Integrand size = 25, antiderivative size = 125 \[ \int \frac {(c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=-\frac {(c-d) (2 c+5 d) \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}-\frac {\left (2 c^2+6 c d+7 d^2\right ) \cos (e+f x)}{15 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{5 f (a+a \sin (e+f x))^3} \] Output:
-1/15*(c-d)*(2*c+5*d)*cos(f*x+e)/a/f/(a+a*sin(f*x+e))^2-1/15*(2*c^2+6*c*d+ 7*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))/f/(a+a*sin(f*x+e))^3
Time = 0.16 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.67 \[ \int \frac {(c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=-\frac {\cos (e+f x) \left (7 c^2+6 c d+2 d^2+6 \left (c^2+3 c d+d^2\right ) \sin (e+f x)+\left (2 c^2+6 c d+7 d^2\right ) \sin ^2(e+f x)\right )}{15 a^3 f (1+\sin (e+f x))^3} \] Input:
Integrate[(c + d*Sin[e + f*x])^2/(a + a*Sin[e + f*x])^3,x]
Output:
-1/15*(Cos[e + f*x]*(7*c^2 + 6*c*d + 2*d^2 + 6*(c^2 + 3*c*d + d^2)*Sin[e + f*x] + (2*c^2 + 6*c*d + 7*d^2)*Sin[e + f*x]^2))/(a^3*f*(1 + Sin[e + f*x]) ^3)
Time = 0.52 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 3239, 25, 3042, 3229, 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.
| \(\displaystyle \int \frac {(c+d \sin (e+f x))^2}{(a \sin (e+f x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(c+d \sin (e+f x))^2}{(a \sin (e+f x)+a)^3}dx\) |
| \(\Big \downarrow \) 3239 |
| \(\displaystyle -\frac {\int -\frac {a \left (2 c^2+4 d c-d^2\right )+a d (c+4 d) \sin (e+f x)}{(\sin (e+f x) a+a)^2}dx}{5 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{5 f (a \sin (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {a \left (2 c^2+4 d c-d^2\right )+a d (c+4 d) \sin (e+f x)}{(\sin (e+f x) a+a)^2}dx}{5 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{5 f (a \sin (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a \left (2 c^2+4 d c-d^2\right )+a d (c+4 d) \sin (e+f x)}{(\sin (e+f x) a+a)^2}dx}{5 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{5 f (a \sin (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 3229 |
| \(\displaystyle \frac {\frac {1}{3} \left (2 c^2+6 c d+7 d^2\right ) \int \frac {1}{\sin (e+f x) a+a}dx-\frac {a (c-d) (2 c+5 d) \cos (e+f x)}{3 f (a \sin (e+f x)+a)^2}}{5 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{5 f (a \sin (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (2 c^2+6 c d+7 d^2\right ) \int \frac {1}{\sin (e+f x) a+a}dx-\frac {a (c-d) (2 c+5 d) \cos (e+f x)}{3 f (a \sin (e+f x)+a)^2}}{5 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{5 f (a \sin (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 3127 |
| \(\displaystyle \frac {-\frac {\left (2 c^2+6 c d+7 d^2\right ) \cos (e+f x)}{3 f (a \sin (e+f x)+a)}-\frac {a (c-d) (2 c+5 d) \cos (e+f x)}{3 f (a \sin (e+f x)+a)^2}}{5 a^2}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{5 f (a \sin (e+f x)+a)^3}\) |
Input:
Int[(c + d*Sin[e + f*x])^2/(a + a*Sin[e + f*x])^3,x]
Output:
-1/5*((c - d)*Cos[e + f*x]*(c + d*Sin[e + f*x]))/(f*(a + a*Sin[e + f*x])^3 ) + (-1/3*(a*(c - d)*(2*c + 5*d)*Cos[e + f*x])/(f*(a + a*Sin[e + f*x])^2) - ((2*c^2 + 6*c*d + 7*d^2)*Cos[e + f*x])/(3*f*(a + a*Sin[e + f*x])))/(5*a^ 2)
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_)])^(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/(a*f*(2*m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)) I nt[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^2, x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f* x])^m*((c + d*Sin[e + f*x])/(a*f*(2*m + 1))), x] + Simp[1/(a*b*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*c*d*(m - 1) + b*(d^2 + c^2*(m + 1) ) + d*(a*d*(m - 1) + b*c*(m + 2))*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] && LtQ[m, -1]
Time = 0.73 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.93
| method | result | size |
| parallelrisch | \(\frac {-30 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4} c^{2}-60 c \left (c +d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+\left (-80 c^{2}-60 c d -40 d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-40 \left (c +\frac {d}{2}\right ) \left (c +d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-14 c^{2}-12 c d -4 d^{2}}{15 f \,a^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) | \(116\) |
| derivativedivides | \(\frac {-\frac {2 \left (8 c^{2}-12 c d +4 d^{2}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {-8 c^{2}+16 c d -8 d^{2}}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}+\frac {4 c \left (c -d \right )}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 c^{2}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {2 \left (4 c^{2}-8 c d +4 d^{2}\right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}}{a^{3} f}\) | \(139\) |
| default | \(\frac {-\frac {2 \left (8 c^{2}-12 c d +4 d^{2}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {-8 c^{2}+16 c d -8 d^{2}}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}+\frac {4 c \left (c -d \right )}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 c^{2}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {2 \left (4 c^{2}-8 c d +4 d^{2}\right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}}{a^{3} f}\) | \(139\) |
| risch | \(-\frac {2 \left (6 c d +7 d^{2}-30 i c d \,{\mathrm e}^{i \left (f x +e \right )}+30 i c d \,{\mathrm e}^{3 i \left (f x +e \right )}+30 i d^{2} {\mathrm e}^{3 i \left (f x +e \right )}-10 i c^{2} {\mathrm e}^{i \left (f x +e \right )}+15 d^{2} {\mathrm e}^{4 i \left (f x +e \right )}-20 c^{2} {\mathrm e}^{2 i \left (f x +e \right )}-40 d^{2} {\mathrm e}^{2 i \left (f x +e \right )}-20 i d^{2} {\mathrm e}^{i \left (f x +e \right )}-30 c d \,{\mathrm e}^{2 i \left (f x +e \right )}+2 c^{2}\right )}{15 f \,a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{5}}\) | \(166\) |
| norman | \(\frac {-\frac {2 c^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{f a}+\frac {-14 c^{2}-12 c d -4 d^{2}}{15 f a}+\frac {4 \left (-c^{2}-c d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{f a}+\frac {4 \left (-9 c^{2}-7 c d -4 d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{5 f a}+\frac {\left (-8 c^{2}-12 c d -4 d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3 f a}+\frac {4 \left (-7 c^{2}-9 c d -2 d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3 f a}+\frac {4 \left (-7 c^{2}-3 c d -2 d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{3 f a}+\frac {2 \left (-16 c^{2}-18 c d -2 d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{3 f a}+\frac {2 \left (-34 c^{2}-22 c d -14 d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{5 f a}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2} a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) | \(310\) |
Input:
int((c+d*sin(f*x+e))^2/(a+sin(f*x+e)*a)^3,x,method=_RETURNVERBOSE)
Output:
1/15*(-30*tan(1/2*f*x+1/2*e)^4*c^2-60*c*(c+d)*tan(1/2*f*x+1/2*e)^3+(-80*c^ 2-60*c*d-40*d^2)*tan(1/2*f*x+1/2*e)^2-40*(c+1/2*d)*(c+d)*tan(1/2*f*x+1/2*e )-14*c^2-12*c*d-4*d^2)/f/a^3/(tan(1/2*f*x+1/2*e)+1)^5
Leaf count of result is larger than twice the leaf count of optimal. 242 vs. \(2 (119) = 238\).
Time = 0.08 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.94 \[ \int \frac {(c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=-\frac {{\left (2 \, c^{2} + 6 \, c d + 7 \, d^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (4 \, c^{2} + 12 \, c d - d^{2}\right )} \cos \left (f x + e\right )^{2} - 3 \, c^{2} + 6 \, c d - 3 \, d^{2} - 3 \, {\left (3 \, c^{2} + 4 \, c d + 3 \, d^{2}\right )} \cos \left (f x + e\right ) - {\left ({\left (2 \, c^{2} + 6 \, c d + 7 \, d^{2}\right )} \cos \left (f x + e\right )^{2} - 3 \, c^{2} + 6 \, c d - 3 \, d^{2} + 6 \, {\left (c^{2} + 3 \, c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \] Input:
integrate((c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^3,x, algorithm="fricas")
Output:
-1/15*((2*c^2 + 6*c*d + 7*d^2)*cos(f*x + e)^3 - (4*c^2 + 12*c*d - d^2)*cos (f*x + e)^2 - 3*c^2 + 6*c*d - 3*d^2 - 3*(3*c^2 + 4*c*d + 3*d^2)*cos(f*x + e) - ((2*c^2 + 6*c*d + 7*d^2)*cos(f*x + e)^2 - 3*c^2 + 6*c*d - 3*d^2 + 6*( c^2 + 3*c*d + d^2)*cos(f*x + e))*sin(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. 1365 vs. \(2 (114) = 228\).
Time = 5.11 (sec) , antiderivative size = 1365, normalized size of antiderivative = 10.92 \[ \int \frac {(c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:
integrate((c+d*sin(f*x+e))**2/(a+a*sin(f*x+e))**3,x)
Output:
Piecewise((-30*c**2*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**2*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**2*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**2*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**2/(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) - 60*c*d*tan(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) - 60*c*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*ta n(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 60*c*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 + f*x/2)...
Leaf count of result is larger than twice the leaf count of optimal. 553 vs. \(2 (119) = 238\).
Time = 0.05 (sec) , antiderivative size = 553, normalized size of antiderivative = 4.42 \[ \int \frac {(c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx =\text {Too large to display} \] Input:
integrate((c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^3,x, algorithm="maxima")
Output:
-2/15*(c^2*(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) + 2*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) + 6*c*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) + 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))/f
Time = 0.15 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.37 \[ \int \frac {(c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=-\frac {2 \, {\left (15 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 30 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 30 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 40 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 30 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 20 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 20 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 30 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 10 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 7 \, c^{2} + 6 \, c d + 2 \, d^{2}\right )}}{15 \, a^{3} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}} \] Input:
integrate((c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^3,x, algorithm="giac")
Output:
-2/15*(15*c^2*tan(1/2*f*x + 1/2*e)^4 + 30*c^2*tan(1/2*f*x + 1/2*e)^3 + 30* c*d*tan(1/2*f*x + 1/2*e)^3 + 40*c^2*tan(1/2*f*x + 1/2*e)^2 + 30*c*d*tan(1/ 2*f*x + 1/2*e)^2 + 20*d^2*tan(1/2*f*x + 1/2*e)^2 + 20*c^2*tan(1/2*f*x + 1/ 2*e) + 30*c*d*tan(1/2*f*x + 1/2*e) + 10*d^2*tan(1/2*f*x + 1/2*e) + 7*c^2 + 6*c*d + 2*d^2)/(a^3*f*(tan(1/2*f*x + 1/2*e) + 1)^5)
Time = 16.84 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.74 \[ \int \frac {(c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=\frac {2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (6\,c\,d-4\,c^2\,\cos \left (e+f\,x\right )+d^2\,\cos \left (e+f\,x\right )+\frac {25\,c^2\,\sin \left (e+f\,x\right )}{2}+\frac {5\,d^2\,\sin \left (e+f\,x\right )}{2}+\frac {53\,c^2}{4}+\frac {13\,d^2}{4}-\frac {9\,c^2\,\cos \left (2\,e+2\,f\,x\right )}{4}-\frac {9\,d^2\,\cos \left (2\,e+2\,f\,x\right )}{4}-\frac {5\,c^2\,\sin \left (2\,e+2\,f\,x\right )}{4}+\frac {5\,d^2\,\sin \left (2\,e+2\,f\,x\right )}{4}+3\,c\,d\,\cos \left (e+f\,x\right )+15\,c\,d\,\sin \left (e+f\,x\right )-3\,c\,d\,\cos \left (2\,e+2\,f\,x\right )\right )}{15\,a^3\,f\,\left (\frac {5\,\sqrt {2}\,\cos \left (\frac {3\,e}{2}+\frac {\pi }{4}+\frac {3\,f\,x}{2}\right )}{4}-\frac {5\,\sqrt {2}\,\cos \left (\frac {e}{2}-\frac {\pi }{4}+\frac {f\,x}{2}\right )}{2}+\frac {\sqrt {2}\,\cos \left (\frac {5\,e}{2}-\frac {\pi }{4}+\frac {5\,f\,x}{2}\right )}{4}\right )} \] Input:
int((c + d*sin(e + f*x))^2/(a + a*sin(e + f*x))^3,x)
Output:
(2*cos(e/2 + (f*x)/2)*(6*c*d - 4*c^2*cos(e + f*x) + d^2*cos(e + f*x) + (25 *c^2*sin(e + f*x))/2 + (5*d^2*sin(e + f*x))/2 + (53*c^2)/4 + (13*d^2)/4 - (9*c^2*cos(2*e + 2*f*x))/4 - (9*d^2*cos(2*e + 2*f*x))/4 - (5*c^2*sin(2*e + 2*f*x))/4 + (5*d^2*sin(2*e + 2*f*x))/4 + 3*c*d*cos(e + f*x) + 15*c*d*sin( e + f*x) - 3*c*d*cos(2*e + 2*f*x)))/(15*a^3*f*((5*2^(1/2)*cos((3*e)/2 + pi /4 + (3*f*x)/2))/4 - (5*2^(1/2)*cos(e/2 - pi/4 + (f*x)/2))/2 + (2^(1/2)*co s((5*e)/2 - pi/4 + (5*f*x)/2))/4))
Time = 0.17 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.66 \[ \int \frac {(c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=\frac {\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} c^{2}}{5}-4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} c d -\frac {4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c^{2}}{3}-4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c d -\frac {8 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} d^{2}}{3}-\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c^{2}}{3}-4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c d -\frac {4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d^{2}}{3}-\frac {8 c^{2}}{15}-\frac {4 c d}{5}-\frac {4 d^{2}}{15}}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )} \] Input:
int((c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^3,x)
Output:
(2*(3*tan((e + f*x)/2)**5*c**2 - 30*tan((e + f*x)/2)**3*c*d - 10*tan((e + f*x)/2)**2*c**2 - 30*tan((e + f*x)/2)**2*c*d - 20*tan((e + f*x)/2)**2*d**2 - 5*tan((e + f*x)/2)*c**2 - 30*tan((e + f*x)/2)*c*d - 10*tan((e + f*x)/2) *d**2 - 4*c**2 - 6*c*d - 2*d**2))/(15*a**3*f*(tan((e + f*x)/2)**5 + 5*tan( (e + f*x)/2)**4 + 10*tan((e + f*x)/2)**3 + 10*tan((e + f*x)/2)**2 + 5*tan( (e + f*x)/2) + 1))