3.5.66 \(\int \frac {1}{(a+a \sin (e+f x))^2} \, dx\) [466]

Optimal. Leaf size=55 \[ -\frac {\cos (e+f x)}{3 f (a+a \sin (e+f x))^2}-\frac {\cos (e+f x)}{3 f \left (a^2+a^2 \sin (e+f x)\right )} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2729, 2727} \begin {gather*} -\frac {\cos (e+f x)}{3 f \left (a^2 \sin (e+f x)+a^2\right )}-\frac {\cos (e+f x)}{3 f (a \sin (e+f x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^(-2),x]

[Out]

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

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]

Rule 2729

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

Rubi steps

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

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Mathematica [A]
time = 0.07, size = 54, normalized size = 0.98 \begin {gather*} -\frac {-3+4 \cos (e+f x)+\cos (2 (e+f x))-4 \sin (e+f x)+\sin (2 (e+f x))}{6 a^2 f (1+\sin (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^(-2),x]

[Out]

-1/6*(-3 + 4*Cos[e + f*x] + Cos[2*(e + f*x)] - 4*Sin[e + f*x] + Sin[2*(e + f*x)])/(a^2*f*(1 + Sin[e + f*x])^2)

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Maple [A]
time = 0.20, size = 53, normalized size = 0.96

method	result	size
risch	\(-\frac {2 i \left (i+3 \,{\mathrm e}^{i \left (f x +e \right )}\right )}{3 f \,a^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{3}}\)	\(38\)
derivativedivides	\(\frac {-\frac {4}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {2}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}}{a^{2} f}\)	\(53\)
default	\(\frac {-\frac {4}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {2}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}}{a^{2} f}\)	\(53\)
norman	\(\frac {-\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a f}-\frac {2 \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a f}-\frac {4}{3 a f}}{a \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sin(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

2/f/a^2*(-2/3/(tan(1/2*f*x+1/2*e)+1)^3+1/(tan(1/2*f*x+1/2*e)+1)^2-1/(tan(1/2*f*x+1/2*e)+1))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 127 vs. \(2 (55) = 110\).
time = 0.28, size = 127, normalized size = 2.31 \begin {gather*} -\frac {2 \, {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 2\right )}}{3 \, {\left (a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.34, size = 103, normalized size = 1.87 \begin {gather*} \frac {\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) - 1\right )} \sin \left (f x + e\right ) + 2 \, \cos \left (f x + e\right ) + 1}{3 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f \cos \left (f x + e\right ) - 2 \, a^{2} f - {\left (a^{2} f \cos \left (f x + e\right ) + 2 \, a^{2} f\right )} \sin \left (f x + e\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (46) = 92\).
time = 0.73, size = 221, normalized size = 4.02 \begin {gather*} \begin {cases} - \frac {6 \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 3 a^{2} f} - \frac {6 \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 3 a^{2} f} - \frac {4}{3 a^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 3 a^{2} f} & \text {for}\: f \neq 0 \\\frac {x}{\left (a \sin {\left (e \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))**2,x)

[Out]

Piecewise((-6*tan(e/2 + f*x/2)**2/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(
e/2 + f*x/2) + 3*a**2*f) - 6*tan(e/2 + f*x/2)/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 + f*x/2)**2 + 9
*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) - 4/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2
*f*tan(e/2 + f*x/2) + 3*a**2*f), Ne(f, 0)), (x/(a*sin(e) + a)**2, True))

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Giac [A]
time = 0.44, size = 50, normalized size = 0.91 \begin {gather*} -\frac {2 \, {\left (3 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2\right )}}{3 \, a^{2} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^2,x, algorithm="giac")

[Out]

-2/3*(3*tan(1/2*f*x + 1/2*e)^2 + 3*tan(1/2*f*x + 1/2*e) + 2)/(a^2*f*(tan(1/2*f*x + 1/2*e) + 1)^3)

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Mupad [B]
time = 6.98, size = 76, normalized size = 1.38 \begin {gather*} -\frac {2\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )-\frac {2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left ({\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-3\right )}{3}}{a^2\,f\,{\left (\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )+\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + a*sin(e + f*x))^2,x)

[Out]

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

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