3.79 \(\int \frac{A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^5} \, dx\)

Optimal. Leaf size=162 \[ \frac{2 (7 A-2 B) \tan ^5(e+f x)}{105 a^3 c^5 f}+\frac{4 (7 A-2 B) \tan ^3(e+f x)}{63 a^3 c^5 f}+\frac{2 (7 A-2 B) \tan (e+f x)}{21 a^3 c^5 f}+\frac{(7 A-2 B) \sec ^5(e+f x)}{63 a^3 f \left (c^5-c^5 \sin (e+f x)\right )}+\frac{(A+B) \sec ^5(e+f x)}{9 a^3 c^3 f (c-c \sin (e+f x))^2} \]

[Out]

((A + B)*Sec[e + f*x]^5)/(9*a^3*c^3*f*(c - c*Sin[e + f*x])^2) + ((7*A - 2*B)*Sec[e + f*x]^5)/(63*a^3*f*(c^5 -
c^5*Sin[e + f*x])) + (2*(7*A - 2*B)*Tan[e + f*x])/(21*a^3*c^5*f) + (4*(7*A - 2*B)*Tan[e + f*x]^3)/(63*a^3*c^5*
f) + (2*(7*A - 2*B)*Tan[e + f*x]^5)/(105*a^3*c^5*f)

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Rubi [A]  time = 0.288791, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2967, 2859, 2672, 3767} \[ \frac{2 (7 A-2 B) \tan ^5(e+f x)}{105 a^3 c^5 f}+\frac{4 (7 A-2 B) \tan ^3(e+f x)}{63 a^3 c^5 f}+\frac{2 (7 A-2 B) \tan (e+f x)}{21 a^3 c^5 f}+\frac{(7 A-2 B) \sec ^5(e+f x)}{63 a^3 f \left (c^5-c^5 \sin (e+f x)\right )}+\frac{(A+B) \sec ^5(e+f x)}{9 a^3 c^3 f (c-c \sin (e+f x))^2} \]

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])^5),x]

[Out]

((A + B)*Sec[e + f*x]^5)/(9*a^3*c^3*f*(c - c*Sin[e + f*x])^2) + ((7*A - 2*B)*Sec[e + f*x]^5)/(63*a^3*f*(c^5 -
c^5*Sin[e + f*x])) + (2*(7*A - 2*B)*Tan[e + f*x])/(21*a^3*c^5*f) + (4*(7*A - 2*B)*Tan[e + f*x]^3)/(63*a^3*c^5*
f) + (2*(7*A - 2*B)*Tan[e + f*x]^5)/(105*a^3*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))^3 (c-c \sin (e+f x))^5} \, dx &=\frac{\int \frac{\sec ^6(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx}{a^3 c^3}\\ &=\frac{(A+B) \sec ^5(e+f x)}{9 a^3 c^3 f (c-c \sin (e+f x))^2}+\frac{(7 A-2 B) \int \frac{\sec ^6(e+f x)}{c-c \sin (e+f x)} \, dx}{9 a^3 c^4}\\ &=\frac{(A+B) \sec ^5(e+f x)}{9 a^3 c^3 f (c-c \sin (e+f x))^2}+\frac{(7 A-2 B) \sec ^5(e+f x)}{63 a^3 f \left (c^5-c^5 \sin (e+f x)\right )}+\frac{(2 (7 A-2 B)) \int \sec ^6(e+f x) \, dx}{21 a^3 c^5}\\ &=\frac{(A+B) \sec ^5(e+f x)}{9 a^3 c^3 f (c-c \sin (e+f x))^2}+\frac{(7 A-2 B) \sec ^5(e+f x)}{63 a^3 f \left (c^5-c^5 \sin (e+f x)\right )}-\frac{(2 (7 A-2 B)) \operatorname{Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (e+f x)\right )}{21 a^3 c^5 f}\\ &=\frac{(A+B) \sec ^5(e+f x)}{9 a^3 c^3 f (c-c \sin (e+f x))^2}+\frac{(7 A-2 B) \sec ^5(e+f x)}{63 a^3 f \left (c^5-c^5 \sin (e+f x)\right )}+\frac{2 (7 A-2 B) \tan (e+f x)}{21 a^3 c^5 f}+\frac{4 (7 A-2 B) \tan ^3(e+f x)}{63 a^3 c^5 f}+\frac{2 (7 A-2 B) \tan ^5(e+f x)}{105 a^3 c^5 f}\\ \end{align*}

Mathematica [B]  time = 1.3207, size = 373, normalized size = 2.3 \[ \frac{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (1125 (49 A+13 B) \cos (e+f x)-20480 (7 A-2 B) \cos (2 (e+f x))-322560 A \sin (e+f x)-24500 A \sin (2 (e+f x))-136192 A \sin (3 (e+f x))-19600 A \sin (4 (e+f x))-7168 A \sin (5 (e+f x))-4900 A \sin (6 (e+f x))+7168 A \sin (7 (e+f x))+23275 A \cos (3 (e+f x))-114688 A \cos (4 (e+f x))+1225 A \cos (5 (e+f x))-28672 A \cos (6 (e+f x))-1225 A \cos (7 (e+f x))+92160 B \sin (e+f x)-6500 B \sin (2 (e+f x))+38912 B \sin (3 (e+f x))-5200 B \sin (4 (e+f x))+2048 B \sin (5 (e+f x))-1300 B \sin (6 (e+f x))-2048 B \sin (7 (e+f x))+6175 B \cos (3 (e+f x))+32768 B \cos (4 (e+f x))+325 B \cos (5 (e+f x))+8192 B \cos (6 (e+f x))-325 B \cos (7 (e+f x))-184320 B)}{1290240 a^3 c^5 f (\sin (e+f x)-1)^5 (\sin (e+f x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^3*(c - 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])*(-184320*B + 1125*(49*A + 13*B)*C
os[e + f*x] - 20480*(7*A - 2*B)*Cos[2*(e + f*x)] + 23275*A*Cos[3*(e + f*x)] + 6175*B*Cos[3*(e + f*x)] - 114688
*A*Cos[4*(e + f*x)] + 32768*B*Cos[4*(e + f*x)] + 1225*A*Cos[5*(e + f*x)] + 325*B*Cos[5*(e + f*x)] - 28672*A*Co
s[6*(e + f*x)] + 8192*B*Cos[6*(e + f*x)] - 1225*A*Cos[7*(e + f*x)] - 325*B*Cos[7*(e + f*x)] - 322560*A*Sin[e +
f*x] + 92160*B*Sin[e + f*x] - 24500*A*Sin[2*(e + f*x)] - 6500*B*Sin[2*(e + f*x)] - 136192*A*Sin[3*(e + f*x)]
+ 38912*B*Sin[3*(e + f*x)] - 19600*A*Sin[4*(e + f*x)] - 5200*B*Sin[4*(e + f*x)] - 7168*A*Sin[5*(e + f*x)] + 20
48*B*Sin[5*(e + f*x)] - 4900*A*Sin[6*(e + f*x)] - 1300*B*Sin[6*(e + f*x)] + 7168*A*Sin[7*(e + f*x)] - 2048*B*S
in[7*(e + f*x)]))/(1290240*a^3*c^5*f*(-1 + Sin[e + f*x])^5*(1 + Sin[e + f*x])^3)

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Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \text{hanged} \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))^3/(c-c*sin(f*x+e))^5,x)

[Out]

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^5,x)

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Maxima [B]  time = 1.18959, size = 1621, normalized size = 10.01 \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))^3/(c-c*sin(f*x+e))^5,x, algorithm="maxima")

[Out]

-2/315*(B*(100*sin(f*x + e)/(cos(f*x + e) + 1) - 340*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 20*sin(f*x + e)^3/(
cos(f*x + e) + 1)^3 + 55*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 88*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 1608*s
in(f*x + e)^6/(cos(f*x + e) + 1)^6 + 1032*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 483*sin(f*x + e)^8/(cos(f*x +
e) + 1)^8 - 588*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 - 420*sin(f*x + e)^10/(cos(f*x + e) + 1)^10 + 420*sin(f*x
+ e)^11/(cos(f*x + e) + 1)^11 - 315*sin(f*x + e)^12/(cos(f*x + e) + 1)^12 - 25)/(a^3*c^5 - 4*a^3*c^5*sin(f*x +
e)/(cos(f*x + e) + 1) + a^3*c^5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 16*a^3*c^5*sin(f*x + e)^3/(cos(f*x + e)
+ 1)^3 - 19*a^3*c^5*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 20*a^3*c^5*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 45
*a^3*c^5*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 45*a^3*c^5*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 20*a^3*c^5*sin
(f*x + e)^9/(cos(f*x + e) + 1)^9 + 19*a^3*c^5*sin(f*x + e)^10/(cos(f*x + e) + 1)^10 - 16*a^3*c^5*sin(f*x + e)^
11/(cos(f*x + e) + 1)^11 - a^3*c^5*sin(f*x + e)^12/(cos(f*x + e) + 1)^12 + 4*a^3*c^5*sin(f*x + e)^13/(cos(f*x
+ e) + 1)^13 - a^3*c^5*sin(f*x + e)^14/(cos(f*x + e) + 1)^14) - 7*A*(5*sin(f*x + e)/(cos(f*x + e) + 1) - 80*si
n(f*x + e)^2/(cos(f*x + e) + 1)^2 + 190*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 50*sin(f*x + e)^4/(cos(f*x + e)
+ 1)^4 - 269*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 96*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 516*sin(f*x + e)^7
/(cos(f*x + e) + 1)^7 - 354*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 - 69*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + 240
*sin(f*x + e)^10/(cos(f*x + e) + 1)^10 + 30*sin(f*x + e)^11/(cos(f*x + e) + 1)^11 - 90*sin(f*x + e)^12/(cos(f*
x + e) + 1)^12 + 45*sin(f*x + e)^13/(cos(f*x + e) + 1)^13 + 10)/(a^3*c^5 - 4*a^3*c^5*sin(f*x + e)/(cos(f*x + e
) + 1) + a^3*c^5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 16*a^3*c^5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 19*a^3
*c^5*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 20*a^3*c^5*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 45*a^3*c^5*sin(f*x
+ e)^6/(cos(f*x + e) + 1)^6 - 45*a^3*c^5*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 20*a^3*c^5*sin(f*x + e)^9/(cos
(f*x + e) + 1)^9 + 19*a^3*c^5*sin(f*x + e)^10/(cos(f*x + e) + 1)^10 - 16*a^3*c^5*sin(f*x + e)^11/(cos(f*x + e)
+ 1)^11 - a^3*c^5*sin(f*x + e)^12/(cos(f*x + e) + 1)^12 + 4*a^3*c^5*sin(f*x + e)^13/(cos(f*x + e) + 1)^13 - a
^3*c^5*sin(f*x + e)^14/(cos(f*x + e) + 1)^14))/f

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Fricas [A]  time = 2.03789, size = 459, normalized size = 2.83 \begin{align*} -\frac{32 \,{\left (7 \, A - 2 \, B\right )} \cos \left (f x + e\right )^{6} - 16 \,{\left (7 \, A - 2 \, B\right )} \cos \left (f x + e\right )^{4} - 4 \,{\left (7 \, A - 2 \, B\right )} \cos \left (f x + e\right )^{2} -{\left (16 \,{\left (7 \, A - 2 \, B\right )} \cos \left (f x + e\right )^{6} - 24 \,{\left (7 \, A - 2 \, B\right )} \cos \left (f x + e\right )^{4} - 10 \,{\left (7 \, A - 2 \, B\right )} \cos \left (f x + e\right )^{2} - 49 \, A + 14 \, B\right )} \sin \left (f x + e\right ) - 14 \, A + 49 \, B}{315 \,{\left (a^{3} c^{5} f \cos \left (f x + e\right )^{7} + 2 \, a^{3} c^{5} f \cos \left (f x + e\right )^{5} \sin \left (f x + e\right ) - 2 \, a^{3} c^{5} f \cos \left (f x + e\right )^{5}\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))^3/(c-c*sin(f*x+e))^5,x, algorithm="fricas")

[Out]

-1/315*(32*(7*A - 2*B)*cos(f*x + e)^6 - 16*(7*A - 2*B)*cos(f*x + e)^4 - 4*(7*A - 2*B)*cos(f*x + e)^2 - (16*(7*
A - 2*B)*cos(f*x + e)^6 - 24*(7*A - 2*B)*cos(f*x + e)^4 - 10*(7*A - 2*B)*cos(f*x + e)^2 - 49*A + 14*B)*sin(f*x
+ e) - 14*A + 49*B)/(a^3*c^5*f*cos(f*x + e)^7 + 2*a^3*c^5*f*cos(f*x + e)^5*sin(f*x + e) - 2*a^3*c^5*f*cos(f*x
+ e)^5)

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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))**3/(c-c*sin(f*x+e))**5,x)

[Out]

Timed out

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Giac [B]  time = 1.25432, size = 560, normalized size = 3.46 \begin{align*} -\frac{\frac{21 \,{\left (435 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 225 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 1470 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 690 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 2060 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 940 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1330 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 590 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 353 \, A - 163 \, B\right )}}{a^{3} c^{5}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{5}} + \frac{31185 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{8} + 4725 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{8} - 185220 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} - 11340 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} + 546840 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} + 15120 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 961380 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 3780 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 1101618 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 24318 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 828492 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 33852 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 404208 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 19368 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 116172 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 6732 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 16373 \, A - 223 \, B}{a^{3} c^{5}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}^{9}}}{20160 \, 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))^3/(c-c*sin(f*x+e))^5,x, algorithm="giac")

[Out]

-1/20160*(21*(435*A*tan(1/2*f*x + 1/2*e)^4 - 225*B*tan(1/2*f*x + 1/2*e)^4 + 1470*A*tan(1/2*f*x + 1/2*e)^3 - 69
0*B*tan(1/2*f*x + 1/2*e)^3 + 2060*A*tan(1/2*f*x + 1/2*e)^2 - 940*B*tan(1/2*f*x + 1/2*e)^2 + 1330*A*tan(1/2*f*x
+ 1/2*e) - 590*B*tan(1/2*f*x + 1/2*e) + 353*A - 163*B)/(a^3*c^5*(tan(1/2*f*x + 1/2*e) + 1)^5) + (31185*A*tan(
1/2*f*x + 1/2*e)^8 + 4725*B*tan(1/2*f*x + 1/2*e)^8 - 185220*A*tan(1/2*f*x + 1/2*e)^7 - 11340*B*tan(1/2*f*x + 1
/2*e)^7 + 546840*A*tan(1/2*f*x + 1/2*e)^6 + 15120*B*tan(1/2*f*x + 1/2*e)^6 - 961380*A*tan(1/2*f*x + 1/2*e)^5 +
3780*B*tan(1/2*f*x + 1/2*e)^5 + 1101618*A*tan(1/2*f*x + 1/2*e)^4 - 24318*B*tan(1/2*f*x + 1/2*e)^4 - 828492*A*
tan(1/2*f*x + 1/2*e)^3 + 33852*B*tan(1/2*f*x + 1/2*e)^3 + 404208*A*tan(1/2*f*x + 1/2*e)^2 - 19368*B*tan(1/2*f*
x + 1/2*e)^2 - 116172*A*tan(1/2*f*x + 1/2*e) + 6732*B*tan(1/2*f*x + 1/2*e) + 16373*A - 223*B)/(a^3*c^5*(tan(1/
2*f*x + 1/2*e) - 1)^9))/f