3.80 \(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^6} \, dx\)

Optimal. Leaf size=205 \[ \frac {2 (8 A-3 B) \tan ^5(e+f x)}{165 a^3 c^6 f}+\frac {4 (8 A-3 B) \tan ^3(e+f x)}{99 a^3 c^6 f}+\frac {2 (8 A-3 B) \tan (e+f x)}{33 a^3 c^6 f}+\frac {(8 A-3 B) \sec ^5(e+f x)}{99 a^3 f \left (c^6-c^6 \sin (e+f x)\right )}+\frac {(8 A-3 B) \sec ^5(e+f x)}{99 a^3 f \left (c^3-c^3 \sin (e+f x)\right )^2}+\frac {(A+B) \sec ^5(e+f x)}{11 a^3 f \left (c^2-c^2 \sin (e+f x)\right )^3} \]

[Out]

1/11*(A+B)*sec(f*x+e)^5/a^3/f/(c^2-c^2*sin(f*x+e))^3+1/99*(8*A-3*B)*sec(f*x+e)^5/a^3/f/(c^3-c^3*sin(f*x+e))^2+
1/99*(8*A-3*B)*sec(f*x+e)^5/a^3/f/(c^6-c^6*sin(f*x+e))+2/33*(8*A-3*B)*tan(f*x+e)/a^3/c^6/f+4/99*(8*A-3*B)*tan(
f*x+e)^3/a^3/c^6/f+2/165*(8*A-3*B)*tan(f*x+e)^5/a^3/c^6/f

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Rubi [A]  time = 0.35, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2967, 2859, 2672, 3767} \[ \frac {2 (8 A-3 B) \tan ^5(e+f x)}{165 a^3 c^6 f}+\frac {4 (8 A-3 B) \tan ^3(e+f x)}{99 a^3 c^6 f}+\frac {2 (8 A-3 B) \tan (e+f x)}{33 a^3 c^6 f}+\frac {(8 A-3 B) \sec ^5(e+f x)}{99 a^3 f \left (c^6-c^6 \sin (e+f x)\right )}+\frac {(8 A-3 B) \sec ^5(e+f x)}{99 a^3 f \left (c^3-c^3 \sin (e+f x)\right )^2}+\frac {(A+B) \sec ^5(e+f x)}{11 a^3 f \left (c^2-c^2 \sin (e+f x)\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^6),x]

[Out]

((A + B)*Sec[e + f*x]^5)/(11*a^3*f*(c^2 - c^2*Sin[e + f*x])^3) + ((8*A - 3*B)*Sec[e + f*x]^5)/(99*a^3*f*(c^3 -
c^3*Sin[e + f*x])^2) + ((8*A - 3*B)*Sec[e + f*x]^5)/(99*a^3*f*(c^6 - c^6*Sin[e + f*x])) + (2*(8*A - 3*B)*Tan[
e + f*x])/(33*a^3*c^6*f) + (4*(8*A - 3*B)*Tan[e + f*x]^3)/(99*a^3*c^6*f) + (2*(8*A - 3*B)*Tan[e + f*x]^5)/(165
*a^3*c^6*f)

Rule 2672

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*Simplify[2*m + p + 1]), x] + Dist[Simplify[m + p + 1]/(a*
Simplify[2*m + p + 1]), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g, m
, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + p + 1], 0] && NeQ[2*m + p + 1, 0] &&  !IGtQ[m, 0]

Rule 2859

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

Rule 2967

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I
ntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin {align*} \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^6} \, dx &=\frac {\int \frac {\sec ^6(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx}{a^3 c^3}\\ &=\frac {(A+B) \sec ^5(e+f x)}{11 a^3 f \left (c^2-c^2 \sin (e+f x)\right )^3}+\frac {(8 A-3 B) \int \frac {\sec ^6(e+f x)}{(c-c \sin (e+f x))^2} \, dx}{11 a^3 c^4}\\ &=\frac {(A+B) \sec ^5(e+f x)}{11 a^3 f \left (c^2-c^2 \sin (e+f x)\right )^3}+\frac {(8 A-3 B) \sec ^5(e+f x)}{99 a^3 f \left (c^3-c^3 \sin (e+f x)\right )^2}+\frac {(7 (8 A-3 B)) \int \frac {\sec ^6(e+f x)}{c-c \sin (e+f x)} \, dx}{99 a^3 c^5}\\ &=\frac {(A+B) \sec ^5(e+f x)}{11 a^3 f \left (c^2-c^2 \sin (e+f x)\right )^3}+\frac {(8 A-3 B) \sec ^5(e+f x)}{99 a^3 f \left (c^3-c^3 \sin (e+f x)\right )^2}+\frac {(8 A-3 B) \sec ^5(e+f x)}{99 a^3 f \left (c^6-c^6 \sin (e+f x)\right )}+\frac {(2 (8 A-3 B)) \int \sec ^6(e+f x) \, dx}{33 a^3 c^6}\\ &=\frac {(A+B) \sec ^5(e+f x)}{11 a^3 f \left (c^2-c^2 \sin (e+f x)\right )^3}+\frac {(8 A-3 B) \sec ^5(e+f x)}{99 a^3 f \left (c^3-c^3 \sin (e+f x)\right )^2}+\frac {(8 A-3 B) \sec ^5(e+f x)}{99 a^3 f \left (c^6-c^6 \sin (e+f x)\right )}-\frac {(2 (8 A-3 B)) \operatorname {Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (e+f x)\right )}{33 a^3 c^6 f}\\ &=\frac {(A+B) \sec ^5(e+f x)}{11 a^3 f \left (c^2-c^2 \sin (e+f x)\right )^3}+\frac {(8 A-3 B) \sec ^5(e+f x)}{99 a^3 f \left (c^3-c^3 \sin (e+f x)\right )^2}+\frac {(8 A-3 B) \sec ^5(e+f x)}{99 a^3 f \left (c^6-c^6 \sin (e+f x)\right )}+\frac {2 (8 A-3 B) \tan (e+f x)}{33 a^3 c^6 f}+\frac {4 (8 A-3 B) \tan ^3(e+f x)}{99 a^3 c^6 f}+\frac {2 (8 A-3 B) \tan ^5(e+f x)}{165 a^3 c^6 f}\\ \end {align*}

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Mathematica [A]  time = 3.55, size = 401, normalized size = 1.96 \[ \frac {-3850 (107 A-3 B) \cos (e+f x)+135168 (8 A-3 B) \cos (2 (e+f x))+1802240 A \sin (e+f x)+247170 A \sin (2 (e+f x))+557056 A \sin (3 (e+f x))+187250 A \sin (4 (e+f x))-163840 A \sin (5 (e+f x))+37450 A \sin (6 (e+f x))-98304 A \sin (7 (e+f x))-3745 A \sin (8 (e+f x))-127330 A \cos (3 (e+f x))+819200 A \cos (4 (e+f x))+37450 A \cos (5 (e+f x))+163840 A \cos (6 (e+f x))+22470 A \cos (7 (e+f x))-16384 A \cos (8 (e+f x))-675840 B \sin (e+f x)-6930 B \sin (2 (e+f x))-208896 B \sin (3 (e+f x))-5250 B \sin (4 (e+f x))+61440 B \sin (5 (e+f x))-1050 B \sin (6 (e+f x))+36864 B \sin (7 (e+f x))+105 B \sin (8 (e+f x))+3570 B \cos (3 (e+f x))-307200 B \cos (4 (e+f x))-1050 B \cos (5 (e+f x))-61440 B \cos (6 (e+f x))-630 B \cos (7 (e+f x))+6144 B \cos (8 (e+f x))+1013760 B}{8110080 a^3 c^6 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^{11} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^6),x]

[Out]

(1013760*B - 3850*(107*A - 3*B)*Cos[e + f*x] + 135168*(8*A - 3*B)*Cos[2*(e + f*x)] - 127330*A*Cos[3*(e + f*x)]
+ 3570*B*Cos[3*(e + f*x)] + 819200*A*Cos[4*(e + f*x)] - 307200*B*Cos[4*(e + f*x)] + 37450*A*Cos[5*(e + f*x)]
- 1050*B*Cos[5*(e + f*x)] + 163840*A*Cos[6*(e + f*x)] - 61440*B*Cos[6*(e + f*x)] + 22470*A*Cos[7*(e + f*x)] -
630*B*Cos[7*(e + f*x)] - 16384*A*Cos[8*(e + f*x)] + 6144*B*Cos[8*(e + f*x)] + 1802240*A*Sin[e + f*x] - 675840*
B*Sin[e + f*x] + 247170*A*Sin[2*(e + f*x)] - 6930*B*Sin[2*(e + f*x)] + 557056*A*Sin[3*(e + f*x)] - 208896*B*Si
n[3*(e + f*x)] + 187250*A*Sin[4*(e + f*x)] - 5250*B*Sin[4*(e + f*x)] - 163840*A*Sin[5*(e + f*x)] + 61440*B*Sin
[5*(e + f*x)] + 37450*A*Sin[6*(e + f*x)] - 1050*B*Sin[6*(e + f*x)] - 98304*A*Sin[7*(e + f*x)] + 36864*B*Sin[7*
(e + f*x)] - 3745*A*Sin[8*(e + f*x)] + 105*B*Sin[8*(e + f*x)])/(8110080*a^3*c^6*f*(Cos[(e + f*x)/2] - Sin[(e +
f*x)/2])^11*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5)

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fricas [A]  time = 0.46, size = 221, normalized size = 1.08 \[ \frac {16 \, {\left (8 \, A - 3 \, B\right )} \cos \left (f x + e\right )^{8} - 72 \, {\left (8 \, A - 3 \, B\right )} \cos \left (f x + e\right )^{6} + 30 \, {\left (8 \, A - 3 \, B\right )} \cos \left (f x + e\right )^{4} + 7 \, {\left (8 \, A - 3 \, B\right )} \cos \left (f x + e\right )^{2} + {\left (48 \, {\left (8 \, A - 3 \, B\right )} \cos \left (f x + e\right )^{6} - 40 \, {\left (8 \, A - 3 \, B\right )} \cos \left (f x + e\right )^{4} - 14 \, {\left (8 \, A - 3 \, B\right )} \cos \left (f x + e\right )^{2} - 72 \, A + 27 \, B\right )} \sin \left (f x + e\right ) + 27 \, A - 72 \, B}{495 \, {\left (3 \, a^{3} c^{6} f \cos \left (f x + e\right )^{7} - 4 \, a^{3} c^{6} f \cos \left (f x + e\right )^{5} - {\left (a^{3} c^{6} f \cos \left (f x + e\right )^{7} - 4 \, a^{3} c^{6} f \cos \left (f x + e\right )^{5}\right )} \sin \left (f x + e\right )\right )}} \]

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-c*sin(f*x+e))^6,x, algorithm="fricas")

[Out]

1/495*(16*(8*A - 3*B)*cos(f*x + e)^8 - 72*(8*A - 3*B)*cos(f*x + e)^6 + 30*(8*A - 3*B)*cos(f*x + e)^4 + 7*(8*A
- 3*B)*cos(f*x + e)^2 + (48*(8*A - 3*B)*cos(f*x + e)^6 - 40*(8*A - 3*B)*cos(f*x + e)^4 - 14*(8*A - 3*B)*cos(f*
x + e)^2 - 72*A + 27*B)*sin(f*x + e) + 27*A - 72*B)/(3*a^3*c^6*f*cos(f*x + e)^7 - 4*a^3*c^6*f*cos(f*x + e)^5 -
(a^3*c^6*f*cos(f*x + e)^7 - 4*a^3*c^6*f*cos(f*x + e)^5)*sin(f*x + e))

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giac [B]  time = 0.30, size = 475, normalized size = 2.32 \[ -\frac {\frac {33 \, {\left (555 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 315 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 1920 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 1020 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2710 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1410 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1760 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 900 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 463 \, A - 243 \, B\right )}}{a^{3} c^{6} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}} + \frac {108405 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} + 10395 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} - 784080 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9} - 5940 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{9} + 2901195 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 79695 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 6652800 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} + 388080 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7} + 10407474 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 816354 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 11435424 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 1114344 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 8949270 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 990990 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 4899840 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 609840 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 1816265 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 235785 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 411664 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 56364 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 47279 \, A - 4179 \, B}{a^{3} c^{6} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{11}}}{63360 \, 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-c*sin(f*x+e))^6,x, algorithm="giac")

[Out]

-1/63360*(33*(555*A*tan(1/2*f*x + 1/2*e)^4 - 315*B*tan(1/2*f*x + 1/2*e)^4 + 1920*A*tan(1/2*f*x + 1/2*e)^3 - 10
20*B*tan(1/2*f*x + 1/2*e)^3 + 2710*A*tan(1/2*f*x + 1/2*e)^2 - 1410*B*tan(1/2*f*x + 1/2*e)^2 + 1760*A*tan(1/2*f
*x + 1/2*e) - 900*B*tan(1/2*f*x + 1/2*e) + 463*A - 243*B)/(a^3*c^6*(tan(1/2*f*x + 1/2*e) + 1)^5) + (108405*A*t
an(1/2*f*x + 1/2*e)^10 + 10395*B*tan(1/2*f*x + 1/2*e)^10 - 784080*A*tan(1/2*f*x + 1/2*e)^9 - 5940*B*tan(1/2*f*
x + 1/2*e)^9 + 2901195*A*tan(1/2*f*x + 1/2*e)^8 - 79695*B*tan(1/2*f*x + 1/2*e)^8 - 6652800*A*tan(1/2*f*x + 1/2
*e)^7 + 388080*B*tan(1/2*f*x + 1/2*e)^7 + 10407474*A*tan(1/2*f*x + 1/2*e)^6 - 816354*B*tan(1/2*f*x + 1/2*e)^6
- 11435424*A*tan(1/2*f*x + 1/2*e)^5 + 1114344*B*tan(1/2*f*x + 1/2*e)^5 + 8949270*A*tan(1/2*f*x + 1/2*e)^4 - 99
0990*B*tan(1/2*f*x + 1/2*e)^4 - 4899840*A*tan(1/2*f*x + 1/2*e)^3 + 609840*B*tan(1/2*f*x + 1/2*e)^3 + 1816265*A
*tan(1/2*f*x + 1/2*e)^2 - 235785*B*tan(1/2*f*x + 1/2*e)^2 - 411664*A*tan(1/2*f*x + 1/2*e) + 56364*B*tan(1/2*f*
x + 1/2*e) + 47279*A - 4179*B)/(a^3*c^6*(tan(1/2*f*x + 1/2*e) - 1)^11))/f

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maple [A]  time = 0.51, size = 365, normalized size = 1.78 \[ \frac {-\frac {2 \left (4 A +4 B \right )}{11 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{11}}-\frac {20 A +20 B}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{10}}-\frac {2 \left (53 A +51 B \right )}{9 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{9}}-\frac {92 A +84 B}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{8}}-\frac {\frac {169 A}{4}+\frac {99 B}{4}}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {\frac {217 A}{2}+84 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{6}}-\frac {2 \left (\frac {219 A}{256}+\frac {21 B}{256}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {2 \left (\frac {231 A}{2}+98 B \right )}{7 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}-\frac {\frac {303 A}{64}+\frac {99 B}{64}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {2 \left (\frac {623 A}{8}+\frac {427 B}{8}\right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {2 \left (\frac {1095 A}{64}+\frac {507 B}{64}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {-\frac {5 A}{32}+\frac {B}{8}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {-\frac {A}{8}+\frac {B}{8}}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 \left (\frac {A}{16}-\frac {B}{16}\right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {2 \left (\frac {7 A}{32}-\frac {3 B}{16}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (\frac {37 A}{256}-\frac {21 B}{256}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}}{f \,a^{3} c^{6}} \]

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-c*sin(f*x+e))^6,x)

[Out]

2/f/a^3/c^6*(-1/11*(4*A+4*B)/(tan(1/2*f*x+1/2*e)-1)^11-1/10*(20*A+20*B)/(tan(1/2*f*x+1/2*e)-1)^10-1/9*(53*A+51
*B)/(tan(1/2*f*x+1/2*e)-1)^9-1/8*(92*A+84*B)/(tan(1/2*f*x+1/2*e)-1)^8-1/4*(169/4*A+99/4*B)/(tan(1/2*f*x+1/2*e)
-1)^4-1/6*(217/2*A+84*B)/(tan(1/2*f*x+1/2*e)-1)^6-(219/256*A+21/256*B)/(tan(1/2*f*x+1/2*e)-1)-1/7*(231/2*A+98*
B)/(tan(1/2*f*x+1/2*e)-1)^7-1/2*(303/64*A+99/64*B)/(tan(1/2*f*x+1/2*e)-1)^2-1/5*(623/8*A+427/8*B)/(tan(1/2*f*x
+1/2*e)-1)^5-1/3*(1095/64*A+507/64*B)/(tan(1/2*f*x+1/2*e)-1)^3-1/2*(-5/32*A+1/8*B)/(tan(1/2*f*x+1/2*e)+1)^2-1/
4*(-1/8*A+1/8*B)/(tan(1/2*f*x+1/2*e)+1)^4-1/5*(1/16*A-1/16*B)/(tan(1/2*f*x+1/2*e)+1)^5-1/3*(7/32*A-3/16*B)/(ta
n(1/2*f*x+1/2*e)+1)^3-(37/256*A-21/256*B)/(tan(1/2*f*x+1/2*e)+1))

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maxima [B]  time = 0.58, size = 1387, normalized size = 6.77 \[ \text {result too large to display} \]

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-c*sin(f*x+e))^6,x, algorithm="maxima")

[Out]

-2/495*(A*(255*sin(f*x + e)/(cos(f*x + e) + 1) + 235*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 3065*sin(f*x + e)^3
/(cos(f*x + e) + 1)^3 + 3775*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 667*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 8
217*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 2035*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 8745*sin(f*x + e)^8/(cos(
f*x + e) + 1)^8 - 11715*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + 33*sin(f*x + e)^10/(cos(f*x + e) + 1)^10 + 4917*
sin(f*x + e)^11/(cos(f*x + e) + 1)^11 - 2475*sin(f*x + e)^12/(cos(f*x + e) + 1)^12 - 1815*sin(f*x + e)^13/(cos
(f*x + e) + 1)^13 + 1485*sin(f*x + e)^14/(cos(f*x + e) + 1)^14 - 495*sin(f*x + e)^15/(cos(f*x + e) + 1)^15 - 1
25)/(a^3*c^6 - 6*a^3*c^6*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*c^6*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10
*a^3*c^6*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 50*a^3*c^6*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 34*a^3*c^6*sin
(f*x + e)^5/(cos(f*x + e) + 1)^5 + 66*a^3*c^6*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 110*a^3*c^6*sin(f*x + e)^7
/(cos(f*x + e) + 1)^7 + 110*a^3*c^6*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 - 66*a^3*c^6*sin(f*x + e)^10/(cos(f*x
+ e) + 1)^10 - 34*a^3*c^6*sin(f*x + e)^11/(cos(f*x + e) + 1)^11 + 50*a^3*c^6*sin(f*x + e)^12/(cos(f*x + e) + 1
)^12 - 10*a^3*c^6*sin(f*x + e)^13/(cos(f*x + e) + 1)^13 - 10*a^3*c^6*sin(f*x + e)^14/(cos(f*x + e) + 1)^14 + 6
*a^3*c^6*sin(f*x + e)^15/(cos(f*x + e) + 1)^15 - a^3*c^6*sin(f*x + e)^16/(cos(f*x + e) + 1)^16) + 3*B*(30*sin(
f*x + e)/(cos(f*x + e) + 1) - 215*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 280*sin(f*x + e)^3/(cos(f*x + e) + 1)^
3 - 245*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 434*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 231*sin(f*x + e)^6/(co
s(f*x + e) + 1)^6 + 880*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 1815*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 330*s
in(f*x + e)^9/(cos(f*x + e) + 1)^9 + 99*sin(f*x + e)^10/(cos(f*x + e) + 1)^10 - 264*sin(f*x + e)^11/(cos(f*x +
e) + 1)^11 - 495*sin(f*x + e)^12/(cos(f*x + e) + 1)^12 + 330*sin(f*x + e)^13/(cos(f*x + e) + 1)^13 - 165*sin(
f*x + e)^14/(cos(f*x + e) + 1)^14 - 5)/(a^3*c^6 - 6*a^3*c^6*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*c^6*sin(f
*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*c^6*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 50*a^3*c^6*sin(f*x + e)^4/(c
os(f*x + e) + 1)^4 + 34*a^3*c^6*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 66*a^3*c^6*sin(f*x + e)^6/(cos(f*x + e)
+ 1)^6 - 110*a^3*c^6*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 110*a^3*c^6*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 - 6
6*a^3*c^6*sin(f*x + e)^10/(cos(f*x + e) + 1)^10 - 34*a^3*c^6*sin(f*x + e)^11/(cos(f*x + e) + 1)^11 + 50*a^3*c^
6*sin(f*x + e)^12/(cos(f*x + e) + 1)^12 - 10*a^3*c^6*sin(f*x + e)^13/(cos(f*x + e) + 1)^13 - 10*a^3*c^6*sin(f*
x + e)^14/(cos(f*x + e) + 1)^14 + 6*a^3*c^6*sin(f*x + e)^15/(cos(f*x + e) + 1)^15 - a^3*c^6*sin(f*x + e)^16/(c
os(f*x + e) + 1)^16))/f

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mupad [B]  time = 14.52, size = 474, normalized size = 2.31 \[ \frac {2\,\left (\frac {165\,B\,\sin \left (e+f\,x\right )}{4}-\frac {6875\,A\,\cos \left (e+f\,x\right )}{64}-\frac {825\,B\,\cos \left (e+f\,x\right )}{64}-110\,A\,\sin \left (e+f\,x\right )-\frac {495\,B}{8}-66\,A\,\cos \left (2\,e+2\,f\,x\right )-\frac {2125\,A\,\cos \left (3\,e+3\,f\,x\right )}{64}-50\,A\,\cos \left (4\,e+4\,f\,x\right )+\frac {625\,A\,\cos \left (5\,e+5\,f\,x\right )}{64}-10\,A\,\cos \left (6\,e+6\,f\,x\right )+\frac {375\,A\,\cos \left (7\,e+7\,f\,x\right )}{64}+A\,\cos \left (8\,e+8\,f\,x\right )+\frac {99\,B\,\cos \left (2\,e+2\,f\,x\right )}{4}-\frac {255\,B\,\cos \left (3\,e+3\,f\,x\right )}{64}+\frac {75\,B\,\cos \left (4\,e+4\,f\,x\right )}{4}+\frac {75\,B\,\cos \left (5\,e+5\,f\,x\right )}{64}+\frac {15\,B\,\cos \left (6\,e+6\,f\,x\right )}{4}+\frac {45\,B\,\cos \left (7\,e+7\,f\,x\right )}{64}-\frac {3\,B\,\cos \left (8\,e+8\,f\,x\right )}{8}+\frac {4125\,A\,\sin \left (2\,e+2\,f\,x\right )}{64}-34\,A\,\sin \left (3\,e+3\,f\,x\right )+\frac {3125\,A\,\sin \left (4\,e+4\,f\,x\right )}{64}+10\,A\,\sin \left (5\,e+5\,f\,x\right )+\frac {625\,A\,\sin \left (6\,e+6\,f\,x\right )}{64}+6\,A\,\sin \left (7\,e+7\,f\,x\right )-\frac {125\,A\,\sin \left (8\,e+8\,f\,x\right )}{128}+\frac {495\,B\,\sin \left (2\,e+2\,f\,x\right )}{64}+\frac {51\,B\,\sin \left (3\,e+3\,f\,x\right )}{4}+\frac {375\,B\,\sin \left (4\,e+4\,f\,x\right )}{64}-\frac {15\,B\,\sin \left (5\,e+5\,f\,x\right )}{4}+\frac {75\,B\,\sin \left (6\,e+6\,f\,x\right )}{64}-\frac {9\,B\,\sin \left (7\,e+7\,f\,x\right )}{4}-\frac {15\,B\,\sin \left (8\,e+8\,f\,x\right )}{128}\right )}{495\,a^3\,c^6\,f\,\left (\frac {5\,\cos \left (5\,e+5\,f\,x\right )}{32}-\frac {17\,\cos \left (3\,e+3\,f\,x\right )}{32}-\frac {55\,\cos \left (e+f\,x\right )}{32}+\frac {3\,\cos \left (7\,e+7\,f\,x\right )}{32}+\frac {33\,\sin \left (2\,e+2\,f\,x\right )}{32}+\frac {25\,\sin \left (4\,e+4\,f\,x\right )}{32}+\frac {5\,\sin \left (6\,e+6\,f\,x\right )}{32}-\frac {\sin \left (8\,e+8\,f\,x\right )}{64}\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 - c*sin(e + f*x))^6),x)

[Out]

(2*((165*B*sin(e + f*x))/4 - (6875*A*cos(e + f*x))/64 - (825*B*cos(e + f*x))/64 - 110*A*sin(e + f*x) - (495*B)
/8 - 66*A*cos(2*e + 2*f*x) - (2125*A*cos(3*e + 3*f*x))/64 - 50*A*cos(4*e + 4*f*x) + (625*A*cos(5*e + 5*f*x))/6
4 - 10*A*cos(6*e + 6*f*x) + (375*A*cos(7*e + 7*f*x))/64 + A*cos(8*e + 8*f*x) + (99*B*cos(2*e + 2*f*x))/4 - (25
5*B*cos(3*e + 3*f*x))/64 + (75*B*cos(4*e + 4*f*x))/4 + (75*B*cos(5*e + 5*f*x))/64 + (15*B*cos(6*e + 6*f*x))/4
+ (45*B*cos(7*e + 7*f*x))/64 - (3*B*cos(8*e + 8*f*x))/8 + (4125*A*sin(2*e + 2*f*x))/64 - 34*A*sin(3*e + 3*f*x)
+ (3125*A*sin(4*e + 4*f*x))/64 + 10*A*sin(5*e + 5*f*x) + (625*A*sin(6*e + 6*f*x))/64 + 6*A*sin(7*e + 7*f*x) -
(125*A*sin(8*e + 8*f*x))/128 + (495*B*sin(2*e + 2*f*x))/64 + (51*B*sin(3*e + 3*f*x))/4 + (375*B*sin(4*e + 4*f
*x))/64 - (15*B*sin(5*e + 5*f*x))/4 + (75*B*sin(6*e + 6*f*x))/64 - (9*B*sin(7*e + 7*f*x))/4 - (15*B*sin(8*e +
8*f*x))/128))/(495*a^3*c^6*f*((5*cos(5*e + 5*f*x))/32 - (17*cos(3*e + 3*f*x))/32 - (55*cos(e + f*x))/32 + (3*c
os(7*e + 7*f*x))/32 + (33*sin(2*e + 2*f*x))/32 + (25*sin(4*e + 4*f*x))/32 + (5*sin(6*e + 6*f*x))/32 - sin(8*e
+ 8*f*x)/64))

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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-c*sin(f*x+e))**6,x)

[Out]

Timed out

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