3.283 \(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))} \, dx\)

Optimal. Leaf size=229 \[ \frac {2 d^2 (B c-A d) \tan ^{-1}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{a^3 f (c-d)^3 \sqrt {c^2-d^2}}-\frac {\left (A \left (2 c^2-9 c d+22 d^2\right )+B \left (3 c^2-16 c d-2 d^2\right )\right ) \cos (e+f x)}{15 f (c-d)^3 \left (a^3 \sin (e+f x)+a^3\right )}-\frac {(2 A c-7 A d+3 B c+2 B d) \cos (e+f x)}{15 a f (c-d)^2 (a \sin (e+f x)+a)^2}-\frac {(A-B) \cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3} \]

[Out]

-1/5*(A-B)*cos(f*x+e)/(c-d)/f/(a+a*sin(f*x+e))^3-1/15*(2*A*c-7*A*d+3*B*c+2*B*d)*cos(f*x+e)/a/(c-d)^2/f/(a+a*si
n(f*x+e))^2-1/15*(B*(3*c^2-16*c*d-2*d^2)+A*(2*c^2-9*c*d+22*d^2))*cos(f*x+e)/(c-d)^3/f/(a^3+a^3*sin(f*x+e))+2*d
^2*(-A*d+B*c)*arctan((d+c*tan(1/2*f*x+1/2*e))/(c^2-d^2)^(1/2))/a^3/(c-d)^3/f/(c^2-d^2)^(1/2)

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Rubi [A]  time = 0.72, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ \frac {2 d^2 (B c-A d) \tan ^{-1}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{a^3 f (c-d)^3 \sqrt {c^2-d^2}}-\frac {\left (A \left (2 c^2-9 c d+22 d^2\right )+B \left (3 c^2-16 c d-2 d^2\right )\right ) \cos (e+f x)}{15 f (c-d)^3 \left (a^3 \sin (e+f x)+a^3\right )}-\frac {(2 A c-7 A d+3 B c+2 B d) \cos (e+f x)}{15 a f (c-d)^2 (a \sin (e+f x)+a)^2}-\frac {(A-B) \cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^3*(c + d*Sin[e + f*x])),x]

[Out]

(2*d^2*(B*c - A*d)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(a^3*(c - d)^3*Sqrt[c^2 - d^2]*f) - ((A -
B)*Cos[e + f*x])/(5*(c - d)*f*(a + a*Sin[e + f*x])^3) - ((2*A*c + 3*B*c - 7*A*d + 2*B*d)*Cos[e + f*x])/(15*a*
(c - d)^2*f*(a + a*Sin[e + f*x])^2) - ((B*(3*c^2 - 16*c*d - 2*d^2) + A*(2*c^2 - 9*c*d + 22*d^2))*Cos[e + f*x])
/(15*(c - d)^3*f*(a^3 + a^3*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 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))^3 (c+d \sin (e+f x))} \, dx &=-\frac {(A-B) \cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {\int \frac {-a (2 A c+3 B c-5 A d)-2 a (A-B) d \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx}{5 a^2 (c-d)}\\ &=-\frac {(A-B) \cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {(2 A c+3 B c-7 A d+2 B d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}+\frac {\int \frac {a^2 \left (B c (3 c-13 d)+A \left (2 c^2-7 c d+15 d^2\right )\right )+a^2 d (2 A c+3 B c-7 A d+2 B d) \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))} \, dx}{15 a^4 (c-d)^2}\\ &=-\frac {(A-B) \cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {(2 A c+3 B c-7 A d+2 B d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}-\frac {\left (B \left (3 c^2-16 c d-2 d^2\right )+A \left (2 c^2-9 c d+22 d^2\right )\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac {\int -\frac {15 a^3 d^2 (B c-A d)}{c+d \sin (e+f x)} \, dx}{15 a^6 (c-d)^3}\\ &=-\frac {(A-B) \cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {(2 A c+3 B c-7 A d+2 B d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}-\frac {\left (B \left (3 c^2-16 c d-2 d^2\right )+A \left (2 c^2-9 c d+22 d^2\right )\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right )}+\frac {\left (d^2 (B c-A d)\right ) \int \frac {1}{c+d \sin (e+f x)} \, dx}{a^3 (c-d)^3}\\ &=-\frac {(A-B) \cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {(2 A c+3 B c-7 A d+2 B d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}-\frac {\left (B \left (3 c^2-16 c d-2 d^2\right )+A \left (2 c^2-9 c d+22 d^2\right )\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right )}+\frac {\left (2 d^2 (B c-A d)\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^3 (c-d)^3 f}\\ &=-\frac {(A-B) \cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {(2 A c+3 B c-7 A d+2 B d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}-\frac {\left (B \left (3 c^2-16 c d-2 d^2\right )+A \left (2 c^2-9 c d+22 d^2\right )\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac {\left (4 d^2 (B c-A d)\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^3 (c-d)^3 f}\\ &=\frac {2 d^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^3 (c-d)^3 \sqrt {c^2-d^2} f}-\frac {(A-B) \cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {(2 A c+3 B c-7 A d+2 B d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}-\frac {\left (B \left (3 c^2-16 c d-2 d^2\right )+A \left (2 c^2-9 c d+22 d^2\right )\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right )}\\ \end {align*}

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Mathematica [B]  time = 1.25, size = 502, normalized size = 2.19 \[ \frac {\left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (-\frac {60 d^2 (A d-B c) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5 \tan ^{-1}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{\sqrt {c^2-d^2}}+20 A c^2 \sin \left (\frac {1}{2} (e+f x)\right )-2 A c^2 \sin \left (\frac {5}{2} (e+f x)\right )-10 A c^2 \cos \left (\frac {3}{2} (e+f x)\right )-75 A c d \sin \left (\frac {1}{2} (e+f x)\right )+9 A c d \sin \left (\frac {5}{2} (e+f x)\right )-15 A c d \cos \left (\frac {1}{2} (e+f x)\right )+45 A c d \cos \left (\frac {3}{2} (e+f x)\right )+145 A d^2 \sin \left (\frac {1}{2} (e+f x)\right )+15 A d^2 \sin \left (\frac {3}{2} (e+f x)\right )-22 A d^2 \sin \left (\frac {5}{2} (e+f x)\right )+75 A d^2 \cos \left (\frac {1}{2} (e+f x)\right )-95 A d^2 \cos \left (\frac {3}{2} (e+f x)\right )+15 B c^2 \sin \left (\frac {1}{2} (e+f x)\right )-3 B c^2 \sin \left (\frac {5}{2} (e+f x)\right )+15 B c^2 \cos \left (\frac {1}{2} (e+f x)\right )-15 B c^2 \cos \left (\frac {3}{2} (e+f x)\right )-85 B c d \sin \left (\frac {1}{2} (e+f x)\right )-15 B c d \sin \left (\frac {3}{2} (e+f x)\right )+16 B c d \sin \left (\frac {5}{2} (e+f x)\right )-75 B c d \cos \left (\frac {1}{2} (e+f x)\right )+65 B c d \cos \left (\frac {3}{2} (e+f x)\right )-20 B d^2 \sin \left (\frac {1}{2} (e+f x)\right )+2 B d^2 \sin \left (\frac {5}{2} (e+f x)\right )+10 B d^2 \cos \left (\frac {3}{2} (e+f x)\right )\right )}{30 a^3 f (c-d)^3 (\sin (e+f x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^3*(c + d*Sin[e + f*x])),x]

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(15*B*c^2*Cos[(e + f*x)/2] - 15*A*c*d*Cos[(e + f*x)/2] - 75*B*c*d*Cos[(
e + f*x)/2] + 75*A*d^2*Cos[(e + f*x)/2] - 10*A*c^2*Cos[(3*(e + f*x))/2] - 15*B*c^2*Cos[(3*(e + f*x))/2] + 45*A
*c*d*Cos[(3*(e + f*x))/2] + 65*B*c*d*Cos[(3*(e + f*x))/2] - 95*A*d^2*Cos[(3*(e + f*x))/2] + 10*B*d^2*Cos[(3*(e
+ f*x))/2] + 20*A*c^2*Sin[(e + f*x)/2] + 15*B*c^2*Sin[(e + f*x)/2] - 75*A*c*d*Sin[(e + f*x)/2] - 85*B*c*d*Sin
[(e + f*x)/2] + 145*A*d^2*Sin[(e + f*x)/2] - 20*B*d^2*Sin[(e + f*x)/2] - (60*d^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])^5)/Sqrt[c^2 - d^2] - 15*B*c*d*Sin[(3*
(e + f*x))/2] + 15*A*d^2*Sin[(3*(e + f*x))/2] - 2*A*c^2*Sin[(5*(e + f*x))/2] - 3*B*c^2*Sin[(5*(e + f*x))/2] +
9*A*c*d*Sin[(5*(e + f*x))/2] + 16*B*c*d*Sin[(5*(e + f*x))/2] - 22*A*d^2*Sin[(5*(e + f*x))/2] + 2*B*d^2*Sin[(5*
(e + f*x))/2]))/(30*a^3*(c - d)^3*f*(1 + Sin[e + f*x])^3)

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fricas [B]  time = 0.54, size = 2292, normalized size = 10.01 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c+d*sin(f*x+e)),x, algorithm="fricas")

[Out]

[1/30*(6*(A - B)*c^4 - 12*(A - B)*c^3*d + 12*(A - B)*c*d^3 - 6*(A - B)*d^4 - 2*((2*A + 3*B)*c^4 - (9*A + 16*B)
*c^3*d + 5*(4*A - B)*c^2*d^2 + (9*A + 16*B)*c*d^3 - 2*(11*A - B)*d^4)*cos(f*x + e)^3 + 2*(2*(2*A + 3*B)*c^4 -
(18*A + 17*B)*c^3*d + 5*(5*A - 2*B)*c^2*d^2 + (18*A + 17*B)*c*d^3 - (29*A - 4*B)*d^4)*cos(f*x + e)^2 + 15*(4*B
*c*d^2 - 4*A*d^3 - (B*c*d^2 - A*d^3)*cos(f*x + e)^3 - 3*(B*c*d^2 - A*d^3)*cos(f*x + e)^2 + 2*(B*c*d^2 - A*d^3)
*cos(f*x + e) + (4*B*c*d^2 - 4*A*d^3 - (B*c*d^2 - A*d^3)*cos(f*x + e)^2 + 2*(B*c*d^2 - A*d^3)*cos(f*x + e))*si
n(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))
+ 6*((3*A + 2*B)*c^4 - (11*A + 9*B)*c^3*d + 5*(3*A - B)*c^2*d^2 + (11*A + 9*B)*c*d^3 - 3*(6*A - B)*d^4)*cos(f*
x + e) - 2*(3*(A - B)*c^4 - 6*(A - B)*c^3*d + 6*(A - B)*c*d^3 - 3*(A - B)*d^4 - ((2*A + 3*B)*c^4 - (9*A + 16*B
)*c^3*d + 5*(4*A - B)*c^2*d^2 + (9*A + 16*B)*c*d^3 - 2*(11*A - B)*d^4)*cos(f*x + e)^2 - 3*((2*A + 3*B)*c^4 - (
9*A + 11*B)*c^3*d + 5*(3*A - B)*c^2*d^2 + (9*A + 11*B)*c*d^3 - (17*A - 2*B)*d^4)*cos(f*x + e))*sin(f*x + e))/(
(a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*f*cos(f*x + e)^3 + 3*(a^3*c^5
- 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*f*cos(f*x + e)^2 - 2*(a^3*c^5 - 3*a^3*c
^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*f*cos(f*x + e) - 4*(a^3*c^5 - 3*a^3*c^4*d + 2*a^
3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*f + ((a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3
- 3*a^3*c*d^4 + a^3*d^5)*f*cos(f*x + e)^2 - 2*(a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*
c*d^4 + a^3*d^5)*f*cos(f*x + e) - 4*(a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3
*d^5)*f)*sin(f*x + e)), 1/15*(3*(A - B)*c^4 - 6*(A - B)*c^3*d + 6*(A - B)*c*d^3 - 3*(A - B)*d^4 - ((2*A + 3*B)
*c^4 - (9*A + 16*B)*c^3*d + 5*(4*A - B)*c^2*d^2 + (9*A + 16*B)*c*d^3 - 2*(11*A - B)*d^4)*cos(f*x + e)^3 + (2*(
2*A + 3*B)*c^4 - (18*A + 17*B)*c^3*d + 5*(5*A - 2*B)*c^2*d^2 + (18*A + 17*B)*c*d^3 - (29*A - 4*B)*d^4)*cos(f*x
+ e)^2 + 15*(4*B*c*d^2 - 4*A*d^3 - (B*c*d^2 - A*d^3)*cos(f*x + e)^3 - 3*(B*c*d^2 - A*d^3)*cos(f*x + e)^2 + 2*
(B*c*d^2 - A*d^3)*cos(f*x + e) + (4*B*c*d^2 - 4*A*d^3 - (B*c*d^2 - A*d^3)*cos(f*x + e)^2 + 2*(B*c*d^2 - A*d^3)
*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))) + 3*
((3*A + 2*B)*c^4 - (11*A + 9*B)*c^3*d + 5*(3*A - B)*c^2*d^2 + (11*A + 9*B)*c*d^3 - 3*(6*A - B)*d^4)*cos(f*x +
e) - (3*(A - B)*c^4 - 6*(A - B)*c^3*d + 6*(A - B)*c*d^3 - 3*(A - B)*d^4 - ((2*A + 3*B)*c^4 - (9*A + 16*B)*c^3*
d + 5*(4*A - B)*c^2*d^2 + (9*A + 16*B)*c*d^3 - 2*(11*A - B)*d^4)*cos(f*x + e)^2 - 3*((2*A + 3*B)*c^4 - (9*A +
11*B)*c^3*d + 5*(3*A - B)*c^2*d^2 + (9*A + 11*B)*c*d^3 - (17*A - 2*B)*d^4)*cos(f*x + e))*sin(f*x + e))/((a^3*c
^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*f*cos(f*x + e)^3 + 3*(a^3*c^5 - 3*a^
3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*f*cos(f*x + e)^2 - 2*(a^3*c^5 - 3*a^3*c^4*d +
2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*f*cos(f*x + e) - 4*(a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*
d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*f + ((a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a
^3*c*d^4 + a^3*d^5)*f*cos(f*x + e)^2 - 2*(a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4
+ a^3*d^5)*f*cos(f*x + e) - 4*(a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*
f)*sin(f*x + e))]

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giac [B]  time = 0.26, size = 578, normalized size = 2.52 \[ \frac {2 \, {\left (\frac {15 \, {\left (B c d^{2} - A d^{3}\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^{3} c^{3} - 3 \, a^{3} c^{2} d + 3 \, a^{3} c d^{2} - a^{3} d^{3}\right )} \sqrt {c^{2} - d^{2}}} - \frac {15 \, A c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 45 \, A c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 45 \, A d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 15 \, B d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 30 \, A c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 15 \, B c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 105 \, A c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 45 \, B c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 135 \, A d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 30 \, B d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 40 \, A c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 15 \, B c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 135 \, A c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 65 \, B c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 185 \, A d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 40 \, B d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 20 \, A c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 15 \, B c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 75 \, A c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 55 \, B c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 115 \, A d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 20 \, B d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 7 \, A c^{2} + 3 \, B c^{2} - 24 \, A c d - 11 \, B c d + 32 \, A d^{2} - 7 \, B d^{2}}{{\left (a^{3} c^{3} - 3 \, a^{3} c^{2} d + 3 \, a^{3} c d^{2} - a^{3} d^{3}\right )} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}}\right )}}{15 \, 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))^3/(c+d*sin(f*x+e)),x, algorithm="giac")

[Out]

2/15*(15*(B*c*d^2 - A*d^3)*(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^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*sqrt(c^2 - d^2)) - (15*A*c^2*tan(1/2*f*x + 1/2*
e)^4 - 45*A*c*d*tan(1/2*f*x + 1/2*e)^4 + 45*A*d^2*tan(1/2*f*x + 1/2*e)^4 - 15*B*d^2*tan(1/2*f*x + 1/2*e)^4 + 3
0*A*c^2*tan(1/2*f*x + 1/2*e)^3 + 15*B*c^2*tan(1/2*f*x + 1/2*e)^3 - 105*A*c*d*tan(1/2*f*x + 1/2*e)^3 - 45*B*c*d
*tan(1/2*f*x + 1/2*e)^3 + 135*A*d^2*tan(1/2*f*x + 1/2*e)^3 - 30*B*d^2*tan(1/2*f*x + 1/2*e)^3 + 40*A*c^2*tan(1/
2*f*x + 1/2*e)^2 + 15*B*c^2*tan(1/2*f*x + 1/2*e)^2 - 135*A*c*d*tan(1/2*f*x + 1/2*e)^2 - 65*B*c*d*tan(1/2*f*x +
1/2*e)^2 + 185*A*d^2*tan(1/2*f*x + 1/2*e)^2 - 40*B*d^2*tan(1/2*f*x + 1/2*e)^2 + 20*A*c^2*tan(1/2*f*x + 1/2*e)
+ 15*B*c^2*tan(1/2*f*x + 1/2*e) - 75*A*c*d*tan(1/2*f*x + 1/2*e) - 55*B*c*d*tan(1/2*f*x + 1/2*e) + 115*A*d^2*t
an(1/2*f*x + 1/2*e) - 20*B*d^2*tan(1/2*f*x + 1/2*e) + 7*A*c^2 + 3*B*c^2 - 24*A*c*d - 11*B*c*d + 32*A*d^2 - 7*B
*d^2)/((a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*(tan(1/2*f*x + 1/2*e) + 1)^5))/f

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maple [B]  time = 0.54, size = 606, normalized size = 2.65 \[ -\frac {2 d^{3} \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^{3} \left (c -d \right )^{3} \sqrt {c^{2}-d^{2}}}+\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 ) B c}{f \,a^{3} \left (c -d \right )^{3} \sqrt {c^{2}-d^{2}}}+\frac {4 A}{f \,a^{3} \left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {4 B}{f \,a^{3} \left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {8 A}{5 f \,a^{3} \left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}+\frac {8 B}{5 f \,a^{3} \left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}+\frac {4 A c}{f \,a^{3} \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {6 A d}{f \,a^{3} \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 B c}{f \,a^{3} \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}+\frac {4 B d}{f \,a^{3} \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {16 A c}{3 f \,a^{3} \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {20 A d}{3 f \,a^{3} \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {4 B c}{f \,a^{3} \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {16 B d}{3 f \,a^{3} \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 A \,c^{2}}{f \,a^{3} \left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {6 A c d}{f \,a^{3} \left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {6 A \,d^{2}}{f \,a^{3} \left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {2 B \,d^{2}}{f \,a^{3} \left (c -d \right )^{3} \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))^3/(c+d*sin(f*x+e)),x)

[Out]

-2/f/a^3*d^3/(c-d)^3/(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^3*d^2/(c
-d)^3/(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+4/f/a^3/(c-d)/(tan(1/2*f*x+
1/2*e)+1)^4*A-4/f/a^3/(c-d)/(tan(1/2*f*x+1/2*e)+1)^4*B-8/5/f/a^3/(c-d)/(tan(1/2*f*x+1/2*e)+1)^5*A+8/5/f/a^3/(c
-d)/(tan(1/2*f*x+1/2*e)+1)^5*B+4/f/a^3/(c-d)^2/(tan(1/2*f*x+1/2*e)+1)^2*A*c-6/f/a^3/(c-d)^2/(tan(1/2*f*x+1/2*e
)+1)^2*A*d-2/f/a^3/(c-d)^2/(tan(1/2*f*x+1/2*e)+1)^2*B*c+4/f/a^3/(c-d)^2/(tan(1/2*f*x+1/2*e)+1)^2*B*d-16/3/f/a^
3/(c-d)^2/(tan(1/2*f*x+1/2*e)+1)^3*A*c+20/3/f/a^3/(c-d)^2/(tan(1/2*f*x+1/2*e)+1)^3*A*d+4/f/a^3/(c-d)^2/(tan(1/
2*f*x+1/2*e)+1)^3*B*c-16/3/f/a^3/(c-d)^2/(tan(1/2*f*x+1/2*e)+1)^3*B*d-2/f/a^3/(c-d)^3/(tan(1/2*f*x+1/2*e)+1)*A
*c^2+6/f/a^3/(c-d)^3/(tan(1/2*f*x+1/2*e)+1)*A*c*d-6/f/a^3/(c-d)^3/(tan(1/2*f*x+1/2*e)+1)*A*d^2+2/f/a^3/(c-d)^3
/(tan(1/2*f*x+1/2*e)+1)*B*d^2

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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))^3/(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?

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mupad [B]  time = 17.18, size = 591, normalized size = 2.58 \[ \frac {2\,d^2\,\mathrm {atan}\left (\frac {\frac {d^2\,\left (A\,d-B\,c\right )\,\left (-2\,a^3\,c^3\,d+6\,a^3\,c^2\,d^2-6\,a^3\,c\,d^3+2\,a^3\,d^4\right )}{a^3\,\sqrt {c+d}\,{\left (c-d\right )}^{7/2}}-\frac {2\,c\,d^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (A\,d-B\,c\right )\,\left (a^3\,c^3-3\,a^3\,c^2\,d+3\,a^3\,c\,d^2-a^3\,d^3\right )}{a^3\,\sqrt {c+d}\,{\left (c-d\right )}^{7/2}}}{2\,A\,d^3-2\,B\,c\,d^2}\right )\,\left (A\,d-B\,c\right )}{a^3\,f\,\sqrt {c+d}\,{\left (c-d\right )}^{7/2}}-\frac {\frac {2\,\left (7\,A\,c^2+32\,A\,d^2+3\,B\,c^2-7\,B\,d^2-24\,A\,c\,d-11\,B\,c\,d\right )}{15\,\left (c-d\right )\,\left (c^2-2\,c\,d+d^2\right )}+\frac {2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (4\,A\,c^2+23\,A\,d^2+3\,B\,c^2-4\,B\,d^2-15\,A\,c\,d-11\,B\,c\,d\right )}{3\,\left (c-d\right )\,\left (c^2-2\,c\,d+d^2\right )}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (2\,A\,c^2+9\,A\,d^2+B\,c^2-2\,B\,d^2-7\,A\,c\,d-3\,B\,c\,d\right )}{\left (c-d\right )\,\left (c^2-2\,c\,d+d^2\right )}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (8\,A\,c^2+37\,A\,d^2+3\,B\,c^2-8\,B\,d^2-27\,A\,c\,d-13\,B\,c\,d\right )}{3\,\left (c-d\right )\,\left (c^2-2\,c\,d+d^2\right )}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (A\,c^2+3\,A\,d^2-B\,d^2-3\,A\,c\,d\right )}{\left (c-d\right )\,\left (c^2-2\,c\,d+d^2\right )}}{f\,\left (a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+5\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+10\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+10\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+5\,a^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a^3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*sin(e + f*x))/((a + a*sin(e + f*x))^3*(c + d*sin(e + f*x))),x)

[Out]

(2*d^2*atan(((d^2*(A*d - B*c)*(2*a^3*d^4 - 6*a^3*c*d^3 - 2*a^3*c^3*d + 6*a^3*c^2*d^2))/(a^3*(c + d)^(1/2)*(c -
d)^(7/2)) - (2*c*d^2*tan(e/2 + (f*x)/2)*(A*d - B*c)*(a^3*c^3 - a^3*d^3 + 3*a^3*c*d^2 - 3*a^3*c^2*d))/(a^3*(c
+ d)^(1/2)*(c - d)^(7/2)))/(2*A*d^3 - 2*B*c*d^2))*(A*d - B*c))/(a^3*f*(c + d)^(1/2)*(c - d)^(7/2)) - ((2*(7*A*
c^2 + 32*A*d^2 + 3*B*c^2 - 7*B*d^2 - 24*A*c*d - 11*B*c*d))/(15*(c - d)*(c^2 - 2*c*d + d^2)) + (2*tan(e/2 + (f*
x)/2)*(4*A*c^2 + 23*A*d^2 + 3*B*c^2 - 4*B*d^2 - 15*A*c*d - 11*B*c*d))/(3*(c - d)*(c^2 - 2*c*d + d^2)) + (2*tan
(e/2 + (f*x)/2)^3*(2*A*c^2 + 9*A*d^2 + B*c^2 - 2*B*d^2 - 7*A*c*d - 3*B*c*d))/((c - d)*(c^2 - 2*c*d + d^2)) + (
2*tan(e/2 + (f*x)/2)^2*(8*A*c^2 + 37*A*d^2 + 3*B*c^2 - 8*B*d^2 - 27*A*c*d - 13*B*c*d))/(3*(c - d)*(c^2 - 2*c*d
+ d^2)) + (2*tan(e/2 + (f*x)/2)^4*(A*c^2 + 3*A*d^2 - B*d^2 - 3*A*c*d))/((c - d)*(c^2 - 2*c*d + d^2)))/(f*(10*
a^3*tan(e/2 + (f*x)/2)^2 + 10*a^3*tan(e/2 + (f*x)/2)^3 + 5*a^3*tan(e/2 + (f*x)/2)^4 + a^3*tan(e/2 + (f*x)/2)^5
+ a^3 + 5*a^3*tan(e/2 + (f*x)/2)))

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

[Out]

Timed out

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