\(\int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx\) [208]

Optimal result

Integrand size = 38, antiderivative size = 118 \[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=-\frac {2 B \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+2 m) \sqrt {c-c \sin (e+f x)}}+\frac {(A+B) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^m}{f (1+2 m) \sqrt {c-c \sin (e+f x)}} \]

[Out]

-2*B*cos(f*x+e)*(a+a*sin(f*x+e))^m/f/(1+2*m)/(c-c*sin(f*x+e))^(1/2)+(A+B)*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)

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3052, 2824, 2746, 70} \[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (1,m+\frac {1}{2},m+\frac {3}{2},\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1) \sqrt {c-c \sin (e+f x)}}-\frac {2 B \cos (e+f x) (a \sin (e+f x)+a)^m}{f (2 m+1) \sqrt {c-c \sin (e+f x)}} \]

[In]

Int[((a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]))/Sqrt[c - c*Sin[e + f*x]],x]

[Out]

(-2*B*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(f*(1 + 2*m)*Sqrt[c - c*Sin[e + f*x]]) + ((A + B)*Cos[e + f*x]*Hype
rgeometric2F1[1, 1/2 + m, 3/2 + m, (1 + Sin[e + f*x])/2]*(a + a*Sin[e + f*x])^m)/(f*(1 + 2*m)*Sqrt[c - c*Sin[e
+ f*x]])

Rule 70

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] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 2746

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[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]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 2824

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist
[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*
FracPart[m])), Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{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])

Rule 3052

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*
(m + n + 1))), x] - Dist[(B*c*(m - n) - A*d*(m + n + 1))/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Si
n[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] &&
!LtQ[m, -2^(-1)] && NeQ[m + n + 1, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 B \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+2 m) \sqrt {c-c \sin (e+f x)}}+(A+B) \int \frac {(a+a \sin (e+f x))^m}{\sqrt {c-c \sin (e+f x)}} \, dx \\ & = -\frac {2 B \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+2 m) \sqrt {c-c \sin (e+f x)}}+\frac {((A+B) \cos (e+f x)) \int \sec (e+f x) (a+a \sin (e+f x))^{\frac {1}{2}+m} \, dx}{\sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \\ & = -\frac {2 B \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+2 m) \sqrt {c-c \sin (e+f x)}}+\frac {(a (A+B) \cos (e+f x)) \text {Subst}\left (\int \frac {(a+x)^{-\frac {1}{2}+m}}{a-x} \, dx,x,a \sin (e+f x)\right )}{f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \\ & = -\frac {2 B \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+2 m) \sqrt {c-c \sin (e+f x)}}+\frac {(A+B) \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)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.12 \[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {\cos (e+f x) (a (1+\sin (e+f x)))^m \left (2 A (3+2 m) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right )+B \left (-6-4 m+(1+2 m) \operatorname {Hypergeometric2F1}\left (1,\frac {3}{2}+m,\frac {5}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right ) (1+\sin (e+f x))\right )\right )}{2 f (1+2 m) (3+2 m) \sqrt {c-c \sin (e+f x)}} \]

[In]

Integrate[((a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]))/Sqrt[c - c*Sin[e + f*x]],x]

[Out]

(Cos[e + f*x]*(a*(1 + Sin[e + f*x]))^m*(2*A*(3 + 2*m)*Hypergeometric2F1[1, 1/2 + m, 3/2 + m, (1 + Sin[e + f*x]
)/2] + B*(-6 - 4*m + (1 + 2*m)*Hypergeometric2F1[1, 3/2 + m, 5/2 + m, (1 + Sin[e + f*x])/2]*(1 + Sin[e + f*x])
)))/(2*f*(1 + 2*m)*(3 + 2*m)*Sqrt[c - c*Sin[e + f*x]])

Maple [F]

\[\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right )}{\sqrt {c -c \sin \left (f x +e \right )}}d x\]

[In]

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x)

[Out]

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x)

Fricas [F]

\[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{\sqrt {-c \sin \left (f x + e\right ) + c}} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral(-(B*sin(f*x + e) + A)*sqrt(-c*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m/(c*sin(f*x + e) - c), x)

Sympy [F]

\[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\int \frac {\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (A + B \sin {\left (e + f x \right )}\right )}{\sqrt {- c \left (\sin {\left (e + f x \right )} - 1\right )}}\, dx \]

[In]

integrate((a+a*sin(f*x+e))**m*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(1/2),x)

[Out]

Integral((a*(sin(e + f*x) + 1))**m*(A + B*sin(e + f*x))/sqrt(-c*(sin(e + f*x) - 1)), x)

Maxima [F]

\[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{\sqrt {-c \sin \left (f x + e\right ) + c}} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m/sqrt(-c*sin(f*x + e) + c), x)

Giac [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{\sqrt {c-c\,\sin \left (e+f\,x\right )}} \,d x \]

[In]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m)/(c - c*sin(e + f*x))^(1/2),x)

[Out]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m)/(c - c*sin(e + f*x))^(1/2), x)