3.2.67 \(\int \sin ^2(e+f x) (a+b \sin (e+f x))^3 \, dx\) [167]

Optimal. Leaf size=160 \[ \frac {1}{8} a \left (4 a^2+9 b^2\right ) x-\frac {b \left (15 a^2+4 b^2\right ) \cos (e+f x)}{5 f}+\frac {b \left (15 a^2+4 b^2\right ) \cos ^3(e+f x)}{15 f}-\frac {a \left (4 a^2+9 b^2\right ) \cos (e+f x) \sin (e+f x)}{8 f}-\frac {11 a b^2 \cos (e+f x) \sin ^3(e+f x)}{20 f}-\frac {b^2 \cos (e+f x) \sin ^3(e+f x) (a+b \sin (e+f x))}{5 f} \]

[Out]

1/8*a*(4*a^2+9*b^2)*x-1/5*b*(15*a^2+4*b^2)*cos(f*x+e)/f+1/15*b*(15*a^2+4*b^2)*cos(f*x+e)^3/f-1/8*a*(4*a^2+9*b^
2)*cos(f*x+e)*sin(f*x+e)/f-11/20*a*b^2*cos(f*x+e)*sin(f*x+e)^3/f-1/5*b^2*cos(f*x+e)*sin(f*x+e)^3*(a+b*sin(f*x+
e))/f

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Rubi [A]
time = 0.16, antiderivative size = 180, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2870, 2832, 2813} \begin {gather*} \frac {\left (3 a^2-16 b^2\right ) \cos (e+f x) (a+b \sin (e+f x))^2}{60 b f}+\frac {a \left (6 a^2-71 b^2\right ) \sin (e+f x) \cos (e+f x)}{120 f}+\frac {1}{8} a x \left (4 a^2+9 b^2\right )+\frac {\left (3 a^4-52 a^2 b^2-16 b^4\right ) \cos (e+f x)}{30 b f}-\frac {\cos (e+f x) (a+b \sin (e+f x))^4}{5 b f}+\frac {a \cos (e+f x) (a+b \sin (e+f x))^3}{20 b f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^2*(a + b*Sin[e + f*x])^3,x]

[Out]

(a*(4*a^2 + 9*b^2)*x)/8 + ((3*a^4 - 52*a^2*b^2 - 16*b^4)*Cos[e + f*x])/(30*b*f) + (a*(6*a^2 - 71*b^2)*Cos[e +
f*x]*Sin[e + f*x])/(120*f) + ((3*a^2 - 16*b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^2)/(60*b*f) + (a*Cos[e + f*x]
*(a + b*Sin[e + f*x])^3)/(20*b*f) - (Cos[e + f*x]*(a + b*Sin[e + f*x])^4)/(5*b*f)

Rule 2813

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(2*a*c +
b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Cos[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rule 2832

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d
)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(f*(m + 1))), x] + Dist[1/(m + 1), Int[(a + b*Sin[e + f*x])^(m - 1)*Sim
p[b*d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], 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] && IntegerQ[2*m]

Rule 2870

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

Rubi steps

\begin {align*} \int \sin ^2(e+f x) (a+b \sin (e+f x))^3 \, dx &=-\frac {\cos (e+f x) (a+b \sin (e+f x))^4}{5 b f}+\frac {\int (4 b-a \sin (e+f x)) (a+b \sin (e+f x))^3 \, dx}{5 b}\\ &=\frac {a \cos (e+f x) (a+b \sin (e+f x))^3}{20 b f}-\frac {\cos (e+f x) (a+b \sin (e+f x))^4}{5 b f}+\frac {\int (a+b \sin (e+f x))^2 \left (13 a b-\left (3 a^2-16 b^2\right ) \sin (e+f x)\right ) \, dx}{20 b}\\ &=\frac {\left (3 a^2-16 b^2\right ) \cos (e+f x) (a+b \sin (e+f x))^2}{60 b f}+\frac {a \cos (e+f x) (a+b \sin (e+f x))^3}{20 b f}-\frac {\cos (e+f x) (a+b \sin (e+f x))^4}{5 b f}+\frac {\int (a+b \sin (e+f x)) \left (b \left (33 a^2+32 b^2\right )-a \left (6 a^2-71 b^2\right ) \sin (e+f x)\right ) \, dx}{60 b}\\ &=\frac {1}{8} a \left (4 a^2+9 b^2\right ) x+\frac {\left (3 a^4-52 a^2 b^2-16 b^4\right ) \cos (e+f x)}{30 b f}+\frac {a \left (6 a^2-71 b^2\right ) \cos (e+f x) \sin (e+f x)}{120 f}+\frac {\left (3 a^2-16 b^2\right ) \cos (e+f x) (a+b \sin (e+f x))^2}{60 b f}+\frac {a \cos (e+f x) (a+b \sin (e+f x))^3}{20 b f}-\frac {\cos (e+f x) (a+b \sin (e+f x))^4}{5 b f}\\ \end {align*}

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Mathematica [A]
time = 0.45, size = 117, normalized size = 0.73 \begin {gather*} \frac {-60 b \left (18 a^2+5 b^2\right ) \cos (e+f x)+10 \left (12 a^2 b+5 b^3\right ) \cos (3 (e+f x))-6 b^3 \cos (5 (e+f x))+15 a \left (4 \left (4 a^2+9 b^2\right ) (e+f x)-8 \left (a^2+3 b^2\right ) \sin (2 (e+f x))+3 b^2 \sin (4 (e+f x))\right )}{480 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^2*(a + b*Sin[e + f*x])^3,x]

[Out]

(-60*b*(18*a^2 + 5*b^2)*Cos[e + f*x] + 10*(12*a^2*b + 5*b^3)*Cos[3*(e + f*x)] - 6*b^3*Cos[5*(e + f*x)] + 15*a*
(4*(4*a^2 + 9*b^2)*(e + f*x) - 8*(a^2 + 3*b^2)*Sin[2*(e + f*x)] + 3*b^2*Sin[4*(e + f*x)]))/(480*f)

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Maple [A]
time = 0.33, size = 124, normalized size = 0.78

method	result	size
derivativedivides	\(\frac {-\frac {b^{3} \left (\frac {8}{3}+\sin ^{4}\left (f x +e \right )+\frac {4 \left (\sin ^{2}\left (f x +e \right )\right )}{3}\right ) \cos \left (f x +e \right )}{5}+3 a \,b^{2} \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-a^{2} b \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )+a^{3} \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f}\)	\(124\)
default	\(\frac {-\frac {b^{3} \left (\frac {8}{3}+\sin ^{4}\left (f x +e \right )+\frac {4 \left (\sin ^{2}\left (f x +e \right )\right )}{3}\right ) \cos \left (f x +e \right )}{5}+3 a \,b^{2} \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-a^{2} b \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )+a^{3} \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f}\)	\(124\)
risch	\(\frac {a^{3} x}{2}+\frac {9 a \,b^{2} x}{8}-\frac {9 b \cos \left (f x +e \right ) a^{2}}{4 f}-\frac {5 b^{3} \cos \left (f x +e \right )}{8 f}-\frac {b^{3} \cos \left (5 f x +5 e \right )}{80 f}+\frac {3 a \,b^{2} \sin \left (4 f x +4 e \right )}{32 f}+\frac {b \cos \left (3 f x +3 e \right ) a^{2}}{4 f}+\frac {5 b^{3} \cos \left (3 f x +3 e \right )}{48 f}-\frac {\sin \left (2 f x +2 e \right ) a^{3}}{4 f}-\frac {3 \sin \left (2 f x +2 e \right ) a \,b^{2}}{4 f}\)	\(149\)
norman	\(\frac {-\frac {60 a^{2} b +16 b^{3}}{15 f}+\frac {a \left (4 a^{2}+9 b^{2}\right ) x}{8}-\frac {12 a^{2} b \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}-\frac {2 \left (42 a^{2} b +16 b^{3}\right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f}-\frac {\left (60 a^{2} b +16 b^{3}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f}-\frac {a \left (4 a^{2}+9 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{4 f}+\frac {a \left (4 a^{2}+9 b^{2}\right ) \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4 f}+\frac {5 a \left (4 a^{2}+9 b^{2}\right ) x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{8}+\frac {5 a \left (4 a^{2}+9 b^{2}\right ) x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4}+\frac {5 a \left (4 a^{2}+9 b^{2}\right ) x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4}+\frac {5 a \left (4 a^{2}+9 b^{2}\right ) x \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{8}+\frac {a \left (4 a^{2}+9 b^{2}\right ) x \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{8}-\frac {a \left (4 a^{2}+21 b^{2}\right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 f}+\frac {a \left (4 a^{2}+21 b^{2}\right ) \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{5}}\)	\(366\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^2*(a+b*sin(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

1/f*(-1/5*b^3*(8/3+sin(f*x+e)^4+4/3*sin(f*x+e)^2)*cos(f*x+e)+3*a*b^2*(-1/4*(sin(f*x+e)^3+3/2*sin(f*x+e))*cos(f
*x+e)+3/8*f*x+3/8*e)-a^2*b*(2+sin(f*x+e)^2)*cos(f*x+e)+a^3*(-1/2*cos(f*x+e)*sin(f*x+e)+1/2*f*x+1/2*e))

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Maxima [A]
time = 0.28, size = 131, normalized size = 0.82 \begin {gather*} \frac {120 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{3} + 480 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{2} b + 45 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} a b^{2} - 32 \, {\left (3 \, \cos \left (f x + e\right )^{5} - 10 \, \cos \left (f x + e\right )^{3} + 15 \, \cos \left (f x + e\right )\right )} b^{3}}{480 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^2*(a+b*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

1/480*(120*(2*f*x + 2*e - sin(2*f*x + 2*e))*a^3 + 480*(cos(f*x + e)^3 - 3*cos(f*x + e))*a^2*b + 45*(12*f*x + 1
2*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*a*b^2 - 32*(3*cos(f*x + e)^5 - 10*cos(f*x + e)^3 + 15*cos(f*x + e
))*b^3)/f

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Fricas [A]
time = 0.45, size = 124, normalized size = 0.78 \begin {gather*} -\frac {24 \, b^{3} \cos \left (f x + e\right )^{5} - 40 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left (4 \, a^{3} + 9 \, a b^{2}\right )} f x + 120 \, {\left (3 \, a^{2} b + b^{3}\right )} \cos \left (f x + e\right ) - 15 \, {\left (6 \, a b^{2} \cos \left (f x + e\right )^{3} - {\left (4 \, a^{3} + 15 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{120 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^2*(a+b*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

-1/120*(24*b^3*cos(f*x + e)^5 - 40*(3*a^2*b + 2*b^3)*cos(f*x + e)^3 - 15*(4*a^3 + 9*a*b^2)*f*x + 120*(3*a^2*b
+ b^3)*cos(f*x + e) - 15*(6*a*b^2*cos(f*x + e)^3 - (4*a^3 + 15*a*b^2)*cos(f*x + e))*sin(f*x + e))/f

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Sympy [A]
time = 0.37, size = 284, normalized size = 1.78 \begin {gather*} \begin {cases} \frac {a^{3} x \sin ^{2}{\left (e + f x \right )}}{2} + \frac {a^{3} x \cos ^{2}{\left (e + f x \right )}}{2} - \frac {a^{3} \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} - \frac {3 a^{2} b \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {2 a^{2} b \cos ^{3}{\left (e + f x \right )}}{f} + \frac {9 a b^{2} x \sin ^{4}{\left (e + f x \right )}}{8} + \frac {9 a b^{2} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac {9 a b^{2} x \cos ^{4}{\left (e + f x \right )}}{8} - \frac {15 a b^{2} \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{8 f} - \frac {9 a b^{2} \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} - \frac {b^{3} \sin ^{4}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {4 b^{3} \sin ^{2}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac {8 b^{3} \cos ^{5}{\left (e + f x \right )}}{15 f} & \text {for}\: f \neq 0 \\x \left (a + b \sin {\left (e \right )}\right )^{3} \sin ^{2}{\left (e \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)**2*(a+b*sin(f*x+e))**3,x)

[Out]

Piecewise((a**3*x*sin(e + f*x)**2/2 + a**3*x*cos(e + f*x)**2/2 - a**3*sin(e + f*x)*cos(e + f*x)/(2*f) - 3*a**2
*b*sin(e + f*x)**2*cos(e + f*x)/f - 2*a**2*b*cos(e + f*x)**3/f + 9*a*b**2*x*sin(e + f*x)**4/8 + 9*a*b**2*x*sin
(e + f*x)**2*cos(e + f*x)**2/4 + 9*a*b**2*x*cos(e + f*x)**4/8 - 15*a*b**2*sin(e + f*x)**3*cos(e + f*x)/(8*f) -
9*a*b**2*sin(e + f*x)*cos(e + f*x)**3/(8*f) - b**3*sin(e + f*x)**4*cos(e + f*x)/f - 4*b**3*sin(e + f*x)**2*co
s(e + f*x)**3/(3*f) - 8*b**3*cos(e + f*x)**5/(15*f), Ne(f, 0)), (x*(a + b*sin(e))**3*sin(e)**2, True))

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Giac [A]
time = 0.46, size = 129, normalized size = 0.81 \begin {gather*} -\frac {b^{3} \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} + \frac {3 \, a b^{2} \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} + \frac {1}{8} \, {\left (4 \, a^{3} + 9 \, a b^{2}\right )} x + \frac {{\left (12 \, a^{2} b + 5 \, b^{3}\right )} \cos \left (3 \, f x + 3 \, e\right )}{48 \, f} - \frac {{\left (18 \, a^{2} b + 5 \, b^{3}\right )} \cos \left (f x + e\right )}{8 \, f} - \frac {{\left (a^{3} + 3 \, a b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^2*(a+b*sin(f*x+e))^3,x, algorithm="giac")

[Out]

-1/80*b^3*cos(5*f*x + 5*e)/f + 3/32*a*b^2*sin(4*f*x + 4*e)/f + 1/8*(4*a^3 + 9*a*b^2)*x + 1/48*(12*a^2*b + 5*b^
3)*cos(3*f*x + 3*e)/f - 1/8*(18*a^2*b + 5*b^3)*cos(f*x + e)/f - 1/4*(a^3 + 3*a*b^2)*sin(2*f*x + 2*e)/f

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Mupad [B]
time = 8.17, size = 328, normalized size = 2.05 \begin {gather*} \frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (4\,a^2+9\,b^2\right )}{4\,\left (a^3+\frac {9\,a\,b^2}{4}\right )}\right )\,\left (4\,a^2+9\,b^2\right )}{4\,f}-\frac {4\,a^2\,b-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9\,\left (a^3+\frac {9\,a\,b^2}{4}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (2\,a^3+\frac {21\,a\,b^2}{2}\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7\,\left (2\,a^3+\frac {21\,a\,b^2}{2}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (20\,a^2\,b+\frac {16\,b^3}{3}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (28\,a^2\,b+\frac {32\,b^3}{3}\right )+\frac {16\,b^3}{15}+\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (a^3+\frac {9\,a\,b^2}{4}\right )+12\,a^2\,b\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}-\frac {a\,\left (4\,a^2+9\,b^2\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )-\frac {f\,x}{2}\right )}{4\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^2*(a + b*sin(e + f*x))^3,x)

[Out]

(a*atan((a*tan(e/2 + (f*x)/2)*(4*a^2 + 9*b^2))/(4*((9*a*b^2)/4 + a^3)))*(4*a^2 + 9*b^2))/(4*f) - (4*a^2*b - ta
n(e/2 + (f*x)/2)^9*((9*a*b^2)/4 + a^3) + tan(e/2 + (f*x)/2)^3*((21*a*b^2)/2 + 2*a^3) - tan(e/2 + (f*x)/2)^7*((
21*a*b^2)/2 + 2*a^3) + tan(e/2 + (f*x)/2)^2*(20*a^2*b + (16*b^3)/3) + tan(e/2 + (f*x)/2)^4*(28*a^2*b + (32*b^3
)/3) + (16*b^3)/15 + tan(e/2 + (f*x)/2)*((9*a*b^2)/4 + a^3) + 12*a^2*b*tan(e/2 + (f*x)/2)^6)/(f*(5*tan(e/2 + (
f*x)/2)^2 + 10*tan(e/2 + (f*x)/2)^4 + 10*tan(e/2 + (f*x)/2)^6 + 5*tan(e/2 + (f*x)/2)^8 + tan(e/2 + (f*x)/2)^10
+ 1)) - (a*(4*a^2 + 9*b^2)*(atan(tan(e/2 + (f*x)/2)) - (f*x)/2))/(4*f)

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