3.7.33 \(\int (3+3 \sin (e+f x))^{-1-m} (a+a \sin (e+f x))^m \, dx\) [633]

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

[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 = 0.01, 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, 2727} \begin {gather*} -\frac {\cos (e+f x) (3 \sin (e+f x)+3)^{-m-1} (a \sin (e+f x)+a)^m}{f} \end {gather*}

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 2727

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] Leaf count is larger than twice the leaf count of optimal. \(104\) vs. \(2(39)=78\).
time = 5.22, size = 104, normalized size = 2.67 \begin {gather*} -\frac {2^{-m} 3^{-1-m} \cos \left (\frac {1}{4} (2 e+\pi +2 f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^{2 (1+m)} (1+\sin (e+f x))^{-1-m} (a (1+\sin (e+f x)))^m \sin ^{-1-2 m}\left (\frac {1}{4} (2 e+\pi +2 f x)\right )}{f} \end {gather*}

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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Maple [F]
time = 0.14, 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 = 0.51, size = 40, normalized size = 1.03 \begin {gather*} -\frac {2 \, a^{m}}{{\left (3^{m + 1} + \frac {3^{m + 1} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )} f} \end {gather*}

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

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Fricas [A]
time = 0.35, size = 47, normalized size = 1.21 \begin {gather*} -\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 )}} \end {gather*}

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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 3^{- m - 1} \int \frac {\left (a \sin {\left (e + f x \right )} + a\right )^{m}}{\left (\sin {\left (e + f x \right )} + 1\right )^{m} \sin {\left (e + f x \right )} + \left (\sin {\left (e + f x \right )} + 1\right )^{m}}\, dx \end {gather*}

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]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 815 vs. \(2 (42) = 84\).
time = 0.72, size = 815, normalized size = 20.90 \begin {gather*} \text {Too large to display} \end {gather*}

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(3/4*pi + pi*m*floor(-1/4*sgn(a) + 1/2) + 1/4*pi*m*sgn(a) + 3/4*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(3/4*pi + pi*m*flo
or(-1/4*sgn(a) + 1/2) + 1/4*pi*m*sgn(a) + 3/4*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(3/4*pi + pi*m*floor(-1/4*sgn(a) + 1/2) + 1/4*pi*m*sgn(a) + 3/4*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(3/4*pi + pi*m*floor(-1/4*sgn(a)
+ 1/2) + 1/4*pi*m*sgn(a) + 3/4*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(abs(a)
) - log(3))*tan(3/4*pi + pi*m*floor(-1/4*sgn(a) + 1/2) + 1/4*pi*m*sgn(a) + 3/4*pi*m - 1/2*f*x - 1/2*e)*tan(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(abs(a))
- log(3))*tan(3/4*pi + pi*m*floor(-1/4*sgn(a) + 1/2) + 1/4*pi*m*sgn(a) + 3/4*pi*m - 1/2*f*x - 1/2*e)^2 - 6*e^
(-m*log(3) + m*log(abs(a)) - log(3))*tan(3/4*pi + pi*m*floor(-1/4*sgn(a) + 1/2) + 1/4*pi*m*sgn(a) + 3/4*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(3/4*pi + pi*m*floor(-1/4*sgn(a) + 1/2) + 1/4*pi*m*sgn(a) + 3/4*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(3/4*pi + pi*m*floor(-1/4*sgn(a) + 1/2) + 1/4*pi*m*sgn(a) + 3/4*pi*m - 1/2*f*x - 1/2*e)^2
*tan(1/2*f*x + 1/2*e)^3 + f*tan(3/4*pi + pi*m*floor(-1/4*sgn(a) + 1/2) + 1/4*pi*m*sgn(a) + 3/4*pi*m - 1/2*f*x
- 1/2*e)^2*tan(1/2*f*x + 1/2*e)^2 + f*tan(3/4*pi + pi*m*floor(-1/4*sgn(a) + 1/2) + 1/4*pi*m*sgn(a) + 3/4*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(3/4*pi + pi*m*floor(-1/4*sgn(a) +
1/2) + 1/4*pi*m*sgn(a) + 3/4*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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Mupad [B]
time = 0.43, size = 52, normalized size = 1.33 \begin {gather*} \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}} \end {gather*}

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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