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

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

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2967, 2859, 3767, 8} \[ \frac {(2 A+B) \tan (e+f x)}{3 a^2 c f}-\frac {(A-B) \sec (e+f x)}{3 c f \left (a^2 \sin (e+f x)+a^2\right )} \]

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])),x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2859

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[((b*c - a*d)*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(2*m +
p + 1)), x] + Dist[(a*d*m + b*c*(m + p + 1))/(a*b*(2*m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^
(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && (LtQ[m, -1] || ILtQ[Simplify[
m + p], 0]) && NeQ[2*m + p + 1, 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))} \, dx &=\frac {\int \frac {\sec ^2(e+f x) (A+B \sin (e+f x))}{a+a \sin (e+f x)} \, dx}{a c}\\ &=-\frac {(A-B) \sec (e+f x)}{3 c f \left (a^2+a^2 \sin (e+f x)\right )}+\frac {(2 A+B) \int \sec ^2(e+f x) \, dx}{3 a^2 c}\\ &=-\frac {(A-B) \sec (e+f x)}{3 c f \left (a^2+a^2 \sin (e+f x)\right )}-\frac {(2 A+B) \operatorname {Subst}(\int 1 \, dx,x,-\tan (e+f x))}{3 a^2 c f}\\ &=-\frac {(A-B) \sec (e+f x)}{3 c f \left (a^2+a^2 \sin (e+f x)\right )}+\frac {(2 A+B) \tan (e+f x)}{3 a^2 c f}\\ \end {align*}

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Mathematica [A]  time = 0.51, size = 110, normalized size = 1.77 \[ \frac {\cos (e+f x) (-2 (A-B) \cos (e+f x)+2 (2 A+B) \cos (2 (e+f x))-8 A \sin (e+f x)-A \sin (2 (e+f x))-4 B \sin (e+f x)+B \sin (2 (e+f x))-6 B)}{12 a^2 c f (\sin (e+f x)-1) (\sin (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[e + f*x]*(-6*B - 2*(A - B)*Cos[e + f*x] + 2*(2*A + B)*Cos[2*(e + f*x)] - 8*A*Sin[e + f*x] - 4*B*Sin[e + f
*x] - A*Sin[2*(e + f*x)] + B*Sin[2*(e + f*x)]))/(12*a^2*c*f*(-1 + Sin[e + f*x])*(1 + Sin[e + f*x])^2)

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fricas [A]  time = 0.41, size = 69, normalized size = 1.11 \[ -\frac {{\left (2 \, A + B\right )} \cos \left (f x + e\right )^{2} - {\left (2 \, A + B\right )} \sin \left (f x + e\right ) - A - 2 \, B}{3 \, {\left (a^{2} c f \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a^{2} c f \cos \left (f x + e\right )\right )}} \]

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

[Out]

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

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giac [A]  time = 0.18, size = 102, normalized size = 1.65 \[ -\frac {\frac {3 \, {\left (A + B\right )}}{a^{2} c {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}} + \frac {9 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 3 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 12 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 7 \, A - B}{a^{2} c {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}}}{6 \, f} \]

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)),x, algorithm="giac")

[Out]

-1/6*(3*(A + B)/(a^2*c*(tan(1/2*f*x + 1/2*e) - 1)) + (9*A*tan(1/2*f*x + 1/2*e)^2 - 3*B*tan(1/2*f*x + 1/2*e)^2
+ 12*A*tan(1/2*f*x + 1/2*e) + 7*A - B)/(a^2*c*(tan(1/2*f*x + 1/2*e) + 1)^3))/f

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maple [A]  time = 0.44, size = 97, normalized size = 1.56 \[ \frac {-\frac {2 \left (\frac {A}{4}+\frac {B}{4}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {-A +B}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (A -B \right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (\frac {3 A}{4}-\frac {B}{4}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}}{a^{2} c f} \]

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)),x)

[Out]

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

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maxima [B]  time = 0.35, size = 265, normalized size = 4.27 \[ \frac {2 \, {\left (\frac {B {\left (\frac {2 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}}{a^{2} c + \frac {2 \, a^{2} c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {2 \, a^{2} c \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {a^{2} c \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}} + \frac {A {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {3 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - 1\right )}}{a^{2} c + \frac {2 \, a^{2} c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {2 \, a^{2} c \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {a^{2} c \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}}\right )}}{3 \, f} \]

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

[Out]

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

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mupad [B]  time = 12.29, size = 117, normalized size = 1.89 \[ \frac {2\,\left (\frac {3\,B}{2}-A\,\cos \left (e+f\,x\right )+B\,\cos \left (e+f\,x\right )+2\,A\,\sin \left (e+f\,x\right )+B\,\sin \left (e+f\,x\right )-A\,\cos \left (2\,e+2\,f\,x\right )-\frac {B\,\cos \left (2\,e+2\,f\,x\right )}{2}-\frac {A\,\sin \left (2\,e+2\,f\,x\right )}{2}+\frac {B\,\sin \left (2\,e+2\,f\,x\right )}{2}\right )}{3\,a^2\,c\,f\,\left (2\,\cos \left (e+f\,x\right )+\sin \left (2\,e+2\,f\,x\right )\right )} \]

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*(c - c*sin(e + f*x))),x)

[Out]

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

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sympy [A]  time = 7.23, size = 578, normalized size = 9.32 \[ \begin {cases} - \frac {6 A \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} c f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 6 a^{2} c f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 6 a^{2} c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 a^{2} c f} - \frac {6 A \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} c f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 6 a^{2} c f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 6 a^{2} c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 a^{2} c f} - \frac {2 A \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} c f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 6 a^{2} c f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 6 a^{2} c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 a^{2} c f} + \frac {2 A}{3 a^{2} c f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 6 a^{2} c f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 6 a^{2} c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 a^{2} c f} - \frac {6 B \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} c f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 6 a^{2} c f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 6 a^{2} c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 a^{2} c f} - \frac {4 B \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} c f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 6 a^{2} c f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 6 a^{2} c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 a^{2} c f} - \frac {2 B}{3 a^{2} c f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 6 a^{2} c f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 6 a^{2} c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 3 a^{2} c 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 )} & \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)),x)

[Out]

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

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