3.12.34 \(\int \frac {1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))^{5/2}} \, dx\) [1134]

Optimal. Leaf size=267 \[ -\frac {i \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{2 a (c-i d)^{5/2} f}+\frac {(i c-6 d) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{2 a (c+i d)^{7/2} f}+\frac {d (3 i c+7 d)}{6 a (i c-d) \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}-\frac {1}{2 (i c-d) f (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}+\frac {d \left (c^2-14 i c d-5 d^2\right )}{2 a (c-i d)^2 (c+i d)^3 f \sqrt {c+d \tan (e+f x)}} \]

[Out]

-1/2*I*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/a/(c-I*d)^(5/2)/f+1/2*(I*c-6*d)*arctanh((c+d*tan(f*x+e))^
(1/2)/(c+I*d)^(1/2))/a/(c+I*d)^(7/2)/f+1/2*d*(c^2-14*I*c*d-5*d^2)/a/(c-I*d)^2/(c+I*d)^3/f/(c+d*tan(f*x+e))^(1/
2)+1/6*d*(3*I*c+7*d)/a/(I*c-d)/(c^2+d^2)/f/(c+d*tan(f*x+e))^(3/2)-1/2/(I*c-d)/f/(a+I*a*tan(f*x+e))/(c+d*tan(f*
x+e))^(3/2)

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Rubi [A]
time = 0.45, antiderivative size = 267, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3633, 3610, 3620, 3618, 65, 214} \begin {gather*} \frac {d (7 d+3 i c)}{6 a f (-d+i c) \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {d \left (c^2-14 i c d-5 d^2\right )}{2 a f (c-i d)^2 (c+i d)^3 \sqrt {c+d \tan (e+f x)}}-\frac {1}{2 f (-d+i c) (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}-\frac {i \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{2 a f (c-i d)^{5/2}}+\frac {(-6 d+i c) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{2 a f (c+i d)^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])*(c + d*Tan[e + f*x])^(5/2)),x]

[Out]

((-1/2*I)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(a*(c - I*d)^(5/2)*f) + ((I*c - 6*d)*ArcTanh[Sqrt[c
+ d*Tan[e + f*x]]/Sqrt[c + I*d]])/(2*a*(c + I*d)^(7/2)*f) + (d*((3*I)*c + 7*d))/(6*a*(I*c - d)*(c^2 + d^2)*f*
(c + d*Tan[e + f*x])^(3/2)) - 1/(2*(I*c - d)*f*(a + I*a*Tan[e + f*x])*(c + d*Tan[e + f*x])^(3/2)) + (d*(c^2 -
(14*I)*c*d - 5*d^2))/(2*a*(c - I*d)^2*(c + I*d)^3*f*Sqrt[c + d*Tan[e + f*x]])

Rule 65

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 3610

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 3618

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c*(
d/f), Subst[Int[(a + (b/d)*x)^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 3620

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 3633

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-a
)*((c + d*Tan[e + f*x])^(n + 1)/(2*f*(b*c - a*d)*(a + b*Tan[e + f*x]))), x] + Dist[1/(2*a*(b*c - a*d)), Int[(c
+ d*Tan[e + f*x])^n*Simp[b*c + a*d*(n - 1) - b*d*n*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] &&  !GtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {1}{(a+i a \tan (e+f x)) (c+d \tan (e+f x))^{5/2}} \, dx &=-\frac {1}{2 (i c-d) f (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}+\frac {\int \frac {\frac {1}{2} a (2 i c-7 d)+\frac {5}{2} i a d \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx}{2 a^2 (i c-d)}\\ &=\frac {(3 c-7 i d) d}{6 a (c-i d) (c+i d)^2 f (c+d \tan (e+f x))^{3/2}}-\frac {1}{2 (i c-d) f (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}+\frac {\int \frac {-\frac {1}{2} a \left (7 c d-i \left (2 c^2+5 d^2\right )\right )+\frac {1}{2} a d (3 i c+7 d) \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}} \, dx}{2 a^2 (i c-d) \left (c^2+d^2\right )}\\ &=\frac {(3 c-7 i d) d}{6 a (c-i d) (c+i d)^2 f (c+d \tan (e+f x))^{3/2}}-\frac {1}{2 (i c-d) f (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}+\frac {d \left (c^2-14 i c d-5 d^2\right )}{2 a (c-i d)^2 (c+i d)^3 f \sqrt {c+d \tan (e+f x)}}-\frac {\int \frac {\frac {1}{2} a \left (2 i c^3-7 c^2 d+8 i c d^2+7 d^3\right )+\frac {1}{2} a d \left (i c^2+14 c d-5 i d^2\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{2 a^2 (i c-d)^3 (c-i d)^2}\\ &=\frac {(3 c-7 i d) d}{6 a (c-i d) (c+i d)^2 f (c+d \tan (e+f x))^{3/2}}-\frac {1}{2 (i c-d) f (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}+\frac {d \left (c^2-14 i c d-5 d^2\right )}{2 a (c-i d)^2 (c+i d)^3 f \sqrt {c+d \tan (e+f x)}}+\frac {\int \frac {1+i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{4 a (c-i d)^2}+\frac {(c+6 i d) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{4 a (c+i d)^3}\\ &=\frac {(3 c-7 i d) d}{6 a (c-i d) (c+i d)^2 f (c+d \tan (e+f x))^{3/2}}-\frac {1}{2 (i c-d) f (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}+\frac {d \left (c^2-14 i c d-5 d^2\right )}{2 a (c-i d)^2 (c+i d)^3 f \sqrt {c+d \tan (e+f x)}}+\frac {i \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {c-i d x}} \, dx,x,i \tan (e+f x)\right )}{4 a (c-i d)^2 f}-\frac {(c+6 i d) \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{4 a (i c-d)^3 f}\\ &=\frac {(3 c-7 i d) d}{6 a (c-i d) (c+i d)^2 f (c+d \tan (e+f x))^{3/2}}-\frac {1}{2 (i c-d) f (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}+\frac {d \left (c^2-14 i c d-5 d^2\right )}{2 a (c-i d)^2 (c+i d)^3 f \sqrt {c+d \tan (e+f x)}}-\frac {\text {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 )}{2 a (c-i d)^2 d f}-\frac {(c+6 i d) \text {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 )}{2 a (c+i d)^3 d f}\\ &=-\frac {i \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{2 a (c-i d)^{5/2} f}+\frac {(i c-6 d) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{2 a (c+i d)^{7/2} f}+\frac {(3 c-7 i d) d}{6 a (c-i d) (c+i d)^2 f (c+d \tan (e+f x))^{3/2}}-\frac {1}{2 (i c-d) f (a+i a \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}+\frac {d \left (c^2-14 i c d-5 d^2\right )}{2 a (c-i d)^2 (c+i d)^3 f \sqrt {c+d \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 6.82, size = 371, normalized size = 1.39 \begin {gather*} \frac {\sec (e+f x) (\cos (f x)+i \sin (f x)) \left (-\frac {2 \left ((-c+i d)^{5/2} (-i c+6 d) \text {ArcTan}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {-c-i d}}\right )-i (-c-i d)^{7/2} \text {ArcTan}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {-c+i d}}\right )\right ) (\cos (e)+i \sin (e))}{(-c-i d)^{7/2} (-c+i d)^{5/2}}+\frac {(\cos (f x)-i \sin (f x)) \left (3 \left (3 i c^4+6 c^3 d-42 i c^2 d^2+2 c d^3-9 i d^4\right ) \cos (e+f x)+(i c+d) \left (\left (3 c^3-3 i c^2 d-43 c d^2+11 i d^3\right ) \cos (3 (e+f x))-8 d \left (-3 c^2+23 i c d+4 d^2\right ) \cos ^2(e+f x) \sin (e+f x)\right )\right ) \sqrt {c+d \tan (e+f x)}}{6 (c-i d)^2 (c+i d)^3 (c \cos (e+f x)+d \sin (e+f x))^2}\right )}{4 f (a+i a \tan (e+f x))} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])*(c + d*Tan[e + f*x])^(5/2)),x]

[Out]

(Sec[e + f*x]*(Cos[f*x] + I*Sin[f*x])*((-2*((-c + I*d)^(5/2)*((-I)*c + 6*d)*ArcTan[Sqrt[c + d*Tan[e + f*x]]/Sq
rt[-c - I*d]] - I*(-c - I*d)^(7/2)*ArcTan[Sqrt[c + d*Tan[e + f*x]]/Sqrt[-c + I*d]])*(Cos[e] + I*Sin[e]))/((-c
- I*d)^(7/2)*(-c + I*d)^(5/2)) + ((Cos[f*x] - I*Sin[f*x])*(3*((3*I)*c^4 + 6*c^3*d - (42*I)*c^2*d^2 + 2*c*d^3 -
(9*I)*d^4)*Cos[e + f*x] + (I*c + d)*((3*c^3 - (3*I)*c^2*d - 43*c*d^2 + (11*I)*d^3)*Cos[3*(e + f*x)] - 8*d*(-3
*c^2 + (23*I)*c*d + 4*d^2)*Cos[e + f*x]^2*Sin[e + f*x]))*Sqrt[c + d*Tan[e + f*x]])/(6*(c - I*d)^2*(c + I*d)^3*
(c*Cos[e + f*x] + d*Sin[e + f*x])^2)))/(4*f*(a + I*a*Tan[e + f*x]))

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Maple [A]
time = 0.47, size = 364, normalized size = 1.36

method	result	size
derivativedivides	\(\frac {2 d^{2} \left (\frac {-\frac {\left (c^{4}+2 c^{2} d^{2}+d^{4}\right ) d \sqrt {c +d \tan \left (f x +e \right )}}{\left (i d +c \right ) \left (-d \tan \left (f x +e \right )+i d \right )}-\frac {\left (i c^{5}+2 i c^{3} d^{2}+i c \,d^{4}-6 c^{4} d -12 c^{2} d^{3}-6 d^{5}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{\left (i d +c \right ) \sqrt {-i d -c}}}{4 d^{2} \left (i d +c \right )^{4} \left (i d -c \right )^{2}}+\frac {\left (i c^{4}-6 i c^{2} d^{2}+i d^{4}-4 c^{3} d +4 c \,d^{3}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{4 \left (i d -c \right )^{\frac {5}{2}} \left (i d +c \right )^{4} d^{2}}-\frac {i c^{3}+i c \,d^{2}-c^{2} d -d^{3}}{3 \left (i d -c \right )^{2} \left (i d +c \right )^{4} \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {3 i c^{2}+i d^{2}-2 c d}{\left (i d -c \right )^{2} \left (i d +c \right )^{4} \sqrt {c +d \tan \left (f x +e \right )}}\right )}{f a}\)	\(364\)
default	\(\frac {2 d^{2} \left (\frac {-\frac {\left (c^{4}+2 c^{2} d^{2}+d^{4}\right ) d \sqrt {c +d \tan \left (f x +e \right )}}{\left (i d +c \right ) \left (-d \tan \left (f x +e \right )+i d \right )}-\frac {\left (i c^{5}+2 i c^{3} d^{2}+i c \,d^{4}-6 c^{4} d -12 c^{2} d^{3}-6 d^{5}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{\left (i d +c \right ) \sqrt {-i d -c}}}{4 d^{2} \left (i d +c \right )^{4} \left (i d -c \right )^{2}}+\frac {\left (i c^{4}-6 i c^{2} d^{2}+i d^{4}-4 c^{3} d +4 c \,d^{3}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{4 \left (i d -c \right )^{\frac {5}{2}} \left (i d +c \right )^{4} d^{2}}-\frac {i c^{3}+i c \,d^{2}-c^{2} d -d^{3}}{3 \left (i d -c \right )^{2} \left (i d +c \right )^{4} \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {3 i c^{2}+i d^{2}-2 c d}{\left (i d -c \right )^{2} \left (i d +c \right )^{4} \sqrt {c +d \tan \left (f x +e \right )}}\right )}{f a}\)	\(364\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+I*a*tan(f*x+e))/(c+d*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/f/a*d^2*(1/4/d^2/(c+I*d)^4/(I*d-c)^2*(-(c^4+2*c^2*d^2+d^4)*d/(c+I*d)*(c+d*tan(f*x+e))^(1/2)/(-d*tan(f*x+e)+I
*d)-(-6*c^4*d-12*c^2*d^3-6*d^5+I*c^5+2*I*c^3*d^2+I*c*d^4)/(c+I*d)/(-I*d-c)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)
/(-I*d-c)^(1/2)))+1/4/(I*d-c)^(5/2)/(c+I*d)^4*(-6*I*c^2*d^2+I*d^4-4*c^3*d+4*c*d^3+I*c^4)/d^2*arctan((c+d*tan(f
*x+e))^(1/2)/(I*d-c)^(1/2))-1/3/(I*d-c)^2/(c+I*d)^4*(I*c^3+I*c*d^2-c^2*d-d^3)/(c+d*tan(f*x+e))^(3/2)-1/(I*d-c)
^2/(c+I*d)^4*(3*I*c^2+I*d^2-2*c*d)/(c+d*tan(f*x+e))^(1/2))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))/(c+d*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2699 vs. \(2 (219) = 438\).
time = 4.03, size = 2699, normalized size = 10.11 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))/(c+d*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/24*(6*((I*a*c^7 + a*c^6*d + 3*I*a*c^5*d^2 + 3*a*c^4*d^3 + 3*I*a*c^3*d^4 + 3*a*c^2*d^5 + I*a*c*d^6 + a*d^7)*f
*e^(6*I*f*x + 6*I*e) + 2*(I*a*c^7 - a*c^6*d + 3*I*a*c^5*d^2 - 3*a*c^4*d^3 + 3*I*a*c^3*d^4 - 3*a*c^2*d^5 + I*a*
c*d^6 - a*d^7)*f*e^(4*I*f*x + 4*I*e) + (I*a*c^7 - 3*a*c^6*d - I*a*c^5*d^2 - 5*a*c^4*d^3 - 5*I*a*c^3*d^4 - a*c^
2*d^5 - 3*I*a*c*d^6 + a*d^7)*f*e^(2*I*f*x + 2*I*e))*sqrt(1/4*I/((-I*a^2*c^5 - 5*a^2*c^4*d + 10*I*a^2*c^3*d^2 +
10*a^2*c^2*d^3 - 5*I*a^2*c*d^4 - a^2*d^5)*f^2))*log(-2*(2*((I*a*c^3 + 3*a*c^2*d - 3*I*a*c*d^2 - a*d^3)*f*e^(2
*I*f*x + 2*I*e) + (I*a*c^3 + 3*a*c^2*d - 3*I*a*c*d^2 - a*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(1/4*I/((-I*a^2*c^5 - 5*a^2*c^4*d + 10*I*a^2*c^3*d^2 + 10*a^2*c^2*d^3 - 5*I*a
^2*c*d^4 - a^2*d^5)*f^2)) - (c - I*d)*e^(2*I*f*x + 2*I*e) - c)*e^(-2*I*f*x - 2*I*e)) + 6*((-I*a*c^7 - a*c^6*d
- 3*I*a*c^5*d^2 - 3*a*c^4*d^3 - 3*I*a*c^3*d^4 - 3*a*c^2*d^5 - I*a*c*d^6 - a*d^7)*f*e^(6*I*f*x + 6*I*e) + 2*(-I
*a*c^7 + a*c^6*d - 3*I*a*c^5*d^2 + 3*a*c^4*d^3 - 3*I*a*c^3*d^4 + 3*a*c^2*d^5 - I*a*c*d^6 + a*d^7)*f*e^(4*I*f*x
+ 4*I*e) + (-I*a*c^7 + 3*a*c^6*d + I*a*c^5*d^2 + 5*a*c^4*d^3 + 5*I*a*c^3*d^4 + a*c^2*d^5 + 3*I*a*c*d^6 - a*d^
7)*f*e^(2*I*f*x + 2*I*e))*sqrt(1/4*I/((-I*a^2*c^5 - 5*a^2*c^4*d + 10*I*a^2*c^3*d^2 + 10*a^2*c^2*d^3 - 5*I*a^2*
c*d^4 - a^2*d^5)*f^2))*log(-2*(2*((-I*a*c^3 - 3*a*c^2*d + 3*I*a*c*d^2 + a*d^3)*f*e^(2*I*f*x + 2*I*e) + (-I*a*c
^3 - 3*a*c^2*d + 3*I*a*c*d^2 + a*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(1/4*I/((-I*a^2*c^5 - 5*a^2*c^4*d + 10*I*a^2*c^3*d^2 + 10*a^2*c^2*d^3 - 5*I*a^2*c*d^4 - a^2*d^5)*f^2)
) - (c - I*d)*e^(2*I*f*x + 2*I*e) - c)*e^(-2*I*f*x - 2*I*e)) + 3*((-I*a*c^7 - a*c^6*d - 3*I*a*c^5*d^2 - 3*a*c^
4*d^3 - 3*I*a*c^3*d^4 - 3*a*c^2*d^5 - I*a*c*d^6 - a*d^7)*f*e^(6*I*f*x + 6*I*e) + 2*(-I*a*c^7 + a*c^6*d - 3*I*a
*c^5*d^2 + 3*a*c^4*d^3 - 3*I*a*c^3*d^4 + 3*a*c^2*d^5 - I*a*c*d^6 + a*d^7)*f*e^(4*I*f*x + 4*I*e) + (-I*a*c^7 +
3*a*c^6*d + I*a*c^5*d^2 + 5*a*c^4*d^3 + 5*I*a*c^3*d^4 + a*c^2*d^5 + 3*I*a*c*d^6 - a*d^7)*f*e^(2*I*f*x + 2*I*e)
)*sqrt(-(-I*c^2 + 12*c*d + 36*I*d^2)/((-I*a^2*c^7 + 7*a^2*c^6*d + 21*I*a^2*c^5*d^2 - 35*a^2*c^4*d^3 - 35*I*a^2
*c^3*d^4 + 21*a^2*c^2*d^5 + 7*I*a^2*c*d^6 - a^2*d^7)*f^2))*log(1/2*(I*c^2 - 7*c*d - 6*I*d^2 + ((a*c^4 + 4*I*a*
c^3*d - 6*a*c^2*d^2 - 4*I*a*c*d^3 + a*d^4)*f*e^(2*I*f*x + 2*I*e) + (a*c^4 + 4*I*a*c^3*d - 6*a*c^2*d^2 - 4*I*a*
c*d^3 + a*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(-(-I*c^2 + 12
*c*d + 36*I*d^2)/((-I*a^2*c^7 + 7*a^2*c^6*d + 21*I*a^2*c^5*d^2 - 35*a^2*c^4*d^3 - 35*I*a^2*c^3*d^4 + 21*a^2*c^
2*d^5 + 7*I*a^2*c*d^6 - a^2*d^7)*f^2)) + (I*c^2 - 6*c*d)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/((a*c^4 + 4
*I*a*c^3*d - 6*a*c^2*d^2 - 4*I*a*c*d^3 + a*d^4)*f)) + 3*((I*a*c^7 + a*c^6*d + 3*I*a*c^5*d^2 + 3*a*c^4*d^3 + 3*
I*a*c^3*d^4 + 3*a*c^2*d^5 + I*a*c*d^6 + a*d^7)*f*e^(6*I*f*x + 6*I*e) + 2*(I*a*c^7 - a*c^6*d + 3*I*a*c^5*d^2 -
3*a*c^4*d^3 + 3*I*a*c^3*d^4 - 3*a*c^2*d^5 + I*a*c*d^6 - a*d^7)*f*e^(4*I*f*x + 4*I*e) + (I*a*c^7 - 3*a*c^6*d -
I*a*c^5*d^2 - 5*a*c^4*d^3 - 5*I*a*c^3*d^4 - a*c^2*d^5 - 3*I*a*c*d^6 + a*d^7)*f*e^(2*I*f*x + 2*I*e))*sqrt(-(-I*
c^2 + 12*c*d + 36*I*d^2)/((-I*a^2*c^7 + 7*a^2*c^6*d + 21*I*a^2*c^5*d^2 - 35*a^2*c^4*d^3 - 35*I*a^2*c^3*d^4 + 2
1*a^2*c^2*d^5 + 7*I*a^2*c*d^6 - a^2*d^7)*f^2))*log(1/2*(I*c^2 - 7*c*d - 6*I*d^2 - ((a*c^4 + 4*I*a*c^3*d - 6*a*
c^2*d^2 - 4*I*a*c*d^3 + a*d^4)*f*e^(2*I*f*x + 2*I*e) + (a*c^4 + 4*I*a*c^3*d - 6*a*c^2*d^2 - 4*I*a*c*d^3 + a*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(-(-I*c^2 + 12*c*d + 36*I*
d^2)/((-I*a^2*c^7 + 7*a^2*c^6*d + 21*I*a^2*c^5*d^2 - 35*a^2*c^4*d^3 - 35*I*a^2*c^3*d^4 + 21*a^2*c^2*d^5 + 7*I*
a^2*c*d^6 - a^2*d^7)*f^2)) + (I*c^2 - 6*c*d)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/((a*c^4 + 4*I*a*c^3*d -
6*a*c^2*d^2 - 4*I*a*c*d^3 + a*d^4)*f)) + 2*(3*c^4 + 6*c^2*d^2 + 3*d^4 + (3*c^4 - 12*I*c^3*d - 98*c^2*d^2 + 10
8*I*c*d^3 + 19*d^4)*e^(6*I*f*x + 6*I*e) + (9*c^4 - 24*I*c^3*d - 178*c^2*d^2 + 48*I*c*d^3 - 19*d^4)*e^(4*I*f*x
+ 4*I*e) + (9*c^4 - 12*I*c^3*d - 74*c^2*d^2 - 60*I*c*d^3 - 35*d^4)*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)))/((-I*a*c^7 - a*c^6*d - 3*I*a*c^5*d^2 - 3*a*c^4*d^3 - 3*I*
a*c^3*d^4 - 3*a*c^2*d^5 - I*a*c*d^6 - a*d^7)*f*e^(6*I*f*x + 6*I*e) + 2*(-I*a*c^7 + a*c^6*d - 3*I*a*c^5*d^2 + 3
*a*c^4*d^3 - 3*I*a*c^3*d^4 + 3*a*c^2*d^5 - I*a*c*d^6 + a*d^7)*f*e^(4*I*f*x + 4*I*e) + (-I*a*c^7 + 3*a*c^6*d +
I*a*c^5*d^2 + 5*a*c^4*d^3 + 5*I*a*c^3*d^4 + a*c^2*d^5 + 3*I*a*c*d^6 - a*d^7)*f*e^(2*I*f*x + 2*I*e))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {i \int \frac {1}{c^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )} - i c^{2} \sqrt {c + d \tan {\left (e + f x \right )}} + 2 c d \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )} - 2 i c d \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )} + d^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{3}{\left (e + f x \right )} - i d^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}}\, dx}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))/(c+d*tan(f*x+e))**(5/2),x)

[Out]

-I*Integral(1/(c**2*sqrt(c + d*tan(e + f*x))*tan(e + f*x) - I*c**2*sqrt(c + d*tan(e + f*x)) + 2*c*d*sqrt(c + d
*tan(e + f*x))*tan(e + f*x)**2 - 2*I*c*d*sqrt(c + d*tan(e + f*x))*tan(e + f*x) + d**2*sqrt(c + d*tan(e + f*x))
*tan(e + f*x)**3 - I*d**2*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2), x)/a

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Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 547 vs. \(2 (219) = 438\).
time = 1.09, size = 547, normalized size = 2.05 \begin {gather*} \frac {2 \, \sqrt {d \tan \left (f x + e\right ) + c} d}{-4 \, {\left (i \, a c^{3} f - 3 \, a c^{2} d f - 3 i \, a c d^{2} f + a d^{3} f\right )} {\left (i \, d \tan \left (f x + e\right ) + d\right )}} + \frac {{\left (-i \, c + 6 \, d\right )} \arctan \left (\frac {2 \, {\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 {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} + i \, \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}}}\right )}{{\left (a c^{3} f + 3 i \, a c^{2} d f - 3 \, a c d^{2} f - i \, a d^{3} f\right )} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} {\left (\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} + \frac {2 \, {\left (9 \, {\left (d \tan \left (f x + e\right ) + c\right )} c d^{2} + c^{2} d^{2} - 3 \, {\left (i \, d \tan \left (f x + e\right ) + i \, c\right )} d^{3} + d^{4}\right )}}{-3 \, {\left (-i \, a c^{5} f + a c^{4} d f - 2 i \, a c^{3} d^{2} f + 2 \, a c^{2} d^{3} f - i \, a c d^{4} f + a d^{5} f\right )} {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} + \frac {i \, \arctan \left (\frac {2 \, {\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 {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} - i \, \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}}}\right )}{{\left (a c^{2} f - 2 i \, a c d f - a d^{2} f\right )} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+I*a*tan(f*x+e))/(c+d*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

2*sqrt(d*tan(f*x + e) + c)*d/((-4*I*a*c^3*f + 12*a*c^2*d*f + 12*I*a*c*d^2*f - 4*a*d^3*f)*(I*d*tan(f*x + e) + d
)) + (-I*c + 6*d)*arctan(2*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c))/(c*sqrt(-2*
c + 2*sqrt(c^2 + d^2)) + I*sqrt(-2*c + 2*sqrt(c^2 + d^2))*d - sqrt(c^2 + d^2)*sqrt(-2*c + 2*sqrt(c^2 + d^2))))
/((a*c^3*f + 3*I*a*c^2*d*f - 3*a*c*d^2*f - I*a*d^3*f)*sqrt(-2*c + 2*sqrt(c^2 + d^2))*(I*d/(c - sqrt(c^2 + d^2)
) + 1)) + 2*(9*(d*tan(f*x + e) + c)*c*d^2 + c^2*d^2 - 3*(I*d*tan(f*x + e) + I*c)*d^3 + d^4)/((3*I*a*c^5*f - 3*
a*c^4*d*f + 6*I*a*c^3*d^2*f - 6*a*c^2*d^3*f + 3*I*a*c*d^4*f - 3*a*d^5*f)*(d*tan(f*x + e) + c)^(3/2)) + I*arcta
n(2*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c))/(c*sqrt(-2*c + 2*sqrt(c^2 + d^2))
- I*sqrt(-2*c + 2*sqrt(c^2 + d^2))*d - sqrt(c^2 + d^2)*sqrt(-2*c + 2*sqrt(c^2 + d^2))))/((a*c^2*f - 2*I*a*c*d*
f - a*d^2*f)*sqrt(-2*c + 2*sqrt(c^2 + d^2))*(-I*d/(c - sqrt(c^2 + d^2)) + 1))

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Mupad [B]
time = 93.43, size = 2500, normalized size = 9.36 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + a*tan(e + f*x)*1i)*(c + d*tan(e + f*x))^(5/2)),x)

[Out]

log((a*f*(139*c*d^7 - d^8*30i + c^2*d^6*180i - 62*c^3*d^5 + c^4*d^4*10i - c^5*d^3))/2 - (((a*f*(208*a^2*c^2*d^
11*f^2 - a^2*c*d^12*f^2*320i - 112*a^2*d^13*f^2 - a^2*c^3*d^10*f^2*640i + 1312*a^2*c^4*d^9*f^2 + 1568*a^2*c^6*
d^7*f^2 + a^2*c^7*d^6*f^2*640i + 592*a^2*c^8*d^5*f^2 + a^2*c^9*d^4*f^2*320i + 16*a^2*c^10*d^3*f^2))/2 - 2*(c +
d*tan(e + f*x))^(1/2)*(a^2*d^2*f^2 - a^2*c^2*f^2 + a^2*c*d*f^2*2i)*((4480*c^2*d^9 - 560*d^11 - c*d^10*2800i +
c^3*d^8*4480i - 560*c^4*d^7 - c^5*d^6*112i - 224*c^6*d^5 + c^7*d^4*32i - a^2*c^10*f^2*(((3920*c*d^12 - 16240*
c^3*d^10 + 5712*c^5*d^8 + 304*c^7*d^6 + 32*c^9*d^4)/(a^2*c^12*f^2 + a^2*d^12*f^2 + 6*a^2*c^2*d^10*f^2 + 15*a^2
*c^4*d^8*f^2 + 20*a^2*c^6*d^6*f^2 + 15*a^2*c^8*d^4*f^2 + 6*a^2*c^10*d^2*f^2) + ((10640*c^2*d^11 - 560*d^13 - 1
4000*c^4*d^9 + 560*c^6*d^7 + 160*c^8*d^5)*1i)/(a^2*c^12*f^2 + a^2*d^12*f^2 + 6*a^2*c^2*d^10*f^2 + 15*a^2*c^4*d
^8*f^2 + 20*a^2*c^6*d^6*f^2 + 15*a^2*c^8*d^4*f^2 + 6*a^2*c^10*d^2*f^2))^2 - 4*(256*d^6 + 256*c^2*d^4)*(((60*c*
d^7 + 10*c^3*d^5)*1i)/(a^4*c^12*f^4 + a^4*d^12*f^4 + 6*a^4*c^2*d^10*f^4 + 15*a^4*c^4*d^8*f^4 + 20*a^4*c^6*d^6*
f^4 + 15*a^4*c^8*d^4*f^4 + 6*a^4*c^10*d^2*f^4) + (36*d^8 - 13*c^2*d^6 + c^4*d^4)/(a^4*c^12*f^4 + a^4*d^12*f^4
+ 6*a^4*c^2*d^10*f^4 + 15*a^4*c^4*d^8*f^4 + 20*a^4*c^6*d^6*f^4 + 15*a^4*c^8*d^4*f^4 + 6*a^4*c^10*d^2*f^4)))^(1
/2)*1i + a^2*d^10*f^2*(((3920*c*d^12 - 16240*c^3*d^10 + 5712*c^5*d^8 + 304*c^7*d^6 + 32*c^9*d^4)/(a^2*c^12*f^2
+ a^2*d^12*f^2 + 6*a^2*c^2*d^10*f^2 + 15*a^2*c^4*d^8*f^2 + 20*a^2*c^6*d^6*f^2 + 15*a^2*c^8*d^4*f^2 + 6*a^2*c^
10*d^2*f^2) + ((10640*c^2*d^11 - 560*d^13 - 14000*c^4*d^9 + 560*c^6*d^7 + 160*c^8*d^5)*1i)/(a^2*c^12*f^2 + a^2
*d^12*f^2 + 6*a^2*c^2*d^10*f^2 + 15*a^2*c^4*d^8*f^2 + 20*a^2*c^6*d^6*f^2 + 15*a^2*c^8*d^4*f^2 + 6*a^2*c^10*d^2
*f^2))^2 - 4*(256*d^6 + 256*c^2*d^4)*(((60*c*d^7 + 10*c^3*d^5)*1i)/(a^4*c^12*f^4 + a^4*d^12*f^4 + 6*a^4*c^2*d^
10*f^4 + 15*a^4*c^4*d^8*f^4 + 20*a^4*c^6*d^6*f^4 + 15*a^4*c^8*d^4*f^4 + 6*a^4*c^10*d^2*f^4) + (36*d^8 - 13*c^2
*d^6 + c^4*d^4)/(a^4*c^12*f^4 + a^4*d^12*f^4 + 6*a^4*c^2*d^10*f^4 + 15*a^4*c^4*d^8*f^4 + 20*a^4*c^6*d^6*f^4 +
15*a^4*c^8*d^4*f^4 + 6*a^4*c^10*d^2*f^4)))^(1/2)*1i + 2*a^2*c*d^9*f^2*(((3920*c*d^12 - 16240*c^3*d^10 + 5712*c
^5*d^8 + 304*c^7*d^6 + 32*c^9*d^4)/(a^2*c^12*f^2 + a^2*d^12*f^2 + 6*a^2*c^2*d^10*f^2 + 15*a^2*c^4*d^8*f^2 + 20
*a^2*c^6*d^6*f^2 + 15*a^2*c^8*d^4*f^2 + 6*a^2*c^10*d^2*f^2) + ((10640*c^2*d^11 - 560*d^13 - 14000*c^4*d^9 + 56
0*c^6*d^7 + 160*c^8*d^5)*1i)/(a^2*c^12*f^2 + a^2*d^12*f^2 + 6*a^2*c^2*d^10*f^2 + 15*a^2*c^4*d^8*f^2 + 20*a^2*c
^6*d^6*f^2 + 15*a^2*c^8*d^4*f^2 + 6*a^2*c^10*d^2*f^2))^2 - 4*(256*d^6 + 256*c^2*d^4)*(((60*c*d^7 + 10*c^3*d^5)
*1i)/(a^4*c^12*f^4 + a^4*d^12*f^4 + 6*a^4*c^2*d^10*f^4 + 15*a^4*c^4*d^8*f^4 + 20*a^4*c^6*d^6*f^4 + 15*a^4*c^8*
d^4*f^4 + 6*a^4*c^10*d^2*f^4) + (36*d^8 - 13*c^2*d^6 + c^4*d^4)/(a^4*c^12*f^4 + a^4*d^12*f^4 + 6*a^4*c^2*d^10*
f^4 + 15*a^4*c^4*d^8*f^4 + 20*a^4*c^6*d^6*f^4 + 15*a^4*c^8*d^4*f^4 + 6*a^4*c^10*d^2*f^4)))^(1/2) + 2*a^2*c^9*d
*f^2*(((3920*c*d^12 - 16240*c^3*d^10 + 5712*c^5*d^8 + 304*c^7*d^6 + 32*c^9*d^4)/(a^2*c^12*f^2 + a^2*d^12*f^2 +
6*a^2*c^2*d^10*f^2 + 15*a^2*c^4*d^8*f^2 + 20*a^2*c^6*d^6*f^2 + 15*a^2*c^8*d^4*f^2 + 6*a^2*c^10*d^2*f^2) + ((1
0640*c^2*d^11 - 560*d^13 - 14000*c^4*d^9 + 560*c^6*d^7 + 160*c^8*d^5)*1i)/(a^2*c^12*f^2 + a^2*d^12*f^2 + 6*a^2
*c^2*d^10*f^2 + 15*a^2*c^4*d^8*f^2 + 20*a^2*c^6*d^6*f^2 + 15*a^2*c^8*d^4*f^2 + 6*a^2*c^10*d^2*f^2))^2 - 4*(256
*d^6 + 256*c^2*d^4)*(((60*c*d^7 + 10*c^3*d^5)*1i)/(a^4*c^12*f^4 + a^4*d^12*f^4 + 6*a^4*c^2*d^10*f^4 + 15*a^4*c
^4*d^8*f^4 + 20*a^4*c^6*d^6*f^4 + 15*a^4*c^8*d^4*f^4 + 6*a^4*c^10*d^2*f^4) + (36*d^8 - 13*c^2*d^6 + c^4*d^4)/(
a^4*c^12*f^4 + a^4*d^12*f^4 + 6*a^4*c^2*d^10*f^4 + 15*a^4*c^4*d^8*f^4 + 20*a^4*c^6*d^6*f^4 + 15*a^4*c^8*d^4*f^
4 + 6*a^4*c^10*d^2*f^4)))^(1/2) + a^2*c^2*d^8*f^2*(((3920*c*d^12 - 16240*c^3*d^10 + 5712*c^5*d^8 + 304*c^7*d^6
+ 32*c^9*d^4)/(a^2*c^12*f^2 + a^2*d^12*f^2 + 6*a^2*c^2*d^10*f^2 + 15*a^2*c^4*d^8*f^2 + 20*a^2*c^6*d^6*f^2 + 1
5*a^2*c^8*d^4*f^2 + 6*a^2*c^10*d^2*f^2) + ((10640*c^2*d^11 - 560*d^13 - 14000*c^4*d^9 + 560*c^6*d^7 + 160*c^8*
d^5)*1i)/(a^2*c^12*f^2 + a^2*d^12*f^2 + 6*a^2*c^2*d^10*f^2 + 15*a^2*c^4*d^8*f^2 + 20*a^2*c^6*d^6*f^2 + 15*a^2*
c^8*d^4*f^2 + 6*a^2*c^10*d^2*f^2))^2 - 4*(256*d^6 + 256*c^2*d^4)*(((60*c*d^7 + 10*c^3*d^5)*1i)/(a^4*c^12*f^4 +
a^4*d^12*f^4 + 6*a^4*c^2*d^10*f^4 + 15*a^4*c^4*d^8*f^4 + 20*a^4*c^6*d^6*f^4 + 15*a^4*c^8*d^4*f^4 + 6*a^4*c^10
*d^2*f^4) + (36*d^8 - 13*c^2*d^6 + c^4*d^4)/(a^4*c^12*f^4 + a^4*d^12*f^4 + 6*a^4*c^2*d^10*f^4 + 15*a^4*c^4*d^8
*f^4 + 20*a^4*c^6*d^6*f^4 + 15*a^4*c^8*d^4*f^4 + 6*a^4*c^10*d^2*f^4)))^(1/2)*3i + 8*a^2*c^3*d^7*f^2*(((3920*c*
d^12 - 16240*c^3*d^10 + 5712*c^5*d^8 + 304*c^7*d^6 + 32*c^9*d^4)/(a^2*c^12*f^2 + a^2*d^12*f^2 + 6*a^2*c^2*d^10
*f^2 + 15*a^2*c^4*d^8*f^2 + 20*a^2*c^6*d^6*f^2 + 15*a^2*c^8*d^4*f^2 + 6*a^2*c^10*d^2*f^2) + ((10640*c^2*d^11 -
560*d^13 - 14000*c^4*d^9 + 560*c^6*d^7 + 160*c^8*d^5)*1i)/(a^2*c^12*f^2 + a^2*d^12*f^2 + 6*a^2*c^2*d^10*f^2 +
15*a^2*c^4*d^8*f^2 + 20*a^2*c^6*d^6*f^2 + 15*a...

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