3.361 \(\int (e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^m \, dx\)

Optimal. Leaf size=142 \[ -\frac{2 (a \sin (c+d x)+a)^{m+2} (e \cos (c+d x))^{-m-2}}{a^2 d e m \left (4-m^2\right )}-\frac{(a \sin (c+d x)+a)^m (e \cos (c+d x))^{-m-2}}{d e (2-m)}+\frac{2 (a \sin (c+d x)+a)^{m+1} (e \cos (c+d x))^{-m-2}}{a d e (2-m) m} \]

[Out]

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

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Rubi [A]  time = 0.222483, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.074, Rules used = {2672, 2671} \[ -\frac{2 (a \sin (c+d x)+a)^{m+2} (e \cos (c+d x))^{-m-2}}{a^2 d e m \left (4-m^2\right )}-\frac{(a \sin (c+d x)+a)^m (e \cos (c+d x))^{-m-2}}{d e (2-m)}+\frac{2 (a \sin (c+d x)+a)^{m+1} (e \cos (c+d x))^{-m-2}}{a d e (2-m) m} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(-3 - m)*(a + a*Sin[c + d*x])^m,x]

[Out]

-(((e*Cos[c + d*x])^(-2 - m)*(a + a*Sin[c + d*x])^m)/(d*e*(2 - m))) + (2*(e*Cos[c + d*x])^(-2 - m)*(a + a*Sin[
c + d*x])^(1 + m))/(a*d*e*(2 - m)*m) - (2*(e*Cos[c + d*x])^(-2 - m)*(a + a*Sin[c + d*x])^(2 + m))/(a^2*d*e*m*(
4 - 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))^{-3-m} (a+a \sin (c+d x))^m \, dx &=-\frac{(e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^m}{d e (2-m)}+\frac{2 \int (e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^{1+m} \, dx}{a (2-m)}\\ &=-\frac{(e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^m}{d e (2-m)}+\frac{2 (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^{1+m}}{a d e (2-m) m}-\frac{2 \int (e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^{2+m} \, dx}{a^2 (2-m) m}\\ &=-\frac{(e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^m}{d e (2-m)}+\frac{2 (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^{1+m}}{a d e (2-m) m}-\frac{2 (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^{2+m}}{a^2 d e m \left (4-m^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.136861, size = 76, normalized size = 0.54 \[ \frac{\sec ^2(c+d x) \left (-2 m \sin (c+d x)+2 \sin ^2(c+d x)+m^2-2\right ) (a (\sin (c+d x)+1))^m (e \cos (c+d x))^{-m}}{d e^3 (m-2) m (m+2)} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(-3 - m)*(a + a*Sin[c + d*x])^m,x]

[Out]

(Sec[c + d*x]^2*(a*(1 + Sin[c + d*x]))^m*(-2 + m^2 - 2*m*Sin[c + d*x] + 2*Sin[c + d*x]^2))/(d*e^3*(-2 + m)*m*(
2 + m)*(e*Cos[c + d*x])^m)

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Maple [F]  time = 0.157, size = 0, normalized size = 0. \begin{align*} \int \left ( e\cos \left ( dx+c \right ) \right ) ^{-3-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))^(-3-m)*(a+a*sin(d*x+c))^m,x)

[Out]

int((e*cos(d*x+c))^(-3-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 - 3}{\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))^(-3-m)*(a+a*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(-m - 3)*(a*sin(d*x + c) + a)^m, x)

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Fricas [A]  time = 2.33089, size = 184, normalized size = 1.3 \begin{align*} \frac{{\left (m^{2} \cos \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{3} - 2 \, m \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \left (e \cos \left (d x + c\right )\right )^{-m - 3}{\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{d m^{3} - 4 \, d m} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(-3-m)*(a+a*sin(d*x+c))^m,x, algorithm="fricas")

[Out]

(m^2*cos(d*x + c) - 2*cos(d*x + c)^3 - 2*m*cos(d*x + c)*sin(d*x + c))*(e*cos(d*x + c))^(-m - 3)*(a*sin(d*x + c
) + a)^m/(d*m^3 - 4*d*m)

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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))**(-3-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 - 3}{\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))^(-3-m)*(a+a*sin(d*x+c))^m,x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^(-m - 3)*(a*sin(d*x + c) + a)^m, x)