Integrand size = 46, 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 = 2.29 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.85 \[ \int (a+a \sin (e+f x))^3 (c-c \sin (e+f x))^n (B (3-n)-B (4+n) \sin (e+f x)) \, dx=\frac {a^3 B (c-c \sin (e+f x))^n (14 \cos (e+f x)-6 \cos (3 (e+f x))+14 \sin (2 (e+f x))-\sin (4 (e+f x)))}{8 f} \] 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*(c - c*Sin[e + f*x])^n*(14*Cos[e + f*x] - 6*Cos[3*(e + f*x)] + 14*S in[2*(e + f*x)] - Sin[4*(e + f*x)]))/(8*f)
Time = 0.51 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, 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.
\(\displaystyle \int (a \sin (e+f x)+a)^3 (B (3-n)-B (n+4) \sin (e+f x)) (c-c \sin (e+f x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \sin (e+f x)+a)^3 (B (3-n)-B (n+4) \sin (e+f x)) (c-c \sin (e+f x))^ndx\) |
\(\Big \downarrow \) 3446 |
\(\displaystyle a^3 c^3 \int \cos ^6(e+f x) (c-c \sin (e+f x))^{n-3} (B (3-n)-B (n+4) \sin (e+f x))dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a^3 c^3 \int \cos (e+f x)^6 (c-c \sin (e+f x))^{n-3} (B (3-n)-B (n+4) \sin (e+f x))dx\) |
\(\Big \downarrow \) 3333 |
\(\displaystyle \frac {a^3 B c^3 \cos ^7(e+f x) (c-c \sin (e+f x))^{n-3}}{f}\) |
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 = 8.72 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.82
method | result | size |
parallelrisch | \(-\frac {a^{3} B \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{n} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{7}}{f \left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{4}}\) | \(62\) |
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*(sin(f*x+e)-1))^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. 78 vs. \(2 (34) = 68\).
Time = 0.09 (sec) , antiderivative size = 78, normalized size of antiderivative = 2.29 \[ \int (a+a \sin (e+f x))^3 (c-c \sin (e+f x))^n (B (3-n)-B (4+n) \sin (e+f x)) \, dx=-\frac {{\left (3 \, B a^{3} \cos \left (f x + e\right )^{3} - 4 \, B a^{3} \cos \left (f x + e\right ) + {\left (B a^{3} \cos \left (f x + e\right )^{3} - 4 \, B a^{3} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n}}{f} \] 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. \(2 (31) = 62\).
Time = 73.19 (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*t an(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*t an(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*t an(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. 6973 vs. \(2 (34) = 68\).
Time = 12.27 (sec) , antiderivative size = 6973, normalized size of antiderivative = 205.09 \[ \int (a+a \sin (e+f x))^3 (c-c \sin (e+f x))^n (B (3-n)-B (4+n) \sin (e+f x)) \, dx=\text {Too large to display} \] 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*ta n(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 = 37.51 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.88 \[ \int (a+a \sin (e+f x))^3 (c-c \sin (e+f x))^n (B (3-n)-B (4+n) \sin (e+f x)) \, dx=\frac {B\,a^3\,{\left (-c\,\left (\sin \left (e+f\,x\right )-1\right )\right )}^n\,\left (14\,\cos \left (e+f\,x\right )-6\,\cos \left (3\,e+3\,f\,x\right )+14\,\sin \left (2\,e+2\,f\,x\right )-\sin \left (4\,e+4\,f\,x\right )\right )}{8\,f} \] 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(( - sin(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*int(( - sin(e + f*x)*c + c)**n*sin(e + f*x),x))