3.103 \(\int \frac{\tan ^4(e+f x)}{\sqrt{a+a \sin (e+f x)}} \, dx\)

Optimal. Leaf size=150 \[ \frac{\tan ^3(e+f x)}{3 f \sqrt{a \sin (e+f x)+a}}+\frac{a \sin (e+f x) \tan (e+f x)}{24 f (a \sin (e+f x)+a)^{3/2}}-\frac{(127 \sin (e+f x)+53) \sec (e+f x)}{192 f \sqrt{a \sin (e+f x)+a}}-\frac{67 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{64 \sqrt{2} \sqrt{a} f} \]

[Out]

(-67*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(64*Sqrt[2]*Sqrt[a]*f) - (Sec[e + f*x
]*(53 + 127*Sin[e + f*x]))/(192*f*Sqrt[a + a*Sin[e + f*x]]) + (a*Sin[e + f*x]*Tan[e + f*x])/(24*f*(a + a*Sin[e
+ f*x])^(3/2)) + Tan[e + f*x]^3/(3*f*Sqrt[a + a*Sin[e + f*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.933351, antiderivative size = 241, normalized size of antiderivative = 1.61, number of steps used = 17, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {2714, 2649, 206, 4401, 2687, 2681, 2650, 2877, 2855} \[ \frac{61 a \cos (e+f x)}{64 f (a \sin (e+f x)+a)^{3/2}}+\frac{7 \sec ^3(e+f x) \sqrt{a \sin (e+f x)+a}}{12 a f}-\frac{5 \sec ^3(e+f x)}{6 f \sqrt{a \sin (e+f x)+a}}-\frac{61 \sec (e+f x)}{48 f \sqrt{a \sin (e+f x)+a}}+\frac{7 a \sec (e+f x)}{24 f (a \sin (e+f x)+a)^{3/2}}-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{\sqrt{a} f}+\frac{61 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{64 \sqrt{2} \sqrt{a} f} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^4/Sqrt[a + a*Sin[e + f*x]],x]

[Out]

(61*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(64*Sqrt[2]*Sqrt[a]*f) - (Sqrt[2]*ArcT
anh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(Sqrt[a]*f) + (61*a*Cos[e + f*x])/(64*f*(a + a
*Sin[e + f*x])^(3/2)) + (7*a*Sec[e + f*x])/(24*f*(a + a*Sin[e + f*x])^(3/2)) - (61*Sec[e + f*x])/(48*f*Sqrt[a
+ a*Sin[e + f*x]]) - (5*Sec[e + f*x]^3)/(6*f*Sqrt[a + a*Sin[e + f*x]]) + (7*Sec[e + f*x]^3*Sqrt[a + a*Sin[e +
f*x]])/(12*a*f)

Rule 2714

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^4, x_Symbol] :> Int[(a + b*Sin[e + f*x
])^m, x] - Int[((a + b*Sin[e + f*x])^m*(1 - 2*Sin[e + f*x]^2))/Cos[e + f*x]^4, x] /; FreeQ[{a, b, e, f, m}, x]
&& EqQ[a^2 - b^2, 0] && IntegerQ[m - 1/2]

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 4401

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

Rule 2687

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> -Simp[(b*(g*
Cos[e + f*x])^(p + 1))/(a*f*g*(p + 1)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(a*(2*p + 1))/(2*g^2*(p + 1)), Int[
(g*Cos[e + f*x])^(p + 2)/(a + b*Sin[e + f*x])^(3/2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& LtQ[p, -1] && IntegerQ[2*p]

Rule 2681

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(2*m + p + 1)), x] + Dist[(m + p + 1)/(a*(2*m + p + 1)),
Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^
2, 0] && LtQ[m, -1] && NeQ[2*m + p + 1, 0] && IntegersQ[2*m, 2*p]

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2877

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[(b*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(2*m + p + 1)), x] - Dist[1/(a^
2*(2*m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1)*(a*m - b*(2*m + p + 1)*Sin[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && LeQ[m, -2^(-1)] && NeQ[2*m + p + 1, 0]

Rule 2855

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[((b*c + a*d)*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(p +
1)), x] + Dist[(b*(a*d*m + b*c*(m + p + 1)))/(a*g^2*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x]
)^(m - 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, -1] && LtQ[p, -1]

Rubi steps

\begin{align*} \int \frac{\tan ^4(e+f x)}{\sqrt{a+a \sin (e+f x)}} \, dx &=\int \frac{1}{\sqrt{a+a \sin (e+f x)}} \, dx-\int \frac{\sec ^4(e+f x) \left (1-2 \sin ^2(e+f x)\right )}{\sqrt{a+a \sin (e+f x)}} \, dx\\ &=-\frac{2 \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cos (e+f x)}{\sqrt{a+a \sin (e+f x)}}\right )}{f}-\int \left (\frac{\sec ^4(e+f x)}{\sqrt{a (1+\sin (e+f x))}}-\frac{2 \sec ^2(e+f x) \tan ^2(e+f x)}{\sqrt{a (1+\sin (e+f x))}}\right ) \, dx\\ &=-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a+a \sin (e+f x)}}\right )}{\sqrt{a} f}+2 \int \frac{\sec ^2(e+f x) \tan ^2(e+f x)}{\sqrt{a (1+\sin (e+f x))}} \, dx-\int \frac{\sec ^4(e+f x)}{\sqrt{a (1+\sin (e+f x))}} \, dx\\ &=-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a+a \sin (e+f x)}}\right )}{\sqrt{a} f}-\frac{5 \sec ^3(e+f x)}{6 f \sqrt{a+a \sin (e+f x)}}+\frac{\int \sec ^4(e+f x) \sqrt{a+a \sin (e+f x)} \left (-\frac{a}{2}+4 a \sin (e+f x)\right ) \, dx}{2 a^2}-\frac{1}{6} (7 a) \int \frac{\sec ^2(e+f x)}{(a+a \sin (e+f x))^{3/2}} \, dx\\ &=-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a+a \sin (e+f x)}}\right )}{\sqrt{a} f}+\frac{7 a \sec (e+f x)}{24 f (a+a \sin (e+f x))^{3/2}}-\frac{5 \sec ^3(e+f x)}{6 f \sqrt{a+a \sin (e+f x)}}+\frac{7 \sec ^3(e+f x) \sqrt{a+a \sin (e+f x)}}{12 a f}-\frac{13}{24} \int \frac{\sec ^2(e+f x)}{\sqrt{a+a \sin (e+f x)}} \, dx-\frac{35}{48} \int \frac{\sec ^2(e+f x)}{\sqrt{a+a \sin (e+f x)}} \, dx\\ &=-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a+a \sin (e+f x)}}\right )}{\sqrt{a} f}+\frac{7 a \sec (e+f x)}{24 f (a+a \sin (e+f x))^{3/2}}-\frac{61 \sec (e+f x)}{48 f \sqrt{a+a \sin (e+f x)}}-\frac{5 \sec ^3(e+f x)}{6 f \sqrt{a+a \sin (e+f x)}}+\frac{7 \sec ^3(e+f x) \sqrt{a+a \sin (e+f x)}}{12 a f}-\frac{1}{16} (13 a) \int \frac{1}{(a+a \sin (e+f x))^{3/2}} \, dx-\frac{1}{32} (35 a) \int \frac{1}{(a+a \sin (e+f x))^{3/2}} \, dx\\ &=-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a+a \sin (e+f x)}}\right )}{\sqrt{a} f}+\frac{61 a \cos (e+f x)}{64 f (a+a \sin (e+f x))^{3/2}}+\frac{7 a \sec (e+f x)}{24 f (a+a \sin (e+f x))^{3/2}}-\frac{61 \sec (e+f x)}{48 f \sqrt{a+a \sin (e+f x)}}-\frac{5 \sec ^3(e+f x)}{6 f \sqrt{a+a \sin (e+f x)}}+\frac{7 \sec ^3(e+f x) \sqrt{a+a \sin (e+f x)}}{12 a f}-\frac{13}{64} \int \frac{1}{\sqrt{a+a \sin (e+f x)}} \, dx-\frac{35}{128} \int \frac{1}{\sqrt{a+a \sin (e+f x)}} \, dx\\ &=-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a+a \sin (e+f x)}}\right )}{\sqrt{a} f}+\frac{61 a \cos (e+f x)}{64 f (a+a \sin (e+f x))^{3/2}}+\frac{7 a \sec (e+f x)}{24 f (a+a \sin (e+f x))^{3/2}}-\frac{61 \sec (e+f x)}{48 f \sqrt{a+a \sin (e+f x)}}-\frac{5 \sec ^3(e+f x)}{6 f \sqrt{a+a \sin (e+f x)}}+\frac{7 \sec ^3(e+f x) \sqrt{a+a \sin (e+f x)}}{12 a f}+\frac{13 \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cos (e+f x)}{\sqrt{a+a \sin (e+f x)}}\right )}{32 f}+\frac{35 \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cos (e+f x)}{\sqrt{a+a \sin (e+f x)}}\right )}{64 f}\\ &=\frac{61 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a+a \sin (e+f x)}}\right )}{64 \sqrt{2} \sqrt{a} f}-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a+a \sin (e+f x)}}\right )}{\sqrt{a} f}+\frac{61 a \cos (e+f x)}{64 f (a+a \sin (e+f x))^{3/2}}+\frac{7 a \sec (e+f x)}{24 f (a+a \sin (e+f x))^{3/2}}-\frac{61 \sec (e+f x)}{48 f \sqrt{a+a \sin (e+f x)}}-\frac{5 \sec ^3(e+f x)}{6 f \sqrt{a+a \sin (e+f x)}}+\frac{7 \sec ^3(e+f x) \sqrt{a+a \sin (e+f x)}}{12 a f}\\ \end{align*}

Mathematica [C]  time = 0.673361, size = 118, normalized size = 0.79 \[ \frac{-\sec ^3(e+f x) (-41 \sin (e+f x)+183 \sin (3 (e+f x))+122 \cos (2 (e+f x))+90)+(804+804 i) (-1)^{3/4} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (e+f x)\right )-1\right )\right )}{768 f \sqrt{a (\sin (e+f x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]^4/Sqrt[a + a*Sin[e + f*x]],x]

[Out]

((804 + 804*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] + Sin[(e +
f*x)/2]) - Sec[e + f*x]^3*(90 + 122*Cos[2*(e + f*x)] - 41*Sin[e + f*x] + 183*Sin[3*(e + f*x)]))/(768*f*Sqrt[a
*(1 + Sin[e + f*x])])

________________________________________________________________________________________

Maple [A]  time = 0.766, size = 231, normalized size = 1.5 \begin{align*}{\frac{1}{ \left ( -384+384\,\sin \left ( fx+e \right ) \right ) \left ( 1+\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) f} \left ( 366\,{a}^{7/2}\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+ \left ( 402\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2} \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}-112\,{a}^{7/2} \right ) \sin \left ( fx+e \right ) + \left ( -201\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2} \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}+122\,{a}^{7/2} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+402\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{a}}} \right ){a}^{2} \left ( a-a\sin \left ( fx+e \right ) \right ) ^{3/2}-16\,{a}^{7/2} \right ){a}^{-{\frac{7}{2}}}{\frac{1}{\sqrt{a+a\sin \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(f*x+e)^4/(a+a*sin(f*x+e))^(1/2),x)

[Out]

1/384*(366*a^(7/2)*sin(f*x+e)*cos(f*x+e)^2+(402*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^
2*(a-a*sin(f*x+e))^(3/2)-112*a^(7/2))*sin(f*x+e)+(-201*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1
/2))*a^2*(a-a*sin(f*x+e))^(3/2)+122*a^(7/2))*cos(f*x+e)^2+402*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/
2)/a^(1/2))*a^2*(a-a*sin(f*x+e))^(3/2)-16*a^(7/2))/a^(7/2)/(-1+sin(f*x+e))/(1+sin(f*x+e))/cos(f*x+e)/(a+a*sin(
f*x+e))^(1/2)/f

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^4/(a+a*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [A]  time = 1.63851, size = 624, normalized size = 4.16 \begin{align*} \frac{201 \, \sqrt{2}{\left (\cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + \cos \left (f x + e\right )^{3}\right )} \sqrt{a} \log \left (-\frac{a \cos \left (f x + e\right )^{2} - 2 \, \sqrt{2} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{a}{\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )} + 3 \, a \cos \left (f x + e\right ) -{\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\cos \left (f x + e\right )^{2} -{\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \,{\left (61 \, \cos \left (f x + e\right )^{2} +{\left (183 \, \cos \left (f x + e\right )^{2} - 56\right )} \sin \left (f x + e\right ) - 8\right )} \sqrt{a \sin \left (f x + e\right ) + a}}{768 \,{\left (a f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + a f \cos \left (f x + e\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^4/(a+a*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

1/768*(201*sqrt(2)*(cos(f*x + e)^3*sin(f*x + e) + cos(f*x + e)^3)*sqrt(a)*log(-(a*cos(f*x + e)^2 - 2*sqrt(2)*s
qrt(a*sin(f*x + e) + a)*sqrt(a)*(cos(f*x + e) - sin(f*x + e) + 1) + 3*a*cos(f*x + e) - (a*cos(f*x + e) - 2*a)*
sin(f*x + e) + 2*a)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) - 4*(61*cos(f*x + e
)^2 + (183*cos(f*x + e)^2 - 56)*sin(f*x + e) - 8)*sqrt(a*sin(f*x + e) + a))/(a*f*cos(f*x + e)^3*sin(f*x + e) +
a*f*cos(f*x + e)^3)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{4}{\left (e + f x \right )}}{\sqrt{a \left (\sin{\left (e + f x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)**4/(a+a*sin(f*x+e))**(1/2),x)

[Out]

Integral(tan(e + f*x)**4/sqrt(a*(sin(e + f*x) + 1)), x)

________________________________________________________________________________________

Giac [B]  time = 5.88912, size = 1056, normalized size = 7.04 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^4/(a+a*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

1/192*(201*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a) + sq
rt(a))/sqrt(-a))/(sqrt(-a)*sgn(tan(1/2*f*x + 1/2*e) + 1)) - 16*(9*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1
/2*f*x + 1/2*e)^2 + a))^5 - 39*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^4*sqrt(a) -
26*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^3*a + 78*(sqrt(a)*tan(1/2*f*x + 1/2*e)
- sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^2*a^(3/2) + 69*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2
*e)^2 + a))*a^2 + 13*a^(5/2))/(((sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^2 - 2*(sqr
t(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))*sqrt(a) - a)^3*sgn(tan(1/2*f*x + 1/2*e) + 1))
- 6*(43*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^7 + 237*(sqrt(a)*tan(1/2*f*x + 1/2
*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^6*sqrt(a) + 161*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x +
1/2*e)^2 + a))^5*a - 221*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^4*a^(3/2) + 25*(
sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^3*a^2 + 103*(sqrt(a)*tan(1/2*f*x + 1/2*e) -
sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^2*a^(5/2) - 93*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e
)^2 + a))*a^3 + 17*a^(7/2))/(((sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))^2 + 2*(sqrt(
a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))*sqrt(a) - a)^4*sgn(tan(1/2*f*x + 1/2*e) + 1)))/f