\(\int \sec ^5(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx\) [1026]

Optimal result

Integrand size = 31, antiderivative size = 104 \[ \int \sec ^5(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=-\frac {a^2 (A (4-m)-B m) \operatorname {Hypergeometric2F1}\left (2,-2+m,-1+m,\frac {1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^{-2+m}}{16 f (2-m)}+\frac {a^4 (A+B) (a+a \sin (e+f x))^{-2+m}}{4 f (a-a \sin (e+f x))^2} \]

[Out]

-1/16*a^2*(A*(4-m)-B*m)*hypergeom([2, -2+m],[-1+m],1/2+1/2*sin(f*x+e))*(a+a*sin(f*x+e))^(-2+m)/f/(2-m)+1/4*a^4
*(A+B)*(a+a*sin(f*x+e))^(-2+m)/f/(a-a*sin(f*x+e))^2

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {2915, 79, 70} \[ \int \sec ^5(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\frac {a^4 (A+B) (a \sin (e+f x)+a)^{m-2}}{4 f (a-a \sin (e+f x))^2}-\frac {a^2 (A (4-m)-B m) (a \sin (e+f x)+a)^{m-2} \operatorname {Hypergeometric2F1}\left (2,m-2,m-1,\frac {1}{2} (\sin (e+f x)+1)\right )}{16 f (2-m)} \]

[In]

Int[Sec[e + f*x]^5*(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]),x]

[Out]

-1/16*(a^2*(A*(4 - m) - B*m)*Hypergeometric2F1[2, -2 + m, -1 + m, (1 + Sin[e + f*x])/2]*(a + a*Sin[e + f*x])^(
-2 + m))/(f*(2 - m)) + (a^4*(A + B)*(a + a*Sin[e + f*x])^(-2 + m))/(4*f*(a - a*Sin[e + f*x])^2)

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 79

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)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

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^5 \text {Subst}\left (\int \frac {(a+x)^{-3+m} \left (A+\frac {B x}{a}\right )}{(a-x)^3} \, dx,x,a \sin (e+f x)\right )}{f} \\ & = \frac {a^4 (A+B) (a+a \sin (e+f x))^{-2+m}}{4 f (a-a \sin (e+f x))^2}+\frac {\left (a^4 (A (4-m)-B m)\right ) \text {Subst}\left (\int \frac {(a+x)^{-3+m}}{(a-x)^2} \, dx,x,a \sin (e+f x)\right )}{4 f} \\ & = -\frac {a^2 (A (4-m)-B m) \operatorname {Hypergeometric2F1}\left (2,-2+m,-1+m,\frac {1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^{-2+m}}{16 f (2-m)}+\frac {a^4 (A+B) (a+a \sin (e+f x))^{-2+m}}{4 f (a-a \sin (e+f x))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.73 \[ \int \sec ^5(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\frac {a^2 \left (-\frac {(A (-4+m)+B m) \operatorname {Hypergeometric2F1}\left (2,-2+m,-1+m,\frac {1}{2} (1+\sin (e+f x))\right )}{-2+m}+\frac {4 (A+B)}{(-1+\sin (e+f x))^2}\right ) (a (1+\sin (e+f x)))^{-2+m}}{16 f} \]

[In]

Integrate[Sec[e + f*x]^5*(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]),x]

[Out]

(a^2*(-(((A*(-4 + m) + B*m)*Hypergeometric2F1[2, -2 + m, -1 + m, (1 + Sin[e + f*x])/2])/(-2 + m)) + (4*(A + B)
)/(-1 + Sin[e + f*x])^2)*(a*(1 + Sin[e + f*x]))^(-2 + m))/(16*f)

Maple [F]

\[\int \left (\sec ^{5}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right )d x\]

[In]

int(sec(f*x+e)^5*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x)

[Out]

int(sec(f*x+e)^5*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x)

Fricas [F]

\[ \int \sec ^5(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{5} \,d x } \]

[In]

integrate(sec(f*x+e)^5*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x, algorithm="fricas")

[Out]

integral((B*sec(f*x + e)^5*sin(f*x + e) + A*sec(f*x + e)^5)*(a*sin(f*x + e) + a)^m, x)

Sympy [F(-1)]

Timed out. \[ \int \sec ^5(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\text {Timed out} \]

[In]

integrate(sec(f*x+e)**5*(a+a*sin(f*x+e))**m*(A+B*sin(f*x+e)),x)

[Out]

Timed out

Maxima [F]

\[ \int \sec ^5(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{5} \,d x } \]

[In]

integrate(sec(f*x+e)^5*(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)^5, x)

Giac [F]

\[ \int \sec ^5(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{5} \,d x } \]

[In]

integrate(sec(f*x+e)^5*(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)^5, x)

Mupad [F(-1)]

Timed out. \[ \int \sec ^5(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\cos \left (e+f\,x\right )}^5} \,d x \]

[In]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m)/cos(e + f*x)^5,x)

[Out]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m)/cos(e + f*x)^5, x)