3.369 \(\int (e \cos (c+d x))^{1-2 m} (a+a \sin (c+d x))^m \, dx\)

Optimal. Leaf size=44 \[ -\frac{a (a \sin (c+d x)+a)^{m-1} (e \cos (c+d x))^{2-2 m}}{d e (1-m)} \]

[Out]

-((a*(e*Cos[c + d*x])^(2 - 2*m)*(a + a*Sin[c + d*x])^(-1 + m))/(d*e*(1 - m)))

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Rubi [A]  time = 0.0554556, antiderivative size = 44, 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 = {2673} \[ -\frac{a (a \sin (c+d x)+a)^{m-1} (e \cos (c+d x))^{2-2 m}}{d e (1-m)} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(1 - 2*m)*(a + a*Sin[c + d*x])^m,x]

[Out]

-((a*(e*Cos[c + d*x])^(2 - 2*m)*(a + a*Sin[c + d*x])^(-1 + m))/(d*e*(1 - m)))

Rule 2673

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 - 1))/(f*g*(m - 1)), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

Rubi steps

\begin{align*} \int (e \cos (c+d x))^{1-2 m} (a+a \sin (c+d x))^m \, dx &=-\frac{a (e \cos (c+d x))^{2-2 m} (a+a \sin (c+d x))^{-1+m}}{d e (1-m)}\\ \end{align*}

Mathematica [A]  time = 0.153695, size = 43, normalized size = 0.98 \[ -\frac{e (\sin (c+d x)-1) (a (\sin (c+d x)+1))^m (e \cos (c+d x))^{-2 m}}{d (m-1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(1 - 2*m)*(a + a*Sin[c + d*x])^m,x]

[Out]

-((e*(-1 + Sin[c + d*x])*(a*(1 + Sin[c + d*x]))^m)/(d*(-1 + m)*(e*Cos[c + d*x])^(2*m)))

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Maple [F]  time = 1.25, size = 0, normalized size = 0. \begin{align*} \int \left ( e\cos \left ( dx+c \right ) \right ) ^{1-2\,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))^(1-2*m)*(a+a*sin(d*x+c))^m,x)

[Out]

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

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Maxima [B]  time = 1.50738, size = 194, normalized size = 4.41 \begin{align*} \frac{{\left (a^{m} e - \frac{2 \, a^{m} e \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{a^{m} e \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} e^{\left (-2 \, 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 )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )\right )}}{{\left (e^{2 \, m}{\left (m - 1\right )} + \frac{e^{2 \, m}{\left (m - 1\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(1-2*m)*(a+a*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.29197, size = 197, normalized size = 4.48 \begin{align*} \frac{\left (e \cos \left (d x + c\right )\right )^{-2 \, m + 1}{\left (a \sin \left (d x + c\right ) + a\right )}^{m}{\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{d m +{\left (d m - d\right )} \cos \left (d x + c\right ) +{\left (d m - d\right )} \sin \left (d x + c\right ) - d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(1-2*m)*(a+a*sin(d*x+c))^m,x, algorithm="fricas")

[Out]

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

[Out]

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