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

Optimal. Leaf size=62 \[ \frac {A \tan ^3(e+f x)}{3 a^2 c^2 f}+\frac {A \tan (e+f x)}{a^2 c^2 f}+\frac {B \sec ^3(e+f x)}{3 a^2 c^2 f} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2967, 2669, 3767} \[ \frac {A \tan ^3(e+f x)}{3 a^2 c^2 f}+\frac {A \tan (e+f x)}{a^2 c^2 f}+\frac {B \sec ^3(e+f x)}{3 a^2 c^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*Sec[e + f*x]^3)/(3*a^2*c^2*f) + (A*Tan[e + f*x])/(a^2*c^2*f) + (A*Tan[e + f*x]^3)/(3*a^2*c^2*f)

Rule 2669

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b*(g*Cos[
e + f*x])^(p + 1))/(f*g*(p + 1)), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x]
&& (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

Rule 2967

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I
ntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin {align*} \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^2} \, dx &=\frac {\int \sec ^4(e+f x) (A+B \sin (e+f x)) \, dx}{a^2 c^2}\\ &=\frac {B \sec ^3(e+f x)}{3 a^2 c^2 f}+\frac {A \int \sec ^4(e+f x) \, dx}{a^2 c^2}\\ &=\frac {B \sec ^3(e+f x)}{3 a^2 c^2 f}-\frac {A \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (e+f x)\right )}{a^2 c^2 f}\\ &=\frac {B \sec ^3(e+f x)}{3 a^2 c^2 f}+\frac {A \tan (e+f x)}{a^2 c^2 f}+\frac {A \tan ^3(e+f x)}{3 a^2 c^2 f}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 53, normalized size = 0.85 \[ \frac {A \left (\frac {1}{3} \tan ^3(e+f x)+\tan (e+f x)\right )}{a^2 c^2 f}+\frac {B \sec ^3(e+f x)}{3 a^2 c^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*Sec[e + f*x]^3)/(3*a^2*c^2*f) + (A*(Tan[e + f*x] + Tan[e + f*x]^3/3))/(a^2*c^2*f)

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fricas [A]  time = 0.42, size = 41, normalized size = 0.66 \[ \frac {{\left (2 \, A \cos \left (f x + e\right )^{2} + A\right )} \sin \left (f x + e\right ) + B}{3 \, a^{2} c^{2} f \cos \left (f x + e\right )^{3}} \]

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/(c-c*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.20, size = 87, normalized size = 1.40 \[ -\frac {2 \, {\left (3 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 3 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 2 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 3 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + B\right )}}{3 \, {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{3} a^{2} c^{2} f} \]

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/(c-c*sin(f*x+e))^2,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.35, size = 145, normalized size = 2.34 \[ \frac {-\frac {2 \left (\frac {A}{2}+\frac {B}{2}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {\frac {A}{2}+\frac {B}{2}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {2 \left (\frac {A}{2}+\frac {B}{4}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {-\frac {A}{2}+\frac {B}{2}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (\frac {A}{2}-\frac {B}{2}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (\frac {A}{2}-\frac {B}{4}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}}{f \,a^{2} c^{2}} \]

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/(c-c*sin(f*x+e))^2,x)

[Out]

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

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maxima [A]  time = 0.39, size = 47, normalized size = 0.76 \[ \frac {\frac {{\left (\tan \left (f x + e\right )^{3} + 3 \, \tan \left (f x + e\right )\right )} A}{a^{2} c^{2}} + \frac {B}{a^{2} c^{2} \cos \left (f x + e\right )^{3}}}{3 \, f} \]

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/(c-c*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 12.38, size = 82, normalized size = 1.32 \[ -\frac {2\,\left (3\,A\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+3\,B\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-2\,A\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+3\,A\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+B\right )}{3\,a^2\,c^2\,f\,{\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )}^3} \]

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*(c - c*sin(e + f*x))^2),x)

[Out]

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

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sympy [A]  time = 7.63, size = 469, normalized size = 7.56 \[ \begin {cases} - \frac {6 A \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} c^{2} f \tan ^{6}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 9 a^{2} c^{2} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} c^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 a^{2} c^{2} f} + \frac {4 A \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} c^{2} f \tan ^{6}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 9 a^{2} c^{2} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} c^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 a^{2} c^{2} f} - \frac {6 A \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} c^{2} f \tan ^{6}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 9 a^{2} c^{2} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} c^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 a^{2} c^{2} f} - \frac {6 B \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} c^{2} f \tan ^{6}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 9 a^{2} c^{2} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} c^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 a^{2} c^{2} f} - \frac {2 B}{3 a^{2} c^{2} f \tan ^{6}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 9 a^{2} c^{2} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} c^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 a^{2} c^{2} f} & \text {for}\: f \neq 0 \\\frac {x \left (A + B \sin {\relax (e )}\right )}{\left (a \sin {\relax (e )} + a\right )^{2} \left (- c \sin {\relax (e )} + c\right )^{2}} & \text {otherwise} \end {cases} \]

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/(c-c*sin(f*x+e))**2,x)

[Out]

Piecewise((-6*A*tan(e/2 + f*x/2)**5/(3*a**2*c**2*f*tan(e/2 + f*x/2)**6 - 9*a**2*c**2*f*tan(e/2 + f*x/2)**4 + 9
*a**2*c**2*f*tan(e/2 + f*x/2)**2 - 3*a**2*c**2*f) + 4*A*tan(e/2 + f*x/2)**3/(3*a**2*c**2*f*tan(e/2 + f*x/2)**6
- 9*a**2*c**2*f*tan(e/2 + f*x/2)**4 + 9*a**2*c**2*f*tan(e/2 + f*x/2)**2 - 3*a**2*c**2*f) - 6*A*tan(e/2 + f*x/
2)/(3*a**2*c**2*f*tan(e/2 + f*x/2)**6 - 9*a**2*c**2*f*tan(e/2 + f*x/2)**4 + 9*a**2*c**2*f*tan(e/2 + f*x/2)**2
- 3*a**2*c**2*f) - 6*B*tan(e/2 + f*x/2)**4/(3*a**2*c**2*f*tan(e/2 + f*x/2)**6 - 9*a**2*c**2*f*tan(e/2 + f*x/2)
**4 + 9*a**2*c**2*f*tan(e/2 + f*x/2)**2 - 3*a**2*c**2*f) - 2*B/(3*a**2*c**2*f*tan(e/2 + f*x/2)**6 - 9*a**2*c**
2*f*tan(e/2 + f*x/2)**4 + 9*a**2*c**2*f*tan(e/2 + f*x/2)**2 - 3*a**2*c**2*f), Ne(f, 0)), (x*(A + B*sin(e))/((a
*sin(e) + a)**2*(-c*sin(e) + c)**2), True))

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