3.1118 \(\int \frac{(c+d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=285 \[ \frac{\left (2 i c^2+5 c d-4 i d^2\right ) \sqrt{c+d \tan (e+f x)}}{16 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac{\left (4 c^2 d+2 i c^3-i c d^2+2 d^3\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{16 a^3 f \sqrt{c+i d}}-\frac{i (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{8 a^3 f}+\frac{(-d+i c) (c+d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac{(c+i d) (2 d+i c) \sqrt{c+d \tan (e+f x)}}{8 a f (a+i a \tan (e+f x))^2} \]

[Out]

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

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Rubi [A]  time = 1.02819, antiderivative size = 285, 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 = {3558, 3595, 3596, 3539, 3537, 63, 208} \[ \frac{\left (2 i c^2+5 c d-4 i d^2\right ) \sqrt{c+d \tan (e+f x)}}{16 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac{\left (4 c^2 d+2 i c^3-i c d^2+2 d^3\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{16 a^3 f \sqrt{c+i d}}-\frac{i (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{8 a^3 f}+\frac{(-d+i c) (c+d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac{(c+i d) (2 d+i c) \sqrt{c+d \tan (e+f x)}}{8 a f (a+i a \tan (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*Tan[e + f*x])^(5/2)/(a + I*a*Tan[e + f*x])^3,x]

[Out]

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

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 3595

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*b - a*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(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 - 1)*Simp[A*(a*c*m + b*d*n
) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a*A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A,
B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && GtQ[n, 0]

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))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx &=\frac{(i c-d) (c+d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}-\frac{\int \frac{\sqrt{c+d \tan (e+f x)} \left (-\frac{3}{2} a \left (2 c^2-3 i c d+d^2\right )-\frac{3}{2} a (c-3 i d) d \tan (e+f x)\right )}{(a+i a \tan (e+f x))^2} \, dx}{6 a^2}\\ &=\frac{(c+i d) (i c+2 d) \sqrt{c+d \tan (e+f x)}}{8 a f (a+i a \tan (e+f x))^2}+\frac{(i c-d) (c+d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac{\int \frac{\frac{3}{2} a^2 \left (4 c^3-9 i c^2 d-5 c d^2-2 i d^3\right )+\frac{3}{2} a^2 d \left (3 c^2-7 i c d-6 d^2\right ) \tan (e+f x)}{(a+i a \tan (e+f x)) \sqrt{c+d \tan (e+f x)}} \, dx}{24 a^4}\\ &=\frac{(c+i d) (i c+2 d) \sqrt{c+d \tan (e+f x)}}{8 a f (a+i a \tan (e+f x))^2}+\frac{\left (2 i c^2+5 c d-4 i d^2\right ) \sqrt{c+d \tan (e+f x)}}{16 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac{(i c-d) (c+d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}-\frac{\int \frac{-\frac{3}{2} a^3 c (i c-d) \left (4 c^2-10 i c d-7 d^2\right )-\frac{3}{2} a^3 (i c-d) d \left (2 c^2-5 i c d-4 d^2\right ) \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{48 a^6 (i c-d)}\\ &=\frac{(c+i d) (i c+2 d) \sqrt{c+d \tan (e+f x)}}{8 a f (a+i a \tan (e+f x))^2}+\frac{\left (2 i c^2+5 c d-4 i d^2\right ) \sqrt{c+d \tan (e+f x)}}{16 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac{(i c-d) (c+d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}+\frac{(c-i d)^3 \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}\\ &=\frac{(c+i d) (i c+2 d) \sqrt{c+d \tan (e+f x)}}{8 a f (a+i a \tan (e+f x))^2}+\frac{\left (2 i c^2+5 c d-4 i d^2\right ) \sqrt{c+d \tan (e+f x)}}{16 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac{(i c-d) (c+d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}-\frac{(i c+d)^3 \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 (2 i c^3+4 c^2 d-i c d^2+2 d^3\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 f}\\ &=\frac{(c+i d) (i c+2 d) \sqrt{c+d \tan (e+f x)}}{8 a f (a+i a \tan (e+f x))^2}+\frac{\left (2 i c^2+5 c d-4 i d^2\right ) \sqrt{c+d \tan (e+f x)}}{16 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac{(i c-d) (c+d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}-\frac{(c-i d)^3 \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 d f}\\ &=-\frac{i (c-i d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{8 a^3 f}+\frac{\left (2 i c^3+4 c^2 d-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 \sqrt{c+i d} f}+\frac{(c+i d) (i c+2 d) \sqrt{c+d \tan (e+f x)}}{8 a f (a+i a \tan (e+f x))^2}+\frac{\left (2 i c^2+5 c d-4 i d^2\right ) \sqrt{c+d \tan (e+f x)}}{16 f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac{(i c-d) (c+d \tan (e+f x))^{3/2}}{6 f (a+i a \tan (e+f x))^3}\\ \end{align*}

Mathematica [A]  time = 2.58806, size = 324, normalized size = 1.14 \[ \frac{\sec ^3(e+f x) (\cos (f x)+i \sin (f x))^3 \left (\frac{2 (\cos (3 e)+i \sin (3 e)) \left (-i \sqrt{-c+i d} \left (-4 i c^2 d+2 c^3-c d^2-2 i d^3\right ) \tan ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{-c-i d}}\right )-2 \sqrt{-c-i d} (d+i c)^3 \tan ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{-c+i d}}\right )\right )}{\sqrt{-c-i d} \sqrt{-c+i d}}+\frac{2}{3} \cos (e+f x) (\sin (3 f x)+i \cos (3 f x)) \sqrt{c+d \tan (e+f x)} \left (\left (9 i c^2+22 c d-2 i d^2\right ) \sin (2 (e+f x))+\left (13 c^2-14 i c d-6 d^2\right ) \cos (2 (e+f x))+7 c^2+i c d+6 d^2\right )\right )}{32 f (a+i a \tan (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

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

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Maple [B]  time = 0.072, size = 1861, normalized size = 6.5 \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))^(5/2)/(a+I*a*tan(f*x+e))^3,x)

[Out]

1/8/f/a^3*d/(-I*d+d*tan(f*x+e))^3*c^7/(-I*d^3-3*c*d^2+3*I*c^2*d+c^3)*(c+d*tan(f*x+e))^(1/2)-5/16/f/a^3*d^3/(-I
*d+d*tan(f*x+e))^3*c^5/(-I*d^3-3*c*d^2+3*I*c^2*d+c^3)*(c+d*tan(f*x+e))^(1/2)-5/4/f/a^3*d^5/(-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))^(1/2)+3/16/f/a^3*d^7/(-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))^(1/2)-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^6+1/8*I/f/a^3*d^6/(-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))^(5/2)-1/8*I/f/a^3*(I*d-c)^(5/2)*arctan((c+d*tan(f*x+e))^(1/2)/(I*d-c)^(1/2))-5/16*I/f/a^3*d^4/
(-I*d^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^2-5/3*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^3-1/2*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^5+1/6*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))^(3/2)*c+7/16*I/f/a^3*d^2/(-I*d+d*tan(f*x+e))^3*c^6/(-I*d^3-3*c*d^2
+3*I*c^2*d+c^3)*(c+d*tan(f*x+e))^(1/2)+5/8*I/f/a^3*d^4/(-I*d+d*tan(f*x+e))^3*c^4/(-I*d^3-3*c*d^2+3*I*c^2*d+c^3
)*(c+d*tan(f*x+e))^(1/2)-13/16*I/f/a^3*d^6/(-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))^(1/2)-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*tan(f*x+e))^(1/2)/(-I*
d-c)^(1/2))*c^4+1/16*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
^4+1/16*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^2+1/12/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))^(3/2)+1/8/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^5+5/16/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^3+7/16/f/a^3*d^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))*c+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^5+5/16/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))^(5/2)*c^3+7/16/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))^(5/2)*c-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^6-5/12/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^4+5/4/
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)*c^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))^(5/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 6.21167, size = 3124, normalized size = 10.96 \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))^(5/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

-1/192*(6*a^3*f*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 + 5*c*d^4 - I*d^5)/(a^6*f^2))*e^(6*I*f*x +
6*I*e)*log(1/8*(16*c^3 - 32*I*c^2*d - 16*c*d^2 - (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^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3
+ 5*c*d^4 - I*d^5)/(a^6*f^2)) + 8*(2*c^3 - 6*I*c^2*d - 6*c*d^2 + 2*I*d^3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2
*I*e)/(c^2 - 2*I*c*d - d^2)) - 6*a^3*f*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 + 5*c*d^4 - I*d^5)/(
a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(1/8*(16*c^3 - 32*I*c^2*d - 16*c*d^2 - (-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^5 - 5*I*c^4*d - 10
*c^3*d^2 + 10*I*c^2*d^3 + 5*c*d^4 - I*d^5)/(a^6*f^2)) + 8*(2*c^3 - 6*I*c^2*d - 6*c*d^2 + 2*I*d^3)*e^(2*I*f*x +
2*I*e))*e^(-2*I*f*x - 2*I*e)/(c^2 - 2*I*c*d - d^2)) - 3*a^3*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)/((I*a^6*c - a^6*d)*f^2))*e^(6*I*f*x + 6*I*e)*log(-1/16*(2*c^4 - 2*I*c^3*d + 3*c^
2*d^2 - 3*I*c*d^3 + 2*d^4 - ((I*a^3*c - a^3*d)*f*e^(2*I*f*x + 2*I*e) + (I*a^3*c - a^3*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(-(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)/((I*a^6*c - a^6*d)*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)/((I*a^3*c - a^3*d)*f)) + 3*a^3*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)/((I*a^6*c - a^6*d)*f^2))*e^(6*I*f*x + 6*I*e)*log(-1/16*(2*c^4 - 2*I*c^3*d + 3*c^2*d^2
- 3*I*c*d^3 + 2*d^4 - ((-I*a^3*c + a^3*d)*f*e^(2*I*f*x + 2*I*e) + (-I*a^3*c + a^3*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(-(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)/((I*a^6*c - a^6*d)*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)/((I*a^3*c - a^3*d)*f)) - 2*(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 + 17*c*d + 2*I*d^2)*e^(4*I*f*x + 4*I*e) + (9*I*c^2 - 5*c*d + 4*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)/(a^3*
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))**(5/2)/(a+I*a*tan(f*x+e))**3,x)

[Out]

Exception raised: AttributeError

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Giac [B]  time = 1.71169, size = 819, normalized size = 2.87 \begin{align*} -\frac{1}{48} \, d^{4}{\left (\frac{12 \,{\left (2 i \, c^{3} + 4 \, c^{2} d - i \, c d^{2} + 2 \, 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 )}{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 )}} + \frac{24 \,{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + 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 )}{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 )}} - \frac{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} - 15 i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}} c d + 12 i \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} c^{2} d + 3 i \, \sqrt{d \tan \left (f x + e\right ) + c} c^{3} d - 12 \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{5}{2}} d^{2} - 20 \,{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}} c d^{2} + 12 \, \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} + 9 i \, \sqrt{d \tan \left (f x + e\right ) + c} c d^{3}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{3} a^{3} d^{3} f}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="giac")

[Out]

-1/48*d^4*(12*(2*I*c^3 + 4*c^2*d - I*c*d^2 + 2*d^3)*arctan(4*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqr
t(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))
+ 24*(-I*c^3 - 3*c^2*d + 3*I*c*d^2 + 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)*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)) - (6*(d*t
an(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 - 15*I*(d*tan(
f*x + e) + c)^(5/2)*c*d + 12*I*(d*tan(f*x + e) + c)^(3/2)*c^2*d + 3*I*sqrt(d*tan(f*x + e) + c)*c^3*d - 12*(d*t
an(f*x + e) + c)^(5/2)*d^2 - 20*(d*tan(f*x + e) + c)^(3/2)*c*d^2 + 12*sqrt(d*tan(f*x + e) + c)*c^2*d^2 + 4*I*(
d*tan(f*x + e) + c)^(3/2)*d^3 + 9*I*sqrt(d*tan(f*x + e) + c)*c*d^3)/((d*tan(f*x + e) - I*d)^3*a^3*d^3*f))