3.69 \(\int \frac{A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^5} \, dx\)

Optimal. Leaf size=175 \[ \frac{4 (2 A-B) \tan ^3(e+f x)}{63 a^2 c^5 f}+\frac{4 (2 A-B) \tan (e+f x)}{21 a^2 c^5 f}+\frac{(2 A-B) \sec ^3(e+f x)}{21 a^2 f \left (c^5-c^5 \sin (e+f x)\right )}+\frac{(2 A-B) \sec ^3(e+f x)}{21 a^2 c^3 f (c-c \sin (e+f x))^2}+\frac{(A+B) \sec ^3(e+f x)}{9 a^2 c^2 f (c-c \sin (e+f x))^3} \]

[Out]

((A + B)*Sec[e + f*x]^3)/(9*a^2*c^2*f*(c - c*Sin[e + f*x])^3) + ((2*A - B)*Sec[e + f*x]^3)/(21*a^2*c^3*f*(c -
c*Sin[e + f*x])^2) + ((2*A - B)*Sec[e + f*x]^3)/(21*a^2*f*(c^5 - c^5*Sin[e + f*x])) + (4*(2*A - B)*Tan[e + f*x
])/(21*a^2*c^5*f) + (4*(2*A - B)*Tan[e + f*x]^3)/(63*a^2*c^5*f)

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Rubi [A]  time = 0.324576, antiderivative size = 175, normalized size of antiderivative = 1., 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} \[ \frac{4 (2 A-B) \tan ^3(e+f x)}{63 a^2 c^5 f}+\frac{4 (2 A-B) \tan (e+f x)}{21 a^2 c^5 f}+\frac{(2 A-B) \sec ^3(e+f x)}{21 a^2 f \left (c^5-c^5 \sin (e+f x)\right )}+\frac{(2 A-B) \sec ^3(e+f x)}{21 a^2 c^3 f (c-c \sin (e+f x))^2}+\frac{(A+B) \sec ^3(e+f x)}{9 a^2 c^2 f (c-c \sin (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^5),x]

[Out]

((A + B)*Sec[e + f*x]^3)/(9*a^2*c^2*f*(c - c*Sin[e + f*x])^3) + ((2*A - B)*Sec[e + f*x]^3)/(21*a^2*c^3*f*(c -
c*Sin[e + f*x])^2) + ((2*A - B)*Sec[e + f*x]^3)/(21*a^2*f*(c^5 - c^5*Sin[e + f*x])) + (4*(2*A - B)*Tan[e + f*x
])/(21*a^2*c^5*f) + (4*(2*A - B)*Tan[e + f*x]^3)/(63*a^2*c^5*f)

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 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 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 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))^2 (c-c \sin (e+f x))^5} \, dx &=\frac{\int \frac{\sec ^4(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx}{a^2 c^2}\\ &=\frac{(A+B) \sec ^3(e+f x)}{9 a^2 c^2 f (c-c \sin (e+f x))^3}+\frac{(2 A-B) \int \frac{\sec ^4(e+f x)}{(c-c \sin (e+f x))^2} \, dx}{3 a^2 c^3}\\ &=\frac{(A+B) \sec ^3(e+f x)}{9 a^2 c^2 f (c-c \sin (e+f x))^3}+\frac{(2 A-B) \sec ^3(e+f x)}{21 a^2 c^3 f (c-c \sin (e+f x))^2}+\frac{(5 (2 A-B)) \int \frac{\sec ^4(e+f x)}{c-c \sin (e+f x)} \, dx}{21 a^2 c^4}\\ &=\frac{(A+B) \sec ^3(e+f x)}{9 a^2 c^2 f (c-c \sin (e+f x))^3}+\frac{(2 A-B) \sec ^3(e+f x)}{21 a^2 c^3 f (c-c \sin (e+f x))^2}+\frac{(2 A-B) \sec ^3(e+f x)}{21 a^2 f \left (c^5-c^5 \sin (e+f x)\right )}+\frac{(4 (2 A-B)) \int \sec ^4(e+f x) \, dx}{21 a^2 c^5}\\ &=\frac{(A+B) \sec ^3(e+f x)}{9 a^2 c^2 f (c-c \sin (e+f x))^3}+\frac{(2 A-B) \sec ^3(e+f x)}{21 a^2 c^3 f (c-c \sin (e+f x))^2}+\frac{(2 A-B) \sec ^3(e+f x)}{21 a^2 f \left (c^5-c^5 \sin (e+f x)\right )}-\frac{(4 (2 A-B)) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (e+f x)\right )}{21 a^2 c^5 f}\\ &=\frac{(A+B) \sec ^3(e+f x)}{9 a^2 c^2 f (c-c \sin (e+f x))^3}+\frac{(2 A-B) \sec ^3(e+f x)}{21 a^2 c^3 f (c-c \sin (e+f x))^2}+\frac{(2 A-B) \sec ^3(e+f x)}{21 a^2 f \left (c^5-c^5 \sin (e+f x)\right )}+\frac{4 (2 A-B) \tan (e+f x)}{21 a^2 c^5 f}+\frac{4 (2 A-B) \tan ^3(e+f x)}{63 a^2 c^5 f}\\ \end{align*}

Mathematica [A]  time = 1.10471, size = 329, normalized size = 1.88 \[ \frac{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (180 (31 A-5 B) \cos (e+f x)-6912 (2 A-B) \cos (2 (e+f x))-18432 A \sin (e+f x)-4185 A \sin (2 (e+f x))-1024 A \sin (3 (e+f x))-1860 A \sin (4 (e+f x))+3072 A \sin (5 (e+f x))+155 A \sin (6 (e+f x))+310 A \cos (3 (e+f x))-6144 A \cos (4 (e+f x))-930 A \cos (5 (e+f x))+512 A \cos (6 (e+f x))+9216 B \sin (e+f x)+675 B \sin (2 (e+f x))+512 B \sin (3 (e+f x))+300 B \sin (4 (e+f x))-1536 B \sin (5 (e+f x))-25 B \sin (6 (e+f x))-50 B \cos (3 (e+f x))+3072 B \cos (4 (e+f x))+150 B \cos (5 (e+f x))-256 B \cos (6 (e+f x))-10752 B)}{64512 a^2 c^5 f (\sin (e+f x)-1)^5 (\sin (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^5),x]

[Out]

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-10752*B + 180*(31*A - 5*B)*Cos[
e + f*x] - 6912*(2*A - B)*Cos[2*(e + f*x)] + 310*A*Cos[3*(e + f*x)] - 50*B*Cos[3*(e + f*x)] - 6144*A*Cos[4*(e
+ f*x)] + 3072*B*Cos[4*(e + f*x)] - 930*A*Cos[5*(e + f*x)] + 150*B*Cos[5*(e + f*x)] + 512*A*Cos[6*(e + f*x)] -
256*B*Cos[6*(e + f*x)] - 18432*A*Sin[e + f*x] + 9216*B*Sin[e + f*x] - 4185*A*Sin[2*(e + f*x)] + 675*B*Sin[2*(
e + f*x)] - 1024*A*Sin[3*(e + f*x)] + 512*B*Sin[3*(e + f*x)] - 1860*A*Sin[4*(e + f*x)] + 300*B*Sin[4*(e + f*x)
] + 3072*A*Sin[5*(e + f*x)] - 1536*B*Sin[5*(e + f*x)] + 155*A*Sin[6*(e + f*x)] - 25*B*Sin[6*(e + f*x)]))/(6451
2*a^2*c^5*f*(-1 + Sin[e + f*x])^5*(1 + Sin[e + f*x])^2)

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Maple [A]  time = 0.118, size = 277, normalized size = 1.6 \begin{align*} 2\,{\frac{1}{{a}^{2}f{c}^{5}} \left ( -1/9\,{\frac{4\,A+4\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{9}}}-1/8\,{\frac{16\,A+16\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{8}}}-1/7\,{\frac{34\,A+32\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{7}}}-1/6\,{\frac{46\,A+40\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{6}}}-1/2\,{\frac{1}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{2}} \left ( 9/2\,A+{\frac{13\,B}{8}} \right ) }-{\frac{1}{\tan \left ( 1/2\,fx+e/2 \right ) -1} \left ({\frac{57\,A}{64}}+{\frac{5\,B}{64}} \right ) }-1/4\,{\frac{1}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{4}} \left ({\frac{59\,A}{2}}+{\frac{39\,B}{2}} \right ) }-1/3\,{\frac{1}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{3}} \left ({\frac{57\,A}{4}}+{\frac{59\,B}{8}} \right ) }-1/5\,{\frac{1}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{5}} \left ({\frac{175\,A}{4}}+{\frac{135\,B}{4}} \right ) }-1/2\,{\frac{-A/16+B/16}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}-1/3\,{\frac{A/16-B/16}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{3}}}-{\frac{1}{\tan \left ( 1/2\,fx+e/2 \right ) +1} \left ({\frac{7\,A}{64}}-{\frac{5\,B}{64}} \right ) } \right ) } \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-c*sin(f*x+e))^5,x)

[Out]

2/f/a^2/c^5*(-1/9*(4*A+4*B)/(tan(1/2*f*x+1/2*e)-1)^9-1/8*(16*A+16*B)/(tan(1/2*f*x+1/2*e)-1)^8-1/7*(34*A+32*B)/
(tan(1/2*f*x+1/2*e)-1)^7-1/6*(46*A+40*B)/(tan(1/2*f*x+1/2*e)-1)^6-1/2*(9/2*A+13/8*B)/(tan(1/2*f*x+1/2*e)-1)^2-
(57/64*A+5/64*B)/(tan(1/2*f*x+1/2*e)-1)-1/4*(59/2*A+39/2*B)/(tan(1/2*f*x+1/2*e)-1)^4-1/3*(57/4*A+59/8*B)/(tan(
1/2*f*x+1/2*e)-1)^3-1/5*(175/4*A+135/4*B)/(tan(1/2*f*x+1/2*e)-1)^5-1/2*(-1/16*A+1/16*B)/(tan(1/2*f*x+1/2*e)+1)
^2-1/3*(1/16*A-1/16*B)/(tan(1/2*f*x+1/2*e)+1)^3-(7/64*A-5/64*B)/(tan(1/2*f*x+1/2*e)+1))

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Maxima [B]  time = 1.14971, size = 1347, normalized size = 7.7 \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-c*sin(f*x+e))^5,x, algorithm="maxima")

[Out]

-2/63*(A*(51*sin(f*x + e)/(cos(f*x + e) + 1) - 39*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 235*sin(f*x + e)^3/(co
s(f*x + e) + 1)^3 + 450*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 306*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 294*si
n(f*x + e)^6/(cos(f*x + e) + 1)^6 + 378*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 63*sin(f*x + e)^8/(cos(f*x + e)
+ 1)^8 - 273*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + 189*sin(f*x + e)^10/(cos(f*x + e) + 1)^10 - 63*sin(f*x + e)
^11/(cos(f*x + e) + 1)^11 - 19)/(a^2*c^5 - 6*a^2*c^5*sin(f*x + e)/(cos(f*x + e) + 1) + 12*a^2*c^5*sin(f*x + e)
^2/(cos(f*x + e) + 1)^2 - 2*a^2*c^5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 27*a^2*c^5*sin(f*x + e)^4/(cos(f*x +
e) + 1)^4 + 36*a^2*c^5*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 36*a^2*c^5*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 +
27*a^2*c^5*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 2*a^2*c^5*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 - 12*a^2*c^5*s
in(f*x + e)^10/(cos(f*x + e) + 1)^10 + 6*a^2*c^5*sin(f*x + e)^11/(cos(f*x + e) + 1)^11 - a^2*c^5*sin(f*x + e)^
12/(cos(f*x + e) + 1)^12) + B*(6*sin(f*x + e)/(cos(f*x + e) + 1) - 75*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 12
8*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 162*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 36*sin(f*x + e)^5/(cos(f*x +
e) + 1)^5 + 42*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 189*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 126*sin(f*x +
e)^9/(cos(f*x + e) + 1)^9 - 63*sin(f*x + e)^10/(cos(f*x + e) + 1)^10 - 1)/(a^2*c^5 - 6*a^2*c^5*sin(f*x + e)/(c
os(f*x + e) + 1) + 12*a^2*c^5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 2*a^2*c^5*sin(f*x + e)^3/(cos(f*x + e) + 1
)^3 - 27*a^2*c^5*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 36*a^2*c^5*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 36*a^2
*c^5*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 27*a^2*c^5*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 2*a^2*c^5*sin(f*x
+ e)^9/(cos(f*x + e) + 1)^9 - 12*a^2*c^5*sin(f*x + e)^10/(cos(f*x + e) + 1)^10 + 6*a^2*c^5*sin(f*x + e)^11/(co
s(f*x + e) + 1)^11 - a^2*c^5*sin(f*x + e)^12/(cos(f*x + e) + 1)^12))/f

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Fricas [A]  time = 1.77727, size = 439, normalized size = 2.51 \begin{align*} \frac{8 \,{\left (2 \, A - B\right )} \cos \left (f x + e\right )^{6} - 36 \,{\left (2 \, A - B\right )} \cos \left (f x + e\right )^{4} + 15 \,{\left (2 \, A - B\right )} \cos \left (f x + e\right )^{2} +{\left (24 \,{\left (2 \, A - B\right )} \cos \left (f x + e\right )^{4} - 20 \,{\left (2 \, A - B\right )} \cos \left (f x + e\right )^{2} - 14 \, A + 7 \, B\right )} \sin \left (f x + e\right ) + 7 \, A - 14 \, B}{63 \,{\left (3 \, a^{2} c^{5} f \cos \left (f x + e\right )^{5} - 4 \, a^{2} c^{5} f \cos \left (f x + e\right )^{3} -{\left (a^{2} c^{5} f \cos \left (f x + e\right )^{5} - 4 \, a^{2} c^{5} f \cos \left (f x + e\right )^{3}\right )} \sin \left (f x + e\right )\right )}} \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-c*sin(f*x+e))^5,x, algorithm="fricas")

[Out]

1/63*(8*(2*A - B)*cos(f*x + e)^6 - 36*(2*A - B)*cos(f*x + e)^4 + 15*(2*A - B)*cos(f*x + e)^2 + (24*(2*A - B)*c
os(f*x + e)^4 - 20*(2*A - B)*cos(f*x + e)^2 - 14*A + 7*B)*sin(f*x + e) + 7*A - 14*B)/(3*a^2*c^5*f*cos(f*x + e)
^5 - 4*a^2*c^5*f*cos(f*x + e)^3 - (a^2*c^5*f*cos(f*x + e)^5 - 4*a^2*c^5*f*cos(f*x + e)^3)*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-c*sin(f*x+e))**5,x)

[Out]

Timed out

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Giac [B]  time = 1.27133, size = 479, normalized size = 2.74 \begin{align*} -\frac{\frac{21 \,{\left (21 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 15 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 36 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 24 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 19 \, A - 13 \, B\right )}}{a^{2} c^{5}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{3}} + \frac{3591 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{8} + 315 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{8} - 19656 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} + 756 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} + 56196 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 4200 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 95760 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 11340 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 107730 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 14994 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 79464 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 13356 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 38484 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 6768 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 10944 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 2196 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1615 \, A - 209 \, B}{a^{2} c^{5}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}^{9}}}{2016 \, 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-c*sin(f*x+e))^5,x, algorithm="giac")

[Out]

-1/2016*(21*(21*A*tan(1/2*f*x + 1/2*e)^2 - 15*B*tan(1/2*f*x + 1/2*e)^2 + 36*A*tan(1/2*f*x + 1/2*e) - 24*B*tan(
1/2*f*x + 1/2*e) + 19*A - 13*B)/(a^2*c^5*(tan(1/2*f*x + 1/2*e) + 1)^3) + (3591*A*tan(1/2*f*x + 1/2*e)^8 + 315*
B*tan(1/2*f*x + 1/2*e)^8 - 19656*A*tan(1/2*f*x + 1/2*e)^7 + 756*B*tan(1/2*f*x + 1/2*e)^7 + 56196*A*tan(1/2*f*x
+ 1/2*e)^6 - 4200*B*tan(1/2*f*x + 1/2*e)^6 - 95760*A*tan(1/2*f*x + 1/2*e)^5 + 11340*B*tan(1/2*f*x + 1/2*e)^5
+ 107730*A*tan(1/2*f*x + 1/2*e)^4 - 14994*B*tan(1/2*f*x + 1/2*e)^4 - 79464*A*tan(1/2*f*x + 1/2*e)^3 + 13356*B*
tan(1/2*f*x + 1/2*e)^3 + 38484*A*tan(1/2*f*x + 1/2*e)^2 - 6768*B*tan(1/2*f*x + 1/2*e)^2 - 10944*A*tan(1/2*f*x
+ 1/2*e) + 2196*B*tan(1/2*f*x + 1/2*e) + 1615*A - 209*B)/(a^2*c^5*(tan(1/2*f*x + 1/2*e) - 1)^9))/f