3.331 \(\int \frac{(A+B \sin (e+f x)) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx\)

Optimal. Leaf size=223 \[ -\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c+d}\right )^{-n} F_1\left (\frac{1}{2};\frac{5}{2},-n;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{d (1-\sin (e+f x))}{c+d}\right )}{2 \sqrt{2} a^2 f \sqrt{\sin (e+f x)+1}}-\frac{B \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c+d}\right )^{-n} F_1\left (\frac{1}{2};\frac{3}{2},-n;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{d (1-\sin (e+f x))}{c+d}\right )}{\sqrt{2} a^2 f \sqrt{\sin (e+f x)+1}} \]

[Out]

-((B*AppellF1[1/2, 3/2, -n, 3/2, (1 - Sin[e + f*x])/2, (d*(1 - Sin[e + f*x]))/(c + d)]*Cos[e + f*x]*(c + d*Sin
[e + f*x])^n)/(Sqrt[2]*a^2*f*Sqrt[1 + Sin[e + f*x]]*((c + d*Sin[e + f*x])/(c + d))^n)) - ((A - B)*AppellF1[1/2
, 5/2, -n, 3/2, (1 - Sin[e + f*x])/2, (d*(1 - Sin[e + f*x]))/(c + d)]*Cos[e + f*x]*(c + d*Sin[e + f*x])^n)/(2*
Sqrt[2]*a^2*f*Sqrt[1 + Sin[e + f*x]]*((c + d*Sin[e + f*x])/(c + d))^n)

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Rubi [A]  time = 0.341262, antiderivative size = 223, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.114, Rules used = {2987, 2784, 139, 138} \[ -\frac{(A-B) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c+d}\right )^{-n} F_1\left (\frac{1}{2};\frac{5}{2},-n;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{d (1-\sin (e+f x))}{c+d}\right )}{2 \sqrt{2} a^2 f \sqrt{\sin (e+f x)+1}}-\frac{B \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c+d}\right )^{-n} F_1\left (\frac{1}{2};\frac{3}{2},-n;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{d (1-\sin (e+f x))}{c+d}\right )}{\sqrt{2} a^2 f \sqrt{\sin (e+f x)+1}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^n)/(a + a*Sin[e + f*x])^2,x]

[Out]

-((B*AppellF1[1/2, 3/2, -n, 3/2, (1 - Sin[e + f*x])/2, (d*(1 - Sin[e + f*x]))/(c + d)]*Cos[e + f*x]*(c + d*Sin
[e + f*x])^n)/(Sqrt[2]*a^2*f*Sqrt[1 + Sin[e + f*x]]*((c + d*Sin[e + f*x])/(c + d))^n)) - ((A - B)*AppellF1[1/2
, 5/2, -n, 3/2, (1 - Sin[e + f*x])/2, (d*(1 - Sin[e + f*x]))/(c + d)]*Cos[e + f*x]*(c + d*Sin[e + f*x])^n)/(2*
Sqrt[2]*a^2*f*Sqrt[1 + Sin[e + f*x]]*((c + d*Sin[e + f*x])/(c + d))^n)

Rule 2987

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

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]

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 \frac{(A+B \sin (e+f x)) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx &=(A-B) \int \frac{(c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx+\frac{B \int \frac{(c+d \sin (e+f x))^n}{a+a \sin (e+f x)} \, dx}{a}\\ &=\frac{((A-B) \cos (e+f x)) \operatorname{Subst}\left (\int \frac{(c+d x)^n}{\sqrt{1-x} (1+x)^{5/2}} \, dx,x,\sin (e+f x)\right )}{a^2 f \sqrt{1-\sin (e+f x)} \sqrt{1+\sin (e+f x)}}+\frac{(B \cos (e+f x)) \operatorname{Subst}\left (\int \frac{(c+d x)^n}{\sqrt{1-x} (1+x)^{3/2}} \, dx,x,\sin (e+f x)\right )}{a^2 f \sqrt{1-\sin (e+f x)} \sqrt{1+\sin (e+f x)}}\\ &=\frac{\left ((A-B) \cos (e+f x) (c+d \sin (e+f x))^n \left (-\frac{c+d \sin (e+f x)}{-c-d}\right )^{-n}\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{c}{-c-d}-\frac{d x}{-c-d}\right )^n}{\sqrt{1-x} (1+x)^{5/2}} \, dx,x,\sin (e+f x)\right )}{a^2 f \sqrt{1-\sin (e+f x)} \sqrt{1+\sin (e+f x)}}+\frac{\left (B \cos (e+f x) (c+d \sin (e+f x))^n \left (-\frac{c+d \sin (e+f x)}{-c-d}\right )^{-n}\right ) \operatorname{Subst}\left (\int \frac{\left (-\frac{c}{-c-d}-\frac{d x}{-c-d}\right )^n}{\sqrt{1-x} (1+x)^{3/2}} \, dx,x,\sin (e+f x)\right )}{a^2 f \sqrt{1-\sin (e+f x)} \sqrt{1+\sin (e+f x)}}\\ &=-\frac{B F_1\left (\frac{1}{2};\frac{3}{2},-n;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c+d}\right )^{-n}}{\sqrt{2} a^2 f \sqrt{1+\sin (e+f x)}}-\frac{(A-B) F_1\left (\frac{1}{2};\frac{5}{2},-n;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x)),\frac{d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c+d}\right )^{-n}}{2 \sqrt{2} a^2 f \sqrt{1+\sin (e+f x)}}\\ \end{align*}

Mathematica [F]  time = 8.67493, size = 0, normalized size = 0. \[ \int \frac{(A+B \sin (e+f x)) (c+d \sin (e+f x))^n}{(a+a \sin (e+f x))^2} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^n)/(a + a*Sin[e + f*x])^2,x]

[Out]

Integrate[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^n)/(a + a*Sin[e + f*x])^2, x]

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Maple [F]  time = 0.714, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( A+B\sin \left ( fx+e \right ) \right ) \left ( c+d\sin \left ( fx+e \right ) \right ) ^{n}}{ \left ( a+a\sin \left ( fx+e \right ) \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^2,x)

[Out]

int((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^2,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(d*sin(f*x + e) + c)^n/(a*sin(f*x + e) + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

integral(-(B*sin(f*x + e) + A)*(d*sin(f*x + e) + c)^n/(a^2*cos(f*x + e)^2 - 2*a^2*sin(f*x + e) - 2*a^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))*(c+d*sin(f*x+e))**n/(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 \frac{{\left (B \sin \left (f x + e\right ) + A\right )}{\left (d \sin \left (f x + e\right ) + c\right )}^{n}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^n/(a+a*sin(f*x+e))^2,x, algorithm="giac")

[Out]

integrate((B*sin(f*x + e) + A)*(d*sin(f*x + e) + c)^n/(a*sin(f*x + e) + a)^2, x)