\(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2} \, dx\) [277]

Optimal result

Integrand size = 35, antiderivative size = 275 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2} \, dx=\frac {2 d \left (A d (3 c+2 d)-B \left (2 c^2+2 c d+d^2\right )\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{a^2 (c-d)^3 (c+d) \sqrt {c^2-d^2} f}-\frac {d \left (A \left (c^2-6 c d-10 d^2\right )+B \left (2 c^2+9 c d+4 d^2\right )\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f (c+d \sin (e+f x))}-\frac {(A c+2 B c-6 A d+3 B d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) (c+d \sin (e+f x))}-\frac {(A-B) \cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))} \]

[Out]

-1/3*d*(A*(c^2-6*c*d-10*d^2)+B*(2*c^2+9*c*d+4*d^2))*cos(f*x+e)/a^2/(c-d)^3/(c+d)/f/(c+d*sin(f*x+e))-1/3*(A*c-6
*A*d+2*B*c+3*B*d)*cos(f*x+e)/a^2/(c-d)^2/f/(1+sin(f*x+e))/(c+d*sin(f*x+e))-1/3*(A-B)*cos(f*x+e)/(c-d)/f/(a+a*s
in(f*x+e))^2/(c+d*sin(f*x+e))+2*d*(A*d*(3*c+2*d)-B*(2*c^2+2*c*d+d^2))*arctan((d+c*tan(1/2*f*x+1/2*e))/(c^2-d^2
)^(1/2))/a^2/(c-d)^3/(c+d)/f/(c^2-d^2)^(1/2)

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3057, 2833, 12, 2739, 632, 210} \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2} \, dx=\frac {2 d \left (A d (3 c+2 d)-B \left (2 c^2+2 c d+d^2\right )\right ) \arctan \left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{a^2 f (c-d)^3 (c+d) \sqrt {c^2-d^2}}-\frac {d \left (A \left (c^2-6 c d-10 d^2\right )+B \left (2 c^2+9 c d+4 d^2\right )\right ) \cos (e+f x)}{3 a^2 f (c-d)^3 (c+d) (c+d \sin (e+f x))}-\frac {(A c-6 A d+2 B c+3 B d) \cos (e+f x)}{3 a^2 f (c-d)^2 (\sin (e+f x)+1) (c+d \sin (e+f x))}-\frac {(A-B) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))} \]

[In]

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

[Out]

(2*d*(A*d*(3*c + 2*d) - B*(2*c^2 + 2*c*d + d^2))*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(a^2*(c - d
)^3*(c + d)*Sqrt[c^2 - d^2]*f) - (d*(A*(c^2 - 6*c*d - 10*d^2) + B*(2*c^2 + 9*c*d + 4*d^2))*Cos[e + f*x])/(3*a^
2*(c - d)^3*(c + d)*f*(c + d*Sin[e + f*x])) - ((A*c + 2*B*c - 6*A*d + 3*B*d)*Cos[e + f*x])/(3*a^2*(c - d)^2*f*
(1 + Sin[e + f*x])*(c + d*Sin[e + f*x])) - ((A - B)*Cos[e + f*x])/(3*(c - d)*f*(a + a*Sin[e + f*x])^2*(c + d*S
in[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 2833

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

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)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))}-\frac {\int \frac {-a (A (c-4 d)+B (2 c+d))-2 a (A-B) d \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx}{3 a^2 (c-d)} \\ & = -\frac {(A c+2 B c-6 A d+3 B d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) (c+d \sin (e+f x))}-\frac {(A-B) \cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))}+\frac {\int \frac {-2 a^2 d (3 B c-5 A d+2 B d)+a^2 d (A c+2 B c-6 A d+3 B d) \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx}{3 a^4 (c-d)^2} \\ & = -\frac {d \left (A \left (c^2-6 c d-10 d^2\right )+B \left (2 c^2+9 c d+4 d^2\right )\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f (c+d \sin (e+f x))}-\frac {(A c+2 B c-6 A d+3 B d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) (c+d \sin (e+f x))}-\frac {(A-B) \cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))}-\frac {\int -\frac {3 a^2 d \left (A d (3 c+2 d)-B \left (2 c^2+2 c d+d^2\right )\right )}{c+d \sin (e+f x)} \, dx}{3 a^4 (c-d)^3 (c+d)} \\ & = -\frac {d \left (A \left (c^2-6 c d-10 d^2\right )+B \left (2 c^2+9 c d+4 d^2\right )\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f (c+d \sin (e+f x))}-\frac {(A c+2 B c-6 A d+3 B d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) (c+d \sin (e+f x))}-\frac {(A-B) \cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))}+\frac {\left (d \left (A d (3 c+2 d)-B \left (2 c^2+2 c d+d^2\right )\right )\right ) \int \frac {1}{c+d \sin (e+f x)} \, dx}{a^2 (c-d)^3 (c+d)} \\ & = -\frac {d \left (A \left (c^2-6 c d-10 d^2\right )+B \left (2 c^2+9 c d+4 d^2\right )\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f (c+d \sin (e+f x))}-\frac {(A c+2 B c-6 A d+3 B d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) (c+d \sin (e+f x))}-\frac {(A-B) \cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))}+\frac {\left (2 d \left (A d (3 c+2 d)-B \left (2 c^2+2 c d+d^2\right )\right )\right ) \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^2 (c-d)^3 (c+d) f} \\ & = -\frac {d \left (A \left (c^2-6 c d-10 d^2\right )+B \left (2 c^2+9 c d+4 d^2\right )\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f (c+d \sin (e+f x))}-\frac {(A c+2 B c-6 A d+3 B d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) (c+d \sin (e+f x))}-\frac {(A-B) \cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))}-\frac {\left (4 d \left (A d (3 c+2 d)-B \left (2 c^2+2 c d+d^2\right )\right )\right ) \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^2 (c-d)^3 (c+d) f} \\ & = \frac {2 d \left (A d (3 c+2 d)-B \left (2 c^2+2 c d+d^2\right )\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{a^2 (c-d)^3 (c+d) \sqrt {c^2-d^2} f}-\frac {d \left (A \left (c^2-6 c d-10 d^2\right )+B \left (2 c^2+9 c d+4 d^2\right )\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f (c+d \sin (e+f x))}-\frac {(A c+2 B c-6 A d+3 B d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) (c+d \sin (e+f x))}-\frac {(A-B) \cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 7.00 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.14 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (2 (A-B) (c-d) \sin \left (\frac {1}{2} (e+f x)\right )+(-A+B) (c-d) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+2 (A (c-7 d)+2 B (c+2 d)) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-\frac {6 d \left (-A d (3 c+2 d)+B \left (2 c^2+2 c d+d^2\right )\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}{(c+d) \sqrt {c^2-d^2}}+\frac {3 d^2 (-B c+A d) \cos (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}{(c+d) (c+d \sin (e+f x))}\right )}{3 a^2 (c-d)^3 f (1+\sin (e+f x))^2} \]

[In]

Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^2*(c + d*Sin[e + f*x])^2),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 - 7*d) + 2*B*(c + 2*d))*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])
^2 - (6*d*(-(A*d*(3*c + 2*d)) + B*(2*c^2 + 2*c*d + d^2))*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]]*(Cos
[(e + f*x)/2] + Sin[(e + f*x)/2])^3)/((c + d)*Sqrt[c^2 - d^2]) + (3*d^2*(-(B*c) + A*d)*Cos[e + f*x]*(Cos[(e +
f*x)/2] + Sin[(e + f*x)/2])^3)/((c + d)*(c + d*Sin[e + f*x]))))/(3*a^2*(c - d)^3*f*(1 + Sin[e + f*x])^2)

Maple [A] (verified)

Time = 1.87 (sec) , antiderivative size = 263, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(\frac {\frac {2 d \left (\frac {\frac {d^{2} \left (d A -B c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c +d \right ) c}+\frac {d \left (d A -B c \right )}{c +d}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {\left (3 A c d +2 A \,d^{2}-2 B \,c^{2}-2 c d B -d^{2} B \right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c +d \right ) \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{3}}-\frac {2 \left (-2 B +2 A \right )}{3 \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 B -2 A}{\left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (A c -3 d A +2 d B \right )}{\left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}}{a^{2} f}\)	\(263\)
default	\(\frac {\frac {2 d \left (\frac {\frac {d^{2} \left (d A -B c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c +d \right ) c}+\frac {d \left (d A -B c \right )}{c +d}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {\left (3 A c d +2 A \,d^{2}-2 B \,c^{2}-2 c d B -d^{2} B \right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c +d \right ) \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{3}}-\frac {2 \left (-2 B +2 A \right )}{3 \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 B -2 A}{\left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (A c -3 d A +2 d B \right )}{\left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}}{a^{2} f}\)	\(263\)
risch	\(\text {Expression too large to display}\)	\(1411\)

[In]

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1518 vs. \(2 (264) = 528\).

Time = 0.36 (sec) , antiderivative size = 3123, normalized size of antiderivative = 11.36 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2} \, dx=\text {Too large to display} \]

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

[1/6*(2*(A - B)*c^5 - 2*(A - B)*c^4*d - 4*(A - B)*c^3*d^2 + 4*(A - B)*c^2*d^3 + 2*(A - B)*c*d^4 - 2*(A - B)*d^
5 - 2*((A + 2*B)*c^4*d - 3*(2*A - 3*B)*c^3*d^2 - (11*A - 2*B)*c^2*d^3 + 3*(2*A - 3*B)*c*d^4 + 2*(5*A - 2*B)*d^
5)*cos(f*x + e)^3 + 2*((A + 2*B)*c^5 - 5*(A - B)*c^4*d - (8*A - 5*B)*c^3*d^2 + (A - 4*B)*c^2*d^3 + 7*(A - B)*c
*d^4 + (4*A - B)*d^5)*cos(f*x + e)^2 - 3*(4*B*c^3*d - 2*(3*A - 4*B)*c^2*d^2 - 2*(5*A - 3*B)*c*d^3 - 2*(2*A - B
)*d^4 - (2*B*c^2*d^2 - (3*A - 2*B)*c*d^3 - (2*A - B)*d^4)*cos(f*x + e)^3 - (2*B*c^3*d - 3*(A - 2*B)*c^2*d^2 -
(8*A - 5*B)*c*d^3 - 2*(2*A - B)*d^4)*cos(f*x + e)^2 + (2*B*c^3*d - (3*A - 4*B)*c^2*d^2 - (5*A - 3*B)*c*d^3 - (
2*A - B)*d^4)*cos(f*x + e) + (4*B*c^3*d - 2*(3*A - 4*B)*c^2*d^2 - 2*(5*A - 3*B)*c*d^3 - 2*(2*A - B)*d^4 - (2*B
*c^2*d^2 - (3*A - 2*B)*c*d^3 - (2*A - B)*d^4)*cos(f*x + e)^2 + (2*B*c^3*d - (3*A - 4*B)*c^2*d^2 - (5*A - 3*B)*
c*d^3 - (2*A - B)*d^4)*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^5 - (5*A - 8*B)*c^4*d - 16*(A - B)*c^3*d^2 - 4*(2*A +
B)*c^2*d^3 + (14*A - 17*B)*c*d^4 + (13*A - 4*B)*d^5)*cos(f*x + e) - 2*((A - B)*c^5 - (A - B)*c^4*d - 2*(A - B)
*c^3*d^2 + 2*(A - B)*c^2*d^3 + (A - B)*c*d^4 - (A - B)*d^5 - ((A + 2*B)*c^4*d - 3*(2*A - 3*B)*c^3*d^2 - (11*A
- 2*B)*c^2*d^3 + 3*(2*A - 3*B)*c*d^4 + 2*(5*A - 2*B)*d^5)*cos(f*x + e)^2 - ((A + 2*B)*c^5 - (4*A - 7*B)*c^4*d
- 14*(A - B)*c^3*d^2 - 2*(5*A + B)*c^2*d^3 + (13*A - 16*B)*c*d^4 + (14*A - 5*B)*d^5)*cos(f*x + e))*sin(f*x + e
))/((a^2*c^6*d - 2*a^2*c^5*d^2 - a^2*c^4*d^3 + 4*a^2*c^3*d^4 - a^2*c^2*d^5 - 2*a^2*c*d^6 + a^2*d^7)*f*cos(f*x
+ e)^3 + (a^2*c^7 - 5*a^2*c^5*d^2 + 2*a^2*c^4*d^3 + 7*a^2*c^3*d^4 - 4*a^2*c^2*d^5 - 3*a^2*c*d^6 + 2*a^2*d^7)*f
*cos(f*x + e)^2 - (a^2*c^7 - a^2*c^6*d - 3*a^2*c^5*d^2 + 3*a^2*c^4*d^3 + 3*a^2*c^3*d^4 - 3*a^2*c^2*d^5 - a^2*c
*d^6 + a^2*d^7)*f*cos(f*x + e) - 2*(a^2*c^7 - a^2*c^6*d - 3*a^2*c^5*d^2 + 3*a^2*c^4*d^3 + 3*a^2*c^3*d^4 - 3*a^
2*c^2*d^5 - a^2*c*d^6 + a^2*d^7)*f + ((a^2*c^6*d - 2*a^2*c^5*d^2 - a^2*c^4*d^3 + 4*a^2*c^3*d^4 - a^2*c^2*d^5 -
2*a^2*c*d^6 + a^2*d^7)*f*cos(f*x + e)^2 - (a^2*c^7 - a^2*c^6*d - 3*a^2*c^5*d^2 + 3*a^2*c^4*d^3 + 3*a^2*c^3*d^
4 - 3*a^2*c^2*d^5 - a^2*c*d^6 + a^2*d^7)*f*cos(f*x + e) - 2*(a^2*c^7 - a^2*c^6*d - 3*a^2*c^5*d^2 + 3*a^2*c^4*d
^3 + 3*a^2*c^3*d^4 - 3*a^2*c^2*d^5 - a^2*c*d^6 + a^2*d^7)*f)*sin(f*x + e)), 1/3*((A - B)*c^5 - (A - B)*c^4*d -
2*(A - B)*c^3*d^2 + 2*(A - B)*c^2*d^3 + (A - B)*c*d^4 - (A - B)*d^5 - ((A + 2*B)*c^4*d - 3*(2*A - 3*B)*c^3*d^
2 - (11*A - 2*B)*c^2*d^3 + 3*(2*A - 3*B)*c*d^4 + 2*(5*A - 2*B)*d^5)*cos(f*x + e)^3 + ((A + 2*B)*c^5 - 5*(A - B
)*c^4*d - (8*A - 5*B)*c^3*d^2 + (A - 4*B)*c^2*d^3 + 7*(A - B)*c*d^4 + (4*A - B)*d^5)*cos(f*x + e)^2 - 3*(4*B*c
^3*d - 2*(3*A - 4*B)*c^2*d^2 - 2*(5*A - 3*B)*c*d^3 - 2*(2*A - B)*d^4 - (2*B*c^2*d^2 - (3*A - 2*B)*c*d^3 - (2*A
- B)*d^4)*cos(f*x + e)^3 - (2*B*c^3*d - 3*(A - 2*B)*c^2*d^2 - (8*A - 5*B)*c*d^3 - 2*(2*A - B)*d^4)*cos(f*x +
e)^2 + (2*B*c^3*d - (3*A - 4*B)*c^2*d^2 - (5*A - 3*B)*c*d^3 - (2*A - B)*d^4)*cos(f*x + e) + (4*B*c^3*d - 2*(3*
A - 4*B)*c^2*d^2 - 2*(5*A - 3*B)*c*d^3 - 2*(2*A - B)*d^4 - (2*B*c^2*d^2 - (3*A - 2*B)*c*d^3 - (2*A - B)*d^4)*c
os(f*x + e)^2 + (2*B*c^3*d - (3*A - 4*B)*c^2*d^2 - (5*A - 3*B)*c*d^3 - (2*A - B)*d^4)*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^5 - (5*A - 8*B
)*c^4*d - 16*(A - B)*c^3*d^2 - 4*(2*A + B)*c^2*d^3 + (14*A - 17*B)*c*d^4 + (13*A - 4*B)*d^5)*cos(f*x + e) - ((
A - B)*c^5 - (A - B)*c^4*d - 2*(A - B)*c^3*d^2 + 2*(A - B)*c^2*d^3 + (A - B)*c*d^4 - (A - B)*d^5 - ((A + 2*B)*
c^4*d - 3*(2*A - 3*B)*c^3*d^2 - (11*A - 2*B)*c^2*d^3 + 3*(2*A - 3*B)*c*d^4 + 2*(5*A - 2*B)*d^5)*cos(f*x + e)^2
- ((A + 2*B)*c^5 - (4*A - 7*B)*c^4*d - 14*(A - B)*c^3*d^2 - 2*(5*A + B)*c^2*d^3 + (13*A - 16*B)*c*d^4 + (14*A
- 5*B)*d^5)*cos(f*x + e))*sin(f*x + e))/((a^2*c^6*d - 2*a^2*c^5*d^2 - a^2*c^4*d^3 + 4*a^2*c^3*d^4 - a^2*c^2*d
^5 - 2*a^2*c*d^6 + a^2*d^7)*f*cos(f*x + e)^3 + (a^2*c^7 - 5*a^2*c^5*d^2 + 2*a^2*c^4*d^3 + 7*a^2*c^3*d^4 - 4*a^
2*c^2*d^5 - 3*a^2*c*d^6 + 2*a^2*d^7)*f*cos(f*x + e)^2 - (a^2*c^7 - a^2*c^6*d - 3*a^2*c^5*d^2 + 3*a^2*c^4*d^3 +
3*a^2*c^3*d^4 - 3*a^2*c^2*d^5 - a^2*c*d^6 + a^2*d^7)*f*cos(f*x + e) - 2*(a^2*c^7 - a^2*c^6*d - 3*a^2*c^5*d^2
+ 3*a^2*c^4*d^3 + 3*a^2*c^3*d^4 - 3*a^2*c^2*d^5 - a^2*c*d^6 + a^2*d^7)*f + ((a^2*c^6*d - 2*a^2*c^5*d^2 - a^2*c
^4*d^3 + 4*a^2*c^3*d^4 - a^2*c^2*d^5 - 2*a^2*c*d^6 + a^2*d^7)*f*cos(f*x + e)^2 - (a^2*c^7 - a^2*c^6*d - 3*a^2*
c^5*d^2 + 3*a^2*c^4*d^3 + 3*a^2*c^3*d^4 - 3*a^2*c^2*d^5 - a^2*c*d^6 + a^2*d^7)*f*cos(f*x + e) - 2*(a^2*c^7 - a
^2*c^6*d - 3*a^2*c^5*d^2 + 3*a^2*c^4*d^3 + 3*a^2*c^3*d^4 - 3*a^2*c^2*d^5 - a^2*c*d^6 + a^2*d^7)*f)*sin(f*x + e
))]

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2} \, dx=\text {Timed out} \]

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))**2/(c+d*sin(f*x+e))**2,x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^2,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 = 0.34 (sec) , antiderivative size = 411, normalized size of antiderivative = 1.49 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2} \, dx=-\frac {2 \, {\left (\frac {3 \, {\left (2 \, B c^{2} d - 3 \, A c d^{2} + 2 \, B c d^{2} - 2 \, A d^{3} + B d^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )}}{{\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} \sqrt {c^{2} - d^{2}}} + \frac {3 \, {\left (B c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - A d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + B c^{2} d^{2} - A c d^{3}\right )}}{{\left (a^{2} c^{5} - 2 \, a^{2} c^{4} d + 2 \, a^{2} c^{2} d^{3} - a^{2} c d^{4}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c\right )}} + \frac {3 \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 9 \, A d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 6 \, B d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, B c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 15 \, A d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 9 \, B d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, A c + B c - 8 \, A d + 5 \, B d}{{\left (a^{2} c^{3} - 3 \, a^{2} c^{2} d + 3 \, a^{2} c d^{2} - a^{2} d^{3}\right )} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}}\right )}}{3 \, f} \]

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^2,x, algorithm="giac")

[Out]

-2/3*(3*(2*B*c^2*d - 3*A*c*d^2 + 2*B*c*d^2 - 2*A*d^3 + B*d^3)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(c) + arcta
n((c*tan(1/2*f*x + 1/2*e) + d)/sqrt(c^2 - d^2)))/((a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*sqrt(c^2 - d
^2)) + 3*(B*c*d^3*tan(1/2*f*x + 1/2*e) - A*d^4*tan(1/2*f*x + 1/2*e) + B*c^2*d^2 - A*c*d^3)/((a^2*c^5 - 2*a^2*c
^4*d + 2*a^2*c^2*d^3 - a^2*c*d^4)*(c*tan(1/2*f*x + 1/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*e) + c)) + (3*A*c*tan(1/2*
f*x + 1/2*e)^2 - 9*A*d*tan(1/2*f*x + 1/2*e)^2 + 6*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) - 15*A*d*tan(1/2*f*x + 1/2*e) + 9*B*d*tan(1/2*f*x + 1/2*e) + 2*A*c + B*c - 8*A*d + 5*
B*d)/((a^2*c^3 - 3*a^2*c^2*d + 3*a^2*c*d^2 - a^2*d^3)*(tan(1/2*f*x + 1/2*e) + 1)^3))/f

Mupad [B] (verification not implemented)

Time = 16.73 (sec) , antiderivative size = 844, normalized size of antiderivative = 3.07 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2} \, dx=\frac {2\,d\,\mathrm {atan}\left (\frac {\frac {d\,\left (-2\,a^2\,c^4\,d+4\,a^2\,c^3\,d^2-4\,a^2\,c\,d^4+2\,a^2\,d^5\right )\,\left (2\,B\,c^2-2\,A\,d^2+B\,d^2-3\,A\,c\,d+2\,B\,c\,d\right )}{a^2\,{\left (c+d\right )}^{3/2}\,{\left (c-d\right )}^{7/2}}-\frac {2\,c\,d\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (a^2\,c^4-2\,a^2\,c^3\,d+2\,a^2\,c\,d^3-a^2\,d^4\right )\,\left (2\,B\,c^2-2\,A\,d^2+B\,d^2-3\,A\,c\,d+2\,B\,c\,d\right )}{a^2\,{\left (c+d\right )}^{3/2}\,{\left (c-d\right )}^{7/2}}}{2\,B\,d^3-4\,A\,d^3-6\,A\,c\,d^2+4\,B\,c\,d^2+4\,B\,c^2\,d}\right )\,\left (2\,B\,c^2-2\,A\,d^2+B\,d^2-3\,A\,c\,d+2\,B\,c\,d\right )}{a^2\,f\,{\left (c+d\right )}^{3/2}\,{\left (c-d\right )}^{7/2}}-\frac {\frac {2\,\left (2\,A\,c^3-3\,A\,d^3+B\,c^3-8\,A\,c\,d^2-6\,A\,c^2\,d+8\,B\,c\,d^2+6\,B\,c^2\,d\right )}{3\,\left (c+d\right )\,\left (c-d\right )\,\left (c^2-2\,c\,d+d^2\right )}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (5\,A\,c^3-9\,A\,d^3+B\,c^3-30\,A\,c\,d^2-11\,A\,c^2\,d+27\,B\,c\,d^2+17\,B\,c^2\,d\right )}{3\,c\,\left (c-d\right )\,\left (c^2-2\,c\,d+d^2\right )}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (A\,c^4-3\,A\,d^4+B\,c^4-9\,A\,c^2\,d^2+8\,B\,c^2\,d^2-7\,A\,c\,d^3-2\,A\,c^3\,d+7\,B\,c\,d^3+4\,B\,c^3\,d\right )}{c\,\left (c+d\right )\,\left (c-d\right )\,\left (c^2-2\,c\,d+d^2\right )}+\frac {2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (3\,A\,c^4-3\,A\,d^4+3\,B\,c^4-27\,A\,c^2\,d^2+30\,B\,c^2\,d^2-25\,A\,c\,d^3-8\,A\,c^3\,d+13\,B\,c\,d^3+14\,B\,c^3\,d\right )}{3\,c\,\left (c+d\right )\,\left (c-d\right )\,\left (c^2-2\,c\,d+d^2\right )}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (A\,c^4-A\,d^4-3\,A\,c^2\,d^2+2\,B\,c^2\,d^2-2\,A\,c^3\,d+B\,c\,d^3+2\,B\,c^3\,d\right )}{c\,\left (c+d\right )\,\left (c-d\right )\,\left (c^2-2\,c\,d+d^2\right )}}{f\,\left (a^2\,c+\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (3\,a^2\,c+2\,a^2\,d\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (3\,a^2\,c+2\,a^2\,d\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (4\,a^2\,c+6\,a^2\,d\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (4\,a^2\,c+6\,a^2\,d\right )+a^2\,c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\right )} \]

[In]

int((A + B*sin(e + f*x))/((a + a*sin(e + f*x))^2*(c + d*sin(e + f*x))^2),x)

[Out]

(2*d*atan(((d*(2*a^2*d^5 - 4*a^2*c*d^4 - 2*a^2*c^4*d + 4*a^2*c^3*d^2)*(2*B*c^2 - 2*A*d^2 + B*d^2 - 3*A*c*d + 2
*B*c*d))/(a^2*(c + d)^(3/2)*(c - d)^(7/2)) - (2*c*d*tan(e/2 + (f*x)/2)*(a^2*c^4 - a^2*d^4 + 2*a^2*c*d^3 - 2*a^
2*c^3*d)*(2*B*c^2 - 2*A*d^2 + B*d^2 - 3*A*c*d + 2*B*c*d))/(a^2*(c + d)^(3/2)*(c - d)^(7/2)))/(2*B*d^3 - 4*A*d^
3 - 6*A*c*d^2 + 4*B*c*d^2 + 4*B*c^2*d))*(2*B*c^2 - 2*A*d^2 + B*d^2 - 3*A*c*d + 2*B*c*d))/(a^2*f*(c + d)^(3/2)*
(c - d)^(7/2)) - ((2*(2*A*c^3 - 3*A*d^3 + B*c^3 - 8*A*c*d^2 - 6*A*c^2*d + 8*B*c*d^2 + 6*B*c^2*d))/(3*(c + d)*(
c - d)*(c^2 - 2*c*d + d^2)) + (2*tan(e/2 + (f*x)/2)^2*(5*A*c^3 - 9*A*d^3 + B*c^3 - 30*A*c*d^2 - 11*A*c^2*d + 2
7*B*c*d^2 + 17*B*c^2*d))/(3*c*(c - d)*(c^2 - 2*c*d + d^2)) + (2*tan(e/2 + (f*x)/2)^3*(A*c^4 - 3*A*d^4 + B*c^4
- 9*A*c^2*d^2 + 8*B*c^2*d^2 - 7*A*c*d^3 - 2*A*c^3*d + 7*B*c*d^3 + 4*B*c^3*d))/(c*(c + d)*(c - d)*(c^2 - 2*c*d
+ d^2)) + (2*tan(e/2 + (f*x)/2)*(3*A*c^4 - 3*A*d^4 + 3*B*c^4 - 27*A*c^2*d^2 + 30*B*c^2*d^2 - 25*A*c*d^3 - 8*A*
c^3*d + 13*B*c*d^3 + 14*B*c^3*d))/(3*c*(c + d)*(c - d)*(c^2 - 2*c*d + d^2)) + (2*tan(e/2 + (f*x)/2)^4*(A*c^4 -
A*d^4 - 3*A*c^2*d^2 + 2*B*c^2*d^2 - 2*A*c^3*d + B*c*d^3 + 2*B*c^3*d))/(c*(c + d)*(c - d)*(c^2 - 2*c*d + d^2))
)/(f*(a^2*c + tan(e/2 + (f*x)/2)*(3*a^2*c + 2*a^2*d) + tan(e/2 + (f*x)/2)^4*(3*a^2*c + 2*a^2*d) + tan(e/2 + (f
*x)/2)^2*(4*a^2*c + 6*a^2*d) + tan(e/2 + (f*x)/2)^3*(4*a^2*c + 6*a^2*d) + a^2*c*tan(e/2 + (f*x)/2)^5))