3.342 \(\int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=284 \[ \frac {\sqrt {2} (A-B) (c-d) \cos (e+f x) (a \sin (e+f x)+a)^m \sqrt {c+d \sin (e+f x)} F_1\left (m+\frac {1}{2};\frac {1}{2},-\frac {3}{2};m+\frac {3}{2};\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{f (2 m+1) \sqrt {1-\sin (e+f x)} \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}+\frac {\sqrt {2} B (c-d) \cos (e+f x) (a \sin (e+f x)+a)^{m+1} \sqrt {c+d \sin (e+f x)} F_1\left (m+\frac {3}{2};\frac {1}{2},-\frac {3}{2};m+\frac {5}{2};\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{a f (2 m+3) \sqrt {1-\sin (e+f x)} \sqrt {\frac {c+d \sin (e+f x)}{c-d}}} \]

[Out]

(A-B)*(c-d)*AppellF1(1/2+m,-3/2,1/2,3/2+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)/f/(1+2*m)/(1-sin(f*x+e))^(1/2)/((c+d*sin(f*x+e))/(c-d))^(1/2)+B*(c-d)*App
ellF1(3/2+m,-3/2,1/2,5/2+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/f/(3+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 = 0.63, antiderivative size = 284, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {2987, 2788, 140, 139, 138} \[ \frac {\sqrt {2} (A-B) (c-d) \cos (e+f x) (a \sin (e+f x)+a)^m \sqrt {c+d \sin (e+f x)} F_1\left (m+\frac {1}{2};\frac {1}{2},-\frac {3}{2};m+\frac {3}{2};\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{f (2 m+1) \sqrt {1-\sin (e+f x)} \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}+\frac {\sqrt {2} B (c-d) \cos (e+f x) (a \sin (e+f x)+a)^{m+1} \sqrt {c+d \sin (e+f x)} F_1\left (m+\frac {3}{2};\frac {1}{2},-\frac {3}{2};m+\frac {5}{2};\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{a f (2 m+3) \sqrt {1-\sin (e+f x)} \sqrt {\frac {c+d \sin (e+f x)}{c-d}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*(A - B)*(c - d)*AppellF1[1/2 + m, 1/2, -3/2, 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]])/(f*(1 + 2*m)*Sqrt[1 - Sin[e + f*x]]*Sq
rt[(c + d*Sin[e + f*x])/(c - d)]) + (Sqrt[2]*B*(c - d)*AppellF1[3/2 + m, 1/2, -3/2, 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
*f*(3 + 2*m)*Sqrt[1 - Sin[e + f*x]]*Sqrt[(c + d*Sin[e + f*x])/(c - d)])

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])

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 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 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 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]

Rubi steps

\begin {align*} \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^{3/2} \, dx &=(A-B) \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{3/2} \, dx+\frac {B \int (a+a \sin (e+f x))^{1+m} (c+d \sin (e+f x))^{3/2} \, dx}{a}\\ &=\frac {\left (a^2 (A-B) \cos (e+f x)\right ) \operatorname {Subst}\left (\int \frac {(a+a x)^{-\frac {1}{2}+m} (c+d x)^{3/2}}{\sqrt {a-a x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}+\frac {(a B \cos (e+f x)) \operatorname {Subst}\left (\int \frac {(a+a x)^{\frac {1}{2}+m} (c+d x)^{3/2}}{\sqrt {a-a x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}\\ &=\frac {\left (a^2 (A-B) \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)^{3/2}}{\sqrt {\frac {1}{2}-\frac {x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt {2} f (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}}+\frac {\left (a B \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)^{3/2}}{\sqrt {\frac {1}{2}-\frac {x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt {2} f (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}}\\ &=\frac {\left (a (A-B) (a c-a d) \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} \left (\frac {a c}{a c-a d}+\frac {a d x}{a c-a d}\right )^{3/2}}{\sqrt {\frac {1}{2}-\frac {x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt {2} f (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 (B (a c-a d) \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} \left (\frac {a c}{a c-a d}+\frac {a d x}{a c-a d}\right )^{3/2}}{\sqrt {\frac {1}{2}-\frac {x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt {2} f (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 {\sqrt {2} (A-B) (c-d) F_1\left (\frac {1}{2}+m;\frac {1}{2},-\frac {3}{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)}}{f (1+2 m) \sqrt {1-\sin (e+f x)} \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}+\frac {\sqrt {2} B (c-d) F_1\left (\frac {3}{2}+m;\frac {1}{2},-\frac {3}{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)}}{f (3+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]  time = 8.14, size = 3281, normalized size = 11.55 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((((-2*B*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])/(3*Sqrt[(c + d - 2*d*Sin[(-e + Pi/2 - f*x)/2]^2)/(c + d)]) - (2*A*d*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)]) + (B*d*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)]) - (2*B*d*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)] - (3*B*d*(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) - (6*B*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)*(C
os[(-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)*AppellF1[1/2, -1/2 - m, -1/2, 3/2, Sin[(-e + P
i/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, 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) - (6*A*
d*(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)*Appe
llF1[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*Ap
pellF1[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) + (6*A*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*S
in[(-e + Pi/2 - f*x)/2]^2)/(c + d)])*Sin[(-e + Pi/2 - f*x)/2]^2) + (3*B*d*(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)/(f*Cos[(-e + Pi/2 - f*x)/2]^(2*m)))

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (B d \cos \left (f x + e\right )^{2} - A c - B d - {\left (B c + A d\right )} \sin \left (f x + e\right )\right )} \sqrt {d \sin \left (f x + e\right ) + c} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sin \left (f x + e\right ) + A\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.20, size = 0, normalized size = 0.00 \[ \int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right ) \left (c +d \sin \left (f x +e \right )\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sin \left (f x + e\right ) + A\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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