3.282 \(\int \frac{A+B \sin (e+f x)}{(a+a \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=102 \[ -\frac{(2 A+3 B) \cos (e+f x)}{15 f \left (a^3 \sin (e+f x)+a^3\right )}-\frac{(2 A+3 B) \cos (e+f x)}{15 a f (a \sin (e+f x)+a)^2}-\frac{(A-B) \cos (e+f x)}{5 f (a \sin (e+f x)+a)^3} \]

[Out]

-((A - B)*Cos[e + f*x])/(5*f*(a + a*Sin[e + f*x])^3) - ((2*A + 3*B)*Cos[e + f*x])/(15*a*f*(a + a*Sin[e + f*x])
^2) - ((2*A + 3*B)*Cos[e + f*x])/(15*f*(a^3 + a^3*Sin[e + f*x]))

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Rubi [A]  time = 0.0750472, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {2750, 2650, 2648} \[ -\frac{(2 A+3 B) \cos (e+f x)}{15 f \left (a^3 \sin (e+f x)+a^3\right )}-\frac{(2 A+3 B) \cos (e+f x)}{15 a f (a \sin (e+f x)+a)^2}-\frac{(A-B) \cos (e+f x)}{5 f (a \sin (e+f x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sin[e + f*x])/(a + a*Sin[e + f*x])^3,x]

[Out]

-((A - B)*Cos[e + f*x])/(5*f*(a + a*Sin[e + f*x])^3) - ((2*A + 3*B)*Cos[e + f*x])/(15*a*f*(a + a*Sin[e + f*x])
^2) - ((2*A + 3*B)*Cos[e + f*x])/(15*f*(a^3 + a^3*Sin[e + f*x]))

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)]

Rule 2650

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]

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 \frac{A+B \sin (e+f x)}{(a+a \sin (e+f x))^3} \, dx &=-\frac{(A-B) \cos (e+f x)}{5 f (a+a \sin (e+f x))^3}+\frac{(2 A+3 B) \int \frac{1}{(a+a \sin (e+f x))^2} \, dx}{5 a}\\ &=-\frac{(A-B) \cos (e+f x)}{5 f (a+a \sin (e+f x))^3}-\frac{(2 A+3 B) \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}+\frac{(2 A+3 B) \int \frac{1}{a+a \sin (e+f x)} \, dx}{15 a^2}\\ &=-\frac{(A-B) \cos (e+f x)}{5 f (a+a \sin (e+f x))^3}-\frac{(2 A+3 B) \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}-\frac{(2 A+3 B) \cos (e+f x)}{15 f \left (a^3+a^3 \sin (e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.0784365, size = 63, normalized size = 0.62 \[ -\frac{\cos (e+f x) \left ((2 A+3 B) \sin ^2(e+f x)+(6 A+9 B) \sin (e+f x)+7 A+3 B\right )}{15 a^3 f (\sin (e+f x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sin[e + f*x])/(a + a*Sin[e + f*x])^3,x]

[Out]

-(Cos[e + f*x]*(7*A + 3*B + (6*A + 9*B)*Sin[e + f*x] + (2*A + 3*B)*Sin[e + f*x]^2))/(15*a^3*f*(1 + Sin[e + f*x
])^3)

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Maple [A]  time = 0.064, size = 114, normalized size = 1.1 \begin{align*} 2\,{\frac{1}{f{a}^{3}} \left ( -1/4\,{\frac{-8\,A+8\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{4}}}-1/5\,{\frac{4\,A-4\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{5}}}-{\frac{A}{\tan \left ( 1/2\,fx+e/2 \right ) +1}}-1/3\,{\frac{8\,A-6\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{3}}}-1/2\,{\frac{-4\,A+2\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3,x)

[Out]

2/f/a^3*(-1/4*(-8*A+8*B)/(tan(1/2*f*x+1/2*e)+1)^4-1/5*(4*A-4*B)/(tan(1/2*f*x+1/2*e)+1)^5-A/(tan(1/2*f*x+1/2*e)
+1)-1/3*(8*A-6*B)/(tan(1/2*f*x+1/2*e)+1)^3-1/2*(-4*A+2*B)/(tan(1/2*f*x+1/2*e)+1)^2)

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Maxima [B]  time = 1.01367, size = 522, normalized size = 5.12 \begin{align*} -\frac{2 \,{\left (\frac{A{\left (\frac{20 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{40 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{30 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 7\right )}}{a^{3} + \frac{5 \, a^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{10 \, a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{10 \, a^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{5 \, a^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{a^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} + \frac{3 \, B{\left (\frac{5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{5 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{5 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + 1\right )}}{a^{3} + \frac{5 \, a^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{10 \, a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{10 \, a^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{5 \, a^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{a^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}\right )}}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

-2/15*(A*(20*sin(f*x + e)/(cos(f*x + e) + 1) + 40*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 30*sin(f*x + e)^3/(cos
(f*x + e) + 1)^3 + 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 7)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) +
10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4
/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 3*B*(5*sin(f*x + e)/(cos(f*x + e) + 1) + 5*
sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(co
s(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*
a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5))/f

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Fricas [A]  time = 1.77057, size = 466, normalized size = 4.57 \begin{align*} -\frac{{\left (2 \, A + 3 \, B\right )} \cos \left (f x + e\right )^{3} - 2 \,{\left (2 \, A + 3 \, B\right )} \cos \left (f x + e\right )^{2} - 3 \,{\left (3 \, A + 2 \, B\right )} \cos \left (f x + e\right ) -{\left ({\left (2 \, A + 3 \, B\right )} \cos \left (f x + e\right )^{2} + 3 \,{\left (2 \, A + 3 \, B\right )} \cos \left (f x + e\right ) - 3 \, A + 3 \, B\right )} \sin \left (f x + e\right ) - 3 \, A + 3 \, B}{15 \,{\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f +{\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

-1/15*((2*A + 3*B)*cos(f*x + e)^3 - 2*(2*A + 3*B)*cos(f*x + e)^2 - 3*(3*A + 2*B)*cos(f*x + e) - ((2*A + 3*B)*c
os(f*x + e)^2 + 3*(2*A + 3*B)*cos(f*x + e) - 3*A + 3*B)*sin(f*x + e) - 3*A + 3*B)/(a^3*f*cos(f*x + e)^3 + 3*a^
3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f + (a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f)*
sin(f*x + e))

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Sympy [A]  time = 11.3529, size = 899, normalized size = 8.81 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))**3,x)

[Out]

Piecewise((6*A*tan(e/2 + f*x/2)**5/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f
*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 20*A*tan(e/2
+ f*x/2)**2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 +
150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 10*A*tan(e/2 + f*x/2)/(15*a**3*f*t
an(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x
/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 8*A/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*
x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a*
*3*f) - 30*B*tan(e/2 + f*x/2)**3/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*t
an(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 30*B*tan(e/2 +
f*x/2)**2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 1
50*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 30*B*tan(e/2 + f*x/2)/(15*a**3*f*tan
(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2
)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 6*B/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/
2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3
*f), Ne(f, 0)), (x*(A + B*sin(e))/(a*sin(e) + a)**3, True))

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Giac [A]  time = 1.29214, size = 176, normalized size = 1.73 \begin{align*} -\frac{2 \,{\left (15 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 30 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 15 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 40 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 15 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 20 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 15 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 7 \, A + 3 \, B\right )}}{15 \, a^{3} f{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3,x, algorithm="giac")

[Out]

-2/15*(15*A*tan(1/2*f*x + 1/2*e)^4 + 30*A*tan(1/2*f*x + 1/2*e)^3 + 15*B*tan(1/2*f*x + 1/2*e)^3 + 40*A*tan(1/2*
f*x + 1/2*e)^2 + 15*B*tan(1/2*f*x + 1/2*e)^2 + 20*A*tan(1/2*f*x + 1/2*e) + 15*B*tan(1/2*f*x + 1/2*e) + 7*A + 3
*B)/(a^3*f*(tan(1/2*f*x + 1/2*e) + 1)^5)