Integrand size = 25, antiderivative size = 135 \[ \int \frac {(c+d \sin (e+f x))^3}{(3+3 \sin (e+f x))^3} \, dx=\frac {d^3 x}{27}-\frac {(c-d)^2 (2 c+7 d) \cos (e+f x)}{45 f (3+3 \sin (e+f x))^2}-\frac {(c-d) \left (2 c^2+11 c d+29 d^2\right ) \cos (e+f x)}{15 f (27+27 \sin (e+f x))}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (3+3 \sin (e+f x))^3} \]
d^3*x/a^3-1/15*(c-d)^2*(2*c+7*d)*cos(f*x+e)/a/f/(a+a*sin(f*x+e))^2-1/15*(c -d)*(2*c^2+11*c*d+29*d^2)*cos(f*x+e)/f/(a^3+a^3*sin(f*x+e))-1/5*(c-d)*cos( f*x+e)*(c+d*sin(f*x+e))^2/f/(a+a*sin(f*x+e))^3
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 8.26 (sec) , antiderivative size = 1363, normalized size of antiderivative = 10.10 \[ \int \frac {(c+d \sin (e+f x))^3}{(3+3 \sin (e+f x))^3} \, dx =\text {Too large to display} \]
-1/58320*(Cos[e + f*x]*(9450*Sqrt[2]*(c + d)^3*ArcSin[Sqrt[1 - Sin[e + f*x ]]/Sqrt[2]] - 4725*(c + d)^3*Sqrt[1 + Cos[2*e + 2*f*x]] + 142*(c + d)^3*Hy pergeometric2F1[3/2, 9/2, 11/2, (1 - Sin[e + f*x])/2]*(1 - Sin[e + f*x])^( 9/2) + 60*(c + d)^3*HypergeometricPFQ[{3/2, 2, 9/2}, {1, 11/2}, (1 - Sin[e + f*x])/2]*(1 - Sin[e + f*x])^(9/2) + 8*(c + d)^3*HypergeometricPFQ[{3/2, 2, 2, 9/2}, {1, 1, 11/2}, (1 - Sin[e + f*x])/2]*(1 - Sin[e + f*x])^(9/2) + 9450*Sqrt[2]*d*(c + d)^2*ArcSin[Sqrt[1 - Sin[e + f*x]]/Sqrt[2]]*(-1 + Si n[e + f*x]) + 5670*Sqrt[2]*d^2*(c + d)*ArcSin[Sqrt[1 - Sin[e + f*x]]/Sqrt[ 2]]*(-1 + Sin[e + f*x])^2 + 1350*Sqrt[2]*d^3*ArcSin[Sqrt[1 - Sin[e + f*x]] /Sqrt[2]]*(-1 + Sin[e + f*x])^3 - 1575*Sqrt[2]*(c + d)^3*(1 - Sin[e + f*x] )^(3/2)*Sqrt[1 + Sin[e + f*x]] - 630*Sqrt[2]*(c + d)^3*(1 - Sin[e + f*x])^ (5/2)*Sqrt[1 + Sin[e + f*x]] + 4725*Sqrt[2]*(c + d)^2*Sqrt[Cos[e + f*x]^2] *(d - d*Sin[e + f*x]) - 282*(c + d)^2*Hypergeometric2F1[3/2, 9/2, 11/2, (1 - Sin[e + f*x])/2]*(1 - Sin[e + f*x])^(9/2)*(d - d*Sin[e + f*x]) - 156*(c + d)^2*HypergeometricPFQ[{3/2, 2, 9/2}, {1, 11/2}, (1 - Sin[e + f*x])/2]* (1 - Sin[e + f*x])^(9/2)*(d - d*Sin[e + f*x]) - 24*(c + d)^2*Hypergeometri cPFQ[{3/2, 2, 2, 9/2}, {1, 1, 11/2}, (1 - Sin[e + f*x])/2]*(1 - Sin[e + f* x])^(9/2)*(d - d*Sin[e + f*x]) + 1575*Sqrt[2]*(c + d)^2*(1 - Sin[e + f*x]) ^(3/2)*Sqrt[1 + Sin[e + f*x]]*(d - d*Sin[e + f*x]) + 630*Sqrt[2]*(c + d)^2 *(1 - Sin[e + f*x])^(5/2)*Sqrt[1 + Sin[e + f*x]]*(d - d*Sin[e + f*x]) -...
Time = 0.89 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.13, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 3244, 25, 3042, 3447, 3042, 3498, 25, 3042, 3214, 3042, 3127}
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.
\(\displaystyle \int \frac {(c+d \sin (e+f x))^3}{(a \sin (e+f x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(c+d \sin (e+f x))^3}{(a \sin (e+f x)+a)^3}dx\) |
\(\Big \downarrow \) 3244 |
\(\displaystyle -\frac {\int -\frac {(c+d \sin (e+f x)) \left (5 a \sin (e+f x) d^2+a \left (2 c^2+5 d c-2 d^2\right )\right )}{(\sin (e+f x) a+a)^2}dx}{5 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a \sin (e+f x)+a)^3}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {(c+d \sin (e+f x)) \left (5 a \sin (e+f x) d^2+a \left (2 c^2+5 d c-2 d^2\right )\right )}{(\sin (e+f x) a+a)^2}dx}{5 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a \sin (e+f x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {(c+d \sin (e+f x)) \left (5 a \sin (e+f x) d^2+a \left (2 c^2+5 d c-2 d^2\right )\right )}{(\sin (e+f x) a+a)^2}dx}{5 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a \sin (e+f x)+a)^3}\) |
\(\Big \downarrow \) 3447 |
\(\displaystyle \frac {\int \frac {5 a \sin ^2(e+f x) d^3+a c \left (2 c^2+5 d c-2 d^2\right )+\left (5 a c d^2+a \left (2 c^2+5 d c-2 d^2\right ) d\right ) \sin (e+f x)}{(\sin (e+f x) a+a)^2}dx}{5 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a \sin (e+f x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {5 a \sin (e+f x)^2 d^3+a c \left (2 c^2+5 d c-2 d^2\right )+\left (5 a c d^2+a \left (2 c^2+5 d c-2 d^2\right ) d\right ) \sin (e+f x)}{(\sin (e+f x) a+a)^2}dx}{5 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a \sin (e+f x)+a)^3}\) |
\(\Big \downarrow \) 3498 |
\(\displaystyle \frac {-\frac {\int -\frac {15 a^2 \sin (e+f x) d^3+a^2 \left (2 c^3+9 d c^2+18 d^2 c-14 d^3\right )}{\sin (e+f x) a+a}dx}{3 a^2}-\frac {a (2 c+7 d) (c-d)^2 \cos (e+f x)}{3 f (a \sin (e+f x)+a)^2}}{5 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a \sin (e+f x)+a)^3}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {\int \frac {15 a^2 \sin (e+f x) d^3+a^2 \left (2 c^3+9 d c^2+18 d^2 c-14 d^3\right )}{\sin (e+f x) a+a}dx}{3 a^2}-\frac {a (c-d)^2 (2 c+7 d) \cos (e+f x)}{3 f (a \sin (e+f x)+a)^2}}{5 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a \sin (e+f x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {15 a^2 \sin (e+f x) d^3+a^2 \left (2 c^3+9 d c^2+18 d^2 c-14 d^3\right )}{\sin (e+f x) a+a}dx}{3 a^2}-\frac {a (c-d)^2 (2 c+7 d) \cos (e+f x)}{3 f (a \sin (e+f x)+a)^2}}{5 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a \sin (e+f x)+a)^3}\) |
\(\Big \downarrow \) 3214 |
\(\displaystyle \frac {\frac {a^2 (c-d) \left (2 c^2+11 c d+29 d^2\right ) \int \frac {1}{\sin (e+f x) a+a}dx+15 a d^3 x}{3 a^2}-\frac {a (c-d)^2 (2 c+7 d) \cos (e+f x)}{3 f (a \sin (e+f x)+a)^2}}{5 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a \sin (e+f x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {a^2 (c-d) \left (2 c^2+11 c d+29 d^2\right ) \int \frac {1}{\sin (e+f x) a+a}dx+15 a d^3 x}{3 a^2}-\frac {a (c-d)^2 (2 c+7 d) \cos (e+f x)}{3 f (a \sin (e+f x)+a)^2}}{5 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a \sin (e+f x)+a)^3}\) |
\(\Big \downarrow \) 3127 |
\(\displaystyle \frac {\frac {15 a d^3 x-\frac {a^2 (c-d) \left (2 c^2+11 c d+29 d^2\right ) \cos (e+f x)}{f (a \sin (e+f x)+a)}}{3 a^2}-\frac {a (c-d)^2 (2 c+7 d) \cos (e+f x)}{3 f (a \sin (e+f x)+a)^2}}{5 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a \sin (e+f x)+a)^3}\) |
-1/5*((c - d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^2)/(f*(a + a*Sin[e + f*x]) ^3) + (-1/3*(a*(c - d)^2*(2*c + 7*d)*Cos[e + f*x])/(f*(a + a*Sin[e + f*x]) ^2) + (15*a*d^3*x - (a^2*(c - d)*(2*c^2 + 11*c*d + 29*d^2)*Cos[e + f*x])/( f*(a + a*Sin[e + f*x])))/(3*a^2))/(5*a^2)
3.5.73.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x] /; FreeQ[{a, b, c, d}, x] && EqQ[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*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n - 1)/(a*f*(2*m + 1))), x] + Simp[1/(a*b* (2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 2)* Simp[b*(c^2*(m + 1) + d^2*(n - 1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) + b*c*(m + n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b - a* B + b*C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] + Simp[1 /(a^2*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b *B - a*C) + b*C*(2*m + 1)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]
Time = 0.98 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.44
method | result | size |
derivativedivides | \(\frac {2 d^{3} \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {2 \left (c^{3}-d^{3}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {-4 c^{3}+6 c^{2} d -2 d^{3}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {-8 c^{3}+24 c^{2} d -24 c \,d^{2}+8 d^{3}}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 \left (4 c^{3}-12 c^{2} d +12 c \,d^{2}-4 d^{3}\right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {2 \left (8 c^{3}-18 c^{2} d +12 c \,d^{2}-2 d^{3}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}}{a^{3} f}\) | \(194\) |
default | \(\frac {2 d^{3} \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {2 \left (c^{3}-d^{3}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {-4 c^{3}+6 c^{2} d -2 d^{3}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {-8 c^{3}+24 c^{2} d -24 c \,d^{2}+8 d^{3}}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 \left (4 c^{3}-12 c^{2} d +12 c \,d^{2}-4 d^{3}\right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {2 \left (8 c^{3}-18 c^{2} d +12 c \,d^{2}-2 d^{3}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}}{a^{3} f}\) | \(194\) |
parallelrisch | \(\frac {15 d^{3} \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) x f +\left (\left (75 f x +30\right ) d^{3}-30 c^{3}\right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (\left (150 f x +150\right ) d^{3}-90 c^{2} d -60 c^{3}\right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (\left (150 f x +290\right ) d^{3}-120 c \,d^{2}-90 c^{2} d -80 c^{3}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (\left (75 f x +190\right ) d^{3}-60 c \,d^{2}-90 c^{2} d -40 c^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\left (15 f x +44\right ) d^{3}-12 c \,d^{2}-18 c^{2} d -14 c^{3}}{15 f \,a^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) | \(208\) |
risch | \(\frac {d^{3} x}{a^{3}}-\frac {2 \left (-120 d^{2} {\mathrm e}^{2 i \left (f x +e \right )} c +45 i c^{2} d \,{\mathrm e}^{3 i \left (f x +e \right )}-135 i d^{3} {\mathrm e}^{3 i \left (f x +e \right )}-45 d^{3} {\mathrm e}^{4 i \left (f x +e \right )}+185 d^{3} {\mathrm e}^{2 i \left (f x +e \right )}-32 d^{3}+2 c^{3}+115 i d^{3} {\mathrm e}^{i \left (f x +e \right )}-60 i c \,d^{2} {\mathrm e}^{i \left (f x +e \right )}-45 i c^{2} d \,{\mathrm e}^{i \left (f x +e \right )}+9 c^{2} d +21 c \,d^{2}-45 d \,c^{2} {\mathrm e}^{2 i \left (f x +e \right )}+45 c \,d^{2} {\mathrm e}^{4 i \left (f x +e \right )}+90 i c \,d^{2} {\mathrm e}^{3 i \left (f x +e \right )}-10 i c^{3} {\mathrm e}^{i \left (f x +e \right )}-20 c^{3} {\mathrm e}^{2 i \left (f x +e \right )}\right )}{15 f \,a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{5}}\) | \(251\) |
norman | \(\frac {\frac {d^{3} x}{a}+\frac {d^{3} x \left (\tan ^{11}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}+\frac {\left (-2 c^{3}+2 d^{3}\right ) \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a}+\frac {\left (-4 c^{3}-6 c^{2} d +10 d^{3}\right ) \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a}+\frac {\left (-12 c^{3}-24 c^{2} d -12 c \,d^{2}+48 d^{3}\right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a}+\frac {-14 c^{3}-18 c^{2} d -12 c \,d^{2}+44 d^{3}}{15 f a}+\frac {5 d^{3} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a}+\frac {13 d^{3} x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}+\frac {25 d^{3} x \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}+\frac {38 d^{3} x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}+\frac {46 d^{3} x \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}+\frac {46 d^{3} x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}+\frac {38 d^{3} x \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}+\frac {25 d^{3} x \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}+\frac {13 d^{3} x \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}+\frac {5 d^{3} x \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a}+\frac {\left (-8 c^{3}-18 c^{2} d -12 c \,d^{2}+38 d^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3 f a}+\frac {2 \left (-10 c^{3}-18 c^{2} d -6 c \,d^{2}+34 d^{3}\right ) \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a}+\frac {2 \left (-22 c^{3}-36 c^{2} d -6 c \,d^{2}+64 d^{3}\right ) \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f a}+\frac {\left (-34 c^{3}-18 c^{2} d -24 c \,d^{2}+76 d^{3}\right ) \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f a}+\frac {2 \left (-52 c^{3}-54 c^{2} d -66 c \,d^{2}+172 d^{3}\right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 f a}+\frac {\left (-122 c^{3}-144 c^{2} d -156 c \,d^{2}+422 d^{3}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15 f a}+\frac {2 \left (-172 c^{3}-144 c^{2} d -186 c \,d^{2}+502 d^{3}\right ) \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15 f a}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{3} a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) | \(687\) |
2/f/a^3*(d^3*arctan(tan(1/2*f*x+1/2*e))-(c^3-d^3)/(tan(1/2*f*x+1/2*e)+1)-1 /2*(-4*c^3+6*c^2*d-2*d^3)/(tan(1/2*f*x+1/2*e)+1)^2-1/4*(-8*c^3+24*c^2*d-24 *c*d^2+8*d^3)/(tan(1/2*f*x+1/2*e)+1)^4-1/5*(4*c^3-12*c^2*d+12*c*d^2-4*d^3) /(tan(1/2*f*x+1/2*e)+1)^5-1/3*(8*c^3-18*c^2*d+12*c*d^2-2*d^3)/(tan(1/2*f*x +1/2*e)+1)^3)
Leaf count of result is larger than twice the leaf count of optimal. 352 vs. \(2 (136) = 272\).
Time = 0.28 (sec) , antiderivative size = 352, normalized size of antiderivative = 2.61 \[ \int \frac {(c+d \sin (e+f x))^3}{(3+3 \sin (e+f x))^3} \, dx=-\frac {60 \, d^{3} f x - {\left (15 \, d^{3} f x - 2 \, c^{3} - 9 \, c^{2} d - 21 \, c d^{2} + 32 \, d^{3}\right )} \cos \left (f x + e\right )^{3} - 3 \, c^{3} + 9 \, c^{2} d - 9 \, c d^{2} + 3 \, d^{3} - {\left (45 \, d^{3} f x + 4 \, c^{3} + 18 \, c^{2} d - 3 \, c d^{2} - 19 \, d^{3}\right )} \cos \left (f x + e\right )^{2} + 3 \, {\left (10 \, d^{3} f x - 3 \, c^{3} - 6 \, c^{2} d - 9 \, c d^{2} + 18 \, d^{3}\right )} \cos \left (f x + e\right ) + {\left (60 \, d^{3} f x + 3 \, c^{3} - 9 \, c^{2} d + 9 \, c d^{2} - 3 \, d^{3} - {\left (15 \, d^{3} f x + 2 \, c^{3} + 9 \, c^{2} d + 21 \, c d^{2} - 32 \, d^{3}\right )} \cos \left (f x + e\right )^{2} + 3 \, {\left (10 \, d^{3} f x - 2 \, c^{3} - 9 \, c^{2} d - 6 \, c d^{2} + 17 \, d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \]
-1/15*(60*d^3*f*x - (15*d^3*f*x - 2*c^3 - 9*c^2*d - 21*c*d^2 + 32*d^3)*cos (f*x + e)^3 - 3*c^3 + 9*c^2*d - 9*c*d^2 + 3*d^3 - (45*d^3*f*x + 4*c^3 + 18 *c^2*d - 3*c*d^2 - 19*d^3)*cos(f*x + e)^2 + 3*(10*d^3*f*x - 3*c^3 - 6*c^2* d - 9*c*d^2 + 18*d^3)*cos(f*x + e) + (60*d^3*f*x + 3*c^3 - 9*c^2*d + 9*c*d ^2 - 3*d^3 - (15*d^3*f*x + 2*c^3 + 9*c^2*d + 21*c*d^2 - 32*d^3)*cos(f*x + e)^2 + 3*(10*d^3*f*x - 2*c^3 - 9*c^2*d - 6*c*d^2 + 17*d^3)*cos(f*x + e))*s in(f*x + e))/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos( f*x + e) - 4*a^3*f + (a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3* f)*sin(f*x + e))
Leaf count of result is larger than twice the leaf count of optimal. 2640 vs. \(2 (126) = 252\).
Time = 8.04 (sec) , antiderivative size = 2640, normalized size of antiderivative = 19.56 \[ \int \frac {(c+d \sin (e+f x))^3}{(3+3 \sin (e+f x))^3} \, dx=\text {Too large to display} \]
Piecewise((-30*c**3*tan(e/2 + f*x/2)**4/(15*a**3*f*tan(e/2 + f*x/2)**5 + 7 5*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f *tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 60*c**3*t an(e/2 + f*x/2)**3/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f* x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 80*c**3*tan(e/2 + f*x/2)**2/(1 5*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f* tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 40*c**3*tan(e/2 + f*x/2)/(15*a**3*f*tan(e/2 + f*x/2 )**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 15 0*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 1 4*c**3/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 15 0*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f* tan(e/2 + f*x/2) + 15*a**3*f) - 90*c**2*d*tan(e/2 + f*x/2)**3/(15*a**3*f*t an(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 90*c**2*d*tan(e/2 + f*x/2)**2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a** 3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 90*c** 2*d*tan(e/2 + f*x/2)/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2...
Leaf count of result is larger than twice the leaf count of optimal. 784 vs. \(2 (136) = 272\).
Time = 0.30 (sec) , antiderivative size = 784, normalized size of antiderivative = 5.81 \[ \int \frac {(c+d \sin (e+f x))^3}{(3+3 \sin (e+f x))^3} \, dx=\text {Too large to display} \]
2/15*(d^3*((95*sin(f*x + e)/(cos(f*x + e) + 1) + 145*sin(f*x + e)^2/(cos(f *x + e) + 1)^2 + 75*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 15*sin(f*x + e)^ 4/(cos(f*x + e) + 1)^4 + 22)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos( f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 15*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/ a^3) - c^3*(20*sin(f*x + e)/(cos(f*x + e) + 1) + 40*sin(f*x + e)^2/(cos(f* x + e) + 1)^2 + 30*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 15*sin(f*x + e)^4 /(cos(f*x + e) + 1)^4 + 7)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f* x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 6*c*d^2*(5*sin(f*x + e)/(cos(f*x + e) + 1) + 10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(cos (f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f *x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 9*c^2*d*(5*sin(f*x + e)/(cos (f*x + e) + 1) + 5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 5*sin(f*x + e)^3/ (cos(f*x + e) + 1)^3 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 1 0*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x ...
Time = 0.48 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.96 \[ \int \frac {(c+d \sin (e+f x))^3}{(3+3 \sin (e+f x))^3} \, dx=\frac {\frac {15 \, {\left (f x + e\right )} d^{3}}{a^{3}} - \frac {2 \, {\left (15 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 15 \, d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 30 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 45 \, c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 75 \, d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 40 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 45 \, c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 60 \, c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 145 \, d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 20 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 45 \, c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 30 \, c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 95 \, d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 7 \, c^{3} + 9 \, c^{2} d + 6 \, c d^{2} - 22 \, d^{3}\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}}}{15 \, f} \]
1/15*(15*(f*x + e)*d^3/a^3 - 2*(15*c^3*tan(1/2*f*x + 1/2*e)^4 - 15*d^3*tan (1/2*f*x + 1/2*e)^4 + 30*c^3*tan(1/2*f*x + 1/2*e)^3 + 45*c^2*d*tan(1/2*f*x + 1/2*e)^3 - 75*d^3*tan(1/2*f*x + 1/2*e)^3 + 40*c^3*tan(1/2*f*x + 1/2*e)^ 2 + 45*c^2*d*tan(1/2*f*x + 1/2*e)^2 + 60*c*d^2*tan(1/2*f*x + 1/2*e)^2 - 14 5*d^3*tan(1/2*f*x + 1/2*e)^2 + 20*c^3*tan(1/2*f*x + 1/2*e) + 45*c^2*d*tan( 1/2*f*x + 1/2*e) + 30*c*d^2*tan(1/2*f*x + 1/2*e) - 95*d^3*tan(1/2*f*x + 1/ 2*e) + 7*c^3 + 9*c^2*d + 6*c*d^2 - 22*d^3)/(a^3*(tan(1/2*f*x + 1/2*e) + 1) ^5))/f
Time = 10.37 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.78 \[ \int \frac {(c+d \sin (e+f x))^3}{(3+3 \sin (e+f x))^3} \, dx=\frac {d^3\,x}{a^3}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (2\,c^3-2\,d^3\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {16\,c^3}{3}+6\,c^2\,d+8\,c\,d^2-\frac {58\,d^3}{3}\right )+\frac {4\,c\,d^2}{5}+\frac {6\,c^2\,d}{5}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (4\,c^3+6\,c^2\,d-10\,d^3\right )+\frac {14\,c^3}{15}-\frac {44\,d^3}{15}+\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {8\,c^3}{3}+6\,c^2\,d+4\,c\,d^2-\frac {38\,d^3}{3}\right )}{f\,\left (a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+5\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+10\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+10\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+5\,a^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a^3\right )} \]
(d^3*x)/a^3 - (tan(e/2 + (f*x)/2)^4*(2*c^3 - 2*d^3) + tan(e/2 + (f*x)/2)^2 *(8*c*d^2 + 6*c^2*d + (16*c^3)/3 - (58*d^3)/3) + (4*c*d^2)/5 + (6*c^2*d)/5 + tan(e/2 + (f*x)/2)^3*(6*c^2*d + 4*c^3 - 10*d^3) + (14*c^3)/15 - (44*d^3 )/15 + tan(e/2 + (f*x)/2)*(4*c*d^2 + 6*c^2*d + (8*c^3)/3 - (38*d^3)/3))/(f *(10*a^3*tan(e/2 + (f*x)/2)^2 + 10*a^3*tan(e/2 + (f*x)/2)^3 + 5*a^3*tan(e/ 2 + (f*x)/2)^4 + a^3*tan(e/2 + (f*x)/2)^5 + a^3 + 5*a^3*tan(e/2 + (f*x)/2) ))