Integrand size = 18, antiderivative size = 94 \[ \int \frac {x^2}{\left (a+b \sin \left (c+d x^3\right )\right )^2} \, dx=\frac {2 a \arctan \left (\frac {b+a \tan \left (\frac {1}{2} \left (c+d x^3\right )\right )}{\sqrt {a^2-b^2}}\right )}{3 \left (a^2-b^2\right )^{3/2} d}+\frac {b \cos \left (c+d x^3\right )}{3 \left (a^2-b^2\right ) d \left (a+b \sin \left (c+d x^3\right )\right )} \] Output:
2/3*a*arctan((b+a*tan(1/2*d*x^3+1/2*c))/(a^2-b^2)^(1/2))/(a^2-b^2)^(3/2)/d +1/3*b*cos(d*x^3+c)/(a^2-b^2)/d/(a+b*sin(d*x^3+c))
Time = 0.24 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.97 \[ \int \frac {x^2}{\left (a+b \sin \left (c+d x^3\right )\right )^2} \, dx=\frac {\frac {2 a \arctan \left (\frac {b+a \tan \left (\frac {1}{2} \left (c+d x^3\right )\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {b \cos \left (c+d x^3\right )}{a+b \sin \left (c+d x^3\right )}}{3 (a-b) (a+b) d} \] Input:
Integrate[x^2/(a + b*Sin[c + d*x^3])^2,x]
Output:
((2*a*ArcTan[(b + a*Tan[(c + d*x^3)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + (b*Cos[c + d*x^3])/(a + b*Sin[c + d*x^3]))/(3*(a - b)*(a + b)*d)
Time = 0.39 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3860, 3042, 3143, 25, 27, 3042, 3139, 1083, 217}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {x^2}{\left (a+b \sin \left (c+d x^3\right )\right )^2} \, dx\) |
| \(\Big \downarrow \) 3860 |
| \(\displaystyle \frac {1}{3} \int \frac {1}{\left (a+b \sin \left (d x^3+c\right )\right )^2}dx^3\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \frac {1}{\left (a+b \sin \left (d x^3+c\right )\right )^2}dx^3\) |
| \(\Big \downarrow \) 3143 |
| \(\displaystyle \frac {1}{3} \left (\frac {b \cos \left (c+d x^3\right )}{d \left (a^2-b^2\right ) \left (a+b \sin \left (c+d x^3\right )\right )}-\frac {\int -\frac {a}{a+b \sin \left (d x^3+c\right )}dx^3}{a^2-b^2}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{3} \left (\frac {\int \frac {a}{a+b \sin \left (d x^3+c\right )}dx^3}{a^2-b^2}+\frac {b \cos \left (c+d x^3\right )}{d \left (a^2-b^2\right ) \left (a+b \sin \left (c+d x^3\right )\right )}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\frac {a \int \frac {1}{a+b \sin \left (d x^3+c\right )}dx^3}{a^2-b^2}+\frac {b \cos \left (c+d x^3\right )}{d \left (a^2-b^2\right ) \left (a+b \sin \left (c+d x^3\right )\right )}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {a \int \frac {1}{a+b \sin \left (d x^3+c\right )}dx^3}{a^2-b^2}+\frac {b \cos \left (c+d x^3\right )}{d \left (a^2-b^2\right ) \left (a+b \sin \left (c+d x^3\right )\right )}\right )\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle \frac {1}{3} \left (\frac {2 a \int \frac {1}{a x^6+a+2 b \tan \left (\frac {1}{2} \left (d x^3+c\right )\right )}d\tan \left (\frac {1}{2} \left (d x^3+c\right )\right )}{d \left (a^2-b^2\right )}+\frac {b \cos \left (c+d x^3\right )}{d \left (a^2-b^2\right ) \left (a+b \sin \left (c+d x^3\right )\right )}\right )\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {1}{3} \left (\frac {b \cos \left (c+d x^3\right )}{d \left (a^2-b^2\right ) \left (a+b \sin \left (c+d x^3\right )\right )}-\frac {4 a \int \frac {1}{-x^6-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {1}{2} \left (d x^3+c\right )\right )\right )}{d \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{3} \left (\frac {2 a \arctan \left (\frac {2 a \tan \left (\frac {1}{2} \left (c+d x^3\right )\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{3/2}}+\frac {b \cos \left (c+d x^3\right )}{d \left (a^2-b^2\right ) \left (a+b \sin \left (c+d x^3\right )\right )}\right )\) |
Input:
Int[x^2/(a + b*Sin[c + d*x^3])^2,x]
Output:
((2*a*ArcTan[(2*b + 2*a*Tan[(c + d*x^3)/2])/(2*Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*d) + (b*Cos[c + d*x^3])/((a^2 - b^2)*d*(a + b*Sin[c + d*x^3]))) /3
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[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]
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] + Simp [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] && IntegerQ[2*n]
Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol ] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^ p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simplify[ (m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[ (m + 1)/n], 0]))
Time = 0.56 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.39
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {2 b^{2} \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )}{a \left (a^{2}-b^{2}\right )}+\frac {2 b}{a^{2}-b^{2}}}{\tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )+a}+\frac {2 a \arctan \left (\frac {2 a \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{2}-b^{2}\right )^{\frac {3}{2}}}}{3 d}\) | \(131\) |
| default | \(\frac {\frac {\frac {2 b^{2} \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )}{a \left (a^{2}-b^{2}\right )}+\frac {2 b}{a^{2}-b^{2}}}{\tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )+a}+\frac {2 a \arctan \left (\frac {2 a \tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{2}-b^{2}\right )^{\frac {3}{2}}}}{3 d}\) | \(131\) |
| risch | \(\frac {\frac {2 i b}{3}+\frac {2 a \,{\mathrm e}^{i \left (d \,x^{3}+c \right )}}{3}}{\left (a^{2}-b^{2}\right ) d \left (b \,{\mathrm e}^{2 i \left (d \,x^{3}+c \right )}-b +2 i a \,{\mathrm e}^{i \left (d \,x^{3}+c \right )}\right )}-\frac {a \ln \left ({\mathrm e}^{i \left (d \,x^{3}+c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{3 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {a \ln \left ({\mathrm e}^{i \left (d \,x^{3}+c \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{3 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}\) | \(232\) |
Input:
int(x^2/(a+b*sin(d*x^3+c))^2,x,method=_RETURNVERBOSE)
Output:
1/3/d*(2*(b^2/a/(a^2-b^2)*tan(1/2*d*x^3+1/2*c)+b/(a^2-b^2))/(tan(1/2*d*x^3 +1/2*c)^2*a+2*b*tan(1/2*d*x^3+1/2*c)+a)+2*a/(a^2-b^2)^(3/2)*arctan(1/2*(2* a*tan(1/2*d*x^3+1/2*c)+2*b)/(a^2-b^2)^(1/2)))
Time = 0.10 (sec) , antiderivative size = 366, normalized size of antiderivative = 3.89 \[ \int \frac {x^2}{\left (a+b \sin \left (c+d x^3\right )\right )^2} \, dx=\left [\frac {{\left (a b \sin \left (d x^{3} + c\right ) + a^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x^{3} + c\right )^{2} - 2 \, a b \sin \left (d x^{3} + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x^{3} + c\right ) \sin \left (d x^{3} + c\right ) + b \cos \left (d x^{3} + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x^{3} + c\right )^{2} - 2 \, a b \sin \left (d x^{3} + c\right ) - a^{2} - b^{2}}\right ) + 2 \, {\left (a^{2} b - b^{3}\right )} \cos \left (d x^{3} + c\right )}{6 \, {\left ({\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \sin \left (d x^{3} + c\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}}, -\frac {{\left (a b \sin \left (d x^{3} + c\right ) + a^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x^{3} + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x^{3} + c\right )}\right ) - {\left (a^{2} b - b^{3}\right )} \cos \left (d x^{3} + c\right )}{3 \, {\left ({\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \sin \left (d x^{3} + c\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}}\right ] \] Input:
integrate(x^2/(a+b*sin(d*x^3+c))^2,x, algorithm="fricas")
Output:
[1/6*((a*b*sin(d*x^3 + c) + a^2)*sqrt(-a^2 + b^2)*log(-((2*a^2 - b^2)*cos( d*x^3 + c)^2 - 2*a*b*sin(d*x^3 + c) - a^2 - b^2 - 2*(a*cos(d*x^3 + c)*sin( d*x^3 + c) + b*cos(d*x^3 + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x^3 + c)^2 - 2 *a*b*sin(d*x^3 + c) - a^2 - b^2)) + 2*(a^2*b - b^3)*cos(d*x^3 + c))/((a^4* b - 2*a^2*b^3 + b^5)*d*sin(d*x^3 + c) + (a^5 - 2*a^3*b^2 + a*b^4)*d), -1/3 *((a*b*sin(d*x^3 + c) + a^2)*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x^3 + c) + b )/(sqrt(a^2 - b^2)*cos(d*x^3 + c))) - (a^2*b - b^3)*cos(d*x^3 + c))/((a^4* b - 2*a^2*b^3 + b^5)*d*sin(d*x^3 + c) + (a^5 - 2*a^3*b^2 + a*b^4)*d)]
Leaf count of result is larger than twice the leaf count of optimal. 2116 vs. \(2 (75) = 150\).
Time = 68.84 (sec) , antiderivative size = 2116, normalized size of antiderivative = 22.51 \[ \int \frac {x^2}{\left (a+b \sin \left (c+d x^3\right )\right )^2} \, dx=\text {Too large to display} \] Input:
integrate(x**2/(a+b*sin(d*x**3+c))**2,x)
Output:
Piecewise((zoo*x**3/sin(c)**2, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), ((tan(c/2 + d*x**3/2)/(6*d) - 1/(6*d*tan(c/2 + d*x**3/2)))/b**2, Eq(a, 0)), (-6*tan( c/2 + d*x**3/2)**2/(9*b**2*d*tan(c/2 + d*x**3/2)**3 - 27*b**2*d*tan(c/2 + d*x**3/2)**2 + 27*b**2*d*tan(c/2 + d*x**3/2) - 9*b**2*d) + 6*tan(c/2 + d*x **3/2)/(9*b**2*d*tan(c/2 + d*x**3/2)**3 - 27*b**2*d*tan(c/2 + d*x**3/2)**2 + 27*b**2*d*tan(c/2 + d*x**3/2) - 9*b**2*d) - 4/(9*b**2*d*tan(c/2 + d*x** 3/2)**3 - 27*b**2*d*tan(c/2 + d*x**3/2)**2 + 27*b**2*d*tan(c/2 + d*x**3/2) - 9*b**2*d), Eq(a, -b)), (-6*tan(c/2 + d*x**3/2)**2/(9*b**2*d*tan(c/2 + d *x**3/2)**3 + 27*b**2*d*tan(c/2 + d*x**3/2)**2 + 27*b**2*d*tan(c/2 + d*x** 3/2) + 9*b**2*d) - 6*tan(c/2 + d*x**3/2)/(9*b**2*d*tan(c/2 + d*x**3/2)**3 + 27*b**2*d*tan(c/2 + d*x**3/2)**2 + 27*b**2*d*tan(c/2 + d*x**3/2) + 9*b** 2*d) - 4/(9*b**2*d*tan(c/2 + d*x**3/2)**3 + 27*b**2*d*tan(c/2 + d*x**3/2)* *2 + 27*b**2*d*tan(c/2 + d*x**3/2) + 9*b**2*d), Eq(a, b)), (x**3/(3*(a + b *sin(c))**2), Eq(d, 0)), (a**3*log(tan(c/2 + d*x**3/2) + b/a - sqrt(-a**2 + b**2)/a)*tan(c/2 + d*x**3/2)**2/(3*a**4*d*sqrt(-a**2 + b**2)*tan(c/2 + d *x**3/2)**2 + 3*a**4*d*sqrt(-a**2 + b**2) + 6*a**3*b*d*sqrt(-a**2 + b**2)* tan(c/2 + d*x**3/2) - 3*a**2*b**2*d*sqrt(-a**2 + b**2)*tan(c/2 + d*x**3/2) **2 - 3*a**2*b**2*d*sqrt(-a**2 + b**2) - 6*a*b**3*d*sqrt(-a**2 + b**2)*tan (c/2 + d*x**3/2)) + a**3*log(tan(c/2 + d*x**3/2) + b/a - sqrt(-a**2 + b**2 )/a)/(3*a**4*d*sqrt(-a**2 + b**2)*tan(c/2 + d*x**3/2)**2 + 3*a**4*d*sqr...
Timed out. \[ \int \frac {x^2}{\left (a+b \sin \left (c+d x^3\right )\right )^2} \, dx=\text {Timed out} \] Input:
integrate(x^2/(a+b*sin(d*x^3+c))^2,x, algorithm="maxima")
Output:
Timed out
Time = 0.13 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.55 \[ \int \frac {x^2}{\left (a+b \sin \left (c+d x^3\right )\right )^2} \, dx=\frac {2 \, {\left (\pi \left \lfloor \frac {d x^{3} + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x^{3} + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a}{3 \, {\left (a^{2} d - b^{2} d\right )} \sqrt {a^{2} - b^{2}}} + \frac {2 \, {\left (b^{2} \tan \left (\frac {1}{2} \, d x^{3} + \frac {1}{2} \, c\right ) + a b\right )}}{3 \, {\left (a^{3} d - a b^{2} d\right )} {\left (a \tan \left (\frac {1}{2} \, d x^{3} + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x^{3} + \frac {1}{2} \, c\right ) + a\right )}} \] Input:
integrate(x^2/(a+b*sin(d*x^3+c))^2,x, algorithm="giac")
Output:
2/3*(pi*floor(1/2*(d*x^3 + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x^3 + 1/2*c) + b)/sqrt(a^2 - b^2)))*a/((a^2*d - b^2*d)*sqrt(a^2 - b^2)) + 2/3*( b^2*tan(1/2*d*x^3 + 1/2*c) + a*b)/((a^3*d - a*b^2*d)*(a*tan(1/2*d*x^3 + 1/ 2*c)^2 + 2*b*tan(1/2*d*x^3 + 1/2*c) + a))
Time = 41.31 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.98 \[ \int \frac {x^2}{\left (a+b \sin \left (c+d x^3\right )\right )^2} \, dx=\frac {\frac {2\,b}{a^2-b^2}+\frac {2\,b^2\,\mathrm {tan}\left (\frac {d\,x^3}{2}+\frac {c}{2}\right )}{a\,\left (a^2-b^2\right )}}{d\,\left (3\,a\,{\mathrm {tan}\left (\frac {d\,x^3}{2}+\frac {c}{2}\right )}^2+6\,b\,\mathrm {tan}\left (\frac {d\,x^3}{2}+\frac {c}{2}\right )+3\,a\right )}+\frac {2\,a\,\mathrm {atan}\left (\frac {3\,\left (a^2-b^2\right )\,\left (\frac {2\,a^2\,\mathrm {tan}\left (\frac {d\,x^3}{2}+\frac {c}{2}\right )}{3\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}+\frac {2\,a\,\left (3\,a^2\,b-3\,b^3\right )}{9\,{\left (a+b\right )}^{3/2}\,\left (a^2-b^2\right )\,{\left (a-b\right )}^{3/2}}\right )}{2\,a}\right )}{3\,d\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}} \] Input:
int(x^2/(a + b*sin(c + d*x^3))^2,x)
Output:
((2*b)/(a^2 - b^2) + (2*b^2*tan(c/2 + (d*x^3)/2))/(a*(a^2 - b^2)))/(d*(3*a + 3*a*tan(c/2 + (d*x^3)/2)^2 + 6*b*tan(c/2 + (d*x^3)/2))) + (2*a*atan((3* (a^2 - b^2)*((2*a^2*tan(c/2 + (d*x^3)/2))/(3*(a + b)^(3/2)*(a - b)^(3/2)) + (2*a*(3*a^2*b - 3*b^3))/(9*(a + b)^(3/2)*(a^2 - b^2)*(a - b)^(3/2))))/(2 *a)))/(3*d*(a + b)^(3/2)*(a - b)^(3/2))
Time = 0.16 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.99 \[ \int \frac {x^2}{\left (a+b \sin \left (c+d x^3\right )\right )^2} \, dx=\frac {2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (d \,x^{3}+c \right ) a b +2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d \,x^{3}}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a^{2}+\cos \left (d \,x^{3}+c \right ) a^{2} b -\cos \left (d \,x^{3}+c \right ) b^{3}}{3 d \left (\sin \left (d \,x^{3}+c \right ) a^{4} b -2 \sin \left (d \,x^{3}+c \right ) a^{2} b^{3}+\sin \left (d \,x^{3}+c \right ) b^{5}+a^{5}-2 a^{3} b^{2}+a \,b^{4}\right )} \] Input:
int(x^2/(a+b*sin(d*x^3+c))^2,x)
Output:
(2*sqrt(a**2 - b**2)*atan((tan((c + d*x**3)/2)*a + b)/sqrt(a**2 - b**2))*s in(c + d*x**3)*a*b + 2*sqrt(a**2 - b**2)*atan((tan((c + d*x**3)/2)*a + b)/ sqrt(a**2 - b**2))*a**2 + cos(c + d*x**3)*a**2*b - cos(c + d*x**3)*b**3)/( 3*d*(sin(c + d*x**3)*a**4*b - 2*sin(c + d*x**3)*a**2*b**3 + sin(c + d*x**3 )*b**5 + a**5 - 2*a**3*b**2 + a*b**4))