Integrand size = 30, antiderivative size = 93 \[ \int \frac {(c-c \sin (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)}} \, dx=\frac {2 c^2 \cos (e+f x) \log (1+\sin (e+f x))}{f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a+a \sin (e+f x)}} \] Output:
2*c^2*cos(f*x+e)*ln(1+sin(f*x+e))/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e) )^(1/2)+c*cos(f*x+e)*(c-c*sin(f*x+e))^(1/2)/f/(a+a*sin(f*x+e))^(1/2)
Time = 3.43 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.28 \[ \int \frac {(c-c \sin (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)}} \, dx=\frac {c \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x)) \left (-4 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+\sin (e+f x)\right ) \sqrt {c-c \sin (e+f x)}}{f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \sqrt {a (1+\sin (e+f x))}} \] Input:
Integrate[(c - c*Sin[e + f*x])^(3/2)/Sqrt[a + a*Sin[e + f*x]],x]
Output:
(c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])*(-4*Log[Cos[( e + f*x)/2] + Sin[(e + f*x)/2]] + Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/ (f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*Sqrt[a*(1 + Sin[e + f*x])])
Time = 0.56 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {3042, 3219, 3042, 3216, 3042, 3146, 16}
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.
\(\displaystyle \int \frac {(c-c \sin (e+f x))^{3/2}}{\sqrt {a \sin (e+f x)+a}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(c-c \sin (e+f x))^{3/2}}{\sqrt {a \sin (e+f x)+a}}dx\) |
\(\Big \downarrow \) 3219 |
\(\displaystyle 2 c \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {\sin (e+f x) a+a}}dx+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 c \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {\sin (e+f x) a+a}}dx+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 3216 |
\(\displaystyle \frac {2 a c^2 \cos (e+f x) \int \frac {\cos (e+f x)}{\sin (e+f x) a+a}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 a c^2 \cos (e+f x) \int \frac {\cos (e+f x)}{\sin (e+f x) a+a}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 3146 |
\(\displaystyle \frac {2 c^2 \cos (e+f x) \int \frac {1}{\sin (e+f x) a+a}d(a \sin (e+f x))}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {2 c^2 \cos (e+f x) \log (a \sin (e+f x)+a)}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\) |
Input:
Int[(c - c*Sin[e + f*x])^(3/2)/Sqrt[a + a*Sin[e + f*x]],x]
Output:
(2*c^2*Cos[e + f*x]*Log[a + a*Sin[e + f*x]])/(f*Sqrt[a + a*Sin[e + f*x]]*S qrt[c - c*Sin[e + f*x]]) + (c*Cos[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(f*Sq rt[a + a*Sin[e + f*x]])
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x )^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/ 2])
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[a*c*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x ]]*Sqrt[c + d*Sin[e + f*x]])) Int[Cos[e + f*x]/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0 ]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[e + f*x]*(a + b*Sin[e + f*x])^ (m - 1)*((c + d*Sin[e + f*x])^n/(f*(m + n))), x] + Simp[a*((2*m - 1)/(m + n )) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; Fre eQ[{a, b, c, d, e, f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I GtQ[m - 1/2, 0] && !LtQ[n, -1] && !(IGtQ[n - 1/2, 0] && LtQ[n, m]) && !( ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])
Time = 1.79 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.33
method | result | size |
default | \(-\frac {\tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right ) \left (\sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}-2 \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )\right )+2 \ln \left (\frac {2}{\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+1}\right )\right ) \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, c \sqrt {4}}{f \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}}\) | \(124\) |
Input:
int((c-c*sin(f*x+e))^(3/2)/(a+sin(f*x+e)*a)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/f*tan(1/4*Pi+1/2*f*x+1/2*e)*(sin(1/4*Pi+1/2*f*x+1/2*e)^2-2*ln(-cot(1/4* Pi+1/2*f*x+1/2*e)+csc(1/4*Pi+1/2*f*x+1/2*e))+2*ln(2/(cos(1/4*Pi+1/2*f*x+1/ 2*e)+1)))*(c*cos(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)*c/(a*sin(1/4*Pi+1/2*f*x+1/ 2*e)^2)^(1/2)*4^(1/2)
\[ \int \frac {(c-c \sin (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)}} \, dx=\int { \frac {{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}{\sqrt {a \sin \left (f x + e\right ) + a}} \,d x } \] Input:
integrate((c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="fric as")
Output:
integral((-c*sin(f*x + e) + c)^(3/2)/sqrt(a*sin(f*x + e) + a), x)
\[ \int \frac {(c-c \sin (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)}} \, dx=\int \frac {\left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{\frac {3}{2}}}{\sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )}}\, dx \] Input:
integrate((c-c*sin(f*x+e))**(3/2)/(a+a*sin(f*x+e))**(1/2),x)
Output:
Integral((-c*(sin(e + f*x) - 1))**(3/2)/sqrt(a*(sin(e + f*x) + 1)), x)
\[ \int \frac {(c-c \sin (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)}} \, dx=\int { \frac {{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}{\sqrt {a \sin \left (f x + e\right ) + a}} \,d x } \] Input:
integrate((c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="maxi ma")
Output:
integrate((-c*sin(f*x + e) + c)^(3/2)/sqrt(a*sin(f*x + e) + a), x)
Time = 0.15 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.03 \[ \int \frac {(c-c \sin (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)}} \, dx=-\frac {2 \, \sqrt {a} c^{\frac {3}{2}} {\left (\frac {\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {\log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}{a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{f} \] Input:
integrate((c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="giac ")
Output:
-2*sqrt(a)*c^(3/2)*(sin(-1/4*pi + 1/2*f*x + 1/2*e)^2/(a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) + log(-sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 1)/(a*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
Timed out. \[ \int \frac {(c-c \sin (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)}} \, dx=\int \frac {{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}}{\sqrt {a+a\,\sin \left (e+f\,x\right )}} \,d x \] Input:
int((c - c*sin(e + f*x))^(3/2)/(a + a*sin(e + f*x))^(1/2),x)
Output:
int((c - c*sin(e + f*x))^(3/2)/(a + a*sin(e + f*x))^(1/2), x)
\[ \int \frac {(c-c \sin (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)}} \, dx=\frac {\sqrt {c}\, \sqrt {a}\, c \left (-\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )+1}d x \right )+\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )+1}d x \right )}{a} \] Input:
int((c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(1/2),x)
Output:
(sqrt(c)*sqrt(a)*c*( - int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x) + 1),x) + int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1))/(sin(e + f*x) + 1),x)))/a