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

Optimal. Leaf size=51 \[ \frac {\tanh ^{-1}(\sin (e+f x)) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m (g \cos (e+f x))^{-2 m}}{f g} \]

[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]  time = 0.17, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {2847, 12, 3770} \[ \frac {\tanh ^{-1}(\sin (e+f x)) (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^m (g \cos (e+f x))^{-2 m}}{f g} \]

Antiderivative was successfully verified.

[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 2847

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 3770

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

Rubi steps

\begin {align*} \int (g \cos (e+f x))^{-1-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m \, dx &=\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 {\tanh ^{-1}(\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]  time = 1.00, size = 94, normalized size = 1.84 \[ \frac {\sqrt {-\tan ^2(e+f x)} \csc (e+f x) \cos ^{2 (m+1)}(e+f x) \sin ^{-1}(\sec (e+f x)) (g \cos (e+f x))^{-2 m-1} \exp (m (\log (a (\sin (e+f x)+1))+\log (c-c \sin (e+f x))-2 \log (\cos (e+f x))))}{f} \]

Antiderivative was successfully verified.

[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

________________________________________________________________________________________

fricas [A]  time = 0.52, size = 48, normalized size = 0.94 \[ \frac {\left (\frac {a c}{g^{2}}\right )^{m} \log \left (\sin \left (f x + e\right ) + 1\right ) - \left (\frac {a c}{g^{2}}\right )^{m} \log \left (-\sin \left (f x + e\right ) + 1\right )}{2 \, f g} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (g \cos \left (f x + e\right )\right )^{-2 \, m - 1} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [F]  time = 0.96, size = 0, normalized size = 0.00 \[ \int \left (g \cos \left (f x +e \right )\right )^{-1-2 m} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (g \cos \left (f x + e\right )\right )^{-2 \, m - 1} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^m}{{\left (g\,\cos \left (e+f\,x\right )\right )}^{2\,m+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________