\(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))} \, dx\) [269]

Optimal result

Integrand size = 35, antiderivative size = 101 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))} \, dx=\frac {2 (B c-A d) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{a (c-d) \sqrt {c^2-d^2} f}-\frac {(A-B) \cos (e+f x)}{(c-d) f (a+a \sin (e+f x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3057, 12, 2739, 632, 210} \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))} \, dx=\frac {2 (B c-A d) \arctan \left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{a f (c-d) \sqrt {c^2-d^2}}-\frac {(A-B) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a)} \]

[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 12

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

Rule 210

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

Rule 632

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 2739

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 3057

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])

Rubi steps \begin{align*} \text {integral}& = -\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)) \text {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)) \text {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) \arctan \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] (verified)

Time = 4.39 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.47 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))} \, dx=\frac {2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left ((A-B) \sqrt {c^2-d^2} \sin \left (\frac {1}{2} (e+f x)\right )+(B c-A d) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )}{a (c-d) \sqrt {c^2-d^2} f (1+\sin (e+f x))} \]

[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]))

Maple [A] (verified)

Time = 0.77 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.93

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (A -B \right )}{\left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {2 \left (-d A +B c \right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right ) \sqrt {c^{2}-d^{2}}}}{a f}\)	\(94\)
default	\(\frac {-\frac {2 \left (A -B \right )}{\left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {2 \left (-d A +B c \right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right ) \sqrt {c^{2}-d^{2}}}}{a f}\)	\(94\)
risch	\(-\frac {2 A}{f \left (c -d \right ) a \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}+\frac {2 B}{f \left (c -d \right ) a \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}+c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) d A}{\sqrt {-c^{2}+d^{2}}\, \left (c -d \right ) f a}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}+c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) B c}{\sqrt {-c^{2}+d^{2}}\, \left (c -d \right ) f a}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}-c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) d A}{\sqrt {-c^{2}+d^{2}}\, \left (c -d \right ) f a}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}-c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) B c}{\sqrt {-c^{2}+d^{2}}\, \left (c -d \right ) f a}\)	\(372\)

[In]

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))/(c+d*sin(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 255 vs. \(2 (96) = 192\).

Time = 0.28 (sec) , antiderivative size = 595, normalized size of antiderivative = 5.89 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))} \, dx=\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 ] \]

[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)]

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))} \, dx=\text {Exception raised: ValueError} \]

[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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*d^2-4*c^2>0)', see `assume?`
for more de

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.09 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))} \, dx=\frac {2 \, {\left (\frac {{\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )} {\left (B c - A d\right )}}{{\left (a c - a d\right )} \sqrt {c^{2} - d^{2}}} - \frac {A - B}{{\left (a c - a d\right )} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}}\right )}}{f} \]

[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

Mupad [B] (verification not implemented)

Time = 13.83 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.52 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))} \, dx=\frac {2\,\mathrm {atan}\left (\frac {\frac {\left (A\,d-B\,c\right )\,\left (2\,a\,d^2-2\,a\,c\,d\right )}{a\,\sqrt {c+d}\,{\left (c-d\right )}^{3/2}}-\frac {2\,c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (A\,d-B\,c\right )\,\left (a\,c-a\,d\right )}{a\,\sqrt {c+d}\,{\left (c-d\right )}^{3/2}}}{2\,A\,d-2\,B\,c}\right )\,\left (A\,d-B\,c\right )}{a\,f\,\sqrt {c+d}\,{\left (c-d\right )}^{3/2}}-\frac {2\,\left (A-B\right )}{f\,\left (a+a\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )\,\left (c-d\right )} \]

[In]

int((A + B*sin(e + f*x))/((a + a*sin(e + f*x))*(c + d*sin(e + f*x))),x)

[Out]

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