3.4.62 \(\int (e \cos (c+d x))^{-2-m} (a+a \sin (c+d x))^m \, dx\) [362]

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

[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 = 0.09, 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 = {2751, 2750} \begin {gather*} \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)} \end {gather*}

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 2750

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[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 2751

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[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 \begin {gather*} \frac {(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)} \end {gather*}

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 = 0.19, 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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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Fricas [A]
time = 0.35, size = 62, normalized size = 0.70 \begin {gather*} \frac {{\left (m \cos \left (d x + c\right ) - \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \left (\cos \left (d x + c\right ) e\right )^{-m - 2} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{d m^{2} - d} \end {gather*}

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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((cos(d*x + c)*e)^(-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 \begin {gather*} -\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 )} \end {gather*}

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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