Integrand size = 33, antiderivative size = 118 \[ \int x^2 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \, dx=\frac {2 x \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{f^2}-\frac {2 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \tan (e+f x)}{f^3}+\frac {x^2 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \tan (e+f x)}{f} \] Output:
2*x*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2)/f^2-2*(a-a*sin(f*x+e))^( 1/2)*(c+c*sin(f*x+e))^(1/2)*tan(f*x+e)/f^3+x^2*(a-a*sin(f*x+e))^(1/2)*(c+c *sin(f*x+e))^(1/2)*tan(f*x+e)/f
Time = 1.32 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.46 \[ \int x^2 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \, dx=\frac {\sqrt {c (1+\sin (e+f x))} \sqrt {a-a \sin (e+f x)} \left (2 f x+\left (-2+f^2 x^2\right ) \tan (e+f x)\right )}{f^3} \] Input:
Integrate[x^2*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]],x]
Output:
(Sqrt[c*(1 + Sin[e + f*x])]*Sqrt[a - a*Sin[e + f*x]]*(2*f*x + (-2 + f^2*x^ 2)*Tan[e + f*x]))/f^3
Time = 0.51 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.66, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {5115, 3042, 3777, 25, 3042, 3777, 3042, 3117}
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.
\(\displaystyle \int x^2 \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \, dx\) |
\(\Big \downarrow \) 5115 |
\(\displaystyle \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \int x^2 \cos (e+f x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \int x^2 \sin \left (e+f x+\frac {\pi }{2}\right )dx\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \left (\frac {2 \int -x \sin (e+f x)dx}{f}+\frac {x^2 \sin (e+f x)}{f}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \left (\frac {x^2 \sin (e+f x)}{f}-\frac {2 \int x \sin (e+f x)dx}{f}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \left (\frac {x^2 \sin (e+f x)}{f}-\frac {2 \int x \sin (e+f x)dx}{f}\right )\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \left (\frac {x^2 \sin (e+f x)}{f}-\frac {2 \left (\frac {\int \cos (e+f x)dx}{f}-\frac {x \cos (e+f x)}{f}\right )}{f}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \left (\frac {x^2 \sin (e+f x)}{f}-\frac {2 \left (\frac {\int \sin \left (e+f x+\frac {\pi }{2}\right )dx}{f}-\frac {x \cos (e+f x)}{f}\right )}{f}\right )\) |
\(\Big \downarrow \) 3117 |
\(\displaystyle \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \left (\frac {x^2 \sin (e+f x)}{f}-\frac {2 \left (\frac {\sin (e+f x)}{f^2}-\frac {x \cos (e+f x)}{f}\right )}{f}\right )\) |
Input:
Int[x^2*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]],x]
Output:
Sec[e + f*x]*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]]*((x^2*Sin[e + f*x])/f - (2*(-((x*Cos[e + f*x])/f) + Sin[e + f*x]/f^2))/f)
Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]
Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[( -(c + d*x)^m)*(Cos[e + f*x]/f), x] + Simp[d*(m/f) Int[(c + d*x)^(m - 1)*C os[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
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]
\[\int x^{2} \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {c +c \sin \left (f x +e \right )}d x\]
Input:
int(x^2*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2),x)
Output:
int(x^2*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2),x)
Exception generated. \[ \int x^2 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(x^2*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2),x, algorithm=" fricas")
Output:
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (has polynomial part)
\[ \int x^2 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \, dx=\int x^{2} \sqrt {c \left (\sin {\left (e + f x \right )} + 1\right )} \sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right )}\, dx \] Input:
integrate(x**2*(a-a*sin(f*x+e))**(1/2)*(c+c*sin(f*x+e))**(1/2),x)
Output:
Integral(x**2*sqrt(c*(sin(e + f*x) + 1))*sqrt(-a*(sin(e + f*x) - 1)), x)
\[ \int x^2 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \, dx=\int { \sqrt {-a \sin \left (f x + e\right ) + a} \sqrt {c \sin \left (f x + e\right ) + c} x^{2} \,d x } \] Input:
integrate(x^2*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2),x, algorithm=" maxima")
Output:
integrate(sqrt(-a*sin(f*x + e) + a)*sqrt(c*sin(f*x + e) + c)*x^2, x)
Time = 0.15 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.01 \[ \int x^2 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \, dx=-{\left (\frac {2 \, x \cos \left (f x + e\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{f^{2}} + \frac {{\left (f^{2} x^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 2 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (f x + e\right )}{f^{3}}\right )} \sqrt {a} \sqrt {c} \] Input:
integrate(x^2*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2),x, algorithm=" giac")
Output:
-(2*x*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^2 + (f^2*x^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(si n(-1/4*pi + 1/2*f*x + 1/2*e)) - 2*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(f*x + e)/f^3)*sqrt(a)*sqrt(c)
Time = 15.82 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.73 \[ \int x^2 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \, dx=\frac {\sqrt {-a\,\left (\sin \left (e+f\,x\right )-1\right )}\,\sqrt {c\,\left (\sin \left (e+f\,x\right )+1\right )}\,\left (2\,f\,x-2\,\sin \left (2\,e+2\,f\,x\right )+2\,f\,x\,\left (2\,{\cos \left (e+f\,x\right )}^2-1\right )+f^2\,x^2\,\sin \left (2\,e+2\,f\,x\right )\right )}{2\,f^3\,{\cos \left (e+f\,x\right )}^2} \] Input:
int(x^2*(a - a*sin(e + f*x))^(1/2)*(c + c*sin(e + f*x))^(1/2),x)
Output:
((-a*(sin(e + f*x) - 1))^(1/2)*(c*(sin(e + f*x) + 1))^(1/2)*(2*f*x - 2*sin (2*e + 2*f*x) + 2*f*x*(2*cos(e + f*x)^2 - 1) + f^2*x^2*sin(2*e + 2*f*x)))/ (2*f^3*cos(e + f*x)^2)
\[ \int x^2 \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \, dx=\sqrt {c}\, \sqrt {a}\, \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, x^{2}d x \right ) \] Input:
int(x^2*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2),x)
Output:
sqrt(c)*sqrt(a)*int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*x**2, x)