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

Optimal. Leaf size=101 \[ \frac{2 (B c-A d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{a f (c-d) \sqrt{c^2-d^2}}-\frac{(A-B) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a)} \]

[Out]

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

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Rubi [A]  time = 0.170218, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 5, 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} \[ \frac{2 (B c-A d) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{a f (c-d) \sqrt{c^2-d^2}}-\frac{(A-B) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.325058, size = 148, normalized size = 1.47 \[ \frac{2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left ((A-B) \sqrt{c^2-d^2} \sin \left (\frac{1}{2} (e+f x)\right )+(B c-A d) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )\right )}{a f (c-d) \sqrt{c^2-d^2} (\sin (e+f x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.114, size = 176, normalized size = 1.7 \begin{align*} -2\,{\frac{Ad}{af \left ( c-d \right ) \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{Bc}{af \left ( c-d \right ) \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}{af \left ( c-d \right ) \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}+2\,{\frac{B}{af \left ( c-d \right ) \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))/(c+d*sin(f*x+e)),x)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.01083, size = 1297, normalized size = 12.84 \begin{align*} \left [-\frac{2 \,{\left (A - B\right )} c^{2} - 2 \,{\left (A - B\right )} d^{2} +{\left (B c - A d +{\left (B c - A d\right )} \cos \left (f x + e\right ) +{\left (B c - A d\right )} \sin \left (f x + e\right )\right )} \sqrt{-c^{2} + d^{2}} \log \left (\frac{{\left (2 \, c^{2} - d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2} + 2 \,{\left (c \cos \left (f x + e\right ) \sin \left (f x + e\right ) + d \cos \left (f x + e\right )\right )} \sqrt{-c^{2} + d^{2}}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}\right ) + 2 \,{\left ({\left (A - B\right )} c^{2} -{\left (A - B\right )} d^{2}\right )} \cos \left (f x + e\right ) - 2 \,{\left ({\left (A - B\right )} c^{2} -{\left (A - B\right )} d^{2}\right )} \sin \left (f x + e\right )}{2 \,{\left ({\left (a c^{3} - a c^{2} d - a c d^{2} + a d^{3}\right )} f \cos \left (f x + e\right ) +{\left (a c^{3} - a c^{2} d - a c d^{2} + a d^{3}\right )} f \sin \left (f x + e\right ) +{\left (a c^{3} - a c^{2} d - a c d^{2} + a d^{3}\right )} f\right )}}, -\frac{{\left (A - B\right )} c^{2} -{\left (A - B\right )} d^{2} +{\left (B c - A d +{\left (B c - A d\right )} \cos \left (f x + e\right ) +{\left (B c - A d\right )} \sin \left (f x + e\right )\right )} \sqrt{c^{2} - d^{2}} \arctan \left (-\frac{c \sin \left (f x + e\right ) + d}{\sqrt{c^{2} - d^{2}} \cos \left (f x + e\right )}\right ) +{\left ({\left (A - B\right )} c^{2} -{\left (A - B\right )} d^{2}\right )} \cos \left (f x + e\right ) -{\left ({\left (A - B\right )} c^{2} -{\left (A - B\right )} d^{2}\right )} \sin \left (f x + e\right )}{{\left (a c^{3} - a c^{2} d - a c d^{2} + a d^{3}\right )} f \cos \left (f x + e\right ) +{\left (a c^{3} - a c^{2} d - a c d^{2} + a d^{3}\right )} f \sin \left (f x + e\right ) +{\left (a c^{3} - a c^{2} d - a c d^{2} + a d^{3}\right )} f}\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))/(c+d*sin(f*x+e)),x, algorithm="fricas")

[Out]

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

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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))/(c+d*sin(f*x+e)),x)

[Out]

Timed out

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Giac [A]  time = 1.24632, size = 153, normalized size = 1.51 \begin{align*} \frac{2 \,{\left (\frac{{\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 (B c - A d\right )}}{{\left (a c - a d\right )} \sqrt{c^{2} - d^{2}}} - \frac{A - B}{{\left (a c - a d\right )}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}}\right )}}{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))/(c+d*sin(f*x+e)),x, algorithm="giac")

[Out]

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