3.13.53 \(\int \frac {(a+b \tan (e+f x))^4}{(c+d \tan (e+f x))^{3/2}} \, dx\) [1253]

Optimal. Leaf size=317 \[ -\frac {i (a-i b)^4 \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{3/2} f}+\frac {i (a+i b)^4 \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(c+i d)^{3/2} f}-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}-\frac {2 b \left (15 a^2 b c d^2-6 a^3 d^3-12 a b^2 d \left (2 c^2+d^2\right )+b^3 \left (8 c^3+5 c d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}-\frac {2 b^2 \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d^2 \left (c^2+d^2\right ) f} \]

[Out]

-I*(a-I*b)^4*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/(c-I*d)^(3/2)/f+I*(a+I*b)^4*arctanh((c+d*tan(f*x+e)
)^(1/2)/(c+I*d)^(1/2))/(c+I*d)^(3/2)/f-2/3*b*(15*a^2*b*c*d^2-6*a^3*d^3-12*a*b^2*d*(2*c^2+d^2)+b^3*(8*c^3+5*c*d
^2))*(c+d*tan(f*x+e))^(1/2)/d^3/(c^2+d^2)/f-2/3*b^2*(3*a*d*(-a*d+2*b*c)-b^2*(4*c^2+d^2))*(c+d*tan(f*x+e))^(1/2
)*tan(f*x+e)/d^2/(c^2+d^2)/f-2*(-a*d+b*c)^2*(a+b*tan(f*x+e))^2/d/(c^2+d^2)/f/(c+d*tan(f*x+e))^(1/2)

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Rubi [A]
time = 0.63, antiderivative size = 317, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {3646, 3718, 3711, 3620, 3618, 65, 214} \begin {gather*} -\frac {2 b \left (-6 a^3 d^3+15 a^2 b c d^2-12 a b^2 d \left (2 c^2+d^2\right )+b^3 \left (8 c^3+5 c d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{3 d^3 f \left (c^2+d^2\right )}-\frac {2 b^2 \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d^2 f \left (c^2+d^2\right )}-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}-\frac {i (a-i b)^4 \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f (c-i d)^{3/2}}+\frac {i (a+i b)^4 \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f (c+i d)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x])^4/(c + d*Tan[e + f*x])^(3/2),x]

[Out]

((-I)*(a - I*b)^4*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((c - I*d)^(3/2)*f) + (I*(a + I*b)^4*ArcTan
h[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((c + I*d)^(3/2)*f) - (2*(b*c - a*d)^2*(a + b*Tan[e + f*x])^2)/(d*(
c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]]) - (2*b*(15*a^2*b*c*d^2 - 6*a^3*d^3 - 12*a*b^2*d*(2*c^2 + d^2) + b^3*(8*
c^3 + 5*c*d^2))*Sqrt[c + d*Tan[e + f*x]])/(3*d^3*(c^2 + d^2)*f) - (2*b^2*(3*a*d*(2*b*c - a*d) - b^2*(4*c^2 + d
^2))*Tan[e + f*x]*Sqrt[c + d*Tan[e + f*x]])/(3*d^2*(c^2 + d^2)*f)

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 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 3646

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

Rule 3711

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[C*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])
^m*Simp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0]
&&  !LeQ[m, -1]

Rule 3718

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*tan[(e
_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[b*C*Tan[e + f*x]*((c + d*Tan[e + f*x])
^(n + 1)/(d*f*(n + 2))), x] - Dist[1/(d*(n + 2)), Int[(c + d*Tan[e + f*x])^n*Simp[b*c*C - a*A*d*(n + 2) - (A*b
+ a*B - b*C)*d*(n + 2)*Tan[e + f*x] - (a*C*d*(n + 2) - b*(c*C - B*d*(n + 2)))*Tan[e + f*x]^2, x], x], x] /; F
reeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[c^2 + d^2, 0] &&  !LtQ[n, -1]

Rubi steps

\begin {align*} \int \frac {(a+b \tan (e+f x))^4}{(c+d \tan (e+f x))^{3/2}} \, dx &=-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}+\frac {2 \int \frac {(a+b \tan (e+f x)) \left (\frac {1}{2} \left (4 b^3 c^2+a^3 c d-9 a b^2 c d+6 a^2 b d^2\right )+\frac {1}{2} d \left (3 a^2 b c-b^3 c-a^3 d+3 a b^2 d\right ) \tan (e+f x)-\frac {1}{2} b \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan ^2(e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx}{d \left (c^2+d^2\right )}\\ &=-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}-\frac {2 b^2 \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d^2 \left (c^2+d^2\right ) f}-\frac {4 \int \frac {\frac {1}{4} \left (-24 a b^3 c^2 d-3 a^4 c d^2+33 a^2 b^2 c d^2-18 a^3 b d^3+2 b^4 c \left (4 c^2+d^2\right )\right )-\frac {3}{4} d^2 \left (4 a^3 b c-4 a b^3 c-a^4 d+6 a^2 b^2 d-b^4 d\right ) \tan (e+f x)+\frac {1}{4} b \left (15 a^2 b c d^2-6 a^3 d^3-12 a b^2 d \left (2 c^2+d^2\right )+b^3 \left (8 c^3+5 c d^2\right )\right ) \tan ^2(e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{3 d^2 \left (c^2+d^2\right )}\\ &=-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}-\frac {2 b \left (15 a^2 b c d^2-6 a^3 d^3-12 a b^2 d \left (2 c^2+d^2\right )+b^3 \left (8 c^3+5 c d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}-\frac {2 b^2 \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d^2 \left (c^2+d^2\right ) f}-\frac {4 \int \frac {-\frac {3}{4} d^2 \left (a^4 c-6 a^2 b^2 c+b^4 c+4 a^3 b d-4 a b^3 d\right )-\frac {3}{4} d^2 \left (4 a^3 b c-4 a b^3 c-a^4 d+6 a^2 b^2 d-b^4 d\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{3 d^2 \left (c^2+d^2\right )}\\ &=-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}-\frac {2 b \left (15 a^2 b c d^2-6 a^3 d^3-12 a b^2 d \left (2 c^2+d^2\right )+b^3 \left (8 c^3+5 c d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}-\frac {2 b^2 \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d^2 \left (c^2+d^2\right ) f}+\frac {(a-i b)^4 \int \frac {1+i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{2 (c-i d)}+\frac {(a+i b)^4 \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx}{2 (c+i d)}\\ &=-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}-\frac {2 b \left (15 a^2 b c d^2-6 a^3 d^3-12 a b^2 d \left (2 c^2+d^2\right )+b^3 \left (8 c^3+5 c d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}-\frac {2 b^2 \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d^2 \left (c^2+d^2\right ) f}+\frac {(a+i b)^4 \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{2 (i c-d) f}-\frac {(a-i b)^4 \text {Subst}\left (\int \frac {1}{(-1+x) \sqrt {c-i d x}} \, dx,x,i \tan (e+f x)\right )}{2 (i c+d) f}\\ &=-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}-\frac {2 b \left (15 a^2 b c d^2-6 a^3 d^3-12 a b^2 d \left (2 c^2+d^2\right )+b^3 \left (8 c^3+5 c d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}-\frac {2 b^2 \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d^2 \left (c^2+d^2\right ) f}-\frac {(a-i b)^4 \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 )}{(c-i d) d f}-\frac {(a+i b)^4 \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 )}{(c+i d) d f}\\ &=-\frac {i (a-i b)^4 \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{3/2} f}+\frac {i (a+i b)^4 \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{(c+i d)^{3/2} f}-\frac {2 (b c-a d)^2 (a+b \tan (e+f x))^2}{d \left (c^2+d^2\right ) f \sqrt {c+d \tan (e+f x)}}-\frac {2 b \left (15 a^2 b c d^2-6 a^3 d^3-12 a b^2 d \left (2 c^2+d^2\right )+b^3 \left (8 c^3+5 c d^2\right )\right ) \sqrt {c+d \tan (e+f x)}}{3 d^3 \left (c^2+d^2\right ) f}-\frac {2 b^2 \left (3 a d (2 b c-a d)-b^2 \left (4 c^2+d^2\right )\right ) \tan (e+f x) \sqrt {c+d \tan (e+f x)}}{3 d^2 \left (c^2+d^2\right ) f}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 6.36, size = 415, normalized size = 1.31 \begin {gather*} \frac {2 b^2 (a+b \tan (e+f x))^2}{3 d f \sqrt {c+d \tan (e+f x)}}+\frac {2 \left (-\frac {2 b^2 (2 b c-5 a d) (a+b \tan (e+f x))}{d f \sqrt {c+d \tan (e+f x)}}+\frac {-\frac {2 \left (8 b^4 c^2-28 a b^3 c d+29 a^2 b^2 d^2-3 b^4 d^2\right )}{d \sqrt {c+d \tan (e+f x)}}+\frac {2 \left (6 a (a-b) b (a+b) d^2 \left (-\frac {i \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{\sqrt {c-i d}}+\frac {i \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{\sqrt {c+i d}}\right )+\frac {\left (-6 a (a-b) b (a+b) c d^3+\frac {3}{2} \left (a^4-6 a^2 b^2+b^4\right ) d^4\right ) \left (-\frac {\, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {c+d \tan (e+f x)}{c-i d}\right )}{(i c+d) \sqrt {c+d \tan (e+f x)}}+\frac {\, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {c+d \tan (e+f x)}{c+i d}\right )}{(i c-d) \sqrt {c+d \tan (e+f x)}}\right )}{d}\right )}{d}}{2 d f}\right )}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[e + f*x])^4/(c + d*Tan[e + f*x])^(3/2),x]

[Out]

(2*b^2*(a + b*Tan[e + f*x])^2)/(3*d*f*Sqrt[c + d*Tan[e + f*x]]) + (2*((-2*b^2*(2*b*c - 5*a*d)*(a + b*Tan[e + f
*x]))/(d*f*Sqrt[c + d*Tan[e + f*x]]) + ((-2*(8*b^4*c^2 - 28*a*b^3*c*d + 29*a^2*b^2*d^2 - 3*b^4*d^2))/(d*Sqrt[c
+ d*Tan[e + f*x]]) + (2*(6*a*(a - b)*b*(a + b)*d^2*(((-I)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/Sq
rt[c - I*d] + (I*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/Sqrt[c + I*d]) + ((-6*a*(a - b)*b*(a + b)*c*
d^3 + (3*(a^4 - 6*a^2*b^2 + b^4)*d^4)/2)*(-(Hypergeometric2F1[-1/2, 1, 1/2, (c + d*Tan[e + f*x])/(c - I*d)]/((
I*c + d)*Sqrt[c + d*Tan[e + f*x]])) + Hypergeometric2F1[-1/2, 1, 1/2, (c + d*Tan[e + f*x])/(c + I*d)]/((I*c -
d)*Sqrt[c + d*Tan[e + f*x]])))/d))/d)/(2*d*f)))/(3*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(5654\) vs. \(2(287)=574\).
time = 0.51, size = 5655, normalized size = 17.84

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(5655\)
default	\(\text {Expression too large to display}\)	\(5655\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(f*x+e))^4/(c+d*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))^4/(c+d*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))^4/(c+d*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{4}}{\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))**4/(c+d*tan(f*x+e))**(3/2),x)

[Out]

Integral((a + b*tan(e + f*x))**4/(c + d*tan(e + f*x))**(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(f*x+e))^4/(c+d*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choi
ce was done

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Mupad [B]
time = 23.38, size = 2500, normalized size = 7.89 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(e + f*x))^4/(c + d*tan(e + f*x))^(3/2),x)

[Out]

(2*b^4*(c + d*tan(e + f*x))^(3/2))/(3*d^3*f) - atan(-(((c + d*tan(e + f*x))^(1/2)*(16*a^8*d^10*f^3 + 16*b^8*d^
10*f^3 - 448*a^2*b^6*d^10*f^3 + 1120*a^4*b^4*d^10*f^3 - 448*a^6*b^2*d^10*f^3 + 32*a^8*c^2*d^8*f^3 - 32*a^8*c^6
*d^4*f^3 - 16*a^8*c^8*d^2*f^3 + 32*b^8*c^2*d^8*f^3 - 32*b^8*c^6*d^4*f^3 - 16*b^8*c^8*d^2*f^3 - 896*a^2*b^6*c^2
*d^8*f^3 + 896*a^2*b^6*c^6*d^4*f^3 + 448*a^2*b^6*c^8*d^2*f^3 - 5376*a^3*b^5*c^3*d^7*f^3 - 5376*a^3*b^5*c^5*d^5
*f^3 - 1792*a^3*b^5*c^7*d^3*f^3 + 2240*a^4*b^4*c^2*d^8*f^3 - 2240*a^4*b^4*c^6*d^4*f^3 - 1120*a^4*b^4*c^8*d^2*f
^3 + 5376*a^5*b^3*c^3*d^7*f^3 + 5376*a^5*b^3*c^5*d^5*f^3 + 1792*a^5*b^3*c^7*d^3*f^3 - 896*a^6*b^2*c^2*d^8*f^3
+ 896*a^6*b^2*c^6*d^4*f^3 + 448*a^6*b^2*c^8*d^2*f^3 + 256*a*b^7*c*d^9*f^3 - 256*a^7*b*c*d^9*f^3 + 768*a*b^7*c^
3*d^7*f^3 + 768*a*b^7*c^5*d^5*f^3 + 256*a*b^7*c^7*d^3*f^3 - 1792*a^3*b^5*c*d^9*f^3 + 1792*a^5*b^3*c*d^9*f^3 -
768*a^7*b*c^3*d^7*f^3 - 768*a^7*b*c^5*d^5*f^3 - 256*a^7*b*c^7*d^3*f^3) + (-(((8*a^8*c^3*f^2 + 8*b^8*c^3*f^2 +
64*a*b^7*d^3*f^2 - 64*a^7*b*d^3*f^2 - 24*a^8*c*d^2*f^2 - 24*b^8*c*d^2*f^2 - 224*a^2*b^6*c^3*f^2 + 560*a^4*b^4*
c^3*f^2 - 224*a^6*b^2*c^3*f^2 - 448*a^3*b^5*d^3*f^2 + 448*a^5*b^3*d^3*f^2 - 192*a*b^7*c^2*d*f^2 + 192*a^7*b*c^
2*d*f^2 + 672*a^2*b^6*c*d^2*f^2 + 1344*a^3*b^5*c^2*d*f^2 - 1680*a^4*b^4*c*d^2*f^2 - 1344*a^5*b^3*c^2*d*f^2 + 6
72*a^6*b^2*c*d^2*f^2)^2/4 - (16*c^6*f^4 + 16*d^6*f^4 + 48*c^2*d^4*f^4 + 48*c^4*d^2*f^4)*(a^16 + b^16 + 8*a^2*b
^14 + 28*a^4*b^12 + 56*a^6*b^10 + 70*a^8*b^8 + 56*a^10*b^6 + 28*a^12*b^4 + 8*a^14*b^2))^(1/2) + 4*a^8*c^3*f^2
+ 4*b^8*c^3*f^2 + 32*a*b^7*d^3*f^2 - 32*a^7*b*d^3*f^2 - 12*a^8*c*d^2*f^2 - 12*b^8*c*d^2*f^2 - 112*a^2*b^6*c^3*
f^2 + 280*a^4*b^4*c^3*f^2 - 112*a^6*b^2*c^3*f^2 - 224*a^3*b^5*d^3*f^2 + 224*a^5*b^3*d^3*f^2 - 96*a*b^7*c^2*d*f
^2 + 96*a^7*b*c^2*d*f^2 + 336*a^2*b^6*c*d^2*f^2 + 672*a^3*b^5*c^2*d*f^2 - 840*a^4*b^4*c*d^2*f^2 - 672*a^5*b^3*
c^2*d*f^2 + 336*a^6*b^2*c*d^2*f^2)/(16*(c^6*f^4 + d^6*f^4 + 3*c^2*d^4*f^4 + 3*c^4*d^2*f^4)))^(1/2)*(128*a^3*b*
d^12*f^4 - 128*a*b^3*d^12*f^4 - (c + d*tan(e + f*x))^(1/2)*(-(((8*a^8*c^3*f^2 + 8*b^8*c^3*f^2 + 64*a*b^7*d^3*f
^2 - 64*a^7*b*d^3*f^2 - 24*a^8*c*d^2*f^2 - 24*b^8*c*d^2*f^2 - 224*a^2*b^6*c^3*f^2 + 560*a^4*b^4*c^3*f^2 - 224*
a^6*b^2*c^3*f^2 - 448*a^3*b^5*d^3*f^2 + 448*a^5*b^3*d^3*f^2 - 192*a*b^7*c^2*d*f^2 + 192*a^7*b*c^2*d*f^2 + 672*
a^2*b^6*c*d^2*f^2 + 1344*a^3*b^5*c^2*d*f^2 - 1680*a^4*b^4*c*d^2*f^2 - 1344*a^5*b^3*c^2*d*f^2 + 672*a^6*b^2*c*d
^2*f^2)^2/4 - (16*c^6*f^4 + 16*d^6*f^4 + 48*c^2*d^4*f^4 + 48*c^4*d^2*f^4)*(a^16 + b^16 + 8*a^2*b^14 + 28*a^4*b
^12 + 56*a^6*b^10 + 70*a^8*b^8 + 56*a^10*b^6 + 28*a^12*b^4 + 8*a^14*b^2))^(1/2) + 4*a^8*c^3*f^2 + 4*b^8*c^3*f^
2 + 32*a*b^7*d^3*f^2 - 32*a^7*b*d^3*f^2 - 12*a^8*c*d^2*f^2 - 12*b^8*c*d^2*f^2 - 112*a^2*b^6*c^3*f^2 + 280*a^4*
b^4*c^3*f^2 - 112*a^6*b^2*c^3*f^2 - 224*a^3*b^5*d^3*f^2 + 224*a^5*b^3*d^3*f^2 - 96*a*b^7*c^2*d*f^2 + 96*a^7*b*
c^2*d*f^2 + 336*a^2*b^6*c*d^2*f^2 + 672*a^3*b^5*c^2*d*f^2 - 840*a^4*b^4*c*d^2*f^2 - 672*a^5*b^3*c^2*d*f^2 + 33
6*a^6*b^2*c*d^2*f^2)/(16*(c^6*f^4 + d^6*f^4 + 3*c^2*d^4*f^4 + 3*c^4*d^2*f^4)))^(1/2)*(64*c*d^12*f^5 + 320*c^3*
d^10*f^5 + 640*c^5*d^8*f^5 + 640*c^7*d^6*f^5 + 320*c^9*d^4*f^5 + 64*c^11*d^2*f^5) + 64*a^4*c*d^11*f^4 + 64*b^4
*c*d^11*f^4 + 256*a^4*c^3*d^9*f^4 + 384*a^4*c^5*d^7*f^4 + 256*a^4*c^7*d^5*f^4 + 64*a^4*c^9*d^3*f^4 + 256*b^4*c
^3*d^9*f^4 + 384*b^4*c^5*d^7*f^4 + 256*b^4*c^7*d^5*f^4 + 64*b^4*c^9*d^3*f^4 - 1536*a^2*b^2*c^3*d^9*f^4 - 2304*
a^2*b^2*c^5*d^7*f^4 - 1536*a^2*b^2*c^7*d^5*f^4 - 384*a^2*b^2*c^9*d^3*f^4 - 384*a*b^3*c^2*d^10*f^4 - 256*a*b^3*
c^4*d^8*f^4 + 256*a*b^3*c^6*d^6*f^4 + 384*a*b^3*c^8*d^4*f^4 + 128*a*b^3*c^10*d^2*f^4 - 384*a^2*b^2*c*d^11*f^4
+ 384*a^3*b*c^2*d^10*f^4 + 256*a^3*b*c^4*d^8*f^4 - 256*a^3*b*c^6*d^6*f^4 - 384*a^3*b*c^8*d^4*f^4 - 128*a^3*b*c
^10*d^2*f^4))*(-(((8*a^8*c^3*f^2 + 8*b^8*c^3*f^2 + 64*a*b^7*d^3*f^2 - 64*a^7*b*d^3*f^2 - 24*a^8*c*d^2*f^2 - 24
*b^8*c*d^2*f^2 - 224*a^2*b^6*c^3*f^2 + 560*a^4*b^4*c^3*f^2 - 224*a^6*b^2*c^3*f^2 - 448*a^3*b^5*d^3*f^2 + 448*a
^5*b^3*d^3*f^2 - 192*a*b^7*c^2*d*f^2 + 192*a^7*b*c^2*d*f^2 + 672*a^2*b^6*c*d^2*f^2 + 1344*a^3*b^5*c^2*d*f^2 -
1680*a^4*b^4*c*d^2*f^2 - 1344*a^5*b^3*c^2*d*f^2 + 672*a^6*b^2*c*d^2*f^2)^2/4 - (16*c^6*f^4 + 16*d^6*f^4 + 48*c
^2*d^4*f^4 + 48*c^4*d^2*f^4)*(a^16 + b^16 + 8*a^2*b^14 + 28*a^4*b^12 + 56*a^6*b^10 + 70*a^8*b^8 + 56*a^10*b^6
+ 28*a^12*b^4 + 8*a^14*b^2))^(1/2) + 4*a^8*c^3*f^2 + 4*b^8*c^3*f^2 + 32*a*b^7*d^3*f^2 - 32*a^7*b*d^3*f^2 - 12*
a^8*c*d^2*f^2 - 12*b^8*c*d^2*f^2 - 112*a^2*b^6*c^3*f^2 + 280*a^4*b^4*c^3*f^2 - 112*a^6*b^2*c^3*f^2 - 224*a^3*b
^5*d^3*f^2 + 224*a^5*b^3*d^3*f^2 - 96*a*b^7*c^2*d*f^2 + 96*a^7*b*c^2*d*f^2 + 336*a^2*b^6*c*d^2*f^2 + 672*a^3*b
^5*c^2*d*f^2 - 840*a^4*b^4*c*d^2*f^2 - 672*a^5*b^3*c^2*d*f^2 + 336*a^6*b^2*c*d^2*f^2)/(16*(c^6*f^4 + d^6*f^4 +
3*c^2*d^4*f^4 + 3*c^4*d^2*f^4)))^(1/2)*1i + ((c + d*tan(e + f*x))^(1/2)*(16*a^8*d^10*f^3 + 16*b^8*d^10*f^3 -
448*a^2*b^6*d^10*f^3 + 1120*a^4*b^4*d^10*f^3 - 448*a^6*b^2*d^10*f^3 + 32*a^8*c^2*d^8*f^3 - 32*a^8*c^6*d^4*f^3
- 16*a^8*c^8*d^2*f^3 + 32*b^8*c^2*d^8*f^3 - 32*...

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