\(\int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx\) [338]

Optimal result

Integrand size = 23, antiderivative size = 117 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=-\frac {B \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m)}-\frac {2^{\frac {1}{2}+m} (A+A m+B m) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m}{f (1+m)} \]

[Out]

-B*cos(f*x+e)*(a+a*sin(f*x+e))^m/f/(1+m)-2^(1/2+m)*(A*m+B*m+A)*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/f/(1+m)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2830, 2731, 2730} \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=-\frac {2^{m+\frac {1}{2}} (A m+A+B m) \cos (e+f x) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {1}{2} (1-\sin (e+f x))\right )}{f (m+1)}-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^m}{f (m+1)} \]

[In]

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

[Out]

-((B*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(f*(1 + m))) - (2^(1/2 + m)*(A + A*m + B*m)*Cos[e + f*x]*Hypergeomet
ric2F1[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)/(f*(1 +
m))

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 2830

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

Rubi steps \begin{align*} \text {integral}& = -\frac {B \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m)}+\frac {(A+A m+B m) \int (a+a \sin (e+f x))^m \, dx}{1+m} \\ & = -\frac {B \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m)}+\frac {\left ((A+A m+B m) (1+\sin (e+f x))^{-m} (a+a \sin (e+f x))^m\right ) \int (1+\sin (e+f x))^m \, dx}{1+m} \\ & = -\frac {B \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+m)}-\frac {2^{\frac {1}{2}+m} (A+A m+B 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}{f (1+m)} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.

Time = 0.17 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.86 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\frac {2^m \left ((A-B) B_{\frac {1}{2} (1+\sin (e+f x))}\left (\frac {1}{2}+m,\frac {1}{2}\right )+2 B B_{\frac {1}{2} (1+\sin (e+f x))}\left (\frac {3}{2}+m,\frac {1}{2}\right )\right ) \sqrt {\cos ^2(e+f x)} \sec (e+f x) (1+\sin (e+f x))^{-m} (a (1+\sin (e+f x)))^m}{f} \]

[In]

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

[Out]

(2^m*((A - B)*Beta[(1 + Sin[e + f*x])/2, 1/2 + m, 1/2] + 2*B*Beta[(1 + Sin[e + f*x])/2, 3/2 + m, 1/2])*Sqrt[Co
s[e + f*x]^2]*Sec[e + f*x]*(a*(1 + Sin[e + f*x]))^m)/(f*(1 + Sin[e + f*x])^m)

Maple [F]

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

[In]

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x)

[Out]

int((a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x)

Fricas [F]

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

[In]

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

[Out]

integral((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m, x)

Sympy [F]

\[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (A + B \sin {\left (e + f x \right )}\right )\, dx \]

[In]

integrate((a+a*sin(f*x+e))**m*(A+B*sin(f*x+e)),x)

[Out]

Integral((a*(sin(e + f*x) + 1))**m*(A + B*sin(e + f*x)), x)

Maxima [F]

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

[In]

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

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m, x)

Giac [F]

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

[In]

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

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^m, x)

Mupad [F(-1)]

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

[In]

int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m,x)

[Out]

int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m, x)