Integrand size = 40, antiderivative size = 123 \[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {(A+C) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^m}{f (1+2 m) \sqrt {c-c \sin (e+f x)}}-\frac {2 C \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (3+2 m) \sqrt {c-c \sin (e+f x)}} \] Output:
(A+C)*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)-2*C*cos(f*x+e)*(a+a*sin(f*x+e) )^(1+m)/a/f/(3+2*m)/(c-c*sin(f*x+e))^(1/2)
Time = 36.80 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.84 \[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{\sqrt {c-c \sin (e+f x)}} \, dx=-\frac {\cos (e+f x) (a (1+\sin (e+f x)))^m \left (-\left ((A+C) (3+2 m) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right )\right )+2 C (1+2 m) (1+\sin (e+f x))\right )}{f (1+2 m) (3+2 m) \sqrt {c-c \sin (e+f x)}} \] Input:
Integrate[((a + a*Sin[e + f*x])^m*(A + C*Sin[e + f*x]^2))/Sqrt[c - c*Sin[e + f*x]],x]
Output:
-((Cos[e + f*x]*(a*(1 + Sin[e + f*x]))^m*(-((A + C)*(3 + 2*m)*Hypergeometr ic2F1[1, 1/2 + m, 3/2 + m, (1 + Sin[e + f*x])/2]) + 2*C*(1 + 2*m)*(1 + Sin [e + f*x])))/(f*(1 + 2*m)*(3 + 2*m)*Sqrt[c - c*Sin[e + f*x]]))
Time = 0.65 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {3042, 3517, 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.
\(\displaystyle \int \frac {(a \sin (e+f x)+a)^m \left (A+C \sin ^2(e+f x)\right )}{\sqrt {c-c \sin (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (e+f x)+a)^m \left (A+C \sin (e+f x)^2\right )}{\sqrt {c-c \sin (e+f x)}}dx\) |
\(\Big \downarrow \) 3517 |
\(\displaystyle (A+C) \int \frac {(\sin (e+f x) a+a)^m}{\sqrt {c-c \sin (e+f x)}}dx-\frac {2 C \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (2 m+3) \sqrt {c-c \sin (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (A+C) \int \frac {(\sin (e+f x) a+a)^m}{\sqrt {c-c \sin (e+f x)}}dx-\frac {2 C \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (2 m+3) \sqrt {c-c \sin (e+f x)}}\) |
\(\Big \downarrow \) 3224 |
\(\displaystyle \frac {(A+C) \cos (e+f x) \int \sec (e+f x) (\sin (e+f x) a+a)^{m+\frac {1}{2}}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {2 C \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (2 m+3) \sqrt {c-c \sin (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(A+C) \cos (e+f x) \int \frac {(\sin (e+f x) a+a)^{m+\frac {1}{2}}}{\cos (e+f x)}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {2 C \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (2 m+3) \sqrt {c-c \sin (e+f x)}}\) |
\(\Big \downarrow \) 3146 |
\(\displaystyle \frac {a (A+C) \cos (e+f x) \int \frac {(\sin (e+f x) a+a)^{m-\frac {1}{2}}}{a-a \sin (e+f x)}d(a \sin (e+f x))}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {2 C \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (2 m+3) \sqrt {c-c \sin (e+f x)}}\) |
\(\Big \downarrow \) 78 |
\(\displaystyle \frac {(A+C) \cos (e+f x) (a \sin (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (1,m+\frac {1}{2},m+\frac {3}{2},\frac {\sin (e+f x) a+a}{2 a}\right )}{f (2 m+1) \sqrt {c-c \sin (e+f x)}}-\frac {2 C \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (2 m+3) \sqrt {c-c \sin (e+f x)}}\) |
Input:
Int[((a + a*Sin[e + f*x])^m*(A + C*Sin[e + f*x]^2))/Sqrt[c - c*Sin[e + f*x ]],x]
Output:
((A + C)*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 ]]) - (2*C*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m))/(a*f*(3 + 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[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2))/Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] : > Simp[-2*C*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(2*m + 3)*Sqrt[ c + d*Sin[e + f*x]])), x] + Simp[(A + C) Int[(a + b*Sin[e + f*x])^m/Sqrt[ c + d*Sin[e + f*x]], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && EqQ [b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
\[\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +C \sin \left (f x +e \right )^{2}\right )}{\sqrt {c -c \sin \left (f x +e \right )}}d x\]
Input:
int((a+a*sin(f*x+e))^m*(A+C*sin(f*x+e)^2)/(c-c*sin(f*x+e))^(1/2),x)
Output:
int((a+a*sin(f*x+e))^m*(A+C*sin(f*x+e)^2)/(c-c*sin(f*x+e))^(1/2),x)
\[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{\sqrt {c-c \sin (e+f x)}} \, dx=\int { \frac {{\left (C \sin \left (f x + e\right )^{2} + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{\sqrt {-c \sin \left (f x + e\right ) + c}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^m*(A+C*sin(f*x+e)^2)/(c-c*sin(f*x+e))^(1/2),x, algorithm="fricas")
Output:
integral((C*cos(f*x + e)^2 - A - C)*sqrt(-c*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m/(c*sin(f*x + e) - c), x)
\[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{\sqrt {c-c \sin (e+f x)}} \, dx=\int \frac {\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (A + C \sin ^{2}{\left (e + f x \right )}\right )}{\sqrt {- c \left (\sin {\left (e + f x \right )} - 1\right )}}\, dx \] Input:
integrate((a+a*sin(f*x+e))**m*(A+C*sin(f*x+e)**2)/(c-c*sin(f*x+e))**(1/2), x)
Output:
Integral((a*(sin(e + f*x) + 1))**m*(A + C*sin(e + f*x)**2)/sqrt(-c*(sin(e + f*x) - 1)), x)
\[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{\sqrt {c-c \sin (e+f x)}} \, dx=\int { \frac {{\left (C \sin \left (f x + e\right )^{2} + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{\sqrt {-c \sin \left (f x + e\right ) + c}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^m*(A+C*sin(f*x+e)^2)/(c-c*sin(f*x+e))^(1/2),x, algorithm="maxima")
Output:
integrate((C*sin(f*x + e)^2 + A)*(a*sin(f*x + e) + a)^m/sqrt(-c*sin(f*x + e) + c), x)
Timed out. \[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{\sqrt {c-c \sin (e+f x)}} \, dx=\text {Timed out} \] Input:
integrate((a+a*sin(f*x+e))^m*(A+C*sin(f*x+e)^2)/(c-c*sin(f*x+e))^(1/2),x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{\sqrt {c-c \sin (e+f x)}} \, dx=\int \frac {\left (C\,{\sin \left (e+f\,x\right )}^2+A\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{\sqrt {c-c\,\sin \left (e+f\,x\right )}} \,d x \] Input:
int(((A + C*sin(e + f*x)^2)*(a + a*sin(e + f*x))^m)/(c - c*sin(e + f*x))^( 1/2),x)
Output:
int(((A + C*sin(e + f*x)^2)*(a + a*sin(e + f*x))^m)/(c - c*sin(e + f*x))^( 1/2), x)
\[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{\sqrt {c-c \sin (e+f x)}} \, dx=-\frac {\sqrt {c}\, \left (\left (\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )-1}d x \right ) c +\left (\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )-1}d x \right ) a \right )}{c} \] Input:
int((a+a*sin(f*x+e))^m*(A+C*sin(f*x+e)^2)/(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)**2)/(sin(e + f*x) - 1),x)*c + int(((sin(e + f*x)*a + a)**m*sqrt( - sin(e + f*x) + 1))/(sin(e + f*x) - 1),x)*a))/c