\(\int (4+3 \sin (c+d x))^n \, dx\) [67]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 12, antiderivative size = 64 \[ \int (4+3 \sin (c+d x))^n \, dx=\frac {\sqrt {2} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-n,\frac {3}{2},\frac {1}{2} (1+\sin (c+d x)),-3 (1+\sin (c+d x))\right ) \cos (c+d x)}{d \sqrt {1-\sin (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2744, 143} \[ \int (4+3 \sin (c+d x))^n \, dx=\frac {\sqrt {2} \cos (c+d x) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-n,\frac {3}{2},\frac {1}{2} (\sin (c+d x)+1),-3 (\sin (c+d x)+1)\right )}{d \sqrt {1-\sin (c+d x)}} \]

[In]

Int[(4 + 3*Sin[c + d*x])^n,x]

[Out]

(Sqrt[2]*AppellF1[1/2, 1/2, -n, 3/2, (1 + Sin[c + d*x])/2, -3*(1 + Sin[c + d*x])]*Cos[c + d*x])/(d*Sqrt[1 - Si
n[c + d*x]])

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 2744

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[Cos[c + d*x]/(d*Sqrt[1 + Sin[c + d*x]]*Sqrt
[1 - Sin[c + d*x]]), Subst[Int[(a + b*x)^n/(Sqrt[1 + x]*Sqrt[1 - x]), x], x, Sin[c + d*x]], x] /; FreeQ[{a, b,
 c, d, n}, x] && NeQ[a^2 - b^2, 0] &&  !IntegerQ[2*n]

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.55 \[ \int (4+3 \sin (c+d x))^n \, dx=\frac {\operatorname {AppellF1}\left (1+n,\frac {1}{2},\frac {1}{2},2+n,4+3 \sin (c+d x),\frac {1}{7} (4+3 \sin (c+d x))\right ) \sec (c+d x) \sqrt {-1-\sin (c+d x)} \sqrt {1-\sin (c+d x)} (4+3 \sin (c+d x))^{1+n}}{\sqrt {7} d (1+n)} \]

[In]

Integrate[(4 + 3*Sin[c + d*x])^n,x]

[Out]

(AppellF1[1 + n, 1/2, 1/2, 2 + n, 4 + 3*Sin[c + d*x], (4 + 3*Sin[c + d*x])/7]*Sec[c + d*x]*Sqrt[-1 - Sin[c + d
*x]]*Sqrt[1 - Sin[c + d*x]]*(4 + 3*Sin[c + d*x])^(1 + n))/(Sqrt[7]*d*(1 + n))

Maple [F]

\[\int \left (4+3 \sin \left (d x +c \right )\right )^{n}d x\]

[In]

int((4+3*sin(d*x+c))^n,x)

[Out]

int((4+3*sin(d*x+c))^n,x)

Fricas [F]

\[ \int (4+3 \sin (c+d x))^n \, dx=\int { {\left (3 \, \sin \left (d x + c\right ) + 4\right )}^{n} \,d x } \]

[In]

integrate((4+3*sin(d*x+c))^n,x, algorithm="fricas")

[Out]

integral((3*sin(d*x + c) + 4)^n, x)

Sympy [F]

\[ \int (4+3 \sin (c+d x))^n \, dx=\int \left (3 \sin {\left (c + d x \right )} + 4\right )^{n}\, dx \]

[In]

integrate((4+3*sin(d*x+c))**n,x)

[Out]

Integral((3*sin(c + d*x) + 4)**n, x)

Maxima [F]

\[ \int (4+3 \sin (c+d x))^n \, dx=\int { {\left (3 \, \sin \left (d x + c\right ) + 4\right )}^{n} \,d x } \]

[In]

integrate((4+3*sin(d*x+c))^n,x, algorithm="maxima")

[Out]

integrate((3*sin(d*x + c) + 4)^n, x)

Giac [F]

\[ \int (4+3 \sin (c+d x))^n \, dx=\int { {\left (3 \, \sin \left (d x + c\right ) + 4\right )}^{n} \,d x } \]

[In]

integrate((4+3*sin(d*x+c))^n,x, algorithm="giac")

[Out]

integrate((3*sin(d*x + c) + 4)^n, x)

Mupad [F(-1)]

Timed out. \[ \int (4+3 \sin (c+d x))^n \, dx=\int {\left (3\,\sin \left (c+d\,x\right )+4\right )}^n \,d x \]

[In]

int((3*sin(c + d*x) + 4)^n,x)

[Out]

int((3*sin(c + d*x) + 4)^n, x)