3.275 \(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2} \, dx\)

Optimal. Leaf size=65 \[ -\frac {(A+2 B) \cos (e+f x)}{3 f \left (a^2 \sin (e+f x)+a^2\right )}-\frac {(A-B) \cos (e+f x)}{3 f (a \sin (e+f x)+a)^2} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2750, 2648} \[ -\frac {(A+2 B) \cos (e+f x)}{3 f \left (a^2 \sin (e+f x)+a^2\right )}-\frac {(A-B) \cos (e+f x)}{3 f (a \sin (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sin[e + f*x])/(a + a*Sin[e + f*x])^2,x]

[Out]

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

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]

Rule 2750

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

Rubi steps

\begin {align*} \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2} \, dx &=-\frac {(A-B) \cos (e+f x)}{3 f (a+a \sin (e+f x))^2}+\frac {(A+2 B) \int \frac {1}{a+a \sin (e+f x)} \, dx}{3 a}\\ &=-\frac {(A-B) \cos (e+f x)}{3 f (a+a \sin (e+f x))^2}-\frac {(A+2 B) \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 = 0.06, size = 43, normalized size = 0.66 \[ -\frac {\cos (e+f x) ((A+2 B) \sin (e+f x)+2 A+B)}{3 a^2 f (\sin (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sin[e + f*x])/(a + a*Sin[e + f*x])^2,x]

[Out]

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

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fricas [A]  time = 0.42, size = 117, normalized size = 1.80 \[ \frac {{\left (A + 2 \, B\right )} \cos \left (f x + e\right )^{2} + {\left (2 \, A + B\right )} \cos \left (f x + e\right ) + {\left ({\left (A + 2 \, B\right )} \cos \left (f x + e\right ) - A + B\right )} \sin \left (f x + e\right ) + A - B}{3 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f \cos \left (f x + e\right ) - 2 \, a^{2} f - {\left (a^{2} f \cos \left (f x + e\right ) + 2 \, a^{2} f\right )} \sin \left (f x + e\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((A + 2*B)*cos(f*x + e)^2 + (2*A + B)*cos(f*x + e) + ((A + 2*B)*cos(f*x + e) - A + B)*sin(f*x + e) + A - B
)/(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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giac [A]  time = 0.19, size = 68, normalized size = 1.05 \[ -\frac {2 \, {\left (3 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, A + B\right )}}{3 \, a^{2} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.33, size = 70, normalized size = 1.08 \[ \frac {-\frac {2 \left (2 A -2 B \right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 A}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {-2 A +2 B}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}}{f \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2,x)

[Out]

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

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maxima [B]  time = 0.35, size = 214, normalized size = 3.29 \[ -\frac {2 \, {\left (\frac {A {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 2\right )}}{a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {B {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}}{a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}\right )}}{3 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(A*(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)
+ B*(3*sin(f*x + e)/(cos(f*x + e) + 1) + 1)/(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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mupad [B]  time = 13.39, size = 97, normalized size = 1.49 \[ -\frac {2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {5\,A}{2}+\frac {B}{2}-\frac {A\,\cos \left (e+f\,x\right )}{2}+\frac {B\,\cos \left (e+f\,x\right )}{2}+\frac {3\,A\,\sin \left (e+f\,x\right )}{2}+\frac {3\,B\,\sin \left (e+f\,x\right )}{2}\right )}{3\,a^2\,f\,\left (\frac {3\,\sqrt {2}\,\cos \left (\frac {e}{2}-\frac {\pi }{4}+\frac {f\,x}{2}\right )}{2}-\frac {\sqrt {2}\,\cos \left (\frac {3\,e}{2}+\frac {\pi }{4}+\frac {3\,f\,x}{2}\right )}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*sin(e + f*x))/(a + a*sin(e + f*x))^2,x)

[Out]

-(2*cos(e/2 + (f*x)/2)*((5*A)/2 + B/2 - (A*cos(e + f*x))/2 + (B*cos(e + f*x))/2 + (3*A*sin(e + f*x))/2 + (3*B*
sin(e + f*x))/2))/(3*a^2*f*((3*2^(1/2)*cos(e/2 - pi/4 + (f*x)/2))/2 - (2^(1/2)*cos((3*e)/2 + pi/4 + (3*f*x)/2)
)/2))

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sympy [A]  time = 4.58, size = 372, normalized size = 5.72 \[ \begin {cases} - \frac {6 A \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 3 a^{2} f} - \frac {6 A \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 3 a^{2} f} - \frac {4 A}{3 a^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 3 a^{2} f} - \frac {6 B \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 3 a^{2} f} - \frac {2 B}{3 a^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 3 a^{2} f} & \text {for}\: f \neq 0 \\\frac {x \left (A + B \sin {\relax (e )}\right )}{\left (a \sin {\relax (e )} + a\right )^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-6*A*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*ta
n(e/2 + f*x/2) + 3*a**2*f) - 6*A*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*A/(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*B*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) - 2*B/(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 + B*sin(e))/(a*sin(e) + a)**2, True))

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