3.1112 \(\int \frac{(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^3} \, dx\)
Optimal. Leaf size=274 \[ -\frac{\left (2 c^2-i c d+2 d^2\right ) \sqrt{c+d \tan (e+f x)}}{16 f (-d+i c) \left (a^3+i a^3 \tan (e+f x)\right )}+\frac{i c \left (2 c^2+3 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{16 a^3 f (c+i d)^{3/2}}-\frac{i (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{8 a^3 f}+\frac{(4 d+3 i c) \sqrt{c+d \tan (e+f x)}}{24 a f (a+i a \tan (e+f x))^2}+\frac{(-d+i c) \sqrt{c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3} \]
[Out]
((-I/8)*(c - I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(a^3*f) + ((I/16)*c*(2*c^2 + 3*d^2)*A
rcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/(a^3*(c + I*d)^(3/2)*f) + ((I*c - d)*Sqrt[c + d*Tan[e + f*x]])
/(6*f*(a + I*a*Tan[e + f*x])^3) + (((3*I)*c + 4*d)*Sqrt[c + d*Tan[e + f*x]])/(24*a*f*(a + I*a*Tan[e + f*x])^2)
- ((2*c^2 - I*c*d + 2*d^2)*Sqrt[c + d*Tan[e + f*x]])/(16*(I*c - d)*f*(a^3 + I*a^3*Tan[e + f*x]))
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Rubi [A] time = 1.05781, antiderivative size = 274, 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 =
{3558, 3596, 3539, 3537, 63, 208} \[ -\frac{\left (2 c^2-i c d+2 d^2\right ) \sqrt{c+d \tan (e+f x)}}{16 f (-d+i c) \left (a^3+i a^3 \tan (e+f x)\right )}+\frac{i c \left (2 c^2+3 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{16 a^3 f (c+i d)^{3/2}}-\frac{i (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{8 a^3 f}+\frac{(4 d+3 i c) \sqrt{c+d \tan (e+f x)}}{24 a f (a+i a \tan (e+f x))^2}+\frac{(-d+i c) \sqrt{c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3} \]
Antiderivative was successfully verified.
[In]
Int[(c + d*Tan[e + f*x])^(3/2)/(a + I*a*Tan[e + f*x])^3,x]
[Out]
((-I/8)*(c - I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(a^3*f) + ((I/16)*c*(2*c^2 + 3*d^2)*A
rcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/(a^3*(c + I*d)^(3/2)*f) + ((I*c - d)*Sqrt[c + d*Tan[e + f*x]])
/(6*f*(a + I*a*Tan[e + f*x])^3) + (((3*I)*c + 4*d)*Sqrt[c + d*Tan[e + f*x]])/(24*a*f*(a + I*a*Tan[e + f*x])^2)
- ((2*c^2 - I*c*d + 2*d^2)*Sqrt[c + d*Tan[e + f*x]])/(16*(I*c - d)*f*(a^3 + I*a^3*Tan[e + f*x]))
Rule 3558
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[((b*c - a*d)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1))/(2*a*f*m), x] + Dist[1/(2*a^2*m), Int[(a
+ b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1))
- d*(b*d*(m - n + 1) - a*c*(m + n - 1))*Tan[e + f*x], 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] && GtQ[n, 1] && (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{(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^3} \, dx &=\frac{(i c-d) \sqrt{c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}-\frac{\int \frac{-\frac{1}{2} a \left (6 c^2-7 i c d+d^2\right )-\frac{1}{2} a (5 c-7 i d) d \tan (e+f x)}{(a+i a \tan (e+f x))^2 \sqrt{c+d \tan (e+f x)}} \, dx}{6 a^2}\\ &=\frac{(i c-d) \sqrt{c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac{(3 i c+4 d) \sqrt{c+d \tan (e+f x)}}{24 a f (a+i a \tan (e+f x))^2}+\frac{\int \frac{\frac{3}{2} a^2 c (c+i d) (4 i c+5 d)+\frac{3}{2} a^2 (i c-d) (3 c-4 i d) d \tan (e+f x)}{(a+i a \tan (e+f x)) \sqrt{c+d \tan (e+f x)}} \, dx}{24 a^4 (i c-d)}\\ &=\frac{(i c-d) \sqrt{c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac{(3 i c+4 d) \sqrt{c+d \tan (e+f x)}}{24 a f (a+i a \tan (e+f x))^2}-\frac{\left (2 c^2-i c d+2 d^2\right ) \sqrt{c+d \tan (e+f x)}}{16 (i c-d) f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac{\int \frac{\frac{3}{2} a^3 (c+i d) \left (4 c^3-2 i c^2 d+5 c d^2-2 i d^3\right )+\frac{3}{2} a^3 (c+i d) d \left (2 c^2-i c d+2 d^2\right ) \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{48 a^6 (c+i d)^2}\\ &=\frac{(i c-d) \sqrt{c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac{(3 i c+4 d) \sqrt{c+d \tan (e+f x)}}{24 a f (a+i a \tan (e+f x))^2}-\frac{\left (2 c^2-i c d+2 d^2\right ) \sqrt{c+d \tan (e+f x)}}{16 (i c-d) f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac{(c-i d)^2 \int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{16 a^3}+\frac{\left (c \left (2 c^2+3 d^2\right )\right ) \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{32 a^3 (c+i d)}\\ &=\frac{(i c-d) \sqrt{c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac{(3 i c+4 d) \sqrt{c+d \tan (e+f x)}}{24 a f (a+i a \tan (e+f x))^2}-\frac{\left (2 c^2-i c d+2 d^2\right ) \sqrt{c+d \tan (e+f x)}}{16 (i c-d) f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac{\left (i (c-i 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^3 f}+\frac{\left (c \left (2 c^2+3 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 )}{32 a^3 (i c-d) f}\\ &=\frac{(i c-d) \sqrt{c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac{(3 i c+4 d) \sqrt{c+d \tan (e+f x)}}{24 a f (a+i a \tan (e+f x))^2}-\frac{\left (2 c^2-i c d+2 d^2\right ) \sqrt{c+d \tan (e+f x)}}{16 (i c-d) f \left (a^3+i a^3 \tan (e+f x)\right )}-\frac{(c-i d)^2 \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 (c \left (2 c^2+3 d^2\right )\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) d f}\\ &=-\frac{i (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{8 a^3 f}+\frac{i c \left (2 c^2+3 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{16 a^3 (c+i d)^{3/2} f}+\frac{(i c-d) \sqrt{c+d \tan (e+f x)}}{6 f (a+i a \tan (e+f x))^3}+\frac{(3 i c+4 d) \sqrt{c+d \tan (e+f x)}}{24 a f (a+i a \tan (e+f x))^2}-\frac{\left (2 c^2-i c d+2 d^2\right ) \sqrt{c+d \tan (e+f x)}}{16 (i c-d) f \left (a^3+i a^3 \tan (e+f x)\right )}\\ \end{align*}
Mathematica [A] time = 2.49677, size = 311, normalized size = 1.14 \[ \frac{\sec ^3(e+f x) (\cos (f x)+i \sin (f x))^3 \left (\frac{2 i (\cos (3 e)+i \sin (3 e)) \left (c \sqrt{-c+i d} \left (2 c^2+3 d^2\right ) \tan ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{-c-i d}}\right )+2 (-c-i d)^{3/2} (c-i d)^2 \tan ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{-c+i d}}\right )\right )}{(-c-i d)^{3/2} \sqrt{-c+i d}}+\frac{2 \cos (e+f x) (\sin (3 f x)+i \cos (3 f x)) \sqrt{c+d \tan (e+f x)} \left (\left (9 i c^2+4 c d+10 i d^2\right ) \sin (2 (e+f x))+\left (13 c^2+4 i c d+6 d^2\right ) \cos (2 (e+f x))+7 c (c+i d)\right )}{3 (c+i d)}\right )}{32 f (a+i a \tan (e+f x))^3} \]
Antiderivative was successfully verified.
[In]
Integrate[(c + d*Tan[e + f*x])^(3/2)/(a + I*a*Tan[e + f*x])^3,x]
[Out]
(Sec[e + f*x]^3*(Cos[f*x] + I*Sin[f*x])^3*(((2*I)*(c*Sqrt[-c + I*d]*(2*c^2 + 3*d^2)*ArcTan[Sqrt[c + d*Tan[e +
f*x]]/Sqrt[-c - I*d]] + 2*(-c - I*d)^(3/2)*(c - I*d)^2*ArcTan[Sqrt[c + d*Tan[e + f*x]]/Sqrt[-c + I*d]])*(Cos[3
*e] + I*Sin[3*e]))/((-c - I*d)^(3/2)*Sqrt[-c + I*d]) + (2*Cos[e + f*x]*(I*Cos[3*f*x] + Sin[3*f*x])*(7*c*(c + I
*d) + (13*c^2 + (4*I)*c*d + 6*d^2)*Cos[2*(e + f*x)] + ((9*I)*c^2 + 4*c*d + (10*I)*d^2)*Sin[2*(e + f*x)])*Sqrt[
c + d*Tan[e + f*x]])/(3*(c + I*d))))/(32*f*(a + I*a*Tan[e + f*x])^3)
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Maple [B] time = 0.068, size = 1266, normalized size = 4.6 \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))^(3/2)/(a+I*a*tan(f*x+e))^3,x)
[Out]
-3/8*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^3+3/16*I/f/a^3*
d^4*c/(-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))+1/8*I/f/a^3*
(I*d-c)^(3/2)*arctan((c+d*tan(f*x+e))^(1/2)/(I*d-c)^(1/2))+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))^(5/2)*c^4+1/8/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))^(5/2)*c^2-1/8/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))^(
5/2)-1/4/f/a^3*d/(-I*d+d*tan(f*x+e))^3*(c+d*tan(f*x+e))^(3/2)*c^2-5/12/f/a^3*d^3/(-I*d+d*tan(f*x+e))^3*(c+d*ta
n(f*x+e))^(3/2)+9/16*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
^5-7/16*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)*c-1/8*I/f/a^3*
c^5/(-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))+1/8/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^6-7/8/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^4-3/8/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^2+1/8/f/a^3*d^7/(-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)-1/16*I/f/a^3*d^2*c^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))+5/16*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))^(5/2)*c+3/16*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))^(5/2)*c^3
+1/4/f/a^3*d*c^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))+3
/8/f/a^3*d^3*c^2/(-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))
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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))^(3/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="maxima")
[Out]
Exception raised: RuntimeError
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Fricas [B] time = 5.8578, size = 3078, normalized size = 11.23 \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))^(3/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="fricas")
[Out]
1/96*(3*(-I*a^3*c + a^3*d)*f*sqrt(-(c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log((2*I
*c^2 + 2*c*d + 2*(a^3*f*e^(2*I*f*x + 2*I*e) + 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^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)/(a^6*f^2)) + (2*I*c^2 + 4*c*d - 2*I*d^2)*e^(2*I*f*x
+ 2*I*e))*e^(-2*I*f*x - 2*I*e)/(I*c + d)) + 3*(I*a^3*c - a^3*d)*f*sqrt(-(c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)/(
a^6*f^2))*e^(6*I*f*x + 6*I*e)*log((2*I*c^2 + 2*c*d - 2*(a^3*f*e^(2*I*f*x + 2*I*e) + 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^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)/(a^6*f^2)) +
(2*I*c^2 + 4*c*d - 2*I*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(I*c + d)) + 24*(-I*a^3*c + a^3*d)*f*sqr
t((-4*I*c^6 - 12*I*c^4*d^2 - 9*I*c^2*d^4)/((256*I*a^6*c^3 - 768*a^6*c^2*d - 768*I*a^6*c*d^2 + 256*a^6*d^3)*f^2
))*e^(6*I*f*x + 6*I*e)*log(-(-2*I*c^4 + 2*c^3*d - 3*I*c^2*d^2 + 3*c*d^3 + ((16*a^3*c^2 + 32*I*a^3*c*d - 16*a^3
*d^2)*f*e^(2*I*f*x + 2*I*e) + (16*a^3*c^2 + 32*I*a^3*c*d - 16*a^3*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^6 - 12*I*c^4*d^2 - 9*I*c^2*d^4)/((256*I*a^6*c^3 - 768*a^6*c
^2*d - 768*I*a^6*c*d^2 + 256*a^6*d^3)*f^2)) + (-2*I*c^4 - 3*I*c^2*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*
e)/((16*a^3*c^2 + 32*I*a^3*c*d - 16*a^3*d^2)*f)) + 24*(I*a^3*c - a^3*d)*f*sqrt((-4*I*c^6 - 12*I*c^4*d^2 - 9*I*
c^2*d^4)/((256*I*a^6*c^3 - 768*a^6*c^2*d - 768*I*a^6*c*d^2 + 256*a^6*d^3)*f^2))*e^(6*I*f*x + 6*I*e)*log(-(-2*I
*c^4 + 2*c^3*d - 3*I*c^2*d^2 + 3*c*d^3 - ((16*a^3*c^2 + 32*I*a^3*c*d - 16*a^3*d^2)*f*e^(2*I*f*x + 2*I*e) + (16
*a^3*c^2 + 32*I*a^3*c*d - 16*a^3*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^6 - 12*I*c^4*d^2 - 9*I*c^2*d^4)/((256*I*a^6*c^3 - 768*a^6*c^2*d - 768*I*a^6*c*d^2 + 256*a^6*
d^3)*f^2)) + (-2*I*c^4 - 3*I*c^2*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/((16*a^3*c^2 + 32*I*a^3*c*d -
16*a^3*d^2)*f)) - (2*c^2 + 4*I*c*d - 2*d^2 + (11*c^2 + 8*d^2)*e^(6*I*f*x + 6*I*e) + (18*c^2 + 7*I*c*d + 8*d^2)
*e^(4*I*f*x + 4*I*e) + (9*c^2 + 11*I*c*d - 2*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)/((I*a^3*c - a^3*d)*f)
________________________________________________________________________________________
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))**(3/2)/(a+I*a*tan(f*x+e))**3,x)
[Out]
Exception raised: AttributeError
________________________________________________________________________________________
Giac [B] time = 1.6268, size = 855, normalized size = 3.12 \begin{align*} -\frac{1}{2} \, d^{4}{\left (\frac{8 \,{\left (2 \, c^{3} + 3 \, c 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 (-16 i \, a^{3} c d^{4} f + 16 \, a^{3} 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{4 \,{\left (-6 i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}} c^{2} + 12 i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} c^{3} - 6 i \, \sqrt{d \tan \left (f x + e\right ) + c} c^{4} - 3 \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}} c d - 12 \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} c^{2} d + 15 \, \sqrt{d \tan \left (f x + e\right ) + c} c^{3} d - 6 i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}} d^{2} + 20 i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} c d^{2} + 6 i \, \sqrt{d \tan \left (f x + e\right ) + c} c^{2} d^{2} - 20 \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} d^{3} + 9 \, \sqrt{d \tan \left (f x + e\right ) + c} c d^{3} + 6 i \, \sqrt{d \tan \left (f x + e\right ) + c} d^{4}\right )}}{{\left (-96 i \, a^{3} c d^{3} f + 96 \, a^{3} d^{4} f\right )}{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{3}} + \frac{{\left (-i \, c^{2} - 2 \, c d + 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 )}{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))^(3/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="giac")
[Out]
-1/2*d^4*(8*(2*c^3 + 3*c*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))))/((-16*I*a^3*c*d^4*f + 16*a^3*d^5*f)*sqrt(-8*c + 8*sqrt(c^2 + d^2))*(I*d/(c - sqrt(c^2 + d^2)) + 1
)) - 4*(-6*I*(d*tan(f*x + e) + c)^(5/2)*c^2 + 12*I*(d*tan(f*x + e) + c)^(3/2)*c^3 - 6*I*sqrt(d*tan(f*x + e) +
c)*c^4 - 3*(d*tan(f*x + e) + c)^(5/2)*c*d - 12*(d*tan(f*x + e) + c)^(3/2)*c^2*d + 15*sqrt(d*tan(f*x + e) + c)*
c^3*d - 6*I*(d*tan(f*x + e) + c)^(5/2)*d^2 + 20*I*(d*tan(f*x + e) + c)^(3/2)*c*d^2 + 6*I*sqrt(d*tan(f*x + e) +
c)*c^2*d^2 - 20*(d*tan(f*x + e) + c)^(3/2)*d^3 + 9*sqrt(d*tan(f*x + e) + c)*c*d^3 + 6*I*sqrt(d*tan(f*x + e) +
c)*d^4)/((-96*I*a^3*c*d^3*f + 96*a^3*d^4*f)*(d*tan(f*x + e) - I*d)^3) + (-I*c^2 - 2*c*d + I*d^2)*arctan(4*(sq
rt(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*sqr
t(-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)))