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

Optimal. Leaf size=131 \[ -\frac{\cos (2 c+2 d x) \left (a+\frac{1}{2} b \sin (2 c+2 d x)\right )^m \left (\frac{2 a+b \sin (2 c+2 d x)}{2 a+b}\right )^{-m} F_1\left (\frac{1}{2};\frac{1}{2},-m;\frac{3}{2};\frac{1}{2} (1-\sin (2 c+2 d x)),\frac{b (1-\sin (2 c+2 d x))}{2 a+b}\right )}{\sqrt{2} d \sqrt{\sin (2 c+2 d x)+1}} \]

[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))

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Rubi [A]  time = 0.112209, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {2666, 2665, 139, 138} \[ -\frac{\cos (2 c+2 d x) \left (a+\frac{1}{2} b \sin (2 c+2 d x)\right )^m \left (\frac{2 a+b \sin (2 c+2 d x)}{2 a+b}\right )^{-m} F_1\left (\frac{1}{2};\frac{1}{2},-m;\frac{3}{2};\frac{1}{2} (1-\sin (2 c+2 d x)),\frac{b (1-\sin (2 c+2 d x))}{2 a+b}\right )}{\sqrt{2} d \sqrt{\sin (2 c+2 d x)+1}} \]

Antiderivative was successfully verified.

[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 2666

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]

Rule 2665

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 139

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 138

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*(m + 1
)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p), 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])

Rubi steps

\begin{align*} \int (a+b \cos (c+d x) \sin (c+d x))^m \, dx &=\int \left (a+\frac{1}{2} b \sin (2 c+2 d x)\right )^m \, dx\\ &=\frac{\cos (2 c+2 d x) \operatorname{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 ) \operatorname{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{F_1\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]  time = 0.610215, size = 145, normalized size = 1.11 \[ \frac{\sec (2 (c+d x)) \sqrt{-\frac{b (\sin (2 (c+d x))-1)}{2 a+b}} \sqrt{\frac{b (\sin (2 (c+d x))+1)}{b-2 a}} \left (a+\frac{1}{2} b \sin (2 (c+d x))\right )^{m+1} F_1\left (m+1;\frac{1}{2},\frac{1}{2};m+2;\frac{2 a+b \sin (2 (c+d x))}{2 a-b},\frac{2 a+b \sin (2 (c+d x))}{2 a+b}\right )}{b d (m+1)} \]

Warning: Unable to verify antiderivative.

[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))

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Maple [F]  time = 0.521, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a\right )}^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a\right )}^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a\right )}^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)