3.1.25 \(\int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n (A+B \sin (e+f x)+C \sin ^2(e+f x)) \, dx\) [25]
Optimal. Leaf size=383 \[ -\frac {C \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{1+n}}{d f (2+m+n)}+\frac {\sqrt {2} (c (C+2 C m)+d (C (1-m+n)+A (2+m+n)-B (2+m+n))) F_1\left (\frac {1}{2}+m;\frac {1}{2},-n;\frac {3}{2}+m;\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c-d}\right )^{-n}}{d f (1+2 m) (2+m+n) \sqrt {1-\sin (e+f x)}}-\frac {\sqrt {2} (c C (1+m)-d (C m+B (2+m+n))) F_1\left (\frac {3}{2}+m;\frac {1}{2},-n;\frac {5}{2}+m;\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) (a+a \sin (e+f x))^{1+m} (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c-d}\right )^{-n}}{a d f (3+2 m) (2+m+n) \sqrt {1-\sin (e+f x)}} \]
[Out]
-C*cos(f*x+e)*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(1+n)/d/f/(2+n+m)+(c*(2*C*m+C)+d*(C*(1-m+n)+A*(2+n+m)-B*(2+n
+m)))*AppellF1(1/2+m,-n,1/2,3/2+m,-d*(1+sin(f*x+e))/(c-d),1/2+1/2*sin(f*x+e))*cos(f*x+e)*(a+a*sin(f*x+e))^m*(c
+d*sin(f*x+e))^n*2^(1/2)/d/f/(1+2*m)/(2+n+m)/(((c+d*sin(f*x+e))/(c-d))^n)/(1-sin(f*x+e))^(1/2)-(c*C*(1+m)-d*(C
*m+B*(2+n+m)))*AppellF1(3/2+m,-n,1/2,5/2+m,-d*(1+sin(f*x+e))/(c-d),1/2+1/2*sin(f*x+e))*cos(f*x+e)*(a+a*sin(f*x
+e))^(1+m)*(c+d*sin(f*x+e))^n*2^(1/2)/a/d/f/(3+2*m)/(2+n+m)/(((c+d*sin(f*x+e))/(c-d))^n)/(1-sin(f*x+e))^(1/2)
________________________________________________________________________________________
Rubi [A]
time = 0.62, 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 = {3124, 3066,
2867, 145, 144, 143} \begin {gather*} \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)} \end {gather*}
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 143
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c
- a*d)), (-f)*((a + b*x)/(b*e - a*f))], 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])
Rule 144
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 145
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 2867
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 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]
Rule 3124
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]
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)) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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] Leaf count is larger than twice the leaf count of optimal. \(2572\) vs. \(2(383)=766\).
time = 7.97, size = 2572, normalized size = 6.72 \begin {gather*} \text {Result too large to show} \end {gather*}
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]
-1/2*(((-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*Sin[(-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*Sin[(-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)*
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) + (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)/(f*Cos[(-e + Pi/2 - f*x
)/2]^(2*m))
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Maple [F]
time = 1.64, size = 0, normalized size = 0.00 \[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c +d \sin \left (f x +e \right )\right )^{n} \left (A +B \sin \left (f x +e \right )+C \left (\sin ^{2}\left (f x +e \right )\right )\right )\, dx\]
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)
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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((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 = 0.00, 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((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)] 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((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]
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((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]
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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^n\,\left (C\,{\sin \left (e+f\,x\right )}^2+B\,\sin \left (e+f\,x\right )+A\right ) \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[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]
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)
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