3.76 \(\int \frac{A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^2} \, dx\)

Optimal. Leaf size=90 \[ \frac{(4 A+B) \tan ^3(e+f x)}{15 a^3 c^2 f}+\frac{(4 A+B) \tan (e+f x)}{5 a^3 c^2 f}-\frac{(A-B) \sec ^3(e+f x)}{5 c^2 f \left (a^3 \sin (e+f x)+a^3\right )} \]

[Out]

-((A - B)*Sec[e + f*x]^3)/(5*c^2*f*(a^3 + a^3*Sin[e + f*x])) + ((4*A + B)*Tan[e + f*x])/(5*a^3*c^2*f) + ((4*A
+ B)*Tan[e + f*x]^3)/(15*a^3*c^2*f)

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Rubi [A]  time = 0.204458, antiderivative size = 90, 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} \[ \frac{(4 A+B) \tan ^3(e+f x)}{15 a^3 c^2 f}+\frac{(4 A+B) \tan (e+f x)}{5 a^3 c^2 f}-\frac{(A-B) \sec ^3(e+f x)}{5 c^2 f \left (a^3 \sin (e+f x)+a^3\right )} \]

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])^2),x]

[Out]

-((A - B)*Sec[e + f*x]^3)/(5*c^2*f*(a^3 + a^3*Sin[e + f*x])) + ((4*A + B)*Tan[e + f*x])/(5*a^3*c^2*f) + ((4*A
+ B)*Tan[e + f*x]^3)/(15*a^3*c^2*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))^2} \, dx &=\frac{\int \frac{\sec ^4(e+f x) (A+B \sin (e+f x))}{a+a \sin (e+f x)} \, dx}{a^2 c^2}\\ &=-\frac{(A-B) \sec ^3(e+f x)}{5 c^2 f \left (a^3+a^3 \sin (e+f x)\right )}+\frac{(4 A+B) \int \sec ^4(e+f x) \, dx}{5 a^3 c^2}\\ &=-\frac{(A-B) \sec ^3(e+f x)}{5 c^2 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac{(4 A+B) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (e+f x)\right )}{5 a^3 c^2 f}\\ &=-\frac{(A-B) \sec ^3(e+f x)}{5 c^2 f \left (a^3+a^3 \sin (e+f x)\right )}+\frac{(4 A+B) \tan (e+f x)}{5 a^3 c^2 f}+\frac{(4 A+B) \tan ^3(e+f x)}{15 a^3 c^2 f}\\ \end{align*}

Mathematica [B]  time = 0.985886, size = 237, normalized size = 2.63 \[ \frac{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (54 (A-B) \cos (e+f x)-32 (4 A+B) \cos (2 (e+f x))+384 A \sin (e+f x)+18 A \sin (2 (e+f x))+128 A \sin (3 (e+f x))+9 A \sin (4 (e+f x))+18 A \cos (3 (e+f x))-64 A \cos (4 (e+f x))+96 B \sin (e+f x)-18 B \sin (2 (e+f x))+32 B \sin (3 (e+f x))-9 B \sin (4 (e+f x))-18 B \cos (3 (e+f x))-16 B \cos (4 (e+f x))+240 B)}{960 a^3 c^2 f (\sin (e+f x)-1)^2 (\sin (e+f x)+1)^3} \]

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])^2),x]

[Out]

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(240*B + 54*(A - B)*Cos[e + f*x]
- 32*(4*A + B)*Cos[2*(e + f*x)] + 18*A*Cos[3*(e + f*x)] - 18*B*Cos[3*(e + f*x)] - 64*A*Cos[4*(e + f*x)] - 16*B
*Cos[4*(e + f*x)] + 384*A*Sin[e + f*x] + 96*B*Sin[e + f*x] + 18*A*Sin[2*(e + f*x)] - 18*B*Sin[2*(e + f*x)] + 1
28*A*Sin[3*(e + f*x)] + 32*B*Sin[3*(e + f*x)] + 9*A*Sin[4*(e + f*x)] - 9*B*Sin[4*(e + f*x)]))/(960*a^3*c^2*f*(
-1 + Sin[e + f*x])^2*(1 + Sin[e + f*x])^3)

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Maple [B]  time = 0.08, size = 185, normalized size = 2.1 \begin{align*} 2\,{\frac{1}{f{a}^{3}{c}^{2}} \left ( -1/3\,{\frac{A/4+B/4}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{3}}}-1/2\,{\frac{A/4+B/4}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{2}}}-{\frac{1}{\tan \left ( 1/2\,fx+e/2 \right ) -1} \left ({\frac{5\,A}{16}}+3/16\,B \right ) }-1/4\,{\frac{-2\,A+2\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{4}}}-1/5\,{\frac{A-B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{5}}}-1/2\,{\frac{-3/2\,A+B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}-1/3\,{\frac{5/2\,A-2\,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}{16}}-3/16\,B \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))^2,x)

[Out]

2/f/a^3/c^2*(-1/3*(1/4*A+1/4*B)/(tan(1/2*f*x+1/2*e)-1)^3-1/2*(1/4*A+1/4*B)/(tan(1/2*f*x+1/2*e)-1)^2-(5/16*A+3/
16*B)/(tan(1/2*f*x+1/2*e)-1)-1/4*(-2*A+2*B)/(tan(1/2*f*x+1/2*e)+1)^4-1/5*(A-B)/(tan(1/2*f*x+1/2*e)+1)^5-1/2*(-
3/2*A+B)/(tan(1/2*f*x+1/2*e)+1)^2-1/3*(5/2*A-2*B)/(tan(1/2*f*x+1/2*e)+1)^3-(11/16*A-3/16*B)/(tan(1/2*f*x+1/2*e
)+1))

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Maxima [B]  time = 1.06267, size = 878, normalized size = 9.76 \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))^2,x, algorithm="maxima")

[Out]

2/15*(A*(9*sin(f*x + e)/(cos(f*x + e) + 1) + 21*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 13*sin(f*x + e)^3/(cos(f
*x + e) + 1)^3 - 25*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 5*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 15*sin(f*x +
e)^6/(cos(f*x + e) + 1)^6 + 15*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 3)/(a^3*c^2 + 2*a^3*c^2*sin(f*x + e)/(co
s(f*x + e) + 1) - 2*a^3*c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 6*a^3*c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^
3 + 6*a^3*c^2*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 2*a^3*c^2*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 2*a^3*c^2*
sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - a^3*c^2*sin(f*x + e)^8/(cos(f*x + e) + 1)^8) + B*(6*sin(f*x + e)/(cos(f*
x + e) + 1) + 9*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 8*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*sin(f*x + e)^4
/(cos(f*x + e) + 1)^4 + 10*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 15*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 3)/(
a^3*c^2 + 2*a^3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) - 2*a^3*c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 6*a^3*c^
2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 6*a^3*c^2*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 2*a^3*c^2*sin(f*x + e)
^6/(cos(f*x + e) + 1)^6 - 2*a^3*c^2*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - a^3*c^2*sin(f*x + e)^8/(cos(f*x + e)
+ 1)^8))/f

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Fricas [A]  time = 1.75245, size = 262, normalized size = 2.91 \begin{align*} -\frac{2 \,{\left (4 \, A + B\right )} \cos \left (f x + e\right )^{4} -{\left (4 \, A + B\right )} \cos \left (f x + e\right )^{2} -{\left (2 \,{\left (4 \, A + B\right )} \cos \left (f x + e\right )^{2} + 4 \, A + B\right )} \sin \left (f x + e\right ) - A - 4 \, B}{15 \,{\left (a^{3} c^{2} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + a^{3} c^{2} 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))^3/(c-c*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

-1/15*(2*(4*A + B)*cos(f*x + e)^4 - (4*A + B)*cos(f*x + e)^2 - (2*(4*A + B)*cos(f*x + e)^2 + 4*A + B)*sin(f*x
+ e) - A - 4*B)/(a^3*c^2*f*cos(f*x + e)^3*sin(f*x + e) + a^3*c^2*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))**3/(c-c*sin(f*x+e))**2,x)

[Out]

Timed out

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Giac [B]  time = 1.20683, size = 317, normalized size = 3.52 \begin{align*} -\frac{\frac{5 \,{\left (15 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 9 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 24 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 12 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 13 \, A + 7 \, B\right )}}{a^{3} c^{2}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}^{3}} + \frac{165 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 45 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 480 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 60 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 650 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 70 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 400 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 20 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 113 \, A - 13 \, B}{a^{3} c^{2}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{5}}}{120 \, 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))^2,x, algorithm="giac")

[Out]

-1/120*(5*(15*A*tan(1/2*f*x + 1/2*e)^2 + 9*B*tan(1/2*f*x + 1/2*e)^2 - 24*A*tan(1/2*f*x + 1/2*e) - 12*B*tan(1/2
*f*x + 1/2*e) + 13*A + 7*B)/(a^3*c^2*(tan(1/2*f*x + 1/2*e) - 1)^3) + (165*A*tan(1/2*f*x + 1/2*e)^4 - 45*B*tan(
1/2*f*x + 1/2*e)^4 + 480*A*tan(1/2*f*x + 1/2*e)^3 - 60*B*tan(1/2*f*x + 1/2*e)^3 + 650*A*tan(1/2*f*x + 1/2*e)^2
- 70*B*tan(1/2*f*x + 1/2*e)^2 + 400*A*tan(1/2*f*x + 1/2*e) - 20*B*tan(1/2*f*x + 1/2*e) + 113*A - 13*B)/(a^3*c
^2*(tan(1/2*f*x + 1/2*e) + 1)^5))/f