Integrand size = 28, antiderivative size = 145 \[ \int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{7/2} \, dx=\frac {256 a^2 c^6 \cos ^5(e+f x)}{1155 f (c-c \sin (e+f x))^{5/2}}+\frac {64 a^2 c^5 \cos ^5(e+f x)}{231 f (c-c \sin (e+f x))^{3/2}}+\frac {8 a^2 c^4 \cos ^5(e+f x)}{33 f \sqrt {c-c \sin (e+f x)}}+\frac {2 a^2 c^3 \cos ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{11 f} \]
256/1155*a^2*c^6*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(5/2)+64/231*a^2*c^5*cos( f*x+e)^5/f/(c-c*sin(f*x+e))^(3/2)+8/33*a^2*c^4*cos(f*x+e)^5/f/(c-c*sin(f*x +e))^(1/2)+2/11*a^2*c^3*cos(f*x+e)^5*(c-c*sin(f*x+e))^(1/2)/f
Leaf count is larger than twice the leaf count of optimal. \(1105\) vs. \(2(145)=290\).
Time = 10.84 (sec) , antiderivative size = 1105, normalized size of antiderivative = 7.62 \[ \int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{7/2} \, dx =\text {Too large to display} \]
(7*Cos[(e + f*x)/2]*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(7/2))/(8* f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)/2] + Sin[(e + f*x )/2])^4) - (Cos[(3*(e + f*x))/2]*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x ])^(7/2))/(8*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4) + (11*Cos[(5*(e + f*x))/2]*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(7/2))/(80*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(C os[(e + f*x)/2] + Sin[(e + f*x)/2])^4) + (Cos[(7*(e + f*x))/2]*(a + a*Sin[ e + f*x])^2*(c - c*Sin[e + f*x])^(7/2))/(112*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4) + (Cos[(9*(e + f*x)) /2]*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(7/2))/(48*f*(Cos[(e + f*x )/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4) + (Cos [(11*(e + f*x))/2]*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(7/2))/(176 *f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)/2] + Sin[(e + f* x)/2])^4) + (7*Sin[(e + f*x)/2]*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x] )^(7/2))/(8*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4) + ((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(7/2)* Sin[(3*(e + f*x))/2])/(8*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4) + (11*(a + a*Sin[e + f*x])^2*(c - c*Sin[ e + f*x])^(7/2)*Sin[(5*(e + f*x))/2])/(80*f*(Cos[(e + f*x)/2] - Sin[(e + f *x)/2])^7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4) - ((a + a*Sin[e + f*...
Time = 0.74 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3042, 3215, 3042, 3153, 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.
| \(\displaystyle \int (a \sin (e+f x)+a)^2 (c-c \sin (e+f x))^{7/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \sin (e+f x)+a)^2 (c-c \sin (e+f x))^{7/2}dx\) |
| \(\Big \downarrow \) 3215 |
| \(\displaystyle a^2 c^2 \int \cos ^4(e+f x) (c-c \sin (e+f x))^{3/2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \int \cos (e+f x)^4 (c-c \sin (e+f x))^{3/2}dx\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle a^2 c^2 \left (\frac {12}{11} c \int \cos ^4(e+f x) \sqrt {c-c \sin (e+f x)}dx+\frac {2 c \cos ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{11 f}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \left (\frac {12}{11} c \int \cos (e+f x)^4 \sqrt {c-c \sin (e+f x)}dx+\frac {2 c \cos ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{11 f}\right )\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle a^2 c^2 \left (\frac {12}{11} c \left (\frac {8}{9} c \int \frac {\cos ^4(e+f x)}{\sqrt {c-c \sin (e+f x)}}dx+\frac {2 c \cos ^5(e+f x)}{9 f \sqrt {c-c \sin (e+f x)}}\right )+\frac {2 c \cos ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{11 f}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \left (\frac {12}{11} c \left (\frac {8}{9} c \int \frac {\cos (e+f x)^4}{\sqrt {c-c \sin (e+f x)}}dx+\frac {2 c \cos ^5(e+f x)}{9 f \sqrt {c-c \sin (e+f x)}}\right )+\frac {2 c \cos ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{11 f}\right )\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle a^2 c^2 \left (\frac {12}{11} c \left (\frac {8}{9} c \left (\frac {4}{7} c \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{3/2}}dx+\frac {2 c \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^{3/2}}\right )+\frac {2 c \cos ^5(e+f x)}{9 f \sqrt {c-c \sin (e+f x)}}\right )+\frac {2 c \cos ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{11 f}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \left (\frac {12}{11} c \left (\frac {8}{9} c \left (\frac {4}{7} c \int \frac {\cos (e+f x)^4}{(c-c \sin (e+f x))^{3/2}}dx+\frac {2 c \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^{3/2}}\right )+\frac {2 c \cos ^5(e+f x)}{9 f \sqrt {c-c \sin (e+f x)}}\right )+\frac {2 c \cos ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{11 f}\right )\) |
| \(\Big \downarrow \) 3152 |
| \(\displaystyle a^2 c^2 \left (\frac {12}{11} c \left (\frac {8}{9} c \left (\frac {8 c^2 \cos ^5(e+f x)}{35 f (c-c \sin (e+f x))^{5/2}}+\frac {2 c \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^{3/2}}\right )+\frac {2 c \cos ^5(e+f x)}{9 f \sqrt {c-c \sin (e+f x)}}\right )+\frac {2 c \cos ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{11 f}\right )\) |
a^2*c^2*((2*c*Cos[e + f*x]^5*Sqrt[c - c*Sin[e + f*x]])/(11*f) + (12*c*((2* c*Cos[e + f*x]^5)/(9*f*Sqrt[c - c*Sin[e + f*x]]) + (8*c*((8*c^2*Cos[e + f* x]^5)/(35*f*(c - c*Sin[e + f*x])^(5/2)) + (2*c*Cos[e + f*x]^5)/(7*f*(c - c *Sin[e + f*x])^(3/2))))/9))/11)
3.3.98.3.1 Defintions of rubi rules used
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 = 7.74 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.56
| method | result | size |
| default | \(\frac {2 \left (\sin \left (f x +e \right )-1\right ) c^{4} \left (\sin \left (f x +e \right )+1\right )^{3} a^{2} \left (105 \left (\sin ^{3}\left (f x +e \right )\right )-455 \left (\sin ^{2}\left (f x +e \right )\right )+755 \sin \left (f x +e \right )-533\right )}{1155 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(81\) |
| parts | \(\frac {2 a^{2} \left (\sin \left (f x +e \right )-1\right ) c^{4} \left (\sin \left (f x +e \right )+1\right ) \left (5 \left (\sin ^{3}\left (f x +e \right )\right )-27 \left (\sin ^{2}\left (f x +e \right )\right )+71 \sin \left (f x +e \right )-177\right )}{35 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {2 a^{2} \left (\sin \left (f x +e \right )-1\right ) c^{4} \left (\sin \left (f x +e \right )+1\right ) \left (315 \left (\sin ^{5}\left (f x +e \right )\right )-1505 \left (\sin ^{4}\left (f x +e \right )\right )+3205 \left (\sin ^{3}\left (f x +e \right )\right )-4539 \left (\sin ^{2}\left (f x +e \right )\right )+6052 \sin \left (f x +e \right )-12104\right )}{3465 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {4 a^{2} \left (\sin \left (f x +e \right )-1\right ) c^{4} \left (\sin \left (f x +e \right )+1\right ) \left (5 \left (\sin ^{4}\left (f x +e \right )\right )-25 \left (\sin ^{3}\left (f x +e \right )\right )+57 \left (\sin ^{2}\left (f x +e \right )\right )-91 \sin \left (f x +e \right )+182\right )}{45 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(266\) |
2/1155*(sin(f*x+e)-1)*c^4*(sin(f*x+e)+1)^3*a^2*(105*sin(f*x+e)^3-455*sin(f *x+e)^2+755*sin(f*x+e)-533)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f
Time = 0.36 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.61 \[ \int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{7/2} \, dx=\frac {2 \, {\left (105 \, a^{2} c^{3} \cos \left (f x + e\right )^{6} + 245 \, a^{2} c^{3} \cos \left (f x + e\right )^{5} - 20 \, a^{2} c^{3} \cos \left (f x + e\right )^{4} + 32 \, a^{2} c^{3} \cos \left (f x + e\right )^{3} - 64 \, a^{2} c^{3} \cos \left (f x + e\right )^{2} + 256 \, a^{2} c^{3} \cos \left (f x + e\right ) + 512 \, a^{2} c^{3} - {\left (105 \, a^{2} c^{3} \cos \left (f x + e\right )^{5} - 140 \, a^{2} c^{3} \cos \left (f x + e\right )^{4} - 160 \, a^{2} c^{3} \cos \left (f x + e\right )^{3} - 192 \, a^{2} c^{3} \cos \left (f x + e\right )^{2} - 256 \, a^{2} c^{3} \cos \left (f x + e\right ) - 512 \, a^{2} c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{1155 \, {\left (f \cos \left (f x + e\right ) - f \sin \left (f x + e\right ) + f\right )}} \]
2/1155*(105*a^2*c^3*cos(f*x + e)^6 + 245*a^2*c^3*cos(f*x + e)^5 - 20*a^2*c ^3*cos(f*x + e)^4 + 32*a^2*c^3*cos(f*x + e)^3 - 64*a^2*c^3*cos(f*x + e)^2 + 256*a^2*c^3*cos(f*x + e) + 512*a^2*c^3 - (105*a^2*c^3*cos(f*x + e)^5 - 1 40*a^2*c^3*cos(f*x + e)^4 - 160*a^2*c^3*cos(f*x + e)^3 - 192*a^2*c^3*cos(f *x + e)^2 - 256*a^2*c^3*cos(f*x + e) - 512*a^2*c^3)*sin(f*x + e))*sqrt(-c* sin(f*x + e) + c)/(f*cos(f*x + e) - f*sin(f*x + e) + f)
Timed out. \[ \int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{7/2} \, dx=\text {Timed out} \]
\[ \int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{7/2} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{2} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}} \,d x } \]
Time = 0.42 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.45 \[ \int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{7/2} \, dx=-\frac {\sqrt {2} {\left (16170 \, a^{2} c^{3} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 2310 \, a^{2} c^{3} \cos \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, f x + \frac {3}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 2541 \, a^{2} c^{3} \cos \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, f x + \frac {5}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 165 \, a^{2} c^{3} \cos \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, f x + \frac {7}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 385 \, a^{2} c^{3} \cos \left (-\frac {9}{4} \, \pi + \frac {9}{2} \, f x + \frac {9}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 105 \, a^{2} c^{3} \cos \left (-\frac {11}{4} \, \pi + \frac {11}{2} \, f x + \frac {11}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {c}}{18480 \, f} \]
-1/18480*sqrt(2)*(16170*a^2*c^3*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^2*c^3*cos(-3/4*pi + 3/2*f*x + 3/2*e)*sgn (sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 2541*a^2*c^3*cos(-5/4*pi + 5/2*f*x + 5/ 2*e)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 165*a^2*c^3*cos(-7/4*pi + 7/2*f *x + 7/2*e)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 385*a^2*c^3*cos(-9/4*pi + 9/2*f*x + 9/2*e)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 105*a^2*c^3*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. \[ \int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{7/2} \, dx=\int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^2\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{7/2} \,d x \]