Integrand size = 45, antiderivative size = 34 \[ \int (a-a \sin (e+f x))^3 (c+c \sin (e+f x))^n (B (3-n)+B (4+n) \sin (e+f x)) \, dx=-\frac {a^3 B c^3 \cos ^7(e+f x) (c+c \sin (e+f x))^{-3+n}}{f} \] Output:
-a^3*B*c^3*cos(f*x+e)^7*(c+c*sin(f*x+e))^(-3+n)/f
Time = 8.99 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.97 \[ \int (a-a \sin (e+f x))^3 (c+c \sin (e+f x))^n (B (3-n)+B (4+n) \sin (e+f x)) \, dx=-\frac {a^3 B \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^7 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (c (1+\sin (e+f x)))^n}{f} \] Input:
Integrate[(a - a*Sin[e + f*x])^3*(c + c*Sin[e + f*x])^n*(B*(3 - n) + B*(4 + n)*Sin[e + f*x]),x]
Output:
-((a^3*B*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)/2] + Sin[( e + f*x)/2])*(c*(1 + Sin[e + f*x]))^n)/f)
Time = 0.47 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {3042, 3446, 3042, 3333}
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.
\(\displaystyle \int (a-a \sin (e+f x))^3 (B (n+4) \sin (e+f x)+B (3-n)) (c \sin (e+f x)+c)^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a-a \sin (e+f x))^3 (B (n+4) \sin (e+f x)+B (3-n)) (c \sin (e+f x)+c)^ndx\) |
\(\Big \downarrow \) 3446 |
\(\displaystyle a^3 c^3 \int \cos ^6(e+f x) (\sin (e+f x) c+c)^{n-3} (B (3-n)+B (n+4) \sin (e+f x))dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a^3 c^3 \int \cos (e+f x)^6 (\sin (e+f x) c+c)^{n-3} (B (3-n)+B (n+4) \sin (e+f x))dx\) |
\(\Big \downarrow \) 3333 |
\(\displaystyle -\frac {a^3 B c^3 \cos ^7(e+f x) (c \sin (e+f x)+c)^{n-3}}{f}\) |
Input:
Int[(a - a*Sin[e + f*x])^3*(c + c*Sin[e + f*x])^n*(B*(3 - n) + B*(4 + n)*S in[e + f*x]),x]
Output:
-((a^3*B*c^3*Cos[e + f*x]^7*(c + c*Sin[e + f*x])^(-3 + n))/f)
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)* (g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(f*g*(m + p + 1))), x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && EqQ[a*d*m + b *c*(m + p + 1), 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Si mp[a^m*c^m Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B*Sin [e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a* d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
Time = 7.87 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.76
method | result | size |
parallelrisch | \(\frac {a^{3} B \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{n} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}{f \left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{4}}\) | \(60\) |
Input:
int((a-a*sin(f*x+e))^3*(c+c*sin(f*x+e))^n*(B*(3-n)+B*(4+n)*sin(f*x+e)),x,m ethod=_RETURNVERBOSE)
Output:
a^3*B*(c*(1+sin(f*x+e)))^n*(tan(1/2*f*x+1/2*e)+1)*(tan(1/2*f*x+1/2*e)-1)^7 /f/(1+tan(1/2*f*x+1/2*e)^2)^4
Leaf count of result is larger than twice the leaf count of optimal. 77 vs. \(2 (34) = 68\).
Time = 0.09 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.26 \[ \int (a-a \sin (e+f x))^3 (c+c \sin (e+f x))^n (B (3-n)+B (4+n) \sin (e+f x)) \, dx=\frac {{\left (3 \, B a^{3} \cos \left (f x + e\right )^{3} - 4 \, B a^{3} \cos \left (f x + e\right ) - {\left (B a^{3} \cos \left (f x + e\right )^{3} - 4 \, B a^{3} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} {\left (c \sin \left (f x + e\right ) + c\right )}^{n}}{f} \] Input:
integrate((a-a*sin(f*x+e))^3*(c+c*sin(f*x+e))^n*(B*(3-n)+B*(4+n)*sin(f*x+e )),x, algorithm="fricas")
Output:
(3*B*a^3*cos(f*x + e)^3 - 4*B*a^3*cos(f*x + e) - (B*a^3*cos(f*x + e)^3 - 4 *B*a^3*cos(f*x + e))*sin(f*x + e))*(c*sin(f*x + e) + c)^n/f
Leaf count of result is larger than twice the leaf count of optimal. 898 vs. \(2 (32) = 64\).
Time = 71.84 (sec) , antiderivative size = 898, normalized size of antiderivative = 26.41 \[ \int (a-a \sin (e+f x))^3 (c+c \sin (e+f x))^n (B (3-n)+B (4+n) \sin (e+f x)) \, dx=\text {Too large to display} \] Input:
integrate((a-a*sin(f*x+e))**3*(c+c*sin(f*x+e))**n*(B*(3-n)+B*(4+n)*sin(f*x +e)),x)
Output:
Piecewise((B*a**3*(c + 2*c*tan(e/2 + f*x/2)/(tan(e/2 + f*x/2)**2 + 1))**n* tan(e/2 + f*x/2)**8/(f*tan(e/2 + f*x/2)**8 + 4*f*tan(e/2 + f*x/2)**6 + 6*f *tan(e/2 + f*x/2)**4 + 4*f*tan(e/2 + f*x/2)**2 + f) - 6*B*a**3*(c + 2*c*ta n(e/2 + f*x/2)/(tan(e/2 + f*x/2)**2 + 1))**n*tan(e/2 + f*x/2)**7/(f*tan(e/ 2 + f*x/2)**8 + 4*f*tan(e/2 + f*x/2)**6 + 6*f*tan(e/2 + f*x/2)**4 + 4*f*ta n(e/2 + f*x/2)**2 + f) + 14*B*a**3*(c + 2*c*tan(e/2 + f*x/2)/(tan(e/2 + f* x/2)**2 + 1))**n*tan(e/2 + f*x/2)**6/(f*tan(e/2 + f*x/2)**8 + 4*f*tan(e/2 + f*x/2)**6 + 6*f*tan(e/2 + f*x/2)**4 + 4*f*tan(e/2 + f*x/2)**2 + f) - 14* B*a**3*(c + 2*c*tan(e/2 + f*x/2)/(tan(e/2 + f*x/2)**2 + 1))**n*tan(e/2 + f *x/2)**5/(f*tan(e/2 + f*x/2)**8 + 4*f*tan(e/2 + f*x/2)**6 + 6*f*tan(e/2 + f*x/2)**4 + 4*f*tan(e/2 + f*x/2)**2 + f) + 14*B*a**3*(c + 2*c*tan(e/2 + f* x/2)/(tan(e/2 + f*x/2)**2 + 1))**n*tan(e/2 + f*x/2)**3/(f*tan(e/2 + f*x/2) **8 + 4*f*tan(e/2 + f*x/2)**6 + 6*f*tan(e/2 + f*x/2)**4 + 4*f*tan(e/2 + f* x/2)**2 + f) - 14*B*a**3*(c + 2*c*tan(e/2 + f*x/2)/(tan(e/2 + f*x/2)**2 + 1))**n*tan(e/2 + f*x/2)**2/(f*tan(e/2 + f*x/2)**8 + 4*f*tan(e/2 + f*x/2)** 6 + 6*f*tan(e/2 + f*x/2)**4 + 4*f*tan(e/2 + f*x/2)**2 + f) + 6*B*a**3*(c + 2*c*tan(e/2 + f*x/2)/(tan(e/2 + f*x/2)**2 + 1))**n*tan(e/2 + f*x/2)/(f*ta n(e/2 + f*x/2)**8 + 4*f*tan(e/2 + f*x/2)**6 + 6*f*tan(e/2 + f*x/2)**4 + 4* f*tan(e/2 + f*x/2)**2 + f) - B*a**3*(c + 2*c*tan(e/2 + f*x/2)/(tan(e/2 + f *x/2)**2 + 1))**n/(f*tan(e/2 + f*x/2)**8 + 4*f*tan(e/2 + f*x/2)**6 + 6*...
\[ \int (a-a \sin (e+f x))^3 (c+c \sin (e+f x))^n (B (3-n)+B (4+n) \sin (e+f x)) \, dx=\int { -{\left (B {\left (n + 4\right )} \sin \left (f x + e\right ) - B {\left (n - 3\right )}\right )} {\left (a \sin \left (f x + e\right ) - a\right )}^{3} {\left (c \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \] Input:
integrate((a-a*sin(f*x+e))^3*(c+c*sin(f*x+e))^n*(B*(3-n)+B*(4+n)*sin(f*x+e )),x, algorithm="maxima")
Output:
-integrate((B*(n + 4)*sin(f*x + e) - B*(n - 3))*(a*sin(f*x + e) - a)^3*(c* sin(f*x + e) + c)^n, x)
Leaf count of result is larger than twice the leaf count of optimal. 6974 vs. \(2 (34) = 68\).
Time = 11.84 (sec) , antiderivative size = 6974, normalized size of antiderivative = 205.12 \[ \int (a-a \sin (e+f x))^3 (c+c \sin (e+f x))^n (B (3-n)+B (4+n) \sin (e+f x)) \, dx=\text {Too large to display} \] Input:
integrate((a-a*sin(f*x+e))^3*(c+c*sin(f*x+e))^n*(B*(3-n)+B*(4+n)*sin(f*x+e )),x, algorithm="giac")
Output:
-(B*a^3*(sqrt(2*tan(f*x + e)^4*tan(1/2*f*x + 1/2*e)^4 + 4*tan(f*x + e)^4*t an(1/2*f*x + 1/2*e)^3 + 4*tan(f*x + e)^4*tan(1/2*f*x + 1/2*e)^2 + 3*tan(f* x + e)^2*tan(1/2*f*x + 1/2*e)^4 + 4*tan(f*x + e)^4*tan(1/2*f*x + 1/2*e) + 8*tan(f*x + e)^2*tan(1/2*f*x + 1/2*e)^3 + 2*tan(f*x + e)^4 + 6*tan(f*x + e )^2*tan(1/2*f*x + 1/2*e)^2 + tan(1/2*f*x + 1/2*e)^4 + 8*tan(f*x + e)^2*tan (1/2*f*x + 1/2*e) + 4*tan(1/2*f*x + 1/2*e)^3 + 3*tan(f*x + e)^2 + 2*tan(1/ 2*f*x + 1/2*e)^2 + 4*tan(1/2*f*x + 1/2*e) + 1)*abs(c)/(tan(f*x + e)^2*tan( 1/2*f*x + 1/2*e)^2 + tan(f*x + e)^2 + tan(1/2*f*x + 1/2*e)^2 + 1))^n*tan(- 1/4*pi*n*sgn(2*c*tan(1/2*f*x + 1/2*e)^4 + 4*c*tan(1/2*f*x + 1/2*e)^3 - 4*c *tan(1/2*f*x + 1/2*e) - 2*c)*sgn(4*c*tan(1/2*f*x + 1/2*e)^3 + 8*c*tan(1/2* f*x + 1/2*e)^2 + 4*c*tan(1/2*f*x + 1/2*e)) + 1/2*pi*n*floor(f*x/pi + e/pi + 1/2) - 1/4*pi*n*sgn(4*c*tan(1/2*f*x + 1/2*e)^3 + 8*c*tan(1/2*f*x + 1/2*e )^2 + 4*c*tan(1/2*f*x + 1/2*e)))^2*tan(1/2*f*x + 1/2*e)^8 - 6*B*a^3*(sqrt( 2*tan(f*x + e)^4*tan(1/2*f*x + 1/2*e)^4 + 4*tan(f*x + e)^4*tan(1/2*f*x + 1 /2*e)^3 + 4*tan(f*x + e)^4*tan(1/2*f*x + 1/2*e)^2 + 3*tan(f*x + e)^2*tan(1 /2*f*x + 1/2*e)^4 + 4*tan(f*x + e)^4*tan(1/2*f*x + 1/2*e) + 8*tan(f*x + e) ^2*tan(1/2*f*x + 1/2*e)^3 + 2*tan(f*x + e)^4 + 6*tan(f*x + e)^2*tan(1/2*f* x + 1/2*e)^2 + tan(1/2*f*x + 1/2*e)^4 + 8*tan(f*x + e)^2*tan(1/2*f*x + 1/2 *e) + 4*tan(1/2*f*x + 1/2*e)^3 + 3*tan(f*x + e)^2 + 2*tan(1/2*f*x + 1/2*e) ^2 + 4*tan(1/2*f*x + 1/2*e) + 1)*abs(c)/(tan(f*x + e)^2*tan(1/2*f*x + 1...
Time = 37.14 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.79 \[ \int (a-a \sin (e+f x))^3 (c+c \sin (e+f x))^n (B (3-n)+B (4+n) \sin (e+f x)) \, dx=-\frac {B\,a^3\,{\left (c\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^n\,\left (14\,\cos \left (e+f\,x\right )-6\,\cos \left (3\,e+3\,f\,x\right )-14\,\sin \left (2\,e+2\,f\,x\right )+\sin \left (4\,e+4\,f\,x\right )\right )}{8\,f} \] Input:
int(-(B*(n - 3) - B*sin(e + f*x)*(n + 4))*(a - a*sin(e + f*x))^3*(c + c*si n(e + f*x))^n,x)
Output:
-(B*a^3*(c*(sin(e + f*x) + 1))^n*(14*cos(e + f*x) - 6*cos(3*e + 3*f*x) - 1 4*sin(2*e + 2*f*x) + sin(4*e + 4*f*x)))/(8*f)
\[ \int (a-a \sin (e+f x))^3 (c+c \sin (e+f x))^n (B (3-n)+B (4+n) \sin (e+f x)) \, dx=a^{3} b \left (-\left (\int \left (\sin \left (f x +e \right ) c +c \right )^{n}d x \right ) n +3 \left (\int \left (\sin \left (f x +e \right ) c +c \right )^{n}d x \right )-\left (\int \left (\sin \left (f x +e \right ) c +c \right )^{n} \sin \left (f x +e \right )^{4}d x \right ) n -4 \left (\int \left (\sin \left (f x +e \right ) c +c \right )^{n} \sin \left (f x +e \right )^{4}d x \right )+4 \left (\int \left (\sin \left (f x +e \right ) c +c \right )^{n} \sin \left (f x +e \right )^{3}d x \right ) n +9 \left (\int \left (\sin \left (f x +e \right ) c +c \right )^{n} \sin \left (f x +e \right )^{3}d x \right )-6 \left (\int \left (\sin \left (f x +e \right ) c +c \right )^{n} \sin \left (f x +e \right )^{2}d x \right ) n -3 \left (\int \left (\sin \left (f x +e \right ) c +c \right )^{n} \sin \left (f x +e \right )^{2}d x \right )+4 \left (\int \left (\sin \left (f x +e \right ) c +c \right )^{n} \sin \left (f x +e \right )d x \right ) n -5 \left (\int \left (\sin \left (f x +e \right ) c +c \right )^{n} \sin \left (f x +e \right )d x \right )\right ) \] Input:
int((a-a*sin(f*x+e))^3*(c+c*sin(f*x+e))^n*(B*(3-n)+B*(4+n)*sin(f*x+e)),x)
Output:
a**3*b*( - int((sin(e + f*x)*c + c)**n,x)*n + 3*int((sin(e + f*x)*c + c)** n,x) - int((sin(e + f*x)*c + c)**n*sin(e + f*x)**4,x)*n - 4*int((sin(e + f *x)*c + c)**n*sin(e + f*x)**4,x) + 4*int((sin(e + f*x)*c + c)**n*sin(e + f *x)**3,x)*n + 9*int((sin(e + f*x)*c + c)**n*sin(e + f*x)**3,x) - 6*int((si n(e + f*x)*c + c)**n*sin(e + f*x)**2,x)*n - 3*int((sin(e + f*x)*c + c)**n* sin(e + f*x)**2,x) + 4*int((sin(e + f*x)*c + c)**n*sin(e + f*x),x)*n - 5*i nt((sin(e + f*x)*c + c)**n*sin(e + f*x),x))