\(\int \frac {(d \sin (e+f x))^n (A+B \sin (e+f x))}{a+a \sin (e+f x)} \, dx\) [4]

Optimal result

Integrand size = 33, antiderivative size = 202 \[ \int \frac {(d \sin (e+f x))^n (A+B \sin (e+f x))}{a+a \sin (e+f x)} \, dx=\frac {(B-A n+B n) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(e+f x)\right ) (d \sin (e+f x))^{1+n}}{a d f (1+n) \sqrt {\cos ^2(e+f x)}}+\frac {(A-B) (1+n) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(e+f x)\right ) (d \sin (e+f x))^{2+n}}{a d^2 f (2+n) \sqrt {\cos ^2(e+f x)}}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (a+a \sin (e+f x))} \]

[Out]

(A-B)*cos(f*x+e)*(d*sin(f*x+e))^(1+n)/d/f/(a+a*sin(f*x+e))+(-A*n+B*n+B)*cos(f*x+e)*hypergeom([1/2, 1/2+1/2*n],
[3/2+1/2*n],sin(f*x+e)^2)*(d*sin(f*x+e))^(1+n)/a/d/f/(1+n)/(cos(f*x+e)^2)^(1/2)+(A-B)*(1+n)*cos(f*x+e)*hyperge
om([1/2, 1+1/2*n],[1/2*n+2],sin(f*x+e)^2)*(d*sin(f*x+e))^(2+n)/a/d^2/f/(2+n)/(cos(f*x+e)^2)^(1/2)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3057, 2827, 2722} \[ \int \frac {(d \sin (e+f x))^n (A+B \sin (e+f x))}{a+a \sin (e+f x)} \, dx=\frac {(n+1) (A-B) \cos (e+f x) (d \sin (e+f x))^{n+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n+2}{2},\frac {n+4}{2},\sin ^2(e+f x)\right )}{a d^2 f (n+2) \sqrt {\cos ^2(e+f x)}}+\frac {(-A n+B n+B) \cos (e+f x) (d \sin (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n+1}{2},\frac {n+3}{2},\sin ^2(e+f x)\right )}{a d f (n+1) \sqrt {\cos ^2(e+f x)}}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{d f (a \sin (e+f x)+a)} \]

[In]

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

[Out]

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

Rule 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 2827

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 3057

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])

Rubi steps \begin{align*} \text {integral}& = \frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (a+a \sin (e+f x))}+\frac {\int (d \sin (e+f x))^n (a d (B-A n+B n)+a (A-B) d (1+n) \sin (e+f x)) \, dx}{a^2 d} \\ & = \frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (a+a \sin (e+f x))}+\frac {((A-B) (1+n)) \int (d \sin (e+f x))^{1+n} \, dx}{a d}+\frac {(B-A n+B n) \int (d \sin (e+f x))^n \, dx}{a} \\ & = \frac {(B-A n+B n) \cos (e+f x) \operatorname {Hypergeometric2F1}\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}}{a d f (1+n) \sqrt {\cos ^2(e+f x)}}+\frac {(A-B) (1+n) \cos (e+f x) \operatorname {Hypergeometric2F1}\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}}{a d^2 f (2+n) \sqrt {\cos ^2(e+f x)}}+\frac {(A-B) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (a+a \sin (e+f x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.79 \[ \int \frac {(d \sin (e+f x))^n (A+B \sin (e+f x))}{a+a \sin (e+f x)} \, dx=\frac {(d \sin (e+f x))^n \left (\frac {(B-A n+B n) \sqrt {\cos ^2(e+f x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(e+f x)\right )}{1+n}+\frac {(A-B) (1+n) \sqrt {\cos ^2(e+f x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(e+f x)\right ) \sin (e+f x)}{2+n}+\frac {(A-B) \cos ^2(e+f x)}{1+\sin (e+f x)}\right ) \tan (e+f x)}{a f} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {\left (d \sin \left (f x +e \right )\right )^{n} \left (A +B \sin \left (f x +e \right )\right )}{a +a \sin \left (f x +e \right )}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {(d \sin (e+f x))^n (A+B \sin (e+f x))}{a+a \sin (e+f x)} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} \left (d \sin \left (f x + e\right )\right )^{n}}{a \sin \left (f x + e\right ) + a} \,d x } \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(d \sin (e+f x))^n (A+B \sin (e+f x))}{a+a \sin (e+f x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {(d \sin (e+f x))^n (A+B \sin (e+f x))}{a+a \sin (e+f x)} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} \left (d \sin \left (f x + e\right )\right )^{n}}{a \sin \left (f x + e\right ) + a} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {(d \sin (e+f x))^n (A+B \sin (e+f x))}{a+a \sin (e+f x)} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} \left (d \sin \left (f x + e\right )\right )^{n}}{a \sin \left (f x + e\right ) + a} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(d \sin (e+f x))^n (A+B \sin (e+f x))}{a+a \sin (e+f x)} \, dx=\int \frac {{\left (d\,\sin \left (e+f\,x\right )\right )}^n\,\left (A+B\,\sin \left (e+f\,x\right )\right )}{a+a\,\sin \left (e+f\,x\right )} \,d x \]

[In]

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

[Out]

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