Integrand size = 25, antiderivative size = 150 \[ \int \frac {(3+b \sin (e+f x))^3}{c+d \sin (e+f x)} \, dx=-\frac {b \left (18 b c d-54 d^2-b^2 \left (2 c^2+d^2\right )\right ) x}{2 d^3}-\frac {2 (b c-3 d)^3 \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{d^3 \sqrt {c^2-d^2} f}+\frac {b^2 (2 b c-15 d) \cos (e+f x)}{2 d^2 f}-\frac {b^2 \cos (e+f x) (3+b \sin (e+f x))}{2 d f} \]
-1/2*b*(6*a*b*c*d-6*a^2*d^2-b^2*(2*c^2+d^2))*x/d^3+1/2*b^2*(-5*a*d+2*b*c)* cos(f*x+e)/d^2/f-1/2*b^2*cos(f*x+e)*(a+b*sin(f*x+e))/d/f-2*(-a*d+b*c)^3*ar ctan((d+c*tan(1/2*f*x+1/2*e))/(c^2-d^2)^(1/2))/d^3/f/(c^2-d^2)^(1/2)
Time = 1.92 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.87 \[ \int \frac {(3+b \sin (e+f x))^3}{c+d \sin (e+f x)} \, dx=\frac {2 b \left (-18 b c d+54 d^2+b^2 \left (2 c^2+d^2\right )\right ) (e+f x)-\frac {8 (b c-3 d)^3 \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\sqrt {c^2-d^2}}+4 b^2 (b c-9 d) d \cos (e+f x)-b^3 d^2 \sin (2 (e+f x))}{4 d^3 f} \]
(2*b*(-18*b*c*d + 54*d^2 + b^2*(2*c^2 + d^2))*(e + f*x) - (8*(b*c - 3*d)^3 *ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/Sqrt[c^2 - d^2] + 4*b^2 *(b*c - 9*d)*d*Cos[e + f*x] - b^3*d^2*Sin[2*(e + f*x)])/(4*d^3*f)
Time = 0.79 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.13, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3272, 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.
| \(\displaystyle \int \frac {(a+b \sin (e+f x))^3}{c+d \sin (e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \sin (e+f x))^3}{c+d \sin (e+f x)}dx\) |
| \(\Big \downarrow \) 3272 |
| \(\displaystyle \frac {\int \frac {2 d a^3-b^2 (2 b c-5 a d) \sin ^2(e+f x)+b^3 c-b \left (-6 d a^2+b c a-b^2 d\right ) \sin (e+f x)}{c+d \sin (e+f x)}dx}{2 d}-\frac {b^2 \cos (e+f x) (a+b \sin (e+f x))}{2 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {2 d a^3-b^2 (2 b c-5 a d) \sin (e+f x)^2+b^3 c-b \left (-6 d a^2+b c a-b^2 d\right ) \sin (e+f x)}{c+d \sin (e+f x)}dx}{2 d}-\frac {b^2 \cos (e+f x) (a+b \sin (e+f x))}{2 d f}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\frac {\int \frac {d \left (2 d a^3+b^3 c\right )-b \left (-\left (\left (2 c^2+d^2\right ) b^2\right )+6 a c d b-6 a^2 d^2\right ) \sin (e+f x)}{c+d \sin (e+f x)}dx}{d}+\frac {b^2 (2 b c-5 a d) \cos (e+f x)}{d f}}{2 d}-\frac {b^2 \cos (e+f x) (a+b \sin (e+f x))}{2 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {d \left (2 d a^3+b^3 c\right )-b \left (-\left (\left (2 c^2+d^2\right ) b^2\right )+6 a c d b-6 a^2 d^2\right ) \sin (e+f x)}{c+d \sin (e+f x)}dx}{d}+\frac {b^2 (2 b c-5 a d) \cos (e+f x)}{d f}}{2 d}-\frac {b^2 \cos (e+f x) (a+b \sin (e+f x))}{2 d f}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {\frac {-\frac {2 (b c-a d)^3 \int \frac {1}{c+d \sin (e+f x)}dx}{d}-\frac {b x \left (-6 a^2 d^2+6 a b c d-\left (b^2 \left (2 c^2+d^2\right )\right )\right )}{d}}{d}+\frac {b^2 (2 b c-5 a d) \cos (e+f x)}{d f}}{2 d}-\frac {b^2 \cos (e+f x) (a+b \sin (e+f x))}{2 d f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {-\frac {2 (b c-a d)^3 \int \frac {1}{c+d \sin (e+f x)}dx}{d}-\frac {b x \left (-6 a^2 d^2+6 a b c d-\left (b^2 \left (2 c^2+d^2\right )\right )\right )}{d}}{d}+\frac {b^2 (2 b c-5 a d) \cos (e+f x)}{d f}}{2 d}-\frac {b^2 \cos (e+f x) (a+b \sin (e+f x))}{2 d f}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle \frac {\frac {-\frac {4 (b c-a d)^3 \int \frac {1}{c \tan ^2\left (\frac {1}{2} (e+f x)\right )+2 d \tan \left (\frac {1}{2} (e+f x)\right )+c}d\tan \left (\frac {1}{2} (e+f x)\right )}{d f}-\frac {b x \left (-6 a^2 d^2+6 a b c d-\left (b^2 \left (2 c^2+d^2\right )\right )\right )}{d}}{d}+\frac {b^2 (2 b c-5 a d) \cos (e+f x)}{d f}}{2 d}-\frac {b^2 \cos (e+f x) (a+b \sin (e+f x))}{2 d f}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {\frac {\frac {8 (b c-a d)^3 \int \frac {1}{-\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )^2-4 \left (c^2-d^2\right )}d\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )}{d f}-\frac {b x \left (-6 a^2 d^2+6 a b c d-\left (b^2 \left (2 c^2+d^2\right )\right )\right )}{d}}{d}+\frac {b^2 (2 b c-5 a d) \cos (e+f x)}{d f}}{2 d}-\frac {b^2 \cos (e+f x) (a+b \sin (e+f x))}{2 d f}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\frac {-\frac {b x \left (-6 a^2 d^2+6 a b c d-\left (b^2 \left (2 c^2+d^2\right )\right )\right )}{d}-\frac {4 (b c-a d)^3 \arctan \left (\frac {2 c \tan \left (\frac {1}{2} (e+f x)\right )+2 d}{2 \sqrt {c^2-d^2}}\right )}{d f \sqrt {c^2-d^2}}}{d}+\frac {b^2 (2 b c-5 a d) \cos (e+f x)}{d f}}{2 d}-\frac {b^2 \cos (e+f x) (a+b \sin (e+f x))}{2 d f}\) |
((-((b*(6*a*b*c*d - 6*a^2*d^2 - b^2*(2*c^2 + d^2))*x)/d) - (4*(b*c - a*d)^ 3*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*(2*b*c - 5*a*d)*Cos[e + f*x])/(d*f))/(2*d) - (b^2*Cos[e + f*x]*(a + b*Sin[e + f*x]))/(2*d*f)
3.7.90.3.1 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)*Cos[e + f*x]*(a + b*Sin[e + f* x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d *(m + n) + b^2*(b*c*(m - 2) + a*d*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Si n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m ] || IntegersQ[2*m, 2*n]) && !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
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 = 1.32 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.53
| method | result | size |
| derivativedivides | \(\frac {\frac {2 \left (d^{3} a^{3}-3 c \,d^{2} a^{2} b +3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{d^{3} \sqrt {c^{2}-d^{2}}}+\frac {2 b \left (\frac {\frac {\left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) b^{2} d^{2}}{2}+\left (-3 a b \,d^{2}+b^{2} c d \right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) b^{2} d^{2}}{2}-3 a b \,d^{2}+b^{2} c d}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}+\frac {\left (6 d^{2} a^{2}-6 a b c d +2 b^{2} c^{2}+d^{2} b^{2}\right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}\right )}{d^{3}}}{f}\) | \(229\) |
| default | \(\frac {\frac {2 \left (d^{3} a^{3}-3 c \,d^{2} a^{2} b +3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{d^{3} \sqrt {c^{2}-d^{2}}}+\frac {2 b \left (\frac {\frac {\left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) b^{2} d^{2}}{2}+\left (-3 a b \,d^{2}+b^{2} c d \right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) b^{2} d^{2}}{2}-3 a b \,d^{2}+b^{2} c d}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}+\frac {\left (6 d^{2} a^{2}-6 a b c d +2 b^{2} c^{2}+d^{2} b^{2}\right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}\right )}{d^{3}}}{f}\) | \(229\) |
| risch | \(\frac {3 b x \,a^{2}}{d}-\frac {3 b^{2} x a c}{d^{2}}+\frac {b^{3} x \,c^{2}}{d^{3}}+\frac {b^{3} x}{2 d}-\frac {3 b^{2} {\mathrm e}^{i \left (f x +e \right )} a}{2 d f}+\frac {b^{3} {\mathrm e}^{i \left (f x +e \right )} c}{2 d^{2} f}-\frac {3 b^{2} {\mathrm e}^{-i \left (f x +e \right )} a}{2 d f}+\frac {b^{3} {\mathrm e}^{-i \left (f x +e \right )} c}{2 d^{2} f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}-c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) a^{3}}{\sqrt {-c^{2}+d^{2}}\, f}+\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}-c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) c \,a^{2} b}{\sqrt {-c^{2}+d^{2}}\, f d}-\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}-c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) a \,b^{2} c^{2}}{\sqrt {-c^{2}+d^{2}}\, f \,d^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}-c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) b^{3} c^{3}}{\sqrt {-c^{2}+d^{2}}\, f \,d^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}+c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) a^{3}}{\sqrt {-c^{2}+d^{2}}\, f}-\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}+c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) c \,a^{2} b}{\sqrt {-c^{2}+d^{2}}\, f d}+\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}+c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) a \,b^{2} c^{2}}{\sqrt {-c^{2}+d^{2}}\, f \,d^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}+c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) b^{3} c^{3}}{\sqrt {-c^{2}+d^{2}}\, f \,d^{3}}-\frac {b^{3} \sin \left (2 f x +2 e \right )}{4 d f}\) | \(733\) |
1/f*(2*(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)/d^3/(c^2-d^2)^(1/2)*a rctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1/2))+2*b/d^3*((1/2*tan( 1/2*f*x+1/2*e)^3*b^2*d^2+(-3*a*b*d^2+b^2*c*d)*tan(1/2*f*x+1/2*e)^2-1/2*tan (1/2*f*x+1/2*e)*b^2*d^2-3*a*b*d^2+b^2*c*d)/(1+tan(1/2*f*x+1/2*e)^2)^2+1/2* (6*a^2*d^2-6*a*b*c*d+2*b^2*c^2+b^2*d^2)*arctan(tan(1/2*f*x+1/2*e))))
Time = 0.30 (sec) , antiderivative size = 578, normalized size of antiderivative = 3.85 \[ \int \frac {(3+b \sin (e+f x))^3}{c+d \sin (e+f x)} \, dx=\left [\frac {{\left (2 \, b^{3} c^{4} - 6 \, a b^{2} c^{3} d + 6 \, a b^{2} c d^{3} + {\left (6 \, a^{2} b - b^{3}\right )} c^{2} d^{2} - {\left (6 \, a^{2} b + b^{3}\right )} d^{4}\right )} f x - {\left (b^{3} c^{2} d^{2} - b^{3} d^{4}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {-c^{2} + d^{2}} \log \left (\frac {{\left (2 \, c^{2} - d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2} + 2 \, {\left (c \cos \left (f x + e\right ) \sin \left (f x + e\right ) + d \cos \left (f x + e\right )\right )} \sqrt {-c^{2} + d^{2}}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}\right ) + 2 \, {\left (b^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} - b^{3} c d^{3} + 3 \, a b^{2} d^{4}\right )} \cos \left (f x + e\right )}{2 \, {\left (c^{2} d^{3} - d^{5}\right )} f}, \frac {{\left (2 \, b^{3} c^{4} - 6 \, a b^{2} c^{3} d + 6 \, a b^{2} c d^{3} + {\left (6 \, a^{2} b - b^{3}\right )} c^{2} d^{2} - {\left (6 \, a^{2} b + b^{3}\right )} d^{4}\right )} f x - {\left (b^{3} c^{2} d^{2} - b^{3} d^{4}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 2 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {c^{2} - d^{2}} \arctan \left (-\frac {c \sin \left (f x + e\right ) + d}{\sqrt {c^{2} - d^{2}} \cos \left (f x + e\right )}\right ) + 2 \, {\left (b^{3} c^{3} d - 3 \, a b^{2} c^{2} d^{2} - b^{3} c d^{3} + 3 \, a b^{2} d^{4}\right )} \cos \left (f x + e\right )}{2 \, {\left (c^{2} d^{3} - d^{5}\right )} f}\right ] \]
[1/2*((2*b^3*c^4 - 6*a*b^2*c^3*d + 6*a*b^2*c*d^3 + (6*a^2*b - b^3)*c^2*d^2 - (6*a^2*b + b^3)*d^4)*f*x - (b^3*c^2*d^2 - b^3*d^4)*cos(f*x + e)*sin(f*x + e) + (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*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*(b^3*c^3*d - 3*a*b^2*c^2 *d^2 - b^3*c*d^3 + 3*a*b^2*d^4)*cos(f*x + e))/((c^2*d^3 - d^5)*f), 1/2*((2 *b^3*c^4 - 6*a*b^2*c^3*d + 6*a*b^2*c*d^3 + (6*a^2*b - b^3)*c^2*d^2 - (6*a^ 2*b + b^3)*d^4)*f*x - (b^3*c^2*d^2 - b^3*d^4)*cos(f*x + e)*sin(f*x + e) + 2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(c^2 - d^2)*arct an(-(c*sin(f*x + e) + d)/(sqrt(c^2 - d^2)*cos(f*x + e))) + 2*(b^3*c^3*d - 3*a*b^2*c^2*d^2 - b^3*c*d^3 + 3*a*b^2*d^4)*cos(f*x + e))/((c^2*d^3 - d^5)* f)]
Timed out. \[ \int \frac {(3+b \sin (e+f x))^3}{c+d \sin (e+f x)} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(3+b \sin (e+f x))^3}{c+d \sin (e+f x)} \, dx=\text {Exception raised: ValueError} \]
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
Time = 0.30 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.63 \[ \int \frac {(3+b \sin (e+f x))^3}{c+d \sin (e+f x)} \, dx=\frac {\frac {{\left (2 \, b^{3} c^{2} - 6 \, a b^{2} c d + 6 \, a^{2} b d^{2} + b^{3} d^{2}\right )} {\left (f x + e\right )}}{d^{3}} - \frac {4 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )}}{\sqrt {c^{2} - d^{2}} d^{3}} + \frac {2 \, {\left (b^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2 \, b^{3} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 6 \, a b^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - b^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, b^{3} c - 6 \, a b^{2} d\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}^{2} d^{2}}}{2 \, f} \]
1/2*((2*b^3*c^2 - 6*a*b^2*c*d + 6*a^2*b*d^2 + b^3*d^2)*(f*x + e)/d^3 - 4*( b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*(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)))/ (sqrt(c^2 - d^2)*d^3) + 2*(b^3*d*tan(1/2*f*x + 1/2*e)^3 + 2*b^3*c*tan(1/2* f*x + 1/2*e)^2 - 6*a*b^2*d*tan(1/2*f*x + 1/2*e)^2 - b^3*d*tan(1/2*f*x + 1/ 2*e) + 2*b^3*c - 6*a*b^2*d)/((tan(1/2*f*x + 1/2*e)^2 + 1)^2*d^2))/f
Time = 15.71 (sec) , antiderivative size = 5902, normalized size of antiderivative = 39.35 \[ \int \frac {(3+b \sin (e+f x))^3}{c+d \sin (e+f x)} \, dx=\text {Too large to display} \]
((2*(b^3*c - 3*a*b^2*d))/d^2 + (b^3*tan(e/2 + (f*x)/2)^3)/d + (2*tan(e/2 + (f*x)/2)^2*(b^3*c - 3*a*b^2*d))/d^2 - (b^3*tan(e/2 + (f*x)/2))/d)/(f*(2*t an(e/2 + (f*x)/2)^2 + tan(e/2 + (f*x)/2)^4 + 1)) + (atan((((b^3*c^2*1i + ( b*d^2*(6*a^2 + b^2)*1i)/2 - a*b^2*c*d*3i)*((8*(b^6*c^2*d^6 + 4*b^6*c^4*d^4 + 4*b^6*c^6*d^2 - 12*a*b^5*c^3*d^5 - 24*a*b^5*c^5*d^3 + 12*a^2*b^4*c^2*d^ 6 + 60*a^2*b^4*c^4*d^4 - 72*a^3*b^3*c^3*d^5 + 36*a^4*b^2*c^2*d^6))/d^5 + ( 8*tan(e/2 + (f*x)/2)*(2*b^6*c*d^8 - 4*a^6*c*d^8 + 7*b^6*c^3*d^6 + 4*b^6*c^ 5*d^4 - 8*b^6*c^7*d^2 - 24*a*b^5*c^2*d^7 - 36*a*b^5*c^4*d^5 + 48*a*b^5*c^6 *d^3 + 24*a^2*b^4*c*d^8 + 72*a^4*b^2*c*d^8 + 24*a^5*b*c^2*d^7 + 108*a^2*b^ 4*c^3*d^6 - 120*a^2*b^4*c^5*d^4 - 144*a^3*b^3*c^2*d^7 + 152*a^3*b^3*c^4*d^ 5 - 96*a^4*b^2*c^3*d^6))/d^6 + ((b^3*c^2*1i + (b*d^2*(6*a^2 + b^2)*1i)/2 - a*b^2*c*d*3i)*((8*tan(e/2 + (f*x)/2)*(8*a^3*c*d^9 - 8*b^3*c^4*d^6 + 24*a* b^2*c^3*d^7 - 24*a^2*b*c^2*d^8))/d^6 - (8*(2*b^3*c*d^8 - 4*a^3*c^2*d^7 + 2 *b^3*c^3*d^6 - 12*a*b^2*c^2*d^7 + 12*a^2*b*c*d^8))/d^5 + ((32*c^2*d^3 + (8 *tan(e/2 + (f*x)/2)*(12*c*d^10 - 8*c^3*d^8))/d^6)*(b^3*c^2*1i + (b*d^2*(6* a^2 + b^2)*1i)/2 - a*b^2*c*d*3i))/d^3))/d^3)*1i)/d^3 + ((b^3*c^2*1i + (b*d ^2*(6*a^2 + b^2)*1i)/2 - a*b^2*c*d*3i)*((8*(b^6*c^2*d^6 + 4*b^6*c^4*d^4 + 4*b^6*c^6*d^2 - 12*a*b^5*c^3*d^5 - 24*a*b^5*c^5*d^3 + 12*a^2*b^4*c^2*d^6 + 60*a^2*b^4*c^4*d^4 - 72*a^3*b^3*c^3*d^5 + 36*a^4*b^2*c^2*d^6))/d^5 + (8*t an(e/2 + (f*x)/2)*(2*b^6*c*d^8 - 4*a^6*c*d^8 + 7*b^6*c^3*d^6 + 4*b^6*c^...