3.360 \(\int (e \cos (c+d x))^{-4-m} (a+a \sin (c+d x))^m \, dx\)

Optimal. Leaf size=201 \[ -\frac {6 (a \sin (c+d x)+a)^{m+3} (e \cos (c+d x))^{-m-3}}{a^3 d e \left (m^4-10 m^2+9\right )}+\frac {6 (a \sin (c+d x)+a)^{m+2} (e \cos (c+d x))^{-m-3}}{a^2 d e (3-m) \left (1-m^2\right )}-\frac {(a \sin (c+d x)+a)^m (e \cos (c+d x))^{-m-3}}{d e (3-m)}-\frac {3 (a \sin (c+d x)+a)^{m+1} (e \cos (c+d x))^{-m-3}}{a d e (1-m) (3-m)} \]

[Out]

-(e*cos(d*x+c))^(-3-m)*(a+a*sin(d*x+c))^m/d/e/(3-m)-3*(e*cos(d*x+c))^(-3-m)*(a+a*sin(d*x+c))^(1+m)/a/d/e/(1-m)
/(3-m)+6*(e*cos(d*x+c))^(-3-m)*(a+a*sin(d*x+c))^(2+m)/a^2/d/e/(3-m)/(-m^2+1)-6*(e*cos(d*x+c))^(-3-m)*(a+a*sin(
d*x+c))^(3+m)/a^3/d/e/(m^4-10*m^2+9)

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Rubi [A]  time = 0.32, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2672, 2671} \[ \frac {6 (a \sin (c+d x)+a)^{m+2} (e \cos (c+d x))^{-m-3}}{a^2 d e (3-m) \left (1-m^2\right )}-\frac {6 (a \sin (c+d x)+a)^{m+3} (e \cos (c+d x))^{-m-3}}{a^3 d e \left (m^4-10 m^2+9\right )}-\frac {(a \sin (c+d x)+a)^m (e \cos (c+d x))^{-m-3}}{d e (3-m)}-\frac {3 (a \sin (c+d x)+a)^{m+1} (e \cos (c+d x))^{-m-3}}{a d e (1-m) (3-m)} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cos[c + d*x])^(-4 - m)*(a + a*Sin[c + d*x])^m,x]

[Out]

-(((e*Cos[c + d*x])^(-3 - m)*(a + a*Sin[c + d*x])^m)/(d*e*(3 - m))) - (3*(e*Cos[c + d*x])^(-3 - m)*(a + a*Sin[
c + d*x])^(1 + m))/(a*d*e*(1 - m)*(3 - m)) + (6*(e*Cos[c + d*x])^(-3 - m)*(a + a*Sin[c + d*x])^(2 + m))/(a^2*d
*e*(3 - m)*(1 - m^2)) - (6*(e*Cos[c + d*x])^(-3 - m)*(a + a*Sin[c + d*x])^(3 + m))/(a^3*d*e*(9 - 10*m^2 + m^4)
)

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]

Rule 2672

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

Rubi steps

\begin {align*} \int (e \cos (c+d x))^{-4-m} (a+a \sin (c+d x))^m \, dx &=-\frac {(e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^m}{d e (3-m)}+\frac {3 \int (e \cos (c+d x))^{-4-m} (a+a \sin (c+d x))^{1+m} \, dx}{a (3-m)}\\ &=-\frac {(e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^m}{d e (3-m)}-\frac {3 (e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^{1+m}}{a d e (1-m) (3-m)}+\frac {6 \int (e \cos (c+d x))^{-4-m} (a+a \sin (c+d x))^{2+m} \, dx}{a^2 (1-m) (3-m)}\\ &=-\frac {(e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^m}{d e (3-m)}-\frac {3 (e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^{1+m}}{a d e (1-m) (3-m)}+\frac {6 (e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^{2+m}}{a^2 d e (1-m) (3-m) (1+m)}-\frac {6 \int (e \cos (c+d x))^{-4-m} (a+a \sin (c+d x))^{3+m} \, dx}{a^3 (1-m) (3-m) (1+m)}\\ &=-\frac {(e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^m}{d e (3-m)}-\frac {3 (e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^{1+m}}{a d e (1-m) (3-m)}+\frac {6 (e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^{2+m}}{a^2 d e (1-m) (3-m) (1+m)}-\frac {6 (e \cos (c+d x))^{-3-m} (a+a \sin (c+d x))^{3+m}}{a^3 d e \left (9-10 m^2+m^4\right )}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 101, normalized size = 0.50 \[ \frac {\sec ^3(c+d x) \left (-3 \left (m^2-3\right ) \sin (c+d x)+6 m \sin ^2(c+d x)-6 \sin ^3(c+d x)+m \left (m^2-7\right )\right ) (a (\sin (c+d x)+1))^m (e \cos (c+d x))^{-m}}{d e^4 (m-3) (m-1) (m+1) (m+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Cos[c + d*x])^(-4 - m)*(a + a*Sin[c + d*x])^m,x]

[Out]

(Sec[c + d*x]^3*(a*(1 + Sin[c + d*x]))^m*(m*(-7 + m^2) - 3*(-3 + m^2)*Sin[c + d*x] + 6*m*Sin[c + d*x]^2 - 6*Si
n[c + d*x]^3))/(d*e^4*(-3 + m)*(-1 + m)*(1 + m)*(3 + m)*(e*Cos[c + d*x])^m)

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fricas [A]  time = 0.48, size = 104, normalized size = 0.52 \[ -\frac {{\left (6 \, m \cos \left (d x + c\right )^{3} - {\left (m^{3} - m\right )} \cos \left (d x + c\right ) - 3 \, {\left (2 \, \cos \left (d x + c\right )^{3} - {\left (m^{2} - 1\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )} \left (e \cos \left (d x + c\right )\right )^{-m - 4} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{d m^{4} - 10 \, d m^{2} + 9 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(-4-m)*(a+a*sin(d*x+c))^m,x, algorithm="fricas")

[Out]

-(6*m*cos(d*x + c)^3 - (m^3 - m)*cos(d*x + c) - 3*(2*cos(d*x + c)^3 - (m^2 - 1)*cos(d*x + c))*sin(d*x + c))*(e
*cos(d*x + c))^(-m - 4)*(a*sin(d*x + c) + a)^m/(d*m^4 - 10*d*m^2 + 9*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x + c\right )\right )^{-m - 4} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(-4-m)*(a+a*sin(d*x+c))^m,x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^(-m - 4)*(a*sin(d*x + c) + a)^m, x)

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maple [F]  time = 0.34, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x +c \right )\right )^{-4-m} \left (a +a \sin \left (d x +c \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*cos(d*x+c))^(-4-m)*(a+a*sin(d*x+c))^m,x)

[Out]

int((e*cos(d*x+c))^(-4-m)*(a+a*sin(d*x+c))^m,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x + c\right )\right )^{-m - 4} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(-4-m)*(a+a*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

integrate((e*cos(d*x + c))^(-m - 4)*(a*sin(d*x + c) + a)^m, x)

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mupad [B]  time = 6.83, size = 137, normalized size = 0.68 \[ \frac {2\,{\left (a\,\left (\sin \left (c+d\,x\right )+1\right )\right )}^m\,\left (12\,\sin \left (2\,c+2\,d\,x\right )+3\,\sin \left (4\,c+4\,d\,x\right )-22\,m\,\cos \left (c+d\,x\right )-6\,m\,\cos \left (3\,c+3\,d\,x\right )+4\,m^3\,\cos \left (c+d\,x\right )-6\,m^2\,\sin \left (2\,c+2\,d\,x\right )\right )}{d\,e^4\,{\left (e\,\cos \left (c+d\,x\right )\right )}^m\,\left (4\,\cos \left (2\,c+2\,d\,x\right )+\cos \left (4\,c+4\,d\,x\right )+3\right )\,\left (m^4-10\,m^2+9\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^m/(e*cos(c + d*x))^(m + 4),x)

[Out]

(2*(a*(sin(c + d*x) + 1))^m*(12*sin(2*c + 2*d*x) + 3*sin(4*c + 4*d*x) - 22*m*cos(c + d*x) - 6*m*cos(3*c + 3*d*
x) + 4*m^3*cos(c + d*x) - 6*m^2*sin(2*c + 2*d*x)))/(d*e^4*(e*cos(c + d*x))^m*(4*cos(2*c + 2*d*x) + cos(4*c + 4
*d*x) + 3)*(m^4 - 10*m^2 + 9))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))**(-4-m)*(a+a*sin(d*x+c))**m,x)

[Out]

Timed out

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