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

Optimal. Leaf size=275 \[ \frac{2 d \left (A d (3 c+2 d)-B \left (2 c^2+2 c d+d^2\right )\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{a^2 f (c-d)^3 (c+d) \sqrt{c^2-d^2}}-\frac{d \left (A \left (c^2-6 c d-10 d^2\right )+B \left (2 c^2+9 c d+4 d^2\right )\right ) \cos (e+f x)}{3 a^2 f (c-d)^3 (c+d) (c+d \sin (e+f x))}-\frac{(A c-6 A d+2 B c+3 B d) \cos (e+f x)}{3 a^2 f (c-d)^2 (\sin (e+f x)+1) (c+d \sin (e+f x))}-\frac{(A-B) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))} \]

[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]))

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Rubi [A]  time = 0.672215, antiderivative size = 275, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {2978, 2754, 12, 2660, 618, 204} \[ \frac{2 d \left (A d (3 c+2 d)-B \left (2 c^2+2 c d+d^2\right )\right ) \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{a^2 f (c-d)^3 (c+d) \sqrt{c^2-d^2}}-\frac{d \left (A \left (c^2-6 c d-10 d^2\right )+B \left (2 c^2+9 c d+4 d^2\right )\right ) \cos (e+f x)}{3 a^2 f (c-d)^3 (c+d) (c+d \sin (e+f x))}-\frac{(A c-6 A d+2 B c+3 B d) \cos (e+f x)}{3 a^2 f (c-d)^2 (\sin (e+f x)+1) (c+d \sin (e+f x))}-\frac{(A-B) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))} \]

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])^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 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 2754

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 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))^2} \, dx &=-\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 ) \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)^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 ) \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)^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 ) \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)^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]  time = 2.88501, size = 313, normalized size = 1.14 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (-\frac{6 d \left (B \left (2 c^2+2 c d+d^2\right )-A d (3 c+2 d)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3 \tan ^{-1}\left (\frac{c \tan \left (\frac{1}{2} (e+f x)\right )+d}{\sqrt{c^2-d^2}}\right )}{(c+d) \sqrt{c^2-d^2}}+\frac{3 d^2 (A d-B c) \cos (e+f x) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3}{(c+d) (c+d \sin (e+f x))}+2 (A-B) (c-d) \sin \left (\frac{1}{2} (e+f x)\right )+2 (A (c-7 d)+2 B (c+2 d)) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2+(B-A) (c-d) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )}{3 a^2 f (c-d)^3 (\sin (e+f x)+1)^2} \]

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])^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)

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Maple [B]  time = 0.16, size = 770, normalized size = 2.8 \begin{align*} \text{result too large to display} \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))^2,x)

[Out]

2/f/a^2/(c-d)^3*d^4/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)/(c+d)/c*tan(1/2*f*x+1/2*e)*A-2/f/a^2/(c-
d)^3*d^3/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)/(c+d)*tan(1/2*f*x+1/2*e)*B+2/f/a^2/(c-d)^3*d^3/(c*t
an(1/2*f*x+1/2*e)^2+2*tan(1/2*f*x+1/2*e)*d+c)/(c+d)*A-2/f/a^2/(c-d)^3*d^2/(c*tan(1/2*f*x+1/2*e)^2+2*tan(1/2*f*
x+1/2*e)*d+c)/(c+d)*B*c+6/f/a^2/(c-d)^3*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))*A*c+4/f/a^2/(c-d)^3*d^3/(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-4/f/a^2/(c-d)^3*d/(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
-4/f/a^2/(c-d)^3*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))*B*c-2/f/a^
2/(c-d)^3*d^3/(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+2/f/a^2/(c-d)^2
/(tan(1/2*f*x+1/2*e)+1)^2*A-2/f/a^2/(c-d)^2/(tan(1/2*f*x+1/2*e)+1)^2*B-4/3/f/a^2/(c-d)^2/(tan(1/2*f*x+1/2*e)+1
)^3*A+4/3/f/a^2/(c-d)^2/(tan(1/2*f*x+1/2*e)+1)^3*B-2/f/a^2/(c-d)^3/(tan(1/2*f*x+1/2*e)+1)*A*c+6/f/a^2/(c-d)^3/
(tan(1/2*f*x+1/2*e)+1)*A*d-4/f/a^2/(c-d)^3/(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))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.02852, size = 6543, normalized size = 23.79 \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))^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
))]

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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))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.2982, size = 574, normalized size = 2.09 \begin{align*} -\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} \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))^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