3.51 \(\int (a+a \sin (e+f x))^m (A+B \sin (e+f x))^p (c-c \sin (e+f x))^n \, dx\)

Optimal. Leaf size=157 \[ \frac{2^{n+\frac{1}{2}} \sec (e+f x) (1-\sin (e+f x))^{\frac{1}{2}-n} (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^n (A+B \sin (e+f x))^p \left (\frac{A+B \sin (e+f x)}{A-B}\right )^{-p} F_1\left (m+\frac{1}{2};\frac{1}{2}-n,-p;m+\frac{3}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{B (\sin (e+f x)+1)}{A-B}\right )}{a f (2 m+1)} \]

[Out]

(2^(1/2 + n)*AppellF1[1/2 + m, 1/2 - n, -p, 3/2 + m, (1 + Sin[e + f*x])/2, -((B*(1 + Sin[e + f*x]))/(A - B))]*
Sec[e + f*x]*(1 - Sin[e + f*x])^(1/2 - n)*(a + a*Sin[e + f*x])^(1 + m)*(A + B*Sin[e + f*x])^p*(c - c*Sin[e + f
*x])^n)/(a*f*(1 + 2*m)*((A + B*Sin[e + f*x])/(A - B))^p)

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Rubi [A]  time = 0.278196, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {3008, 140, 139, 138} \[ \frac{2^{n+\frac{1}{2}} \sec (e+f x) (1-\sin (e+f x))^{\frac{1}{2}-n} (a \sin (e+f x)+a)^{m+1} (c-c \sin (e+f x))^n (A+B \sin (e+f x))^p \left (\frac{A+B \sin (e+f x)}{A-B}\right )^{-p} F_1\left (m+\frac{1}{2};\frac{1}{2}-n,-p;m+\frac{3}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{B (\sin (e+f x)+1)}{A-B}\right )}{a f (2 m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x])^p*(c - c*Sin[e + f*x])^n,x]

[Out]

(2^(1/2 + n)*AppellF1[1/2 + m, 1/2 - n, -p, 3/2 + m, (1 + Sin[e + f*x])/2, -((B*(1 + Sin[e + f*x]))/(A - B))]*
Sec[e + f*x]*(1 - Sin[e + f*x])^(1/2 - n)*(a + a*Sin[e + f*x])^(1 + m)*(A + B*Sin[e + f*x])^p*(c - c*Sin[e + f
*x])^n)/(a*f*(1 + 2*m)*((A + B*Sin[e + f*x])/(A - B))^p)

Rule 3008

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])^(p_)*((c_) + (d_.)*si
n[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])/(f*Cos[e +
f*x]), Subst[Int[(a + b*x)^(m - 1/2)*(c + d*x)^(n - 1/2)*(A + B*x)^p, x], x, Sin[e + f*x]], x] /; FreeQ[{a, b
, c, d, e, f, A, B, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]

Rule 140

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c + d*x)^
FracPart[n]/((b/(b*c - a*d))^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*((b*c)/(b*c
- a*d) + (b*d*x)/(b*c - a*d))^n*(e + f*x)^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] &&  !SimplerQ[c + d*x, a + b*x] &&  !SimplerQ[e +
f*x, a + b*x]

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+a \sin (e+f x))^m (A+B \sin (e+f x))^p (c-c \sin (e+f x))^n \, dx &=\frac{\left (\sec (e+f x) \sqrt{a+a \sin (e+f x)} \sqrt{c-c \sin (e+f x)}\right ) \operatorname{Subst}\left (\int (a+a x)^{-\frac{1}{2}+m} (A+B x)^p (c-c x)^{-\frac{1}{2}+n} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\left (\sec (e+f x) \sqrt{a+a \sin (e+f x)} (A+B \sin (e+f x))^p \left (\frac{a (A+B \sin (e+f x))}{a A-a B}\right )^{-p} \sqrt{c-c \sin (e+f x)}\right ) \operatorname{Subst}\left (\int (a+a x)^{-\frac{1}{2}+m} \left (\frac{a A}{a A-a B}+\frac{a B x}{a A-a B}\right )^p (c-c x)^{-\frac{1}{2}+n} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\left (2^{-\frac{1}{2}+n} \sec (e+f x) \sqrt{a+a \sin (e+f x)} (A+B \sin (e+f x))^p \left (\frac{a (A+B \sin (e+f x))}{a A-a B}\right )^{-p} (c-c \sin (e+f x))^n \left (\frac{c-c \sin (e+f x)}{c}\right )^{\frac{1}{2}-n}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}-\frac{x}{2}\right )^{-\frac{1}{2}+n} (a+a x)^{-\frac{1}{2}+m} \left (\frac{a A}{a A-a B}+\frac{a B x}{a A-a B}\right )^p \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{2^{\frac{1}{2}+n} F_1\left (\frac{1}{2}+m;\frac{1}{2}-n,-p;\frac{3}{2}+m;\frac{1}{2} (1+\sin (e+f x)),-\frac{B (1+\sin (e+f x))}{A-B}\right ) \sec (e+f x) (1-\sin (e+f x))^{\frac{1}{2}-n} (a+a \sin (e+f x))^{1+m} (A+B \sin (e+f x))^p \left (\frac{A+B \sin (e+f x)}{A-B}\right )^{-p} (c-c \sin (e+f x))^n}{a f (1+2 m)}\\ \end{align*}

Mathematica [A]  time = 1.05879, size = 168, normalized size = 1.07 \[ -\frac{2 \cot \left (\frac{1}{4} (2 e+2 f x+\pi )\right ) \sin ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right )^{\frac{1}{2}-m} (a (\sin (e+f x)+1))^m (c-c \sin (e+f x))^n (A+B \sin (e+f x))^p \left (\frac{A+B \sin (e+f x)}{A+B}\right )^{-p} F_1\left (n+\frac{1}{2};\frac{1}{2}-m,-p;n+\frac{3}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 B \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{A+B}\right )}{2 f n+f} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x])^p*(c - c*Sin[e + f*x])^n,x]

[Out]

(-2*AppellF1[1/2 + n, 1/2 - m, -p, 3/2 + n, Cos[(2*e + Pi + 2*f*x)/4]^2, (2*B*Sin[(2*e - Pi + 2*f*x)/4]^2)/(A
+ B)]*Cot[(2*e + Pi + 2*f*x)/4]*(a*(1 + Sin[e + f*x]))^m*(A + B*Sin[e + f*x])^p*(c - c*Sin[e + f*x])^n*(Sin[(2
*e + Pi + 2*f*x)/4]^2)^(1/2 - m))/((f + 2*f*n)*((A + B*Sin[e + f*x])/(A + B))^p)

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Maple [F]  time = 4.027, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( A+B\sin \left ( fx+e \right ) \right ) ^{p} \left ( c-c\sin \left ( fx+e \right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))^p*(c-c*sin(f*x+e))^n,x)

[Out]

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))^p*(c-c*sin(f*x+e))^n,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))^p*(c-c*sin(f*x+e))^n,x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)^p*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^n, 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 )}^{p}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}{\left (-c \sin \left (f x + e\right ) + c\right )}^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))^p*(c-c*sin(f*x+e))^n,x, algorithm="fricas")

[Out]

integral((B*sin(f*x + e) + A)^p*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^n, 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+a*sin(f*x+e))**m*(A+B*sin(f*x+e))**p*(c-c*sin(f*x+e))**n,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e))^p*(c-c*sin(f*x+e))^n,x, algorithm="giac")

[Out]

integrate((B*sin(f*x + e) + A)^p*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^n, x)