\(\int (3+3 \sin (e+f x))^2 (c+d \sin (e+f x))^n \, dx\) [656]

Optimal result

Integrand size = 25, antiderivative size = 104 \[ \int (3+3 \sin (e+f x))^2 (c+d \sin (e+f x))^n \, dx=-\frac {36 \sqrt {2} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {3}{2},-n,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{f \sqrt {1+\sin (e+f x)}} \]

[Out]

-4*a^2*AppellF1(1/2,-n,-3/2,3/2,d*(1-sin(f*x+e))/(c+d),1/2-1/2*sin(f*x+e))*cos(f*x+e)*(c+d*sin(f*x+e))^n*2^(1/
2)/f/(((c+d*sin(f*x+e))/(c+d))^n)/(1+sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2863, 144, 143} \[ \int (3+3 \sin (e+f x))^2 (c+d \sin (e+f x))^n \, dx=-\frac {4 \sqrt {2} a^2 \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {3}{2},-n,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right )}{f \sqrt {\sin (e+f x)+1}} \]

[In]

Int[(a + a*Sin[e + f*x])^2*(c + d*Sin[e + f*x])^n,x]

[Out]

(-4*Sqrt[2]*a^2*AppellF1[1/2, -3/2, -n, 3/2, (1 - Sin[e + f*x])/2, (d*(1 - Sin[e + f*x]))/(c + d)]*Cos[e + f*x
]*(c + d*Sin[e + f*x])^n)/(f*Sqrt[1 + Sin[e + f*x]]*((c + d*Sin[e + f*x])/(c + d))^n)

Rule 143

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c
- a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !Inte
gerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] &&  !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) &&  !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])

Rule 144

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((b/(b*e - a*f))^IntPart[p]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p]), Int[(a + b*x)^m*(c + d*x)^n*
(b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !GtQ[b/(b*e - a*f), 0]

Rule 2863

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*(Cos[e + f*x]/(f*Sqrt[1 + Sin[e + f*x]]*Sqrt[1 - Sin[e + f*x]])), Subst[Int[(1 + (b/a)*x)^(m - 1/2)*((c
+ d*x)^n/Sqrt[1 - (b/a)*x]), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0]
&& EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (a^2 \cos (e+f x)\right ) \text {Subst}\left (\int \frac {(1+x)^{3/2} (c+d x)^n}{\sqrt {1-x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}} \\ & = \frac {\left (a^2 \cos (e+f x) (c+d \sin (e+f x))^n \left (-\frac {c+d \sin (e+f x)}{-c-d}\right )^{-n}\right ) \text {Subst}\left (\int \frac {(1+x)^{3/2} \left (-\frac {c}{-c-d}-\frac {d x}{-c-d}\right )^n}{\sqrt {1-x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}} \\ & = -\frac {4 \sqrt {2} a^2 \operatorname {AppellF1}\left (\frac {1}{2},-\frac {3}{2},-n,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{f \sqrt {1+\sin (e+f x)}} \\ \end{align*}

Mathematica [F]

\[ \int (3+3 \sin (e+f x))^2 (c+d \sin (e+f x))^n \, dx=\int (3+3 \sin (e+f x))^2 (c+d \sin (e+f x))^n \, dx \]

[In]

Integrate[(3 + 3*Sin[e + f*x])^2*(c + d*Sin[e + f*x])^n,x]

[Out]

Integrate[(3 + 3*Sin[e + f*x])^2*(c + d*Sin[e + f*x])^n, x]

Maple [F]

\[\int \left (a +a \sin \left (f x +e \right )\right )^{2} \left (c +d \sin \left (f x +e \right )\right )^{n}d x\]

[In]

int((a+a*sin(f*x+e))^2*(c+d*sin(f*x+e))^n,x)

[Out]

int((a+a*sin(f*x+e))^2*(c+d*sin(f*x+e))^n,x)

Fricas [F]

\[ \int (3+3 \sin (e+f x))^2 (c+d \sin (e+f x))^n \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{2} {\left (d \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^2*(c+d*sin(f*x+e))^n,x, algorithm="fricas")

[Out]

integral(-(a^2*cos(f*x + e)^2 - 2*a^2*sin(f*x + e) - 2*a^2)*(d*sin(f*x + e) + c)^n, x)

Sympy [F(-1)]

Timed out. \[ \int (3+3 \sin (e+f x))^2 (c+d \sin (e+f x))^n \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(f*x+e))**2*(c+d*sin(f*x+e))**n,x)

[Out]

Timed out

Maxima [F]

\[ \int (3+3 \sin (e+f x))^2 (c+d \sin (e+f x))^n \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{2} {\left (d \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^2*(c+d*sin(f*x+e))^n,x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^2*(d*sin(f*x + e) + c)^n, x)

Giac [F]

\[ \int (3+3 \sin (e+f x))^2 (c+d \sin (e+f x))^n \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{2} {\left (d \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^2*(c+d*sin(f*x+e))^n,x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^2*(d*sin(f*x + e) + c)^n, x)

Mupad [F(-1)]

Timed out. \[ \int (3+3 \sin (e+f x))^2 (c+d \sin (e+f x))^n \, dx=\int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^2\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^n \,d x \]

[In]

int((a + a*sin(e + f*x))^2*(c + d*sin(e + f*x))^n,x)

[Out]

int((a + a*sin(e + f*x))^2*(c + d*sin(e + f*x))^n, x)