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(d*x+c))^(-1-m)*(a+a*sin(d*x+c))^m/d/e/(1-m)+(e*cos(d*x+c))^(-1-m)*(a+a*sin(d*x+c))^(1+m)/a/d/e/(-m^2+1
)

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Rubi [A]  time = 0.12, antiderivative size = 89, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.12, size = 53, normalized size = 0.60 \[ \frac {(m-\sin (c+d x)) (a (\sin (c+d x)+1))^m (e \cos (c+d x))^{-m-1}}{d e (m-1) (m+1)} \]

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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fricas [A]  time = 0.47, size = 61, normalized size = 0.69 \[ \frac {{\left (m \cos \left (d x + c\right ) - \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \left (e \cos \left (d x + c\right )\right )^{-m - 2} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{d m^{2} - d} \]

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

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)

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maple [F]  time = 0.32, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x +c \right )\right )^{-2-m} \left (a +a \sin \left (d x +c \right )\right )^{m}\, dx \]

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

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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mupad [B]  time = 5.60, size = 71, normalized size = 0.80 \[ -\frac {\left (\sin \left (2\,c+2\,d\,x\right )-2\,m\,\cos \left (c+d\,x\right )\right )\,{\left (a\,\left (\sin \left (c+d\,x\right )+1\right )\right )}^m}{d\,e^2\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )\,{\left (e\,\cos \left (c+d\,x\right )\right )}^m\,\left (m^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((sin(2*c + 2*d*x) - 2*m*cos(c + d*x))*(a*(sin(c + d*x) + 1))^m)/(d*e^2*(cos(2*c + 2*d*x) + 1)*(e*cos(c + d*x
))^m*(m^2 - 1))

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

[Out]

Timed out

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