Integrand size = 28, antiderivative size = 68 \[ \int \frac {(a+a \sin (e+f x))^m}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {\cos (e+f x) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^m}{f (1+2 m) \sqrt {c-c \sin (e+f x)}} \] Output:
cos(f*x+e)*hypergeom([1, 1/2+m],[3/2+m],1/2+1/2*sin(f*x+e))*(a+a*sin(f*x+e ))^m/f/(1+2*m)/(c-c*sin(f*x+e))^(1/2)
Time = 0.16 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.97 \[ \int \frac {(a+a \sin (e+f x))^m}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {\cos (e+f x) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right ) (a (1+\sin (e+f x)))^m}{(f+2 f m) \sqrt {c-c \sin (e+f x)}} \] Input:
Integrate[(a + a*Sin[e + f*x])^m/Sqrt[c - c*Sin[e + f*x]],x]
Output:
(Cos[e + f*x]*Hypergeometric2F1[1, 1/2 + m, 3/2 + m, (1 + Sin[e + f*x])/2] *(a*(1 + Sin[e + f*x]))^m)/((f + 2*f*m)*Sqrt[c - c*Sin[e + f*x]])
Time = 0.38 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3042, 3224, 3042, 3146, 78}
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.
\(\displaystyle \int \frac {(a \sin (e+f x)+a)^m}{\sqrt {c-c \sin (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (e+f x)+a)^m}{\sqrt {c-c \sin (e+f x)}}dx\) |
\(\Big \downarrow \) 3224 |
\(\displaystyle \frac {\cos (e+f x) \int \sec (e+f x) (\sin (e+f x) a+a)^{m+\frac {1}{2}}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\cos (e+f x) \int \frac {(\sin (e+f x) a+a)^{m+\frac {1}{2}}}{\cos (e+f x)}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\) |
\(\Big \downarrow \) 3146 |
\(\displaystyle \frac {a \cos (e+f x) \int \frac {(\sin (e+f x) a+a)^{m-\frac {1}{2}}}{a-a \sin (e+f x)}d(a \sin (e+f x))}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\) |
\(\Big \downarrow \) 78 |
\(\displaystyle \frac {\cos (e+f x) (a \sin (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (1,m+\frac {1}{2},m+\frac {3}{2},\frac {\sin (e+f x) a+a}{2 a}\right )}{f (2 m+1) \sqrt {c-c \sin (e+f x)}}\) |
Input:
Int[(a + a*Sin[e + f*x])^m/Sqrt[c - c*Sin[e + f*x]],x]
Output:
(Cos[e + f*x]*Hypergeometric2F1[1, 1/2 + m, 3/2 + m, (a + a*Sin[e + f*x])/ (2*a)]*(a + a*Sin[e + f*x])^m)/(f*(1 + 2*m)*Sqrt[c - c*Sin[e + f*x]])
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b *c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] && !IntegerQ[m] && IntegerQ[n]
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[((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])^FracPart[m]/Cos[e + f*x]^(2*FracP art[m])) Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; F reeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && (FractionQ[m] || !FractionQ[n])
\[\int \frac {\left (a +\sin \left (f x +e \right ) a \right )^{m}}{\sqrt {c -c \sin \left (f x +e \right )}}d x\]
Input:
int((a+sin(f*x+e)*a)^m/(c-c*sin(f*x+e))^(1/2),x)
Output:
int((a+sin(f*x+e)*a)^m/(c-c*sin(f*x+e))^(1/2),x)
\[ \int \frac {(a+a \sin (e+f x))^m}{\sqrt {c-c \sin (e+f x)}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{\sqrt {-c \sin \left (f x + e\right ) + c}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^m/(c-c*sin(f*x+e))^(1/2),x, algorithm="fricas")
Output:
integral(-sqrt(-c*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m/(c*sin(f*x + e) - c), x)
\[ \int \frac {(a+a \sin (e+f x))^m}{\sqrt {c-c \sin (e+f x)}} \, dx=\int \frac {\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m}}{\sqrt {- c \left (\sin {\left (e + f x \right )} - 1\right )}}\, dx \] Input:
integrate((a+a*sin(f*x+e))**m/(c-c*sin(f*x+e))**(1/2),x)
Output:
Integral((a*(sin(e + f*x) + 1))**m/sqrt(-c*(sin(e + f*x) - 1)), x)
\[ \int \frac {(a+a \sin (e+f x))^m}{\sqrt {c-c \sin (e+f x)}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{\sqrt {-c \sin \left (f x + e\right ) + c}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^m/(c-c*sin(f*x+e))^(1/2),x, algorithm="maxima")
Output:
integrate((a*sin(f*x + e) + a)^m/sqrt(-c*sin(f*x + e) + c), x)
\[ \int \frac {(a+a \sin (e+f x))^m}{\sqrt {c-c \sin (e+f x)}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{\sqrt {-c \sin \left (f x + e\right ) + c}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^m/(c-c*sin(f*x+e))^(1/2),x, algorithm="giac")
Output:
integrate((a*sin(f*x + e) + a)^m/sqrt(-c*sin(f*x + e) + c), x)
Timed out. \[ \int \frac {(a+a \sin (e+f x))^m}{\sqrt {c-c \sin (e+f x)}} \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{\sqrt {c-c\,\sin \left (e+f\,x\right )}} \,d x \] Input:
int((a + a*sin(e + f*x))^m/(c - c*sin(e + f*x))^(1/2),x)
Output:
int((a + a*sin(e + f*x))^m/(c - c*sin(e + f*x))^(1/2), x)
\[ \int \frac {(a+a \sin (e+f x))^m}{\sqrt {c-c \sin (e+f x)}} \, dx=-\frac {\sqrt {c}\, \left (\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )-1}d x \right )}{c} \] Input:
int((a+a*sin(f*x+e))^m/(c-c*sin(f*x+e))^(1/2),x)
Output:
( - sqrt(c)*int(((sin(e + f*x)*a + a)**m*sqrt( - sin(e + f*x) + 1))/(sin(e + f*x) - 1),x))/c