3.1.12 \(\int (a+a \sin (e+f x))^m \sqrt {c+d \sin (e+f x)} (A+C \sin ^2(e+f x)) \, dx\) [12]

Optimal. Leaf size=375 \[ -\frac {2 C \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{3/2}}{d f (5+2 m)}+\frac {\sqrt {2} (2 c (C+2 C m)+d (C (3-2 m)+A (5+2 m))) F_1\left (\frac {1}{2}+m;\frac {1}{2},-\frac {1}{2};\frac {3}{2}+m;\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) (a+a \sin (e+f x))^m \sqrt {c+d \sin (e+f x)}}{d f (1+2 m) (5+2 m) \sqrt {1-\sin (e+f x)} \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}+\frac {2 \sqrt {2} C (d m-c (1+m)) F_1\left (\frac {3}{2}+m;\frac {1}{2},-\frac {1}{2};\frac {5}{2}+m;\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) (a+a \sin (e+f x))^{1+m} \sqrt {c+d \sin (e+f x)}}{a d f (3+2 m) (5+2 m) \sqrt {1-\sin (e+f x)} \sqrt {\frac {c+d \sin (e+f x)}{c-d}}} \]

[Out]

-2*C*cos(f*x+e)*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(3/2)/d/f/(5+2*m)+(2*c*(2*C*m+C)+d*(C*(3-2*m)+A*(5+2*m)))*
AppellF1(1/2+m,-1/2,1/2,3/2+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*2^(1/2
)*(c+d*sin(f*x+e))^(1/2)/d/f/(1+2*m)/(5+2*m)/(1-sin(f*x+e))^(1/2)/((c+d*sin(f*x+e))/(c-d))^(1/2)+2*C*(d*m-c*(1
+m))*AppellF1(3/2+m,-1/2,1/2,5/2+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)*2^(1/2)*(c+d*sin(f*x+e))^(1/2)/a/d/f/(3+2*m)/(5+2*m)/(1-sin(f*x+e))^(1/2)/((c+d*sin(f*x+e))/(c-d))^(1/2)

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Rubi [A]
time = 0.61, antiderivative size = 374, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3125, 3066, 2867, 145, 144, 143} \begin {gather*} \frac {\sqrt {2} \cos (e+f x) (A d (2 m+5)+2 c (2 C m+C)+C d (3-2 m)) (a \sin (e+f x)+a)^m \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 {2 \sqrt {2} C (d m-c (m+1)) \cos (e+f x) (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)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^m*Sqrt[c + d*Sin[e + f*x]]*(A + 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]*(C*d*(3 - 2*m
) + A*d*(5 + 2*m) + 2*c*(C + 2*C*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)*S
qrt[1 - Sin[e + f*x]]*Sqrt[(c + d*Sin[e + f*x])/(c - d)]) + (2*Sqrt[2]*C*(d*m - c*(1 + 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 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 3125

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (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*Si
mp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1)) + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a,
b, c, d, e, f, A, 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 \sqrt {c+d \sin (e+f x)} \left (A+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 )+a C (d m-c (1+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-c (1+m))) \int (a+a \sin (e+f x))^{1+m} \sqrt {c+d \sin (e+f x)} \, dx}{a d (5+2 m)}+\frac {(C d (3-2 m)+A d (5+2 m)+2 c (C+2 C 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 {(2 a C (d m-c (1+m)) \cos (e+f x)) \text {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 (C d (3-2 m)+A d (5+2 m)+2 c (C+2 C m)) \cos (e+f x)\right ) \text {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 (\sqrt {2} a C (d m-c (1+m)) \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} \sqrt {c+d x}}{\sqrt {\frac {1}{2}-\frac {x}{2}}} \, dx,x,\sin (e+f x)\right )}{d f (5+2 m) (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}}+\frac {\left (a^2 (C d (3-2 m)+A d (5+2 m)+2 c (C+2 C m)) \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} \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 (\sqrt {2} a C (d m-c (1+m)) \cos (e+f x) \sqrt {\frac {a-a \sin (e+f x)}{a}} \sqrt {c+d \sin (e+f x)}\right ) \text {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 )}{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 (C d (3-2 m)+A d (5+2 m)+2 c (C+2 C m)) \cos (e+f x) \sqrt {\frac {a-a \sin (e+f x)}{a}} \sqrt {c+d \sin (e+f x)}\right ) \text {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} (C d (3-2 m)+A d (5+2 m)+2 c (C+2 C 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 {2 \sqrt {2} C (d m-c (1+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*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(1874\) vs. \(2(375)=750\).
time = 7.26, size = 1874, normalized size = 5.00 \begin {gather*} \text {Too large to display} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Sin[e + f*x])^m*Sqrt[c + d*Sin[e + f*x]]*(A + C*Sin[e + f*x]^2),x]

[Out]

(((-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 + P
i/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*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)*(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))*(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 = 0.60, size = 0, normalized size = 0.00 \[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \sqrt {c +d \sin \left (f x +e \right )}\, \left (A +C \left (\sin ^{2}\left (f x +e \right )\right )\right )\, dx\]

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))^(1/2)*(A+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+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))^(1/2)*(A+C*sin(f*x+e)^2),x, algorithm="maxima")

[Out]

integrate((C*sin(f*x + e)^2 + A)*sqrt(d*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m, 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))^(1/2)*(A+C*sin(f*x+e)^2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (A + C \sin ^{2}{\left (e + f x \right )}\right ) \sqrt {c + d \sin {\left (e + f x \right )}}\, dx \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))**(1/2)*(A+C*sin(f*x+e)**2),x)

[Out]

Integral((a*(sin(e + f*x) + 1))**m*(A + C*sin(e + f*x)**2)*sqrt(c + d*sin(e + f*x)), x)

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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))^(1/2)*(A+C*sin(f*x+e)^2),x, algorithm="giac")

[Out]

integrate((C*sin(f*x + e)^2 + A)*sqrt(d*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \left (C\,{\sin \left (e+f\,x\right )}^2+A\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\sqrt {c+d\,\sin \left (e+f\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C*sin(e + f*x)^2)*(a + a*sin(e + f*x))^m*(c + d*sin(e + f*x))^(1/2),x)

[Out]

int((A + C*sin(e + f*x)^2)*(a + a*sin(e + f*x))^m*(c + d*sin(e + f*x))^(1/2), x)

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