\(\int \sec (e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx\) [1024]

Optimal result

Integrand size = 29, antiderivative size = 80 \[ \int \sec (e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\frac {(A-B) (a+a \sin (e+f x))^m}{2 f m}+\frac {(A+B) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^{1+m}}{4 a f (1+m)} \]

[Out]

1/2*(A-B)*(a+a*sin(f*x+e))^m/f/m+1/4*(A+B)*hypergeom([1, 1+m],[2+m],1/2+1/2*sin(f*x+e))*(a+a*sin(f*x+e))^(1+m)
/a/f/(1+m)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2915, 80, 70} \[ \int \sec (e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\frac {(A+B) (a \sin (e+f x)+a)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {1}{2} (\sin (e+f x)+1)\right )}{4 a f (m+1)}+\frac {(A-B) (a \sin (e+f x)+a)^m}{2 f m} \]

[In]

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

[Out]

((A - B)*(a + a*Sin[e + f*x])^m)/(2*f*m) + ((A + B)*Hypergeometric2F1[1, 1 + m, 2 + m, (1 + Sin[e + f*x])/2]*(
a + a*Sin[e + f*x])^(1 + m))/(4*a*f*(1 + m))

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^Simplify[p + 1], x], x] /; FreeQ[{a, b, c
, d, e, f, n, p}, x] &&  !RationalQ[p] && SumSimplerQ[p, 1]

Rule 2915

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rubi steps \begin{align*} \text {integral}& = \frac {a \text {Subst}\left (\int \frac {(a+x)^{-1+m} \left (A+\frac {B x}{a}\right )}{a-x} \, dx,x,a \sin (e+f x)\right )}{f} \\ & = \frac {(A-B) (a+a \sin (e+f x))^m}{2 f m}+\frac {(A+B) \text {Subst}\left (\int \frac {(a+x)^m}{a-x} \, dx,x,a \sin (e+f x)\right )}{2 f} \\ & = \frac {(A-B) (a+a \sin (e+f x))^m}{2 f m}+\frac {(A+B) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^{1+m}}{4 a f (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.89 \[ \int \sec (e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\frac {(a (1+\sin (e+f x)))^m \left (2 (A-B) (1+m)+(A+B) m \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {1}{2} (1+\sin (e+f x))\right ) (1+\sin (e+f x))\right )}{4 f m (1+m)} \]

[In]

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

[Out]

((a*(1 + Sin[e + f*x]))^m*(2*(A - B)*(1 + m) + (A + B)*m*Hypergeometric2F1[1, 1 + m, 2 + m, (1 + Sin[e + f*x])
/2]*(1 + Sin[e + f*x])))/(4*f*m*(1 + m))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

integrate(sec(f*x+e)*(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))*sec(e + f*x), x)

Maxima [F]

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

[In]

integrate(sec(f*x+e)*(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*sec(f*x + e), x)

Giac [F]

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

[In]

integrate(sec(f*x+e)*(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*sec(f*x + e), x)

Mupad [F(-1)]

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

[In]

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

[Out]

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