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]

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

________________________________________________________________________________________

Rubi [A]  time = 0.42, antiderivative size = 152, normalized size of antiderivative = 1.00, 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 12

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

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])

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 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 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])

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 = 0.63, 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)

________________________________________________________________________________________

fricas [B]  time = 0.50, size = 1285, normalized size = 8.45 \[ \text {result too large to display} \]

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))]

________________________________________________________________________________________

giac [A]  time = 0.22, size = 259, normalized size = 1.70 \[ -\frac {2 \, {\left (\frac {3 \, {\left (B c d - A d^{2}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (c) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )}}{{\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\right )} \sqrt {c^{2} - d^{2}}} + \frac {3 \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 6 \, A d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, B d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, B c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 9 \, A d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, B d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, A c + B c - 5 \, A d + 2 \, B d}{{\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\right )} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}}\right )}}{3 \, f} \]

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

________________________________________________________________________________________

maple [B]  time = 0.56, size = 327, normalized size = 2.15 \[ \frac {2 d^{2} \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right ) A}{f \,a^{2} \left (c -d \right )^{2} \sqrt {c^{2}-d^{2}}}-\frac {2 d \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right ) B c}{f \,a^{2} \left (c -d \right )^{2} \sqrt {c^{2}-d^{2}}}+\frac {2 A}{f \,a^{2} \left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 B}{f \,a^{2} \left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {4 A}{3 f \,a^{2} \left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {4 B}{3 f \,a^{2} \left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 A c}{f \,a^{2} \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {4 A d}{f \,a^{2} \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {2 B d}{f \,a^{2} \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )} \]

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

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> 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 details)Is 4*d^2-4*c^2 positive or negative?

________________________________________________________________________________________

mupad [B]  time = 14.78, size = 302, normalized size = 1.99 \[ \frac {2\,d\,\mathrm {atan}\left (\frac {\frac {d\,\left (A\,d-B\,c\right )\,\left (2\,a^2\,c^2\,d-4\,a^2\,c\,d^2+2\,a^2\,d^3\right )}{a^2\,\sqrt {c+d}\,{\left (c-d\right )}^{5/2}}+\frac {2\,c\,d\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (A\,d-B\,c\right )\,\left (a^2\,c^2-2\,a^2\,c\,d+a^2\,d^2\right )}{a^2\,\sqrt {c+d}\,{\left (c-d\right )}^{5/2}}}{2\,A\,d^2-2\,B\,c\,d}\right )\,\left (A\,d-B\,c\right )}{a^2\,f\,\sqrt {c+d}\,{\left (c-d\right )}^{5/2}}-\frac {\frac {2\,\left (2\,A\,c-5\,A\,d+B\,c+2\,B\,d\right )}{3\,{\left (c-d\right )}^2}+\frac {2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (A\,c-3\,A\,d+B\,c+B\,d\right )}{{\left (c-d\right )}^2}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (A\,c-2\,A\,d+B\,d\right )}{{\left (c-d\right )}^2}}{f\,\left (a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+3\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+3\,a^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a^2\right )} \]

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 + d*sin(e + f*x))),x)

[Out]

(2*d*atan(((d*(A*d - B*c)*(2*a^2*d^3 - 4*a^2*c*d^2 + 2*a^2*c^2*d))/(a^2*(c + d)^(1/2)*(c - d)^(5/2)) + (2*c*d*
tan(e/2 + (f*x)/2)*(A*d - B*c)*(a^2*c^2 + a^2*d^2 - 2*a^2*c*d))/(a^2*(c + d)^(1/2)*(c - d)^(5/2)))/(2*A*d^2 -
2*B*c*d))*(A*d - B*c))/(a^2*f*(c + d)^(1/2)*(c - d)^(5/2)) - ((2*(2*A*c - 5*A*d + B*c + 2*B*d))/(3*(c - d)^2)
+ (2*tan(e/2 + (f*x)/2)*(A*c - 3*A*d + B*c + B*d))/(c - d)^2 + (2*tan(e/2 + (f*x)/2)^2*(A*c - 2*A*d + B*d))/(c
- d)^2)/(f*(3*a^2*tan(e/2 + (f*x)/2)^2 + a^2*tan(e/2 + (f*x)/2)^3 + a^2 + 3*a^2*tan(e/2 + (f*x)/2)))

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

________________________________________________________________________________________