Integrand size = 25, antiderivative size = 138 \[ \int (a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x)) \, dx=-\frac {64 a^3 (7 c+5 d) \cos (e+f x)}{105 f \sqrt {a+a \sin (e+f x)}}-\frac {16 a^2 (7 c+5 d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{105 f}-\frac {2 a (7 c+5 d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 f}-\frac {2 d \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 f} \] Output:
-64/105*a^3*(7*c+5*d)*cos(f*x+e)/f/(a+a*sin(f*x+e))^(1/2)-16/105*a^2*(7*c+ 5*d)*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/f-2/35*a*(7*c+5*d)*cos(f*x+e)*(a+a* sin(f*x+e))^(3/2)/f-2/7*d*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/f
Time = 3.07 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.86 \[ \int (a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x)) \, dx=-\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} (1246 c+1040 d-6 (7 c+20 d) \cos (2 (e+f x))+(392 c+505 d) \sin (e+f x)-15 d \sin (3 (e+f x)))}{210 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )} \] Input:
Integrate[(a + a*Sin[e + f*x])^(5/2)*(c + d*Sin[e + f*x]),x]
Output:
-1/210*(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x] )]*(1246*c + 1040*d - 6*(7*c + 20*d)*Cos[2*(e + f*x)] + (392*c + 505*d)*Si n[e + f*x] - 15*d*Sin[3*(e + f*x)]))/(f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/ 2]))
Time = 0.51 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.96, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3230, 3042, 3126, 3042, 3126, 3042, 3125}
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.
| \(\displaystyle \int (a \sin (e+f x)+a)^{5/2} (c+d \sin (e+f x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \sin (e+f x)+a)^{5/2} (c+d \sin (e+f x))dx\) |
| \(\Big \downarrow \) 3230 |
| \(\displaystyle \frac {1}{7} (7 c+5 d) \int (\sin (e+f x) a+a)^{5/2}dx-\frac {2 d \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} (7 c+5 d) \int (\sin (e+f x) a+a)^{5/2}dx-\frac {2 d \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 f}\) |
| \(\Big \downarrow \) 3126 |
| \(\displaystyle \frac {1}{7} (7 c+5 d) \left (\frac {8}{5} a \int (\sin (e+f x) a+a)^{3/2}dx-\frac {2 a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}\right )-\frac {2 d \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} (7 c+5 d) \left (\frac {8}{5} a \int (\sin (e+f x) a+a)^{3/2}dx-\frac {2 a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}\right )-\frac {2 d \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 f}\) |
| \(\Big \downarrow \) 3126 |
| \(\displaystyle \frac {1}{7} (7 c+5 d) \left (\frac {8}{5} a \left (\frac {4}{3} a \int \sqrt {\sin (e+f x) a+a}dx-\frac {2 a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}\right )-\frac {2 a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}\right )-\frac {2 d \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} (7 c+5 d) \left (\frac {8}{5} a \left (\frac {4}{3} a \int \sqrt {\sin (e+f x) a+a}dx-\frac {2 a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}\right )-\frac {2 a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}\right )-\frac {2 d \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 f}\) |
| \(\Big \downarrow \) 3125 |
| \(\displaystyle \frac {1}{7} (7 c+5 d) \left (\frac {8}{5} a \left (-\frac {8 a^2 \cos (e+f x)}{3 f \sqrt {a \sin (e+f x)+a}}-\frac {2 a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}\right )-\frac {2 a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}\right )-\frac {2 d \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 f}\) |
Input:
Int[(a + a*Sin[e + f*x])^(5/2)*(c + d*Sin[e + f*x]),x]
Output:
(-2*d*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(7*f) + ((7*c + 5*d)*((-2*a *Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(5*f) + (8*a*((-8*a^2*Cos[e + f* x])/(3*f*Sqrt[a + a*Sin[e + f*x]]) - (2*a*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(3*f)))/5))/7
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*b*(Cos [c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]])), x] /; FreeQ[{a, b, c, d}, x] && Eq Q[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)), x] + Simp[a*((2*n - 1)/n) Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[ a^2 - b^2, 0] && IGtQ[n - 1/2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( f*(m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(b*(m + 1)) Int[(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
Time = 1.65 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.72
| method | result | size |
| default | \(\frac {2 \left (1+\sin \left (f x +e \right )\right ) a^{3} \left (-1+\sin \left (f x +e \right )\right ) \left (-15 \cos \left (f x +e \right )^{2} \sin \left (f x +e \right ) d +\left (-21 c -60 d \right ) \cos \left (f x +e \right )^{2}+\left (98 c +130 d \right ) \sin \left (f x +e \right )+322 c +290 d \right )}{105 \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\) | \(99\) |
| parts | \(\frac {2 c \left (1+\sin \left (f x +e \right )\right ) a^{3} \left (-1+\sin \left (f x +e \right )\right ) \left (3 \sin \left (f x +e \right )^{2}+14 \sin \left (f x +e \right )+43\right )}{15 \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}+\frac {2 d \left (1+\sin \left (f x +e \right )\right ) a^{3} \left (-1+\sin \left (f x +e \right )\right ) \left (3 \sin \left (f x +e \right )^{3}+12 \sin \left (f x +e \right )^{2}+23 \sin \left (f x +e \right )+46\right )}{21 \cos \left (f x +e \right ) \sqrt {a +\sin \left (f x +e \right ) a}\, f}\) | \(142\) |
Input:
int((a+sin(f*x+e)*a)^(5/2)*(c+d*sin(f*x+e)),x,method=_RETURNVERBOSE)
Output:
2/105*(1+sin(f*x+e))*a^3*(-1+sin(f*x+e))*(-15*cos(f*x+e)^2*sin(f*x+e)*d+(- 21*c-60*d)*cos(f*x+e)^2+(98*c+130*d)*sin(f*x+e)+322*c+290*d)/cos(f*x+e)/(a +sin(f*x+e)*a)^(1/2)/f
Time = 0.08 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.49 \[ \int (a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x)) \, dx=\frac {2 \, {\left (15 \, a^{2} d \cos \left (f x + e\right )^{4} + 3 \, {\left (7 \, a^{2} c + 20 \, a^{2} d\right )} \cos \left (f x + e\right )^{3} - 224 \, a^{2} c - 160 \, a^{2} d - {\left (77 \, a^{2} c + 85 \, a^{2} d\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (161 \, a^{2} c + 145 \, a^{2} d\right )} \cos \left (f x + e\right ) + {\left (15 \, a^{2} d \cos \left (f x + e\right )^{3} + 224 \, a^{2} c + 160 \, a^{2} d - 3 \, {\left (7 \, a^{2} c + 15 \, a^{2} d\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (49 \, a^{2} c + 65 \, a^{2} d\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{105 \, {\left (f \cos \left (f x + e\right ) + f \sin \left (f x + e\right ) + f\right )}} \] Input:
integrate((a+a*sin(f*x+e))^(5/2)*(c+d*sin(f*x+e)),x, algorithm="fricas")
Output:
2/105*(15*a^2*d*cos(f*x + e)^4 + 3*(7*a^2*c + 20*a^2*d)*cos(f*x + e)^3 - 2 24*a^2*c - 160*a^2*d - (77*a^2*c + 85*a^2*d)*cos(f*x + e)^2 - 2*(161*a^2*c + 145*a^2*d)*cos(f*x + e) + (15*a^2*d*cos(f*x + e)^3 + 224*a^2*c + 160*a^ 2*d - 3*(7*a^2*c + 15*a^2*d)*cos(f*x + e)^2 - 2*(49*a^2*c + 65*a^2*d)*cos( f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)/(f*cos(f*x + e) + f*sin(f *x + e) + f)
\[ \int (a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x)) \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}} \left (c + d \sin {\left (e + f x \right )}\right )\, dx \] Input:
integrate((a+a*sin(f*x+e))**(5/2)*(c+d*sin(f*x+e)),x)
Output:
Integral((a*(sin(e + f*x) + 1))**(5/2)*(c + d*sin(e + f*x)), x)
\[ \int (a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x)) \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (d \sin \left (f x + e\right ) + c\right )} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^(5/2)*(c+d*sin(f*x+e)),x, algorithm="maxima")
Output:
integrate((a*sin(f*x + e) + a)^(5/2)*(d*sin(f*x + e) + c), x)
Time = 0.15 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.46 \[ \int (a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x)) \, dx=\frac {\sqrt {2} {\left (15 \, a^{2} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, f x + \frac {7}{2} \, e\right ) + 525 \, {\left (4 \, a^{2} c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 3 \, a^{2} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 35 \, {\left (10 \, a^{2} c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 11 \, a^{2} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, f x + \frac {3}{2} \, e\right ) + 21 \, {\left (2 \, a^{2} c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 5 \, a^{2} d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, f x + \frac {5}{2} \, e\right )\right )} \sqrt {a}}{420 \, f} \] Input:
integrate((a+a*sin(f*x+e))^(5/2)*(c+d*sin(f*x+e)),x, algorithm="giac")
Output:
1/420*sqrt(2)*(15*a^2*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-7/4*pi + 7/2*f*x + 7/2*e) + 525*(4*a^2*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 3*a^ 2*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sin(-1/4*pi + 1/2*f*x + 1/2*e) + 35*(10*a^2*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 11*a^2*d*sgn(cos(-1/4*p i + 1/2*f*x + 1/2*e)))*sin(-3/4*pi + 3/2*f*x + 3/2*e) + 21*(2*a^2*c*sgn(co s(-1/4*pi + 1/2*f*x + 1/2*e)) + 5*a^2*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e) ))*sin(-5/4*pi + 5/2*f*x + 5/2*e))*sqrt(a)/f
Timed out. \[ \int (a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x)) \, dx=\int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}\,\left (c+d\,\sin \left (e+f\,x\right )\right ) \,d x \] Input:
int((a + a*sin(e + f*x))^(5/2)*(c + d*sin(e + f*x)),x)
Output:
int((a + a*sin(e + f*x))^(5/2)*(c + d*sin(e + f*x)), x)
\[ \int (a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x)) \, dx=\sqrt {a}\, a^{2} \left (\left (\int \sqrt {\sin \left (f x +e \right )+1}d x \right ) c +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}d x \right ) d +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right ) c +2 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right ) d +2 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) c +\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) d \right ) \] Input:
int((a+a*sin(f*x+e))^(5/2)*(c+d*sin(f*x+e)),x)
Output:
sqrt(a)*a**2*(int(sqrt(sin(e + f*x) + 1),x)*c + int(sqrt(sin(e + f*x) + 1) *sin(e + f*x)**3,x)*d + int(sqrt(sin(e + f*x) + 1)*sin(e + f*x)**2,x)*c + 2*int(sqrt(sin(e + f*x) + 1)*sin(e + f*x)**2,x)*d + 2*int(sqrt(sin(e + f*x ) + 1)*sin(e + f*x),x)*c + int(sqrt(sin(e + f*x) + 1)*sin(e + f*x),x)*d)