3.78 \(\int \frac{A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^4} \, dx\)

Optimal. Leaf size=121 \[ \frac{(6 A-B) \tan ^5(e+f x)}{35 a^3 c^4 f}+\frac{2 (6 A-B) \tan ^3(e+f x)}{21 a^3 c^4 f}+\frac{(6 A-B) \tan (e+f x)}{7 a^3 c^4 f}+\frac{(A+B) \sec ^5(e+f x)}{7 a^3 f \left (c^4-c^4 \sin (e+f x)\right )} \]

[Out]

((A + B)*Sec[e + f*x]^5)/(7*a^3*f*(c^4 - c^4*Sin[e + f*x])) + ((6*A - B)*Tan[e + f*x])/(7*a^3*c^4*f) + (2*(6*A
- B)*Tan[e + f*x]^3)/(21*a^3*c^4*f) + ((6*A - B)*Tan[e + f*x]^5)/(35*a^3*c^4*f)

________________________________________________________________________________________

Rubi [A]  time = 0.223278, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {2967, 2859, 3767} \[ \frac{(6 A-B) \tan ^5(e+f x)}{35 a^3 c^4 f}+\frac{2 (6 A-B) \tan ^3(e+f x)}{21 a^3 c^4 f}+\frac{(6 A-B) \tan (e+f x)}{7 a^3 c^4 f}+\frac{(A+B) \sec ^5(e+f x)}{7 a^3 f \left (c^4-c^4 \sin (e+f x)\right )} \]

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])^4),x]

[Out]

((A + B)*Sec[e + f*x]^5)/(7*a^3*f*(c^4 - c^4*Sin[e + f*x])) + ((6*A - B)*Tan[e + f*x])/(7*a^3*c^4*f) + (2*(6*A
- B)*Tan[e + f*x]^3)/(21*a^3*c^4*f) + ((6*A - B)*Tan[e + f*x]^5)/(35*a^3*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 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))^4} \, dx &=\frac{\int \frac{\sec ^6(e+f x) (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx}{a^3 c^3}\\ &=\frac{(A+B) \sec ^5(e+f x)}{7 a^3 f \left (c^4-c^4 \sin (e+f x)\right )}+\frac{(6 A-B) \int \sec ^6(e+f x) \, dx}{7 a^3 c^4}\\ &=\frac{(A+B) \sec ^5(e+f x)}{7 a^3 f \left (c^4-c^4 \sin (e+f x)\right )}-\frac{(6 A-B) \operatorname{Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (e+f x)\right )}{7 a^3 c^4 f}\\ &=\frac{(A+B) \sec ^5(e+f x)}{7 a^3 f \left (c^4-c^4 \sin (e+f x)\right )}+\frac{(6 A-B) \tan (e+f x)}{7 a^3 c^4 f}+\frac{2 (6 A-B) \tan ^3(e+f x)}{21 a^3 c^4 f}+\frac{(6 A-B) \tan ^5(e+f x)}{35 a^3 c^4 f}\\ \end{align*}

Mathematica [B]  time = 1.09089, size = 325, normalized size = 2.69 \[ -\frac{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (1500 (A+B) \cos (e+f x)-640 (6 A-B) \cos (2 (e+f x))-15360 A \sin (e+f x)-375 A \sin (2 (e+f x))-7680 A \sin (3 (e+f x))-300 A \sin (4 (e+f x))-1536 A \sin (5 (e+f x))-75 A \sin (6 (e+f x))+750 A \cos (3 (e+f x))-3072 A \cos (4 (e+f x))+150 A \cos (5 (e+f x))-768 A \cos (6 (e+f x))+2560 B \sin (e+f x)-375 B \sin (2 (e+f x))+1280 B \sin (3 (e+f x))-300 B \sin (4 (e+f x))+256 B \sin (5 (e+f x))-75 B \sin (6 (e+f x))+750 B \cos (3 (e+f x))+512 B \cos (4 (e+f x))+150 B \cos (5 (e+f x))+128 B \cos (6 (e+f x))-8960 B)}{53760 a^3 c^4 f (\sin (e+f x)-1)^4 (\sin (e+f x)+1)^3} \]

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])^4),x]

[Out]

-((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-8960*B + 1500*(A + B)*Cos[e +
f*x] - 640*(6*A - B)*Cos[2*(e + f*x)] + 750*A*Cos[3*(e + f*x)] + 750*B*Cos[3*(e + f*x)] - 3072*A*Cos[4*(e + f*
x)] + 512*B*Cos[4*(e + f*x)] + 150*A*Cos[5*(e + f*x)] + 150*B*Cos[5*(e + f*x)] - 768*A*Cos[6*(e + f*x)] + 128*
B*Cos[6*(e + f*x)] - 15360*A*Sin[e + f*x] + 2560*B*Sin[e + f*x] - 375*A*Sin[2*(e + f*x)] - 375*B*Sin[2*(e + f*
x)] - 7680*A*Sin[3*(e + f*x)] + 1280*B*Sin[3*(e + f*x)] - 300*A*Sin[4*(e + f*x)] - 300*B*Sin[4*(e + f*x)] - 15
36*A*Sin[5*(e + f*x)] + 256*B*Sin[5*(e + f*x)] - 75*A*Sin[6*(e + f*x)] - 75*B*Sin[6*(e + f*x)]))/(53760*a^3*c^
4*f*(-1 + Sin[e + f*x])^4*(1 + Sin[e + f*x])^3)

________________________________________________________________________________________

Maple [B]  time = 0.1, size = 271, normalized size = 2.2 \begin{align*} 2\,{\frac{1}{f{a}^{3}{c}^{4}} \left ( -1/7\,{\frac{A+B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{7}}}-1/6\,{\frac{3\,A+3\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{6}}}-1/4\,{\frac{11/2\,A+9/2\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{4}}}-1/2\,{\frac{1}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{2}} \left ({\frac{15\,A}{8}}+B \right ) }-{\frac{1}{\tan \left ( 1/2\,fx+e/2 \right ) -1} \left ({\frac{21\,A}{32}}+{\frac{5\,B}{32}} \right ) }-1/5\,{\frac{1}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{5}} \left ({\frac{21\,A}{4}}+{\frac{19\,B}{4}} \right ) }-1/3\,{\frac{1}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{3}} \left ({\frac{33\,A}{8}}+11/4\,B \right ) }-1/2\,{\frac{-A/2+3/8\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}-1/4\,{\frac{-A/2+B/2}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{4}}}-1/5\,{\frac{A/4-B/4}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{5}}}-1/3\,{\frac{3/4\,A-5/8\,B}{ \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{11\,A}{32}}-{\frac{5\,B}{32}} \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))^3/(c-c*sin(f*x+e))^4,x)

[Out]

2/f/a^3/c^4*(-1/7*(A+B)/(tan(1/2*f*x+1/2*e)-1)^7-1/6*(3*A+3*B)/(tan(1/2*f*x+1/2*e)-1)^6-1/4*(11/2*A+9/2*B)/(ta
n(1/2*f*x+1/2*e)-1)^4-1/2*(15/8*A+B)/(tan(1/2*f*x+1/2*e)-1)^2-(21/32*A+5/32*B)/(tan(1/2*f*x+1/2*e)-1)-1/5*(21/
4*A+19/4*B)/(tan(1/2*f*x+1/2*e)-1)^5-1/3*(33/8*A+11/4*B)/(tan(1/2*f*x+1/2*e)-1)^3-1/2*(-1/2*A+3/8*B)/(tan(1/2*
f*x+1/2*e)+1)^2-1/4*(-1/2*A+1/2*B)/(tan(1/2*f*x+1/2*e)+1)^4-1/5*(1/4*A-1/4*B)/(tan(1/2*f*x+1/2*e)+1)^5-1/3*(3/
4*A-5/8*B)/(tan(1/2*f*x+1/2*e)+1)^3-(11/32*A-5/32*B)/(tan(1/2*f*x+1/2*e)+1))

________________________________________________________________________________________

Maxima [B]  time = 1.14742, size = 1376, normalized size = 11.37 \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))^4,x, algorithm="maxima")

[Out]

-2/105*(B*(30*sin(f*x + e)/(cos(f*x + e) + 1) - 45*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 80*sin(f*x + e)^3/(co
s(f*x + e) + 1)^3 - 110*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 188*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 266*si
n(f*x + e)^6/(cos(f*x + e) + 1)^6 - 112*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 35*sin(f*x + e)^8/(cos(f*x + e)
+ 1)^8 + 70*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 - 105*sin(f*x + e)^10/(cos(f*x + e) + 1)^10 - 15)/(a^3*c^4 - 2
*a^3*c^4*sin(f*x + e)/(cos(f*x + e) + 1) - 4*a^3*c^4*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*c^4*sin(f*x
+ e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*c^4*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 20*a^3*c^4*sin(f*x + e)^5/(cos(f
*x + e) + 1)^5 + 20*a^3*c^4*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 5*a^3*c^4*sin(f*x + e)^8/(cos(f*x + e) + 1)^
8 - 10*a^3*c^4*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + 4*a^3*c^4*sin(f*x + e)^10/(cos(f*x + e) + 1)^10 + 2*a^3*c
^4*sin(f*x + e)^11/(cos(f*x + e) + 1)^11 - a^3*c^4*sin(f*x + e)^12/(cos(f*x + e) + 1)^12) - 3*A*(25*sin(f*x +
e)/(cos(f*x + e) + 1) - 55*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 15*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 130*
sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 26*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 182*sin(f*x + e)^6/(cos(f*x + e
) + 1)^6 + 126*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 105*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 - 35*sin(f*x + e)
^9/(cos(f*x + e) + 1)^9 - 35*sin(f*x + e)^10/(cos(f*x + e) + 1)^10 + 35*sin(f*x + e)^11/(cos(f*x + e) + 1)^11
+ 5)/(a^3*c^4 - 2*a^3*c^4*sin(f*x + e)/(cos(f*x + e) + 1) - 4*a^3*c^4*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10
*a^3*c^4*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*c^4*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 20*a^3*c^4*sin(
f*x + e)^5/(cos(f*x + e) + 1)^5 + 20*a^3*c^4*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 5*a^3*c^4*sin(f*x + e)^8/(c
os(f*x + e) + 1)^8 - 10*a^3*c^4*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + 4*a^3*c^4*sin(f*x + e)^10/(cos(f*x + e)
+ 1)^10 + 2*a^3*c^4*sin(f*x + e)^11/(cos(f*x + e) + 1)^11 - a^3*c^4*sin(f*x + e)^12/(cos(f*x + e) + 1)^12))/f

________________________________________________________________________________________

Fricas [A]  time = 2.09174, size = 350, normalized size = 2.89 \begin{align*} -\frac{8 \,{\left (6 \, A - B\right )} \cos \left (f x + e\right )^{6} - 4 \,{\left (6 \, A - B\right )} \cos \left (f x + e\right )^{4} -{\left (6 \, A - B\right )} \cos \left (f x + e\right )^{2} +{\left (8 \,{\left (6 \, A - B\right )} \cos \left (f x + e\right )^{4} + 4 \,{\left (6 \, A - B\right )} \cos \left (f x + e\right )^{2} + 18 \, A - 3 \, B\right )} \sin \left (f x + e\right ) - 3 \, A + 18 \, B}{105 \,{\left (a^{3} c^{4} f \cos \left (f x + e\right )^{5} \sin \left (f x + e\right ) - a^{3} c^{4} 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))^4,x, algorithm="fricas")

[Out]

-1/105*(8*(6*A - B)*cos(f*x + e)^6 - 4*(6*A - B)*cos(f*x + e)^4 - (6*A - B)*cos(f*x + e)^2 + (8*(6*A - B)*cos(
f*x + e)^4 + 4*(6*A - B)*cos(f*x + e)^2 + 18*A - 3*B)*sin(f*x + e) - 3*A + 18*B)/(a^3*c^4*f*cos(f*x + e)^5*sin
(f*x + e) - a^3*c^4*f*cos(f*x + e)^5)

________________________________________________________________________________________

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))**4,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.21273, size = 479, normalized size = 3.96 \begin{align*} -\frac{\frac{7 \,{\left (165 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 75 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 540 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 210 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 750 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 280 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 480 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 170 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 129 \, A - 49 \, B\right )}}{a^{3} c^{4}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{5}} + \frac{2205 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} + 525 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 10080 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 1470 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 21945 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 2555 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 26460 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 2240 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 18963 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1407 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 7476 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 434 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1383 \, A + 137 \, B}{a^{3} c^{4}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}^{7}}}{1680 \, 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))^4,x, algorithm="giac")

[Out]

-1/1680*(7*(165*A*tan(1/2*f*x + 1/2*e)^4 - 75*B*tan(1/2*f*x + 1/2*e)^4 + 540*A*tan(1/2*f*x + 1/2*e)^3 - 210*B*
tan(1/2*f*x + 1/2*e)^3 + 750*A*tan(1/2*f*x + 1/2*e)^2 - 280*B*tan(1/2*f*x + 1/2*e)^2 + 480*A*tan(1/2*f*x + 1/2
*e) - 170*B*tan(1/2*f*x + 1/2*e) + 129*A - 49*B)/(a^3*c^4*(tan(1/2*f*x + 1/2*e) + 1)^5) + (2205*A*tan(1/2*f*x
+ 1/2*e)^6 + 525*B*tan(1/2*f*x + 1/2*e)^6 - 10080*A*tan(1/2*f*x + 1/2*e)^5 - 1470*B*tan(1/2*f*x + 1/2*e)^5 + 2
1945*A*tan(1/2*f*x + 1/2*e)^4 + 2555*B*tan(1/2*f*x + 1/2*e)^4 - 26460*A*tan(1/2*f*x + 1/2*e)^3 - 2240*B*tan(1/
2*f*x + 1/2*e)^3 + 18963*A*tan(1/2*f*x + 1/2*e)^2 + 1407*B*tan(1/2*f*x + 1/2*e)^2 - 7476*A*tan(1/2*f*x + 1/2*e
) - 434*B*tan(1/2*f*x + 1/2*e) + 1383*A + 137*B)/(a^3*c^4*(tan(1/2*f*x + 1/2*e) - 1)^7))/f