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

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

[Out]

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

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Rubi [A]  time = 0.0506929, antiderivative size = 34, normalized size of antiderivative = 1., 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 = {2671} \[ \frac{(a \sin (c+d x)+a)^m (e \cos (c+d x))^{-m}}{d e m} \]

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 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))^{-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*}

Mathematica [A]  time = 0.0531585, size = 34, normalized size = 1. \[ \frac{(a (\sin (c+d x)+1))^m (e \cos (c+d x))^{-m}}{d e m} \]

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 = 0.144, size = 0, normalized size = 0. \begin{align*} \int \left ( e\cos \left ( dx+c \right ) \right ) ^{-1-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))^(-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 = 1.57164, size = 88, normalized size = 2.59 \begin{align*} \frac{a^{m} e^{-m - 1} 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 )\right )}}{d m} \end{align*}

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 - 1)*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))/(d
*m)

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Fricas [A]  time = 2.28788, size = 93, normalized size = 2.74 \begin{align*} \frac{\left (e \cos \left (d x + c\right )\right )^{-m - 1}{\left (a \sin \left (d x + c\right ) + a\right )}^{m} \cos \left (d x + c\right )}{d m} \end{align*}

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]

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

[Out]

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