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

Optimal. Leaf size=142 \[ -\frac {2 (a \sin (c+d x)+a)^{m+2} (e \cos (c+d x))^{-m-2}}{a^2 d e m \left (4-m^2\right )}-\frac {(a \sin (c+d x)+a)^m (e \cos (c+d x))^{-m-2}}{d e (2-m)}+\frac {2 (a \sin (c+d x)+a)^{m+1} (e \cos (c+d x))^{-m-2}}{a d e (2-m) m} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 142, normalized size of antiderivative = 1.00, 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 = {2672, 2671} \[ -\frac {2 (a \sin (c+d x)+a)^{m+2} (e \cos (c+d x))^{-m-2}}{a^2 d e m \left (4-m^2\right )}-\frac {(a \sin (c+d x)+a)^m (e \cos (c+d x))^{-m-2}}{d e (2-m)}+\frac {2 (a \sin (c+d x)+a)^{m+1} (e \cos (c+d x))^{-m-2}}{a d e (2-m) m} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.13, size = 76, normalized size = 0.54 \[ \frac {\sec ^2(c+d x) \left (-2 m \sin (c+d x)+2 \sin ^2(c+d x)+m^2-2\right ) (a (\sin (c+d x)+1))^m (e \cos (c+d x))^{-m}}{d e^3 (m-2) m (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 75, normalized size = 0.53 \[ \frac {{\left (m^{2} \cos \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{3} - 2 \, m \cos \left (d x + c\right ) \sin \left (d x + c\right )\right )} \left (e \cos \left (d x + c\right )\right )^{-m - 3} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{d m^{3} - 4 \, d m} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x + c\right )\right )^{-m - 3} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.31, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x +c \right )\right )^{-3-m} \left (a +a \sin \left (d x +c \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x + c\right )\right )^{-m - 3} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 6.13, size = 103, normalized size = 0.73 \[ -\frac {2\,{\left (a\,\left (\sin \left (c+d\,x\right )+1\right )\right )}^m\,\left (-2\,\cos \left (c+d\,x\right )\,m^2+2\,\sin \left (2\,c+2\,d\,x\right )\,m+3\,\cos \left (c+d\,x\right )+\cos \left (3\,c+3\,d\,x\right )\right )}{d\,e^3\,m\,{\left (e\,\cos \left (c+d\,x\right )\right )}^m\,\left (m^2-4\right )\,\left (3\,\cos \left (c+d\,x\right )+\cos \left (3\,c+3\,d\,x\right )\right )} \]

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 + 3),x)

[Out]

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

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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))**(-3-m)*(a+a*sin(d*x+c))**m,x)

[Out]

Timed out

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