d \left (2 c^2+7 i c d+25 d^2\right )}{8 a^2 f (c-i d) (c+i d)^3 \sqrt{c+d \tan (e+f x)}}+\frac{\left (2 i c^2-10 c d-23 i d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{8 a^2 f (c+i d)^{7/2}}+\frac{-7 d+2 i c}{8 a^2 f (c+i d)^2 (1+i \tan (e+f x)) \sqrt{c+d \tan (e+f x)}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{4 a^2 f (c-i d)^{3/2}}-\frac{1}{4 f (-d+i c) (a+i a \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}} \]

[Out]

((-I/4)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(a^2*(c - I*d)^(3/2)*f) + (((2*I)*c^2 - 10*c*d - (23*
I)*d^2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/(8*a^2*(c + I*d)^(7/2)*f) + (d*(2*c^2 + (7*I)*c*d + 2
5*d^2))/(8*a^2*(c - I*d)*(c + I*d)^3*f*Sqrt[c + d*Tan[e + f*x]]) + ((2*I)*c - 7*d)/(8*a^2*(c + I*d)^2*f*(1 + I
*Tan[e + f*x])*Sqrt[c + d*Tan[e + f*x]]) - 1/(4*(I*c - d)*f*(a + I*a*Tan[e + f*x])^2*Sqrt[c + d*Tan[e + f*x]])

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Rubi [A]  time = 0.76385, antiderivative size = 281, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.233, Rules used = {3559, 3596, 3529, 3539, 3537, 63, 208} \[ \frac{d \left (2 c^2+7 i c d+25 d^2\right )}{8 a^2 f (c-i d) (c+i d)^3 \sqrt{c+d \tan (e+f x)}}+\frac{\left (2 i c^2-10 c d-23 i d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{8 a^2 f (c+i d)^{7/2}}+\frac{-7 d+2 i c}{8 a^2 f (c+i d)^2 (1+i \tan (e+f x)) \sqrt{c+d \tan (e+f x)}}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{4 a^2 f (c-i d)^{3/2}}-\frac{1}{4 f (-d+i c) (a+i a \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])^2*(c + d*Tan[e + f*x])^(3/2)),x]

[Out]

((-I/4)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(a^2*(c - I*d)^(3/2)*f) + (((2*I)*c^2 - 10*c*d - (23*
I)*d^2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/(8*a^2*(c + I*d)^(7/2)*f) + (d*(2*c^2 + (7*I)*c*d + 2
5*d^2))/(8*a^2*(c - I*d)*(c + I*d)^3*f*Sqrt[c + d*Tan[e + f*x]]) + ((2*I)*c - 7*d)/(8*a^2*(c + I*d)^2*f*(1 + I
*Tan[e + f*x])*Sqrt[c + d*Tan[e + f*x]]) - 1/(4*(I*c - d)*f*(a + I*a*Tan[e + f*x])^2*Sqrt[c + d*Tan[e + f*x]])

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 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 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 (c+d \tan (e+f x))^{3/2}} \, dx &=-\frac{1}{4 (i c-d) f (a+i a \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}-\frac{\int \frac{-\frac{1}{2} a (4 i c-9 d)-\frac{5}{2} i a d \tan (e+f x)}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2}} \, dx}{4 a^2 (i c-d)}\\ &=\frac{2 i c-7 d}{8 a^2 (c+i d)^2 f (1+i \tan (e+f x)) \sqrt{c+d \tan (e+f x)}}-\frac{1}{4 (i c-d) f (a+i a \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}-\frac{\int \frac{-\frac{1}{2} a^2 \left (4 c^2+14 i c d-25 d^2\right )-\frac{3}{2} a^2 (2 c+7 i d) d \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}} \, dx}{8 a^4 (c+i d)^2}\\ &=\frac{d \left (2 c^2+7 i c d+25 d^2\right )}{8 a^2 (c+i d)^2 \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{2 i c-7 d}{8 a^2 (c+i d)^2 f (1+i \tan (e+f x)) \sqrt{c+d \tan (e+f x)}}-\frac{1}{4 (i c-d) f (a+i a \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}-\frac{\int \frac{-\frac{1}{2} a^2 \left (4 c^3+14 i c^2 d-19 c d^2+21 i d^3\right )-\frac{1}{2} a^2 d \left (2 c^2+7 i c d+25 d^2\right ) \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{8 a^4 (c+i d)^2 \left (c^2+d^2\right )}\\ &=\frac{d \left (2 c^2+7 i c d+25 d^2\right )}{8 a^2 (c+i d)^2 \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{2 i c-7 d}{8 a^2 (c+i d)^2 f (1+i \tan (e+f x)) \sqrt{c+d \tan (e+f x)}}-\frac{1}{4 (i c-d) f (a+i a \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}+\frac{\int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{8 a^2 (c-i d)}+\frac{\left (2 c^2+10 i c d-23 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)^3}\\ &=\frac{d \left (2 c^2+7 i c d+25 d^2\right )}{8 a^2 (c+i d)^2 \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{2 i c-7 d}{8 a^2 (c+i d)^2 f (1+i \tan (e+f x)) \sqrt{c+d \tan (e+f x)}}-\frac{1}{4 (i c-d) f (a+i a \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}-\frac{\operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c-i d x}} \, dx,x,i \tan (e+f x)\right )}{8 a^2 (i c+d) f}-\frac{\left (2 c^2+10 i c d-23 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)^3 f}\\ &=\frac{d \left (2 c^2+7 i c d+25 d^2\right )}{8 a^2 (c+i d)^2 \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{2 i c-7 d}{8 a^2 (c+i d)^2 f (1+i \tan (e+f x)) \sqrt{c+d \tan (e+f x)}}-\frac{1}{4 (i c-d) f (a+i a \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}-\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 (c-i d) d f}-\frac{\left (2 c^2+10 i c d-23 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)^3 d f}\\ &=-\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{4 a^2 (c-i d)^{3/2} f}-\frac{\left (10 c d-i \left (2 c^2-23 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)^{7/2} f}+\frac{d \left (2 c^2+7 i c d+25 d^2\right )}{8 a^2 (c+i d)^2 \left (c^2+d^2\right ) f \sqrt{c+d \tan (e+f x)}}+\frac{2 i c-7 d}{8 a^2 (c+i d)^2 f (1+i \tan (e+f x)) \sqrt{c+d \tan (e+f x)}}-\frac{1}{4 (i c-d) f (a+i a \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 4.37606, size = 388, normalized size = 1.38 \[ \frac{\sec ^2(e+f x) (\cos (f x)+i \sin (f x))^2 \left (\frac{(\cos (2 f x)-i \sin (2 f x)) \sqrt{c+d \tan (e+f x)} \left (\left (-23 c^2 d+12 i c^3+26 i c d^2+23 d^3\right ) \cos (e+f x)+\left (-5 c^2 d+4 i c^3+18 i c d^2+41 d^3\right ) \cos (3 (e+f x))-4 \left (3 i c^2 d+2 c^3+16 c d^2-43 i d^3\right ) \sin (e+f x) \cos ^2(e+f x)\right )}{2 (c-i d) (c+i d)^3 (c \cos (e+f x)+d \sin (e+f x))}+\frac{2 (\cos (2 e)+i \sin (2 e)) \left (\sqrt{-c+i d} \left (8 c^2 d-2 i c^3+13 i c d^2+23 d^3\right ) \tan ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{-c-i d}}\right )-2 i (-c-i d)^{7/2} \tan ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{-c+i d}}\right )\right )}{(-c-i d)^{7/2} (-c+i d)^{3/2}}\right )}{16 f (a+i a \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])^2*(c + d*Tan[e + f*x])^(3/2)),x]

[Out]

(Sec[e + f*x]^2*(Cos[f*x] + I*Sin[f*x])^2*((2*(Sqrt[-c + I*d]*((-2*I)*c^3 + 8*c^2*d + (13*I)*c*d^2 + 23*d^3)*A
rcTan[Sqrt[c + d*Tan[e + f*x]]/Sqrt[-c - I*d]] - (2*I)*(-c - I*d)^(7/2)*ArcTan[Sqrt[c + d*Tan[e + f*x]]/Sqrt[-
c + I*d]])*(Cos[2*e] + I*Sin[2*e]))/((-c - I*d)^(7/2)*(-c + I*d)^(3/2)) + ((Cos[2*f*x] - I*Sin[2*f*x])*(((12*I
)*c^3 - 23*c^2*d + (26*I)*c*d^2 + 23*d^3)*Cos[e + f*x] + ((4*I)*c^3 - 5*c^2*d + (18*I)*c*d^2 + 41*d^3)*Cos[3*(
e + f*x)] - 4*(2*c^3 + (3*I)*c^2*d + 16*c*d^2 - (43*I)*d^3)*Cos[e + f*x]^2*Sin[e + f*x])*Sqrt[c + d*Tan[e + f*
x]])/(2*(c - I*d)*(c + I*d)^3*(c*Cos[e + f*x] + d*Sin[e + f*x]))))/(16*f*(a + I*a*Tan[e + f*x])^2)

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Maple [B]  time = 0.078, size = 1568, normalized size = 5.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+I*a*tan(f*x+e))^2/(c+d*tan(f*x+e))^(3/2),x)

[Out]

3/4/f/a^2*d/(I*d-c)^(3/2)/(c+I*d)^3*arctan((c+d*tan(f*x+e))^(1/2)/(I*d-c)^(1/2))*c^2-1/4/f/a^2*d^3/(I*d-c)^(3/
2)/(c+I*d)^3*arctan((c+d*tan(f*x+e))^(1/2)/(I*d-c)^(1/2))-1/4*I/f/a^2/(I*d-c)^(3/2)/(c+I*d)^3*arctan((c+d*tan(
f*x+e))^(1/2)/(I*d-c)^(1/2))*c^3+1/2*I/f/a^2*d^4/(c+I*d)^3/(I*d-c)/(-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^2-1/4/f/a^2*d/(c+I*d)^3/(I*d-c)/(-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^4+7/8/f/a^2*d^3/(c+I*d)^3/(I*d-c)/(-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
^2+9/8/f/a^2*d^5/(c+I*d)^3/(I*d-c)/(-I*d+d*tan(f*x+e))^2/(-d^2+2*I*c*d+c^2)*(c+d*tan(f*x+e))^(3/2)+3/4*I/f/a^2
*d^2/(I*d-c)^(3/2)/(c+I*d)^3*arctan((c+d*tan(f*x+e))^(1/2)/(I*d-c)^(1/2))*c-33/8*I/f/a^2*d^4/(c+I*d)^3/(I*d-c)
/(-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+1/4/f/a^2*d/(c+I*d)^3/(I*d-
c)/(-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^5-11/4/f/a^2*d^3/(c+I*d)^3/(I*d-c)/(-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^3-3/f/a^2*d^5/(c+I*d)^3/(I*d-c)/(-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*I/f/a^2/(c+I*d)^3/(I*d-c)/(-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^5-31/8*I/f/a^2*d^2/(c+I*d)^3/(I*d-c)/(-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^3+15/8*I/f/a^2*d^2/(c+I*d)^3/(I*d-c)/(-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^4-3/2/f/a^2*d/(c+I*d)^3/(I*d-c)/(-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^4+11/8/f/a^2*d^3/(c+I*d)^3/(I*d-c)/(-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+23/8/f/a^2*d^5/(c+I*d)^3/(I*d-c)/(-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))-11/8*I/f/a^2*d^2/(c+I*d)^3/(I*d-c)/(-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^3-11/8*I/f/a^2*d^4/(c+I*d)^3/(I*d-c)/(-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-11/8*I/f/a^2*d^6/(c+I*d)^3/(I*d-c)/(-I*d+d*tan(f*x+e))^2/(
-d^2+2*I*c*d+c^2)*(c+d*tan(f*x+e))^(1/2)-2/f/a^2*d^3/(I*c-d)/(I*c+d)/(c+I*d)^2/(c+d*tan(f*x+e))^(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/(a+I*a*tan(f*x+e))^2/(c+d*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 12.0133, size = 5484, normalized size = 19.52 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^2/(c+d*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

(((-4*I*a^2*c^5 + 4*a^2*c^4*d - 8*I*a^2*c^3*d^2 + 8*a^2*c^2*d^3 - 4*I*a^2*c*d^4 + 4*a^2*d^5)*f*e^(6*I*f*x + 6*
I*e) + (-4*I*a^2*c^5 + 12*a^2*c^4*d + 8*I*a^2*c^3*d^2 + 8*a^2*c^2*d^3 + 12*I*a^2*c*d^4 - 4*a^2*d^5)*f*e^(4*I*f
*x + 4*I*e))*sqrt(I/((-16*I*a^4*c^3 - 48*a^4*c^2*d + 48*I*a^4*c*d^2 + 16*a^4*d^3)*f^2))*log((((8*I*a^2*c^2 + 1
6*a^2*c*d - 8*I*a^2*d^2)*f*e^(2*I*f*x + 2*I*e) + (8*I*a^2*c^2 + 16*a^2*c*d - 8*I*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(I/((-16*I*a^4*c^3 - 48*a^4*c^2*d + 48*I*a^4*c*d^
2 + 16*a^4*d^3)*f^2)) + 2*(c - I*d)*e^(2*I*f*x + 2*I*e) + 2*c)*e^(-2*I*f*x - 2*I*e)) + ((4*I*a^2*c^5 - 4*a^2*c
^4*d + 8*I*a^2*c^3*d^2 - 8*a^2*c^2*d^3 + 4*I*a^2*c*d^4 - 4*a^2*d^5)*f*e^(6*I*f*x + 6*I*e) + (4*I*a^2*c^5 - 12*
a^2*c^4*d - 8*I*a^2*c^3*d^2 - 8*a^2*c^2*d^3 - 12*I*a^2*c*d^4 + 4*a^2*d^5)*f*e^(4*I*f*x + 4*I*e))*sqrt(I/((-16*
I*a^4*c^3 - 48*a^4*c^2*d + 48*I*a^4*c*d^2 + 16*a^4*d^3)*f^2))*log((((-8*I*a^2*c^2 - 16*a^2*c*d + 8*I*a^2*d^2)*
f*e^(2*I*f*x + 2*I*e) + (-8*I*a^2*c^2 - 16*a^2*c*d + 8*I*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(I/((-16*I*a^4*c^3 - 48*a^4*c^2*d + 48*I*a^4*c*d^2 + 16*a^4*d^3)*f^2)) +
2*(c - I*d)*e^(2*I*f*x + 2*I*e) + 2*c)*e^(-2*I*f*x - 2*I*e)) + ((-4*I*a^2*c^5 + 4*a^2*c^4*d - 8*I*a^2*c^3*d^2
+ 8*a^2*c^2*d^3 - 4*I*a^2*c*d^4 + 4*a^2*d^5)*f*e^(6*I*f*x + 6*I*e) + (-4*I*a^2*c^5 + 12*a^2*c^4*d + 8*I*a^2*c^
3*d^2 + 8*a^2*c^2*d^3 + 12*I*a^2*c*d^4 - 4*a^2*d^5)*f*e^(4*I*f*x + 4*I*e))*sqrt((-4*I*c^4 + 40*c^3*d + 192*I*c
^2*d^2 - 460*c*d^3 - 529*I*d^4)/((64*I*a^4*c^7 - 448*a^4*c^6*d - 1344*I*a^4*c^5*d^2 + 2240*a^4*c^4*d^3 + 2240*
I*a^4*c^3*d^4 - 1344*a^4*c^2*d^5 - 448*I*a^4*c*d^6 + 64*a^4*d^7)*f^2))*log((2*I*c^3 - 12*c^2*d - 33*I*c*d^2 +
23*d^3 + ((8*a^2*c^4 + 32*I*a^2*c^3*d - 48*a^2*c^2*d^2 - 32*I*a^2*c*d^3 + 8*a^2*d^4)*f*e^(2*I*f*x + 2*I*e) + (
8*a^2*c^4 + 32*I*a^2*c^3*d - 48*a^2*c^2*d^2 - 32*I*a^2*c*d^3 + 8*a^2*d^4)*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 + 40*c^3*d + 192*I*c^2*d^2 - 460*c*d^3 - 529*I*d^4)/((
64*I*a^4*c^7 - 448*a^4*c^6*d - 1344*I*a^4*c^5*d^2 + 2240*a^4*c^4*d^3 + 2240*I*a^4*c^3*d^4 - 1344*a^4*c^2*d^5 -
448*I*a^4*c*d^6 + 64*a^4*d^7)*f^2)) + (2*I*c^3 - 10*c^2*d - 23*I*c*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*
I*e)/((8*a^2*c^4 + 32*I*a^2*c^3*d - 48*a^2*c^2*d^2 - 32*I*a^2*c*d^3 + 8*a^2*d^4)*f)) + ((4*I*a^2*c^5 - 4*a^2*c
^4*d + 8*I*a^2*c^3*d^2 - 8*a^2*c^2*d^3 + 4*I*a^2*c*d^4 - 4*a^2*d^5)*f*e^(6*I*f*x + 6*I*e) + (4*I*a^2*c^5 - 12*
a^2*c^4*d - 8*I*a^2*c^3*d^2 - 8*a^2*c^2*d^3 - 12*I*a^2*c*d^4 + 4*a^2*d^5)*f*e^(4*I*f*x + 4*I*e))*sqrt((-4*I*c^
4 + 40*c^3*d + 192*I*c^2*d^2 - 460*c*d^3 - 529*I*d^4)/((64*I*a^4*c^7 - 448*a^4*c^6*d - 1344*I*a^4*c^5*d^2 + 22
40*a^4*c^4*d^3 + 2240*I*a^4*c^3*d^4 - 1344*a^4*c^2*d^5 - 448*I*a^4*c*d^6 + 64*a^4*d^7)*f^2))*log((2*I*c^3 - 12
*c^2*d - 33*I*c*d^2 + 23*d^3 - ((8*a^2*c^4 + 32*I*a^2*c^3*d - 48*a^2*c^2*d^2 - 32*I*a^2*c*d^3 + 8*a^2*d^4)*f*e
^(2*I*f*x + 2*I*e) + (8*a^2*c^4 + 32*I*a^2*c^3*d - 48*a^2*c^2*d^2 - 32*I*a^2*c*d^3 + 8*a^2*d^4)*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 + 40*c^3*d + 192*I*c^2*d^2 - 460
*c*d^3 - 529*I*d^4)/((64*I*a^4*c^7 - 448*a^4*c^6*d - 1344*I*a^4*c^5*d^2 + 2240*a^4*c^4*d^3 + 2240*I*a^4*c^3*d^
4 - 1344*a^4*c^2*d^5 - 448*I*a^4*c*d^6 + 64*a^4*d^7)*f^2)) + (2*I*c^3 - 10*c^2*d - 23*I*c*d^2)*e^(2*I*f*x + 2*
I*e))*e^(-2*I*f*x - 2*I*e)/((8*a^2*c^4 + 32*I*a^2*c^3*d - 48*a^2*c^2*d^2 - 32*I*a^2*c*d^3 + 8*a^2*d^4)*f)) + (
c^3 + I*c^2*d + c*d^2 + I*d^3 + (3*c^3 + 4*I*c^2*d + 17*c*d^2 - 42*I*d^3)*e^(6*I*f*x + 6*I*e) + (7*c^3 + 13*I*
c^2*d + 21*c*d^2 - 33*I*d^3)*e^(4*I*f*x + 4*I*e) + (5*c^3 + 10*I*c^2*d + 5*c*d^2 + 10*I*d^3)*e^(2*I*f*x + 2*I*
e))*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))/((-16*I*a^2*c^5 + 16*a^2*c^4*d
- 32*I*a^2*c^3*d^2 + 32*a^2*c^2*d^3 - 16*I*a^2*c*d^4 + 16*a^2*d^5)*f*e^(6*I*f*x + 6*I*e) + (-16*I*a^2*c^5 + 48
*a^2*c^4*d + 32*I*a^2*c^3*d^2 + 32*a^2*c^2*d^3 + 48*I*a^2*c*d^4 - 16*a^2*d^5)*f*e^(4*I*f*x + 4*I*e))

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

[Out]

Exception raised: AttributeError

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Giac [B]  time = 1.64954, size = 811, normalized size = 2.89 \begin{align*} -2 \, d^{3}{\left (\frac{2 \,{\left (2 i \, c^{2} - 10 \, c d - 23 i \, 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 \, a^{2} c^{3} d^{3} f + 24 i \, a^{2} c^{2} d^{4} f - 24 \, a^{2} c d^{5} f - 8 i \, a^{2} d^{6} f\right )} \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}}{\left (\frac{i \, d}{c - \sqrt{c^{2} + d^{2}}} + 1\right )}} - \frac{1}{{\left (a^{2} c^{4} f + 2 i \, a^{2} c^{3} d f + 2 i \, a^{2} c d^{3} f - a^{2} d^{4} f\right )} \sqrt{d \tan \left (f x + e\right ) + c}} + \frac{2 \, \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 (4 i \, a^{2} c d^{3} f + 4 \, 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} + 9 i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} d - 13 i \, \sqrt{d \tan \left (f x + e\right ) + c} c d + 11 \, \sqrt{d \tan \left (f x + e\right ) + c} d^{2}}{{\left (16 \, a^{2} c^{3} d^{2} f + 48 i \, a^{2} c^{2} d^{3} f - 48 \, a^{2} c d^{4} f - 16 i \, a^{2} d^{5} f\right )}{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))^2/(c+d*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

-2*d^3*(2*(2*I*c^2 - 10*c*d - 23*I*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*a^2*c^3*d^3*f + 24*I*a^2*c^2*d^4*f - 24*a^2*c*d^5*f - 8*I*a^2*d^6*f)*sqrt(-8*c + 8*
sqrt(c^2 + d^2))*(I*d/(c - sqrt(c^2 + d^2)) + 1)) - 1/((a^2*c^4*f + 2*I*a^2*c^3*d*f + 2*I*a^2*c*d^3*f - a^2*d^
4*f)*sqrt(d*tan(f*x + e) + c)) + 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*s
qrt(c^2 + d^2))))/((4*I*a^2*c*d^3*f + 4*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 + 9*I*(d*tan(f*x + e) + c)^(3/2)*d -
13*I*sqrt(d*tan(f*x + e) + c)*c*d + 11*sqrt(d*tan(f*x + e) + c)*d^2)/((16*a^2*c^3*d^2*f + 48*I*a^2*c^2*d^3*f -
48*a^2*c*d^4*f - 16*I*a^2*d^5*f)*(d*tan(f*x + e) - I*d)^2))