Integrand size = 31, antiderivative size = 168 \[ \int x \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=\frac {c \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{f^2}-\frac {3 c x \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{4 f}+\frac {c \sin (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{4 f^2}+\frac {x \sec (e+f x) \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{5/2}}{2 c f} \]
1/2*x*sec(f*x+e)*(c+c*sin(f*x+e))^(5/2)*(a-a*sin(f*x+e))^(1/2)/c/f+c*(a-a* sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2)/f^2-3/4*c*x*sec(f*x+e)*(a-a*sin(f *x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2)/f+1/4*c*sin(f*x+e)*(a-a*sin(f*x+e))^(1 /2)*(c+c*sin(f*x+e))^(1/2)/f^2
Time = 0.89 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.43 \[ \int x \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=\frac {c \sqrt {c (1+\sin (e+f x))} \sqrt {a-a \sin (e+f x)} (4-f x \cos (2 (e+f x)) \sec (e+f x)+\sin (e+f x)+4 f x \tan (e+f x))}{4 f^2} \]
(c*Sqrt[c*(1 + Sin[e + f*x])]*Sqrt[a - a*Sin[e + f*x]]*(4 - f*x*Cos[2*(e + f*x)]*Sec[e + f*x] + Sin[e + f*x] + 4*f*x*Tan[e + f*x]))/(4*f^2)
Time = 0.40 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.68, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {5115, 4922, 3042, 3123}
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.
\(\displaystyle \int x \sqrt {a-a \sin (e+f x)} (c \sin (e+f x)+c)^{3/2} \, dx\) |
\(\Big \downarrow \) 5115 |
\(\displaystyle \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \int x \cos (e+f x) (\sin (e+f x) c+c)dx\) |
\(\Big \downarrow \) 4922 |
\(\displaystyle \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \left (\frac {x (c \sin (e+f x)+c)^2}{2 c f}-\frac {\int (\sin (e+f x) c+c)^2dx}{2 c f}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \left (\frac {x (c \sin (e+f x)+c)^2}{2 c f}-\frac {\int (\sin (e+f x) c+c)^2dx}{2 c f}\right )\) |
\(\Big \downarrow \) 3123 |
\(\displaystyle \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \left (\frac {x (c \sin (e+f x)+c)^2}{2 c f}-\frac {-\frac {2 c^2 \cos (e+f x)}{f}-\frac {c^2 \sin (e+f x) \cos (e+f x)}{2 f}+\frac {3 c^2 x}{2}}{2 c f}\right )\) |
Sec[e + f*x]*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]]*((x*(c + c* Sin[e + f*x])^2)/(2*c*f) - ((3*c^2*x)/2 - (2*c^2*Cos[e + f*x])/f - (c^2*Co s[e + f*x]*Sin[e + f*x])/(2*f))/(2*c*f))
3.2.73.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[(2*a^2 + b^ 2)*(x/2), x] + (-Simp[2*a*b*(Cos[c + d*x]/d), x] - Simp[b^2*Cos[c + d*x]*(S in[c + d*x]/(2*d)), x]) /; FreeQ[{a, b, c, d}, x]
Int[Cos[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.)*((a_) + (b_.)*Sin[(c _.) + (d_.)*(x_)])^(n_.), x_Symbol] :> Simp[(e + f*x)^m*((a + b*Sin[c + d*x ])^(n + 1)/(b*d*(n + 1))), x] - Simp[f*(m/(b*d*(n + 1))) Int[(e + f*x)^(m - 1)*(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && IGtQ[m, 0] && NeQ[n, -1]
Int[((g_.) + (h_.)*(x_))^(p_.)*((a_) + (b_.)*Sin[(e_.) + (f_.)*(x_)])^(m_)* ((c_) + (d_.)*Sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a^IntPart[m] *c^IntPart[m]*(a + b*Sin[e + f*x])^FracPart[m]*((c + d*Sin[e + f*x])^FracPa rt[m]/Cos[e + f*x]^(2*FracPart[m])) Int[(g + h*x)^p*Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[p] && IntegerQ[2*m] && IGeQ[n - m, 0]
\[\int x \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {a -\sin \left (f x +e \right ) a}d x\]
Exception generated. \[ \int x \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (has polynomial part)
\[ \int x \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=\int x \left (c \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right )}\, dx \]
\[ \int x \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=\int { \sqrt {-a \sin \left (f x + e\right ) + a} {\left (c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}} x \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 311 vs. \(2 (146) = 292\).
Time = 0.34 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.85 \[ \int x \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=-\frac {{\left (\frac {8 \, c \cos \left (f x + e\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{f} + \frac {c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (2 \, f x + 2 \, e\right )}{f} - \frac {{\left (\pi c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - {\left (\pi - 2 \, f x - 2 \, e\right )} c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 2 \, c e \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \cos \left (2 \, f x + 2 \, e\right )}{f} + \frac {4 \, {\left (\pi c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - {\left (\pi - 2 \, f x - 2 \, e\right )} c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 2 \, c e \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (f x + e\right )}{f}\right )} \sqrt {a} \sqrt {c}}{8 \, f} \]
-1/8*(8*c*cos(f*x + e)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/f + c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/ 4*pi + 1/2*f*x + 1/2*e))*sin(2*f*x + 2*e)/f - (pi*c*sgn(cos(-1/4*pi + 1/2* f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - (pi - 2*f*x - 2*e)*c*s gn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 2 *c*e*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e )))*cos(2*f*x + 2*e)/f + 4*(pi*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(s in(-1/4*pi + 1/2*f*x + 1/2*e)) - (pi - 2*f*x - 2*e)*c*sgn(cos(-1/4*pi + 1/ 2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 2*c*e*sgn(cos(-1/4*p i + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))*sin(f*x + e)/f) *sqrt(a)*sqrt(c)/f
Time = 29.20 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.73 \[ \int x \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx=-\frac {c\,\sqrt {-a\,\left (\sin \left (e+f\,x\right )-1\right )}\,\sqrt {c\,\left (\sin \left (e+f\,x\right )+1\right )}\,\left (-16\,{\sin \left (e+f\,x\right )}^2+\sin \left (e+f\,x\right )+\sin \left (3\,e+3\,f\,x\right )+8\,f\,x\,\sin \left (2\,e+2\,f\,x\right )+2\,f\,x\,\left (2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )+2\,f\,x\,\left (2\,{\sin \left (\frac {3\,e}{2}+\frac {3\,f\,x}{2}\right )}^2-1\right )+16\right )}{8\,f^2\,\left (2\,{\sin \left (e+f\,x\right )}^2-2\right )} \]