3.4.41 \(\int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx\) [341]

Optimal. Leaf size=467 \[ \frac {\left (B \left (2 d^3 m+c^3 (1-m) m+2 c^2 d (1-m) m-c d^2 \left (3-3 m+m^2\right )\right )-A d \left (2 c d (2-m) m-c^2 \left (2-3 m+m^2\right )-d^2 \left (1-m+m^2\right )\right )\right ) F_1\left (\frac {1}{2}+m;\frac {1}{2},1;\frac {3}{2}+m;\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{\sqrt {2} (c-d)^3 d (c+d)^2 f (1+2 m) \sqrt {1-\sin (e+f x)}}-\frac {2^{-\frac {1}{2}+m} m \left (A d (c (3-m)-d m)-B \left (2 d^2+c^2 (1-m)-c d m\right )\right ) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m}{d \left (c^2-d^2\right )^2 f}-\frac {(B c-A d) \cos (e+f x) (a+a \sin (e+f x))^m}{2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}+\frac {\left (A d (c (3-m)-d m)-B \left (2 d^2+c^2 (1-m)-c d m\right )\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{2 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))} \]

[Out]

-2^(-1/2+m)*m*(A*d*(c*(3-m)-d*m)-B*(2*d^2+c^2*(1-m)-c*d*m))*cos(f*x+e)*hypergeom([1/2, 1/2-m],[3/2],1/2-1/2*si
n(f*x+e))*(1+sin(f*x+e))^(-1/2-m)*(a+a*sin(f*x+e))^m/d/(c^2-d^2)^2/f-1/2*(-A*d+B*c)*cos(f*x+e)*(a+a*sin(f*x+e)
)^m/(c^2-d^2)/f/(c+d*sin(f*x+e))^2+1/2*(A*d*(c*(3-m)-d*m)-B*(2*d^2+c^2*(1-m)-c*d*m))*cos(f*x+e)*(a+a*sin(f*x+e
))^m/(c^2-d^2)^2/f/(c+d*sin(f*x+e))+1/2*(B*(2*d^3*m+c^3*(1-m)*m+2*c^2*d*(1-m)*m-c*d^2*(m^2-3*m+3))-A*d*(2*c*d*
(2-m)*m-c^2*(m^2-3*m+2)-d^2*(m^2-m+1)))*AppellF1(1/2+m,1,1/2,3/2+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)^3/d/(c+d)^2/f/(1+2*m)*2^(1/2)/(1-sin(f*x+e))^(1/2)

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Rubi [A]
time = 0.91, antiderivative size = 467, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3063, 3065, 2731, 2730, 2867, 142, 141} \begin {gather*} \frac {\cos (e+f x) \left (B \left (c^3 (1-m) m+2 c^2 d (1-m) m-c d^2 \left (m^2-3 m+3\right )+2 d^3 m\right )-A d \left (-\left (c^2 \left (m^2-3 m+2\right )\right )+2 c d (2-m) m-d^2 \left (m^2-m+1\right )\right )\right ) (a \sin (e+f x)+a)^m F_1\left (m+\frac {1}{2};\frac {1}{2},1;m+\frac {3}{2};\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{\sqrt {2} d f (2 m+1) (c-d)^3 (c+d)^2 \sqrt {1-\sin (e+f x)}}-\frac {2^{m-\frac {1}{2}} m \cos (e+f x) \left (A d (c (3-m)-d m)-B \left (c^2 (1-m)-c d m+2 d^2\right )\right ) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right )}{d f \left (c^2-d^2\right )^2}+\frac {\cos (e+f x) \left (A d (c (3-m)-d m)-B \left (c^2 (1-m)-c d m+2 d^2\right )\right ) (a \sin (e+f x)+a)^m}{2 f \left (c^2-d^2\right )^2 (c+d \sin (e+f x))}-\frac {(B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^m}{2 f \left (c^2-d^2\right ) (c+d \sin (e+f x))^2} \end {gather*}

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,x]

[Out]

((B*(2*d^3*m + c^3*(1 - m)*m + 2*c^2*d*(1 - m)*m - c*d^2*(3 - 3*m + m^2)) - A*d*(2*c*d*(2 - m)*m - c^2*(2 - 3*
m + m^2) - d^2*(1 - m + m^2)))*AppellF1[1/2 + m, 1/2, 1, 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[2]*(c - d)^3*d*(c + d)^2*f*(1 + 2*m)*Sqrt[1 - Sin[e +
f*x]]) - (2^(-1/2 + m)*m*(A*d*(c*(3 - m) - d*m) - B*(2*d^2 + c^2*(1 - m) - c*d*m))*Cos[e + f*x]*Hypergeometric
2F1[1/2, 1/2 - m, 3/2, (1 - Sin[e + f*x])/2]*(1 + Sin[e + f*x])^(-1/2 - m)*(a + a*Sin[e + f*x])^m)/(d*(c^2 - d
^2)^2*f) - ((B*c - A*d)*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(2*(c^2 - d^2)*f*(c + d*Sin[e + f*x])^2) + ((A*d*
(c*(3 - m) - d*m) - B*(2*d^2 + c^2*(1 - m) - c*d*m))*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(2*(c^2 - d^2)^2*f*(
c + d*Sin[e + f*x]))

Rule 141

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(b*e - a*f
)^p*((a + b*x)^(m + 1)/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n))*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}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] && IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !(GtQ[d/(d*a - c*b), 0] && SimplerQ[c + d*x, a + b*x])

Rule 142

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}, x] &&  !IntegerQ[m] &&
!IntegerQ[n] && IntegerQ[p] &&  !GtQ[b/(b*c - a*d), 0] &&  !SimplerQ[c + d*x, a + b*x]

Rule 2730

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-2^(n + 1/2))*a^(n - 1/2)*b*(Cos[c + d*x]/
(d*Sqrt[a + b*Sin[c + d*x]]))*Hypergeometric2F1[1/2, 1/2 - n, 3/2, (1/2)*(1 - b*(Sin[c + d*x]/a))], x] /; Free
Q[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rule 2731

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[a^IntPart[n]*((a + b*Sin[c + d*x])^FracPart
[n]/(1 + (b/a)*Sin[c + d*x])^FracPart[n]), Int[(1 + (b/a)*Sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, n}, x]
&& EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] &&  !GtQ[a, 0]

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 3063

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x]
)^(n + 1)/(f*(n + 1)*(c^2 - d^2))), x] + Dist[1/(b*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin
[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e +
f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2
- d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 0])

Rule 3065

Int[(((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] &
& EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && NeQ[m + 1/2, 0]

Rubi steps

\begin {align*} \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx &=-\frac {(B c-A d) \cos (e+f x) (a+a \sin (e+f x))^m}{2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}-\frac {\int \frac {(a+a \sin (e+f x))^m (-a (2 A c-2 B d+B c m-A d m)-a (B c-A d) (1-m) \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx}{2 a \left (c^2-d^2\right )}\\ &=-\frac {(B c-A d) \cos (e+f x) (a+a \sin (e+f x))^m}{2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}+\frac {\left (A d (c (3-m)-d m)-B \left (2 d^2+c^2 (1-m)-c d m\right )\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{2 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}+\frac {\int \frac {(a+a \sin (e+f x))^m \left (-a^2 ((B c-A d) (1-m) (d-c m)-(c-d m) (2 A c-2 B d+B c m-A d m))+a^2 m \left (A d (c (3-m)-d m)-B \left (2 d^2+c^2 (1-m)-c d m\right )\right ) \sin (e+f x)\right )}{c+d \sin (e+f x)} \, dx}{2 a^2 \left (c^2-d^2\right )^2}\\ &=-\frac {(B c-A d) \cos (e+f x) (a+a \sin (e+f x))^m}{2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}+\frac {\left (A d (c (3-m)-d m)-B \left (2 d^2+c^2 (1-m)-c d m\right )\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{2 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}+\frac {\left (m \left (A d (c (3-m)-d m)-B \left (2 d^2+c^2 (1-m)-c d m\right )\right )\right ) \int (a+a \sin (e+f x))^m \, dx}{2 d \left (c^2-d^2\right )^2}+\frac {\left (B \left (2 d^3 m+c^3 (1-m) m+2 c^2 d (1-m) m-c d^2 \left (3-3 m+m^2\right )\right )-A d \left (2 c d (2-m) m-c^2 \left (2-3 m+m^2\right )-d^2 \left (1-m+m^2\right )\right )\right ) \int \frac {(a+a \sin (e+f x))^m}{c+d \sin (e+f x)} \, dx}{2 d \left (c^2-d^2\right )^2}\\ &=-\frac {(B c-A d) \cos (e+f x) (a+a \sin (e+f x))^m}{2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}+\frac {\left (A d (c (3-m)-d m)-B \left (2 d^2+c^2 (1-m)-c d m\right )\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{2 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}+\frac {\left (a^2 \left (B \left (2 d^3 m+c^3 (1-m) m+2 c^2 d (1-m) m-c d^2 \left (3-3 m+m^2\right )\right )-A d \left (2 c d (2-m) m-c^2 \left (2-3 m+m^2\right )-d^2 \left (1-m+m^2\right )\right )\right ) \cos (e+f x)\right ) \text {Subst}\left (\int \frac {(a+a x)^{-\frac {1}{2}+m}}{\sqrt {a-a x} (c+d x)} \, dx,x,\sin (e+f x)\right )}{2 d \left (c^2-d^2\right )^2 f \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}+\frac {\left (m \left (A d (c (3-m)-d m)-B \left (2 d^2+c^2 (1-m)-c d m\right )\right ) (1+\sin (e+f x))^{-m} (a+a \sin (e+f x))^m\right ) \int (1+\sin (e+f x))^m \, dx}{2 d \left (c^2-d^2\right )^2}\\ &=-\frac {2^{-\frac {1}{2}+m} m \left (A d (c (3-m)-d m)-B \left (2 d^2+c^2 (1-m)-c d m\right )\right ) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m}{d \left (c^2-d^2\right )^2 f}-\frac {(B c-A d) \cos (e+f x) (a+a \sin (e+f x))^m}{2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}+\frac {\left (A d (c (3-m)-d m)-B \left (2 d^2+c^2 (1-m)-c d m\right )\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{2 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}+\frac {\left (a^2 \left (B \left (2 d^3 m+c^3 (1-m) m+2 c^2 d (1-m) m-c d^2 \left (3-3 m+m^2\right )\right )-A d \left (2 c d (2-m) m-c^2 \left (2-3 m+m^2\right )-d^2 \left (1-m+m^2\right )\right )\right ) \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 {\frac {1}{2}-\frac {x}{2}} (c+d x)} \, dx,x,\sin (e+f x)\right )}{2 \sqrt {2} d \left (c^2-d^2\right )^2 f (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}}\\ &=\frac {\left (B \left (2 d^3 m+c^3 (1-m) m+2 c^2 d (1-m) m-c d^2 \left (3-3 m+m^2\right )\right )-A d \left (2 c d (2-m) m-c^2 \left (2-3 m+m^2\right )-d^2 \left (1-m+m^2\right )\right )\right ) F_1\left (\frac {1}{2}+m;\frac {1}{2},1;\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 {2} (c-d)^3 d (c+d)^2 f (1+2 m) \sqrt {1-\sin (e+f x)}}-\frac {2^{-\frac {1}{2}+m} m \left (A d (c (3-m)-d m)-B \left (2 d^2+c^2 (1-m)-c d m\right )\right ) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m}{d \left (c^2-d^2\right )^2 f}-\frac {(B c-A d) \cos (e+f x) (a+a \sin (e+f x))^m}{2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))^2}+\frac {\left (A d (c (3-m)-d m)-B \left (2 d^2+c^2 (1-m)-c d m\right )\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{2 \left (c^2-d^2\right )^2 f (c+d \sin (e+f x))}\\ \end {align*}

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Mathematica [A]
time = 1.73, size = 651, normalized size = 1.39 \begin {gather*} \frac {6 (c+d) \cos ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )^{-\frac {1}{2}+m} \cot \left (\frac {1}{4} (2 e+\pi +2 f x)\right ) (a (1+\sin (e+f x)))^m \left (\frac {(-B c+A d) F_1\left (\frac {1}{2};\frac {1}{2}-m,3;\frac {3}{2};\cos ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{c+d}\right )}{-3 (c+d) F_1\left (\frac {1}{2};\frac {1}{2}-m,3;\frac {3}{2};\cos ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{c+d}\right )+\left (-12 d F_1\left (\frac {3}{2};\frac {1}{2}-m,4;\frac {5}{2};\cos ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{c+d}\right )+(c+d) (-1+2 m) F_1\left (\frac {3}{2};\frac {3}{2}-m,3;\frac {5}{2};\cos ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{c+d}\right )\right ) \cos ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right )}+\frac {B F_1\left (\frac {1}{2};\frac {1}{2}-m,2;\frac {3}{2};\cos ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{c+d}\right ) (c+d \sin (e+f x))}{-3 (c+d) F_1\left (\frac {1}{2};\frac {1}{2}-m,2;\frac {3}{2};\cos ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{c+d}\right )+\left (-8 d F_1\left (\frac {3}{2};\frac {1}{2}-m,3;\frac {5}{2};\cos ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{c+d}\right )+(c+d) (-1+2 m) F_1\left (\frac {3}{2};\frac {3}{2}-m,2;\frac {5}{2};\cos ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e-\pi +2 f x)\right )}{c+d}\right )\right ) \cos ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right )}\right ) \sin ^2\left (\frac {1}{4} (2 e+\pi +2 f x)\right )^{\frac {1}{2}-m}}{d f (c+d \sin (e+f x))^3} \end {gather*}

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,x]

[Out]

(6*(c + d)*(Cos[(2*e - Pi + 2*f*x)/4]^2)^(-1/2 + m)*Cot[(2*e + Pi + 2*f*x)/4]*(a*(1 + Sin[e + f*x]))^m*(((-(B*
c) + A*d)*AppellF1[1/2, 1/2 - m, 3, 3/2, Cos[(2*e + Pi + 2*f*x)/4]^2, (2*d*Sin[(2*e - Pi + 2*f*x)/4]^2)/(c + d
)])/(-3*(c + d)*AppellF1[1/2, 1/2 - m, 3, 3/2, Cos[(2*e + Pi + 2*f*x)/4]^2, (2*d*Sin[(2*e - Pi + 2*f*x)/4]^2)/
(c + d)] + (-12*d*AppellF1[3/2, 1/2 - m, 4, 5/2, Cos[(2*e + Pi + 2*f*x)/4]^2, (2*d*Sin[(2*e - Pi + 2*f*x)/4]^2
)/(c + d)] + (c + d)*(-1 + 2*m)*AppellF1[3/2, 3/2 - m, 3, 5/2, Cos[(2*e + Pi + 2*f*x)/4]^2, (2*d*Sin[(2*e - Pi
+ 2*f*x)/4]^2)/(c + d)])*Cos[(2*e + Pi + 2*f*x)/4]^2) + (B*AppellF1[1/2, 1/2 - m, 2, 3/2, Cos[(2*e + Pi + 2*f
*x)/4]^2, (2*d*Sin[(2*e - Pi + 2*f*x)/4]^2)/(c + d)]*(c + d*Sin[e + f*x]))/(-3*(c + d)*AppellF1[1/2, 1/2 - m,
2, 3/2, Cos[(2*e + Pi + 2*f*x)/4]^2, (2*d*Sin[(2*e - Pi + 2*f*x)/4]^2)/(c + d)] + (-8*d*AppellF1[3/2, 1/2 - m,
3, 5/2, Cos[(2*e + Pi + 2*f*x)/4]^2, (2*d*Sin[(2*e - Pi + 2*f*x)/4]^2)/(c + d)] + (c + d)*(-1 + 2*m)*AppellF1
[3/2, 3/2 - m, 2, 5/2, Cos[(2*e + Pi + 2*f*x)/4]^2, (2*d*Sin[(2*e - Pi + 2*f*x)/4]^2)/(c + d)])*Cos[(2*e + Pi
+ 2*f*x)/4]^2))*(Sin[(2*e + Pi + 2*f*x)/4]^2)^(1/2 - m))/(d*f*(c + d*Sin[e + f*x])^3)

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Maple [F]
time = 2.16, size = 0, normalized size = 0.00 \[\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right )}{\left (c +d \sin \left (f x +e \right )\right )^{3}}\, dx\]

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,x)

[Out]

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

[Out]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\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} \,d x \end {gather*}

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,x)

[Out]

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

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