3.367 \(\int (e \cos (c+d x))^{5-2 m} (a+a \sin (c+d x))^m \, dx\)

Optimal. Leaf size=150 \[ -\frac{8 a^3 (a \sin (c+d x)+a)^{m-3} (e \cos (c+d x))^{6-2 m}}{d e (5-m) \left (m^2-7 m+12\right )}-\frac{4 a^2 (a \sin (c+d x)+a)^{m-2} (e \cos (c+d x))^{6-2 m}}{d e \left (m^2-9 m+20\right )}-\frac{a (a \sin (c+d x)+a)^{m-1} (e \cos (c+d x))^{6-2 m}}{d e (5-m)} \]

[Out]

(-8*a^3*(e*Cos[c + d*x])^(6 - 2*m)*(a + a*Sin[c + d*x])^(-3 + m))/(d*e*(5 - m)*(12 - 7*m + m^2)) - (4*a^2*(e*C
os[c + d*x])^(6 - 2*m)*(a + a*Sin[c + d*x])^(-2 + m))/(d*e*(20 - 9*m + m^2)) - (a*(e*Cos[c + d*x])^(6 - 2*m)*(
a + a*Sin[c + d*x])^(-1 + m))/(d*e*(5 - m))

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Rubi [A]  time = 0.240436, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.074, Rules used = {2674, 2673} \[ -\frac{8 a^3 (a \sin (c+d x)+a)^{m-3} (e \cos (c+d x))^{6-2 m}}{d e (5-m) \left (m^2-7 m+12\right )}-\frac{4 a^2 (a \sin (c+d x)+a)^{m-2} (e \cos (c+d x))^{6-2 m}}{d e \left (m^2-9 m+20\right )}-\frac{a (a \sin (c+d x)+a)^{m-1} (e \cos (c+d x))^{6-2 m}}{d e (5-m)} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(5 - 2*m)*(a + a*Sin[c + d*x])^m,x]

[Out]

(-8*a^3*(e*Cos[c + d*x])^(6 - 2*m)*(a + a*Sin[c + d*x])^(-3 + m))/(d*e*(5 - m)*(12 - 7*m + m^2)) - (4*a^2*(e*C
os[c + d*x])^(6 - 2*m)*(a + a*Sin[c + d*x])^(-2 + m))/(d*e*(20 - 9*m + m^2)) - (a*(e*Cos[c + d*x])^(6 - 2*m)*(
a + a*Sin[c + d*x])^(-1 + m))/(d*e*(5 - 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))^{5-2 m} (a+a \sin (c+d x))^m \, dx &=-\frac{a (e \cos (c+d x))^{6-2 m} (a+a \sin (c+d x))^{-1+m}}{d e (5-m)}+\frac{(4 a) \int (e \cos (c+d x))^{5-2 m} (a+a \sin (c+d x))^{-1+m} \, dx}{5-m}\\ &=-\frac{4 a^2 (e \cos (c+d x))^{6-2 m} (a+a \sin (c+d x))^{-2+m}}{d e \left (20-9 m+m^2\right )}-\frac{a (e \cos (c+d x))^{6-2 m} (a+a \sin (c+d x))^{-1+m}}{d e (5-m)}+\frac{\left (8 a^2\right ) \int (e \cos (c+d x))^{5-2 m} (a+a \sin (c+d x))^{-2+m} \, dx}{20-9 m+m^2}\\ &=-\frac{8 a^3 (e \cos (c+d x))^{6-2 m} (a+a \sin (c+d x))^{-3+m}}{d e (3-m) \left (20-9 m+m^2\right )}-\frac{4 a^2 (e \cos (c+d x))^{6-2 m} (a+a \sin (c+d x))^{-2+m}}{d e \left (20-9 m+m^2\right )}-\frac{a (e \cos (c+d x))^{6-2 m} (a+a \sin (c+d x))^{-1+m}}{d e (5-m)}\\ \end{align*}

Mathematica [A]  time = 0.434272, size = 105, normalized size = 0.7 \[ \frac{e^5 \cos ^6(c+d x) \left (\left (m^2-7 m+12\right ) \sin ^2(c+d x)+2 \left (m^2-9 m+18\right ) \sin (c+d x)+m^2-11 m+32\right ) (a (\sin (c+d x)+1))^m (e \cos (c+d x))^{-2 m}}{d (m-5) (m-4) (m-3) (\sin (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(5 - 2*m)*(a + a*Sin[c + d*x])^m,x]

[Out]

(e^5*Cos[c + d*x]^6*(a*(1 + Sin[c + d*x]))^m*(32 - 11*m + m^2 + 2*(18 - 9*m + m^2)*Sin[c + d*x] + (12 - 7*m +
m^2)*Sin[c + d*x]^2))/(d*(-5 + m)*(-4 + m)*(-3 + m)*(e*Cos[c + d*x])^(2*m)*(1 + Sin[c + d*x])^3)

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Maple [F]  time = 0.914, size = 0, normalized size = 0. \begin{align*} \int \left ( e\cos \left ( dx+c \right ) \right ) ^{5-2\,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))^(5-2*m)*(a+a*sin(d*x+c))^m,x)

[Out]

int((e*cos(d*x+c))^(5-2*m)*(a+a*sin(d*x+c))^m,x)

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Maxima [B]  time = 1.66588, size = 842, normalized size = 5.61 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(5-2*m)*(a+a*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

((m^2 - 11*m + 32)*a^m*e^5 - 2*(m^2 - 15*m + 60)*a^m*e^5*sin(d*x + c)/(cos(d*x + c) + 1) - (3*m^2 - m - 160)*a
^m*e^5*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 8*(m^2 - 7*m - 20)*a^m*e^5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 +
2*(m^2 + 5*m + 160)*a^m*e^5*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 4*(3*m^2 - 13*m + 116)*a^m*e^5*sin(d*x + c)^
5/(cos(d*x + c) + 1)^5 + 2*(m^2 + 5*m + 160)*a^m*e^5*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 8*(m^2 - 7*m - 20)*
a^m*e^5*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - (3*m^2 - m - 160)*a^m*e^5*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 -
2*(m^2 - 15*m + 60)*a^m*e^5*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + (m^2 - 11*m + 32)*a^m*e^5*sin(d*x + c)^10/(c
os(d*x + c) + 1)^10)*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^3 - 12*m^2 + 47*m - 60)*e^(2*m) + 5*(m^3 - 12*m^2 + 47*m - 60)*e^(2*m)*sin(d*x + c)^2/(cos(d*x
+ c) + 1)^2 + 10*(m^3 - 12*m^2 + 47*m - 60)*e^(2*m)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 10*(m^3 - 12*m^2 +
47*m - 60)*e^(2*m)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 5*(m^3 - 12*m^2 + 47*m - 60)*e^(2*m)*sin(d*x + c)^8/(
cos(d*x + c) + 1)^8 + (m^3 - 12*m^2 + 47*m - 60)*e^(2*m)*sin(d*x + c)^10/(cos(d*x + c) + 1)^10)*d)

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Fricas [B]  time = 2.46202, size = 813, normalized size = 5.42 \begin{align*} -\frac{{\left ({\left (m^{2} - 7 \, m + 12\right )} \cos \left (d x + c\right )^{3} -{\left (m^{2} - 11 \, m + 24\right )} \cos \left (d x + c\right )^{2} - 2 \,{\left (m^{2} - 9 \, m + 22\right )} \cos \left (d x + c\right ) -{\left ({\left (m^{2} - 7 \, m + 12\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (m^{2} - 9 \, m + 18\right )} \cos \left (d x + c\right ) - 8\right )} \sin \left (d x + c\right ) - 8\right )} \left (e \cos \left (d x + c\right )\right )^{-2 \, m + 5}{\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{4 \, d m^{3} -{\left (d m^{3} - 12 \, d m^{2} + 47 \, d m - 60 \, d\right )} \cos \left (d x + c\right )^{3} - 48 \, d m^{2} - 3 \,{\left (d m^{3} - 12 \, d m^{2} + 47 \, d m - 60 \, d\right )} \cos \left (d x + c\right )^{2} + 188 \, d m + 2 \,{\left (d m^{3} - 12 \, d m^{2} + 47 \, d m - 60 \, d\right )} \cos \left (d x + c\right ) +{\left (4 \, d m^{3} - 48 \, d m^{2} -{\left (d m^{3} - 12 \, d m^{2} + 47 \, d m - 60 \, d\right )} \cos \left (d x + c\right )^{2} + 188 \, d m + 2 \,{\left (d m^{3} - 12 \, d m^{2} + 47 \, d m - 60 \, d\right )} \cos \left (d x + c\right ) - 240 \, d\right )} \sin \left (d x + c\right ) - 240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(5-2*m)*(a+a*sin(d*x+c))^m,x, algorithm="fricas")

[Out]

-((m^2 - 7*m + 12)*cos(d*x + c)^3 - (m^2 - 11*m + 24)*cos(d*x + c)^2 - 2*(m^2 - 9*m + 22)*cos(d*x + c) - ((m^2
- 7*m + 12)*cos(d*x + c)^2 + 2*(m^2 - 9*m + 18)*cos(d*x + c) - 8)*sin(d*x + c) - 8)*(e*cos(d*x + c))^(-2*m +
5)*(a*sin(d*x + c) + a)^m/(4*d*m^3 - (d*m^3 - 12*d*m^2 + 47*d*m - 60*d)*cos(d*x + c)^3 - 48*d*m^2 - 3*(d*m^3 -
12*d*m^2 + 47*d*m - 60*d)*cos(d*x + c)^2 + 188*d*m + 2*(d*m^3 - 12*d*m^2 + 47*d*m - 60*d)*cos(d*x + c) + (4*d
*m^3 - 48*d*m^2 - (d*m^3 - 12*d*m^2 + 47*d*m - 60*d)*cos(d*x + c)^2 + 188*d*m + 2*(d*m^3 - 12*d*m^2 + 47*d*m -
60*d)*cos(d*x + c) - 240*d)*sin(d*x + c) - 240*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))**(5-2*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 + 5}{\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))^(5-2*m)*(a+a*sin(d*x+c))^m,x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^(-2*m + 5)*(a*sin(d*x + c) + a)^m, x)