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

Optimal. Leaf size=152 \[ -\frac{2 d (B c-A d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{a^2 f (c-d)^2 \sqrt{c^2-d^2}}-\frac{(A (c-4 d)+B (2 c+d)) \cos (e+f x)}{3 a^2 f (c-d)^2 (\sin (e+f x)+1)}-\frac{(A-B) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2} \]

[Out]

(-2*d*(B*c - A*d)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(a^2*(c - d)^2*Sqrt[c^2 - d^2]*f) - ((A*(c
- 4*d) + B*(2*c + d))*Cos[e + f*x])/(3*a^2*(c - d)^2*f*(1 + Sin[e + f*x])) - ((A - B)*Cos[e + f*x])/(3*(c - d
)*f*(a + a*Sin[e + f*x])^2)

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Rubi [A]  time = 0.419258, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2978, 12, 2660, 618, 204} \[ -\frac{2 d (B c-A d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{a^2 f (c-d)^2 \sqrt{c^2-d^2}}-\frac{(A (c-4 d)+B (2 c+d)) \cos (e+f x)}{3 a^2 f (c-d)^2 (\sin (e+f x)+1)}-\frac{(A-B) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d*(B*c - A*d)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(a^2*(c - d)^2*Sqrt[c^2 - d^2]*f) - ((A*(c
- 4*d) + B*(2*c + d))*Cos[e + f*x])/(3*a^2*(c - d)^2*f*(1 + Sin[e + f*x])) - ((A - B)*Cos[e + f*x])/(3*(c - d
)*f*(a + a*Sin[e + f*x])^2)

Rule 2978

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.638223, size = 229, normalized size = 1.51 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (\frac{6 d (A d-B c) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3 \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{\sqrt{c^2-d^2}}+2 (A-B) (c-d) \sin \left (\frac{1}{2} (e+f x)\right )+2 (A (c-4 d)+B (2 c+d)) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2+(B-A) (c-d) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )}{3 a^2 f (c-d)^2 (\sin (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(2*(A - B)*(c - d)*Sin[(e + f*x)/2] + (-A + B)*(c - d)*(Cos[(e + f*x)/2
] + Sin[(e + f*x)/2]) + 2*(A*(c - 4*d) + B*(2*c + d))*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2
+ (6*d*(-(B*c) + A*d)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^
3)/Sqrt[c^2 - d^2]))/(3*a^2*(c - d)^2*f*(1 + Sin[e + f*x])^2)

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Maple [B]  time = 0.131, size = 327, normalized size = 2.2 \begin{align*} 2\,{\frac{A{d}^{2}}{{a}^{2}f \left ( c-d \right ) ^{2}\sqrt{{c}^{2}-{d}^{2}}}\arctan \left ( 1/2\,{\frac{2\,c\tan \left ( 1/2\,fx+e/2 \right ) +2\,d}{\sqrt{{c}^{2}-{d}^{2}}}} \right ) }-2\,{\frac{Bcd}{{a}^{2}f \left ( c-d \right ) ^{2}\sqrt{{c}^{2}-{d}^{2}}}\arctan \left ( 1/2\,{\frac{2\,c\tan \left ( 1/2\,fx+e/2 \right ) +2\,d}{\sqrt{{c}^{2}-{d}^{2}}}} \right ) }+2\,{\frac{A}{{a}^{2}f \left ( c-d \right ) \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}-2\,{\frac{B}{{a}^{2}f \left ( c-d \right ) \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}-{\frac{4\,A}{3\,{a}^{2}f \left ( c-d \right ) } \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-3}}+{\frac{4\,B}{3\,{a}^{2}f \left ( c-d \right ) } \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-3}}-2\,{\frac{Ac}{{a}^{2}f \left ( c-d \right ) ^{2} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}+4\,{\frac{Ad}{{a}^{2}f \left ( c-d \right ) ^{2} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}-2\,{\frac{Bd}{{a}^{2}f \left ( c-d \right ) ^{2} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \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))^2/(c+d*sin(f*x+e)),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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))^2/(c+d*sin(f*x+e)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.42788, size = 2755, normalized size = 18.12 \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))^2/(c+d*sin(f*x+e)),x, algorithm="fricas")

[Out]

[1/6*(2*(A - B)*c^3 - 2*(A - B)*c^2*d - 2*(A - B)*c*d^2 + 2*(A - B)*d^3 + 2*((A + 2*B)*c^3 - (4*A - B)*c^2*d -
(A + 2*B)*c*d^2 + (4*A - B)*d^3)*cos(f*x + e)^2 - 3*(2*B*c*d - 2*A*d^2 - (B*c*d - A*d^2)*cos(f*x + e)^2 + (B*
c*d - A*d^2)*cos(f*x + e) + (2*B*c*d - 2*A*d^2 + (B*c*d - A*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(-c^2 + d^2)*
log(((2*c^2 - d^2)*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2 + 2*(c*cos(f*x + e)*sin(f*x + e) + d*cos(f*
x + e))*sqrt(-c^2 + d^2))/(d^2*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2)) + 2*((2*A + B)*c^3 - (5*A - 2
*B)*c^2*d - (2*A + B)*c*d^2 + (5*A - 2*B)*d^3)*cos(f*x + e) - 2*((A - B)*c^3 - (A - B)*c^2*d - (A - B)*c*d^2 +
(A - B)*d^3 - ((A + 2*B)*c^3 - (4*A - B)*c^2*d - (A + 2*B)*c*d^2 + (4*A - B)*d^3)*cos(f*x + e))*sin(f*x + e))
/((a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*f*cos(f*x + e)^2 - (a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^
2*d^4)*f*cos(f*x + e) - 2*(a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*f - ((a^2*c^4 - 2*a^2*c^3*d + 2*a^2*
c*d^3 - a^2*d^4)*f*cos(f*x + e) + 2*(a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*f)*sin(f*x + e)), 1/3*((A
- B)*c^3 - (A - B)*c^2*d - (A - B)*c*d^2 + (A - B)*d^3 + ((A + 2*B)*c^3 - (4*A - B)*c^2*d - (A + 2*B)*c*d^2 +
(4*A - B)*d^3)*cos(f*x + e)^2 - 3*(2*B*c*d - 2*A*d^2 - (B*c*d - A*d^2)*cos(f*x + e)^2 + (B*c*d - A*d^2)*cos(f*
x + e) + (2*B*c*d - 2*A*d^2 + (B*c*d - A*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(c^2 - d^2)*arctan(-(c*sin(f*x +
e) + d)/(sqrt(c^2 - d^2)*cos(f*x + e))) + ((2*A + B)*c^3 - (5*A - 2*B)*c^2*d - (2*A + B)*c*d^2 + (5*A - 2*B)*
d^3)*cos(f*x + e) - ((A - B)*c^3 - (A - B)*c^2*d - (A - B)*c*d^2 + (A - B)*d^3 - ((A + 2*B)*c^3 - (4*A - B)*c^
2*d - (A + 2*B)*c*d^2 + (4*A - B)*d^3)*cos(f*x + e))*sin(f*x + e))/((a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2
*d^4)*f*cos(f*x + e)^2 - (a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*f*cos(f*x + e) - 2*(a^2*c^4 - 2*a^2*c
^3*d + 2*a^2*c*d^3 - a^2*d^4)*f - ((a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*f*cos(f*x + e) + 2*(a^2*c^4
- 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*f)*sin(f*x + e))]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))**2/(c+d*sin(f*x+e)),x)

[Out]

Timed out

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Giac [A]  time = 1.2664, size = 350, normalized size = 2.3 \begin{align*} -\frac{2 \,{\left (\frac{3 \,{\left (B c d - A d^{2}\right )}{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (c\right ) + \arctan \left (\frac{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + d}{\sqrt{c^{2} - d^{2}}}\right )\right )}}{{\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\right )} \sqrt{c^{2} - d^{2}}} + \frac{3 \, A c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 6 \, A d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3 \, B d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3 \, A c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 3 \, B c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 9 \, A d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 3 \, B d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 2 \, A c + B c - 5 \, A d + 2 \, B d}{{\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\right )}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{3}}\right )}}{3 \, 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))^2/(c+d*sin(f*x+e)),x, algorithm="giac")

[Out]

-2/3*(3*(B*c*d - A*d^2)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(c) + arctan((c*tan(1/2*f*x + 1/2*e) + d)/sqrt(c^
2 - d^2)))/((a^2*c^2 - 2*a^2*c*d + a^2*d^2)*sqrt(c^2 - d^2)) + (3*A*c*tan(1/2*f*x + 1/2*e)^2 - 6*A*d*tan(1/2*f
*x + 1/2*e)^2 + 3*B*d*tan(1/2*f*x + 1/2*e)^2 + 3*A*c*tan(1/2*f*x + 1/2*e) + 3*B*c*tan(1/2*f*x + 1/2*e) - 9*A*d
*tan(1/2*f*x + 1/2*e) + 3*B*d*tan(1/2*f*x + 1/2*e) + 2*A*c + B*c - 5*A*d + 2*B*d)/((a^2*c^2 - 2*a^2*c*d + a^2*
d^2)*(tan(1/2*f*x + 1/2*e) + 1)^3))/f