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

Optimal. Leaf size=176 \[ -\frac{2 a (A-2 B) \cos (e+f x)}{315 f \left (c^5-c^5 \sin (e+f x)\right )}-\frac{2 a c (A-2 B) \cos (e+f x)}{315 f \left (c^3-c^3 \sin (e+f x)\right )^2}-\frac{a c (A-2 B) \cos (e+f x)}{105 f \left (c^2-c^2 \sin (e+f x)\right )^3}-\frac{a (A+19 B) \cos (e+f x)}{63 c f (c-c \sin (e+f x))^4}+\frac{2 a (A+B) \cos (e+f x)}{9 f (c-c \sin (e+f x))^5} \]

[Out]

(2*a*(A + B)*Cos[e + f*x])/(9*f*(c - c*Sin[e + f*x])^5) - (a*(A + 19*B)*Cos[e + f*x])/(63*c*f*(c - c*Sin[e + f
*x])^4) - (a*(A - 2*B)*c*Cos[e + f*x])/(105*f*(c^2 - c^2*Sin[e + f*x])^3) - (2*a*(A - 2*B)*c*Cos[e + f*x])/(31
5*f*(c^3 - c^3*Sin[e + f*x])^2) - (2*a*(A - 2*B)*Cos[e + f*x])/(315*f*(c^5 - c^5*Sin[e + f*x]))

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Rubi [A]  time = 0.307073, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {2967, 2857, 2750, 2650, 2648} \[ -\frac{2 a (A-2 B) \cos (e+f x)}{315 f \left (c^5-c^5 \sin (e+f x)\right )}-\frac{2 a c (A-2 B) \cos (e+f x)}{315 f \left (c^3-c^3 \sin (e+f x)\right )^2}-\frac{a c (A-2 B) \cos (e+f x)}{105 f \left (c^2-c^2 \sin (e+f x)\right )^3}-\frac{a (A+19 B) \cos (e+f x)}{63 c f (c-c \sin (e+f x))^4}+\frac{2 a (A+B) \cos (e+f x)}{9 f (c-c \sin (e+f x))^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(A + B)*Cos[e + f*x])/(9*f*(c - c*Sin[e + f*x])^5) - (a*(A + 19*B)*Cos[e + f*x])/(63*c*f*(c - c*Sin[e + f
*x])^4) - (a*(A - 2*B)*c*Cos[e + f*x])/(105*f*(c^2 - c^2*Sin[e + f*x])^3) - (2*a*(A - 2*B)*c*Cos[e + f*x])/(31
5*f*(c^3 - c^3*Sin[e + f*x])^2) - (2*a*(A - 2*B)*Cos[e + f*x])/(315*f*(c^5 - c^5*Sin[e + f*x]))

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 2857

Int[cos[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_
)]), x_Symbol] :> Simp[(2*(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(2*m + 3)), x] + Dist[
1/(b^3*(2*m + 3)), Int[(a + b*Sin[e + f*x])^(m + 2)*(b*c + 2*a*d*(m + 1) - b*d*(2*m + 3)*Sin[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -3/2]

Rule 2750

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)/(a*f*(2*m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)
), Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 -
b^2, 0] && LtQ[m, -2^(-1)]

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^5} \, dx &=(a c) \int \frac{\cos ^2(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^6} \, dx\\ &=\frac{2 a (A+B) \cos (e+f x)}{9 f (c-c \sin (e+f x))^5}+\frac{a \int \frac{-A c-10 B c-9 B c \sin (e+f x)}{(c-c \sin (e+f x))^4} \, dx}{9 c^2}\\ &=\frac{2 a (A+B) \cos (e+f x)}{9 f (c-c \sin (e+f x))^5}-\frac{a (A+19 B) \cos (e+f x)}{63 c f (c-c \sin (e+f x))^4}-\frac{(a (A-2 B)) \int \frac{1}{(c-c \sin (e+f x))^3} \, dx}{21 c^2}\\ &=\frac{2 a (A+B) \cos (e+f x)}{9 f (c-c \sin (e+f x))^5}-\frac{a (A+19 B) \cos (e+f x)}{63 c f (c-c \sin (e+f x))^4}-\frac{a (A-2 B) \cos (e+f x)}{105 c^2 f (c-c \sin (e+f x))^3}-\frac{(2 a (A-2 B)) \int \frac{1}{(c-c \sin (e+f x))^2} \, dx}{105 c^3}\\ &=\frac{2 a (A+B) \cos (e+f x)}{9 f (c-c \sin (e+f x))^5}-\frac{a (A+19 B) \cos (e+f x)}{63 c f (c-c \sin (e+f x))^4}-\frac{a (A-2 B) \cos (e+f x)}{105 c^2 f (c-c \sin (e+f x))^3}-\frac{2 a (A-2 B) \cos (e+f x)}{315 c^3 f (c-c \sin (e+f x))^2}-\frac{(2 a (A-2 B)) \int \frac{1}{c-c \sin (e+f x)} \, dx}{315 c^4}\\ &=\frac{2 a (A+B) \cos (e+f x)}{9 f (c-c \sin (e+f x))^5}-\frac{a (A+19 B) \cos (e+f x)}{63 c f (c-c \sin (e+f x))^4}-\frac{a (A-2 B) \cos (e+f x)}{105 c^2 f (c-c \sin (e+f x))^3}-\frac{2 a (A-2 B) \cos (e+f x)}{315 c^3 f (c-c \sin (e+f x))^2}-\frac{2 a (A-2 B) \cos (e+f x)}{315 f \left (c^5-c^5 \sin (e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.824741, size = 200, normalized size = 1.14 \[ \frac{a \left (-42 (2 A+B) \cos \left (e+\frac{3 f x}{2}\right )+36 A \sin \left (2 e+\frac{5 f x}{2}\right )-A \sin \left (4 e+\frac{9 f x}{2}\right )+315 A \cos \left (e+\frac{f x}{2}\right )+9 A \cos \left (3 e+\frac{7 f x}{2}\right )+189 A \sin \left (\frac{f x}{2}\right )+210 B \sin \left (2 e+\frac{3 f x}{2}\right )-72 B \sin \left (2 e+\frac{5 f x}{2}\right )+2 B \sin \left (4 e+\frac{9 f x}{2}\right )-18 B \cos \left (3 e+\frac{7 f x}{2}\right )+252 B \sin \left (\frac{f x}{2}\right )\right )}{1260 c^5 f \left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^9} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(315*A*Cos[e + (f*x)/2] - 42*(2*A + B)*Cos[e + (3*f*x)/2] + 9*A*Cos[3*e + (7*f*x)/2] - 18*B*Cos[3*e + (7*f*
x)/2] + 189*A*Sin[(f*x)/2] + 252*B*Sin[(f*x)/2] + 210*B*Sin[2*e + (3*f*x)/2] + 36*A*Sin[2*e + (5*f*x)/2] - 72*
B*Sin[2*e + (5*f*x)/2] - A*Sin[4*e + (9*f*x)/2] + 2*B*Sin[4*e + (9*f*x)/2]))/(1260*c^5*f*(Cos[e/2] - Sin[e/2])
*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^9)

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Maple [A]  time = 0.136, size = 203, normalized size = 1.2 \begin{align*} 2\,{\frac{a}{f{c}^{5}} \left ( -1/8\,{\frac{128\,A+128\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{8}}}-1/2\,{\frac{10\,A+2\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{2}}}-1/5\,{\frac{236\,A+168\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{5}}}-1/3\,{\frac{46\,A+18\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{3}}}-1/6\,{\frac{296\,A+248\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{6}}}-1/9\,{\frac{32\,A+32\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{9}}}-1/7\,{\frac{248\,A+232\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{7}}}-1/4\,{\frac{128\,A+72\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{4}}}-{\frac{A}{\tan \left ( 1/2\,fx+e/2 \right ) -1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^5,x)

[Out]

2/f*a/c^5*(-1/8*(128*A+128*B)/(tan(1/2*f*x+1/2*e)-1)^8-1/2*(10*A+2*B)/(tan(1/2*f*x+1/2*e)-1)^2-1/5*(236*A+168*
B)/(tan(1/2*f*x+1/2*e)-1)^5-1/3*(46*A+18*B)/(tan(1/2*f*x+1/2*e)-1)^3-1/6*(296*A+248*B)/(tan(1/2*f*x+1/2*e)-1)^
6-1/9*(32*A+32*B)/(tan(1/2*f*x+1/2*e)-1)^9-1/7*(248*A+232*B)/(tan(1/2*f*x+1/2*e)-1)^7-1/4*(128*A+72*B)/(tan(1/
2*f*x+1/2*e)-1)^4-A/(tan(1/2*f*x+1/2*e)-1))

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Maxima [B]  time = 1.155, size = 1924, normalized size = 10.93 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^5,x, algorithm="maxima")

[Out]

-2/315*(A*a*(432*sin(f*x + e)/(cos(f*x + e) + 1) - 1728*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 3612*sin(f*x + e
)^3/(cos(f*x + e) + 1)^3 - 5418*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 5040*sin(f*x + e)^5/(cos(f*x + e) + 1)^5
- 3360*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 1260*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 315*sin(f*x + e)^8/(c
os(f*x + e) + 1)^8 - 83)/(c^5 - 9*c^5*sin(f*x + e)/(cos(f*x + e) + 1) + 36*c^5*sin(f*x + e)^2/(cos(f*x + e) +
1)^2 - 84*c^5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 126*c^5*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 126*c^5*sin(
f*x + e)^5/(cos(f*x + e) + 1)^5 + 84*c^5*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 36*c^5*sin(f*x + e)^7/(cos(f*x
+ e) + 1)^7 + 9*c^5*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 - c^5*sin(f*x + e)^9/(cos(f*x + e) + 1)^9) - 5*A*a*(45
*sin(f*x + e)/(cos(f*x + e) + 1) - 117*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 273*sin(f*x + e)^3/(cos(f*x + e)
+ 1)^3 - 315*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 315*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 147*sin(f*x + e)^
6/(cos(f*x + e) + 1)^6 + 63*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 5)/(c^5 - 9*c^5*sin(f*x + e)/(cos(f*x + e) +
1) + 36*c^5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 84*c^5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 126*c^5*sin(f*
x + e)^4/(cos(f*x + e) + 1)^4 - 126*c^5*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 84*c^5*sin(f*x + e)^6/(cos(f*x +
e) + 1)^6 - 36*c^5*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 9*c^5*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 - c^5*sin(
f*x + e)^9/(cos(f*x + e) + 1)^9) - 5*B*a*(45*sin(f*x + e)/(cos(f*x + e) + 1) - 117*sin(f*x + e)^2/(cos(f*x + e
) + 1)^2 + 273*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 315*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 315*sin(f*x + e
)^5/(cos(f*x + e) + 1)^5 - 147*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 63*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 -
5)/(c^5 - 9*c^5*sin(f*x + e)/(cos(f*x + e) + 1) + 36*c^5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 84*c^5*sin(f*x
+ e)^3/(cos(f*x + e) + 1)^3 + 126*c^5*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 126*c^5*sin(f*x + e)^5/(cos(f*x +
e) + 1)^5 + 84*c^5*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 36*c^5*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 9*c^5*si
n(f*x + e)^8/(cos(f*x + e) + 1)^8 - c^5*sin(f*x + e)^9/(cos(f*x + e) + 1)^9) + 14*B*a*(9*sin(f*x + e)/(cos(f*x
+ e) + 1) - 36*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 54*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 81*sin(f*x + e)
^4/(cos(f*x + e) + 1)^4 + 45*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 30*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 1)
/(c^5 - 9*c^5*sin(f*x + e)/(cos(f*x + e) + 1) + 36*c^5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 84*c^5*sin(f*x +
e)^3/(cos(f*x + e) + 1)^3 + 126*c^5*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 126*c^5*sin(f*x + e)^5/(cos(f*x + e)
+ 1)^5 + 84*c^5*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 36*c^5*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 9*c^5*sin(
f*x + e)^8/(cos(f*x + e) + 1)^8 - c^5*sin(f*x + e)^9/(cos(f*x + e) + 1)^9))/f

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Fricas [A]  time = 1.4781, size = 794, normalized size = 4.51 \begin{align*} -\frac{2 \,{\left (A - 2 \, B\right )} a \cos \left (f x + e\right )^{5} - 8 \,{\left (A - 2 \, B\right )} a \cos \left (f x + e\right )^{4} - 25 \,{\left (A - 2 \, B\right )} a \cos \left (f x + e\right )^{3} + 5 \,{\left (4 \, A + 13 \, B\right )} a \cos \left (f x + e\right )^{2} - 35 \,{\left (A + B\right )} a \cos \left (f x + e\right ) - 70 \,{\left (A + B\right )} a +{\left (2 \,{\left (A - 2 \, B\right )} a \cos \left (f x + e\right )^{4} + 10 \,{\left (A - 2 \, B\right )} a \cos \left (f x + e\right )^{3} - 15 \,{\left (A - 2 \, B\right )} a \cos \left (f x + e\right )^{2} - 35 \,{\left (A + B\right )} a \cos \left (f x + e\right ) - 70 \,{\left (A + B\right )} a\right )} \sin \left (f x + e\right )}{315 \,{\left (c^{5} f \cos \left (f x + e\right )^{5} + 5 \, c^{5} f \cos \left (f x + e\right )^{4} - 8 \, c^{5} f \cos \left (f x + e\right )^{3} - 20 \, c^{5} f \cos \left (f x + e\right )^{2} + 8 \, c^{5} f \cos \left (f x + e\right ) + 16 \, c^{5} f -{\left (c^{5} f \cos \left (f x + e\right )^{4} - 4 \, c^{5} f \cos \left (f x + e\right )^{3} - 12 \, c^{5} f \cos \left (f x + e\right )^{2} + 8 \, c^{5} f \cos \left (f x + e\right ) + 16 \, c^{5} f\right )} \sin \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^5,x, algorithm="fricas")

[Out]

-1/315*(2*(A - 2*B)*a*cos(f*x + e)^5 - 8*(A - 2*B)*a*cos(f*x + e)^4 - 25*(A - 2*B)*a*cos(f*x + e)^3 + 5*(4*A +
13*B)*a*cos(f*x + e)^2 - 35*(A + B)*a*cos(f*x + e) - 70*(A + B)*a + (2*(A - 2*B)*a*cos(f*x + e)^4 + 10*(A - 2
*B)*a*cos(f*x + e)^3 - 15*(A - 2*B)*a*cos(f*x + e)^2 - 35*(A + B)*a*cos(f*x + e) - 70*(A + B)*a)*sin(f*x + e))
/(c^5*f*cos(f*x + e)^5 + 5*c^5*f*cos(f*x + e)^4 - 8*c^5*f*cos(f*x + e)^3 - 20*c^5*f*cos(f*x + e)^2 + 8*c^5*f*c
os(f*x + e) + 16*c^5*f - (c^5*f*cos(f*x + e)^4 - 4*c^5*f*cos(f*x + e)^3 - 12*c^5*f*cos(f*x + e)^2 + 8*c^5*f*co
s(f*x + e) + 16*c^5*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+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**5,x)

[Out]

Timed out

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Giac [A]  time = 1.20943, size = 360, normalized size = 2.05 \begin{align*} -\frac{2 \,{\left (315 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{8} - 945 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} + 315 \, B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} + 2625 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 315 \, B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 3465 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 945 \, B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 3843 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 441 \, B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 2247 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 609 \, B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 1143 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 81 \, B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 207 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 99 \, B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 58 \, A a - 11 \, B a\right )}}{315 \, c^{5} f{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^5,x, algorithm="giac")

[Out]

-2/315*(315*A*a*tan(1/2*f*x + 1/2*e)^8 - 945*A*a*tan(1/2*f*x + 1/2*e)^7 + 315*B*a*tan(1/2*f*x + 1/2*e)^7 + 262
5*A*a*tan(1/2*f*x + 1/2*e)^6 - 315*B*a*tan(1/2*f*x + 1/2*e)^6 - 3465*A*a*tan(1/2*f*x + 1/2*e)^5 + 945*B*a*tan(
1/2*f*x + 1/2*e)^5 + 3843*A*a*tan(1/2*f*x + 1/2*e)^4 - 441*B*a*tan(1/2*f*x + 1/2*e)^4 - 2247*A*a*tan(1/2*f*x +
1/2*e)^3 + 609*B*a*tan(1/2*f*x + 1/2*e)^3 + 1143*A*a*tan(1/2*f*x + 1/2*e)^2 - 81*B*a*tan(1/2*f*x + 1/2*e)^2 -
207*A*a*tan(1/2*f*x + 1/2*e) + 99*B*a*tan(1/2*f*x + 1/2*e) + 58*A*a - 11*B*a)/(c^5*f*(tan(1/2*f*x + 1/2*e) -
1)^9)