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[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))

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Rubi [A]  time = 0.172097, antiderivative size = 51, normalized size of antiderivative = 1., 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 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 12

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

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 = 0.95267, 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

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Maple [F]  time = 0.446, size = 0, normalized size = 0. \begin{align*} \int \left ( g\cos \left ( fx+e \right ) \right ) ^{-1-2\,m} \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( c-c\sin \left ( fx+e \right ) \right ) ^{m}\, dx \end{align*}

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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)

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Fricas [A]  time = 1.77073, size = 113, normalized size = 2.22 \begin{align*} \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} \end{align*}

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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)