3.14 \(\int (a+b \sin (e+f x))^m (A-A \sin ^2(e+f x)) \, dx\)

Optimal. Leaf size=211 \[ \frac{4 \sqrt{2} A \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac{a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac{1}{2};-\frac{3}{2},-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{b (1-\sin (e+f x))}{a+b}\right )}{f \sqrt{\sin (e+f x)+1}}-\frac{4 \sqrt{2} A \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac{a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac{1}{2};-\frac{1}{2},-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{b (1-\sin (e+f x))}{a+b}\right )}{f \sqrt{\sin (e+f x)+1}} \]

[Out]

(4*Sqrt[2]*A*AppellF1[1/2, -3/2, -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) - (4*Sqrt[2]*A*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)

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Rubi [A]  time = 0.227248, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {3018, 2755, 139, 138, 2784} \[ \frac{4 \sqrt{2} A \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac{a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac{1}{2};-\frac{3}{2},-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{b (1-\sin (e+f x))}{a+b}\right )}{f \sqrt{\sin (e+f x)+1}}-\frac{4 \sqrt{2} A \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac{a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac{1}{2};-\frac{1}{2},-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{b (1-\sin (e+f x))}{a+b}\right )}{f \sqrt{\sin (e+f x)+1}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[e + f*x])^m*(A - A*Sin[e + f*x]^2),x]

[Out]

(4*Sqrt[2]*A*AppellF1[1/2, -3/2, -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) - (4*Sqrt[2]*A*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 3018

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[
A - C, Int[(a + b*Sin[e + f*x])^m*(1 + Sin[e + f*x]), x], x] + Dist[C, Int[(a + b*Sin[e + f*x])^m*(1 + Sin[e +
f*x])^2, x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && EqQ[A + C, 0] &&  !IntegerQ[2*m]

Rule 2755

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

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])

Rule 2784

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*x)/a)^(m - 1/2)*(c
+ d*x)^n)/Sqrt[1 - (b*x)/a], 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*} \int (a+b \sin (e+f x))^m \left (A-A \sin ^2(e+f x)\right ) \, dx &=-\left (A \int (1+\sin (e+f x))^2 (a+b \sin (e+f x))^m \, dx\right )+(2 A) \int (1+\sin (e+f x)) (a+b \sin (e+f x))^m \, dx\\ &=-\frac{(A \cos (e+f x)) \operatorname{Subst}\left (\int \frac{(1+x)^{3/2} (a+b x)^m}{\sqrt{1-x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt{1-\sin (e+f x)} \sqrt{1+\sin (e+f x)}}+\frac{(2 A \cos (e+f x)) \operatorname{Subst}\left (\int \frac{\sqrt{1+x} (a+b x)^m}{\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 \cos (e+f x) (a+b \sin (e+f x))^m \left (-\frac{a+b \sin (e+f x)}{-a-b}\right )^{-m}\right ) \operatorname{Subst}\left (\int \frac{(1+x)^{3/2} \left (-\frac{a}{-a-b}-\frac{b x}{-a-b}\right )^m}{\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 (2 A \cos (e+f x) (a+b \sin (e+f x))^m \left (-\frac{a+b \sin (e+f x)}{-a-b}\right )^{-m}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1+x} \left (-\frac{a}{-a-b}-\frac{b x}{-a-b}\right )^m}{\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 F_1\left (\frac{1}{2};-\frac{3}{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)}}-\frac{4 \sqrt{2} A 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*}

Mathematica [F]  time = 3.42141, size = 0, normalized size = 0. \[ \int (a+b \sin (e+f x))^m \left (A-A \sin ^2(e+f x)\right ) \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(a + b*Sin[e + f*x])^m*(A - A*Sin[e + f*x]^2),x]

[Out]

Integrate[(a + b*Sin[e + f*x])^m*(A - A*Sin[e + f*x]^2), x]

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Maple [F]  time = 0.276, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\sin \left ( fx+e \right ) \right ) ^{m} \left ( A-A \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(f*x+e))^m*(A-A*sin(f*x+e)^2),x)

[Out]

int((a+b*sin(f*x+e))^m*(A-A*sin(f*x+e)^2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int{\left (A \sin \left (f x + e\right )^{2} - A\right )}{\left (b \sin \left (f x + e\right ) + a\right )}^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))^m*(A-A*sin(f*x+e)^2),x, algorithm="maxima")

[Out]

-integrate((A*sin(f*x + e)^2 - A)*(b*sin(f*x + e) + a)^m, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \sin \left (f x + e\right ) + a\right )}^{m} A \cos \left (f x + e\right )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))^m*(A-A*sin(f*x+e)^2),x, algorithm="fricas")

[Out]

integral((b*sin(f*x + e) + a)^m*A*cos(f*x + e)^2, 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*sin(f*x+e))**m*(A-A*sin(f*x+e)**2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -{\left (A \sin \left (f x + e\right )^{2} - A\right )}{\left (b \sin \left (f x + e\right ) + a\right )}^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))^m*(A-A*sin(f*x+e)^2),x, algorithm="giac")

[Out]

integrate(-(A*sin(f*x + e)^2 - A)*(b*sin(f*x + e) + a)^m, x)