\(\int (d \sin (e+f x))^n (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) \, dx\) [2]

Optimal result

Integrand size = 33, antiderivative size = 277 \[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) \, dx=-\frac {a^2 (A (3+n)+B (4+n)) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (2+n) (3+n)}+\frac {a^2 (2 B (1+n)+A (3+2 n)) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(e+f x)\right ) (d \sin (e+f x))^{1+n}}{d f (1+n) (2+n) \sqrt {\cos ^2(e+f x)}}+\frac {a^2 (2 A (3+n)+B (5+2 n)) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(e+f x)\right ) (d \sin (e+f x))^{2+n}}{d^2 f (2+n) (3+n) \sqrt {\cos ^2(e+f x)}}-\frac {B \cos (e+f x) (d \sin (e+f x))^{1+n} \left (a^2+a^2 \sin (e+f x)\right )}{d f (3+n)} \]

[Out]

-a^2*(A*(3+n)+B*(4+n))*cos(f*x+e)*(d*sin(f*x+e))^(1+n)/d/f/(2+n)/(3+n)-B*cos(f*x+e)*(d*sin(f*x+e))^(1+n)*(a^2+
a^2*sin(f*x+e))/d/f/(3+n)+a^2*(2*B*(1+n)+A*(3+2*n))*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)/d/f/(1+n)/(2+n)/(cos(f*x+e)^2)^(1/2)+a^2*(2*A*(3+n)+B*(5+2*n))*cos(f*x+e)*hypergeom
([1/2, 1+1/2*n],[1/2*n+2],sin(f*x+e)^2)*(d*sin(f*x+e))^(2+n)/d^2/f/(2+n)/(3+n)/(cos(f*x+e)^2)^(1/2)

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3055, 3047, 3102, 2827, 2722} \[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) \, dx=\frac {a^2 (2 A (n+3)+B (2 n+5)) \cos (e+f x) (d \sin (e+f x))^{n+2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n+2}{2},\frac {n+4}{2},\sin ^2(e+f x)\right )}{d^2 f (n+2) (n+3) \sqrt {\cos ^2(e+f x)}}+\frac {a^2 (A (2 n+3)+2 B (n+1)) \cos (e+f x) (d \sin (e+f x))^{n+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n+1}{2},\frac {n+3}{2},\sin ^2(e+f x)\right )}{d f (n+1) (n+2) \sqrt {\cos ^2(e+f x)}}-\frac {a^2 (A (n+3)+B (n+4)) \cos (e+f x) (d \sin (e+f x))^{n+1}}{d f (n+2) (n+3)}-\frac {B \cos (e+f x) \left (a^2 \sin (e+f x)+a^2\right ) (d \sin (e+f x))^{n+1}}{d f (n+3)} \]

[In]

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

[Out]

-((a^2*(A*(3 + n) + B*(4 + n))*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + n))/(d*f*(2 + n)*(3 + n))) + (a^2*(2*B*(1 +
n) + A*(3 + 2*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))/(d*f*(1 + n)*(2 + n)*Sqrt[Cos[e + f*x]^2]) + (a^2*(2*A*(3 + n) + B*(5 + 2*n))*Cos[e + f*x]*Hypergeomet
ric2F1[1/2, (2 + n)/2, (4 + n)/2, Sin[e + f*x]^2]*(d*Sin[e + f*x])^(2 + n))/(d^2*f*(2 + n)*(3 + n)*Sqrt[Cos[e
+ f*x]^2]) - (B*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + n)*(a^2 + a^2*Sin[e + f*x]))/(d*f*(3 + n))

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 3047

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

Rule 3055

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)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x
])^(n + 1)/(d*f*(m + n + 1))), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*
x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))
*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] && GtQ[m, 1/2] &&  !LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 3102

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

Rubi steps \begin{align*} \text {integral}& = -\frac {B \cos (e+f x) (d \sin (e+f x))^{1+n} \left (a^2+a^2 \sin (e+f x)\right )}{d f (3+n)}+\frac {\int (d \sin (e+f x))^n (a+a \sin (e+f x)) (a d (B (1+n)+A (3+n))+a d (A (3+n)+B (4+n)) \sin (e+f x)) \, dx}{d (3+n)} \\ & = -\frac {B \cos (e+f x) (d \sin (e+f x))^{1+n} \left (a^2+a^2 \sin (e+f x)\right )}{d f (3+n)}+\frac {\int (d \sin (e+f x))^n \left (a^2 d (B (1+n)+A (3+n))+\left (a^2 d (B (1+n)+A (3+n))+a^2 d (A (3+n)+B (4+n))\right ) \sin (e+f x)+a^2 d (A (3+n)+B (4+n)) \sin ^2(e+f x)\right ) \, dx}{d (3+n)} \\ & = -\frac {a^2 (A (3+n)+B (4+n)) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (2+n) (3+n)}-\frac {B \cos (e+f x) (d \sin (e+f x))^{1+n} \left (a^2+a^2 \sin (e+f x)\right )}{d f (3+n)}+\frac {\int (d \sin (e+f x))^n \left (a^2 d^2 (3+n) (2 B (1+n)+A (3+2 n))+a^2 d^2 (2+n) (2 A (3+n)+B (5+2 n)) \sin (e+f x)\right ) \, dx}{d^2 (2+n) (3+n)} \\ & = -\frac {a^2 (A (3+n)+B (4+n)) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (2+n) (3+n)}-\frac {B \cos (e+f x) (d \sin (e+f x))^{1+n} \left (a^2+a^2 \sin (e+f x)\right )}{d f (3+n)}+\frac {\left (a^2 (2 B (1+n)+A (3+2 n))\right ) \int (d \sin (e+f x))^n \, dx}{2+n}+\frac {\left (a^2 (2 A (3+n)+B (5+2 n))\right ) \int (d \sin (e+f x))^{1+n} \, dx}{d (3+n)} \\ & = -\frac {a^2 (A (3+n)+B (4+n)) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (2+n) (3+n)}+\frac {a^2 (2 B (1+n)+A (3+2 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}}{d f (1+n) (2+n) \sqrt {\cos ^2(e+f x)}}+\frac {a^2 (2 A (3+n)+B (5+2 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}}{d^2 f (2+n) (3+n) \sqrt {\cos ^2(e+f x)}}-\frac {B \cos (e+f x) (d \sin (e+f x))^{1+n} \left (a^2+a^2 \sin (e+f x)\right )}{d f (3+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.98 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.74 \[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) \, dx=\frac {a^2 \cos (e+f x) \sin (e+f x) (d \sin (e+f x))^n \left (\frac {A \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\sin ^2(e+f x)\right )}{1+n}+\sin (e+f x) \left (\frac {(2 A+B) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2+n}{2},\frac {4+n}{2},\sin ^2(e+f x)\right )}{2+n}+\sin (e+f x) \left (\frac {(A+2 B) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3+n}{2},\frac {5+n}{2},\sin ^2(e+f x)\right )}{3+n}+\frac {B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4+n}{2},\frac {6+n}{2},\sin ^2(e+f x)\right ) \sin (e+f x)}{4+n}\right )\right )\right )}{f \sqrt {\cos ^2(e+f x)}} \]

[In]

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

[Out]

(a^2*Cos[e + f*x]*Sin[e + f*x]*(d*Sin[e + f*x])^n*((A*Hypergeometric2F1[1/2, (1 + n)/2, (3 + n)/2, Sin[e + f*x
]^2])/(1 + n) + Sin[e + f*x]*(((2*A + B)*Hypergeometric2F1[1/2, (2 + n)/2, (4 + n)/2, Sin[e + f*x]^2])/(2 + n)
+ Sin[e + f*x]*(((A + 2*B)*Hypergeometric2F1[1/2, (3 + n)/2, (5 + n)/2, Sin[e + f*x]^2])/(3 + n) + (B*Hyperge
ometric2F1[1/2, (4 + n)/2, (6 + n)/2, Sin[e + f*x]^2]*Sin[e + f*x])/(4 + n)))))/(f*Sqrt[Cos[e + f*x]^2])

Maple [F]

\[\int \left (d \sin \left (f x +e \right )\right )^{n} \left (a +a \sin \left (f x +e \right )\right )^{2} \left (A +B \sin \left (f x +e \right )\right )d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{2} \left (d \sin \left (f x + e\right )\right )^{n} \,d x } \]

[In]

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

[Out]

integral(-((A + 2*B)*a^2*cos(f*x + e)^2 - 2*(A + B)*a^2 + (B*a^2*cos(f*x + e)^2 - 2*(A + B)*a^2)*sin(f*x + e))
*(d*sin(f*x + e))^n, x)

Sympy [F(-1)]

Timed out. \[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{2} \left (d \sin \left (f x + e\right )\right )^{n} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{2} \left (d \sin \left (f x + e\right )\right )^{n} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (d \sin (e+f x))^n (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) \, dx=\int {\left (d\,\sin \left (e+f\,x\right )\right )}^n\,\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2 \,d x \]

[In]

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

[Out]

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