3.13.29 \(\int (a+b \tan (e+f x))^3 \sqrt {c+d \tan (e+f x)} \, dx\) [1229]

Optimal. Leaf size=209 \[ \frac {(i a+b)^3 \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}-\frac {(i a-b)^3 \sqrt {c+i d} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 b \left (3 a^2-b^2\right ) \sqrt {c+d \tan (e+f x)}}{f}-\frac {4 b^2 (b c-6 a d) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}+\frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}{5 d f} \]

[Out]

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

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Rubi [A]
time = 0.39, antiderivative size = 209, 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 = {3647, 3711, 3609, 3620, 3618, 65, 214} \begin {gather*} \frac {2 b \left (3 a^2-b^2\right ) \sqrt {c+d \tan (e+f x)}}{f}-\frac {4 b^2 (b c-6 a d) (c+d \tan (e+f x))^{3/2}}{15 d^2 f}+\frac {2 b^2 (a+b \tan (e+f x)) (c+d \tan (e+f x))^{3/2}}{5 d f}-\frac {(-b+i a)^3 \sqrt {c+i d} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {(b+i a)^3 \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I*a + b)^3*Sqrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/f - ((I*a - b)^3*Sqrt[c + I*d]*Arc
Tanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/f + (2*b*(3*a^2 - b^2)*Sqrt[c + d*Tan[e + f*x]])/f - (4*b^2*(b*c
- 6*a*d)*(c + d*Tan[e + f*x])^(3/2))/(15*d^2*f) + (2*b^2*(a + b*Tan[e + f*x])*(c + d*Tan[e + f*x])^(3/2))/(5*
d*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 3609

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*
((a + b*Tan[e + f*x])^m/(f*m)), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

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 3647

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

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]

Rubi steps

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

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Mathematica [A]
time = 2.32, size = 194, normalized size = 0.93 \begin {gather*} \frac {-15 i (a-i b)^3 \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+15 i (a+i b)^3 \sqrt {c+i d} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )+\frac {2 b \sqrt {c+d \tan (e+f x)} \left (15 a b c d+45 a^2 d^2-b^2 \left (2 c^2+15 d^2\right )+b d (b c+15 a d) \tan (e+f x)+3 b^2 d^2 \tan ^2(e+f x)\right )}{d^2}}{15 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-15*I)*(a - I*b)^3*Sqrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]] + (15*I)*(a + I*b)^3*Sqrt[c
+ I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]] + (2*b*Sqrt[c + d*Tan[e + f*x]]*(15*a*b*c*d + 45*a^2*d
^2 - b^2*(2*c^2 + 15*d^2) + b*d*(b*c + 15*a*d)*Tan[e + f*x] + 3*b^2*d^2*Tan[e + f*x]^2))/d^2)/(15*f)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1095\) vs. \(2(181)=362\).
time = 0.52, size = 1096, normalized size = 5.24 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d*tan(f*x+e))^(1/2)*(a+b*tan(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(1/2)*(a+b*tan(f*x+e))^3,x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e) + a)^3*sqrt(d*tan(f*x + e) + c), x)

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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((c+d*tan(f*x+e))^(1/2)*(a+b*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))**(1/2)*(a+b*tan(f*x+e))**3,x)

[Out]

Integral((a + b*tan(e + f*x))**3*sqrt(c + d*tan(e + f*x)), 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((c+d*tan(f*x+e))^(1/2)*(a+b*tan(f*x+e))^3,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 = 21.26, size = 2500, normalized size = 11.96 \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))^3*(c + d*tan(e + f*x))^(1/2),x)

[Out]

atan(((((8*(4*b^3*d^4*f^2 - 12*a^2*b*d^4*f^2 + 4*b^3*c^2*d^2*f^2 - 12*a^2*b*c^2*d^2*f^2))/f^3 - 64*c*d^2*(c +
d*tan(e + f*x))^(1/2)*(-(((8*b^6*c*f^2 - 8*a^6*c*f^2 - 120*a^2*b^4*c*f^2 + 120*a^4*b^2*c*f^2 - 160*a^3*b^3*d*f
^2 + 48*a*b^5*d*f^2 + 48*a^5*b*d*f^2)^2/64 - f^4*(a^12*c^2 + a^12*d^2 + b^12*c^2 + b^12*d^2 + 6*a^2*b^10*c^2 +
15*a^4*b^8*c^2 + 20*a^6*b^6*c^2 + 15*a^8*b^4*c^2 + 6*a^10*b^2*c^2 + 6*a^2*b^10*d^2 + 15*a^4*b^8*d^2 + 20*a^6*
b^6*d^2 + 15*a^8*b^4*d^2 + 6*a^10*b^2*d^2))^(1/2) + a^6*c*f^2 - b^6*c*f^2 + 15*a^2*b^4*c*f^2 - 15*a^4*b^2*c*f^
2 + 20*a^3*b^3*d*f^2 - 6*a*b^5*d*f^2 - 6*a^5*b*d*f^2)/(4*f^4))^(1/2))*(-(((8*b^6*c*f^2 - 8*a^6*c*f^2 - 120*a^2
*b^4*c*f^2 + 120*a^4*b^2*c*f^2 - 160*a^3*b^3*d*f^2 + 48*a*b^5*d*f^2 + 48*a^5*b*d*f^2)^2/64 - f^4*(a^12*c^2 + a
^12*d^2 + b^12*c^2 + b^12*d^2 + 6*a^2*b^10*c^2 + 15*a^4*b^8*c^2 + 20*a^6*b^6*c^2 + 15*a^8*b^4*c^2 + 6*a^10*b^2
*c^2 + 6*a^2*b^10*d^2 + 15*a^4*b^8*d^2 + 20*a^6*b^6*d^2 + 15*a^8*b^4*d^2 + 6*a^10*b^2*d^2))^(1/2) + a^6*c*f^2
- b^6*c*f^2 + 15*a^2*b^4*c*f^2 - 15*a^4*b^2*c*f^2 + 20*a^3*b^3*d*f^2 - 6*a*b^5*d*f^2 - 6*a^5*b*d*f^2)/(4*f^4))
^(1/2) + (16*(c + d*tan(e + f*x))^(1/2)*(a^6*d^4 - b^6*d^4 + 15*a^2*b^4*d^4 - 15*a^4*b^2*d^4 - a^6*c^2*d^2 + b
^6*c^2*d^2 - 40*a^3*b^3*c*d^3 - 15*a^2*b^4*c^2*d^2 + 15*a^4*b^2*c^2*d^2 + 12*a*b^5*c*d^3 + 12*a^5*b*c*d^3))/f^
2)*(-(((8*b^6*c*f^2 - 8*a^6*c*f^2 - 120*a^2*b^4*c*f^2 + 120*a^4*b^2*c*f^2 - 160*a^3*b^3*d*f^2 + 48*a*b^5*d*f^2
+ 48*a^5*b*d*f^2)^2/64 - f^4*(a^12*c^2 + a^12*d^2 + b^12*c^2 + b^12*d^2 + 6*a^2*b^10*c^2 + 15*a^4*b^8*c^2 + 2
0*a^6*b^6*c^2 + 15*a^8*b^4*c^2 + 6*a^10*b^2*c^2 + 6*a^2*b^10*d^2 + 15*a^4*b^8*d^2 + 20*a^6*b^6*d^2 + 15*a^8*b^
4*d^2 + 6*a^10*b^2*d^2))^(1/2) + a^6*c*f^2 - b^6*c*f^2 + 15*a^2*b^4*c*f^2 - 15*a^4*b^2*c*f^2 + 20*a^3*b^3*d*f^
2 - 6*a*b^5*d*f^2 - 6*a^5*b*d*f^2)/(4*f^4))^(1/2)*1i - (((8*(4*b^3*d^4*f^2 - 12*a^2*b*d^4*f^2 + 4*b^3*c^2*d^2*
f^2 - 12*a^2*b*c^2*d^2*f^2))/f^3 + 64*c*d^2*(c + d*tan(e + f*x))^(1/2)*(-(((8*b^6*c*f^2 - 8*a^6*c*f^2 - 120*a^
2*b^4*c*f^2 + 120*a^4*b^2*c*f^2 - 160*a^3*b^3*d*f^2 + 48*a*b^5*d*f^2 + 48*a^5*b*d*f^2)^2/64 - f^4*(a^12*c^2 +
a^12*d^2 + b^12*c^2 + b^12*d^2 + 6*a^2*b^10*c^2 + 15*a^4*b^8*c^2 + 20*a^6*b^6*c^2 + 15*a^8*b^4*c^2 + 6*a^10*b^
2*c^2 + 6*a^2*b^10*d^2 + 15*a^4*b^8*d^2 + 20*a^6*b^6*d^2 + 15*a^8*b^4*d^2 + 6*a^10*b^2*d^2))^(1/2) + a^6*c*f^2
- b^6*c*f^2 + 15*a^2*b^4*c*f^2 - 15*a^4*b^2*c*f^2 + 20*a^3*b^3*d*f^2 - 6*a*b^5*d*f^2 - 6*a^5*b*d*f^2)/(4*f^4)
)^(1/2))*(-(((8*b^6*c*f^2 - 8*a^6*c*f^2 - 120*a^2*b^4*c*f^2 + 120*a^4*b^2*c*f^2 - 160*a^3*b^3*d*f^2 + 48*a*b^5
*d*f^2 + 48*a^5*b*d*f^2)^2/64 - f^4*(a^12*c^2 + a^12*d^2 + b^12*c^2 + b^12*d^2 + 6*a^2*b^10*c^2 + 15*a^4*b^8*c
^2 + 20*a^6*b^6*c^2 + 15*a^8*b^4*c^2 + 6*a^10*b^2*c^2 + 6*a^2*b^10*d^2 + 15*a^4*b^8*d^2 + 20*a^6*b^6*d^2 + 15*
a^8*b^4*d^2 + 6*a^10*b^2*d^2))^(1/2) + a^6*c*f^2 - b^6*c*f^2 + 15*a^2*b^4*c*f^2 - 15*a^4*b^2*c*f^2 + 20*a^3*b^
3*d*f^2 - 6*a*b^5*d*f^2 - 6*a^5*b*d*f^2)/(4*f^4))^(1/2) - (16*(c + d*tan(e + f*x))^(1/2)*(a^6*d^4 - b^6*d^4 +
15*a^2*b^4*d^4 - 15*a^4*b^2*d^4 - a^6*c^2*d^2 + b^6*c^2*d^2 - 40*a^3*b^3*c*d^3 - 15*a^2*b^4*c^2*d^2 + 15*a^4*b
^2*c^2*d^2 + 12*a*b^5*c*d^3 + 12*a^5*b*c*d^3))/f^2)*(-(((8*b^6*c*f^2 - 8*a^6*c*f^2 - 120*a^2*b^4*c*f^2 + 120*a
^4*b^2*c*f^2 - 160*a^3*b^3*d*f^2 + 48*a*b^5*d*f^2 + 48*a^5*b*d*f^2)^2/64 - f^4*(a^12*c^2 + a^12*d^2 + b^12*c^2
+ b^12*d^2 + 6*a^2*b^10*c^2 + 15*a^4*b^8*c^2 + 20*a^6*b^6*c^2 + 15*a^8*b^4*c^2 + 6*a^10*b^2*c^2 + 6*a^2*b^10*
d^2 + 15*a^4*b^8*d^2 + 20*a^6*b^6*d^2 + 15*a^8*b^4*d^2 + 6*a^10*b^2*d^2))^(1/2) + a^6*c*f^2 - b^6*c*f^2 + 15*a
^2*b^4*c*f^2 - 15*a^4*b^2*c*f^2 + 20*a^3*b^3*d*f^2 - 6*a*b^5*d*f^2 - 6*a^5*b*d*f^2)/(4*f^4))^(1/2)*1i)/((((8*(
4*b^3*d^4*f^2 - 12*a^2*b*d^4*f^2 + 4*b^3*c^2*d^2*f^2 - 12*a^2*b*c^2*d^2*f^2))/f^3 - 64*c*d^2*(c + d*tan(e + f*
x))^(1/2)*(-(((8*b^6*c*f^2 - 8*a^6*c*f^2 - 120*a^2*b^4*c*f^2 + 120*a^4*b^2*c*f^2 - 160*a^3*b^3*d*f^2 + 48*a*b^
5*d*f^2 + 48*a^5*b*d*f^2)^2/64 - f^4*(a^12*c^2 + a^12*d^2 + b^12*c^2 + b^12*d^2 + 6*a^2*b^10*c^2 + 15*a^4*b^8*
c^2 + 20*a^6*b^6*c^2 + 15*a^8*b^4*c^2 + 6*a^10*b^2*c^2 + 6*a^2*b^10*d^2 + 15*a^4*b^8*d^2 + 20*a^6*b^6*d^2 + 15
*a^8*b^4*d^2 + 6*a^10*b^2*d^2))^(1/2) + a^6*c*f^2 - b^6*c*f^2 + 15*a^2*b^4*c*f^2 - 15*a^4*b^2*c*f^2 + 20*a^3*b
^3*d*f^2 - 6*a*b^5*d*f^2 - 6*a^5*b*d*f^2)/(4*f^4))^(1/2))*(-(((8*b^6*c*f^2 - 8*a^6*c*f^2 - 120*a^2*b^4*c*f^2 +
120*a^4*b^2*c*f^2 - 160*a^3*b^3*d*f^2 + 48*a*b^5*d*f^2 + 48*a^5*b*d*f^2)^2/64 - f^4*(a^12*c^2 + a^12*d^2 + b^
12*c^2 + b^12*d^2 + 6*a^2*b^10*c^2 + 15*a^4*b^8*c^2 + 20*a^6*b^6*c^2 + 15*a^8*b^4*c^2 + 6*a^10*b^2*c^2 + 6*a^2
*b^10*d^2 + 15*a^4*b^8*d^2 + 20*a^6*b^6*d^2 + 15*a^8*b^4*d^2 + 6*a^10*b^2*d^2))^(1/2) + a^6*c*f^2 - b^6*c*f^2
+ 15*a^2*b^4*c*f^2 - 15*a^4*b^2*c*f^2 + 20*a^3*b^3*d*f^2 - 6*a*b^5*d*f^2 - 6*a^5*b*d*f^2)/(4*f^4))^(1/2) + (16
*(c + d*tan(e + f*x))^(1/2)*(a^6*d^4 - b^6*d^4 + 15*a^2*b^4*d^4 - 15*a^4*b^2*d^4 - a^6*c^2*d^2 + b^6*c^2*d^2 -
40*a^3*b^3*c*d^3 - 15*a^2*b^4*c^2*d^2 + 15*a^4*b^2*c^2*d^2 + 12*a*b^5*c*d^3 + 12*a^5*b*c*d^3))/f^2)*(-(((8*b^
6*c*f^2 - 8*a^6*c*f^2 - 120*a^2*b^4*c*f^2 + 120...

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