Integrand size = 25, antiderivative size = 201 \[ \int \frac {(3+b \sin (e+f x))^3}{(c+d \sin (e+f x))^2} \, dx=-\frac {b^2 (2 b c-9 d) x}{d^3}+\frac {2 (b c-3 d)^2 \left (2 b c^2+3 c d-3 b d^2\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{d^3 \left (c^2-d^2\right )^{3/2} f}+\frac {b \left (6 b c d-9 d^2-b^2 \left (2 c^2-d^2\right )\right ) \cos (e+f x)}{d^2 \left (c^2-d^2\right ) f}+\frac {(b c-3 d)^2 \cos (e+f x) (3+b \sin (e+f x))}{d \left (c^2-d^2\right ) f (c+d \sin (e+f x))} \]
-b^2*(-3*a*d+2*b*c)*x/d^3+2*(-a*d+b*c)^2*(a*c*d+2*b*c^2-3*b*d^2)*arctan((d +c*tan(1/2*f*x+1/2*e))/(c^2-d^2)^(1/2))/d^3/(c^2-d^2)^(3/2)/f+b*(2*a*b*c*d -a^2*d^2-b^2*(2*c^2-d^2))*cos(f*x+e)/d^2/(c^2-d^2)/f+(-a*d+b*c)^2*cos(f*x+ e)*(a+b*sin(f*x+e))/d/(c^2-d^2)/f/(c+d*sin(f*x+e))
Time = 2.07 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.75 \[ \int \frac {(3+b \sin (e+f x))^3}{(c+d \sin (e+f x))^2} \, dx=\frac {-b^2 (2 b c-9 d) (e+f x)+\frac {2 (b c-3 d)^2 \left (2 b c^2+3 c d-3 b d^2\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\left (c^2-d^2\right )^{3/2}}-b^3 d \cos (e+f x)+\frac {d (-b c+3 d)^3 \cos (e+f x)}{(c-d) (c+d) (c+d \sin (e+f x))}}{d^3 f} \]
(-(b^2*(2*b*c - 9*d)*(e + f*x)) + (2*(b*c - 3*d)^2*(2*b*c^2 + 3*c*d - 3*b* d^2)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(c^2 - d^2)^(3/2) - b^3*d*Cos[e + f*x] + (d*(-(b*c) + 3*d)^3*Cos[e + f*x])/((c - d)*(c + d)*( c + d*Sin[e + f*x])))/(d^3*f)
Time = 1.00 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.16, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3271, 3042, 3502, 3042, 3214, 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.
| \(\displaystyle \int \frac {(a+b \sin (e+f x))^3}{(c+d \sin (e+f x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \sin (e+f x))^3}{(c+d \sin (e+f x))^2}dx\) |
| \(\Big \downarrow \) 3271 |
| \(\displaystyle \frac {(b c-a d)^2 \cos (e+f x) (a+b \sin (e+f x))}{d f \left (c^2-d^2\right ) (c+d \sin (e+f x))}-\frac {\int \frac {-c d a^3+3 b d^2 a^2-3 b^2 c d a+b^3 c^2+b \left (-\left (\left (2 c^2-d^2\right ) b^2\right )+2 a c d b-a^2 d^2\right ) \sin ^2(e+f x)-b \left (a b c^2+\left (a^2+b^2\right ) d c-3 a b d^2\right ) \sin (e+f x)}{c+d \sin (e+f x)}dx}{d \left (c^2-d^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(b c-a d)^2 \cos (e+f x) (a+b \sin (e+f x))}{d f \left (c^2-d^2\right ) (c+d \sin (e+f x))}-\frac {\int \frac {-c d a^3+3 b d^2 a^2-3 b^2 c d a+b^3 c^2+b \left (-\left (\left (2 c^2-d^2\right ) b^2\right )+2 a c d b-a^2 d^2\right ) \sin (e+f x)^2-b \left (a b c^2+\left (a^2+b^2\right ) d c-3 a b d^2\right ) \sin (e+f x)}{c+d \sin (e+f x)}dx}{d \left (c^2-d^2\right )}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {(b c-a d)^2 \cos (e+f x) (a+b \sin (e+f x))}{d f \left (c^2-d^2\right ) (c+d \sin (e+f x))}-\frac {\frac {\int \frac {(2 b c-3 a d) \left (c^2-d^2\right ) \sin (e+f x) b^2+d \left (-c d a^3+3 b d^2 a^2-3 b^2 c d a+b^3 c^2\right )}{c+d \sin (e+f x)}dx}{d}-\frac {b \left (-a^2 d^2+2 a b c d-\left (b^2 \left (2 c^2-d^2\right )\right )\right ) \cos (e+f x)}{d f}}{d \left (c^2-d^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(b c-a d)^2 \cos (e+f x) (a+b \sin (e+f x))}{d f \left (c^2-d^2\right ) (c+d \sin (e+f x))}-\frac {\frac {\int \frac {(2 b c-3 a d) \left (c^2-d^2\right ) \sin (e+f x) b^2+d \left (-c d a^3+3 b d^2 a^2-3 b^2 c d a+b^3 c^2\right )}{c+d \sin (e+f x)}dx}{d}-\frac {b \left (-a^2 d^2+2 a b c d-\left (b^2 \left (2 c^2-d^2\right )\right )\right ) \cos (e+f x)}{d f}}{d \left (c^2-d^2\right )}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {(b c-a d)^2 \cos (e+f x) (a+b \sin (e+f x))}{d f \left (c^2-d^2\right ) (c+d \sin (e+f x))}-\frac {\frac {\frac {b^2 x \left (c^2-d^2\right ) (2 b c-3 a d)}{d}-\frac {(b c-a d)^2 \left (a c d+2 b c^2-3 b d^2\right ) \int \frac {1}{c+d \sin (e+f x)}dx}{d}}{d}-\frac {b \left (-a^2 d^2+2 a b c d-\left (b^2 \left (2 c^2-d^2\right )\right )\right ) \cos (e+f x)}{d f}}{d \left (c^2-d^2\right )}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(b c-a d)^2 \cos (e+f x) (a+b \sin (e+f x))}{d f \left (c^2-d^2\right ) (c+d \sin (e+f x))}-\frac {\frac {\frac {b^2 x \left (c^2-d^2\right ) (2 b c-3 a d)}{d}-\frac {(b c-a d)^2 \left (a c d+2 b c^2-3 b d^2\right ) \int \frac {1}{c+d \sin (e+f x)}dx}{d}}{d}-\frac {b \left (-a^2 d^2+2 a b c d-\left (b^2 \left (2 c^2-d^2\right )\right )\right ) \cos (e+f x)}{d f}}{d \left (c^2-d^2\right )}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle \frac {(b c-a d)^2 \cos (e+f x) (a+b \sin (e+f x))}{d f \left (c^2-d^2\right ) (c+d \sin (e+f x))}-\frac {\frac {\frac {b^2 x \left (c^2-d^2\right ) (2 b c-3 a d)}{d}-\frac {2 (b c-a d)^2 \left (a c d+2 b c^2-3 b d^2\right ) \int \frac {1}{c \tan ^2\left (\frac {1}{2} (e+f x)\right )+2 d \tan \left (\frac {1}{2} (e+f x)\right )+c}d\tan \left (\frac {1}{2} (e+f x)\right )}{d f}}{d}-\frac {b \left (-a^2 d^2+2 a b c d-\left (b^2 \left (2 c^2-d^2\right )\right )\right ) \cos (e+f x)}{d f}}{d \left (c^2-d^2\right )}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {(b c-a d)^2 \cos (e+f x) (a+b \sin (e+f x))}{d f \left (c^2-d^2\right ) (c+d \sin (e+f x))}-\frac {\frac {\frac {4 (b c-a d)^2 \left (a c d+2 b c^2-3 b d^2\right ) \int \frac {1}{-\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )^2-4 \left (c^2-d^2\right )}d\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )}{d f}+\frac {b^2 x \left (c^2-d^2\right ) (2 b c-3 a d)}{d}}{d}-\frac {b \left (-a^2 d^2+2 a b c d-\left (b^2 \left (2 c^2-d^2\right )\right )\right ) \cos (e+f x)}{d f}}{d \left (c^2-d^2\right )}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {(b c-a d)^2 \cos (e+f x) (a+b \sin (e+f x))}{d f \left (c^2-d^2\right ) (c+d \sin (e+f x))}-\frac {\frac {\frac {b^2 x \left (c^2-d^2\right ) (2 b c-3 a d)}{d}-\frac {2 (b c-a d)^2 \left (a c d+2 b c^2-3 b d^2\right ) \arctan \left (\frac {2 c \tan \left (\frac {1}{2} (e+f x)\right )+2 d}{2 \sqrt {c^2-d^2}}\right )}{d f \sqrt {c^2-d^2}}}{d}-\frac {b \left (-a^2 d^2+2 a b c d-\left (b^2 \left (2 c^2-d^2\right )\right )\right ) \cos (e+f x)}{d f}}{d \left (c^2-d^2\right )}\) |
-((((b^2*(2*b*c - 3*a*d)*(c^2 - d^2)*x)/d - (2*(b*c - a*d)^2*(2*b*c^2 + a* c*d - 3*b*d^2)*ArcTan[(2*d + 2*c*Tan[(e + f*x)/2])/(2*Sqrt[c^2 - d^2])])/( d*Sqrt[c^2 - d^2]*f))/d - (b*(2*a*b*c*d - a^2*d^2 - b^2*(2*c^2 - d^2))*Cos [e + f*x])/(d*f))/(d*(c^2 - d^2))) + ((b*c - a*d)^2*Cos[e + f*x]*(a + b*Si n[e + f*x]))/(d*(c^2 - d^2)*f*(c + d*Sin[e + f*x]))
3.7.91.3.1 Defintions of rubi rules used
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[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(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]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2))*Co s[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] + Simp[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])
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[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]
Time = 1.66 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.50
| method | result | size |
| derivativedivides | \(\frac {\frac {2 b^{2} \left (-\frac {b d}{1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )}+\left (3 d a -2 c b \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{d^{3}}+\frac {\frac {2 \left (\frac {d^{2} \left (d^{3} a^{3}-3 c \,d^{2} a^{2} b +3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c^{2}-d^{2}\right ) c}+\frac {d \left (d^{3} a^{3}-3 c \,d^{2} a^{2} b +3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}{c^{2}-d^{2}}\right )}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {2 \left (a^{3} c \,d^{3}-3 a^{2} b \,d^{4}-3 a \,b^{2} c^{3} d +6 a \,b^{2} c \,d^{3}+2 b^{3} c^{4}-3 b^{3} c^{2} d^{2}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c^{2}-d^{2}\right )^{\frac {3}{2}}}}{d^{3}}}{f}\) | \(302\) |
| default | \(\frac {\frac {2 b^{2} \left (-\frac {b d}{1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )}+\left (3 d a -2 c b \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right )}{d^{3}}+\frac {\frac {2 \left (\frac {d^{2} \left (d^{3} a^{3}-3 c \,d^{2} a^{2} b +3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c^{2}-d^{2}\right ) c}+\frac {d \left (d^{3} a^{3}-3 c \,d^{2} a^{2} b +3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}{c^{2}-d^{2}}\right )}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {2 \left (a^{3} c \,d^{3}-3 a^{2} b \,d^{4}-3 a \,b^{2} c^{3} d +6 a \,b^{2} c \,d^{3}+2 b^{3} c^{4}-3 b^{3} c^{2} d^{2}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c^{2}-d^{2}\right )^{\frac {3}{2}}}}{d^{3}}}{f}\) | \(302\) |
| risch | \(\text {Expression too large to display}\) | \(1185\) |
1/f*(2*b^2/d^3*(-b*d/(1+tan(1/2*f*x+1/2*e)^2)+(3*a*d-2*b*c)*arctan(tan(1/2 *f*x+1/2*e)))+2/d^3*((d^2*(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)/(c ^2-d^2)/c*tan(1/2*f*x+1/2*e)+d*(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^ 3)/(c^2-d^2))/(tan(1/2*f*x+1/2*e)^2*c+2*d*tan(1/2*f*x+1/2*e)+c)+(a^3*c*d^3 -3*a^2*b*d^4-3*a*b^2*c^3*d+6*a*b^2*c*d^3+2*b^3*c^4-3*b^3*c^2*d^2)/(c^2-d^2 )^(3/2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 488 vs. \(2 (203) = 406\).
Time = 0.34 (sec) , antiderivative size = 1062, normalized size of antiderivative = 5.28 \[ \int \frac {(3+b \sin (e+f x))^3}{(c+d \sin (e+f x))^2} \, dx=\left [-\frac {2 \, {\left (2 \, b^{3} c^{6} - 3 \, a b^{2} c^{5} d - 4 \, b^{3} c^{4} d^{2} + 6 \, a b^{2} c^{3} d^{3} + 2 \, b^{3} c^{2} d^{4} - 3 \, a b^{2} c d^{5}\right )} f x + {\left (2 \, b^{3} c^{5} - 3 \, a b^{2} c^{4} d - 3 \, b^{3} c^{3} d^{2} - 3 \, a^{2} b c d^{4} + {\left (a^{3} + 6 \, a b^{2}\right )} c^{2} d^{3} + {\left (2 \, b^{3} c^{4} d - 3 \, a b^{2} c^{3} d^{2} - 3 \, b^{3} c^{2} d^{3} - 3 \, a^{2} b d^{5} + {\left (a^{3} + 6 \, a b^{2}\right )} c d^{4}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c^{2} + d^{2}} \log \left (\frac {{\left (2 \, c^{2} - d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2} + 2 \, {\left (c \cos \left (f x + e\right ) \sin \left (f x + e\right ) + d \cos \left (f x + e\right )\right )} \sqrt {-c^{2} + d^{2}}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}\right ) + 2 \, {\left (2 \, b^{3} c^{5} d - 3 \, a b^{2} c^{4} d^{2} + a^{3} d^{6} + 3 \, {\left (a^{2} b - b^{3}\right )} c^{3} d^{3} - {\left (a^{3} - 3 \, a b^{2}\right )} c^{2} d^{4} - {\left (3 \, a^{2} b - b^{3}\right )} c d^{5}\right )} \cos \left (f x + e\right ) + 2 \, {\left ({\left (2 \, b^{3} c^{5} d - 3 \, a b^{2} c^{4} d^{2} - 4 \, b^{3} c^{3} d^{3} + 6 \, a b^{2} c^{2} d^{4} + 2 \, b^{3} c d^{5} - 3 \, a b^{2} d^{6}\right )} f x + {\left (b^{3} c^{4} d^{2} - 2 \, b^{3} c^{2} d^{4} + b^{3} d^{6}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{2 \, {\left ({\left (c^{4} d^{4} - 2 \, c^{2} d^{6} + d^{8}\right )} f \sin \left (f x + e\right ) + {\left (c^{5} d^{3} - 2 \, c^{3} d^{5} + c d^{7}\right )} f\right )}}, -\frac {{\left (2 \, b^{3} c^{6} - 3 \, a b^{2} c^{5} d - 4 \, b^{3} c^{4} d^{2} + 6 \, a b^{2} c^{3} d^{3} + 2 \, b^{3} c^{2} d^{4} - 3 \, a b^{2} c d^{5}\right )} f x + {\left (2 \, b^{3} c^{5} - 3 \, a b^{2} c^{4} d - 3 \, b^{3} c^{3} d^{2} - 3 \, a^{2} b c d^{4} + {\left (a^{3} + 6 \, a b^{2}\right )} c^{2} d^{3} + {\left (2 \, b^{3} c^{4} d - 3 \, a b^{2} c^{3} d^{2} - 3 \, b^{3} c^{2} d^{3} - 3 \, a^{2} b d^{5} + {\left (a^{3} + 6 \, a b^{2}\right )} c d^{4}\right )} \sin \left (f x + e\right )\right )} \sqrt {c^{2} - d^{2}} \arctan \left (-\frac {c \sin \left (f x + e\right ) + d}{\sqrt {c^{2} - d^{2}} \cos \left (f x + e\right )}\right ) + {\left (2 \, b^{3} c^{5} d - 3 \, a b^{2} c^{4} d^{2} + a^{3} d^{6} + 3 \, {\left (a^{2} b - b^{3}\right )} c^{3} d^{3} - {\left (a^{3} - 3 \, a b^{2}\right )} c^{2} d^{4} - {\left (3 \, a^{2} b - b^{3}\right )} c d^{5}\right )} \cos \left (f x + e\right ) + {\left ({\left (2 \, b^{3} c^{5} d - 3 \, a b^{2} c^{4} d^{2} - 4 \, b^{3} c^{3} d^{3} + 6 \, a b^{2} c^{2} d^{4} + 2 \, b^{3} c d^{5} - 3 \, a b^{2} d^{6}\right )} f x + {\left (b^{3} c^{4} d^{2} - 2 \, b^{3} c^{2} d^{4} + b^{3} d^{6}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{{\left (c^{4} d^{4} - 2 \, c^{2} d^{6} + d^{8}\right )} f \sin \left (f x + e\right ) + {\left (c^{5} d^{3} - 2 \, c^{3} d^{5} + c d^{7}\right )} f}\right ] \]
[-1/2*(2*(2*b^3*c^6 - 3*a*b^2*c^5*d - 4*b^3*c^4*d^2 + 6*a*b^2*c^3*d^3 + 2* b^3*c^2*d^4 - 3*a*b^2*c*d^5)*f*x + (2*b^3*c^5 - 3*a*b^2*c^4*d - 3*b^3*c^3* d^2 - 3*a^2*b*c*d^4 + (a^3 + 6*a*b^2)*c^2*d^3 + (2*b^3*c^4*d - 3*a*b^2*c^3 *d^2 - 3*b^3*c^2*d^3 - 3*a^2*b*d^5 + (a^3 + 6*a*b^2)*c*d^4)*sin(f*x + e))* sqrt(-c^2 + d^2)*log(((2*c^2 - d^2)*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2 + 2*(c*cos(f*x + e)*sin(f*x + e) + d*cos(f*x + e))*sqrt(-c^2 + d ^2))/(d^2*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2)) + 2*(2*b^3*c^5 *d - 3*a*b^2*c^4*d^2 + a^3*d^6 + 3*(a^2*b - b^3)*c^3*d^3 - (a^3 - 3*a*b^2) *c^2*d^4 - (3*a^2*b - b^3)*c*d^5)*cos(f*x + e) + 2*((2*b^3*c^5*d - 3*a*b^2 *c^4*d^2 - 4*b^3*c^3*d^3 + 6*a*b^2*c^2*d^4 + 2*b^3*c*d^5 - 3*a*b^2*d^6)*f* x + (b^3*c^4*d^2 - 2*b^3*c^2*d^4 + b^3*d^6)*cos(f*x + e))*sin(f*x + e))/(( c^4*d^4 - 2*c^2*d^6 + d^8)*f*sin(f*x + e) + (c^5*d^3 - 2*c^3*d^5 + c*d^7)* f), -((2*b^3*c^6 - 3*a*b^2*c^5*d - 4*b^3*c^4*d^2 + 6*a*b^2*c^3*d^3 + 2*b^3 *c^2*d^4 - 3*a*b^2*c*d^5)*f*x + (2*b^3*c^5 - 3*a*b^2*c^4*d - 3*b^3*c^3*d^2 - 3*a^2*b*c*d^4 + (a^3 + 6*a*b^2)*c^2*d^3 + (2*b^3*c^4*d - 3*a*b^2*c^3*d^ 2 - 3*b^3*c^2*d^3 - 3*a^2*b*d^5 + (a^3 + 6*a*b^2)*c*d^4)*sin(f*x + e))*sqr t(c^2 - d^2)*arctan(-(c*sin(f*x + e) + d)/(sqrt(c^2 - d^2)*cos(f*x + e))) + (2*b^3*c^5*d - 3*a*b^2*c^4*d^2 + a^3*d^6 + 3*(a^2*b - b^3)*c^3*d^3 - (a^ 3 - 3*a*b^2)*c^2*d^4 - (3*a^2*b - b^3)*c*d^5)*cos(f*x + e) + ((2*b^3*c^5*d - 3*a*b^2*c^4*d^2 - 4*b^3*c^3*d^3 + 6*a*b^2*c^2*d^4 + 2*b^3*c*d^5 - 3*...
Timed out. \[ \int \frac {(3+b \sin (e+f x))^3}{(c+d \sin (e+f x))^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(3+b \sin (e+f x))^3}{(c+d \sin (e+f x))^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*d^2-4*c^2>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 564 vs. \(2 (203) = 406\).
Time = 0.30 (sec) , antiderivative size = 564, normalized size of antiderivative = 2.81 \[ \int \frac {(3+b \sin (e+f x))^3}{(c+d \sin (e+f x))^2} \, dx=\frac {\frac {2 \, {\left (2 \, b^{3} c^{4} - 3 \, a b^{2} c^{3} d - 3 \, b^{3} c^{2} d^{2} + a^{3} c d^{3} + 6 \, a b^{2} c d^{3} - 3 \, a^{2} b d^{4}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )}}{{\left (c^{2} d^{3} - d^{5}\right )} \sqrt {c^{2} - d^{2}}} - \frac {2 \, {\left (b^{3} c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 3 \, a b^{2} c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 3 \, a^{2} b c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - a^{3} d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2 \, b^{3} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 3 \, a b^{2} c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, a^{2} b c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - b^{3} c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a^{3} c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, b^{3} c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 3 \, a b^{2} c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, a^{2} b c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, b^{3} c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - a^{3} d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, b^{3} c^{4} - 3 \, a b^{2} c^{3} d + 3 \, a^{2} b c^{2} d^{2} - b^{3} c^{2} d^{2} - a^{3} c d^{3}\right )}}{{\left (c^{3} d^{2} - c d^{4}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c\right )}} - \frac {{\left (2 \, b^{3} c - 3 \, a b^{2} d\right )} {\left (f x + e\right )}}{d^{3}}}{f} \]
(2*(2*b^3*c^4 - 3*a*b^2*c^3*d - 3*b^3*c^2*d^2 + a^3*c*d^3 + 6*a*b^2*c*d^3 - 3*a^2*b*d^4)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(c) + arctan((c*tan(1/ 2*f*x + 1/2*e) + d)/sqrt(c^2 - d^2)))/((c^2*d^3 - d^5)*sqrt(c^2 - d^2)) - 2*(b^3*c^3*d*tan(1/2*f*x + 1/2*e)^3 - 3*a*b^2*c^2*d^2*tan(1/2*f*x + 1/2*e) ^3 + 3*a^2*b*c*d^3*tan(1/2*f*x + 1/2*e)^3 - a^3*d^4*tan(1/2*f*x + 1/2*e)^3 + 2*b^3*c^4*tan(1/2*f*x + 1/2*e)^2 - 3*a*b^2*c^3*d*tan(1/2*f*x + 1/2*e)^2 + 3*a^2*b*c^2*d^2*tan(1/2*f*x + 1/2*e)^2 - b^3*c^2*d^2*tan(1/2*f*x + 1/2* e)^2 - a^3*c*d^3*tan(1/2*f*x + 1/2*e)^2 + 3*b^3*c^3*d*tan(1/2*f*x + 1/2*e) - 3*a*b^2*c^2*d^2*tan(1/2*f*x + 1/2*e) + 3*a^2*b*c*d^3*tan(1/2*f*x + 1/2* e) - 2*b^3*c*d^3*tan(1/2*f*x + 1/2*e) - a^3*d^4*tan(1/2*f*x + 1/2*e) + 2*b ^3*c^4 - 3*a*b^2*c^3*d + 3*a^2*b*c^2*d^2 - b^3*c^2*d^2 - a^3*c*d^3)/((c^3* d^2 - c*d^4)*(c*tan(1/2*f*x + 1/2*e)^4 + 2*d*tan(1/2*f*x + 1/2*e)^3 + 2*c* tan(1/2*f*x + 1/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*e) + c)) - (2*b^3*c - 3*a*b ^2*d)*(f*x + e)/d^3)/f
Time = 19.13 (sec) , antiderivative size = 8953, normalized size of antiderivative = 44.54 \[ \int \frac {(3+b \sin (e+f x))^3}{(c+d \sin (e+f x))^2} \, dx=\text {Too large to display} \]
((2*(a^3*d^3 - 2*b^3*c^3 + b^3*c*d^2 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2))/(d^ 2*(c^2 - d^2)) + (2*tan(e/2 + (f*x)/2)^2*(a^3*d^3 - 2*b^3*c^3 + b^3*c*d^2 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2))/(d^2*(c^2 - d^2)) + (2*tan(e/2 + (f*x)/2 )*(a^3*d^3 - 3*b^3*c^3 + 2*b^3*c*d^2 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2))/(c* d*(c^2 - d^2)) + (2*tan(e/2 + (f*x)/2)^3*(a^3*d^3 - b^3*c^3 + 3*a*b^2*c^2* d - 3*a^2*b*c*d^2))/(c*d*(c^2 - d^2)))/(f*(c + 2*d*tan(e/2 + (f*x)/2) + 2* c*tan(e/2 + (f*x)/2)^2 + c*tan(e/2 + (f*x)/2)^4 + 2*d*tan(e/2 + (f*x)/2)^3 )) + (2*b^2*atan(((b^2*(3*a*d - 2*b*c)*((32*(4*b^6*c^4*d^6 - 8*b^6*c^6*d^4 + 4*b^6*c^8*d^2 - 12*a*b^5*c^3*d^7 + 24*a*b^5*c^5*d^5 - 12*a*b^5*c^7*d^3 + 9*a^2*b^4*c^2*d^8 - 18*a^2*b^4*c^4*d^6 + 9*a^2*b^4*c^6*d^4))/(d^9 - 2*c^ 2*d^7 + c^4*d^5) - (32*tan(e/2 + (f*x)/2)*(a^6*c^3*d^8 - 8*b^6*c^3*d^8 + 2 9*b^6*c^5*d^6 - 28*b^6*c^7*d^4 + 8*b^6*c^9*d^2 + 24*a*b^5*c^2*d^9 - 96*a*b ^5*c^4*d^7 + 90*a*b^5*c^6*d^5 - 24*a*b^5*c^8*d^3 - 18*a^2*b^4*c*d^10 + 9*a ^4*b^2*c*d^10 - 6*a^5*b*c^2*d^9 + 99*a^2*b^4*c^3*d^8 - 84*a^2*b^4*c^5*d^6 + 18*a^2*b^4*c^7*d^4 - 36*a^3*b^3*c^2*d^9 + 12*a^3*b^3*c^4*d^7 + 4*a^3*b^3 *c^6*d^5 + 12*a^4*b^2*c^3*d^8 - 6*a^4*b^2*c^5*d^6))/(d^10 - 2*c^2*d^8 + c^ 4*d^6) + (b^2*(3*a*d - 2*b*c)*((32*tan(e/2 + (f*x)/2)*(2*a^3*c^2*d^11 - 2* a^3*c^4*d^9 - 6*b^3*c^3*d^10 + 10*b^3*c^5*d^8 - 4*b^3*c^7*d^6 + 12*a*b^2*c ^2*d^11 - 18*a*b^2*c^4*d^9 + 6*a*b^2*c^6*d^7 + 6*a^2*b*c^3*d^10 - 6*a^2*b* c*d^12))/(d^10 - 2*c^2*d^8 + c^4*d^6) - (32*(a^3*c^5*d^7 - a^3*c^3*d^9 ...