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

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

[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]))

________________________________________________________________________________________

Rubi [A]  time = 1.34696, antiderivative size = 467, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2984, 2986, 2652, 2651, 2788, 137, 136} \[ \frac{\cos (e+f x) \left (B \left (2 c^2 d (1-m) m+c^3 (1-m) m-c d^2 \left (m^2-3 m+3\right )+2 d^3 m\right )-A d \left (c^2 \left (-\left (m^2-3 m+2\right )\right )+2 c d (2-m) m-d^2 \left (m^2-m+1\right )\right )\right ) (a \sin (e+f x)+a)^m F_1\left (m+\frac{1}{2};\frac{1}{2},1;m+\frac{3}{2};\frac{1}{2} (\sin (e+f x)+1),-\frac{d (\sin (e+f x)+1)}{c-d}\right )}{\sqrt{2} d f (2 m+1) (c-d)^3 (c+d)^2 \sqrt{1-\sin (e+f x)}}-\frac{2^{m-\frac{1}{2}} m \cos (e+f x) \left (A d (c (3-m)-d m)-B \left (c^2 (1-m)-c d m+2 d^2\right )\right ) (\sin (e+f x)+1)^{-m-\frac{1}{2}} (a \sin (e+f x)+a)^m \, _2F_1\left (\frac{1}{2},\frac{1}{2}-m;\frac{3}{2};\frac{1}{2} (1-\sin (e+f x))\right )}{d f \left (c^2-d^2\right )^2}+\frac{\cos (e+f x) \left (A d (c (3-m)-d m)-B \left (c^2 (1-m)-c d m+2 d^2\right )\right ) (a \sin (e+f x)+a)^m}{2 f \left (c^2-d^2\right )^2 (c+d \sin (e+f x))}-\frac{(B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^m}{2 f \left (c^2-d^2\right ) (c+d \sin (e+f x))^2} \]

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 2984

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 2986

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]

Rule 2652

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

Rule 2651

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(2^(n + 1/2)*a^(n - 1/2)*b*Cos[c + d*x]*Hy
pergeometric2F1[1/2, 1/2 - n, 3/2, (1*(1 - (b*Sin[c + d*x])/a))/2])/(d*Sqrt[a + b*Sin[c + d*x]]), x] /; FreeQ[
{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 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 137

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 136

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)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f
))])/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n), 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])

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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [A]  time = 6.1267, size = 654, normalized size = 1.4 \[ \frac{6 (c+d) \cot \left (\frac{1}{4} (2 e+2 f x+\pi )\right ) \sin ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right )^{\frac{1}{2}-m} \cos ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )^{m-\frac{1}{2}} (a (\sin (e+f x)+1))^m \left (\frac{(B c-A d) F_1\left (\frac{1}{2};\frac{1}{2}-m,3;\frac{3}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )}{\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ) \left (12 d F_1\left (\frac{3}{2};\frac{1}{2}-m,4;\frac{5}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )-(2 m-1) (c+d) F_1\left (\frac{3}{2};\frac{3}{2}-m,3;\frac{5}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )\right )+3 (c+d) F_1\left (\frac{1}{2};\frac{1}{2}-m,3;\frac{3}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )}-\frac{B (c+d \sin (e+f x)) F_1\left (\frac{1}{2};\frac{1}{2}-m,2;\frac{3}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )}{\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ) \left (8 d F_1\left (\frac{3}{2};\frac{1}{2}-m,3;\frac{5}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )-(2 m-1) (c+d) F_1\left (\frac{3}{2};\frac{3}{2}-m,2;\frac{5}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )\right )+3 (c+d) F_1\left (\frac{1}{2};\frac{1}{2}-m,2;\frac{3}{2};\cos ^2\left (\frac{1}{4} (2 e+2 f x+\pi )\right ),\frac{2 d \sin ^2\left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )}\right )}{d f (c+d \sin (e+f x))^3} \]

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 = 2.198, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( A+B\sin \left ( fx+e \right ) \right ) }{ \left ( c+d\sin \left ( fx+e \right ) \right ) ^{3}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{3}}\,{d x} \end{align*}

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., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{3 \, c d^{2} \cos \left (f x + e\right )^{2} - c^{3} - 3 \, c d^{2} +{\left (d^{3} \cos \left (f x + e\right )^{2} - 3 \, c^{2} d - d^{3}\right )} \sin \left (f x + e\right )}, x\right ) \end{align*}

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)]  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*(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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{3}}\,{d x} \end{align*}

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)