3.368 \(\int (e \cos (c+d x))^{3-2 m} (a+a \sin (c+d x))^m \, dx\)

Optimal. Leaf size=94 \[ -\frac{2 a^2 (a \sin (c+d x)+a)^{m-2} (e \cos (c+d x))^{4-2 m}}{d e \left (m^2-5 m+6\right )}-\frac{a (a \sin (c+d x)+a)^{m-1} (e \cos (c+d x))^{4-2 m}}{d e (3-m)} \]

[Out]

(-2*a^2*(e*Cos[c + d*x])^(4 - 2*m)*(a + a*Sin[c + d*x])^(-2 + m))/(d*e*(6 - 5*m + m^2)) - (a*(e*Cos[c + d*x])^
(4 - 2*m)*(a + a*Sin[c + d*x])^(-1 + m))/(d*e*(3 - m))

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Rubi [A]  time = 0.140641, antiderivative size = 94, normalized size of antiderivative = 1., 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 = {2674, 2673} \[ -\frac{2 a^2 (a \sin (c+d x)+a)^{m-2} (e \cos (c+d x))^{4-2 m}}{d e \left (m^2-5 m+6\right )}-\frac{a (a \sin (c+d x)+a)^{m-1} (e \cos (c+d x))^{4-2 m}}{d e (3-m)} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(3 - 2*m)*(a + a*Sin[c + d*x])^m,x]

[Out]

(-2*a^2*(e*Cos[c + d*x])^(4 - 2*m)*(a + a*Sin[c + d*x])^(-2 + m))/(d*e*(6 - 5*m + m^2)) - (a*(e*Cos[c + d*x])^
(4 - 2*m)*(a + a*Sin[c + d*x])^(-1 + m))/(d*e*(3 - m))

Rule 2674

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 + p)), x] + Dist[(a*(2*m + p - 1))/(m + p), 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]
&& IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]

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))^{3-2 m} (a+a \sin (c+d x))^m \, dx &=-\frac{a (e \cos (c+d x))^{4-2 m} (a+a \sin (c+d x))^{-1+m}}{d e (3-m)}+\frac{(2 a) \int (e \cos (c+d x))^{3-2 m} (a+a \sin (c+d x))^{-1+m} \, dx}{3-m}\\ &=-\frac{2 a^2 (e \cos (c+d x))^{4-2 m} (a+a \sin (c+d x))^{-2+m}}{d e \left (6-5 m+m^2\right )}-\frac{a (e \cos (c+d x))^{4-2 m} (a+a \sin (c+d x))^{-1+m}}{d e (3-m)}\\ \end{align*}

Mathematica [A]  time = 0.219216, size = 72, normalized size = 0.77 \[ \frac{e^3 \cos ^4(c+d x) ((m-2) \sin (c+d x)+m-4) (a (\sin (c+d x)+1))^m (e \cos (c+d x))^{-2 m}}{d (m-3) (m-2) (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(3 - 2*m)*(a + a*Sin[c + d*x])^m,x]

[Out]

(e^3*Cos[c + d*x]^4*(a*(1 + Sin[c + d*x]))^m*(-4 + m + (-2 + m)*Sin[c + d*x]))/(d*(-3 + m)*(-2 + m)*(e*Cos[c +
d*x])^(2*m)*(1 + Sin[c + d*x])^2)

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Maple [F]  time = 0.935, size = 0, normalized size = 0. \begin{align*} \int \left ( e\cos \left ( dx+c \right ) \right ) ^{3-2\,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))^(3-2*m)*(a+a*sin(d*x+c))^m,x)

[Out]

int((e*cos(d*x+c))^(3-2*m)*(a+a*sin(d*x+c))^m,x)

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Maxima [B]  time = 1.5684, size = 474, normalized size = 5.04 \begin{align*} \frac{{\left (a^{m} e^{3}{\left (m - 4\right )} - \frac{2 \, a^{m} e^{3}{\left (m - 6\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{a^{m} e^{3}{\left (m + 12\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{4 \, a^{m} e^{3}{\left (m + 2\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{a^{m} e^{3}{\left (m + 12\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{2 \, a^{m} e^{3}{\left (m - 6\right )} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{a^{m} e^{3}{\left (m - 4\right )} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}\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 ({\left (m^{2} - 5 \, m + 6\right )} e^{2 \, m} + \frac{3 \,{\left (m^{2} - 5 \, m + 6\right )} e^{2 \, m} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{3 \,{\left (m^{2} - 5 \, m + 6\right )} e^{2 \, m} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{{\left (m^{2} - 5 \, m + 6\right )} e^{2 \, m} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(3-2*m)*(a+a*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

(a^m*e^3*(m - 4) - 2*a^m*e^3*(m - 6)*sin(d*x + c)/(cos(d*x + c) + 1) - a^m*e^3*(m + 12)*sin(d*x + c)^2/(cos(d*
x + c) + 1)^2 + 4*a^m*e^3*(m + 2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - a^m*e^3*(m + 12)*sin(d*x + c)^4/(cos(d
*x + c) + 1)^4 - 2*a^m*e^3*(m - 6)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + a^m*e^3*(m - 4)*sin(d*x + c)^6/(cos(d
*x + c) + 1)^6)*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^2 - 5*m + 6)*e^(2*m) + 3*(m^2 - 5*m + 6)*e^(2*m)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*(m^2 - 5*m
+ 6)*e^(2*m)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + (m^2 - 5*m + 6)*e^(2*m)*sin(d*x + c)^6/(cos(d*x + c) + 1)^
6)*d)

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Fricas [A]  time = 2.37375, size = 440, normalized size = 4.68 \begin{align*} \frac{{\left ({\left (m - 2\right )} \cos \left (d x + c\right )^{2} +{\left (m - 4\right )} \cos \left (d x + c\right ) +{\left ({\left (m - 2\right )} \cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - 2\right )} \left (e \cos \left (d x + c\right )\right )^{-2 \, m + 3}{\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{2 \, d m^{2} -{\left (d m^{2} - 5 \, d m + 6 \, d\right )} \cos \left (d x + c\right )^{2} - 10 \, d m +{\left (d m^{2} - 5 \, d m + 6 \, d\right )} \cos \left (d x + c\right ) +{\left (2 \, d m^{2} - 10 \, d m +{\left (d m^{2} - 5 \, d m + 6 \, d\right )} \cos \left (d x + c\right ) + 12 \, d\right )} \sin \left (d x + c\right ) + 12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(3-2*m)*(a+a*sin(d*x+c))^m,x, algorithm="fricas")

[Out]

((m - 2)*cos(d*x + c)^2 + (m - 4)*cos(d*x + c) + ((m - 2)*cos(d*x + c) + 2)*sin(d*x + c) - 2)*(e*cos(d*x + c))
^(-2*m + 3)*(a*sin(d*x + c) + a)^m/(2*d*m^2 - (d*m^2 - 5*d*m + 6*d)*cos(d*x + c)^2 - 10*d*m + (d*m^2 - 5*d*m +
6*d)*cos(d*x + c) + (2*d*m^2 - 10*d*m + (d*m^2 - 5*d*m + 6*d)*cos(d*x + c) + 12*d)*sin(d*x + c) + 12*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))**(3-2*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 + 3}{\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))^(3-2*m)*(a+a*sin(d*x+c))^m,x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^(-2*m + 3)*(a*sin(d*x + c) + a)^m, x)