3.28 \(\int (a+a \sin (e+f x))^m \sqrt{c+d \sin (e+f x)} (A+B \sin (e+f x)+C \sin ^2(e+f x)) \, dx\)
Optimal. Leaf size=396 \[ \frac{\sqrt{2} \cos (e+f x) (a \sin (e+f x)+a)^m (2 c (2 C m+C)-d (-A (2 m+5)+2 B m+5 B+2 C m-3 C)) \sqrt{c+d \sin (e+f x)} F_1\left (m+\frac{1}{2};\frac{1}{2},-\frac{1}{2};m+\frac{3}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{d f (2 m+1) (2 m+5) \sqrt{1-\sin (e+f x)} \sqrt{\frac{c+d \sin (e+f x)}{c-d}}}-\frac{\sqrt{2} \cos (e+f x) (2 c C (m+1)-d (B (2 m+5)+2 C m)) (a \sin (e+f x)+a)^{m+1} \sqrt{c+d \sin (e+f x)} F_1\left (m+\frac{3}{2};\frac{1}{2},-\frac{1}{2};m+\frac{5}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{a d f (2 m+3) (2 m+5) \sqrt{1-\sin (e+f x)} \sqrt{\frac{c+d \sin (e+f x)}{c-d}}}-\frac{2 C \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{3/2}}{d f (2 m+5)} \]
[Out]
(-2*C*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(3/2))/(d*f*(5 + 2*m)) + (Sqrt[2]*(2*c*(C + 2*C
*m) - d*(5*B - 3*C + 2*B*m + 2*C*m - A*(5 + 2*m)))*AppellF1[1/2 + m, 1/2, -1/2, 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*Sqrt[c + d*Sin[e + f*x]])/(d*f*(1 + 2*
m)*(5 + 2*m)*Sqrt[1 - Sin[e + f*x]]*Sqrt[(c + d*Sin[e + f*x])/(c - d)]) - (Sqrt[2]*(2*c*C*(1 + m) - d*(2*C*m +
B*(5 + 2*m)))*AppellF1[3/2 + m, 1/2, -1/2, 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)*Sqrt[c + d*Sin[e + f*x]])/(a*d*f*(3 + 2*m)*(5 + 2*m)*Sqrt[1 - Sin[e
+ f*x]]*Sqrt[(c + d*Sin[e + f*x])/(c - d)])
________________________________________________________________________________________
Rubi [A] time = 0.923472, antiderivative size = 393, normalized size of antiderivative =
0.99, number of steps used = 10, number of rules used = 6, integrand size = 47, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.128, Rules
used = {3045, 2987, 2788, 140, 139, 138} \[ \frac{\sqrt{2} \cos (e+f x) (a \sin (e+f x)+a)^m (2 c (2 C m+C)-d (-A (2 m+5)+2 B m+5 B+2 C m-3 C)) \sqrt{c+d \sin (e+f x)} F_1\left (m+\frac{1}{2};\frac{1}{2},-\frac{1}{2};m+\frac{3}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{d f (2 m+1) (2 m+5) \sqrt{1-\sin (e+f x)} \sqrt{\frac{c+d \sin (e+f x)}{c-d}}}+\frac{\sqrt{2} \cos (e+f x) (B d (2 m+5)-2 c C (m+1)+2 C d m) (a \sin (e+f x)+a)^{m+1} \sqrt{c+d \sin (e+f x)} F_1\left (m+\frac{3}{2};\frac{1}{2},-\frac{1}{2};m+\frac{5}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{a d f (2 m+3) (2 m+5) \sqrt{1-\sin (e+f x)} \sqrt{\frac{c+d \sin (e+f x)}{c-d}}}-\frac{2 C \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{3/2}}{d f (2 m+5)} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Sin[e + f*x])^m*Sqrt[c + d*Sin[e + f*x]]*(A + B*Sin[e + f*x] + C*Sin[e + f*x]^2),x]
[Out]
(-2*C*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(3/2))/(d*f*(5 + 2*m)) + (Sqrt[2]*(2*c*(C + 2*C
*m) - d*(5*B - 3*C + 2*B*m + 2*C*m - A*(5 + 2*m)))*AppellF1[1/2 + m, 1/2, -1/2, 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*Sqrt[c + d*Sin[e + f*x]])/(d*f*(1 + 2*
m)*(5 + 2*m)*Sqrt[1 - Sin[e + f*x]]*Sqrt[(c + d*Sin[e + f*x])/(c - d)]) + (Sqrt[2]*(2*C*d*m - 2*c*C*(1 + m) +
B*d*(5 + 2*m))*AppellF1[3/2 + m, 1/2, -1/2, 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)*Sqrt[c + d*Sin[e + f*x]])/(a*d*f*(3 + 2*m)*(5 + 2*m)*Sqrt[1 - Sin[e
+ f*x]]*Sqrt[(c + d*Sin[e + f*x])/(c - d)])
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 \sqrt{c+d \sin (e+f x)} \left (A+B \sin (e+f x)+C \sin ^2(e+f x)\right ) \, dx &=-\frac{2 C \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{3/2}}{d f (5+2 m)}+\frac{2 \int (a+a \sin (e+f x))^m \sqrt{c+d \sin (e+f x)} \left (\frac{1}{2} a \left (2 A d \left (\frac{5}{2}+m\right )+2 C \left (\frac{3 d}{2}+c m\right )\right )+\frac{1}{2} a (2 C d m-2 c C (1+m)+B d (5+2 m)) \sin (e+f x)\right ) \, dx}{a d (5+2 m)}\\ &=-\frac{2 C \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{3/2}}{d f (5+2 m)}+\frac{(2 C d m-2 c C (1+m)+B d (5+2 m)) \int (a+a \sin (e+f x))^{1+m} \sqrt{c+d \sin (e+f x)} \, dx}{a d (5+2 m)}+\frac{(2 c (C+2 C m)-d (5 B-3 C+2 B m+2 C m-A (5+2 m))) \int (a+a \sin (e+f x))^m \sqrt{c+d \sin (e+f x)} \, dx}{d (5+2 m)}\\ &=-\frac{2 C \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{3/2}}{d f (5+2 m)}+\frac{(a (2 C d m-2 c C (1+m)+B d (5+2 m)) \cos (e+f x)) \operatorname{Subst}\left (\int \frac{(a+a x)^{\frac{1}{2}+m} \sqrt{c+d x}}{\sqrt{a-a x}} \, dx,x,\sin (e+f x)\right )}{d f (5+2 m) \sqrt{a-a \sin (e+f x)} \sqrt{a+a \sin (e+f x)}}+\frac{\left (a^2 (2 c (C+2 C m)-d (5 B-3 C+2 B m+2 C m-A (5+2 m))) \cos (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{-\frac{1}{2}+m} \sqrt{c+d x}}{\sqrt{a-a x}} \, dx,x,\sin (e+f x)\right )}{d f (5+2 m) \sqrt{a-a \sin (e+f x)} \sqrt{a+a \sin (e+f x)}}\\ &=-\frac{2 C \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{3/2}}{d f (5+2 m)}+\frac{\left (a (2 C d m-2 c C (1+m)+B d (5+2 m)) \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} \sqrt{c+d x}}{\sqrt{\frac{1}{2}-\frac{x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt{2} d f (5+2 m) (a-a \sin (e+f x)) \sqrt{a+a \sin (e+f x)}}+\frac{\left (a^2 (2 c (C+2 C m)-d (5 B-3 C+2 B m+2 C m-A (5+2 m))) \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} \sqrt{c+d x}}{\sqrt{\frac{1}{2}-\frac{x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt{2} d f (5+2 m) (a-a \sin (e+f x)) \sqrt{a+a \sin (e+f x)}}\\ &=-\frac{2 C \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{3/2}}{d f (5+2 m)}+\frac{\left (a (2 C d m-2 c C (1+m)+B d (5+2 m)) \cos (e+f x) \sqrt{\frac{a-a \sin (e+f x)}{a}} \sqrt{c+d \sin (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{\frac{1}{2}+m} \sqrt{\frac{a c}{a c-a d}+\frac{a d x}{a c-a d}}}{\sqrt{\frac{1}{2}-\frac{x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt{2} d f (5+2 m) (a-a \sin (e+f x)) \sqrt{a+a \sin (e+f x)} \sqrt{\frac{a (c+d \sin (e+f x))}{a c-a d}}}+\frac{\left (a^2 (2 c (C+2 C m)-d (5 B-3 C+2 B m+2 C m-A (5+2 m))) \cos (e+f x) \sqrt{\frac{a-a \sin (e+f x)}{a}} \sqrt{c+d \sin (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{(a+a x)^{-\frac{1}{2}+m} \sqrt{\frac{a c}{a c-a d}+\frac{a d x}{a c-a d}}}{\sqrt{\frac{1}{2}-\frac{x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt{2} d f (5+2 m) (a-a \sin (e+f x)) \sqrt{a+a \sin (e+f x)} \sqrt{\frac{a (c+d \sin (e+f x))}{a c-a d}}}\\ &=-\frac{2 C \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{3/2}}{d f (5+2 m)}+\frac{\sqrt{2} (2 c (C+2 C m)-d (5 B-3 C+2 B m+2 C m-A (5+2 m))) F_1\left (\frac{1}{2}+m;\frac{1}{2},-\frac{1}{2};\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 \sqrt{c+d \sin (e+f x)}}{d f (1+2 m) (5+2 m) \sqrt{1-\sin (e+f x)} \sqrt{\frac{c+d \sin (e+f x)}{c-d}}}+\frac{\sqrt{2} (2 C d m-2 c C (1+m)+B d (5+2 m)) F_1\left (\frac{3}{2}+m;\frac{1}{2},-\frac{1}{2};\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} \sqrt{c+d \sin (e+f x)}}{d f (3+2 m) (5+2 m) (a-a \sin (e+f x)) \sqrt{\frac{c+d \sin (e+f x)}{c-d}}}\\ \end{align*}
Mathematica [B] time = 8.61626, size = 2574, normalized size = 6.5 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + a*Sin[e + f*x])^m*Sqrt[c + d*Sin[e + f*x]]*(A + B*Sin[e + f*x] + C*Sin[e + f*x]^2),x]
[Out]
(((4*B*AppellF1[3/2, (1 - 2*m)/2, -1/2, 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)*Sqrt[c + d - 2*d*Sin[(-e + Pi/2 - f*x)/2]^2])/(
3*Sqrt[(c + d - 2*d*Sin[(-e + Pi/2 - f*x)/2]^2)/(c + d)]) - (2*C*AppellF1[5/2, (1 - 2*m)/2, -1/2, 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)*Sqrt[c + d - 2*d*Sin[(-e + Pi/2 - f*x)/2]^2])/(5*Sqrt[(c + d - 2*d*Sin[(-e + Pi/2 - f*x)/2]^2)/(c +
d)]) + (4*C*AppellF1[3/2, (-1 - 2*m)/2, -1/2, 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)*Sqrt[c + d - 2*d*Sin[(-e + Pi/2 - f*x)/2
]^2])/Sqrt[(c + d - 2*d*Sin[(-e + Pi/2 - f*x)/2]^2)/(c + d)] + (6*C*(c + d)*AppellF1[1/2, -3/2 - m, -1/2, 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)*Sqrt[c + d - 2*d*Sin[(-e + Pi/2 - f*x)/2]^2])/(-3*(c + d)*AppellF1[1/2, -3/2 - m, -1/2, 3/2, Sin[(-e + Pi/
2 - f*x)/2]^2, (2*d*Sin[(-e + Pi/2 - f*x)/2]^2)/(c + d)] + (2*d*AppellF1[3/2, -3/2 - m, 1/2, 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, -1/2, 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, -1/2, 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)*Sqrt[c + d - 2*d*Sin[(-e + Pi/2 - f*x)/2]^2])/(-3*(c + d)*Appell
F1[1/2, -1/2 - m, -1/2, 3/2, Sin[(-e + Pi/2 - f*x)/2]^2, (2*d*Sin[(-e + Pi/2 - f*x)/2]^2)/(c + d)] + (2*d*Appe
llF1[3/2, -1/2 - m, 1/2, 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, -1/2, 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, -1/2, 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)*Sqrt[c + d - 2*d*Sin[(-e + Pi/2 -
f*x)/2]^2])/(3*(c + d)*AppellF1[1/2, 1/2 - m, -1/2, 3/2, Sin[(-e + Pi/2 - f*x)/2]^2, (2*d*Sin[(-e + Pi/2 - f*
x)/2]^2)/(c + d)] - (2*d*AppellF1[3/2, 1/2 - m, 1/2, 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, -1/2, 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, -1/2, 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)*(Co
s[(-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)*Sqrt[
c + d - 2*d*Sin[(-e + Pi/2 - f*x)/2]^2])/(3*(c + d)*AppellF1[1/2, 1/2 - m, -1/2, 3/2, Sin[(-e + Pi/2 - f*x)/2]
^2, (2*d*Sin[(-e + Pi/2 - f*x)/2]^2)/(c + d)] - (2*d*AppellF1[3/2, 1/2 - m, 1/2, 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, -1/2, 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 = 0.712, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m}\sqrt{c+d\sin \left ( fx+e \right ) } \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))^(1/2)*(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))^(1/2)*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x)
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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 )} \sqrt{d \sin \left (f x + e\right ) + c}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}\,{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))^(1/2)*(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)*sqrt(d*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m, 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 )} \sqrt{d \sin \left (f x + e\right ) + c}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}, 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))^(1/2)*(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)*sqrt(d*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m, 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))**(1/2)*(A+B*sin(f*x+e)+C*sin(f*x+e)**2),x)
[Out]
Timed out
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \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))^(1/2)*(A+B*sin(f*x+e)+C*sin(f*x+e)^2),x, algorithm="giac")
[Out]
Exception raised: AttributeError