3.25 \(\int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n (A+B \sin (e+f x)+C \sin ^2(e+f x)) \, dx\)
Optimal. Leaf size=383 \[ \frac{\sqrt{2} \cos (e+f x) (a \sin (e+f x)+a)^m (d (A (m+n+2)-B (m+n+2)+C (-m+n+1))+c (2 C m+C)) (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c-d}\right )^{-n} F_1\left (m+\frac{1}{2};\frac{1}{2},-n;m+\frac{3}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{d f (2 m+1) (m+n+2) \sqrt{1-\sin (e+f x)}}-\frac{\sqrt{2} \cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c C (m+1)-d (B (m+n+2)+C m)) (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c-d}\right )^{-n} F_1\left (m+\frac{3}{2};\frac{1}{2},-n;m+\frac{5}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{a d f (2 m+3) (m+n+2) \sqrt{1-\sin (e+f x)}}-\frac{C \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{n+1}}{d f (m+n+2)} \]
[Out]
-((C*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(1 + n))/(d*f*(2 + m + n))) + (Sqrt[2]*(c*(C + 2
*C*m) + d*(C*(1 - m + n) + A*(2 + m + n) - B*(2 + m + n)))*AppellF1[1/2 + m, 1/2, -n, 3/2 + m, (1 + Sin[e + f*
x])/2, -((d*(1 + Sin[e + f*x]))/(c - d))]*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n)/(d*f*(1
+ 2*m)*(2 + m + n)*Sqrt[1 - Sin[e + f*x]]*((c + d*Sin[e + f*x])/(c - d))^n) - (Sqrt[2]*(c*C*(1 + m) - d*(C*m +
B*(2 + m + n)))*AppellF1[3/2 + m, 1/2, -n, 5/2 + m, (1 + Sin[e + f*x])/2, -((d*(1 + Sin[e + f*x]))/(c - d))]*
Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m)*(c + d*Sin[e + f*x])^n)/(a*d*f*(3 + 2*m)*(2 + m + n)*Sqrt[1 - Sin[e
+ f*x]]*((c + d*Sin[e + f*x])/(c - d))^n)
________________________________________________________________________________________
Rubi [A] time = 0.903591, antiderivative size = 381, normalized size of antiderivative =
0.99, number of steps used = 10, number of rules used = 6, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules
used = {3045, 2987, 2788, 140, 139, 138} \[ \frac{\sqrt{2} \cos (e+f x) (a \sin (e+f x)+a)^m (d (A (m+n+2)-B (m+n+2)+C (-m+n+1))+c (2 C m+C)) (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c-d}\right )^{-n} F_1\left (m+\frac{1}{2};\frac{1}{2},-n;m+\frac{3}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{d f (2 m+1) (m+n+2) \sqrt{1-\sin (e+f x)}}+\frac{\sqrt{2} \cos (e+f x) (a \sin (e+f x)+a)^{m+1} (B d (m+n+2)-c C (m+1)+C d m) (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c-d}\right )^{-n} F_1\left (m+\frac{3}{2};\frac{1}{2},-n;m+\frac{5}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{a d f (2 m+3) (m+n+2) \sqrt{1-\sin (e+f x)}}-\frac{C \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{n+1}}{d f (m+n+2)} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n*(A + B*Sin[e + f*x] + C*Sin[e + f*x]^2),x]
[Out]
-((C*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(1 + n))/(d*f*(2 + m + n))) + (Sqrt[2]*(c*(C + 2
*C*m) + d*(C*(1 - m + n) + A*(2 + m + n) - B*(2 + m + n)))*AppellF1[1/2 + m, 1/2, -n, 3/2 + m, (1 + Sin[e + f*
x])/2, -((d*(1 + Sin[e + f*x]))/(c - d))]*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n)/(d*f*(1
+ 2*m)*(2 + m + n)*Sqrt[1 - Sin[e + f*x]]*((c + d*Sin[e + f*x])/(c - d))^n) + (Sqrt[2]*(C*d*m - c*C*(1 + m) +
B*d*(2 + m + n))*AppellF1[3/2 + m, 1/2, -n, 5/2 + m, (1 + Sin[e + f*x])/2, -((d*(1 + Sin[e + f*x]))/(c - d))]*
Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m)*(c + d*Sin[e + f*x])^n)/(a*d*f*(3 + 2*m)*(2 + m + n)*Sqrt[1 - Sin[e
+ f*x]]*((c + d*Sin[e + f*x])/(c - d))^n)
Rule 3045
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((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*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(b*d*(m + n + 2)), Int[(a + b*Sin[e + f*x
])^m*(c + d*Sin[e + f*x])^n*Simp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1)) + (C*(a*d*m - b*c*(m + 1)) + b*B*
d*(m + n + 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[m, -2^(-1)] && NeQ[m + n + 2, 0]
Rule 2987
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]
Rule 2788
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dis
t[(a^2*Cos[e + f*x])/(f*Sqrt[a + b*Sin[e + f*x]]*Sqrt[a - b*Sin[e + f*x]]), Subst[Int[((a + b*x)^(m - 1/2)*(c
+ d*x)^n)/Sqrt[a - b*x], x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a*d, 0] &
& EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !IntegerQ[m]
Rule 140
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), 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*((b*c)/(b*c
- a*d) + (b*d*x)/(b*c - a*d))^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m]
&& !IntegerQ[n] && !IntegerQ[p] && !GtQ[b/(b*c - a*d), 0] && !SimplerQ[c + d*x, a + b*x] && !SimplerQ[e +
f*x, a + b*x]
Rule 139
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(e + f*x)^
FracPart[p]/((b/(b*e - a*f))^IntPart[p]*((b*(e + f*x))/(b*e - a*f))^FracPart[p]), Int[(a + b*x)^m*(c + d*x)^n*
((b*e)/(b*e - a*f) + (b*f*x)/(b*e - a*f))^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m]
&& !IntegerQ[n] && !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && !GtQ[b/(b*e - a*f), 0]
Rule 138
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*(m + 1
)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p), x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !Inte
gerQ[n] && !IntegerQ[p] && GtQ[b/(b*c - a*d), 0] && GtQ[b/(b*e - a*f), 0] && !(GtQ[d/(d*a - c*b), 0] && GtQ[
d/(d*e - c*f), 0] && SimplerQ[c + d*x, a + b*x]) && !(GtQ[f/(f*a - e*b), 0] && GtQ[f/(f*c - e*d), 0] && Simpl
erQ[e + f*x, a + b*x])
Rubi steps
\begin{align*} \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx &=-\frac{C \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{1+n}}{d f (2+m+n)}+\frac{\int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n (a (A d (2+m+n)+C (d+c m+d n))+a (C d m-c C (1+m)+B d (2+m+n)) \sin (e+f x)) \, dx}{a d (2+m+n)}\\ &=-\frac{C \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{1+n}}{d f (2+m+n)}+\frac{(C d m-c C (1+m)+B d (2+m+n)) \int (a+a \sin (e+f x))^{1+m} (c+d \sin (e+f x))^n \, dx}{a d (2+m+n)}+\frac{(c (C+2 C m)+d (C (1-m+n)+A (2+m+n)-B (2+m+n))) \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx}{d (2+m+n)}\\ &=-\frac{C \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{1+n}}{d f (2+m+n)}+\frac{(a (C d m-c C (1+m)+B d (2+m+n)) \cos (e+f x)) \operatorname{Subst}\left (\int \frac{(a+a x)^{\frac{1}{2}+m} (c+d x)^n}{\sqrt{a-a x}} \, dx,x,\sin (e+f x)\right )}{d f (2+m+n) \sqrt{a-a \sin (e+f x)} \sqrt{a+a \sin (e+f x)}}+\frac{\left (a^2 (c (C+2 C m)+d (C (1-m+n)+A (2+m+n)-B (2+m+n))) \cos (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{-\frac{1}{2}+m} (c+d x)^n}{\sqrt{a-a x}} \, dx,x,\sin (e+f x)\right )}{d f (2+m+n) \sqrt{a-a \sin (e+f x)} \sqrt{a+a \sin (e+f x)}}\\ &=-\frac{C \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{1+n}}{d f (2+m+n)}+\frac{\left (a (C d m-c C (1+m)+B d (2+m+n)) \cos (e+f x) \sqrt{\frac{a-a \sin (e+f x)}{a}}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{\frac{1}{2}+m} (c+d x)^n}{\sqrt{\frac{1}{2}-\frac{x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt{2} d f (2+m+n) (a-a \sin (e+f x)) \sqrt{a+a \sin (e+f x)}}+\frac{\left (a^2 (c (C+2 C m)+d (C (1-m+n)+A (2+m+n)-B (2+m+n))) \cos (e+f x) \sqrt{\frac{a-a \sin (e+f x)}{a}}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{-\frac{1}{2}+m} (c+d x)^n}{\sqrt{\frac{1}{2}-\frac{x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt{2} d f (2+m+n) (a-a \sin (e+f x)) \sqrt{a+a \sin (e+f x)}}\\ &=-\frac{C \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{1+n}}{d f (2+m+n)}+\frac{\left (a (C d m-c C (1+m)+B d (2+m+n)) \cos (e+f x) \sqrt{\frac{a-a \sin (e+f x)}{a}} (c+d \sin (e+f x))^n \left (\frac{a (c+d \sin (e+f x))}{a c-a d}\right )^{-n}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{\frac{1}{2}+m} \left (\frac{a c}{a c-a d}+\frac{a d x}{a c-a d}\right )^n}{\sqrt{\frac{1}{2}-\frac{x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt{2} d f (2+m+n) (a-a \sin (e+f x)) \sqrt{a+a \sin (e+f x)}}+\frac{\left (a^2 (c (C+2 C m)+d (C (1-m+n)+A (2+m+n)-B (2+m+n))) \cos (e+f x) \sqrt{\frac{a-a \sin (e+f x)}{a}} (c+d \sin (e+f x))^n \left (\frac{a (c+d \sin (e+f x))}{a c-a d}\right )^{-n}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{-\frac{1}{2}+m} \left (\frac{a c}{a c-a d}+\frac{a d x}{a c-a d}\right )^n}{\sqrt{\frac{1}{2}-\frac{x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt{2} d f (2+m+n) (a-a \sin (e+f x)) \sqrt{a+a \sin (e+f x)}}\\ &=-\frac{C \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{1+n}}{d f (2+m+n)}+\frac{\sqrt{2} (c (C+2 C m)+d (C (1-m+n)+A (2+m+n)-B (2+m+n))) F_1\left (\frac{1}{2}+m;\frac{1}{2},-n;\frac{3}{2}+m;\frac{1}{2} (1+\sin (e+f x)),-\frac{d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c-d}\right )^{-n}}{d f (1+2 m) (2+m+n) \sqrt{1-\sin (e+f x)}}+\frac{\sqrt{2} (C d m-c C (1+m)+B d (2+m+n)) F_1\left (\frac{3}{2}+m;\frac{1}{2},-n;\frac{5}{2}+m;\frac{1}{2} (1+\sin (e+f x)),-\frac{d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) \sqrt{1-\sin (e+f x)} (a+a \sin (e+f x))^{1+m} (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c-d}\right )^{-n}}{d f (3+2 m) (2+m+n) (a-a \sin (e+f x))}\\ \end{align*}
Mathematica [B] time = 8.77524, size = 2572, normalized size = 6.72 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n*(A + B*Sin[e + f*x] + C*Sin[e + f*x]^2),x]
[Out]
-(((-4*B*AppellF1[3/2, (1 - 2*m)/2, -n, 5/2, Sin[(-e + Pi/2 - f*x)/2]^2, (2*d*Sin[(-e + Pi/2 - f*x)/2]^2)/(c +
d)]*Cos[(-e + Pi/2 - f*x)/2]^(-1 + 2*m)*(Cos[(-e + Pi/2 - f*x)/2]^2)^((1 - 2*m)/2)*Sin[(-e + Pi/2 - f*x)/2]^3
*(1 - Sin[(-e + Pi/2 - f*x)/2]^2)^((1 - 2*m)/2 + (-1 + 2*m)/2)*(c + d - 2*d*Sin[(-e + Pi/2 - f*x)/2]^2)^n)/(3*
((c + d - 2*d*Sin[(-e + Pi/2 - f*x)/2]^2)/(c + d))^n) + (2*C*AppellF1[5/2, (1 - 2*m)/2, -n, 7/2, Sin[(-e + Pi/
2 - f*x)/2]^2, (2*d*Sin[(-e + Pi/2 - f*x)/2]^2)/(c + d)]*Cos[(-e + Pi/2 - f*x)/2]^(-1 + 2*m)*(Cos[(-e + Pi/2 -
f*x)/2]^2)^((1 - 2*m)/2)*Sin[(-e + Pi/2 - f*x)/2]^5*(1 - Sin[(-e + Pi/2 - f*x)/2]^2)^((1 - 2*m)/2 + (-1 + 2*m
)/2)*(c + d - 2*d*Sin[(-e + Pi/2 - f*x)/2]^2)^n)/(5*((c + d - 2*d*Sin[(-e + Pi/2 - f*x)/2]^2)/(c + d))^n) - (4
*C*AppellF1[3/2, (-1 - 2*m)/2, -n, 5/2, Sin[(-e + Pi/2 - f*x)/2]^2, (2*d*Sin[(-e + Pi/2 - f*x)/2]^2)/(c + d)]*
Cos[(-e + Pi/2 - f*x)/2]^(1 + 2*m)*(Cos[(-e + Pi/2 - f*x)/2]^2)^((-1 - 2*m)/2)*Sin[(-e + Pi/2 - f*x)/2]^3*(1 -
Sin[(-e + Pi/2 - f*x)/2]^2)^((-1 - 2*m)/2 + (1 + 2*m)/2)*(c + d - 2*d*Sin[(-e + Pi/2 - f*x)/2]^2)^n)/((c + d
- 2*d*Sin[(-e + Pi/2 - f*x)/2]^2)/(c + d))^n - (6*C*(c + d)*AppellF1[1/2, -3/2 - m, -n, 3/2, Sin[(-e + Pi/2 -
f*x)/2]^2, (2*d*Sin[(-e + Pi/2 - f*x)/2]^2)/(c + d)]*Cos[(-e + Pi/2 - f*x)/2]^(3 + 2*m)*(Cos[(-e + Pi/2 - f*x)
/2]^2)^(1/2 + (-4 - 2*m)/2)*Sin[(-e + Pi/2 - f*x)/2]*(1 - Sin[(-e + Pi/2 - f*x)/2]^2)^(3/2 + m)*(c + d - 2*d*S
in[(-e + Pi/2 - f*x)/2]^2)^n)/(-3*(c + d)*AppellF1[1/2, -3/2 - m, -n, 3/2, Sin[(-e + Pi/2 - f*x)/2]^2, (2*d*Si
n[(-e + Pi/2 - f*x)/2]^2)/(c + d)] + (4*d*n*AppellF1[3/2, -3/2 - m, 1 - n, 5/2, Sin[(-e + Pi/2 - f*x)/2]^2, (2
*d*Sin[(-e + Pi/2 - f*x)/2]^2)/(c + d)] + (c + d)*(3 + 2*m)*AppellF1[3/2, -1/2 - m, -n, 5/2, Sin[(-e + Pi/2 -
f*x)/2]^2, (2*d*Sin[(-e + Pi/2 - f*x)/2]^2)/(c + d)])*Sin[(-e + Pi/2 - f*x)/2]^2) - (12*B*(c + d)*AppellF1[1/2
, -1/2 - m, -n, 3/2, Sin[(-e + Pi/2 - f*x)/2]^2, (2*d*Sin[(-e + Pi/2 - f*x)/2]^2)/(c + d)]*Cos[(-e + Pi/2 - f*
x)/2]^(1 + 2*m)*(Cos[(-e + Pi/2 - f*x)/2]^2)^(1/2 + (-2 - 2*m)/2)*Sin[(-e + Pi/2 - f*x)/2]*(1 - Sin[(-e + Pi/2
- f*x)/2]^2)^(1/2 + m)*(c + d - 2*d*Sin[(-e + Pi/2 - f*x)/2]^2)^n)/(-3*(c + d)*AppellF1[1/2, -1/2 - m, -n, 3/
2, Sin[(-e + Pi/2 - f*x)/2]^2, (2*d*Sin[(-e + Pi/2 - f*x)/2]^2)/(c + d)] + (4*d*n*AppellF1[3/2, -1/2 - m, 1 -
n, 5/2, Sin[(-e + Pi/2 - f*x)/2]^2, (2*d*Sin[(-e + Pi/2 - f*x)/2]^2)/(c + d)] + (c + d)*(1 + 2*m)*AppellF1[3/2
, 1/2 - m, -n, 5/2, Sin[(-e + Pi/2 - f*x)/2]^2, (2*d*Sin[(-e + Pi/2 - f*x)/2]^2)/(c + d)])*Sin[(-e + Pi/2 - f*
x)/2]^2) + (12*A*(c + d)*AppellF1[1/2, 1/2 - m, -n, 3/2, Sin[(-e + Pi/2 - f*x)/2]^2, (2*d*Sin[(-e + Pi/2 - f*x
)/2]^2)/(c + d)]*Cos[(-e + Pi/2 - f*x)/2]^(-1 + 2*m)*(Cos[(-e + Pi/2 - f*x)/2]^2)^(1/2 - m)*Sin[(-e + Pi/2 - f
*x)/2]*(1 - Sin[(-e + Pi/2 - f*x)/2]^2)^(-1/2 + m)*(c + d - 2*d*Sin[(-e + Pi/2 - f*x)/2]^2)^n)/(3*(c + d)*Appe
llF1[1/2, 1/2 - m, -n, 3/2, Sin[(-e + Pi/2 - f*x)/2]^2, (2*d*Sin[(-e + Pi/2 - f*x)/2]^2)/(c + d)] - (4*d*n*App
ellF1[3/2, 1/2 - m, 1 - n, 5/2, Sin[(-e + Pi/2 - f*x)/2]^2, (2*d*Sin[(-e + Pi/2 - f*x)/2]^2)/(c + d)] + (c + d
)*(-1 + 2*m)*AppellF1[3/2, 3/2 - m, -n, 5/2, Sin[(-e + Pi/2 - f*x)/2]^2, (2*d*Sin[(-e + Pi/2 - f*x)/2]^2)/(c +
d)])*Sin[(-e + Pi/2 - f*x)/2]^2) + (6*C*(c + d)*AppellF1[1/2, 1/2 - m, -n, 3/2, Sin[(-e + Pi/2 - f*x)/2]^2, (
2*d*Sin[(-e + Pi/2 - f*x)/2]^2)/(c + d)]*Cos[(-e + Pi/2 - f*x)/2]^(-1 + 2*m)*(Cos[(-e + Pi/2 - f*x)/2]^2)^(1/2
- m)*Sin[(-e + Pi/2 - f*x)/2]*(1 - Sin[(-e + Pi/2 - f*x)/2]^2)^(-1/2 + m)*(c + d - 2*d*Sin[(-e + Pi/2 - f*x)/
2]^2)^n)/(3*(c + d)*AppellF1[1/2, 1/2 - m, -n, 3/2, Sin[(-e + Pi/2 - f*x)/2]^2, (2*d*Sin[(-e + Pi/2 - f*x)/2]^
2)/(c + d)] - (4*d*n*AppellF1[3/2, 1/2 - m, 1 - n, 5/2, Sin[(-e + Pi/2 - f*x)/2]^2, (2*d*Sin[(-e + Pi/2 - f*x)
/2]^2)/(c + d)] + (c + d)*(-1 + 2*m)*AppellF1[3/2, 3/2 - m, -n, 5/2, Sin[(-e + Pi/2 - f*x)/2]^2, (2*d*Sin[(-e
+ Pi/2 - f*x)/2]^2)/(c + d)])*Sin[(-e + Pi/2 - f*x)/2]^2))*(a + a*Sin[e + f*x])^m)/(2*f*Cos[(-e + Pi/2 - f*x)/
2]^(2*m))
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Maple [F] time = 0.702, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( c+d\sin \left ( fx+e \right ) \right ) ^{n} \left ( A+B\sin \left ( fx+e \right ) +C \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^n*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x)
[Out]
int((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^n*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x)
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \sin \left (f x + e\right )^{2} + B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}{\left (d \sin \left (f x + e\right ) + c\right )}^{n}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^n*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x, algorithm="maxima")
[Out]
integrate((C*sin(f*x + e)^2 + B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m*(d*sin(f*x + e) + c)^n, x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (C \cos \left (f x + e\right )^{2} - B \sin \left (f x + e\right ) - A - C\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}{\left (d \sin \left (f x + e\right ) + c\right )}^{n}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^n*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x, algorithm="fricas")
[Out]
integral(-(C*cos(f*x + e)^2 - B*sin(f*x + e) - A - C)*(a*sin(f*x + e) + a)^m*(d*sin(f*x + e) + c)^n, 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((a+a*sin(f*x+e))**m*(c+d*sin(f*x+e))**n*(A+B*sin(f*x+e)+C*sin(f*x+e)**2),x)
[Out]
Timed out
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Giac [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((a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^n*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x, algorithm="giac")
[Out]
Timed out