\(\int (a+b \cos (c+d x) \sin (c+d x))^m \, dx\) [566]

Optimal result

Integrand size = 18, antiderivative size = 131 \[ \int (a+b \cos (c+d x) \sin (c+d x))^m \, dx=-\frac {\operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (2 c+2 d x)),\frac {b (1-\sin (2 c+2 d x))}{2 a+b}\right ) \cos (2 c+2 d x) \left (a+\frac {1}{2} b \sin (2 c+2 d x)\right )^m \left (\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}\right )^{-m}}{\sqrt {2} d \sqrt {1+\sin (2 c+2 d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2745, 2744, 144, 143} \[ \int (a+b \cos (c+d x) \sin (c+d x))^m \, dx=-\frac {\cos (2 c+2 d x) \left (a+\frac {1}{2} b \sin (2 c+2 d x)\right )^m \left (\frac {2 a+b \sin (2 c+2 d x)}{2 a+b}\right )^{-m} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sin (2 c+2 d x)),\frac {b (1-\sin (2 c+2 d x))}{2 a+b}\right )}{\sqrt {2} d \sqrt {\sin (2 c+2 d x)+1}} \]

[In]

Int[(a + b*Cos[c + d*x]*Sin[c + d*x])^m,x]

[Out]

-((AppellF1[1/2, 1/2, -m, 3/2, (1 - Sin[2*c + 2*d*x])/2, (b*(1 - Sin[2*c + 2*d*x]))/(2*a + b)]*Cos[2*c + 2*d*x
]*(a + (b*Sin[2*c + 2*d*x])/2)^m)/(Sqrt[2]*d*Sqrt[1 + Sin[2*c + 2*d*x]]*((2*a + b*Sin[2*c + 2*d*x])/(2*a + b))
^m))

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 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]

Rule 2745

Int[((a_) + cos[(c_.) + (d_.)*(x_)]*(b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Int[(a + b*(Sin[2*c + 2*
d*x]/2))^n, x] /; FreeQ[{a, b, c, d, n}, x]

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

Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.11 \[ \int (a+b \cos (c+d x) \sin (c+d x))^m \, dx=\frac {\operatorname {AppellF1}\left (1+m,\frac {1}{2},\frac {1}{2},2+m,\frac {2 a+b \sin (2 (c+d x))}{2 a-b},\frac {2 a+b \sin (2 (c+d x))}{2 a+b}\right ) \sec (2 (c+d x)) \sqrt {-\frac {b (-1+\sin (2 (c+d x)))}{2 a+b}} \sqrt {\frac {b (1+\sin (2 (c+d x)))}{-2 a+b}} \left (a+\frac {1}{2} b \sin (2 (c+d x))\right )^{1+m}}{b d (1+m)} \]

[In]

Integrate[(a + b*Cos[c + d*x]*Sin[c + d*x])^m,x]

[Out]

(AppellF1[1 + m, 1/2, 1/2, 2 + m, (2*a + b*Sin[2*(c + d*x)])/(2*a - b), (2*a + b*Sin[2*(c + d*x)])/(2*a + b)]*
Sec[2*(c + d*x)]*Sqrt[-((b*(-1 + Sin[2*(c + d*x)]))/(2*a + b))]*Sqrt[(b*(1 + Sin[2*(c + d*x)]))/(-2*a + b)]*(a
+ (b*Sin[2*(c + d*x)])/2)^(1 + m))/(b*d*(1 + m))

Maple [F]

\[\int \left (a +\cos \left (d x +c \right ) \sin \left (d x +c \right ) b \right )^{m}d x\]

[In]

int((a+cos(d*x+c)*sin(d*x+c)*b)^m,x)

[Out]

int((a+cos(d*x+c)*sin(d*x+c)*b)^m,x)

Fricas [F]

\[ \int (a+b \cos (c+d x) \sin (c+d x))^m \, dx=\int { {\left (b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]

[In]

integrate((a+b*cos(d*x+c)*sin(d*x+c))^m,x, algorithm="fricas")

[Out]

integral((b*cos(d*x + c)*sin(d*x + c) + a)^m, x)

Sympy [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x) \sin (c+d x))^m \, dx=\text {Timed out} \]

[In]

integrate((a+b*cos(d*x+c)*sin(d*x+c))**m,x)

[Out]

Timed out

Maxima [F]

\[ \int (a+b \cos (c+d x) \sin (c+d x))^m \, dx=\int { {\left (b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]

[In]

integrate((a+b*cos(d*x+c)*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

integrate((b*cos(d*x + c)*sin(d*x + c) + a)^m, x)

Giac [F]

\[ \int (a+b \cos (c+d x) \sin (c+d x))^m \, dx=\int { {\left (b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]

[In]

integrate((a+b*cos(d*x+c)*sin(d*x+c))^m,x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c)*sin(d*x + c) + a)^m, x)

Mupad [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x) \sin (c+d x))^m \, dx=\int {\left (a+b\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\right )}^m \,d x \]

[In]

int((a + b*cos(c + d*x)*sin(c + d*x))^m,x)

[Out]

int((a + b*cos(c + d*x)*sin(c + d*x))^m, x)