\(\int \frac {\sin ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx\) [62]
Optimal result
Integrand size = 23, antiderivative size = 129 \[
\int \frac {\sin ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\frac {\left (3 a^2+6 a b-b^2\right ) x}{8 (a-b)^3}-\frac {a^{3/2} \sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{(a-b)^3 f}-\frac {(5 a-b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f}
\]
[Out]
1/8*(3*a^2+6*a*b-b^2)*x/(a-b)^3-1/8*(5*a-b)*cos(f*x+e)*sin(f*x+e)/(a-b)^2/f+1/4*cos(f*x+e)^3*sin(f*x+e)/(a-b)/
f-a^(3/2)*arctan(b^(1/2)*tan(f*x+e)/a^(1/2))*b^(1/2)/(a-b)^3/f
Rubi [A] (verified)
Time = 0.27 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of
steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3744, 481, 541, 536, 209, 211}
\[
\int \frac {\sin ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {a^{3/2} \sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{f (a-b)^3}+\frac {x \left (3 a^2+6 a b-b^2\right )}{8 (a-b)^3}+\frac {\sin (e+f x) \cos ^3(e+f x)}{4 f (a-b)}-\frac {(5 a-b) \sin (e+f x) \cos (e+f x)}{8 f (a-b)^2}
\]
[In]
Int[Sin[e + f*x]^4/(a + b*Tan[e + f*x]^2),x]
[Out]
((3*a^2 + 6*a*b - b^2)*x)/(8*(a - b)^3) - (a^(3/2)*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/((a - b)^3*
f) - ((5*a - b)*Cos[e + f*x]*Sin[e + f*x])/(8*(a - b)^2*f) + (Cos[e + f*x]^3*Sin[e + f*x])/(4*(a - b)*f)
Rule 209
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rule 211
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rule 481
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-a)*e^(
2*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*n*(b*c - a*d)*(p + 1))), x] + Dist[e^
(2*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1)
+ (a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0]
&& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
Rule 536
Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]
Rule 541
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[(
-(b*e - a*f))*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*n*(b*c - a*d)*(p + 1))), x] + Dist[1/(a*n*(b*c - a
*d)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*
f)*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]
Rule 3744
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[c*(ff^(m + 1)/f), Subst[Int[x^m*((a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2)
^(m/2 + 1)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2]
Rubi steps \begin{align*}
\text {integral}& = \frac {\text {Subst}\left (\int \frac {x^4}{\left (1+x^2\right )^3 \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f}-\frac {\text {Subst}\left (\int \frac {a+(-4 a+b) x^2}{\left (1+x^2\right )^2 \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{4 (a-b) f} \\ & = -\frac {(5 a-b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f}+\frac {\text {Subst}\left (\int \frac {a (3 a+b)-(5 a-b) b x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{8 (a-b)^2 f} \\ & = -\frac {(5 a-b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f}-\frac {\left (a^2 b\right ) \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{(a-b)^3 f}+\frac {\left (3 a^2+6 a b-b^2\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{8 (a-b)^3 f} \\ & = \frac {\left (3 a^2+6 a b-b^2\right ) x}{8 (a-b)^3}-\frac {a^{3/2} \sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{(a-b)^3 f}-\frac {(5 a-b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f} \\
\end{align*}
Mathematica [A] (verified)
Time = 0.72 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.77
\[
\int \frac {\sin ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\frac {4 \left (3 a^2+6 a b-b^2\right ) (e+f x)-32 a^{3/2} \sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )-8 a (a-b) \sin (2 (e+f x))+(a-b)^2 \sin (4 (e+f x))}{32 (a-b)^3 f}
\]
[In]
Integrate[Sin[e + f*x]^4/(a + b*Tan[e + f*x]^2),x]
[Out]
(4*(3*a^2 + 6*a*b - b^2)*(e + f*x) - 32*a^(3/2)*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]] - 8*a*(a - b)*S
in[2*(e + f*x)] + (a - b)^2*Sin[4*(e + f*x)])/(32*(a - b)^3*f)
Maple [A] (verified)
Time = 5.17 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.02
| | |
method | result | size |
| | |
derivativedivides |
\(\frac {-\frac {a^{2} b \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{\left (a -b \right )^{3} \sqrt {a b}}+\frac {\frac {\left (-\frac {5}{8} a^{2}+\frac {3}{4} a b -\frac {1}{8} b^{2}\right ) \tan \left (f x +e \right )^{3}+\left (-\frac {3}{8} a^{2}+\frac {1}{4} a b +\frac {1}{8} b^{2}\right ) \tan \left (f x +e \right )}{\left (1+\tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}+6 a b -b^{2}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{8}}{\left (a -b \right )^{3}}}{f}\) |
\(131\) |
default |
\(\frac {-\frac {a^{2} b \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{\left (a -b \right )^{3} \sqrt {a b}}+\frac {\frac {\left (-\frac {5}{8} a^{2}+\frac {3}{4} a b -\frac {1}{8} b^{2}\right ) \tan \left (f x +e \right )^{3}+\left (-\frac {3}{8} a^{2}+\frac {1}{4} a b +\frac {1}{8} b^{2}\right ) \tan \left (f x +e \right )}{\left (1+\tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}+6 a b -b^{2}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{8}}{\left (a -b \right )^{3}}}{f}\) |
\(131\) |
risch |
\(\frac {3 x \,a^{2}}{8 \left (a -b \right )^{3}}+\frac {3 x a b}{4 \left (a -b \right )^{3}}-\frac {x \,b^{2}}{8 \left (a -b \right )^{3}}+\frac {i a \,{\mathrm e}^{2 i \left (f x +e \right )}}{8 \left (a -b \right )^{2} f}-\frac {i a \,{\mathrm e}^{-2 i \left (f x +e \right )}}{8 \left (a^{2}-2 a b +b^{2}\right ) f}+\frac {\sqrt {-a b}\, a \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right )}{2 \left (a -b \right )^{3} f}-\frac {\sqrt {-a b}\, a \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right )}{2 \left (a -b \right )^{3} f}+\frac {\sin \left (4 f x +4 e \right )}{32 \left (a -b \right ) f}\) |
\(218\) |
| | |
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[In]
int(sin(f*x+e)^4/(a+b*tan(f*x+e)^2),x,method=_RETURNVERBOSE)
[Out]
1/f*(-a^2*b/(a-b)^3/(a*b)^(1/2)*arctan(b*tan(f*x+e)/(a*b)^(1/2))+1/(a-b)^3*(((-5/8*a^2+3/4*a*b-1/8*b^2)*tan(f*
x+e)^3+(-3/8*a^2+1/4*a*b+1/8*b^2)*tan(f*x+e))/(1+tan(f*x+e)^2)^2+1/8*(3*a^2+6*a*b-b^2)*arctan(tan(f*x+e))))
Fricas [A] (verification not implemented)
none
Time = 0.32 (sec) , antiderivative size = 383, normalized size of antiderivative = 2.97
\[
\int \frac {\sin ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\left [\frac {{\left (3 \, a^{2} + 6 \, a b - b^{2}\right )} f x - 2 \, \sqrt {-a b} a \log \left (\frac {{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a + b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt {-a b} \sin \left (f x + e\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}}\right ) + {\left (2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (5 \, a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} f}, \frac {{\left (3 \, a^{2} + 6 \, a b - b^{2}\right )} f x + 4 \, \sqrt {a b} a \arctan \left (\frac {{\left ({\left (a + b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {a b}}{2 \, a b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) + {\left (2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (5 \, a^{2} - 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} f}\right ]
\]
[In]
integrate(sin(f*x+e)^4/(a+b*tan(f*x+e)^2),x, algorithm="fricas")
[Out]
[1/8*((3*a^2 + 6*a*b - b^2)*f*x - 2*sqrt(-a*b)*a*log(((a^2 + 6*a*b + b^2)*cos(f*x + e)^4 - 2*(3*a*b + b^2)*cos
(f*x + e)^2 - 4*((a + b)*cos(f*x + e)^3 - b*cos(f*x + e))*sqrt(-a*b)*sin(f*x + e) + b^2)/((a^2 - 2*a*b + b^2)*
cos(f*x + e)^4 + 2*(a*b - b^2)*cos(f*x + e)^2 + b^2)) + (2*(a^2 - 2*a*b + b^2)*cos(f*x + e)^3 - (5*a^2 - 6*a*b
+ b^2)*cos(f*x + e))*sin(f*x + e))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*f), 1/8*((3*a^2 + 6*a*b - b^2)*f*x + 4*sq
rt(a*b)*a*arctan(1/2*((a + b)*cos(f*x + e)^2 - b)*sqrt(a*b)/(a*b*cos(f*x + e)*sin(f*x + e))) + (2*(a^2 - 2*a*b
+ b^2)*cos(f*x + e)^3 - (5*a^2 - 6*a*b + b^2)*cos(f*x + e))*sin(f*x + e))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*f)
]
Sympy [F(-1)]
Timed out. \[
\int \frac {\sin ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\text {Timed out}
\]
[In]
integrate(sin(f*x+e)**4/(a+b*tan(f*x+e)**2),x)
[Out]
Timed out
Maxima [A] (verification not implemented)
none
Time = 0.31 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.42
\[
\int \frac {\sin ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {\frac {8 \, a^{2} b \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt {a b}} - \frac {{\left (3 \, a^{2} + 6 \, a b - b^{2}\right )} {\left (f x + e\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (5 \, a - b\right )} \tan \left (f x + e\right )^{3} + {\left (3 \, a + b\right )} \tan \left (f x + e\right )}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} - 2 \, a b + b^{2}}}{8 \, f}
\]
[In]
integrate(sin(f*x+e)^4/(a+b*tan(f*x+e)^2),x, algorithm="maxima")
[Out]
-1/8*(8*a^2*b*arctan(b*tan(f*x + e)/sqrt(a*b))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*sqrt(a*b)) - (3*a^2 + 6*a*b -
b^2)*(f*x + e)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + ((5*a - b)*tan(f*x + e)^3 + (3*a + b)*tan(f*x + e))/((a^2 - 2
*a*b + b^2)*tan(f*x + e)^4 + 2*(a^2 - 2*a*b + b^2)*tan(f*x + e)^2 + a^2 - 2*a*b + b^2))/f
Giac [A] (verification not implemented)
none
Time = 0.50 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.41
\[
\int \frac {\sin ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {\frac {8 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )\right )} a^{2} b}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt {a b}} - \frac {{\left (3 \, a^{2} + 6 \, a b - b^{2}\right )} {\left (f x + e\right )}}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {5 \, a \tan \left (f x + e\right )^{3} - b \tan \left (f x + e\right )^{3} + 3 \, a \tan \left (f x + e\right ) + b \tan \left (f x + e\right )}{{\left (a^{2} - 2 \, a b + b^{2}\right )} {\left (\tan \left (f x + e\right )^{2} + 1\right )}^{2}}}{8 \, f}
\]
[In]
integrate(sin(f*x+e)^4/(a+b*tan(f*x+e)^2),x, algorithm="giac")
[Out]
-1/8*(8*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b)))*a^2*b/((a^3 - 3*a^2*b + 3*a*b
^2 - b^3)*sqrt(a*b)) - (3*a^2 + 6*a*b - b^2)*(f*x + e)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + (5*a*tan(f*x + e)^3 -
b*tan(f*x + e)^3 + 3*a*tan(f*x + e) + b*tan(f*x + e))/((a^2 - 2*a*b + b^2)*(tan(f*x + e)^2 + 1)^2))/f
Mupad [B] (verification not implemented)
Time = 13.78 (sec) , antiderivative size = 3588, normalized size of antiderivative = 27.81
\[
\int \frac {\sin ^4(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\text {Too large to display}
\]
[In]
int(sin(e + f*x)^4/(a + b*tan(e + f*x)^2),x)
[Out]
(atan(((((tan(e + f*x)*(b^7 - 12*a*b^6 + 30*a^2*b^5 + 36*a^3*b^4 + 73*a^4*b^3))/(32*(a^4 - 4*a^3*b - 4*a*b^3 +
b^4 + 6*a^2*b^2)) + (((32*a*b^9 - 96*a^2*b^8 - 96*a^3*b^7 + 800*a^4*b^6 - 1440*a^5*b^5 + 1248*a^6*b^4 - 544*a
^7*b^3 + 96*a^8*b^2)/(64*(a^6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*a^2*b^4 - 20*a^3*b^3 + 15*a^4*b^2)) - (tan(e + f*
x)*(-a^3*b)^(1/2)*(1280*a*b^8 - 256*b^9 - 2304*a^2*b^7 + 1280*a^3*b^6 + 1280*a^4*b^5 - 2304*a^5*b^4 + 1280*a^6
*b^3 - 256*a^7*b^2))/(64*(3*a*b^2 - 3*a^2*b + a^3 - b^3)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)))*(-a^3*b
)^(1/2))/(2*(3*a*b^2 - 3*a^2*b + a^3 - b^3)))*(-a^3*b)^(1/2)*1i)/(2*(3*a*b^2 - 3*a^2*b + a^3 - b^3)) + (((tan(
e + f*x)*(b^7 - 12*a*b^6 + 30*a^2*b^5 + 36*a^3*b^4 + 73*a^4*b^3))/(32*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b
^2)) - (((32*a*b^9 - 96*a^2*b^8 - 96*a^3*b^7 + 800*a^4*b^6 - 1440*a^5*b^5 + 1248*a^6*b^4 - 544*a^7*b^3 + 96*a^
8*b^2)/(64*(a^6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*a^2*b^4 - 20*a^3*b^3 + 15*a^4*b^2)) + (tan(e + f*x)*(-a^3*b)^(1
/2)*(1280*a*b^8 - 256*b^9 - 2304*a^2*b^7 + 1280*a^3*b^6 + 1280*a^4*b^5 - 2304*a^5*b^4 + 1280*a^6*b^3 - 256*a^7
*b^2))/(64*(3*a*b^2 - 3*a^2*b + a^3 - b^3)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)))*(-a^3*b)^(1/2))/(2*(3
*a*b^2 - 3*a^2*b + a^3 - b^3)))*(-a^3*b)^(1/2)*1i)/(2*(3*a*b^2 - 3*a^2*b + a^3 - b^3)))/((a^2*b^6 - 11*a^3*b^5
+ 27*a^4*b^4 + 15*a^5*b^3)/(32*(a^6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*a^2*b^4 - 20*a^3*b^3 + 15*a^4*b^2)) + (((t
an(e + f*x)*(b^7 - 12*a*b^6 + 30*a^2*b^5 + 36*a^3*b^4 + 73*a^4*b^3))/(32*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^
2*b^2)) + (((32*a*b^9 - 96*a^2*b^8 - 96*a^3*b^7 + 800*a^4*b^6 - 1440*a^5*b^5 + 1248*a^6*b^4 - 544*a^7*b^3 + 96
*a^8*b^2)/(64*(a^6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*a^2*b^4 - 20*a^3*b^3 + 15*a^4*b^2)) - (tan(e + f*x)*(-a^3*b)
^(1/2)*(1280*a*b^8 - 256*b^9 - 2304*a^2*b^7 + 1280*a^3*b^6 + 1280*a^4*b^5 - 2304*a^5*b^4 + 1280*a^6*b^3 - 256*
a^7*b^2))/(64*(3*a*b^2 - 3*a^2*b + a^3 - b^3)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)))*(-a^3*b)^(1/2))/(2
*(3*a*b^2 - 3*a^2*b + a^3 - b^3)))*(-a^3*b)^(1/2))/(2*(3*a*b^2 - 3*a^2*b + a^3 - b^3)) - (((tan(e + f*x)*(b^7
- 12*a*b^6 + 30*a^2*b^5 + 36*a^3*b^4 + 73*a^4*b^3))/(32*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)) - (((32*a
*b^9 - 96*a^2*b^8 - 96*a^3*b^7 + 800*a^4*b^6 - 1440*a^5*b^5 + 1248*a^6*b^4 - 544*a^7*b^3 + 96*a^8*b^2)/(64*(a^
6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*a^2*b^4 - 20*a^3*b^3 + 15*a^4*b^2)) + (tan(e + f*x)*(-a^3*b)^(1/2)*(1280*a*b^
8 - 256*b^9 - 2304*a^2*b^7 + 1280*a^3*b^6 + 1280*a^4*b^5 - 2304*a^5*b^4 + 1280*a^6*b^3 - 256*a^7*b^2))/(64*(3*
a*b^2 - 3*a^2*b + a^3 - b^3)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)))*(-a^3*b)^(1/2))/(2*(3*a*b^2 - 3*a^2
*b + a^3 - b^3)))*(-a^3*b)^(1/2))/(2*(3*a*b^2 - 3*a^2*b + a^3 - b^3))))*(-a^3*b)^(1/2)*1i)/(f*(3*a*b^2 - 3*a^2
*b + a^3 - b^3)) - ((tan(e + f*x)^3*(5*a - b))/(8*(a^2 - 2*a*b + b^2)) + (tan(e + f*x)*(3*a + b))/(8*(a^2 - 2*
a*b + b^2)))/(f*(2*tan(e + f*x)^2 + tan(e + f*x)^4 + 1)) + (atan(((((tan(e + f*x)*(b^7 - 12*a*b^6 + 30*a^2*b^5
+ 36*a^3*b^4 + 73*a^4*b^3))/(32*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)) + (((32*a*b^9 - 96*a^2*b^8 - 96*
a^3*b^7 + 800*a^4*b^6 - 1440*a^5*b^5 + 1248*a^6*b^4 - 544*a^7*b^3 + 96*a^8*b^2)/(64*(a^6 - 6*a^5*b - 6*a*b^5 +
b^6 + 15*a^2*b^4 - 20*a^3*b^3 + 15*a^4*b^2)) - (tan(e + f*x)*(6*a*b + 3*a^2 - b^2)*(1280*a*b^8 - 256*b^9 - 23
04*a^2*b^7 + 1280*a^3*b^6 + 1280*a^4*b^5 - 2304*a^5*b^4 + 1280*a^6*b^3 - 256*a^7*b^2))/(512*(a*b^2*3i - a^2*b*
3i + a^3*1i - b^3*1i)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)))*(6*a*b + 3*a^2 - b^2))/(16*(a*b^2*3i - a^2
*b*3i + a^3*1i - b^3*1i)))*(6*a*b + 3*a^2 - b^2)*1i)/(16*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)) + (((tan(e +
f*x)*(b^7 - 12*a*b^6 + 30*a^2*b^5 + 36*a^3*b^4 + 73*a^4*b^3))/(32*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)
) - (((32*a*b^9 - 96*a^2*b^8 - 96*a^3*b^7 + 800*a^4*b^6 - 1440*a^5*b^5 + 1248*a^6*b^4 - 544*a^7*b^3 + 96*a^8*b
^2)/(64*(a^6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*a^2*b^4 - 20*a^3*b^3 + 15*a^4*b^2)) + (tan(e + f*x)*(6*a*b + 3*a^2
- b^2)*(1280*a*b^8 - 256*b^9 - 2304*a^2*b^7 + 1280*a^3*b^6 + 1280*a^4*b^5 - 2304*a^5*b^4 + 1280*a^6*b^3 - 256
*a^7*b^2))/(512*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)))*(6*a*b +
3*a^2 - b^2))/(16*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)))*(6*a*b + 3*a^2 - b^2)*1i)/(16*(a*b^2*3i - a^2*b*3
i + a^3*1i - b^3*1i)))/((a^2*b^6 - 11*a^3*b^5 + 27*a^4*b^4 + 15*a^5*b^3)/(32*(a^6 - 6*a^5*b - 6*a*b^5 + b^6 +
15*a^2*b^4 - 20*a^3*b^3 + 15*a^4*b^2)) + (((tan(e + f*x)*(b^7 - 12*a*b^6 + 30*a^2*b^5 + 36*a^3*b^4 + 73*a^4*b^
3))/(32*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)) + (((32*a*b^9 - 96*a^2*b^8 - 96*a^3*b^7 + 800*a^4*b^6 - 1
440*a^5*b^5 + 1248*a^6*b^4 - 544*a^7*b^3 + 96*a^8*b^2)/(64*(a^6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*a^2*b^4 - 20*a^
3*b^3 + 15*a^4*b^2)) - (tan(e + f*x)*(6*a*b + 3*a^2 - b^2)*(1280*a*b^8 - 256*b^9 - 2304*a^2*b^7 + 1280*a^3*b^6
+ 1280*a^4*b^5 - 2304*a^5*b^4 + 1280*a^6*b^3 - 256*a^7*b^2))/(512*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)*(a^
4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)))*(6*a*b + 3*a^2 - b^2))/(16*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i))
)*(6*a*b + 3*a^2 - b^2))/(16*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i)) - (((tan(e + f*x)*(b^7 - 12*a*b^6 + 30*a
^2*b^5 + 36*a^3*b^4 + 73*a^4*b^3))/(32*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)) - (((32*a*b^9 - 96*a^2*b^8
- 96*a^3*b^7 + 800*a^4*b^6 - 1440*a^5*b^5 + 1248*a^6*b^4 - 544*a^7*b^3 + 96*a^8*b^2)/(64*(a^6 - 6*a^5*b - 6*a
*b^5 + b^6 + 15*a^2*b^4 - 20*a^3*b^3 + 15*a^4*b^2)) + (tan(e + f*x)*(6*a*b + 3*a^2 - b^2)*(1280*a*b^8 - 256*b^
9 - 2304*a^2*b^7 + 1280*a^3*b^6 + 1280*a^4*b^5 - 2304*a^5*b^4 + 1280*a^6*b^3 - 256*a^7*b^2))/(512*(a*b^2*3i -
a^2*b*3i + a^3*1i - b^3*1i)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)))*(6*a*b + 3*a^2 - b^2))/(16*(a*b^2*3i
- a^2*b*3i + a^3*1i - b^3*1i)))*(6*a*b + 3*a^2 - b^2))/(16*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i))))*(6*a*b
+ 3*a^2 - b^2)*1i)/(8*f*(a*b^2*3i - a^2*b*3i + a^3*1i - b^3*1i))