Integrand size = 28, antiderivative size = 109 \[ \int (a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2} \, dx=\frac {64 a^3 c^6 \cos ^7(e+f x)}{693 f (c-c \sin (e+f x))^{7/2}}+\frac {16 a^3 c^5 \cos ^7(e+f x)}{99 f (c-c \sin (e+f x))^{5/2}}+\frac {2 a^3 c^4 \cos ^7(e+f x)}{11 f (c-c \sin (e+f x))^{3/2}} \] Output:
64/693*a^3*c^6*cos(f*x+e)^7/f/(c-c*sin(f*x+e))^(7/2)+16/99*a^3*c^5*cos(f*x +e)^7/f/(c-c*sin(f*x+e))^(5/2)+2/11*a^3*c^4*cos(f*x+e)^7/f/(c-c*sin(f*x+e) )^(3/2)
Leaf count is larger than twice the leaf count of optimal. \(1105\) vs. \(2(109)=218\).
Time = 13.41 (sec) , antiderivative size = 1105, normalized size of antiderivative = 10.14 \[ \int (a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2} \, dx =\text {Too large to display} \] Input:
Integrate[(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2),x]
Output:
(5*Cos[(e + f*x)/2]*(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2))/(8* f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x )/2])^6) - (5*Cos[(3*(e + f*x))/2]*(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f *x])^(5/2))/(24*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2 ] + Sin[(e + f*x)/2])^6) + (Cos[(5*(e + f*x))/2]*(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2))/(16*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(C os[(e + f*x)/2] + Sin[(e + f*x)/2])^6) - (5*Cos[(7*(e + f*x))/2]*(a + a*Si n[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2))/(112*f*(Cos[(e + f*x)/2] - Sin[( e + f*x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6) + (Cos[(9*(e + f*x ))/2]*(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2))/(144*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6) - ( Cos[(11*(e + f*x))/2]*(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2))/( 176*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6) + (5*Sin[(e + f*x)/2]*(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f *x])^(5/2))/(8*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6) + (5*(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^( 5/2)*Sin[(3*(e + f*x))/2])/(24*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*( Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6) + ((a + a*Sin[e + f*x])^3*(c - c*S in[e + f*x])^(5/2)*Sin[(5*(e + f*x))/2])/(16*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6) + (5*(a + a*Sin[e...
Time = 0.66 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3215, 3042, 3153, 3042, 3153, 3042, 3152}
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.
| \(\displaystyle \int (a \sin (e+f x)+a)^3 (c-c \sin (e+f x))^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \sin (e+f x)+a)^3 (c-c \sin (e+f x))^{5/2}dx\) |
| \(\Big \downarrow \) 3215 |
| \(\displaystyle a^3 c^3 \int \frac {\cos ^6(e+f x)}{\sqrt {c-c \sin (e+f x)}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \int \frac {\cos (e+f x)^6}{\sqrt {c-c \sin (e+f x)}}dx\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle a^3 c^3 \left (\frac {8}{11} c \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^{3/2}}dx+\frac {2 c \cos ^7(e+f x)}{11 f (c-c \sin (e+f x))^{3/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (\frac {8}{11} c \int \frac {\cos (e+f x)^6}{(c-c \sin (e+f x))^{3/2}}dx+\frac {2 c \cos ^7(e+f x)}{11 f (c-c \sin (e+f x))^{3/2}}\right )\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle a^3 c^3 \left (\frac {8}{11} c \left (\frac {4}{9} c \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^{5/2}}dx+\frac {2 c \cos ^7(e+f x)}{9 f (c-c \sin (e+f x))^{5/2}}\right )+\frac {2 c \cos ^7(e+f x)}{11 f (c-c \sin (e+f x))^{3/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \left (\frac {8}{11} c \left (\frac {4}{9} c \int \frac {\cos (e+f x)^6}{(c-c \sin (e+f x))^{5/2}}dx+\frac {2 c \cos ^7(e+f x)}{9 f (c-c \sin (e+f x))^{5/2}}\right )+\frac {2 c \cos ^7(e+f x)}{11 f (c-c \sin (e+f x))^{3/2}}\right )\) |
| \(\Big \downarrow \) 3152 |
| \(\displaystyle a^3 c^3 \left (\frac {8}{11} c \left (\frac {8 c^2 \cos ^7(e+f x)}{63 f (c-c \sin (e+f x))^{7/2}}+\frac {2 c \cos ^7(e+f x)}{9 f (c-c \sin (e+f x))^{5/2}}\right )+\frac {2 c \cos ^7(e+f x)}{11 f (c-c \sin (e+f x))^{3/2}}\right )\) |
Input:
Int[(a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2),x]
Output:
a^3*c^3*((2*c*Cos[e + f*x]^7)/(11*f*(c - c*Sin[e + f*x])^(3/2)) + (8*c*((8 *c^2*Cos[e + f*x]^7)/(63*f*(c - c*Sin[e + f*x])^(7/2)) + (2*c*Cos[e + f*x] ^7)/(9*f*(c - c*Sin[e + f*x])^(5/2))))/11)
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x ])^(m - 1)/(f*g*(m - 1))), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[(-b)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m + p))), x] + Simp[a*((2*m + p - 1)/(m + p)) Int[(g* Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*c^m Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[ b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && !(IntegerQ[n] && ((Lt Q[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
Time = 38.50 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.65
| method | result | size |
| default | \(-\frac {2 \left (-1+\sin \left (f x +e \right )\right ) c^{3} \left (1+\sin \left (f x +e \right )\right )^{4} a^{3} \left (63 \sin \left (f x +e \right )^{2}-182 \sin \left (f x +e \right )+151\right )}{693 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(71\) |
| parts | \(-\frac {2 a^{3} \left (-1+\sin \left (f x +e \right )\right ) c^{3} \left (1+\sin \left (f x +e \right )\right ) \left (3 \sin \left (f x +e \right )^{2}-14 \sin \left (f x +e \right )+43\right )}{15 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}-\frac {2 a^{3} \left (-1+\sin \left (f x +e \right )\right ) c^{3} \left (1+\sin \left (f x +e \right )\right ) \left (63 \sin \left (f x +e \right )^{5}-224 \sin \left (f x +e \right )^{4}+355 \sin \left (f x +e \right )^{3}-426 \sin \left (f x +e \right )^{2}+568 \sin \left (f x +e \right )-1136\right )}{693 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}-\frac {2 a^{3} \left (-1+\sin \left (f x +e \right )\right ) c^{3} \left (1+\sin \left (f x +e \right )\right ) \left (3 \sin \left (f x +e \right )^{3}-12 \sin \left (f x +e \right )^{2}+23 \sin \left (f x +e \right )-46\right )}{7 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}-\frac {2 a^{3} \left (-1+\sin \left (f x +e \right )\right ) c^{3} \left (1+\sin \left (f x +e \right )\right ) \left (35 \sin \left (f x +e \right )^{4}-130 \sin \left (f x +e \right )^{3}+219 \sin \left (f x +e \right )^{2}-292 \sin \left (f x +e \right )+584\right )}{105 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(334\) |
Input:
int((a+sin(f*x+e)*a)^3*(c-c*sin(f*x+e))^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/693*(-1+sin(f*x+e))*c^3*(1+sin(f*x+e))^4*a^3*(63*sin(f*x+e)^2-182*sin(f *x+e)+151)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f
Leaf count of result is larger than twice the leaf count of optimal. 234 vs. \(2 (97) = 194\).
Time = 0.09 (sec) , antiderivative size = 234, normalized size of antiderivative = 2.15 \[ \int (a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2} \, dx=-\frac {2 \, {\left (63 \, a^{3} c^{2} \cos \left (f x + e\right )^{6} - 7 \, a^{3} c^{2} \cos \left (f x + e\right )^{5} + 10 \, a^{3} c^{2} \cos \left (f x + e\right )^{4} - 16 \, a^{3} c^{2} \cos \left (f x + e\right )^{3} + 32 \, a^{3} c^{2} \cos \left (f x + e\right )^{2} - 128 \, a^{3} c^{2} \cos \left (f x + e\right ) - 256 \, a^{3} c^{2} - {\left (63 \, a^{3} c^{2} \cos \left (f x + e\right )^{5} + 70 \, a^{3} c^{2} \cos \left (f x + e\right )^{4} + 80 \, a^{3} c^{2} \cos \left (f x + e\right )^{3} + 96 \, a^{3} c^{2} \cos \left (f x + e\right )^{2} + 128 \, a^{3} c^{2} \cos \left (f x + e\right ) + 256 \, a^{3} c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{693 \, {\left (f \cos \left (f x + e\right ) - f \sin \left (f x + e\right ) + f\right )}} \] Input:
integrate((a+a*sin(f*x+e))^3*(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas")
Output:
-2/693*(63*a^3*c^2*cos(f*x + e)^6 - 7*a^3*c^2*cos(f*x + e)^5 + 10*a^3*c^2* cos(f*x + e)^4 - 16*a^3*c^2*cos(f*x + e)^3 + 32*a^3*c^2*cos(f*x + e)^2 - 1 28*a^3*c^2*cos(f*x + e) - 256*a^3*c^2 - (63*a^3*c^2*cos(f*x + e)^5 + 70*a^ 3*c^2*cos(f*x + e)^4 + 80*a^3*c^2*cos(f*x + e)^3 + 96*a^3*c^2*cos(f*x + e) ^2 + 128*a^3*c^2*cos(f*x + e) + 256*a^3*c^2)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c)/(f*cos(f*x + e) - f*sin(f*x + e) + f)
\[ \int (a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2} \, dx=a^{3} \left (\int c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c}\, dx + \int c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )}\, dx + \int \left (- 2 c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )}\right )\, dx + \int \left (- 2 c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{3}{\left (e + f x \right )}\right )\, dx + \int c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{4}{\left (e + f x \right )}\, dx + \int c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{5}{\left (e + f x \right )}\, dx\right ) \] Input:
integrate((a+a*sin(f*x+e))**3*(c-c*sin(f*x+e))**(5/2),x)
Output:
a**3*(Integral(c**2*sqrt(-c*sin(e + f*x) + c), x) + Integral(c**2*sqrt(-c* sin(e + f*x) + c)*sin(e + f*x), x) + Integral(-2*c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**2, x) + Integral(-2*c**2*sqrt(-c*sin(e + f*x) + c)*sin (e + f*x)**3, x) + Integral(c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**4 , x) + Integral(c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**5, x))
\[ \int (a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{3} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^3*(c-c*sin(f*x+e))^(5/2),x, algorithm="maxima")
Output:
integrate((a*sin(f*x + e) + a)^3*(-c*sin(f*x + e) + c)^(5/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 210 vs. \(2 (97) = 194\).
Time = 0.23 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.93 \[ \int (a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2} \, dx=-\frac {\sqrt {2} {\left (6930 \, a^{3} c^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 2310 \, a^{3} c^{2} \cos \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, f x + \frac {3}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 693 \, a^{3} c^{2} \cos \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, f x + \frac {5}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 495 \, a^{3} c^{2} \cos \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, f x + \frac {7}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 77 \, a^{3} c^{2} \cos \left (-\frac {9}{4} \, \pi + \frac {9}{2} \, f x + \frac {9}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 63 \, a^{3} c^{2} \cos \left (-\frac {11}{4} \, \pi + \frac {11}{2} \, f x + \frac {11}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {c}}{11088 \, f} \] Input:
integrate((a+a*sin(f*x+e))^3*(c-c*sin(f*x+e))^(5/2),x, algorithm="giac")
Output:
-1/11088*sqrt(2)*(6930*a^3*c^2*cos(-1/4*pi + 1/2*f*x + 1/2*e)*sgn(sin(-1/4 *pi + 1/2*f*x + 1/2*e)) + 2310*a^3*c^2*cos(-3/4*pi + 3/2*f*x + 3/2*e)*sgn( sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 693*a^3*c^2*cos(-5/4*pi + 5/2*f*x + 5/2* e)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 495*a^3*c^2*cos(-7/4*pi + 7/2*f*x + 7/2*e)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 77*a^3*c^2*cos(-9/4*pi + 9 /2*f*x + 9/2*e)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 63*a^3*c^2*cos(-11/4 *pi + 11/2*f*x + 11/2*e)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(c)/f
Timed out. \[ \int (a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2} \, dx=\int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^3\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2} \,d x \] Input:
int((a + a*sin(e + f*x))^3*(c - c*sin(e + f*x))^(5/2),x)
Output:
int((a + a*sin(e + f*x))^3*(c - c*sin(e + f*x))^(5/2), x)
\[ \int (a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2} \, dx=\sqrt {c}\, a^{3} c^{2} \left (\int \sqrt {-\sin \left (f x +e \right )+1}d x +\int \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{5}d x +\int \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{4}d x -2 \left (\int \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}d x \right )-2 \left (\int \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right )+\int \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right ) \] Input:
int((a+a*sin(f*x+e))^3*(c-c*sin(f*x+e))^(5/2),x)
Output:
sqrt(c)*a**3*c**2*(int(sqrt( - sin(e + f*x) + 1),x) + int(sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**5,x) + int(sqrt( - sin(e + f*x) + 1)*sin(e + f*x)* *4,x) - 2*int(sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**3,x) - 2*int(sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**2,x) + int(sqrt( - sin(e + f*x) + 1)*sin( e + f*x),x))