\(\int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx\) [340]

Optimal result

Integrand size = 35, antiderivative size = 293 \[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx=\frac {\sqrt {2} \left (A d (c (1-m)-d m)-B \left (d^2-c^2 m-c d m\right )\right ) \operatorname {AppellF1}\left (\frac {1}{2}+m,\frac {1}{2},1,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{(c-d)^2 d (c+d) f (1+2 m) \sqrt {1-\sin (e+f x)}}+\frac {2^{\frac {1}{2}+m} (B c-A d) m \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m}{d \left (c^2-d^2\right ) f}-\frac {(B c-A d) \cos (e+f x) (a+a \sin (e+f x))^m}{\left (c^2-d^2\right ) f (c+d \sin (e+f x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3063, 3065, 2731, 2730, 2867, 142, 141} \[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx=\frac {\sqrt {2} \cos (e+f x) \left (A d (c (1-m)-d m)-B \left (c^2 (-m)-c d m+d^2\right )\right ) (a \sin (e+f x)+a)^m \operatorname {AppellF1}\left (m+\frac {1}{2},\frac {1}{2},1,m+\frac {3}{2},\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{d f (2 m+1) (c-d)^2 (c+d) \sqrt {1-\sin (e+f x)}}+\frac {2^{m+\frac {1}{2}} m (B c-A d) \cos (e+f x) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x))\right )}{d f \left (c^2-d^2\right )}-\frac {(B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^m}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))} \]

[In]

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

[Out]

(Sqrt[2]*(A*d*(c*(1 - m) - d*m) - B*(d^2 - c^2*m - c*d*m))*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)/((c - d)^2*d*(c + d)*f*(1 + 2*m)
*Sqrt[1 - Sin[e + f*x]]) + (2^(1/2 + m)*(B*c - A*d)*m*Cos[e + f*x]*Hypergeometric2F1[1/2, 1/2 - m, 3/2, (1 - S
in[e + f*x])/2]*(1 + Sin[e + f*x])^(-1/2 - m)*(a + a*Sin[e + f*x])^m)/(d*(c^2 - d^2)*f) - ((B*c - A*d)*Cos[e +
f*x]*(a + a*Sin[e + f*x])^m)/((c^2 - d^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*} \text {integral}& = -\frac {(B c-A d) \cos (e+f x) (a+a \sin (e+f x))^m}{\left (c^2-d^2\right ) f (c+d \sin (e+f x))}-\frac {\int \frac {(a+a \sin (e+f x))^m (-a (A c-B d+B c m-A d m)+a (B c-A d) m \sin (e+f x))}{c+d \sin (e+f x)} \, dx}{a \left (c^2-d^2\right )} \\ & = -\frac {(B c-A d) \cos (e+f x) (a+a \sin (e+f x))^m}{\left (c^2-d^2\right ) f (c+d \sin (e+f x))}-\frac {((B c-A d) m) \int (a+a \sin (e+f x))^m \, dx}{d \left (c^2-d^2\right )}+\frac {\left (A d (c (1-m)-d m)-B \left (d^2-c^2 m-c d m\right )\right ) \int \frac {(a+a \sin (e+f x))^m}{c+d \sin (e+f x)} \, dx}{d \left (c^2-d^2\right )} \\ & = -\frac {(B c-A d) \cos (e+f x) (a+a \sin (e+f x))^m}{\left (c^2-d^2\right ) f (c+d \sin (e+f x))}+\frac {\left (a^2 \left (A d (c (1-m)-d m)-B \left (d^2-c^2 m-c d m\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 )}{d \left (c^2-d^2\right ) f \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}-\frac {\left ((B c-A d) m (1+\sin (e+f x))^{-m} (a+a \sin (e+f x))^m\right ) \int (1+\sin (e+f x))^m \, dx}{d \left (c^2-d^2\right )} \\ & = \frac {2^{\frac {1}{2}+m} (B c-A d) m \cos (e+f x) \operatorname {Hypergeometric2F1}\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 ) f}-\frac {(B c-A d) \cos (e+f x) (a+a \sin (e+f x))^m}{\left (c^2-d^2\right ) f (c+d \sin (e+f x))}+\frac {\left (a^2 \left (A d (c (1-m)-d m)-B \left (d^2-c^2 m-c d m\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 )}{\sqrt {2} d \left (c^2-d^2\right ) f (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}} \\ & = \frac {\sqrt {2} \left (A d (c (1-m)-d m)-B \left (d^2-c^2 m-c d m\right )\right ) \operatorname {AppellF1}\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}{(c-d)^2 d (c+d) f (1+2 m) \sqrt {1-\sin (e+f x)}}+\frac {2^{\frac {1}{2}+m} (B c-A d) m \cos (e+f x) \operatorname {Hypergeometric2F1}\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 ) f}-\frac {(B c-A d) \cos (e+f x) (a+a \sin (e+f x))^m}{\left (c^2-d^2\right ) f (c+d \sin (e+f x))} \\ \end{align*}

Mathematica [F]

\[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx=\int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right )}{\left (c +d \sin \left (f x +e \right )\right )^{2}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^2} \,d x \]

[In]

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

[Out]

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