3.1.1 \(\int (d \sin (e+f x))^n (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) \, dx\) [1]

Optimal. Leaf size=373 \[ -\frac {a^3 \left (B \left (27+14 n+2 n^2\right )+A \left (28+15 n+2 n^2\right )\right ) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (2+n) (3+n) (4+n)}+\frac {a^3 \left (B \left (15+19 n+4 n^2\right )+A \left (20+21 n+4 n^2\right )\right ) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1+n}{2};\frac {3+n}{2};\sin ^2(e+f x)\right ) (d \sin (e+f x))^{1+n}}{d f (1+n) (2+n) (4+n) \sqrt {\cos ^2(e+f x)}}+\frac {a^3 (B (9+4 n)+A (11+4 n)) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {2+n}{2};\frac {4+n}{2};\sin ^2(e+f x)\right ) (d \sin (e+f x))^{2+n}}{d^2 f (2+n) (3+n) \sqrt {\cos ^2(e+f x)}}-\frac {a B \cos (e+f x) (d \sin (e+f x))^{1+n} (a+a \sin (e+f x))^2}{d f (4+n)}-\frac {(A (4+n)+B (6+n)) \cos (e+f x) (d \sin (e+f x))^{1+n} \left (a^3+a^3 \sin (e+f x)\right )}{d f (3+n) (4+n)} \]

[Out]

-a^3*(B*(2*n^2+14*n+27)+A*(2*n^2+15*n+28))*cos(f*x+e)*(d*sin(f*x+e))^(1+n)/d/f/(4+n)/(n^2+5*n+6)-a*B*cos(f*x+e
)*(d*sin(f*x+e))^(1+n)*(a+a*sin(f*x+e))^2/d/f/(4+n)-(A*(4+n)+B*(6+n))*cos(f*x+e)*(d*sin(f*x+e))^(1+n)*(a^3+a^3
*sin(f*x+e))/d/f/(3+n)/(4+n)+a^3*(B*(4*n^2+19*n+15)+A*(4*n^2+21*n+20))*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/(4+n)/(n^2+3*n+2)/(cos(f*x+e)^2)^(1/2)+a^3*(B*(9+4*n)+A*(11+
4*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]
time = 0.60, antiderivative size = 373, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3055, 3047, 3102, 2827, 2722} \begin {gather*} \frac {a^3 (A (4 n+11)+B (4 n+9)) \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 )}{d^2 f (n+2) (n+3) \sqrt {\cos ^2(e+f x)}}+\frac {a^3 \left (A \left (4 n^2+21 n+20\right )+B \left (4 n^2+19 n+15\right )\right ) \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 )}{d f (n+1) (n+2) (n+4) \sqrt {\cos ^2(e+f x)}}-\frac {a^3 \left (A \left (2 n^2+15 n+28\right )+B \left (2 n^2+14 n+27\right )\right ) \cos (e+f x) (d \sin (e+f x))^{n+1}}{d f (n+2) (n+3) (n+4)}-\frac {(A (n+4)+B (n+6)) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right ) (d \sin (e+f x))^{n+1}}{d f (n+3) (n+4)}-\frac {a B \cos (e+f x) (a \sin (e+f x)+a)^2 (d \sin (e+f x))^{n+1}}{d f (n+4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*Sin[e + f*x])^n*(a + a*Sin[e + f*x])^3*(A + B*Sin[e + f*x]),x]

[Out]

-((a^3*(B*(27 + 14*n + 2*n^2) + A*(28 + 15*n + 2*n^2))*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + n))/(d*f*(2 + n)*(3
+ n)*(4 + n))) + (a^3*(B*(15 + 19*n + 4*n^2) + A*(20 + 21*n + 4*n^2))*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)*(4 + n)*Sqrt[Cos[e + f*x]^2])
+ (a^3*(B*(9 + 4*n) + A*(11 + 4*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))/(d^2*f*(2 + n)*(3 + n)*Sqrt[Cos[e + f*x]^2]) - (a*B*Cos[e + f*x]*(d*Sin[e + f*x])^(
1 + n)*(a + a*Sin[e + f*x])^2)/(d*f*(4 + n)) - ((A*(4 + n) + B*(6 + n))*Cos[e + f*x]*(d*Sin[e + f*x])^(1 + n)*
(a^3 + a^3*Sin[e + f*x]))/(d*f*(3 + n)*(4 + 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*} \int (d \sin (e+f x))^n (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) \, dx &=-\frac {a B \cos (e+f x) (d \sin (e+f x))^{1+n} (a+a \sin (e+f x))^2}{d f (4+n)}+\frac {\int (d \sin (e+f x))^n (a+a \sin (e+f x))^2 (a d (B (1+n)+A (4+n))+a d (A (4+n)+B (6+n)) \sin (e+f x)) \, dx}{d (4+n)}\\ &=-\frac {a B \cos (e+f x) (d \sin (e+f x))^{1+n} (a+a \sin (e+f x))^2}{d f (4+n)}-\frac {(A (4+n)+B (6+n)) \cos (e+f x) (d \sin (e+f x))^{1+n} \left (a^3+a^3 \sin (e+f x)\right )}{d f (3+n) (4+n)}+\frac {\int (d \sin (e+f x))^n (a+a \sin (e+f x)) \left (a^2 d^2 \left (2 A \left (8+6 n+n^2\right )+B \left (9+11 n+2 n^2\right )\right )+a^2 d^2 \left (B \left (27+14 n+2 n^2\right )+A \left (28+15 n+2 n^2\right )\right ) \sin (e+f x)\right ) \, dx}{d^2 (3+n) (4+n)}\\ &=-\frac {a B \cos (e+f x) (d \sin (e+f x))^{1+n} (a+a \sin (e+f x))^2}{d f (4+n)}-\frac {(A (4+n)+B (6+n)) \cos (e+f x) (d \sin (e+f x))^{1+n} \left (a^3+a^3 \sin (e+f x)\right )}{d f (3+n) (4+n)}+\frac {\int (d \sin (e+f x))^n \left (a^3 d^2 \left (2 A \left (8+6 n+n^2\right )+B \left (9+11 n+2 n^2\right )\right )+\left (a^3 d^2 \left (2 A \left (8+6 n+n^2\right )+B \left (9+11 n+2 n^2\right )\right )+a^3 d^2 \left (B \left (27+14 n+2 n^2\right )+A \left (28+15 n+2 n^2\right )\right )\right ) \sin (e+f x)+a^3 d^2 \left (B \left (27+14 n+2 n^2\right )+A \left (28+15 n+2 n^2\right )\right ) \sin ^2(e+f x)\right ) \, dx}{d^2 (3+n) (4+n)}\\ &=-\frac {a^3 \left (B \left (27+14 n+2 n^2\right )+A \left (28+15 n+2 n^2\right )\right ) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (2+n) (3+n) (4+n)}-\frac {a B \cos (e+f x) (d \sin (e+f x))^{1+n} (a+a \sin (e+f x))^2}{d f (4+n)}-\frac {(A (4+n)+B (6+n)) \cos (e+f x) (d \sin (e+f x))^{1+n} \left (a^3+a^3 \sin (e+f x)\right )}{d f (3+n) (4+n)}+\frac {\int (d \sin (e+f x))^n \left (a^3 d^3 (3+n) \left (B \left (15+19 n+4 n^2\right )+A \left (20+21 n+4 n^2\right )\right )+a^3 d^3 (2+n) (4+n) (B (9+4 n)+A (11+4 n)) \sin (e+f x)\right ) \, dx}{d^3 (2+n) (3+n) (4+n)}\\ &=-\frac {a^3 \left (B \left (27+14 n+2 n^2\right )+A \left (28+15 n+2 n^2\right )\right ) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (2+n) (3+n) (4+n)}-\frac {a B \cos (e+f x) (d \sin (e+f x))^{1+n} (a+a \sin (e+f x))^2}{d f (4+n)}-\frac {(A (4+n)+B (6+n)) \cos (e+f x) (d \sin (e+f x))^{1+n} \left (a^3+a^3 \sin (e+f x)\right )}{d f (3+n) (4+n)}+\frac {\left (a^3 (B (9+4 n)+A (11+4 n))\right ) \int (d \sin (e+f x))^{1+n} \, dx}{d (3+n)}+\frac {\left (a^3 \left (B \left (15+19 n+4 n^2\right )+A \left (20+21 n+4 n^2\right )\right )\right ) \int (d \sin (e+f x))^n \, dx}{(2+n) (4+n)}\\ &=-\frac {a^3 \left (B \left (27+14 n+2 n^2\right )+A \left (28+15 n+2 n^2\right )\right ) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (2+n) (3+n) (4+n)}+\frac {a^3 \left (B \left (15+19 n+4 n^2\right )+A \left (20+21 n+4 n^2\right )\right ) \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}}{d f (1+n) (2+n) (4+n) \sqrt {\cos ^2(e+f x)}}+\frac {a^3 (B (9+4 n)+A (11+4 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}}{d^2 f (2+n) (3+n) \sqrt {\cos ^2(e+f x)}}-\frac {a B \cos (e+f x) (d \sin (e+f x))^{1+n} (a+a \sin (e+f x))^2}{d f (4+n)}-\frac {(A (4+n)+B (6+n)) \cos (e+f x) (d \sin (e+f x))^{1+n} \left (a^3+a^3 \sin (e+f x)\right )}{d f (3+n) (4+n)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 1.61, size = 248, normalized size = 0.66 \begin {gather*} \frac {a^3 \cos (e+f x) \sin (e+f x) (d \sin (e+f x))^n \left (\frac {A \, _2F_1\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 {(3 A+B) \, _2F_1\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 {3 (A+B) \, _2F_1\left (\frac {1}{2},\frac {3+n}{2};\frac {5+n}{2};\sin ^2(e+f x)\right )}{3+n}+\sin (e+f x) \left (\frac {(A+3 B) \, _2F_1\left (\frac {1}{2},\frac {4+n}{2};\frac {6+n}{2};\sin ^2(e+f x)\right )}{4+n}+\frac {B \, _2F_1\left (\frac {1}{2},\frac {5+n}{2};\frac {7+n}{2};\sin ^2(e+f x)\right ) \sin (e+f x)}{5+n}\right )\right )\right )\right )}{f \sqrt {\cos ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d*Sin[e + f*x])^n*(a + a*Sin[e + f*x])^3*(A + B*Sin[e + f*x]),x]

[Out]

(a^3*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]*(((3*A + B)*Hypergeometric2F1[1/2, (2 + n)/2, (4 + n)/2, Sin[e + f*x]^2])/(2 + n)
+ Sin[e + f*x]*((3*(A + B)*Hypergeometric2F1[1/2, (3 + n)/2, (5 + n)/2, Sin[e + f*x]^2])/(3 + n) + Sin[e + f*
x]*(((A + 3*B)*Hypergeometric2F1[1/2, (4 + n)/2, (6 + n)/2, Sin[e + f*x]^2])/(4 + n) + (B*Hypergeometric2F1[1/
2, (5 + n)/2, (7 + n)/2, Sin[e + f*x]^2]*Sin[e + f*x])/(5 + n))))))/(f*Sqrt[Cos[e + f*x]^2])

________________________________________________________________________________________

Maple [F]
time = 0.83, size = 0, normalized size = 0.00 \[\int \left (d \sin \left (f x +e \right )\right )^{n} \left (a +a \sin \left (f x +e \right )\right )^{3} \left (A +B \sin \left (f x +e \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sin(f*x+e))^n*(a+a*sin(f*x+e))^3*(A+B*sin(f*x+e)),x)

[Out]

int((d*sin(f*x+e))^n*(a+a*sin(f*x+e))^3*(A+B*sin(f*x+e)),x)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sin(f*x+e))^n*(a+a*sin(f*x+e))^3*(A+B*sin(f*x+e)),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^3*(d*sin(f*x + e))^n, x)

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sin(f*x+e))^n*(a+a*sin(f*x+e))^3*(A+B*sin(f*x+e)),x, algorithm="fricas")

[Out]

integral((B*a^3*cos(f*x + e)^4 - (3*A + 5*B)*a^3*cos(f*x + e)^2 + 4*(A + B)*a^3 - ((A + 3*B)*a^3*cos(f*x + e)^
2 - 4*(A + B)*a^3)*sin(f*x + e))*(d*sin(f*x + e))^n, x)

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sin(f*x+e))**n*(a+a*sin(f*x+e))**3*(A+B*sin(f*x+e)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sin(f*x+e))^n*(a+a*sin(f*x+e))^3*(A+B*sin(f*x+e)),x, algorithm="giac")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^3*(d*sin(f*x + e))^n, x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \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 )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*sin(e + f*x))^n*(A + B*sin(e + f*x))*(a + a*sin(e + f*x))^3,x)

[Out]

int((d*sin(e + f*x))^n*(A + B*sin(e + f*x))*(a + a*sin(e + f*x))^3, x)

________________________________________________________________________________________