3.202 \(\int \frac{1}{(a+b \sin (c+d x))^4} \, dx\)
Optimal. Leaf size=182 \[ \frac{a \left (2 a^2+3 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{7/2}}+\frac{b \left (11 a^2+4 b^2\right ) \cos (c+d x)}{6 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}+\frac{5 a b \cos (c+d x)}{6 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}+\frac{b \cos (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^3} \]
[Out]
(a*(2*a^2 + 3*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(7/2)*d) + (b*Cos[c + d*x])/
(3*(a^2 - b^2)*d*(a + b*Sin[c + d*x])^3) + (5*a*b*Cos[c + d*x])/(6*(a^2 - b^2)^2*d*(a + b*Sin[c + d*x])^2) + (
b*(11*a^2 + 4*b^2)*Cos[c + d*x])/(6*(a^2 - b^2)^3*d*(a + b*Sin[c + d*x]))
________________________________________________________________________________________
Rubi [A] time = 0.224564, antiderivative size = 182, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used =
{2664, 2754, 12, 2660, 618, 204} \[ \frac{a \left (2 a^2+3 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{7/2}}+\frac{b \left (11 a^2+4 b^2\right ) \cos (c+d x)}{6 d \left (a^2-b^2\right )^3 (a+b \sin (c+d x))}+\frac{5 a b \cos (c+d x)}{6 d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^2}+\frac{b \cos (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sin (c+d x))^3} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*Sin[c + d*x])^(-4),x]
[Out]
(a*(2*a^2 + 3*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(7/2)*d) + (b*Cos[c + d*x])/
(3*(a^2 - b^2)*d*(a + b*Sin[c + d*x])^3) + (5*a*b*Cos[c + d*x])/(6*(a^2 - b^2)^2*d*(a + b*Sin[c + d*x])^2) + (
b*(11*a^2 + 4*b^2)*Cos[c + d*x])/(6*(a^2 - b^2)^3*d*(a + b*Sin[c + d*x]))
Rule 2664
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n +
1))/(d*(n + 1)*(a^2 - b^2)), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1
) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integer
Q[2*n]
Rule 2754
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*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] && LtQ[m, -1] && IntegerQ[2*m]
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
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{1}{(a+b \sin (c+d x))^4} \, dx &=\frac{b \cos (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^3}-\frac{\int \frac{-3 a+2 b \sin (c+d x)}{(a+b \sin (c+d x))^3} \, dx}{3 \left (a^2-b^2\right )}\\ &=\frac{b \cos (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^3}+\frac{5 a b \cos (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}+\frac{\int \frac{2 \left (3 a^2+2 b^2\right )-5 a b \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx}{6 \left (a^2-b^2\right )^2}\\ &=\frac{b \cos (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^3}+\frac{5 a b \cos (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}+\frac{b \left (11 a^2+4 b^2\right ) \cos (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}-\frac{\int -\frac{3 a \left (2 a^2+3 b^2\right )}{a+b \sin (c+d x)} \, dx}{6 \left (a^2-b^2\right )^3}\\ &=\frac{b \cos (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^3}+\frac{5 a b \cos (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}+\frac{b \left (11 a^2+4 b^2\right ) \cos (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac{\left (a \left (2 a^2+3 b^2\right )\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{2 \left (a^2-b^2\right )^3}\\ &=\frac{b \cos (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^3}+\frac{5 a b \cos (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}+\frac{b \left (11 a^2+4 b^2\right ) \cos (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}+\frac{\left (a \left (2 a^2+3 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^3 d}\\ &=\frac{b \cos (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^3}+\frac{5 a b \cos (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}+\frac{b \left (11 a^2+4 b^2\right ) \cos (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}-\frac{\left (2 a \left (2 a^2+3 b^2\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{1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^3 d}\\ &=\frac{a \left (2 a^2+3 b^2\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2} d}+\frac{b \cos (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sin (c+d x))^3}+\frac{5 a b \cos (c+d x)}{6 \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))^2}+\frac{b \left (11 a^2+4 b^2\right ) \cos (c+d x)}{6 \left (a^2-b^2\right )^3 d (a+b \sin (c+d x))}\\ \end{align*}
Mathematica [A] time = 1.06478, size = 157, normalized size = 0.86 \[ \frac{\frac{6 a \left (2 a^2+3 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{7/2}}+\frac{b \cos (c+d x) \left (b^2 \left (11 a^2+4 b^2\right ) \sin ^2(c+d x)+3 a b \left (9 a^2+b^2\right ) \sin (c+d x)-5 a^2 b^2+18 a^4+2 b^4\right )}{(a-b)^3 (a+b)^3 (a+b \sin (c+d x))^3}}{6 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + b*Sin[c + d*x])^(-4),x]
[Out]
((6*a*(2*a^2 + 3*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(7/2) + (b*Cos[c + d*x]*(1
8*a^4 - 5*a^2*b^2 + 2*b^4 + 3*a*b*(9*a^2 + b^2)*Sin[c + d*x] + b^2*(11*a^2 + 4*b^2)*Sin[c + d*x]^2))/((a - b)^
3*(a + b)^3*(a + b*Sin[c + d*x])^3))/(6*d)
________________________________________________________________________________________
Maple [B] time = 0.076, size = 1733, normalized size = 9.5 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(a+b*sin(d*x+c))^4,x)
[Out]
9/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^3*b^2*a^3/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)*tan(1/2*d*x+1/2*
c)^5-6/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^3*b^4*a/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)*tan(1/2*d*x+1
/2*c)^5+2/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^3*b^6/a/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)*tan(1/2*d*
x+1/2*c)^5+6/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^3*b*a^4/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)*tan(1/2
*d*x+1/2*c)^4+27/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^3*b^3*a^2/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)*t
an(1/2*d*x+1/2*c)^4-12/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^3*b^5/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)
*tan(1/2*d*x+1/2*c)^4+4/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^3*b^7/a^2/(a^6-3*a^4*b^2+3*a^2*b^4
-b^6)*tan(1/2*d*x+1/2*c)^4+36/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^3*a^3*b^2/(a^6-3*a^4*b^2+3*a
^2*b^4-b^6)*tan(1/2*d*x+1/2*c)^3+14/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^3*a*b^4/(a^6-3*a^4*b^2
+3*a^2*b^4-b^6)*tan(1/2*d*x+1/2*c)^3-8/3/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^3/a*b^6/(a^6-3*a^
4*b^2+3*a^2*b^4-b^6)*tan(1/2*d*x+1/2*c)^3+8/3/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^3/a^3*b^8/(a
^6-3*a^4*b^2+3*a^2*b^4-b^6)*tan(1/2*d*x+1/2*c)^3+12/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^3*a^4*
b/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)*tan(1/2*d*x+1/2*c)^2+40/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^3*
a^2*b^3/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)*tan(1/2*d*x+1/2*c)^2-6/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+
a)^3*b^5/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)*tan(1/2*d*x+1/2*c)^2+4/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b
+a)^3/a^2*b^7/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)*tan(1/2*d*x+1/2*c)^2+27/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/
2*c)*b+a)^3*b^2*a^3/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)*tan(1/2*d*x+1/2*c)-4/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x
+1/2*c)*b+a)^3*b^4*a/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)*tan(1/2*d*x+1/2*c)+2/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*
x+1/2*c)*b+a)^3*b^6/a/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)*tan(1/2*d*x+1/2*c)+6/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d
*x+1/2*c)*b+a)^3*b/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)*a^4-5/3/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^3
*b^3/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)*a^2+2/3/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^3*b^5/(a^6-3*a^
4*b^2+3*a^2*b^4-b^6)+2/d*a^3/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+
2*b)/(a^2-b^2)^(1/2))+3/d*a/(a^6-3*a^4*b^2+3*a^2*b^4-b^6)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2
*b)/(a^2-b^2)^(1/2))*b^2
________________________________________________________________________________________
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(1/(a+b*sin(d*x+c))^4,x, algorithm="maxima")
[Out]
Exception raised: ValueError
________________________________________________________________________________________
Fricas [B] time = 2.03495, size = 2101, normalized size = 11.54 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*sin(d*x+c))^4,x, algorithm="fricas")
[Out]
[1/12*(2*(11*a^4*b^3 - 7*a^2*b^5 - 4*b^7)*cos(d*x + c)^3 - 6*(9*a^5*b^2 - 8*a^3*b^4 - a*b^6)*cos(d*x + c)*sin(
d*x + c) - 3*(2*a^6 + 9*a^4*b^2 + 9*a^2*b^4 - 3*(2*a^4*b^2 + 3*a^2*b^4)*cos(d*x + c)^2 + (6*a^5*b + 11*a^3*b^3
+ 3*a*b^5 - (2*a^3*b^3 + 3*a*b^5)*cos(d*x + c)^2)*sin(d*x + c))*sqrt(-a^2 + b^2)*log(-((2*a^2 - b^2)*cos(d*x
+ c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 - 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(
b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) - 12*(3*a^6*b - 2*a^4*b^3 - b^7)*cos(d*x + c))/(3*(a^9*b
^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*d*cos(d*x + c)^2 - (a^11 - a^9*b^2 - 6*a^7*b^4 + 14*a^5*b^6 -
11*a^3*b^8 + 3*a*b^10)*d + ((a^8*b^3 - 4*a^6*b^5 + 6*a^4*b^7 - 4*a^2*b^9 + b^11)*d*cos(d*x + c)^2 - (3*a^10*b
- 11*a^8*b^3 + 14*a^6*b^5 - 6*a^4*b^7 - a^2*b^9 + b^11)*d)*sin(d*x + c)), 1/6*((11*a^4*b^3 - 7*a^2*b^5 - 4*b^
7)*cos(d*x + c)^3 - 3*(9*a^5*b^2 - 8*a^3*b^4 - a*b^6)*cos(d*x + c)*sin(d*x + c) + 3*(2*a^6 + 9*a^4*b^2 + 9*a^2
*b^4 - 3*(2*a^4*b^2 + 3*a^2*b^4)*cos(d*x + c)^2 + (6*a^5*b + 11*a^3*b^3 + 3*a*b^5 - (2*a^3*b^3 + 3*a*b^5)*cos(
d*x + c)^2)*sin(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) - 6*(3*
a^6*b - 2*a^4*b^3 - b^7)*cos(d*x + c))/(3*(a^9*b^2 - 4*a^7*b^4 + 6*a^5*b^6 - 4*a^3*b^8 + a*b^10)*d*cos(d*x + c
)^2 - (a^11 - a^9*b^2 - 6*a^7*b^4 + 14*a^5*b^6 - 11*a^3*b^8 + 3*a*b^10)*d + ((a^8*b^3 - 4*a^6*b^5 + 6*a^4*b^7
- 4*a^2*b^9 + b^11)*d*cos(d*x + c)^2 - (3*a^10*b - 11*a^8*b^3 + 14*a^6*b^5 - 6*a^4*b^7 - a^2*b^9 + b^11)*d)*si
n(d*x + c))]
________________________________________________________________________________________
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(1/(a+b*sin(d*x+c))**4,x)
[Out]
Timed out
________________________________________________________________________________________
Giac [B] time = 1.23768, size = 689, normalized size = 3.79 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+b*sin(d*x+c))^4,x, algorithm="giac")
[Out]
1/3*(3*(2*a^3 + 3*a*b^2)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a
^2 - b^2)))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sqrt(a^2 - b^2)) + (27*a^6*b^2*tan(1/2*d*x + 1/2*c)^5 - 18*a^
4*b^4*tan(1/2*d*x + 1/2*c)^5 + 6*a^2*b^6*tan(1/2*d*x + 1/2*c)^5 + 18*a^7*b*tan(1/2*d*x + 1/2*c)^4 + 81*a^5*b^3
*tan(1/2*d*x + 1/2*c)^4 - 36*a^3*b^5*tan(1/2*d*x + 1/2*c)^4 + 12*a*b^7*tan(1/2*d*x + 1/2*c)^4 + 108*a^6*b^2*ta
n(1/2*d*x + 1/2*c)^3 + 42*a^4*b^4*tan(1/2*d*x + 1/2*c)^3 - 8*a^2*b^6*tan(1/2*d*x + 1/2*c)^3 + 8*b^8*tan(1/2*d*
x + 1/2*c)^3 + 36*a^7*b*tan(1/2*d*x + 1/2*c)^2 + 120*a^5*b^3*tan(1/2*d*x + 1/2*c)^2 - 18*a^3*b^5*tan(1/2*d*x +
1/2*c)^2 + 12*a*b^7*tan(1/2*d*x + 1/2*c)^2 + 81*a^6*b^2*tan(1/2*d*x + 1/2*c) - 12*a^4*b^4*tan(1/2*d*x + 1/2*c
) + 6*a^2*b^6*tan(1/2*d*x + 1/2*c) + 18*a^7*b - 5*a^5*b^3 + 2*a^3*b^5)/((a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^3*b^6
)*(a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)^3))/d