\(\int \sec ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx\) [1025]

Optimal result

Integrand size = 31, antiderivative size = 100 \[ \int \sec ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=-\frac {a (A (2-m)-B m) \operatorname {Hypergeometric2F1}\left (1,-1+m,m,\frac {1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^{-1+m}}{4 f (1-m)}+\frac {a^2 (A+B) (a+a \sin (e+f x))^{-1+m}}{2 f (a-a \sin (e+f x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 100, 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} \[ \int \sec ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\frac {a^2 (A+B) (a \sin (e+f x)+a)^{m-1}}{2 f (a-a \sin (e+f x))}-\frac {a (A (2-m)-B m) (a \sin (e+f x)+a)^{m-1} \operatorname {Hypergeometric2F1}\left (1,m-1,m,\frac {1}{2} (\sin (e+f x)+1)\right )}{4 f (1-m)} \]

[In]

Int[Sec[e + f*x]^3*(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.82 \[ \int \sec ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=-\frac {a \left (2 (A+B) (-1+m)+(A (-2+m)+B m) \operatorname {Hypergeometric2F1}\left (1,-1+m,m,\frac {1}{2} (1+\sin (e+f x))\right ) (-1+\sin (e+f x))\right ) (a (1+\sin (e+f x)))^{-1+m}}{4 f (-1+m) (-1+\sin (e+f x))} \]

[In]

Integrate[Sec[e + f*x]^3*(a + a*Sin[e + f*x])^m*(A + B*Sin[e + f*x]),x]

[Out]

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

Maple [F]

\[\int \left (\sec ^{3}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right )d x\]

[In]

int(sec(f*x+e)^3*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x)

[Out]

int(sec(f*x+e)^3*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x)

Fricas [F]

\[ \int \sec ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{3} \,d x } \]

[In]

integrate(sec(f*x+e)^3*(a+a*sin(f*x+e))^m*(A+B*sin(f*x+e)),x, algorithm="fricas")

[Out]

integral((B*sec(f*x + e)^3*sin(f*x + e) + A*sec(f*x + e)^3)*(a*sin(f*x + e) + a)^m, x)

Sympy [F(-1)]

Timed out. \[ \int \sec ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\text {Timed out} \]

[In]

integrate(sec(f*x+e)**3*(a+a*sin(f*x+e))**m*(A+B*sin(f*x+e)),x)

[Out]

Timed out

Maxima [F]

\[ \int \sec ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{3} \,d x } \]

[In]

integrate(sec(f*x+e)^3*(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)^3, x)

Giac [F]

\[ \int \sec ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{3} \,d x } \]

[In]

integrate(sec(f*x+e)^3*(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)^3, x)

Mupad [F(-1)]

Timed out. \[ \int \sec ^3(e+f x) (a+a \sin (e+f x))^m (A+B \sin (e+f x)) \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\cos \left (e+f\,x\right )}^3} \,d x \]

[In]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m)/cos(e + f*x)^3,x)

[Out]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^m)/cos(e + f*x)^3, x)