\(\int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx\) [404]

Optimal result

Integrand size = 26, antiderivative size = 110 \[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\frac {2^{\frac {1}{2}+n} c \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (1+2 m),\frac {1}{2} (1-2 n),\frac {1}{2} (3+2 m),\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1}{2}-n} (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^{-1+n}}{f (1+2 m)} \]

[Out]

2^(1/2+n)*c*cos(f*x+e)*hypergeom([1/2-n, 1/2+m],[3/2+m],1/2+1/2*sin(f*x+e))*(1-sin(f*x+e))^(1/2-n)*(a+a*sin(f*
x+e))^m*(c-c*sin(f*x+e))^(-1+n)/f/(1+2*m)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2824, 2768, 72, 71} \[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=\frac {c 2^{n+\frac {1}{2}} \cos (e+f x) (1-\sin (e+f x))^{\frac {1}{2}-n} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n-1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (2 m+1),\frac {1}{2} (1-2 n),\frac {1}{2} (2 m+3),\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1)} \]

[In]

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

[Out]

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

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 72

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n]), Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c -
a*d)), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 2768

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[a^2*(
(g*Cos[e + f*x])^(p + 1)/(f*g*(a + b*Sin[e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2))), Subst[Int[(
a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&
EqQ[a^2 - b^2, 0] &&  !IntegerQ[m]

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}& = \left (\cos ^{-2 m}(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m\right ) \int \cos ^{2 m}(e+f x) (c-c \sin (e+f x))^{-m+n} \, dx \\ & = \frac {\left (c^2 \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{\frac {1}{2} (-1-2 m)+m} (c+c \sin (e+f x))^{\frac {1}{2} (-1-2 m)}\right ) \text {Subst}\left (\int (c-c x)^{-m+\frac {1}{2} (-1+2 m)+n} (c+c x)^{\frac {1}{2} (-1+2 m)} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\left (2^{-\frac {1}{2}+n} c^2 \cos (e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-\frac {1}{2}+\frac {1}{2} (-1-2 m)+m+n} \left (\frac {c-c \sin (e+f x)}{c}\right )^{\frac {1}{2}-n} (c+c \sin (e+f x))^{\frac {1}{2} (-1-2 m)}\right ) \text {Subst}\left (\int \left (\frac {1}{2}-\frac {x}{2}\right )^{-m+\frac {1}{2} (-1+2 m)+n} (c+c x)^{\frac {1}{2} (-1+2 m)} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {2^{\frac {1}{2}+n} c \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (1+2 m),\frac {1}{2} (1-2 n),\frac {1}{2} (3+2 m),\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1}{2}-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+n}}{f (1+2 m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.38 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.98 \[ \int (3+3 \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx=-\frac {2\ 3^m \cot \left (\frac {1}{4} (2 e+\pi +2 f x)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2}-m,\frac {1}{2}+n,\frac {3}{2}+n,\cos ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right )\right ) (1+\sin (e+f x))^m (c-c \sin (e+f x))^n \sin ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right )^{\frac {1}{2}-m}}{f+2 f n} \]

[In]

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

[Out]

(-2*3^m*Cot[(2*e + Pi + 2*f*x)/4]*Hypergeometric2F1[1/2 - m, 1/2 + n, 3/2 + n, Cos[(2*e + Pi + 2*f*x)/4]^2]*(1
+ Sin[e + f*x])^m*(c - c*Sin[e + f*x])^n*(Sin[(2*e + Pi + 2*f*x)/4]^2)^(1/2 - m))/(f + 2*f*n)

Maple [F]

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

[In]

int((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x)

[Out]

int((a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x)

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

integrate((a+a*sin(f*x+e))**m*(c-c*sin(f*x+e))**n,x)

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((a + a*sin(e + f*x))^m*(c - c*sin(e + f*x))^n,x)

[Out]

int((a + a*sin(e + f*x))^m*(c - c*sin(e + f*x))^n, x)