3.4.63 \(\int (e \cos (c+d x))^{-1-m} (a+a \sin (c+d x))^m \, dx\) [363]

Optimal. Leaf size=34 \[ \frac {(e \cos (c+d x))^{-m} (a+a \sin (c+d x))^m}{d e m} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {2750} \begin {gather*} \frac {(a \sin (c+d x)+a)^m (e \cos (c+d x))^{-m}}{d e m} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rubi steps

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

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Mathematica [A]
time = 0.06, size = 34, normalized size = 1.00 \begin {gather*} \frac {(e \cos (c+d x))^{-m} (a (1+\sin (c+d x)))^m}{d e m} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.20, size = 0, normalized size = 0.00 \[\int \left (e \cos \left (d x +c \right )\right )^{-1-m} \left (a +a \sin \left (d x +c \right )\right )^{m}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(-1-m)*(a+a*sin(d*x+c))^m,x)

[Out]

int((e*cos(d*x+c))^(-1-m)*(a+a*sin(d*x+c))^m,x)

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Maxima [A]
time = 0.49, size = 62, normalized size = 1.82 \begin {gather*} \frac {a^{m} e^{\left (m \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right ) - m \log \left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right ) - m - 1\right )}}{d m} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(-1-m)*(a+a*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.37, size = 40, normalized size = 1.18 \begin {gather*} \frac {\left (\cos \left (d x + c\right ) e\right )^{-m - 1} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} \cos \left (d x + c\right )}{d m} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(-1-m)*(a+a*sin(d*x+c))^m,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{m} \left (e \cos {\left (c + d x \right )}\right )^{- m - 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))**(-1-m)*(a+a*sin(d*x+c))**m,x)

[Out]

Integral((a*(sin(c + d*x) + 1))**m*(e*cos(c + d*x))**(-m - 1), x)

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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))^(-1-m)*(a+a*sin(d*x+c))^m,x, algorithm="giac")

[Out]

integrate((cos(d*x + c)*e)^(-m - 1)*(a*sin(d*x + c) + a)^m, x)

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Mupad [B]
time = 0.29, size = 34, normalized size = 1.00 \begin {gather*} \frac {{\left (a\,\left (\sin \left (c+d\,x\right )+1\right )\right )}^m}{d\,e\,m\,{\left (e\,\cos \left (c+d\,x\right )\right )}^m} \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 + 1),x)

[Out]

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

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