3.68 \(\int \frac{A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^4} \, dx\)

Optimal. Leaf size=135 \[ \frac{4 (5 A-2 B) \tan ^3(e+f x)}{105 a^2 c^4 f}+\frac{4 (5 A-2 B) \tan (e+f x)}{35 a^2 c^4 f}+\frac{(5 A-2 B) \sec ^3(e+f x)}{35 a^2 f \left (c^4-c^4 \sin (e+f x)\right )}+\frac{(A+B) \sec ^3(e+f x)}{7 a^2 f \left (c^2-c^2 \sin (e+f x)\right )^2} \]

[Out]

((A + B)*Sec[e + f*x]^3)/(7*a^2*f*(c^2 - c^2*Sin[e + f*x])^2) + ((5*A - 2*B)*Sec[e + f*x]^3)/(35*a^2*f*(c^4 -
c^4*Sin[e + f*x])) + (4*(5*A - 2*B)*Tan[e + f*x])/(35*a^2*c^4*f) + (4*(5*A - 2*B)*Tan[e + f*x]^3)/(105*a^2*c^4
*f)

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Rubi [A]  time = 0.270265, antiderivative size = 135, 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} \[ \frac{4 (5 A-2 B) \tan ^3(e+f x)}{105 a^2 c^4 f}+\frac{4 (5 A-2 B) \tan (e+f x)}{35 a^2 c^4 f}+\frac{(5 A-2 B) \sec ^3(e+f x)}{35 a^2 f \left (c^4-c^4 \sin (e+f x)\right )}+\frac{(A+B) \sec ^3(e+f x)}{7 a^2 f \left (c^2-c^2 \sin (e+f x)\right )^2} \]

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])^4),x]

[Out]

((A + B)*Sec[e + f*x]^3)/(7*a^2*f*(c^2 - c^2*Sin[e + f*x])^2) + ((5*A - 2*B)*Sec[e + f*x]^3)/(35*a^2*f*(c^4 -
c^4*Sin[e + f*x])) + (4*(5*A - 2*B)*Tan[e + f*x])/(35*a^2*c^4*f) + (4*(5*A - 2*B)*Tan[e + f*x]^3)/(105*a^2*c^4
*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))^4} \, dx &=\frac{\int \frac{\sec ^4(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx}{a^2 c^2}\\ &=\frac{(A+B) \sec ^3(e+f x)}{7 a^2 f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac{(5 A-2 B) \int \frac{\sec ^4(e+f x)}{c-c \sin (e+f x)} \, dx}{7 a^2 c^3}\\ &=\frac{(A+B) \sec ^3(e+f x)}{7 a^2 f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac{(5 A-2 B) \sec ^3(e+f x)}{35 a^2 f \left (c^4-c^4 \sin (e+f x)\right )}+\frac{(4 (5 A-2 B)) \int \sec ^4(e+f x) \, dx}{35 a^2 c^4}\\ &=\frac{(A+B) \sec ^3(e+f x)}{7 a^2 f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac{(5 A-2 B) \sec ^3(e+f x)}{35 a^2 f \left (c^4-c^4 \sin (e+f x)\right )}-\frac{(4 (5 A-2 B)) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (e+f x)\right )}{35 a^2 c^4 f}\\ &=\frac{(A+B) \sec ^3(e+f x)}{7 a^2 f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac{(5 A-2 B) \sec ^3(e+f x)}{35 a^2 f \left (c^4-c^4 \sin (e+f x)\right )}+\frac{4 (5 A-2 B) \tan (e+f x)}{35 a^2 c^4 f}+\frac{4 (5 A-2 B) \tan ^3(e+f x)}{105 a^2 c^4 f}\\ \end{align*}

Mathematica [B]  time = 0.924016, size = 285, normalized size = 2.11 \[ -\frac{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (42 (25 A+4 B) \cos (e+f x)-512 (5 A-2 B) \cos (2 (e+f x))-4480 A \sin (e+f x)-600 A \sin (2 (e+f x))-960 A \sin (3 (e+f x))-300 A \sin (4 (e+f x))+320 A \sin (5 (e+f x))+225 A \cos (3 (e+f x))-1280 A \cos (4 (e+f x))-75 A \cos (5 (e+f x))+1792 B \sin (e+f x)-96 B \sin (2 (e+f x))+384 B \sin (3 (e+f x))-48 B \sin (4 (e+f x))-128 B \sin (5 (e+f x))+36 B \cos (3 (e+f x))+512 B \cos (4 (e+f x))-12 B \cos (5 (e+f x))-2688 B)}{13440 a^2 c^4 f (\sin (e+f x)-1)^4 (\sin (e+f x)+1)^2} \]

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])^4),x]

[Out]

-((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-2688*B + 42*(25*A + 4*B)*Cos[e
+ f*x] - 512*(5*A - 2*B)*Cos[2*(e + f*x)] + 225*A*Cos[3*(e + f*x)] + 36*B*Cos[3*(e + f*x)] - 1280*A*Cos[4*(e
+ f*x)] + 512*B*Cos[4*(e + f*x)] - 75*A*Cos[5*(e + f*x)] - 12*B*Cos[5*(e + f*x)] - 4480*A*Sin[e + f*x] + 1792*
B*Sin[e + f*x] - 600*A*Sin[2*(e + f*x)] - 96*B*Sin[2*(e + f*x)] - 960*A*Sin[3*(e + f*x)] + 384*B*Sin[3*(e + f*
x)] - 300*A*Sin[4*(e + f*x)] - 48*B*Sin[4*(e + f*x)] + 320*A*Sin[5*(e + f*x)] - 128*B*Sin[5*(e + f*x)]))/(1344
0*a^2*c^4*f*(-1 + Sin[e + f*x])^4*(1 + Sin[e + f*x])^2)

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Maple [A]  time = 0.116, size = 233, normalized size = 1.7 \begin{align*} 2\,{\frac{1}{{a}^{2}f{c}^{4}} \left ( -1/7\,{\frac{2\,A+2\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{7}}}-1/6\,{\frac{6\,A+6\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{6}}}-1/4\,{\frac{10\,A+8\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{4}}}-1/5\,{\frac{10\,A+9\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{5}}}-{\frac{1}{\tan \left ( 1/2\,fx+e/2 \right ) -1} \left ({\frac{13\,A}{16}}+B/8 \right ) }-1/2\,{\frac{1}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{2}} \left ({\frac{23\,A}{8}}+{\frac{11\,B}{8}} \right ) }-1/3\,{\frac{1}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{3}} \left ({\frac{55\,A}{8}}+{\frac{35\,B}{8}} \right ) }-1/2\,{\frac{-A/8+B/8}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}-1/3\,{\frac{A/8-B/8}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{3}}}-{\frac{3/16\,A-B/8}{\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+B*sin(f*x+e))/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^4,x)

[Out]

2/f/a^2/c^4*(-1/7*(2*A+2*B)/(tan(1/2*f*x+1/2*e)-1)^7-1/6*(6*A+6*B)/(tan(1/2*f*x+1/2*e)-1)^6-1/4*(10*A+8*B)/(ta
n(1/2*f*x+1/2*e)-1)^4-1/5*(10*A+9*B)/(tan(1/2*f*x+1/2*e)-1)^5-(13/16*A+1/8*B)/(tan(1/2*f*x+1/2*e)-1)-1/2*(23/8
*A+11/8*B)/(tan(1/2*f*x+1/2*e)-1)^2-1/3*(55/8*A+35/8*B)/(tan(1/2*f*x+1/2*e)-1)^3-1/2*(-1/8*A+1/8*B)/(tan(1/2*f
*x+1/2*e)+1)^2-1/3*(1/8*A-1/8*B)/(tan(1/2*f*x+1/2*e)+1)^3-(3/16*A-1/8*B)/(tan(1/2*f*x+1/2*e)+1))

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Maxima [B]  time = 1.09332, size = 1127, normalized size = 8.35 \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))^4,x, algorithm="maxima")

[Out]

-2/105*(B*(36*sin(f*x + e)/(cos(f*x + e) + 1) - 132*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 68*sin(f*x + e)^3/(c
os(f*x + e) + 1)^3 - 14*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 84*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 140*sin
(f*x + e)^6/(cos(f*x + e) + 1)^6 + 140*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 105*sin(f*x + e)^8/(cos(f*x + e)
+ 1)^8 - 9)/(a^2*c^4 - 4*a^2*c^4*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*c^4*sin(f*x + e)^2/(cos(f*x + e) + 1)
^2 + 8*a^2*c^4*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 14*a^2*c^4*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 14*a^2*c
^4*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 8*a^2*c^4*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 3*a^2*c^4*sin(f*x + e
)^8/(cos(f*x + e) + 1)^8 + 4*a^2*c^4*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 - a^2*c^4*sin(f*x + e)^10/(cos(f*x +
e) + 1)^10) + 5*A*(3*sin(f*x + e)/(cos(f*x + e) + 1) + 24*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 76*sin(f*x + e
)^3/(cos(f*x + e) + 1)^3 + 28*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 42*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 5
6*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 28*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 42*sin(f*x + e)^8/(cos(f*x +
e) + 1)^8 - 21*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 - 6)/(a^2*c^4 - 4*a^2*c^4*sin(f*x + e)/(cos(f*x + e) + 1) +
3*a^2*c^4*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 8*a^2*c^4*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 14*a^2*c^4*si
n(f*x + e)^4/(cos(f*x + e) + 1)^4 + 14*a^2*c^4*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 8*a^2*c^4*sin(f*x + e)^7/
(cos(f*x + e) + 1)^7 - 3*a^2*c^4*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 4*a^2*c^4*sin(f*x + e)^9/(cos(f*x + e)
+ 1)^9 - a^2*c^4*sin(f*x + e)^10/(cos(f*x + e) + 1)^10))/f

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Fricas [A]  time = 1.65329, size = 371, normalized size = 2.75 \begin{align*} -\frac{16 \,{\left (5 \, A - 2 \, B\right )} \cos \left (f x + e\right )^{4} - 8 \,{\left (5 \, A - 2 \, B\right )} \cos \left (f x + e\right )^{2} -{\left (8 \,{\left (5 \, A - 2 \, B\right )} \cos \left (f x + e\right )^{4} - 12 \,{\left (5 \, A - 2 \, B\right )} \cos \left (f x + e\right )^{2} - 25 \, A + 10 \, B\right )} \sin \left (f x + e\right ) - 10 \, A + 25 \, B}{105 \,{\left (a^{2} c^{4} f \cos \left (f x + e\right )^{5} + 2 \, a^{2} c^{4} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) - 2 \, a^{2} c^{4} f \cos \left (f x + e\right )^{3}\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))^4,x, algorithm="fricas")

[Out]

-1/105*(16*(5*A - 2*B)*cos(f*x + e)^4 - 8*(5*A - 2*B)*cos(f*x + e)^2 - (8*(5*A - 2*B)*cos(f*x + e)^4 - 12*(5*A
- 2*B)*cos(f*x + e)^2 - 25*A + 10*B)*sin(f*x + e) - 10*A + 25*B)/(a^2*c^4*f*cos(f*x + e)^5 + 2*a^2*c^4*f*cos(
f*x + e)^3*sin(f*x + e) - 2*a^2*c^4*f*cos(f*x + e)^3)

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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))**4,x)

[Out]

Timed out

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Giac [B]  time = 1.23712, size = 398, normalized size = 2.95 \begin{align*} -\frac{\frac{35 \,{\left (9 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 6 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 15 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 9 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 8 \, A - 5 \, B\right )}}{a^{2} c^{4}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{3}} + \frac{1365 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} + 210 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 5775 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 105 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 12250 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 175 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 14350 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 910 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 10185 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 756 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 3955 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 427 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 760 \, A - 31 \, B}{a^{2} c^{4}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}^{7}}}{840 \, 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))^4,x, algorithm="giac")

[Out]

-1/840*(35*(9*A*tan(1/2*f*x + 1/2*e)^2 - 6*B*tan(1/2*f*x + 1/2*e)^2 + 15*A*tan(1/2*f*x + 1/2*e) - 9*B*tan(1/2*
f*x + 1/2*e) + 8*A - 5*B)/(a^2*c^4*(tan(1/2*f*x + 1/2*e) + 1)^3) + (1365*A*tan(1/2*f*x + 1/2*e)^6 + 210*B*tan(
1/2*f*x + 1/2*e)^6 - 5775*A*tan(1/2*f*x + 1/2*e)^5 - 105*B*tan(1/2*f*x + 1/2*e)^5 + 12250*A*tan(1/2*f*x + 1/2*
e)^4 - 175*B*tan(1/2*f*x + 1/2*e)^4 - 14350*A*tan(1/2*f*x + 1/2*e)^3 + 910*B*tan(1/2*f*x + 1/2*e)^3 + 10185*A*
tan(1/2*f*x + 1/2*e)^2 - 756*B*tan(1/2*f*x + 1/2*e)^2 - 3955*A*tan(1/2*f*x + 1/2*e) + 427*B*tan(1/2*f*x + 1/2*
e) + 760*A - 31*B)/(a^2*c^4*(tan(1/2*f*x + 1/2*e) - 1)^7))/f