3.194 \(\int \frac{\sin ^4(x)}{(a+b \sin (x))^3} \, dx\)

Optimal. Leaf size=179 \[ -\frac{\left (3 a^2-2 b^2\right ) \cos (x)}{2 b^3 \left (a^2-b^2\right )}+\frac{3 a^2 \left (-5 a^2 b^2+2 a^4+4 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{b^4 \left (a^2-b^2\right )^{5/2}}+\frac{a^2 \sin ^2(x) \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac{3 a^3 \left (a^2-2 b^2\right ) \cos (x)}{2 b^3 \left (a^2-b^2\right )^2 (a+b \sin (x))}-\frac{3 a x}{b^4} \]

[Out]

(-3*a*x)/b^4 + (3*a^2*(2*a^4 - 5*a^2*b^2 + 4*b^4)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(b^4*(a^2 - b^2)^(
5/2)) - ((3*a^2 - 2*b^2)*Cos[x])/(2*b^3*(a^2 - b^2)) + (a^2*Cos[x]*Sin[x]^2)/(2*b*(a^2 - b^2)*(a + b*Sin[x])^2
) - (3*a^3*(a^2 - 2*b^2)*Cos[x])/(2*b^3*(a^2 - b^2)^2*(a + b*Sin[x]))

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Rubi [A]  time = 0.411073, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.538, Rules used = {2792, 3031, 3023, 2735, 2660, 618, 204} \[ -\frac{\left (3 a^2-2 b^2\right ) \cos (x)}{2 b^3 \left (a^2-b^2\right )}+\frac{3 a^2 \left (-5 a^2 b^2+2 a^4+4 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{b^4 \left (a^2-b^2\right )^{5/2}}+\frac{a^2 \sin ^2(x) \cos (x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac{3 a^3 \left (a^2-2 b^2\right ) \cos (x)}{2 b^3 \left (a^2-b^2\right )^2 (a+b \sin (x))}-\frac{3 a x}{b^4} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^4/(a + b*Sin[x])^3,x]

[Out]

(-3*a*x)/b^4 + (3*a^2*(2*a^4 - 5*a^2*b^2 + 4*b^4)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(b^4*(a^2 - b^2)^(
5/2)) - ((3*a^2 - 2*b^2)*Cos[x])/(2*b^3*(a^2 - b^2)) + (a^2*Cos[x]*Sin[x]^2)/(2*b*(a^2 - b^2)*(a + b*Sin[x])^2
) - (3*a^3*(a^2 - 2*b^2)*Cos[x])/(2*b^3*(a^2 - b^2)^2*(a + b*Sin[x]))

Rule 2792

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(
d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e +
f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*
d*(a^2 + b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - d^2) - m*(b*c - a*d)^2 + d*
n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3031

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

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*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[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2735

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

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rubi steps

\begin{align*} \int \frac{\sin ^4(x)}{(a+b \sin (x))^3} \, dx &=\frac{a^2 \cos (x) \sin ^2(x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac{\int \frac{\sin (x) \left (2 a^2-2 a b \sin (x)-\left (3 a^2-2 b^2\right ) \sin ^2(x)\right )}{(a+b \sin (x))^2} \, dx}{2 b \left (a^2-b^2\right )}\\ &=\frac{a^2 \cos (x) \sin ^2(x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac{3 a^3 \left (a^2-2 b^2\right ) \cos (x)}{2 b^3 \left (a^2-b^2\right )^2 (a+b \sin (x))}-\frac{\int \frac{3 a^2 b \left (a^2-2 b^2\right )+a \left (3 a^2-4 b^2\right ) \left (a^2-b^2\right ) \sin (x)-b \left (3 a^2-2 b^2\right ) \left (a^2-b^2\right ) \sin ^2(x)}{a+b \sin (x)} \, dx}{2 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac{\left (3 a^2-2 b^2\right ) \cos (x)}{2 b^3 \left (a^2-b^2\right )}+\frac{a^2 \cos (x) \sin ^2(x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac{3 a^3 \left (a^2-2 b^2\right ) \cos (x)}{2 b^3 \left (a^2-b^2\right )^2 (a+b \sin (x))}-\frac{\int \frac{3 a^2 b^2 \left (a^2-2 b^2\right )+6 a b \left (a^2-b^2\right )^2 \sin (x)}{a+b \sin (x)} \, dx}{2 b^4 \left (a^2-b^2\right )^2}\\ &=-\frac{3 a x}{b^4}-\frac{\left (3 a^2-2 b^2\right ) \cos (x)}{2 b^3 \left (a^2-b^2\right )}+\frac{a^2 \cos (x) \sin ^2(x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac{3 a^3 \left (a^2-2 b^2\right ) \cos (x)}{2 b^3 \left (a^2-b^2\right )^2 (a+b \sin (x))}+\frac{\left (3 a^2 \left (2 a^4-5 a^2 b^2+4 b^4\right )\right ) \int \frac{1}{a+b \sin (x)} \, dx}{2 b^4 \left (a^2-b^2\right )^2}\\ &=-\frac{3 a x}{b^4}-\frac{\left (3 a^2-2 b^2\right ) \cos (x)}{2 b^3 \left (a^2-b^2\right )}+\frac{a^2 \cos (x) \sin ^2(x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac{3 a^3 \left (a^2-2 b^2\right ) \cos (x)}{2 b^3 \left (a^2-b^2\right )^2 (a+b \sin (x))}+\frac{\left (3 a^2 \left (2 a^4-5 a^2 b^2+4 b^4\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{b^4 \left (a^2-b^2\right )^2}\\ &=-\frac{3 a x}{b^4}-\frac{\left (3 a^2-2 b^2\right ) \cos (x)}{2 b^3 \left (a^2-b^2\right )}+\frac{a^2 \cos (x) \sin ^2(x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac{3 a^3 \left (a^2-2 b^2\right ) \cos (x)}{2 b^3 \left (a^2-b^2\right )^2 (a+b \sin (x))}-\frac{\left (6 a^2 \left (2 a^4-5 a^2 b^2+4 b^4\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac{x}{2}\right )\right )}{b^4 \left (a^2-b^2\right )^2}\\ &=-\frac{3 a x}{b^4}+\frac{3 a^2 \left (2 a^4-5 a^2 b^2+4 b^4\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{b^4 \left (a^2-b^2\right )^{5/2}}-\frac{\left (3 a^2-2 b^2\right ) \cos (x)}{2 b^3 \left (a^2-b^2\right )}+\frac{a^2 \cos (x) \sin ^2(x)}{2 b \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac{3 a^3 \left (a^2-2 b^2\right ) \cos (x)}{2 b^3 \left (a^2-b^2\right )^2 (a+b \sin (x))}\\ \end{align*}

Mathematica [A]  time = 0.756829, size = 144, normalized size = 0.8 \[ \frac{\frac{6 a^2 \left (-5 a^2 b^2+2 a^4+4 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac{a^3 b \left (8 b^2-5 a^2\right ) \cos (x)}{(a-b)^2 (a+b)^2 (a+b \sin (x))}+\frac{a^4 b \cos (x)}{(a-b) (a+b) (a+b \sin (x))^2}-6 a x-2 b \cos (x)}{2 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^4/(a + b*Sin[x])^3,x]

[Out]

(-6*a*x + (6*a^2*(2*a^4 - 5*a^2*b^2 + 4*b^4)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) - 2*b
*Cos[x] + (a^4*b*Cos[x])/((a - b)*(a + b)*(a + b*Sin[x])^2) + (a^3*b*(-5*a^2 + 8*b^2)*Cos[x])/((a - b)^2*(a +
b)^2*(a + b*Sin[x])))/(2*b^4)

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Maple [B]  time = 0.062, size = 634, normalized size = 3.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^4/(a+b*sin(x))^3,x)

[Out]

-2/b^3/(tan(1/2*x)^2+1)-6/b^4*a*arctan(tan(1/2*x))-3*a^5/b^2/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)^2/(a^4-2*a^2*b^
2+b^4)*tan(1/2*x)^3+6*a^3/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)^3-4*a^6/b^3/(tan(
1/2*x)^2*a+2*tan(1/2*x)*b+a)^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)^2-a^4/b/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)^2/(a^4
-2*a^2*b^2+b^4)*tan(1/2*x)^2+14*a^2*b/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)^2-13*
a^5/b^2/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)+22*a^3/(tan(1/2*x)^2*a+2*tan(1/2*x)
*b+a)^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)-4*a^6/b^3/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)^2/(a^4-2*a^2*b^2+b^4)+7*a^4
/b/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)^2/(a^4-2*a^2*b^2+b^4)+6*a^6/b^4/(a^4-2*a^2*b^2+b^4)/(a^2-b^2)^(1/2)*arcta
n(1/2*(2*a*tan(1/2*x)+2*b)/(a^2-b^2)^(1/2))-15*a^4/b^2/(a^4-2*a^2*b^2+b^4)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan
(1/2*x)+2*b)/(a^2-b^2)^(1/2))+12*a^2/(a^4-2*a^2*b^2+b^4)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*x)+2*b)/(a^2-
b^2)^(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(x)^4/(a+b*sin(x))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.28055, size = 2013, normalized size = 11.25 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^4/(a+b*sin(x))^3,x, algorithm="fricas")

[Out]

[1/4*(12*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*x*cos(x)^2 + 4*(a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*cos(
x)^3 - 3*(2*a^8 - 3*a^6*b^2 - a^4*b^4 + 4*a^2*b^6 - (2*a^6*b^2 - 5*a^4*b^4 + 4*a^2*b^6)*cos(x)^2 + 2*(2*a^7*b
- 5*a^5*b^3 + 4*a^3*b^5)*sin(x))*sqrt(-a^2 + b^2)*log(((2*a^2 - b^2)*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2 + 2*(
a*cos(x)*sin(x) + b*cos(x))*sqrt(-a^2 + b^2))/(b^2*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2)) - 12*(a^9 - 2*a^7*b^2
+ 2*a^3*b^6 - a*b^8)*x - 2*(6*a^8*b - 15*a^6*b^3 + 7*a^4*b^5 + 4*a^2*b^7 - 2*b^9)*cos(x) - 2*(12*(a^8*b - 3*a
^6*b^3 + 3*a^4*b^5 - a^2*b^7)*x + (9*a^7*b^2 - 25*a^5*b^4 + 20*a^3*b^6 - 4*a*b^8)*cos(x))*sin(x))/(a^8*b^4 - 2
*a^6*b^6 + 2*a^2*b^10 - b^12 - (a^6*b^6 - 3*a^4*b^8 + 3*a^2*b^10 - b^12)*cos(x)^2 + 2*(a^7*b^5 - 3*a^5*b^7 + 3
*a^3*b^9 - a*b^11)*sin(x)), 1/2*(6*(a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*x*cos(x)^2 + 2*(a^6*b^3 - 3*a^4*b
^5 + 3*a^2*b^7 - b^9)*cos(x)^3 - 3*(2*a^8 - 3*a^6*b^2 - a^4*b^4 + 4*a^2*b^6 - (2*a^6*b^2 - 5*a^4*b^4 + 4*a^2*b
^6)*cos(x)^2 + 2*(2*a^7*b - 5*a^5*b^3 + 4*a^3*b^5)*sin(x))*sqrt(a^2 - b^2)*arctan(-(a*sin(x) + b)/(sqrt(a^2 -
b^2)*cos(x))) - 6*(a^9 - 2*a^7*b^2 + 2*a^3*b^6 - a*b^8)*x - (6*a^8*b - 15*a^6*b^3 + 7*a^4*b^5 + 4*a^2*b^7 - 2*
b^9)*cos(x) - (12*(a^8*b - 3*a^6*b^3 + 3*a^4*b^5 - a^2*b^7)*x + (9*a^7*b^2 - 25*a^5*b^4 + 20*a^3*b^6 - 4*a*b^8
)*cos(x))*sin(x))/(a^8*b^4 - 2*a^6*b^6 + 2*a^2*b^10 - b^12 - (a^6*b^6 - 3*a^4*b^8 + 3*a^2*b^10 - b^12)*cos(x)^
2 + 2*(a^7*b^5 - 3*a^5*b^7 + 3*a^3*b^9 - a*b^11)*sin(x))]

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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(x)**4/(a+b*sin(x))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.98173, size = 346, normalized size = 1.93 \begin{align*} \frac{3 \,{\left (2 \, a^{6} - 5 \, a^{4} b^{2} + 4 \, a^{2} b^{4}\right )}{\left (\pi \left \lfloor \frac{x}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, x\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} b^{4} - 2 \, a^{2} b^{6} + b^{8}\right )} \sqrt{a^{2} - b^{2}}} - \frac{3 \, a^{5} b \tan \left (\frac{1}{2} \, x\right )^{3} - 6 \, a^{3} b^{3} \tan \left (\frac{1}{2} \, x\right )^{3} + 4 \, a^{6} \tan \left (\frac{1}{2} \, x\right )^{2} + a^{4} b^{2} \tan \left (\frac{1}{2} \, x\right )^{2} - 14 \, a^{2} b^{4} \tan \left (\frac{1}{2} \, x\right )^{2} + 13 \, a^{5} b \tan \left (\frac{1}{2} \, x\right ) - 22 \, a^{3} b^{3} \tan \left (\frac{1}{2} \, x\right ) + 4 \, a^{6} - 7 \, a^{4} b^{2}}{{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )}{\left (a \tan \left (\frac{1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, x\right ) + a\right )}^{2}} - \frac{3 \, a x}{b^{4}} - \frac{2}{{\left (\tan \left (\frac{1}{2} \, x\right )^{2} + 1\right )} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^4/(a+b*sin(x))^3,x, algorithm="giac")

[Out]

3*(2*a^6 - 5*a^4*b^2 + 4*a^2*b^4)*(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*x) + b)/sqrt(a^2 - b^2)
))/((a^4*b^4 - 2*a^2*b^6 + b^8)*sqrt(a^2 - b^2)) - (3*a^5*b*tan(1/2*x)^3 - 6*a^3*b^3*tan(1/2*x)^3 + 4*a^6*tan(
1/2*x)^2 + a^4*b^2*tan(1/2*x)^2 - 14*a^2*b^4*tan(1/2*x)^2 + 13*a^5*b*tan(1/2*x) - 22*a^3*b^3*tan(1/2*x) + 4*a^
6 - 7*a^4*b^2)/((a^4*b^3 - 2*a^2*b^5 + b^7)*(a*tan(1/2*x)^2 + 2*b*tan(1/2*x) + a)^2) - 3*a*x/b^4 - 2/((tan(1/2
*x)^2 + 1)*b^3)