\(\int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m \, dx\) [181]

Optimal result

Integrand size = 40, antiderivative size = 51 \[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m \, dx=\frac {\text {arctanh}(\sin (e+f x)) (g \cos (e+f x))^{-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m}{f g} \]

[Out]

arctanh(sin(f*x+e))*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^m/f/g/((g*cos(f*x+e))^(2*m))

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {2926, 12, 3855} \[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m \, dx=\frac {\text {arctanh}(\sin (e+f x)) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m (g \cos (e+f x))^{-2 m}}{f g} \]

[In]

Int[(g*Cos[e + f*x])^(-1 - 2*m)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^m,x]

[Out]

(ArcTanh[Sin[e + f*x]]*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^m)/(f*g*(g*Cos[e + f*x])^(2*m))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2926

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

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

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

Mathematica [A] (verified)

Time = 2.22 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.84 \[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m \, dx=\frac {e^{m (-2 \log (\cos (e+f x))+\log (a (1+\sin (e+f x)))+\log (c-c \sin (e+f x)))} \arcsin (\sec (e+f x)) \cos ^{2 (1+m)}(e+f x) (g \cos (e+f x))^{-1-2 m} \csc (e+f x) \sqrt {-\tan ^2(e+f x)}}{f} \]

[In]

Integrate[(g*Cos[e + f*x])^(-1 - 2*m)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^m,x]

[Out]

(E^(m*(-2*Log[Cos[e + f*x]] + Log[a*(1 + Sin[e + f*x])] + Log[c - c*Sin[e + f*x]]))*ArcSin[Sec[e + f*x]]*Cos[e
+ f*x]^(2*(1 + m))*(g*Cos[e + f*x])^(-1 - 2*m)*Csc[e + f*x]*Sqrt[-Tan[e + f*x]^2])/f

Maple [F]

\[\int \left (g \cos \left (f x +e \right )\right )^{-1-2 m} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{m}d x\]

[In]

int((g*cos(f*x+e))^(-1-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^m,x)

[Out]

int((g*cos(f*x+e))^(-1-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^m,x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.94 \[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m \, dx=\frac {\left (\frac {a c}{g^{2}}\right )^{m} \log \left (\sin \left (f x + e\right ) + 1\right ) - \left (\frac {a c}{g^{2}}\right )^{m} \log \left (-\sin \left (f x + e\right ) + 1\right )}{2 \, f g} \]

[In]

integrate((g*cos(f*x+e))^(-1-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^m,x, algorithm="fricas")

[Out]

1/2*((a*c/g^2)^m*log(sin(f*x + e) + 1) - (a*c/g^2)^m*log(-sin(f*x + e) + 1))/(f*g)

Sympy [F(-1)]

Timed out. \[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m \, dx=\text {Timed out} \]

[In]

integrate((g*cos(f*x+e))**(-1-2*m)*(a+a*sin(f*x+e))**m*(c-c*sin(f*x+e))**m,x)

[Out]

Timed out

Maxima [F]

\[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{-2 \, m - 1} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{m} \,d x } \]

[In]

integrate((g*cos(f*x+e))^(-1-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^m,x, algorithm="maxima")

[Out]

integrate((g*cos(f*x + e))^(-2*m - 1)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^m, x)

Giac [F]

\[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{-2 \, m - 1} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{m} \,d x } \]

[In]

integrate((g*cos(f*x+e))^(-1-2*m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^m,x, algorithm="giac")

[Out]

integrate((g*cos(f*x + e))^(-2*m - 1)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^m, x)

Mupad [F(-1)]

Timed out. \[ \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^m}{{\left (g\,\cos \left (e+f\,x\right )\right )}^{2\,m+1}} \,d x \]

[In]

int(((a + a*sin(e + f*x))^m*(c - c*sin(e + f*x))^m)/(g*cos(e + f*x))^(2*m + 1),x)

[Out]

int(((a + a*sin(e + f*x))^m*(c - c*sin(e + f*x))^m)/(g*cos(e + f*x))^(2*m + 1), x)