3.4.69 \(\int (e \cos (c+d x))^{1-2 m} (a+a \sin (c+d x))^m \, dx\) [369]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 44, 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 = {2752} \begin {gather*} -\frac {a (a \sin (c+d x)+a)^{m-1} (e \cos (c+d x))^{2-2 m}}{d e (1-m)} \end {gather*}

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 2752

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 - 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*}

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Mathematica [A]
time = 0.16, size = 43, normalized size = 0.98 \begin {gather*} -\frac {e (e \cos (c+d x))^{-2 m} (-1+\sin (c+d x)) (a (1+\sin (c+d x)))^m}{d (-1+m)} \end {gather*}

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 = 0.19, size = 0, normalized size = 0.00 \[\int \left (e \cos \left (d x +c \right )\right )^{1-2 m} \left (a +a \sin \left (d x +c \right )\right )^{m}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(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] Leaf count of result is larger than twice the leaf count of optimal. 155 vs. \(2 (41) = 82\).
time = 0.53, size = 155, normalized size = 3.52 \begin {gather*} \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 (m e^{\left (2 \, m\right )} + \frac {{\left (m e^{\left (2 \, m\right )} - e^{\left (2 \, m\right )}\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - e^{\left (2 \, m\right )}\right )} d} \end {gather*}

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

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Fricas [A]
time = 0.35, size = 81, normalized size = 1.84 \begin {gather*} \frac {\left (\cos \left (d x + c\right ) e\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 {gather*}

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]

(cos(d*x + c)*e)^(-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]
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 )^{1 - 2 m}\, dx \end {gather*}

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]

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

[Out]

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

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Mupad [B]
time = 5.58, size = 58, normalized size = 1.32 \begin {gather*} \frac {e\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )\,{\left (a\,\left (\sin \left (c+d\,x\right )+1\right )\right )}^m}{2\,d\,{\left (e\,\cos \left (c+d\,x\right )\right )}^{2\,m}\,\left (m-1\right )\,\left (\sin \left (c+d\,x\right )+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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