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

Optimal. Leaf size=279 \[ \frac{(n+1) (2 A (1-n)+2 B n+B) \cos (e+f x) (d \sin (e+f x))^{n+2} \, _2F_1\left (\frac{1}{2},\frac{n+2}{2};\frac{n+4}{2};\sin ^2(e+f x)\right )}{3 a^2 d^2 f (n+2) \sqrt{\cos ^2(e+f x)}}-\frac{n (-2 A n+A+2 B (n+1)) \cos (e+f x) (d \sin (e+f x))^{n+1} \, _2F_1\left (\frac{1}{2},\frac{n+1}{2};\frac{n+3}{2};\sin ^2(e+f x)\right )}{3 a^2 d f (n+1) \sqrt{\cos ^2(e+f x)}}+\frac{(2 A (1-n)+2 B n+B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{3 a^2 d f (\sin (e+f x)+1)}+\frac{(A-B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{3 d f (a \sin (e+f x)+a)^2} \]

[Out]

-(n*(A - 2*A*n + 2*B*(1 + n))*Cos[e + f*x]*Hypergeometric2F1[1/2, (1 + n)/2, (3 + n)/2, Sin[e + f*x]^2]*(d*Sin
[e + f*x])^(1 + n))/(3*a^2*d*f*(1 + n)*Sqrt[Cos[e + f*x]^2]) + ((1 + n)*(B + 2*A*(1 - n) + 2*B*n)*Cos[e + f*x]
*Hypergeometric2F1[1/2, (2 + n)/2, (4 + n)/2, Sin[e + f*x]^2]*(d*Sin[e + f*x])^(2 + n))/(3*a^2*d^2*f*(2 + n)*S
qrt[Cos[e + f*x]^2]) + ((B + 2*A*(1 - n) + 2*B*n)*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + n))/(3*a^2*d*f*(1 + Sin[e
+ f*x])) + ((A - B)*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + n))/(3*d*f*(a + a*Sin[e + f*x])^2)

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Rubi [A]  time = 0.4879, antiderivative size = 279, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {2978, 2748, 2643} \[ \frac{(n+1) (2 A (1-n)+2 B n+B) \cos (e+f x) (d \sin (e+f x))^{n+2} \, _2F_1\left (\frac{1}{2},\frac{n+2}{2};\frac{n+4}{2};\sin ^2(e+f x)\right )}{3 a^2 d^2 f (n+2) \sqrt{\cos ^2(e+f x)}}-\frac{n (-2 A n+A+2 B (n+1)) \cos (e+f x) (d \sin (e+f x))^{n+1} \, _2F_1\left (\frac{1}{2},\frac{n+1}{2};\frac{n+3}{2};\sin ^2(e+f x)\right )}{3 a^2 d f (n+1) \sqrt{\cos ^2(e+f x)}}+\frac{(2 A (1-n)+2 B n+B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{3 a^2 d f (\sin (e+f x)+1)}+\frac{(A-B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{3 d f (a \sin (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(n*(A - 2*A*n + 2*B*(1 + n))*Cos[e + f*x]*Hypergeometric2F1[1/2, (1 + n)/2, (3 + n)/2, Sin[e + f*x]^2]*(d*Sin
[e + f*x])^(1 + n))/(3*a^2*d*f*(1 + n)*Sqrt[Cos[e + f*x]^2]) + ((1 + n)*(B + 2*A*(1 - n) + 2*B*n)*Cos[e + f*x]
*Hypergeometric2F1[1/2, (2 + n)/2, (4 + n)/2, Sin[e + f*x]^2]*(d*Sin[e + f*x])^(2 + n))/(3*a^2*d^2*f*(2 + n)*S
qrt[Cos[e + f*x]^2]) + ((B + 2*A*(1 - n) + 2*B*n)*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + n))/(3*a^2*d*f*(1 + Sin[e
+ f*x])) + ((A - B)*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + n))/(3*d*f*(a + a*Sin[e + f*x])^2)

Rule 2978

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

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rubi steps

\begin{align*} \int \frac{(d \sin (e+f x))^n (A+B \sin (e+f x))}{(a+a \sin (e+f x))^2} \, dx &=\frac{(A-B) \cos (e+f x) (d \sin (e+f x))^{1+n}}{3 d f (a+a \sin (e+f x))^2}+\frac{\int \frac{(d \sin (e+f x))^n (a d (2 A+B-A n+B n)+a (A-B) d n \sin (e+f x))}{a+a \sin (e+f x)} \, dx}{3 a^2 d}\\ &=\frac{(B+2 A (1-n)+2 B n) \cos (e+f x) (d \sin (e+f x))^{1+n}}{3 a^2 d f (1+\sin (e+f x))}+\frac{(A-B) \cos (e+f x) (d \sin (e+f x))^{1+n}}{3 d f (a+a \sin (e+f x))^2}+\frac{\int (d \sin (e+f x))^n \left (-a^2 d^2 n (A-2 A n+2 B (1+n))+a^2 d^2 (1+n) (2 A (1-n)+B (1+2 n)) \sin (e+f x)\right ) \, dx}{3 a^4 d^2}\\ &=\frac{(B+2 A (1-n)+2 B n) \cos (e+f x) (d \sin (e+f x))^{1+n}}{3 a^2 d f (1+\sin (e+f x))}+\frac{(A-B) \cos (e+f x) (d \sin (e+f x))^{1+n}}{3 d f (a+a \sin (e+f x))^2}+\frac{((1+n) (B+2 A (1-n)+2 B n)) \int (d \sin (e+f x))^{1+n} \, dx}{3 a^2 d}-\frac{(n (A-2 A n+2 B (1+n))) \int (d \sin (e+f x))^n \, dx}{3 a^2}\\ &=-\frac{n (A-2 A n+2 B (1+n)) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1+n}{2};\frac{3+n}{2};\sin ^2(e+f x)\right ) (d \sin (e+f x))^{1+n}}{3 a^2 d f (1+n) \sqrt{\cos ^2(e+f x)}}+\frac{(1+n) (B+2 A (1-n)+2 B n) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{2+n}{2};\frac{4+n}{2};\sin ^2(e+f x)\right ) (d \sin (e+f x))^{2+n}}{3 a^2 d^2 f (2+n) \sqrt{\cos ^2(e+f x)}}+\frac{(B+2 A (1-n)+2 B n) \cos (e+f x) (d \sin (e+f x))^{1+n}}{3 a^2 d f (1+\sin (e+f x))}+\frac{(A-B) \cos (e+f x) (d \sin (e+f x))^{1+n}}{3 d f (a+a \sin (e+f x))^2}\\ \end{align*}

Mathematica [A]  time = 1.27717, size = 212, normalized size = 0.76 \[ \frac{\sin (e+f x) \cos (e+f x) (d \sin (e+f x))^n \left (-\frac{n (-2 A n+A+2 B (n+1)) \, _2F_1\left (\frac{1}{2},\frac{n+1}{2};\frac{n+3}{2};\sin ^2(e+f x)\right )}{(n+1) \sqrt{\cos ^2(e+f x)}}+\frac{(n+1) (-2 A (n-1)+2 B n+B) \sin (e+f x) \, _2F_1\left (\frac{1}{2},\frac{n+2}{2};\frac{n+4}{2};\sin ^2(e+f x)\right )}{(n+2) \sqrt{\cos ^2(e+f x)}}+\frac{n (B-A)}{\sin (e+f x)+1}+\frac{A (-n)+2 A+B n+B}{\sin (e+f x)+1}+\frac{A-B}{(\sin (e+f x)+1)^2}\right )}{3 a^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[e + f*x]*Sin[e + f*x]*(d*Sin[e + f*x])^n*(-((n*(A - 2*A*n + 2*B*(1 + n))*Hypergeometric2F1[1/2, (1 + n)/2
, (3 + n)/2, Sin[e + f*x]^2])/((1 + n)*Sqrt[Cos[e + f*x]^2])) + ((1 + n)*(B - 2*A*(-1 + n) + 2*B*n)*Hypergeome
tric2F1[1/2, (2 + n)/2, (4 + n)/2, Sin[e + f*x]^2]*Sin[e + f*x])/((2 + n)*Sqrt[Cos[e + f*x]^2]) + (A - B)/(1 +
Sin[e + f*x])^2 + ((-A + B)*n)/(1 + Sin[e + f*x]) + (2*A + B - A*n + B*n)/(1 + Sin[e + f*x])))/(3*a^2*f)

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Maple [F]  time = 1.539, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( d\sin \left ( fx+e \right ) \right ) ^{n} \left ( A+B\sin \left ( fx+e \right ) \right ) }{ \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((d*sin(f*x+e))^n*(A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2,x)

[Out]

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

[Out]

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

[Out]

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

[Out]

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