\(\int (1+\sin (e+f x))^m (3+5 \sin (e+f x))^{-1-m} \, dx\) [620]

Optimal result

Integrand size = 27, antiderivative size = 62 \[ \int (1+\sin (e+f x))^m (3+5 \sin (e+f x))^{-1-m} \, dx=-\frac {4^{-1-m} \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1+m,\frac {3}{2},\frac {1-\sin (e+f x)}{4 (1+\sin (e+f x))}\right )}{f (1+\sin (e+f x))} \]

[Out]

-4^(-1-m)*cos(f*x+e)*hypergeom([1/2, 1+m],[3/2],1/4*(1-sin(f*x+e))/(1+sin(f*x+e)))/f/(1+sin(f*x+e))

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.79, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2867, 134} \[ \int (1+\sin (e+f x))^m (3+5 \sin (e+f x))^{-1-m} \, dx=-\frac {2^{-2 m-1} \cos (e+f x) (\sin (e+f x)+1)^{m-1} \left (\frac {\sin (e+f x)+1}{5 \sin (e+f x)+3}\right )^{\frac {1}{2}-m} (5 \sin (e+f x)+3)^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},-\frac {1-\sin (e+f x)}{5 \sin (e+f x)+3}\right )}{f} \]

[In]

Int[(1 + Sin[e + f*x])^m*(3 + 5*Sin[e + f*x])^(-1 - m),x]

[Out]

-((2^(-1 - 2*m)*Cos[e + f*x]*Hypergeometric2F1[1/2, 1/2 - m, 3/2, -((1 - Sin[e + f*x])/(3 + 5*Sin[e + f*x]))]*
(1 + Sin[e + f*x])^(-1 + m)*((1 + Sin[e + f*x])/(3 + 5*Sin[e + f*x]))^(1/2 - m))/(f*(3 + 5*Sin[e + f*x])^m))

Rule 134

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

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\cos (e+f x) \text {Subst}\left (\int \frac {(1+x)^{-\frac {1}{2}+m} (3+5 x)^{-1-m}}{\sqrt {1-x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}} \\ & = -\frac {2^{-1-2 m} \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},-\frac {1-\sin (e+f x)}{3+5 \sin (e+f x)}\right ) (1+\sin (e+f x))^{-1+m} \left (\frac {1+\sin (e+f x)}{3+5 \sin (e+f x)}\right )^{\frac {1}{2}-m} (3+5 \sin (e+f x))^{-m}}{f} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(196\) vs. \(2(62)=124\).

Time = 7.67 (sec) , antiderivative size = 196, normalized size of antiderivative = 3.16 \[ \int (1+\sin (e+f x))^m (3+5 \sin (e+f x))^{-1-m} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (1+m,1+2 m,2 (1+m),\frac {4 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )}{3 \cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )}\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (-\frac {\cos \left (\frac {1}{2} (e+f x)\right )+3 \sin \left (\frac {1}{2} (e+f x)\right )}{3 \cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )}\right )^m (1+\sin (e+f x))^m (3+5 \sin (e+f x))^{-m}}{f (1+2 m) \left (3 \cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

[In]

Integrate[(1 + Sin[e + f*x])^m*(3 + 5*Sin[e + f*x])^(-1 - m),x]

[Out]

-((Hypergeometric2F1[1 + m, 1 + 2*m, 2*(1 + m), (4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))/(3*Cos[(e + f*x)/2]
+ Sin[(e + f*x)/2])]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-((Cos[(e + f*x)/2] + 3*Sin[(e + f*x)/2])/(3*Cos[(
e + f*x)/2] + Sin[(e + f*x)/2])))^m*(1 + Sin[e + f*x])^m)/(f*(1 + 2*m)*(3*Cos[(e + f*x)/2] + Sin[(e + f*x)/2])
*(3 + 5*Sin[e + f*x])^m))

Maple [F]

\[\int \left (\sin \left (f x +e \right )+1\right )^{m} \left (3+5 \sin \left (f x +e \right )\right )^{-1-m}d x\]

[In]

int((sin(f*x+e)+1)^m*(3+5*sin(f*x+e))^(-1-m),x)

[Out]

int((sin(f*x+e)+1)^m*(3+5*sin(f*x+e))^(-1-m),x)

Fricas [F]

\[ \int (1+\sin (e+f x))^m (3+5 \sin (e+f x))^{-1-m} \, dx=\int { {\left (5 \, \sin \left (f x + e\right ) + 3\right )}^{-m - 1} {\left (\sin \left (f x + e\right ) + 1\right )}^{m} \,d x } \]

[In]

integrate((1+sin(f*x+e))^m*(3+5*sin(f*x+e))^(-1-m),x, algorithm="fricas")

[Out]

integral((5*sin(f*x + e) + 3)^(-m - 1)*(sin(f*x + e) + 1)^m, x)

Sympy [F]

\[ \int (1+\sin (e+f x))^m (3+5 \sin (e+f x))^{-1-m} \, dx=\int \left (\sin {\left (e + f x \right )} + 1\right )^{m} \left (5 \sin {\left (e + f x \right )} + 3\right )^{- m - 1}\, dx \]

[In]

integrate((1+sin(f*x+e))**m*(3+5*sin(f*x+e))**(-1-m),x)

[Out]

Integral((sin(e + f*x) + 1)**m*(5*sin(e + f*x) + 3)**(-m - 1), x)

Maxima [F]

\[ \int (1+\sin (e+f x))^m (3+5 \sin (e+f x))^{-1-m} \, dx=\int { {\left (5 \, \sin \left (f x + e\right ) + 3\right )}^{-m - 1} {\left (\sin \left (f x + e\right ) + 1\right )}^{m} \,d x } \]

[In]

integrate((1+sin(f*x+e))^m*(3+5*sin(f*x+e))^(-1-m),x, algorithm="maxima")

[Out]

integrate((5*sin(f*x + e) + 3)^(-m - 1)*(sin(f*x + e) + 1)^m, x)

Giac [F]

\[ \int (1+\sin (e+f x))^m (3+5 \sin (e+f x))^{-1-m} \, dx=\int { {\left (5 \, \sin \left (f x + e\right ) + 3\right )}^{-m - 1} {\left (\sin \left (f x + e\right ) + 1\right )}^{m} \,d x } \]

[In]

integrate((1+sin(f*x+e))^m*(3+5*sin(f*x+e))^(-1-m),x, algorithm="giac")

[Out]

integrate((5*sin(f*x + e) + 3)^(-m - 1)*(sin(f*x + e) + 1)^m, x)

Mupad [F(-1)]

Timed out. \[ \int (1+\sin (e+f x))^m (3+5 \sin (e+f x))^{-1-m} \, dx=\int \frac {{\left (\sin \left (e+f\,x\right )+1\right )}^m}{{\left (5\,\sin \left (e+f\,x\right )+3\right )}^{m+1}} \,d x \]

[In]

int((sin(e + f*x) + 1)^m/(5*sin(e + f*x) + 3)^(m + 1),x)

[Out]

int((sin(e + f*x) + 1)^m/(5*sin(e + f*x) + 3)^(m + 1), x)