3.362 \(\int (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^m \, dx\)

Optimal. Leaf size=89 \[ \frac{(a \sin (c+d x)+a)^{m+1} (e \cos (c+d x))^{-m-1}}{a d e \left (1-m^2\right )}-\frac{(a \sin (c+d x)+a)^m (e \cos (c+d x))^{-m-1}}{d e (1-m)} \]

[Out]

-(((e*Cos[c + d*x])^(-1 - m)*(a + a*Sin[c + d*x])^m)/(d*e*(1 - m))) + ((e*Cos[c + d*x])^(-1 - m)*(a + a*Sin[c
+ d*x])^(1 + m))/(a*d*e*(1 - m^2))

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Rubi [A]  time = 0.124096, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.074, Rules used = {2672, 2671} \[ \frac{(a \sin (c+d x)+a)^{m+1} (e \cos (c+d x))^{-m-1}}{a d e \left (1-m^2\right )}-\frac{(a \sin (c+d x)+a)^m (e \cos (c+d x))^{-m-1}}{d e (1-m)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((e*Cos[c + d*x])^(-1 - m)*(a + a*Sin[c + d*x])^m)/(d*e*(1 - m))) + ((e*Cos[c + d*x])^(-1 - m)*(a + a*Sin[c
+ d*x])^(1 + m))/(a*d*e*(1 - m^2))

Rule 2672

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*Simplify[2*m + p + 1]), x] + Dist[Simplify[m + p + 1]/(a*
Simplify[2*m + p + 1]), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g, m
, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + p + 1], 0] && NeQ[2*m + p + 1, 0] &&  !IGtQ[m, 0]

Rule 2671

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

Rubi steps

\begin{align*} \int (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^m \, dx &=-\frac{(e \cos (c+d x))^{-1-m} (a+a \sin (c+d x))^m}{d e (1-m)}+\frac{\int (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^{1+m} \, dx}{a (1-m)}\\ &=-\frac{(e \cos (c+d x))^{-1-m} (a+a \sin (c+d x))^m}{d e (1-m)}+\frac{(e \cos (c+d x))^{-1-m} (a+a \sin (c+d x))^{1+m}}{a d e \left (1-m^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.120738, size = 53, normalized size = 0.6 \[ \frac{(m-\sin (c+d x)) (a (\sin (c+d x)+1))^m (e \cos (c+d x))^{-m-1}}{d e (m-1) (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*Cos[c + d*x])^(-1 - m)*(m - Sin[c + d*x])*(a*(1 + Sin[c + d*x]))^m)/(d*e*(-1 + m)*(1 + m))

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Maple [F]  time = 0.148, size = 0, normalized size = 0. \begin{align*} \int \left ( e\cos \left ( dx+c \right ) \right ) ^{-2-m} \left ( a+a\sin \left ( dx+c \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e \cos \left (d x + c\right )\right )^{-m - 2}{\left (a \sin \left (d x + c\right ) + a\right )}^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.37061, size = 144, normalized size = 1.62 \begin{align*} \frac{{\left (m \cos \left (d x + c\right ) - \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \left (e \cos \left (d x + c\right )\right )^{-m - 2}{\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{d m^{2} - d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(m*cos(d*x + c) - cos(d*x + c)*sin(d*x + c))*(e*cos(d*x + c))^(-m - 2)*(a*sin(d*x + c) + a)^m/(d*m^2 - d)

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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((e*cos(d*x+c))**(-2-m)*(a+a*sin(d*x+c))**m,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e \cos \left (d x + c\right )\right )^{-m - 2}{\left (a \sin \left (d x + c\right ) + a\right )}^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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