3.1106 \(\int \frac{\sqrt{c+d \tan (e+f x)}}{(a+i a \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=280 \[ \frac{\left (-4 c^2 d+2 i c^3-i c d^2-2 d^3\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{16 a^3 f (c+i d)^{5/2}}+\frac{c (-3 d+2 i c) \sqrt{c+d \tan (e+f x)}}{16 f (c+i d)^2 \left (a^3+i a^3 \tan (e+f x)\right )}-\frac{i \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{8 a^3 f}+\frac{(-2 d+3 i c) \sqrt{c+d \tan (e+f x)}}{24 a f (c+i d) (a+i a \tan (e+f x))^2}+\frac{i \sqrt{c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3} \]

[Out]

((-I/8)*Sqrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(a^3*f) + (((2*I)*c^3 - 4*c^2*d - I*c*d
^2 - 2*d^3)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/(16*a^3*(c + I*d)^(5/2)*f) + ((I/6)*Sqrt[c + d*Ta
n[e + f*x]])/(f*(a + I*a*Tan[e + f*x])^3) + (((3*I)*c - 2*d)*Sqrt[c + d*Tan[e + f*x]])/(24*a*(c + I*d)*f*(a +
I*a*Tan[e + f*x])^2) + (c*((2*I)*c - 3*d)*Sqrt[c + d*Tan[e + f*x]])/(16*(c + I*d)^2*f*(a^3 + I*a^3*Tan[e + f*x
]))

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Rubi [A]  time = 0.981864, antiderivative size = 280, normalized size of antiderivative = 1., number of steps used = 10, 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} \[ \frac{\left (-4 c^2 d+2 i c^3-i c d^2-2 d^3\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{16 a^3 f (c+i d)^{5/2}}+\frac{c (-3 d+2 i c) \sqrt{c+d \tan (e+f x)}}{16 f (c+i d)^2 \left (a^3+i a^3 \tan (e+f x)\right )}-\frac{i \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{8 a^3 f}+\frac{(-2 d+3 i c) \sqrt{c+d \tan (e+f x)}}{24 a f (c+i d) (a+i a \tan (e+f x))^2}+\frac{i \sqrt{c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/8)*Sqrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(a^3*f) + (((2*I)*c^3 - 4*c^2*d - I*c*d
^2 - 2*d^3)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/(16*a^3*(c + I*d)^(5/2)*f) + ((I/6)*Sqrt[c + d*Ta
n[e + f*x]])/(f*(a + I*a*Tan[e + f*x])^3) + (((3*I)*c - 2*d)*Sqrt[c + d*Tan[e + f*x]])/(24*a*(c + I*d)*f*(a +
I*a*Tan[e + f*x])^2) + (c*((2*I)*c - 3*d)*Sqrt[c + d*Tan[e + f*x]])/(16*(c + I*d)^2*f*(a^3 + I*a^3*Tan[e + f*x
]))

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))^3} \, dx &=\frac{i \sqrt{c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac{\int \frac{-a (6 c-i d)-5 a d \tan (e+f x)}{(a+i a \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}} \, dx}{12 a^2}\\ &=\frac{i \sqrt{c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac{(3 i c-2 d) \sqrt{c+d \tan (e+f x)}}{24 a (c+i d) f (a+i a \tan (e+f x))^2}+\frac{\int \frac{-3 a^2 \left (3 c d-i \left (4 c^2+2 d^2\right )\right )+3 a^2 (3 i c-2 d) d \tan (e+f x)}{(a+i a \tan (e+f x)) \sqrt{c+d \tan (e+f x)}} \, dx}{48 a^4 (i c-d)}\\ &=\frac{i \sqrt{c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac{(3 i c-2 d) \sqrt{c+d \tan (e+f x)}}{24 a (c+i d) f (a+i a \tan (e+f x))^2}+\frac{c (2 i c-3 d) \sqrt{c+d \tan (e+f x)}}{16 (c+i d)^2 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac{\int \frac{3 a^3 \left (4 c^3+6 i c^2 d+c d^2+4 i d^3\right )+3 a^3 c (2 c+3 i d) d \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{96 a^6 (c+i d)^2}\\ &=\frac{i \sqrt{c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac{(3 i c-2 d) \sqrt{c+d \tan (e+f x)}}{24 a (c+i d) f (a+i a \tan (e+f x))^2}+\frac{c (2 i c-3 d) \sqrt{c+d \tan (e+f x)}}{16 (c+i d)^2 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac{(c-i d) \int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{16 a^3}+\frac{\left (2 c^3+4 i c^2 d-c d^2+2 i d^3\right ) \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{32 a^3 (c+i d)^2}\\ &=\frac{i \sqrt{c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac{(3 i c-2 d) \sqrt{c+d \tan (e+f x)}}{24 a (c+i d) f (a+i a \tan (e+f x))^2}+\frac{c (2 i c-3 d) \sqrt{c+d \tan (e+f x)}}{16 (c+i d)^2 f \left (a^3+i a^3 \tan (e+f x)\right )}+\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 )}{16 a^3 f}-\frac{\left (i \left (2 c^3+4 i c^2 d-c d^2+2 i d^3\right )\right ) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{32 a^3 (c+i d)^2 f}\\ &=\frac{i \sqrt{c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac{(3 i c-2 d) \sqrt{c+d \tan (e+f x)}}{24 a (c+i d) f (a+i a \tan (e+f x))^2}+\frac{c (2 i c-3 d) \sqrt{c+d \tan (e+f x)}}{16 (c+i d)^2 f \left (a^3+i a^3 \tan (e+f x)\right )}-\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 )}{8 a^3 d f}-\frac{\left (2 c^3+4 i c^2 d-c d^2+2 i d^3\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 )}{16 a^3 (c+i d)^2 d f}\\ &=-\frac{i \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{8 a^3 f}-\frac{\left (4 c^2 d-i \left (2 c^3-c d^2+2 i d^3\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{16 a^3 (c+i d)^{5/2} f}+\frac{i \sqrt{c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac{(3 i c-2 d) \sqrt{c+d \tan (e+f x)}}{24 a (c+i d) f (a+i a \tan (e+f x))^2}+\frac{c (2 i c-3 d) \sqrt{c+d \tan (e+f x)}}{16 (c+i d)^2 f \left (a^3+i a^3 \tan (e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 2.56476, size = 329, normalized size = 1.18 \[ \frac{\sec ^3(e+f x) (\cos (f x)+i \sin (f x))^3 \left (\frac{2 (\cos (3 e)+i \sin (3 e)) \left (\sqrt{-c+i d} \left (4 c^2 d-2 i c^3+i c d^2+2 d^3\right ) \tan ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{-c-i d}}\right )+2 (d+i c) (-c-i d)^{5/2} \tan ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{-c+i d}}\right )\right )}{(-c-i d)^{5/2} \sqrt{-c+i d}}+\frac{2 \cos (e+f x) (\sin (3 f x)+i \cos (3 f x)) \sqrt{c+d \tan (e+f x)} \left (i \left (9 c^2+14 i c d-2 d^2\right ) \sin (2 (e+f x))+\left (13 c^2+22 i c d-6 d^2\right ) \cos (2 (e+f x))+7 c^2+13 i c d-6 d^2\right )}{3 (c+i d)^2}\right )}{32 f (a+i a \tan (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[e + f*x]^3*(Cos[f*x] + I*Sin[f*x])^3*((2*(Sqrt[-c + I*d]*((-2*I)*c^3 + 4*c^2*d + I*c*d^2 + 2*d^3)*ArcTan[
Sqrt[c + d*Tan[e + f*x]]/Sqrt[-c - I*d]] + 2*(-c - I*d)^(5/2)*(I*c + d)*ArcTan[Sqrt[c + d*Tan[e + f*x]]/Sqrt[-
c + I*d]])*(Cos[3*e] + I*Sin[3*e]))/((-c - I*d)^(5/2)*Sqrt[-c + I*d]) + (2*Cos[e + f*x]*(I*Cos[3*f*x] + Sin[3*
f*x])*(7*c^2 + (13*I)*c*d - 6*d^2 + (13*c^2 + (22*I)*c*d - 6*d^2)*Cos[2*(e + f*x)] + I*(9*c^2 + (14*I)*c*d - 2
*d^2)*Sin[2*(e + f*x)])*Sqrt[c + d*Tan[e + f*x]])/(3*(c + I*d)^2)))/(32*f*(a + I*a*Tan[e + f*x])^3)

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Maple [B]  time = 0.091, size = 1315, normalized size = 4.7 \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))^3,x)

[Out]

2/3*I/f/a^3*d^4/(-I*d+d*tan(f*x+e))^3/(-I*d^3-3*c*d^2+3*I*c^2*d+c^3)*(c+d*tan(f*x+e))^(3/2)*c+5/16*I/f/a^3*d^2
/(-I*d+d*tan(f*x+e))^3*c^2/(-I*d^3-3*c*d^2+3*I*c^2*d+c^3)*(c+d*tan(f*x+e))^(5/2)+1/8/f/a^3*d/(-I*d+d*tan(f*x+e
))^3*c^3/(-I*d^3-3*c*d^2+3*I*c^2*d+c^3)*(c+d*tan(f*x+e))^(5/2)-3/16/f/a^3*d^3/(-I*d+d*tan(f*x+e))^3*c/(-I*d^3-
3*c*d^2+3*I*c^2*d+c^3)*(c+d*tan(f*x+e))^(5/2)+11/16*I/f/a^3*d^2/(-I*d+d*tan(f*x+e))^3/(-I*d^3-3*c*d^2+3*I*c^2*
d+c^3)*(c+d*tan(f*x+e))^(1/2)*c^4+5/16*I/f/a^3*d^2/(-I*d^3-3*c*d^2+3*I*c^2*d+c^3)/(-I*d-c)^(1/2)*arctan((c+d*t
an(f*x+e))^(1/2)/(-I*d-c)^(1/2))*c^2-1/4/f/a^3*d/(-I*d+d*tan(f*x+e))^3/(-I*d^3-3*c*d^2+3*I*c^2*d+c^3)*(c+d*tan
(f*x+e))^(3/2)*c^4+4/3/f/a^3*d^3/(-I*d+d*tan(f*x+e))^3/(-I*d^3-3*c*d^2+3*I*c^2*d+c^3)*(c+d*tan(f*x+e))^(3/2)*c
^2-1/12/f/a^3*d^5/(-I*d+d*tan(f*x+e))^3/(-I*d^3-3*c*d^2+3*I*c^2*d+c^3)*(c+d*tan(f*x+e))^(3/2)-I/f/a^3*d^2/(-I*
d+d*tan(f*x+e))^3/(-I*d^3-3*c*d^2+3*I*c^2*d+c^3)*(c+d*tan(f*x+e))^(3/2)*c^3-1/8*I/f/a^3*(I*d-c)^(1/2)*arctan((
c+d*tan(f*x+e))^(1/2)/(I*d-c)^(1/2))-29/16*I/f/a^3*d^4/(-I*d+d*tan(f*x+e))^3/(-I*d^3-3*c*d^2+3*I*c^2*d+c^3)*(c
+d*tan(f*x+e))^(1/2)*c^2+1/8/f/a^3*d/(-I*d+d*tan(f*x+e))^3/(-I*d^3-3*c*d^2+3*I*c^2*d+c^3)*(c+d*tan(f*x+e))^(1/
2)*c^5-25/16/f/a^3*d^3/(-I*d+d*tan(f*x+e))^3/(-I*d^3-3*c*d^2+3*I*c^2*d+c^3)*(c+d*tan(f*x+e))^(1/2)*c^3+17/16/f
/a^3*d^5/(-I*d+d*tan(f*x+e))^3/(-I*d^3-3*c*d^2+3*I*c^2*d+c^3)*(c+d*tan(f*x+e))^(1/2)*c-1/8*I/f/a^3/(-I*d^3-3*c
*d^2+3*I*c^2*d+c^3)/(-I*d-c)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)/(-I*d-c)^(1/2))*c^4+1/8*I/f/a^3*d^4/(-I*d^3-3
*c*d^2+3*I*c^2*d+c^3)/(-I*d-c)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)/(-I*d-c)^(1/2))+1/4*I/f/a^3*d^6/(-I*d+d*tan
(f*x+e))^3/(-I*d^3-3*c*d^2+3*I*c^2*d+c^3)*(c+d*tan(f*x+e))^(1/2)+3/8/f/a^3*d/(-I*d^3-3*c*d^2+3*I*c^2*d+c^3)/(-
I*d-c)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)/(-I*d-c)^(1/2))*c^3+1/16/f/a^3*d^3/(-I*d^3-3*c*d^2+3*I*c^2*d+c^3)/(
-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((c+d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 6.79283, size = 3671, normalized size = 13.11 \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))^3,x, algorithm="fricas")

[Out]

1/8*((24*a^3*c^2 + 48*I*a^3*c*d - 24*a^3*d^2)*f*sqrt(-(c - I*d)/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(1/8*((16*I*
a^3*f*e^(2*I*f*x + 2*I*e) + 16*I*a^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(-(c - I*d)/(a^6*f^2)) + 16*(c - I*d)*e^(2*I*f*x + 2*I*e) + 16*c)*e^(-2*I*f*x - 2*I*e)) - (24*a^3*c^2
+ 48*I*a^3*c*d - 24*a^3*d^2)*f*sqrt(-(c - I*d)/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(1/8*((-16*I*a^3*f*e^(2*I*f*x
+ 2*I*e) - 16*I*a^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(-(c - I
*d)/(a^6*f^2)) + 16*(c - I*d)*e^(2*I*f*x + 2*I*e) + 16*c)*e^(-2*I*f*x - 2*I*e)) + 8*(24*a^3*c^2 + 48*I*a^3*c*d
- 24*a^3*d^2)*f*sqrt((4*I*c^6 - 16*c^5*d - 20*I*c^4*d^2 - 15*I*c^2*d^4 + 4*c*d^5 - 4*I*d^6)/((-256*I*a^6*c^5
+ 1280*a^6*c^4*d + 2560*I*a^6*c^3*d^2 - 2560*a^6*c^2*d^3 - 1280*I*a^6*c*d^4 + 256*a^6*d^5)*f^2))*e^(6*I*f*x +
6*I*e)*log(-(2*c^4 + 6*I*c^3*d - 5*c^2*d^2 + I*c*d^3 - 2*d^4 - ((16*I*a^3*c^3 - 48*a^3*c^2*d - 48*I*a^3*c*d^2
+ 16*a^3*d^3)*f*e^(2*I*f*x + 2*I*e) + (16*I*a^3*c^3 - 48*a^3*c^2*d - 48*I*a^3*c*d^2 + 16*a^3*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^6 - 16*c^5*d - 20*I*c^4*d^2 - 15*
I*c^2*d^4 + 4*c*d^5 - 4*I*d^6)/((-256*I*a^6*c^5 + 1280*a^6*c^4*d + 2560*I*a^6*c^3*d^2 - 2560*a^6*c^2*d^3 - 128
0*I*a^6*c*d^4 + 256*a^6*d^5)*f^2)) + (2*c^4 + 4*I*c^3*d - c^2*d^2 + 2*I*c*d^3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*
x - 2*I*e)/((16*I*a^3*c^3 - 48*a^3*c^2*d - 48*I*a^3*c*d^2 + 16*a^3*d^3)*f)) - 8*(24*a^3*c^2 + 48*I*a^3*c*d - 2
4*a^3*d^2)*f*sqrt((4*I*c^6 - 16*c^5*d - 20*I*c^4*d^2 - 15*I*c^2*d^4 + 4*c*d^5 - 4*I*d^6)/((-256*I*a^6*c^5 + 12
80*a^6*c^4*d + 2560*I*a^6*c^3*d^2 - 2560*a^6*c^2*d^3 - 1280*I*a^6*c*d^4 + 256*a^6*d^5)*f^2))*e^(6*I*f*x + 6*I*
e)*log(-(2*c^4 + 6*I*c^3*d - 5*c^2*d^2 + I*c*d^3 - 2*d^4 - ((-16*I*a^3*c^3 + 48*a^3*c^2*d + 48*I*a^3*c*d^2 - 1
6*a^3*d^3)*f*e^(2*I*f*x + 2*I*e) + (-16*I*a^3*c^3 + 48*a^3*c^2*d + 48*I*a^3*c*d^2 - 16*a^3*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^6 - 16*c^5*d - 20*I*c^4*d^2 - 15*I*
c^2*d^4 + 4*c*d^5 - 4*I*d^6)/((-256*I*a^6*c^5 + 1280*a^6*c^4*d + 2560*I*a^6*c^3*d^2 - 2560*a^6*c^2*d^3 - 1280*
I*a^6*c*d^4 + 256*a^6*d^5)*f^2)) + (2*c^4 + 4*I*c^3*d - c^2*d^2 + 2*I*c*d^3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x
- 2*I*e)/((16*I*a^3*c^3 - 48*a^3*c^2*d - 48*I*a^3*c*d^2 + 16*a^3*d^3)*f)) - 8*(-2*I*c^2 + 4*c*d + 2*I*d^2 + (-
11*I*c^2 + 18*c*d + 4*I*d^2)*e^(6*I*f*x + 6*I*e) + (-18*I*c^2 + 31*c*d + 10*I*d^2)*e^(4*I*f*x + 4*I*e) + (-9*I
*c^2 + 17*c*d + 8*I*d^2)*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)))*e^(-6*I*f*x - 6*I*e)/((96*a^3*c^2 + 192*I*a^3*c*d - 96*a^3*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((c+d*tan(f*x+e))**(1/2)/(a+I*a*tan(f*x+e))**3,x)

[Out]

Exception raised: AttributeError

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Giac [B]  time = 1.66205, size = 868, normalized size = 3.1 \begin{align*} \frac{1}{2} \, d^{4}{\left (\frac{8 \,{\left (2 \, c^{3} + 4 i \, c^{2} d - c d^{2} + 2 i \, d^{3}\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 (16 i \, a^{3} c^{2} d^{4} f - 32 \, a^{3} c d^{5} f - 16 i \, a^{3} 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{4 \,{\left (6 \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}} c^{2} - 12 \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} c^{3} + 6 \, \sqrt{d \tan \left (f x + e\right ) + c} c^{4} + 9 i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}} c d - 36 i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} c^{2} d + 27 i \, \sqrt{d \tan \left (f x + e\right ) + c} c^{3} d + 28 \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} c d^{2} - 48 \, \sqrt{d \tan \left (f x + e\right ) + c} c^{2} d^{2} + 4 i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} d^{3} - 39 i \, \sqrt{d \tan \left (f x + e\right ) + c} c d^{3} + 12 \, \sqrt{d \tan \left (f x + e\right ) + c} d^{4}\right )}}{{\left (96 \, a^{3} c^{2} d^{3} f + 192 i \, a^{3} c d^{4} f - 96 \, a^{3} d^{5} f\right )}{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{3}} - \frac{{\left (-i \, c - d\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 )}{a^{3} \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}} d^{4} 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))^3,x, algorithm="giac")

[Out]

1/2*d^4*(8*(2*c^3 + 4*I*c^2*d - c*d^2 + 2*I*d^3)*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)*s
qrt(-8*c + 8*sqrt(c^2 + d^2))))/((16*I*a^3*c^2*d^4*f - 32*a^3*c*d^5*f - 16*I*a^3*d^6*f)*sqrt(-8*c + 8*sqrt(c^2
+ d^2))*(I*d/(c - sqrt(c^2 + d^2)) + 1)) + 4*(6*(d*tan(f*x + e) + c)^(5/2)*c^2 - 12*(d*tan(f*x + e) + c)^(3/2
)*c^3 + 6*sqrt(d*tan(f*x + e) + c)*c^4 + 9*I*(d*tan(f*x + e) + c)^(5/2)*c*d - 36*I*(d*tan(f*x + e) + c)^(3/2)*
c^2*d + 27*I*sqrt(d*tan(f*x + e) + c)*c^3*d + 28*(d*tan(f*x + e) + c)^(3/2)*c*d^2 - 48*sqrt(d*tan(f*x + e) + c
)*c^2*d^2 + 4*I*(d*tan(f*x + e) + c)^(3/2)*d^3 - 39*I*sqrt(d*tan(f*x + e) + c)*c*d^3 + 12*sqrt(d*tan(f*x + e)
+ c)*d^4)/((96*a^3*c^2*d^3*f + 192*I*a^3*c*d^4*f - 96*a^3*d^5*f)*(d*tan(f*x + e) - I*d)^3) - (-I*c - d)*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))))/(a^3*sqrt(-8*c + 8*sqrt
(c^2 + d^2))*d^4*f*(-I*d/(c - sqrt(c^2 + d^2)) + 1)))