3.1.100 \(\int (c+d \tan (e+f x))^{3/2} (A+B \tan (e+f x)+C \tan ^2(e+f x)) \, dx\) [100]

Optimal. Leaf size=187 \[ -\frac {(i A+B-i C) (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}-\frac {(B-i (A-C)) (c+i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 (B c+(A-C) d) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 B (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 d f} \]

[Out]

-(I*A+B-I*C)*(c-I*d)^(3/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/f-(B-I*(A-C))*(c+I*d)^(3/2)*arctanh((
c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/f+2*(B*c+(A-C)*d)*(c+d*tan(f*x+e))^(1/2)/f+2/3*B*(c+d*tan(f*x+e))^(3/2)/f
+2/5*C*(c+d*tan(f*x+e))^(5/2)/d/f

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Rubi [A]
time = 0.29, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3711, 3609, 3620, 3618, 65, 214} \begin {gather*} \frac {2 (d (A-C)+B c) \sqrt {c+d \tan (e+f x)}}{f}-\frac {(c-i d)^{3/2} (i A+B-i C) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}-\frac {(c+i d)^{3/2} (B-i (A-C)) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 B (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 d f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*Tan[e + f*x])^(3/2)*(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2),x]

[Out]

-(((I*A + B - I*C)*(c - I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/f) - ((B - I*(A - C))*(c +
I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/f + (2*(B*c + (A - C)*d)*Sqrt[c + d*Tan[e + f*x]]
)/f + (2*B*(c + d*Tan[e + f*x])^(3/2))/(3*f) + (2*C*(c + d*Tan[e + f*x])^(5/2))/(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 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 (c+d \tan (e+f x))^{3/2} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right ) \, dx &=\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 d f}+\int (A-C+B \tan (e+f x)) (c+d \tan (e+f x))^{3/2} \, dx\\ &=\frac {2 B (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 d f}+\int \sqrt {c+d \tan (e+f x)} (A c-c C-B d+(B c+(A-C) d) \tan (e+f x)) \, dx\\ &=\frac {2 (B c+(A-C) d) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 B (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 d f}+\int \frac {-c^2 C-2 B c d+C d^2+A \left (c^2-d^2\right )+\left (2 c (A-C) d+B \left (c^2-d^2\right )\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx\\ &=\frac {2 (B c+(A-C) d) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 B (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 d f}+\frac {1}{2} \left ((A-i B-C) (c-i d)^2\right ) \int \frac {1+i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx+\frac {1}{2} \left ((A+i B-C) (c+i d)^2\right ) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}} \, dx\\ &=\frac {2 (B c+(A-C) d) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 B (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 d f}+\frac {\left ((i A+B-i C) (c-i d)^2\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 (i (A+i B-C) (c+i d)^2\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 c+(A-C) d) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 B (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 d f}-\frac {\left ((A-i B-C) (c-i d)^2\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 ((A+i B-C) (c+i d)^2\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-i C) (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f}-\frac {(B-i (A-C)) (c+i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{f}+\frac {2 (B c+(A-C) d) \sqrt {c+d \tan (e+f x)}}{f}+\frac {2 B (c+d \tan (e+f x))^{3/2}}{3 f}+\frac {2 C (c+d \tan (e+f x))^{5/2}}{5 d f}\\ \end {align*}

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Mathematica [A]
time = 0.87, size = 202, normalized size = 1.08 \begin {gather*} \frac {\frac {6 C (c+d \tan (e+f x))^{5/2}}{d}+5 (i A+B-i C) \left (-3 (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+\sqrt {c+d \tan (e+f x)} (4 c-3 i d+d \tan (e+f x))\right )+5 (-i A+B+i C) \left (-3 (c+i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )+\sqrt {c+d \tan (e+f x)} (4 c+3 i d+d \tan (e+f x))\right )}{15 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*Tan[e + f*x])^(3/2)*(A + B*Tan[e + f*x] + C*Tan[e + f*x]^2),x]

[Out]

((6*C*(c + d*Tan[e + f*x])^(5/2))/d + 5*(I*A + B - I*C)*(-3*(c - I*d)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/S
qrt[c - I*d]] + Sqrt[c + d*Tan[e + f*x]]*(4*c - (3*I)*d + d*Tan[e + f*x])) + 5*((-I)*A + B + I*C)*(-3*(c + I*d
)^(3/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]] + Sqrt[c + d*Tan[e + f*x]]*(4*c + (3*I)*d + d*Tan[e +
f*x])))/(15*f)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1292\) vs. \(2(158)=316\).
time = 0.50, size = 1293, normalized size = 6.91 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x,method=_RETURNVERBOSE)

[Out]

2/f/d*(1/5*C*(c+d*tan(f*x+e))^(5/2)+1/3*B*d*(c+d*tan(f*x+e))^(3/2)+A*d^2*(c+d*tan(f*x+e))^(1/2)+B*c*d*(c+d*tan
(f*x+e))^(1/2)-C*d^2*(c+d*tan(f*x+e))^(1/2)-d*(1/4/d*(-1/2*(A*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*c-
A*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2+A*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*d^2-B*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d
^2)^(1/2)*d+2*B*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c*d-C*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*c+C*(2*(c^2+
d^2)^(1/2)+2*c)^(1/2)*c^2-C*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*d^2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*
c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))+2*(-2*A*(c^2+d^2)^(1/2)*d^2-2*B*(c^2+d^2)^(1/2)*c*d+2*C*(c^2+d^2)^(1/
2)*d^2+1/2*(A*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*c-A*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2+A*(2*(c^2+d^
2)^(1/2)+2*c)^(1/2)*d^2-B*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*d+2*B*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c*
d-C*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*c+C*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2-C*(2*(c^2+d^2)^(1/2)+2
*c)^(1/2)*d^2)*(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^2+d^2)^(1/2)+2*c)^(1
/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)))+1/4/d*(1/2*(A*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2
+d^2)^(1/2)*c-A*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2+A*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*d^2-B*(2*(c^2+d^2)^(1/2)+2*c
)^(1/2)*(c^2+d^2)^(1/2)*d+2*B*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c*d-C*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2
)*c+C*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2-C*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*d^2)*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*(2*A*(c^2+d^2)^(1/2)*d^2+2*B*(c^2+d^2)^(1/2)*c*d-2*C*
(c^2+d^2)^(1/2)*d^2-1/2*(A*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*c-A*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2
+A*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*d^2-B*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*d+2*B*(2*(c^2+d^2)^(1/2)+
2*c)^(1/2)*c*d-C*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*(c^2+d^2)^(1/2)*c+C*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*c^2-C*(2*(c^2
+d^2)^(1/2)+2*c)^(1/2)*d^2)*(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))^(3/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2),x, algorithm="maxima")

[Out]

integrate((C*tan(f*x + e)^2 + B*tan(f*x + e) + A)*(d*tan(f*x + e) + c)^(3/2), 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))^(3/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^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 \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \left (A + B \tan {\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))**(3/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)**2),x)

[Out]

Integral((c + d*tan(e + f*x))**(3/2)*(A + B*tan(e + f*x) + C*tan(e + f*x)**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((c+d*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^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 = 44.87, size = 2500, normalized size = 13.37 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*tan(e + f*x))^(3/2)*(A + B*tan(e + f*x) + C*tan(e + f*x)^2),x)

[Out]

((2*C*c^2)/(d*f) - (2*C*(d^3*f + c^2*d*f))/(d^2*f^2))*(c + d*tan(e + f*x))^(1/2) - log((((16*c*d^2*(((-B^4*d^2
*f^4*(3*c^2 - d^2)^2)^(1/2) + B^2*c^3*f^2 - 3*B^2*c*d^2*f^2)/f^4)^(1/2)*(B*c^2 + B*d^2 + f*(((-B^4*d^2*f^4*(3*
c^2 - d^2)^2)^(1/2) + B^2*c^3*f^2 - 3*B^2*c*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/f - (16*B^2*d^2*(
c + d*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6*c^2*d^2))/f^2)*(((-B^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + B^2*c^3*f^2 -
3*B^2*c*d^2*f^2)/f^4)^(1/2))/2 - (8*B^3*d^2*(c^2 - d^2)*(c^2 + d^2)^2)/f^3)*(((6*B^4*c^2*d^4*f^4 - B^4*d^6*f^
4 - 9*B^4*c^4*d^2*f^4)^(1/2) + B^2*c^3*f^2 - 3*B^2*c*d^2*f^2)/(4*f^4))^(1/2) - log((((16*c*d^2*(-((-B^4*d^2*f^
4*(3*c^2 - d^2)^2)^(1/2) - B^2*c^3*f^2 + 3*B^2*c*d^2*f^2)/f^4)^(1/2)*(B*c^2 + B*d^2 + f*(-((-B^4*d^2*f^4*(3*c^
2 - d^2)^2)^(1/2) - B^2*c^3*f^2 + 3*B^2*c*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/f - (16*B^2*d^2*(c
+ d*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6*c^2*d^2))/f^2)*(-((-B^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - B^2*c^3*f^2 +
3*B^2*c*d^2*f^2)/f^4)^(1/2))/2 - (8*B^3*d^2*(c^2 - d^2)*(c^2 + d^2)^2)/f^3)*(-((6*B^4*c^2*d^4*f^4 - B^4*d^6*f^
4 - 9*B^4*c^4*d^2*f^4)^(1/2) - B^2*c^3*f^2 + 3*B^2*c*d^2*f^2)/(4*f^4))^(1/2) + log((((16*c*d^2*(((-B^4*d^2*f^4
*(3*c^2 - d^2)^2)^(1/2) + B^2*c^3*f^2 - 3*B^2*c*d^2*f^2)/f^4)^(1/2)*(B*c^2 + B*d^2 - f*(((-B^4*d^2*f^4*(3*c^2
- d^2)^2)^(1/2) + B^2*c^3*f^2 - 3*B^2*c*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/f + (16*B^2*d^2*(c +
d*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6*c^2*d^2))/f^2)*(((-B^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + B^2*c^3*f^2 - 3*B
^2*c*d^2*f^2)/f^4)^(1/2))/2 - (8*B^3*d^2*(c^2 - d^2)*(c^2 + d^2)^2)/f^3)*((6*B^4*c^2*d^4*f^4 - B^4*d^6*f^4 - 9
*B^4*c^4*d^2*f^4)^(1/2)/(4*f^4) + (B^2*c^3)/(4*f^2) - (3*B^2*c*d^2)/(4*f^2))^(1/2) + log((((16*c*d^2*(-((-B^4*
d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - B^2*c^3*f^2 + 3*B^2*c*d^2*f^2)/f^4)^(1/2)*(B*c^2 + B*d^2 - f*(-((-B^4*d^2*f^4
*(3*c^2 - d^2)^2)^(1/2) - B^2*c^3*f^2 + 3*B^2*c*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/f + (16*B^2*d
^2*(c + d*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6*c^2*d^2))/f^2)*(-((-B^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - B^2*c^3*
f^2 + 3*B^2*c*d^2*f^2)/f^4)^(1/2))/2 - (8*B^3*d^2*(c^2 - d^2)*(c^2 + d^2)^2)/f^3)*((B^2*c^3)/(4*f^2) - (6*B^4*
c^2*d^4*f^4 - B^4*d^6*f^4 - 9*B^4*c^4*d^2*f^4)^(1/2)/(4*f^4) - (3*B^2*c*d^2)/(4*f^2))^(1/2) - log((((16*d^2*(-
((-A^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + A^2*c^3*f^2 - 3*A^2*c*d^2*f^2)/f^4)^(1/2)*(A*d^3 + A*c^2*d + c*f*(-((-
A^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + A^2*c^3*f^2 - 3*A^2*c*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/f
+ (16*A^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6*c^2*d^2))/f^2)*(-((-A^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2)
+ A^2*c^3*f^2 - 3*A^2*c*d^2*f^2)/f^4)^(1/2))/2 - (16*A^3*c*d^3*(c^2 + d^2)^2)/f^3)*(-((6*A^4*c^2*d^4*f^4 - A^
4*d^6*f^4 - 9*A^4*c^4*d^2*f^4)^(1/2) + A^2*c^3*f^2 - 3*A^2*c*d^2*f^2)/(4*f^4))^(1/2) - log((((16*d^2*(((-A^4*d
^2*f^4*(3*c^2 - d^2)^2)^(1/2) - A^2*c^3*f^2 + 3*A^2*c*d^2*f^2)/f^4)^(1/2)*(A*d^3 + A*c^2*d + c*f*(((-A^4*d^2*f
^4*(3*c^2 - d^2)^2)^(1/2) - A^2*c^3*f^2 + 3*A^2*c*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/f + (16*A^2
*d^2*(c + d*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6*c^2*d^2))/f^2)*(((-A^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - A^2*c^3
*f^2 + 3*A^2*c*d^2*f^2)/f^4)^(1/2))/2 - (16*A^3*c*d^3*(c^2 + d^2)^2)/f^3)*(((6*A^4*c^2*d^4*f^4 - A^4*d^6*f^4 -
9*A^4*c^4*d^2*f^4)^(1/2) - A^2*c^3*f^2 + 3*A^2*c*d^2*f^2)/(4*f^4))^(1/2) + log((((16*d^2*(((-A^4*d^2*f^4*(3*c
^2 - d^2)^2)^(1/2) - A^2*c^3*f^2 + 3*A^2*c*d^2*f^2)/f^4)^(1/2)*(A*d^3 + A*c^2*d - c*f*(((-A^4*d^2*f^4*(3*c^2 -
d^2)^2)^(1/2) - A^2*c^3*f^2 + 3*A^2*c*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/f - (16*A^2*d^2*(c + d
*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6*c^2*d^2))/f^2)*(((-A^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - A^2*c^3*f^2 + 3*A^
2*c*d^2*f^2)/f^4)^(1/2))/2 - (16*A^3*c*d^3*(c^2 + d^2)^2)/f^3)*((6*A^4*c^2*d^4*f^4 - A^4*d^6*f^4 - 9*A^4*c^4*d
^2*f^4)^(1/2)/(4*f^4) - (A^2*c^3)/(4*f^2) + (3*A^2*c*d^2)/(4*f^2))^(1/2) + log((((16*d^2*(-((-A^4*d^2*f^4*(3*c
^2 - d^2)^2)^(1/2) + A^2*c^3*f^2 - 3*A^2*c*d^2*f^2)/f^4)^(1/2)*(A*d^3 + A*c^2*d - c*f*(-((-A^4*d^2*f^4*(3*c^2
- d^2)^2)^(1/2) + A^2*c^3*f^2 - 3*A^2*c*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e + f*x))^(1/2)))/f - (16*A^2*d^2*(c +
d*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6*c^2*d^2))/f^2)*(-((-A^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + A^2*c^3*f^2 - 3*
A^2*c*d^2*f^2)/f^4)^(1/2))/2 - (16*A^3*c*d^3*(c^2 + d^2)^2)/f^3)*((3*A^2*c*d^2)/(4*f^2) - (A^2*c^3)/(4*f^2) -
(6*A^4*c^2*d^4*f^4 - A^4*d^6*f^4 - 9*A^4*c^4*d^2*f^4)^(1/2)/(4*f^4))^(1/2) - log((16*C^3*c*d^3*(c^2 + d^2)^2)/
f^3 - (((16*d^2*(-((-C^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + C^2*c^3*f^2 - 3*C^2*c*d^2*f^2)/f^4)^(1/2)*(C*d^3 + C
*c^2*d - c*f*(-((-C^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + C^2*c^3*f^2 - 3*C^2*c*d^2*f^2)/f^4)^(1/2)*(c + d*tan(e
+ f*x))^(1/2)))/f - (16*C^2*d^2*(c + d*tan(e + f*x))^(1/2)*(c^4 + d^4 - 6*c^2*d^2))/f^2)*(-((-C^4*d^2*f^4*(3*c
^2 - d^2)^2)^(1/2) + C^2*c^3*f^2 - 3*C^2*c*d^2*f^2)/f^4)^(1/2))/2)*(-((6*C^4*c^2*d^4*f^4 - C^4*d^6*f^4 - 9*C^4
*c^4*d^2*f^4)^(1/2) + C^2*c^3*f^2 - 3*C^2*c*d^2...

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