Integrand size = 44, antiderivative size = 33 \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^3 (B (-3+m)-B (4+m) \sin (e+f x)) \, dx=\frac {a^3 B c^3 \cos ^7(e+f x) (a+a \sin (e+f x))^{-3+m}}{f} \] Output:
a^3*B*c^3*cos(f*x+e)^7*(a+a*sin(f*x+e))^(-3+m)/f
Time = 8.73 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.00 \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^3 (B (-3+m)-B (4+m) \sin (e+f x)) \, dx=\frac {B c^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^7 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (a (1+\sin (e+f x)))^m}{f} \] Input:
Integrate[(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^3*(B*(-3 + m) - B*(4 + m)*Sin[e + f*x]),x]
Output:
(B*c^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^m)/f
Time = 0.48 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {3042, 3446, 25, 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.
\(\displaystyle \int (c-c \sin (e+f x))^3 (a \sin (e+f x)+a)^m (B (m-3)-B (m+4) \sin (e+f x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (c-c \sin (e+f x))^3 (a \sin (e+f x)+a)^m (B (m-3)-B (m+4) \sin (e+f x))dx\) |
\(\Big \downarrow \) 3446 |
\(\displaystyle a^3 c^3 \int -\cos ^6(e+f x) (\sin (e+f x) a+a)^{m-3} (B (3-m)+B (m+4) \sin (e+f x))dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -a^3 c^3 \int \cos ^6(e+f x) (\sin (e+f x) a+a)^{m-3} (B (3-m)+B (m+4) \sin (e+f x))dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -a^3 c^3 \int \cos (e+f x)^6 (\sin (e+f x) a+a)^{m-3} (B (3-m)+B (m+4) \sin (e+f x))dx\) |
\(\Big \downarrow \) 3333 |
\(\displaystyle \frac {a^3 B c^3 \cos ^7(e+f x) (a \sin (e+f x)+a)^{m-3}}{f}\) |
Input:
Int[(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^3*(B*(-3 + m) - B*(4 + m)* Sin[e + f*x]),x]
Output:
(a^3*B*c^3*Cos[e + f*x]^7*(a + a*Sin[e + f*x])^(-3 + m))/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 = 7.45 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.85
method | result | size |
parallelrisch | \(-\frac {c^{3} B \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{m} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}{f \left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{4}}\) | \(61\) |
Input:
int((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^3*(B*(-3+m)-B*(4+m)*sin(f*x+e)),x, method=_RETURNVERBOSE)
Output:
-c^3*B*(a*(1+sin(f*x+e)))^m*(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. \(2 (33) = 66\).
Time = 0.09 (sec) , antiderivative size = 78, normalized size of antiderivative = 2.36 \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^3 (B (-3+m)-B (4+m) \sin (e+f x)) \, dx=-\frac {{\left (3 \, B c^{3} \cos \left (f x + e\right )^{3} - 4 \, B c^{3} \cos \left (f x + e\right ) - {\left (B c^{3} \cos \left (f x + e\right )^{3} - 4 \, B c^{3} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{f} \] Input:
integrate((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^3*(B*(-3+m)-B*(4+m)*sin(f*x+ e)),x, algorithm="fricas")
Output:
-(3*B*c^3*cos(f*x + e)^3 - 4*B*c^3*cos(f*x + e) - (B*c^3*cos(f*x + e)^3 - 4*B*c^3*cos(f*x + e))*sin(f*x + e))*(a*sin(f*x + e) + a)^m/f
Leaf count of result is larger than twice the leaf count of optimal. 898 vs. \(2 (31) = 62\).
Time = 71.04 (sec) , antiderivative size = 898, normalized size of antiderivative = 27.21 \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^3 (B (-3+m)-B (4+m) \sin (e+f x)) \, dx=\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))**m*(c-c*sin(f*x+e))**3*(B*(-3+m)-B*(4+m)*sin(f* x+e)),x)
Output:
Piecewise((-B*c**3*(a + 2*a*tan(e/2 + f*x/2)/(tan(e/2 + f*x/2)**2 + 1))**m *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*c**3*(a + 2*a*t an(e/2 + f*x/2)/(tan(e/2 + f*x/2)**2 + 1))**m*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*c**3*(a + 2*a*tan(e/2 + f*x/2)/(tan(e/2 + f *x/2)**2 + 1))**m*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*c**3*(a + 2*a*tan(e/2 + f*x/2)/(tan(e/2 + f*x/2)**2 + 1))**m*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*c**3*(a + 2*a*tan(e/2 + f *x/2)/(tan(e/2 + f*x/2)**2 + 1))**m*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*c**3*(a + 2*a*tan(e/2 + f*x/2)/(tan(e/2 + f*x/2)**2 + 1))**m*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*c**3*(a + 2*a*tan(e/2 + f*x/2)/(tan(e/2 + f*x/2)**2 + 1))**m*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*c**3*(a + 2*a*tan(e/2 + f*x/2)/(tan(e/2 + f*x/2)**2 + 1))**m/(f*tan(e/2 + f*x/2)**8 + 4*f*tan(e/2 + f*x/2)**6 + 6...
\[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^3 (B (-3+m)-B (4+m) \sin (e+f x)) \, dx=\int { {\left (B {\left (m + 4\right )} \sin \left (f x + e\right ) - B {\left (m - 3\right )}\right )} {\left (c \sin \left (f x + e\right ) - c\right )}^{3} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^3*(B*(-3+m)-B*(4+m)*sin(f*x+ e)),x, algorithm="maxima")
Output:
integrate((B*(m + 4)*sin(f*x + e) - B*(m - 3))*(c*sin(f*x + e) - c)^3*(a*s in(f*x + e) + a)^m, x)
Leaf count of result is larger than twice the leaf count of optimal. 6973 vs. \(2 (33) = 66\).
Time = 11.88 (sec) , antiderivative size = 6973, normalized size of antiderivative = 211.30 \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^3 (B (-3+m)-B (4+m) \sin (e+f x)) \, dx=\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^3*(B*(-3+m)-B*(4+m)*sin(f*x+ e)),x, algorithm="giac")
Output:
(B*c^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(a)/(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))^m*tan(-1 /4*pi*m*sgn(2*a*tan(1/2*f*x + 1/2*e)^4 + 4*a*tan(1/2*f*x + 1/2*e)^3 - 4*a* tan(1/2*f*x + 1/2*e) - 2*a)*sgn(4*a*tan(1/2*f*x + 1/2*e)^3 + 8*a*tan(1/2*f *x + 1/2*e)^2 + 4*a*tan(1/2*f*x + 1/2*e)) + 1/2*pi*m*floor(f*x/pi + e/pi + 1/2) - 1/4*pi*m*sgn(4*a*tan(1/2*f*x + 1/2*e)^3 + 8*a*tan(1/2*f*x + 1/2*e) ^2 + 4*a*tan(1/2*f*x + 1/2*e)))^2*tan(1/2*f*x + 1/2*e)^8 - 6*B*c^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(a)/(tan(f*x + e)^2*tan(1/2*f*x + 1/...
Time = 36.93 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.85 \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^3 (B (-3+m)-B (4+m) \sin (e+f x)) \, dx=\frac {B\,c^3\,{\left (a\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,\left (14\,\cos \left (e+f\,x\right )-6\,\cos \left (3\,e+3\,f\,x\right )-14\,\sin \left (2\,e+2\,f\,x\right )+\sin \left (4\,e+4\,f\,x\right )\right )}{8\,f} \] Input:
int((B*(m - 3) - B*sin(e + f*x)*(m + 4))*(a + a*sin(e + f*x))^m*(c - c*sin (e + f*x))^3,x)
Output:
(B*c^3*(a*(sin(e + f*x) + 1))^m*(14*cos(e + f*x) - 6*cos(3*e + 3*f*x) - 14 *sin(2*e + 2*f*x) + sin(4*e + 4*f*x)))/(8*f)
\[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^3 (B (-3+m)-B (4+m) \sin (e+f x)) \, dx=b \,c^{3} \left (\left (\int \left (a +a \sin \left (f x +e \right )\right )^{m}d x \right ) m -3 \left (\int \left (a +a \sin \left (f x +e \right )\right )^{m}d x \right )+\left (\int \left (a +a \sin \left (f x +e \right )\right )^{m} \sin \left (f x +e \right )^{4}d x \right ) m +4 \left (\int \left (a +a \sin \left (f x +e \right )\right )^{m} \sin \left (f x +e \right )^{4}d x \right )-4 \left (\int \left (a +a \sin \left (f x +e \right )\right )^{m} \sin \left (f x +e \right )^{3}d x \right ) m -9 \left (\int \left (a +a \sin \left (f x +e \right )\right )^{m} \sin \left (f x +e \right )^{3}d x \right )+6 \left (\int \left (a +a \sin \left (f x +e \right )\right )^{m} \sin \left (f x +e \right )^{2}d x \right ) m +3 \left (\int \left (a +a \sin \left (f x +e \right )\right )^{m} \sin \left (f x +e \right )^{2}d x \right )-4 \left (\int \left (a +a \sin \left (f x +e \right )\right )^{m} \sin \left (f x +e \right )d x \right ) m +5 \left (\int \left (a +a \sin \left (f x +e \right )\right )^{m} \sin \left (f x +e \right )d x \right )\right ) \] Input:
int((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^3*(B*(-3+m)-B*(4+m)*sin(f*x+e)),x)
Output:
b*c**3*(int((sin(e + f*x)*a + a)**m,x)*m - 3*int((sin(e + f*x)*a + a)**m,x ) + int((sin(e + f*x)*a + a)**m*sin(e + f*x)**4,x)*m + 4*int((sin(e + f*x) *a + a)**m*sin(e + f*x)**4,x) - 4*int((sin(e + f*x)*a + a)**m*sin(e + f*x) **3,x)*m - 9*int((sin(e + f*x)*a + a)**m*sin(e + f*x)**3,x) + 6*int((sin(e + f*x)*a + a)**m*sin(e + f*x)**2,x)*m + 3*int((sin(e + f*x)*a + a)**m*sin (e + f*x)**2,x) - 4*int((sin(e + f*x)*a + a)**m*sin(e + f*x),x)*m + 5*int( (sin(e + f*x)*a + a)**m*sin(e + f*x),x))