3.360 \(\int (e \cos (c+d x))^{-4-m} (a+a \sin (c+d x))^m \, dx\)

Optimal. Leaf size=201 \[ \frac{6 (a \sin (c+d x)+a)^{m+2} (e \cos (c+d x))^{-m-3}}{a^2 d e (3-m) \left (1-m^2\right )}-\frac{6 (a \sin (c+d x)+a)^{m+3} (e \cos (c+d x))^{-m-3}}{a^3 d e \left (m^4-10 m^2+9\right )}-\frac{(a \sin (c+d x)+a)^m (e \cos (c+d x))^{-m-3}}{d e (3-m)}-\frac{3 (a \sin (c+d x)+a)^{m+1} (e \cos (c+d x))^{-m-3}}{a d e (1-m) (3-m)} \]

[Out]

-(((e*Cos[c + d*x])^(-3 - m)*(a + a*Sin[c + d*x])^m)/(d*e*(3 - m))) - (3*(e*Cos[c + d*x])^(-3 - m)*(a + a*Sin[
c + d*x])^(1 + m))/(a*d*e*(1 - m)*(3 - m)) + (6*(e*Cos[c + d*x])^(-3 - m)*(a + a*Sin[c + d*x])^(2 + m))/(a^2*d
*e*(3 - m)*(1 - m^2)) - (6*(e*Cos[c + d*x])^(-3 - m)*(a + a*Sin[c + d*x])^(3 + m))/(a^3*d*e*(9 - 10*m^2 + m^4)
)

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Rubi [A]  time = 0.321228, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.074, Rules used = {2672, 2671} \[ \frac{6 (a \sin (c+d x)+a)^{m+2} (e \cos (c+d x))^{-m-3}}{a^2 d e (3-m) \left (1-m^2\right )}-\frac{6 (a \sin (c+d x)+a)^{m+3} (e \cos (c+d x))^{-m-3}}{a^3 d e \left (m^4-10 m^2+9\right )}-\frac{(a \sin (c+d x)+a)^m (e \cos (c+d x))^{-m-3}}{d e (3-m)}-\frac{3 (a \sin (c+d x)+a)^{m+1} (e \cos (c+d x))^{-m-3}}{a d e (1-m) (3-m)} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(-4 - m)*(a + a*Sin[c + d*x])^m,x]

[Out]

-(((e*Cos[c + d*x])^(-3 - m)*(a + a*Sin[c + d*x])^m)/(d*e*(3 - m))) - (3*(e*Cos[c + d*x])^(-3 - m)*(a + a*Sin[
c + d*x])^(1 + m))/(a*d*e*(1 - m)*(3 - m)) + (6*(e*Cos[c + d*x])^(-3 - m)*(a + a*Sin[c + d*x])^(2 + m))/(a^2*d
*e*(3 - m)*(1 - m^2)) - (6*(e*Cos[c + d*x])^(-3 - m)*(a + a*Sin[c + d*x])^(3 + m))/(a^3*d*e*(9 - 10*m^2 + m^4)
)

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))^{-4-m} (a+a \sin (c+d x))^m \, dx &=-\frac{(e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^m}{d e (3-m)}+\frac{3 \int (e \cos (c+d x))^{-4-m} (a+a \sin (c+d x))^{1+m} \, dx}{a (3-m)}\\ &=-\frac{(e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^m}{d e (3-m)}-\frac{3 (e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^{1+m}}{a d e (1-m) (3-m)}+\frac{6 \int (e \cos (c+d x))^{-4-m} (a+a \sin (c+d x))^{2+m} \, dx}{a^2 (1-m) (3-m)}\\ &=-\frac{(e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^m}{d e (3-m)}-\frac{3 (e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^{1+m}}{a d e (1-m) (3-m)}+\frac{6 (e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^{2+m}}{a^2 d e (1-m) (3-m) (1+m)}-\frac{6 \int (e \cos (c+d x))^{-4-m} (a+a \sin (c+d x))^{3+m} \, dx}{a^3 (1-m) (3-m) (1+m)}\\ &=-\frac{(e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^m}{d e (3-m)}-\frac{3 (e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^{1+m}}{a d e (1-m) (3-m)}+\frac{6 (e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^{2+m}}{a^2 d e (1-m) (3-m) (1+m)}-\frac{6 (e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^{3+m}}{a^3 d e \left (9-10 m^2+m^4\right )}\\ \end{align*}

Mathematica [A]  time = 0.196414, size = 101, normalized size = 0.5 \[ \frac{\sec ^3(c+d x) \left (-3 \left (m^2-3\right ) \sin (c+d x)+6 m \sin ^2(c+d x)-6 \sin ^3(c+d x)+m \left (m^2-7\right )\right ) (a (\sin (c+d x)+1))^m (e \cos (c+d x))^{-m}}{d e^4 (m-3) (m-1) (m+1) (m+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(-4 - m)*(a + a*Sin[c + d*x])^m,x]

[Out]

(Sec[c + d*x]^3*(a*(1 + Sin[c + d*x]))^m*(m*(-7 + m^2) - 3*(-3 + m^2)*Sin[c + d*x] + 6*m*Sin[c + d*x]^2 - 6*Si
n[c + d*x]^3))/(d*e^4*(-3 + m)*(-1 + m)*(1 + m)*(3 + m)*(e*Cos[c + d*x])^m)

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Maple [F]  time = 0.171, size = 0, normalized size = 0. \begin{align*} \int \left ( e\cos \left ( dx+c \right ) \right ) ^{-4-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))^(-4-m)*(a+a*sin(d*x+c))^m,x)

[Out]

int((e*cos(d*x+c))^(-4-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 - 4}{\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))^(-4-m)*(a+a*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(-m - 4)*(a*sin(d*x + c) + a)^m, x)

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Fricas [A]  time = 2.37926, size = 247, normalized size = 1.23 \begin{align*} -\frac{{\left (6 \, m \cos \left (d x + c\right )^{3} -{\left (m^{3} - m\right )} \cos \left (d x + c\right ) - 3 \,{\left (2 \, \cos \left (d x + c\right )^{3} -{\left (m^{2} - 1\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )} \left (e \cos \left (d x + c\right )\right )^{-m - 4}{\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{d m^{4} - 10 \, d m^{2} + 9 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(-4-m)*(a+a*sin(d*x+c))^m,x, algorithm="fricas")

[Out]

-(6*m*cos(d*x + c)^3 - (m^3 - m)*cos(d*x + c) - 3*(2*cos(d*x + c)^3 - (m^2 - 1)*cos(d*x + c))*sin(d*x + c))*(e
*cos(d*x + c))^(-m - 4)*(a*sin(d*x + c) + a)^m/(d*m^4 - 10*d*m^2 + 9*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))**(-4-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 - 4}{\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))^(-4-m)*(a+a*sin(d*x+c))^m,x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^(-m - 4)*(a*sin(d*x + c) + a)^m, x)