\(\int \frac {(3+3 \sin (e+f x))^m}{\sqrt {c-c \sin (e+f x)}} \, dx\) [417]

Optimal result

Integrand size = 28, antiderivative size = 68 \[ \int \frac {(3+3 \sin (e+f x))^m}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {\cos (e+f x) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right ) (3+3 \sin (e+f x))^m}{f (1+2 m) \sqrt {c-c \sin (e+f x)}} \]

[Out]

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 = 0.10 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {2824, 2746, 70} \[ \int \frac {(3+3 \sin (e+f x))^m}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {\cos (e+f x) (a \sin (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (1,m+\frac {1}{2},m+\frac {3}{2},\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1) \sqrt {c-c \sin (e+f x)}} \]

[In]

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

[Out]

(Cos[e + f*x]*Hypergeometric2F1[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])

Rubi steps \begin{align*} \text {integral}& = \frac {\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 {(a \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 {\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 = 0.10 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.99 \[ \int \frac {(3+3 \sin (e+f x))^m}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {3^m \cos (e+f x) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right ) (1+\sin (e+f x))^m}{(f+2 f m) \sqrt {c-c \sin (e+f x)}} \]

[In]

Integrate[(3 + 3*Sin[e + f*x])^m/Sqrt[c - c*Sin[e + f*x]],x]

[Out]

(3^m*Cos[e + f*x]*Hypergeometric2F1[1, 1/2 + m, 3/2 + m, (1 + Sin[e + f*x])/2]*(1 + Sin[e + f*x])^m)/((f + 2*f
*m)*Sqrt[c - c*Sin[e + f*x]])

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {(3+3 \sin (e+f x))^m}{\sqrt {c-c \sin (e+f x)}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{\sqrt {-c \sin \left (f x + e\right ) + c}} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {(3+3 \sin (e+f x))^m}{\sqrt {c-c \sin (e+f x)}} \, dx=\int \frac {\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m}}{\sqrt {- c \left (\sin {\left (e + f x \right )} - 1\right )}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {(3+3 \sin (e+f x))^m}{\sqrt {c-c \sin (e+f x)}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{\sqrt {-c \sin \left (f x + e\right ) + c}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {(3+3 \sin (e+f x))^m}{\sqrt {c-c \sin (e+f x)}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{\sqrt {-c \sin \left (f x + e\right ) + c}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(3+3 \sin (e+f x))^m}{\sqrt {c-c \sin (e+f x)}} \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{\sqrt {c-c\,\sin \left (e+f\,x\right )}} \,d x \]

[In]

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

[Out]

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