3.81 \(\int \frac{\sin ^3(e+f x)}{(a+b \tan ^2(e+f x))^3} \, dx\)

Optimal. Leaf size=180 \[ \frac{\cos ^3(e+f x)}{3 f (a-b)^3}-\frac{(a+2 b) \cos (e+f x)}{f (a-b)^4}-\frac{b (7 a+4 b) \sec (e+f x)}{8 f (a-b)^4 \left (a+b \sec ^2(e+f x)-b\right )}-\frac{a b \sec (e+f x)}{4 f (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )^2}-\frac{5 \sqrt{b} (3 a+4 b) \tan ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a-b}}\right )}{8 f (a-b)^{9/2}} \]

[Out]

(-5*Sqrt[b]*(3*a + 4*b)*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/(8*(a - b)^(9/2)*f) - ((a + 2*b)*Cos[e + f
*x])/((a - b)^4*f) + Cos[e + f*x]^3/(3*(a - b)^3*f) - (a*b*Sec[e + f*x])/(4*(a - b)^3*f*(a - b + b*Sec[e + f*x
]^2)^2) - (b*(7*a + 4*b)*Sec[e + f*x])/(8*(a - b)^4*f*(a - b + b*Sec[e + f*x]^2))

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Rubi [A]  time = 0.246349, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3664, 456, 1259, 1261, 205} \[ \frac{\cos ^3(e+f x)}{3 f (a-b)^3}-\frac{(a+2 b) \cos (e+f x)}{f (a-b)^4}-\frac{b (7 a+4 b) \sec (e+f x)}{8 f (a-b)^4 \left (a+b \sec ^2(e+f x)-b\right )}-\frac{a b \sec (e+f x)}{4 f (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )^2}-\frac{5 \sqrt{b} (3 a+4 b) \tan ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a-b}}\right )}{8 f (a-b)^{9/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*Sqrt[b]*(3*a + 4*b)*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/(8*(a - b)^(9/2)*f) - ((a + 2*b)*Cos[e + f
*x])/((a - b)^4*f) + Cos[e + f*x]^3/(3*(a - b)^3*f) - (a*b*Sec[e + f*x])/(4*(a - b)^3*f*(a - b + b*Sec[e + f*x
]^2)^2) - (b*(7*a + 4*b)*Sec[e + f*x])/(8*(a - b)^4*f*(a - b + b*Sec[e + f*x]^2))

Rule 3664

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a - b + b*ff^2*x^2)^p)/x^(
m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 456

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1259

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((-d)^(
m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*(d + e*x^2)^(q + 1))/(2*e^(2*p + m/2)*(q + 1)), x] + Dist[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*(-d)^(-(m/2) + 1)*e^(2*p)*(q + 1)*
(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2)))/(d + e*x^2)], x], x]
, x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]

Rule 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{-1+x^2}{x^4 \left (a-b+b x^2\right )^3} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{a b \sec (e+f x)}{4 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac{b \operatorname{Subst}\left (\int \frac{\frac{4}{(a-b) b}-\frac{4 a x^2}{(a-b)^2 b}+\frac{3 a x^4}{(a-b)^3}}{x^4 \left (a-b+b x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{4 f}\\ &=-\frac{a b \sec (e+f x)}{4 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac{b (7 a+4 b) \sec (e+f x)}{8 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \frac{8 (a-b) b-8 b (a+b) x^2+\frac{b^2 (7 a+4 b) x^4}{a-b}}{x^4 \left (a-b+b x^2\right )} \, dx,x,\sec (e+f x)\right )}{8 (a-b)^3 b f}\\ &=-\frac{a b \sec (e+f x)}{4 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac{b (7 a+4 b) \sec (e+f x)}{8 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{\operatorname{Subst}\left (\int \left (\frac{8 b}{x^4}-\frac{8 b (a+2 b)}{(a-b) x^2}+\frac{5 b^2 (3 a+4 b)}{(a-b) \left (a-b+b x^2\right )}\right ) \, dx,x,\sec (e+f x)\right )}{8 (a-b)^3 b f}\\ &=-\frac{(a+2 b) \cos (e+f x)}{(a-b)^4 f}+\frac{\cos ^3(e+f x)}{3 (a-b)^3 f}-\frac{a b \sec (e+f x)}{4 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac{b (7 a+4 b) \sec (e+f x)}{8 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac{(5 b (3 a+4 b)) \operatorname{Subst}\left (\int \frac{1}{a-b+b x^2} \, dx,x,\sec (e+f x)\right )}{8 (a-b)^4 f}\\ &=-\frac{5 \sqrt{b} (3 a+4 b) \tan ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a-b}}\right )}{8 (a-b)^{9/2} f}-\frac{(a+2 b) \cos (e+f x)}{(a-b)^4 f}+\frac{\cos ^3(e+f x)}{3 (a-b)^3 f}-\frac{a b \sec (e+f x)}{4 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac{b (7 a+4 b) \sec (e+f x)}{8 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 5.43852, size = 230, normalized size = 1.28 \[ \frac{\frac{2 \left (3 \cos (e+f x) \left (a \left (\frac{4 b^2}{((a-b) \cos (2 (e+f x))+a+b)^2}-\frac{9 b}{(a-b) \cos (2 (e+f x))+a+b}-3\right )+b \left (-\frac{4 b}{(a-b) \cos (2 (e+f x))+a+b}-9\right )\right )+(a-b) \cos (3 (e+f x))\right )}{(a-b)^4}+\frac{15 \sqrt{b} (3 a+4 b) \tan ^{-1}\left (\frac{\sqrt{a-b}-\sqrt{a} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{b}}\right )}{(a-b)^{9/2}}+\frac{15 \sqrt{b} (3 a+4 b) \tan ^{-1}\left (\frac{\sqrt{a-b}+\sqrt{a} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{b}}\right )}{(a-b)^{9/2}}}{24 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((15*Sqrt[b]*(3*a + 4*b)*ArcTan[(Sqrt[a - b] - Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]])/(a - b)^(9/2) + (15*Sqrt[b]
*(3*a + 4*b)*ArcTan[(Sqrt[a - b] + Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]])/(a - b)^(9/2) + (2*(3*Cos[e + f*x]*(a*(
-3 + (4*b^2)/(a + b + (a - b)*Cos[2*(e + f*x)])^2 - (9*b)/(a + b + (a - b)*Cos[2*(e + f*x)])) + b*(-9 - (4*b)/
(a + b + (a - b)*Cos[2*(e + f*x)]))) + (a - b)*Cos[3*(e + f*x)]))/(a - b)^4)/(24*f)

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Maple [B]  time = 0.082, size = 504, normalized size = 2.8 \begin{align*}{\frac{a \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{3\,f \left ({a}^{3}-3\,{a}^{2}b+3\,a{b}^{2}-{b}^{3} \right ) \left ( a-b \right ) }}-{\frac{b \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{3\,f \left ({a}^{3}-3\,{a}^{2}b+3\,a{b}^{2}-{b}^{3} \right ) \left ( a-b \right ) }}-{\frac{\cos \left ( fx+e \right ) a}{f \left ({a}^{3}-3\,{a}^{2}b+3\,a{b}^{2}-{b}^{3} \right ) \left ( a-b \right ) }}-2\,{\frac{\cos \left ( fx+e \right ) b}{f \left ({a}^{3}-3\,{a}^{2}b+3\,a{b}^{2}-{b}^{3} \right ) \left ( a-b \right ) }}-{\frac{9\,{a}^{2}b \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{8\,f \left ( a-b \right ) ^{4} \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) ^{2}}}+{\frac{5\,{b}^{2} \left ( \cos \left ( fx+e \right ) \right ) ^{3}a}{8\,f \left ( a-b \right ) ^{4} \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) ^{2}}}+{\frac{{b}^{3} \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{2\,f \left ( a-b \right ) ^{4} \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) ^{2}}}-{\frac{7\,a{b}^{2}\cos \left ( fx+e \right ) }{8\,f \left ( a-b \right ) ^{4} \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) ^{2}}}-{\frac{{b}^{3}\cos \left ( fx+e \right ) }{2\,f \left ( a-b \right ) ^{4} \left ( a \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}b+b \right ) ^{2}}}+{\frac{15\,ab}{8\,f \left ( a-b \right ) ^{4}}\arctan \left ({ \left ( a-b \right ) \cos \left ( fx+e \right ){\frac{1}{\sqrt{b \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{b \left ( a-b \right ) }}}}+{\frac{5\,{b}^{2}}{2\,f \left ( a-b \right ) ^{4}}\arctan \left ({ \left ( a-b \right ) \cos \left ( fx+e \right ){\frac{1}{\sqrt{b \left ( a-b \right ) }}}} \right ){\frac{1}{\sqrt{b \left ( a-b \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/f/(a^3-3*a^2*b+3*a*b^2-b^3)/(a-b)*a*cos(f*x+e)^3-1/3/f/(a^3-3*a^2*b+3*a*b^2-b^3)/(a-b)*b*cos(f*x+e)^3-1/f/
(a^3-3*a^2*b+3*a*b^2-b^3)/(a-b)*a*cos(f*x+e)-2/f/(a^3-3*a^2*b+3*a*b^2-b^3)/(a-b)*cos(f*x+e)*b-9/8/f*b/(a-b)^4/
(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^2*a^2*cos(f*x+e)^3+5/8/f*b^2/(a-b)^4/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^2*cos
(f*x+e)^3*a+1/2/f*b^3/(a-b)^4/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^2*cos(f*x+e)^3-7/8/f*b^2/(a-b)^4/(a*cos(f*x+e)
^2-cos(f*x+e)^2*b+b)^2*a*cos(f*x+e)-1/2/f*b^3/(a-b)^4/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^2*cos(f*x+e)+15/8/f*b/
(a-b)^4/(b*(a-b))^(1/2)*arctan((a-b)*cos(f*x+e)/(b*(a-b))^(1/2))*a+5/2/f*b^2/(a-b)^4/(b*(a-b))^(1/2)*arctan((a
-b)*cos(f*x+e)/(b*(a-b))^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.96241, size = 1723, normalized size = 9.57 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(16*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^7 - 16*(3*a^3 - 2*a^2*b - 5*a*b^2 + 4*b^3)*cos(f*x + e)
^5 - 50*(3*a^2*b + a*b^2 - 4*b^3)*cos(f*x + e)^3 + 15*((3*a^3 - 2*a^2*b - 5*a*b^2 + 4*b^3)*cos(f*x + e)^4 + 3*
a*b^2 + 4*b^3 + 2*(3*a^2*b + a*b^2 - 4*b^3)*cos(f*x + e)^2)*sqrt(-b/(a - b))*log(((a - b)*cos(f*x + e)^2 + 2*(
a - b)*sqrt(-b/(a - b))*cos(f*x + e) - b)/((a - b)*cos(f*x + e)^2 + b)) - 30*(3*a*b^2 + 4*b^3)*cos(f*x + e))/(
(a^6 - 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^3 + 15*a^2*b^4 - 6*a*b^5 + b^6)*f*cos(f*x + e)^4 + 2*(a^5*b - 5*a^4*b^2
+ 10*a^3*b^3 - 10*a^2*b^4 + 5*a*b^5 - b^6)*f*cos(f*x + e)^2 + (a^4*b^2 - 4*a^3*b^3 + 6*a^2*b^4 - 4*a*b^5 + b^
6)*f), 1/24*(8*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^7 - 8*(3*a^3 - 2*a^2*b - 5*a*b^2 + 4*b^3)*cos(f*x
+ e)^5 - 25*(3*a^2*b + a*b^2 - 4*b^3)*cos(f*x + e)^3 - 15*((3*a^3 - 2*a^2*b - 5*a*b^2 + 4*b^3)*cos(f*x + e)^4
+ 3*a*b^2 + 4*b^3 + 2*(3*a^2*b + a*b^2 - 4*b^3)*cos(f*x + e)^2)*sqrt(b/(a - b))*arctan(-(a - b)*sqrt(b/(a - b)
)*cos(f*x + e)/b) - 15*(3*a*b^2 + 4*b^3)*cos(f*x + e))/((a^6 - 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^3 + 15*a^2*b^4
- 6*a*b^5 + b^6)*f*cos(f*x + e)^4 + 2*(a^5*b - 5*a^4*b^2 + 10*a^3*b^3 - 10*a^2*b^4 + 5*a*b^5 - b^6)*f*cos(f*x
+ e)^2 + (a^4*b^2 - 4*a^3*b^3 + 6*a^2*b^4 - 4*a*b^5 + b^6)*f)]

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Sympy [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(sin(f*x+e)**3/(a+b*tan(f*x+e)**2)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.60447, size = 760, normalized size = 4.22 \begin{align*} \frac{a^{6} f^{17} \cos \left (f x + e\right )^{3} - 6 \, a^{5} b f^{17} \cos \left (f x + e\right )^{3} + 15 \, a^{4} b^{2} f^{17} \cos \left (f x + e\right )^{3} - 20 \, a^{3} b^{3} f^{17} \cos \left (f x + e\right )^{3} + 15 \, a^{2} b^{4} f^{17} \cos \left (f x + e\right )^{3} - 6 \, a b^{5} f^{17} \cos \left (f x + e\right )^{3} + b^{6} f^{17} \cos \left (f x + e\right )^{3} - 3 \, a^{6} f^{17} \cos \left (f x + e\right ) + 9 \, a^{5} b f^{17} \cos \left (f x + e\right ) - 30 \, a^{3} b^{3} f^{17} \cos \left (f x + e\right ) + 45 \, a^{2} b^{4} f^{17} \cos \left (f x + e\right ) - 27 \, a b^{5} f^{17} \cos \left (f x + e\right ) + 6 \, b^{6} f^{17} \cos \left (f x + e\right )}{3 \,{\left (a^{9} f^{18} - 9 \, a^{8} b f^{18} + 36 \, a^{7} b^{2} f^{18} - 84 \, a^{6} b^{3} f^{18} + 126 \, a^{5} b^{4} f^{18} - 126 \, a^{4} b^{5} f^{18} + 84 \, a^{3} b^{6} f^{18} - 36 \, a^{2} b^{7} f^{18} + 9 \, a b^{8} f^{18} - b^{9} f^{18}\right )}} + \frac{5 \,{\left (3 \, a b + 4 \, b^{2}\right )} \arctan \left (\frac{a \cos \left (f x + e\right ) - b \cos \left (f x + e\right )}{\sqrt{a b - b^{2}}}\right )}{8 \,{\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} \sqrt{a b - b^{2}} f} - \frac{\frac{9 \, a^{2} b \cos \left (f x + e\right )^{3}}{f} - \frac{5 \, a b^{2} \cos \left (f x + e\right )^{3}}{f} - \frac{4 \, b^{3} \cos \left (f x + e\right )^{3}}{f} + \frac{7 \, a b^{2} \cos \left (f x + e\right )}{f} + \frac{4 \, b^{3} \cos \left (f x + e\right )}{f}}{8 \,{\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )}{\left (a \cos \left (f x + e\right )^{2} - b \cos \left (f x + e\right )^{2} + b\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(a^6*f^17*cos(f*x + e)^3 - 6*a^5*b*f^17*cos(f*x + e)^3 + 15*a^4*b^2*f^17*cos(f*x + e)^3 - 20*a^3*b^3*f^17*
cos(f*x + e)^3 + 15*a^2*b^4*f^17*cos(f*x + e)^3 - 6*a*b^5*f^17*cos(f*x + e)^3 + b^6*f^17*cos(f*x + e)^3 - 3*a^
6*f^17*cos(f*x + e) + 9*a^5*b*f^17*cos(f*x + e) - 30*a^3*b^3*f^17*cos(f*x + e) + 45*a^2*b^4*f^17*cos(f*x + e)
- 27*a*b^5*f^17*cos(f*x + e) + 6*b^6*f^17*cos(f*x + e))/(a^9*f^18 - 9*a^8*b*f^18 + 36*a^7*b^2*f^18 - 84*a^6*b^
3*f^18 + 126*a^5*b^4*f^18 - 126*a^4*b^5*f^18 + 84*a^3*b^6*f^18 - 36*a^2*b^7*f^18 + 9*a*b^8*f^18 - b^9*f^18) +
5/8*(3*a*b + 4*b^2)*arctan((a*cos(f*x + e) - b*cos(f*x + e))/sqrt(a*b - b^2))/((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*
a*b^3 + b^4)*sqrt(a*b - b^2)*f) - 1/8*(9*a^2*b*cos(f*x + e)^3/f - 5*a*b^2*cos(f*x + e)^3/f - 4*b^3*cos(f*x + e
)^3/f + 7*a*b^2*cos(f*x + e)/f + 4*b^3*cos(f*x + e)/f)/((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*(a*cos(f*x
+ e)^2 - b*cos(f*x + e)^2 + b)^2)