3.650 \(\int (-3-3 \sin (e+f x))^{-1-m} (a+a \sin (e+f x))^m \, dx\)

Optimal. Leaf size=39 \[ -\frac {\cos (e+f x) (-3 \sin (e+f x)-3)^{-m-1} (a \sin (e+f x)+a)^m}{f} \]

[Out]

-cos(f*x+e)*(-3-3*sin(f*x+e))^(-1-m)*(a+a*sin(f*x+e))^m/f

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Rubi [A]  time = 0.02, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {23, 2648} \[ -\frac {\cos (e+f x) (-3 \sin (e+f x)-3)^{-m-1} (a \sin (e+f x)+a)^m}{f} \]

Antiderivative was successfully verified.

[In]

Int[(-3 - 3*Sin[e + f*x])^(-1 - m)*(a + a*Sin[e + f*x])^m,x]

[Out]

-((Cos[e + f*x]*(-3 - 3*Sin[e + f*x])^(-1 - m)*(a + a*Sin[e + f*x])^m)/f)

Rule 23

Int[(u_.)*((a_) + (b_.)*(v_))^(m_)*((c_) + (d_.)*(v_))^(n_), x_Symbol] :> Dist[(a + b*v)^m/(c + d*v)^m, Int[u*
(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[b*c - a*d, 0] &&  !(IntegerQ[m] || IntegerQ[n
] || GtQ[b/d, 0])

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

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

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Mathematica [B]  time = 0.52, size = 106, normalized size = 2.72 \[ -\frac {2^{-m} 3^{-m-1} \cos \left (\frac {1}{4} (2 e+2 f x+\pi )\right ) (-\sin (e+f x)-1)^{-m-1} \sin ^{-2 m-1}\left (\frac {1}{4} (2 e+2 f x+\pi )\right ) (a (\sin (e+f x)+1))^m \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^{2 (m+1)}}{f} \]

Antiderivative was successfully verified.

[In]

Integrate[(-3 - 3*Sin[e + f*x])^(-1 - m)*(a + a*Sin[e + f*x])^m,x]

[Out]

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

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fricas [A]  time = 0.48, size = 43, normalized size = 1.10 \[ \frac {\left (-\frac {1}{3} \, a\right )^{m} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )}}{3 \, {\left (f \cos \left (f x + e\right ) + f \sin \left (f x + e\right ) + f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3-3*sin(f*x+e))^(-1-m)*(a+a*sin(f*x+e))^m,x, algorithm="fricas")

[Out]

1/3*(-1/3*a)^m*(cos(f*x + e) - sin(f*x + e) + 1)/(f*cos(f*x + e) + f*sin(f*x + e) + f)

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giac [B]  time = 1.26, size = 827, normalized size = 21.21 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3-3*sin(f*x+e))^(-1-m)*(a+a*sin(f*x+e))^m,x, algorithm="giac")

[Out]

(e^(-m*log(3) + m*log(abs(a)) - log(3))*tan(-5/4*pi - pi*m*floor(-1/4*sgn(a) + 1/2) - 1/4*pi*m*sgn(a) - 5/4*pi
*m + 1/2*f*x + 1/2*e)^2*tan(1/2*f*x + 1/2*e)^3 + 3*e^(-m*log(3) + m*log(abs(a)) - log(3))*tan(-5/4*pi - pi*m*f
loor(-1/4*sgn(a) + 1/2) - 1/4*pi*m*sgn(a) - 5/4*pi*m + 1/2*f*x + 1/2*e)^2*tan(1/2*f*x + 1/2*e)^2 - 2*e^(-m*log
(3) + m*log(abs(a)) - log(3))*tan(-5/4*pi - pi*m*floor(-1/4*sgn(a) + 1/2) - 1/4*pi*m*sgn(a) - 5/4*pi*m + 1/2*f
*x + 1/2*e)*tan(1/2*f*x + 1/2*e)^3 - 3*e^(-m*log(3) + m*log(abs(a)) - log(3))*tan(-5/4*pi - pi*m*floor(-1/4*sg
n(a) + 1/2) - 1/4*pi*m*sgn(a) - 5/4*pi*m + 1/2*f*x + 1/2*e)^2*tan(1/2*f*x + 1/2*e) + 6*e^(-m*log(3) + m*log(ab
s(a)) - log(3))*tan(-5/4*pi - pi*m*floor(-1/4*sgn(a) + 1/2) - 1/4*pi*m*sgn(a) - 5/4*pi*m + 1/2*f*x + 1/2*e)*ta
n(1/2*f*x + 1/2*e)^2 - e^(-m*log(3) + m*log(abs(a)) - log(3))*tan(1/2*f*x + 1/2*e)^3 - e^(-m*log(3) + m*log(ab
s(a)) - log(3))*tan(-5/4*pi - pi*m*floor(-1/4*sgn(a) + 1/2) - 1/4*pi*m*sgn(a) - 5/4*pi*m + 1/2*f*x + 1/2*e)^2
+ 6*e^(-m*log(3) + m*log(abs(a)) - log(3))*tan(-5/4*pi - pi*m*floor(-1/4*sgn(a) + 1/2) - 1/4*pi*m*sgn(a) - 5/4
*pi*m + 1/2*f*x + 1/2*e)*tan(1/2*f*x + 1/2*e) - 3*e^(-m*log(3) + m*log(abs(a)) - log(3))*tan(1/2*f*x + 1/2*e)^
2 - 2*e^(-m*log(3) + m*log(abs(a)) - log(3))*tan(-5/4*pi - pi*m*floor(-1/4*sgn(a) + 1/2) - 1/4*pi*m*sgn(a) - 5
/4*pi*m + 1/2*f*x + 1/2*e) + 3*e^(-m*log(3) + m*log(abs(a)) - log(3))*tan(1/2*f*x + 1/2*e) + e^(-m*log(3) + m*
log(abs(a)) - log(3)))/(f*tan(-5/4*pi - pi*m*floor(-1/4*sgn(a) + 1/2) - 1/4*pi*m*sgn(a) - 5/4*pi*m + 1/2*f*x +
1/2*e)^2*tan(1/2*f*x + 1/2*e)^3 + f*tan(-5/4*pi - pi*m*floor(-1/4*sgn(a) + 1/2) - 1/4*pi*m*sgn(a) - 5/4*pi*m
+ 1/2*f*x + 1/2*e)^2*tan(1/2*f*x + 1/2*e)^2 + f*tan(-5/4*pi - pi*m*floor(-1/4*sgn(a) + 1/2) - 1/4*pi*m*sgn(a)
- 5/4*pi*m + 1/2*f*x + 1/2*e)^2*tan(1/2*f*x + 1/2*e) + f*tan(1/2*f*x + 1/2*e)^3 + f*tan(-5/4*pi - pi*m*floor(-
1/4*sgn(a) + 1/2) - 1/4*pi*m*sgn(a) - 5/4*pi*m + 1/2*f*x + 1/2*e)^2 + f*tan(1/2*f*x + 1/2*e)^2 + f*tan(1/2*f*x
+ 1/2*e) + f)

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maple [F]  time = 0.64, size = 0, normalized size = 0.00 \[ \int \left (-3-3 \sin \left (f x +e \right )\right )^{-1-m} \left (a +a \sin \left (f x +e \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3-3*sin(f*x+e))^(-1-m)*(a+a*sin(f*x+e))^m,x)

[Out]

int((-3-3*sin(f*x+e))^(-1-m)*(a+a*sin(f*x+e))^m,x)

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maxima [A]  time = 1.23, size = 45, normalized size = 1.15 \[ \frac {2 \, a^{m}}{{\left (3^{m + 1} \left (-1\right )^{m} + \frac {3^{m + 1} \left (-1\right )^{m} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3-3*sin(f*x+e))^(-1-m)*(a+a*sin(f*x+e))^m,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.39, size = 52, normalized size = 1.33 \[ \frac {{\left (a\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,\left (-\cos \left (e+f\,x\right )+\sin \left (e+f\,x\right )\,1{}\mathrm {i}+1{}\mathrm {i}\right )}{f\,{\left (-3\,\sin \left (e+f\,x\right )-3\right )}^{m+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(e + f*x))^m/(- 3*sin(e + f*x) - 3)^(m + 1),x)

[Out]

((a*(sin(e + f*x) + 1))^m*(sin(e + f*x)*1i - cos(e + f*x) + 1i))/(f*(- 3*sin(e + f*x) - 3)^(m + 1))

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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((-3-3*sin(f*x+e))**(-1-m)*(a+a*sin(f*x+e))**m,x)

[Out]

Timed out

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