\(\int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c+d \sin (e+f x))^n \, dx\) [332]

Optimal result

Integrand size = 37, antiderivative size = 427 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=-\frac {2 a^2 (A-B) \cos (e+f x) (c+d \sin (e+f x))^{1+n}}{d f (3+2 n) \sqrt {a+a \sin (e+f x)}}+\frac {2 a^2 B (3 c-d (11+4 n)) \cos (e+f x) (c+d \sin (e+f x))^{1+n}}{d^2 f (3+2 n) (5+2 n) \sqrt {a+a \sin (e+f x)}}-\frac {2 a B \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{1+n}}{d f (5+2 n)}+\frac {2 a^2 (A-B) (c-d (5+4 n)) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},\frac {d (1-\sin (e+f x))}{c+d}\right ) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{d f (3+2 n) \sqrt {a+a \sin (e+f x)}}-\frac {2 a^2 B \left (3 c^2-2 c d (7+4 n)+d^2 \left (43+56 n+16 n^2\right )\right ) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},\frac {d (1-\sin (e+f x))}{c+d}\right ) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{d^2 f (3+2 n) (5+2 n) \sqrt {a+a \sin (e+f x)}} \]

[Out]

-2*a^2*(A-B)*cos(f*x+e)*(c+d*sin(f*x+e))^(1+n)/d/f/(3+2*n)/(a+a*sin(f*x+e))^(1/2)+2*a^2*B*(3*c-d*(11+4*n))*cos
(f*x+e)*(c+d*sin(f*x+e))^(1+n)/d^2/f/(4*n^2+16*n+15)/(a+a*sin(f*x+e))^(1/2)+2*a^2*(A-B)*(c-d*(5+4*n))*cos(f*x+
e)*hypergeom([1/2, -n],[3/2],d*(1-sin(f*x+e))/(c+d))*(c+d*sin(f*x+e))^n/d/f/(3+2*n)/(((c+d*sin(f*x+e))/(c+d))^
n)/(a+a*sin(f*x+e))^(1/2)-2*a^2*B*(3*c^2-2*c*d*(7+4*n)+d^2*(16*n^2+56*n+43))*cos(f*x+e)*hypergeom([1/2, -n],[3
/2],d*(1-sin(f*x+e))/(c+d))*(c+d*sin(f*x+e))^n/d^2/f/(4*n^2+16*n+15)/(((c+d*sin(f*x+e))/(c+d))^n)/(a+a*sin(f*x
+e))^(1/2)-2*a*B*cos(f*x+e)*(c+d*sin(f*x+e))^(1+n)*(a+a*sin(f*x+e))^(1/2)/d/f/(5+2*n)

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 427, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.189, Rules used = {3066, 2842, 21, 2855, 72, 71, 3060} \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=\frac {2 a^2 (A-B) (c-d (4 n+5)) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},\frac {d (1-\sin (e+f x))}{c+d}\right )}{d f (2 n+3) \sqrt {a \sin (e+f x)+a}}-\frac {2 a^2 (A-B) \cos (e+f x) (c+d \sin (e+f x))^{n+1}}{d f (2 n+3) \sqrt {a \sin (e+f x)+a}}-\frac {2 a^2 B \left (3 c^2-2 c d (4 n+7)+d^2 \left (16 n^2+56 n+43\right )\right ) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},\frac {d (1-\sin (e+f x))}{c+d}\right )}{d^2 f (2 n+3) (2 n+5) \sqrt {a \sin (e+f x)+a}}+\frac {2 a^2 B (3 c-d (4 n+11)) \cos (e+f x) (c+d \sin (e+f x))^{n+1}}{d^2 f (2 n+3) (2 n+5) \sqrt {a \sin (e+f x)+a}}-\frac {2 a B \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{n+1}}{d f (2 n+5)} \]

[In]

Int[(a + a*Sin[e + f*x])^(3/2)*(A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^n,x]

[Out]

(-2*a^2*(A - B)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(1 + n))/(d*f*(3 + 2*n)*Sqrt[a + a*Sin[e + f*x]]) + (2*a^2*B
*(3*c - d*(11 + 4*n))*Cos[e + f*x]*(c + d*Sin[e + f*x])^(1 + n))/(d^2*f*(3 + 2*n)*(5 + 2*n)*Sqrt[a + a*Sin[e +
f*x]]) - (2*a*B*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(1 + n))/(d*f*(5 + 2*n)) + (2*a^2*
(A - B)*(c - d*(5 + 4*n))*Cos[e + f*x]*Hypergeometric2F1[1/2, -n, 3/2, (d*(1 - Sin[e + f*x]))/(c + d)]*(c + d*
Sin[e + f*x])^n)/(d*f*(3 + 2*n)*Sqrt[a + a*Sin[e + f*x]]*((c + d*Sin[e + f*x])/(c + d))^n) - (2*a^2*B*(3*c^2 -
2*c*d*(7 + 4*n) + d^2*(43 + 56*n + 16*n^2))*Cos[e + f*x]*Hypergeometric2F1[1/2, -n, 3/2, (d*(1 - Sin[e + f*x]
))/(c + d)]*(c + d*Sin[e + f*x])^n)/(d^2*f*(3 + 2*n)*(5 + 2*n)*Sqrt[a + a*Sin[e + f*x]]*((c + d*Sin[e + f*x])/
(c + d))^n)

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 72

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n]), Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c -
a*d)), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 2842

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Dist[1/(
d*(m + n)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c*(m - 2) + b^2*d*(n + 1) + a^2*d
*(m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Sin[e + f*x], x], 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] && GtQ[m, 1] &&  !LtQ[n, -1] && (IntegersQ[2*m,
2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))

Rule 2855

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist
[a^2*(Cos[e + f*x]/(f*Sqrt[a + b*Sin[e + f*x]]*Sqrt[a - b*Sin[e + f*x]])), Subst[Int[(c + d*x)^n/Sqrt[a - b*x]
, 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[2*n]

Rule 3060

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt
[a + b*Sin[e + f*x]])), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b*d*(2*n + 3)), Int[Sqrt[a + b*
Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, 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[n, -1]

Rule 3066

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]

Rubi steps \begin{align*} \text {integral}& = (A-B) \int (a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^n \, dx+\frac {B \int (a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x))^n \, dx}{a} \\ & = -\frac {2 a^2 (A-B) \cos (e+f x) (c+d \sin (e+f x))^{1+n}}{d f (3+2 n) \sqrt {a+a \sin (e+f x)}}-\frac {2 a B \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{1+n}}{d f (5+2 n)}+\frac {(2 (A-B)) \int \frac {(c+d \sin (e+f x))^n \left (-\frac {1}{2} a^2 (c-5 d-4 d n)-\frac {1}{2} a^2 (c-5 d-4 d n) \sin (e+f x)\right )}{\sqrt {a+a \sin (e+f x)}} \, dx}{d (3+2 n)}+\frac {(2 B) \int \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^n \left (\frac {1}{2} a^2 (c+d (7+4 n))-\frac {1}{2} a^2 (3 c-11 d-4 d n) \sin (e+f x)\right ) \, dx}{a d (5+2 n)} \\ & = -\frac {2 a^2 (A-B) \cos (e+f x) (c+d \sin (e+f x))^{1+n}}{d f (3+2 n) \sqrt {a+a \sin (e+f x)}}+\frac {2 a^2 B (3 c-d (11+4 n)) \cos (e+f x) (c+d \sin (e+f x))^{1+n}}{d^2 f (3+2 n) (5+2 n) \sqrt {a+a \sin (e+f x)}}-\frac {2 a B \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{1+n}}{d f (5+2 n)}-\frac {(a (A-B) (c-d (5+4 n))) \int \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^n \, dx}{d (3+2 n)}+\frac {\left (a B \left (3 c^2-2 c d (7+4 n)+d^2 \left (43+56 n+16 n^2\right )\right )\right ) \int \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^n \, dx}{d^2 (3+2 n) (5+2 n)} \\ & = -\frac {2 a^2 (A-B) \cos (e+f x) (c+d \sin (e+f x))^{1+n}}{d f (3+2 n) \sqrt {a+a \sin (e+f x)}}+\frac {2 a^2 B (3 c-d (11+4 n)) \cos (e+f x) (c+d \sin (e+f x))^{1+n}}{d^2 f (3+2 n) (5+2 n) \sqrt {a+a \sin (e+f x)}}-\frac {2 a B \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{1+n}}{d f (5+2 n)}-\frac {\left (a^3 (A-B) (c-d (5+4 n)) \cos (e+f x)\right ) \text {Subst}\left (\int \frac {(c+d x)^n}{\sqrt {a-a x}} \, dx,x,\sin (e+f x)\right )}{d f (3+2 n) \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}+\frac {\left (a^3 B \left (3 c^2-2 c d (7+4 n)+d^2 \left (43+56 n+16 n^2\right )\right ) \cos (e+f x)\right ) \text {Subst}\left (\int \frac {(c+d x)^n}{\sqrt {a-a x}} \, dx,x,\sin (e+f x)\right )}{d^2 f (3+2 n) (5+2 n) \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}} \\ & = -\frac {2 a^2 (A-B) \cos (e+f x) (c+d \sin (e+f x))^{1+n}}{d f (3+2 n) \sqrt {a+a \sin (e+f x)}}+\frac {2 a^2 B (3 c-d (11+4 n)) \cos (e+f x) (c+d \sin (e+f x))^{1+n}}{d^2 f (3+2 n) (5+2 n) \sqrt {a+a \sin (e+f x)}}-\frac {2 a B \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{1+n}}{d f (5+2 n)}-\frac {\left (a^3 (A-B) (c-d (5+4 n)) \cos (e+f x) (c+d \sin (e+f x))^n \left (-\frac {a (c+d \sin (e+f x))}{-a c-a d}\right )^{-n}\right ) \text {Subst}\left (\int \frac {\left (\frac {c}{c+d}+\frac {d x}{c+d}\right )^n}{\sqrt {a-a x}} \, dx,x,\sin (e+f x)\right )}{d f (3+2 n) \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}+\frac {\left (a^3 B \left (3 c^2-2 c d (7+4 n)+d^2 \left (43+56 n+16 n^2\right )\right ) \cos (e+f x) (c+d \sin (e+f x))^n \left (-\frac {a (c+d \sin (e+f x))}{-a c-a d}\right )^{-n}\right ) \text {Subst}\left (\int \frac {\left (\frac {c}{c+d}+\frac {d x}{c+d}\right )^n}{\sqrt {a-a x}} \, dx,x,\sin (e+f x)\right )}{d^2 f (3+2 n) (5+2 n) \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}} \\ & = -\frac {2 a^2 (A-B) \cos (e+f x) (c+d \sin (e+f x))^{1+n}}{d f (3+2 n) \sqrt {a+a \sin (e+f x)}}+\frac {2 a^2 B (3 c-d (11+4 n)) \cos (e+f x) (c+d \sin (e+f x))^{1+n}}{d^2 f (3+2 n) (5+2 n) \sqrt {a+a \sin (e+f x)}}-\frac {2 a B \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{1+n}}{d f (5+2 n)}+\frac {2 a^2 (A-B) (c-d (5+4 n)) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},\frac {d (1-\sin (e+f x))}{c+d}\right ) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{d f (3+2 n) \sqrt {a+a \sin (e+f x)}}-\frac {2 a^2 B \left (3 c^2-2 c d (7+4 n)+d^2 \left (43+56 n+16 n^2\right )\right ) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},\frac {d (1-\sin (e+f x))}{c+d}\right ) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n}}{d^2 f (3+2 n) (5+2 n) \sqrt {a+a \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 11.79 (sec) , antiderivative size = 245, normalized size of antiderivative = 0.57 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{-n} \left (-30 (A+B) (c-d (5+4 n)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-n,\frac {3}{2},-\frac {d (-1+\sin (e+f x))}{c+d}\right )+6 B d (3+2 n) \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-n,\frac {7}{2},-\frac {d (-1+\sin (e+f x))}{c+d}\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4+20 B d (3+2 n) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-n,\frac {5}{2},-\frac {d (-1+\sin (e+f x))}{c+d}\right ) (-1+\sin (e+f x))+30 (A+B) (c+d) \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{1+n}\right )}{15 d f (3+2 n) \sqrt {a (1+\sin (e+f x))}} \]

[In]

Integrate[(a + a*Sin[e + f*x])^(3/2)*(A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^n,x]

[Out]

-1/15*(a^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^n*(-30*(A + B)*(c - d*(5 + 4*n))*Hypergeometric2F1[1/2, -n, 3/2,
-((d*(-1 + Sin[e + f*x]))/(c + d))] + 6*B*d*(3 + 2*n)*Hypergeometric2F1[5/2, -n, 7/2, -((d*(-1 + Sin[e + f*x])
)/(c + d))]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4 + 20*B*d*(3 + 2*n)*Hypergeometric2F1[3/2, -n, 5/2, -((d*(-
1 + Sin[e + f*x]))/(c + d))]*(-1 + Sin[e + f*x]) + 30*(A + B)*(c + d)*((c + d*Sin[e + f*x])/(c + d))^(1 + n)))
/(d*f*(3 + 2*n)*Sqrt[a*(1 + Sin[e + f*x])]*((c + d*Sin[e + f*x])/(c + d))^n)

Maple [F]

\[\int \left (a +a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \left (A +B \sin \left (f x +e \right )\right ) \left (c +d \sin \left (f x +e \right )\right )^{n}d x\]

[In]

int((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c+d*sin(f*x+e))^n,x)

[Out]

int((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c+d*sin(f*x+e))^n,x)

Fricas [F]

\[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (d \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c+d*sin(f*x+e))^n,x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(f*x+e))**(3/2)*(A+B*sin(f*x+e))*(c+d*sin(f*x+e))**n,x)

[Out]

Timed out

Maxima [F]

\[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (d \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c+d*sin(f*x+e))^n,x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^(3/2)*(d*sin(f*x + e) + c)^n, x)

Giac [F]

\[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (d \sin \left (f x + e\right ) + c\right )}^{n} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))*(c+d*sin(f*x+e))^n,x, algorithm="giac")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^(3/2)*(d*sin(f*x + e) + c)^n, x)

Mupad [F(-1)]

Timed out. \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) (c+d \sin (e+f x))^n \, dx=\int \left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^n \,d x \]

[In]

int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(3/2)*(c + d*sin(e + f*x))^n,x)

[Out]

int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(3/2)*(c + d*sin(e + f*x))^n, x)