3.9.4 \(\int (a+b \sin (e+f x))^m \, dx\) [804]

Optimal. Leaf size=104 \[ -\frac {\sqrt {2} F_1\left (\frac {1}{2};\frac {1}{2},-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right ) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m}}{f \sqrt {1+\sin (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2744, 144, 143} \begin {gather*} -\frac {\sqrt {2} \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac {1}{2};\frac {1}{2},-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{f \sqrt {\sin (e+f x)+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[e + f*x])^m,x]

[Out]

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

Rubi steps

\begin {align*} \int (a+b \sin (e+f x))^m \, dx &=\frac {\cos (e+f x) \text {Subst}\left (\int \frac {(a+b x)^m}{\sqrt {1-x} \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 (\cos (e+f x) (a+b \sin (e+f x))^m \left (-\frac {a+b \sin (e+f x)}{-a-b}\right )^{-m}\right ) \text {Subst}\left (\int \frac {\left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^m}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}}\\ &=-\frac {\sqrt {2} F_1\left (\frac {1}{2};\frac {1}{2},-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right ) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m}}{f \sqrt {1+\sin (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 0.16, size = 120, normalized size = 1.15 \begin {gather*} \frac {F_1\left (1+m;\frac {1}{2},\frac {1}{2};2+m;\frac {a+b \sin (e+f x)}{a-b},\frac {a+b \sin (e+f x)}{a+b}\right ) \sec (e+f x) \sqrt {-\frac {b (-1+\sin (e+f x))}{a+b}} \sqrt {\frac {b (1+\sin (e+f x))}{-a+b}} (a+b \sin (e+f x))^{1+m}}{b f (1+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[e + f*x])^m,x]

[Out]

(AppellF1[1 + m, 1/2, 1/2, 2 + m, (a + b*Sin[e + f*x])/(a - b), (a + b*Sin[e + f*x])/(a + b)]*Sec[e + f*x]*Sqr
t[-((b*(-1 + Sin[e + f*x]))/(a + b))]*Sqrt[(b*(1 + Sin[e + f*x]))/(-a + b)]*(a + b*Sin[e + f*x])^(1 + m))/(b*f
*(1 + m))

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \left (a +b \sin \left (f x +e \right )\right )^{m}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(f*x+e))^m,x)

[Out]

int((a+b*sin(f*x+e))^m,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sin(f*x + e) + a)^m, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*sin(f*x + e) + a)^m, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sin {\left (e + f x \right )}\right )^{m}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))**m,x)

[Out]

Integral((a + b*sin(e + f*x))**m, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sin(f*x + e) + a)^m, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (a+b\,\sin \left (e+f\,x\right )\right )}^m \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(e + f*x))^m,x)

[Out]

int((a + b*sin(e + f*x))^m, x)

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