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

Optimal. Leaf size=142 \[ \frac{2 (4 A-3 B) \tan (e+f x)}{35 a c^4 f}+\frac{(4 A-3 B) \sec (e+f x)}{35 a f \left (c^4-c^4 \sin (e+f x)\right )}+\frac{(4 A-3 B) \sec (e+f x)}{35 a f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac{(A+B) \sec (e+f x)}{7 a c f (c-c \sin (e+f x))^3} \]

[Out]

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

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Rubi [A]  time = 0.306909, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.139, Rules used = {2967, 2859, 2672, 3767, 8} \[ \frac{2 (4 A-3 B) \tan (e+f x)}{35 a c^4 f}+\frac{(4 A-3 B) \sec (e+f x)}{35 a f \left (c^4-c^4 \sin (e+f x)\right )}+\frac{(4 A-3 B) \sec (e+f x)}{35 a f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac{(A+B) \sec (e+f x)}{7 a c f (c-c \sin (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])*(c - c*Sin[e + f*x])^4),x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [A]  time = 1.09769, size = 240, normalized size = 1.69 \[ \frac{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) ((182 B-406 A) \cos (e+f x)+224 (4 A-3 B) \cos (2 (e+f x))+896 A \sin (e+f x)+406 A \sin (2 (e+f x))-384 A \sin (3 (e+f x))-29 A \sin (4 (e+f x))+174 A \cos (3 (e+f x))-64 A \cos (4 (e+f x))-672 B \sin (e+f x)-182 B \sin (2 (e+f x))+288 B \sin (3 (e+f x))+13 B \sin (4 (e+f x))-78 B \cos (3 (e+f x))+48 B \cos (4 (e+f x))+560 B)}{2240 a c^4 f (\sin (e+f x)-1)^4 (\sin (e+f x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])*(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])*(560*B + (-406*A + 182*B)*Cos[e +
f*x] + 224*(4*A - 3*B)*Cos[2*(e + f*x)] + 174*A*Cos[3*(e + f*x)] - 78*B*Cos[3*(e + f*x)] - 64*A*Cos[4*(e + f*
x)] + 48*B*Cos[4*(e + f*x)] + 896*A*Sin[e + f*x] - 672*B*Sin[e + f*x] + 406*A*Sin[2*(e + f*x)] - 182*B*Sin[2*(
e + f*x)] - 384*A*Sin[3*(e + f*x)] + 288*B*Sin[3*(e + f*x)] - 29*A*Sin[4*(e + f*x)] + 13*B*Sin[4*(e + f*x)]))/
(2240*a*c^4*f*(-1 + Sin[e + f*x])^4*(1 + Sin[e + f*x]))

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Maple [A]  time = 0.087, size = 189, normalized size = 1.3 \begin{align*} 2\,{\frac{1}{af{c}^{4}} \left ( -1/7\,{\frac{4\,A+4\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{7}}}-1/6\,{\frac{12\,A+12\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{6}}}-1/4\,{\frac{18\,A+14\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{4}}}-1/5\,{\frac{19\,A+17\,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{15\,A}{16}}+B/16 \right ) }-1/2\,{\frac{1}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{2}} \left ({\frac{17\,A}{4}}+7/4\,B \right ) }-1/3\,{\frac{1}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{3}} \left ({\frac{45\,A}{4}}+{\frac{27\,B}{4}} \right ) }-{\frac{A/16-B/16}{\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))/(c-c*sin(f*x+e))^4,x)

[Out]

2/f/a/c^4*(-1/7*(4*A+4*B)/(tan(1/2*f*x+1/2*e)-1)^7-1/6*(12*A+12*B)/(tan(1/2*f*x+1/2*e)-1)^6-1/4*(18*A+14*B)/(t
an(1/2*f*x+1/2*e)-1)^4-1/5*(19*A+17*B)/(tan(1/2*f*x+1/2*e)-1)^5-(15/16*A+1/16*B)/(tan(1/2*f*x+1/2*e)-1)-1/2*(1
7/4*A+7/4*B)/(tan(1/2*f*x+1/2*e)-1)^2-1/3*(45/4*A+27/4*B)/(tan(1/2*f*x+1/2*e)-1)^3-(1/16*A-1/16*B)/(tan(1/2*f*
x+1/2*e)+1))

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Maxima [B]  time = 1.07238, size = 836, normalized size = 5.89 \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))/(c-c*sin(f*x+e))^4,x, algorithm="maxima")

[Out]

-2/35*(A*(43*sin(f*x + e)/(cos(f*x + e) + 1) - 77*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 7*sin(f*x + e)^3/(cos(
f*x + e) + 1)^3 + 105*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 175*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 105*sin(
f*x + e)^6/(cos(f*x + e) + 1)^6 - 35*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 13)/(a*c^4 - 6*a*c^4*sin(f*x + e)/(
cos(f*x + e) + 1) + 14*a*c^4*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 14*a*c^4*sin(f*x + e)^3/(cos(f*x + e) + 1)^
3 + 14*a*c^4*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 14*a*c^4*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 6*a*c^4*sin(
f*x + e)^7/(cos(f*x + e) + 1)^7 - a*c^4*sin(f*x + e)^8/(cos(f*x + e) + 1)^8) - B*(6*sin(f*x + e)/(cos(f*x + e)
+ 1) + 21*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 56*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 105*sin(f*x + e)^4/(
cos(f*x + e) + 1)^4 - 70*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 35*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 1)/(a*
c^4 - 6*a*c^4*sin(f*x + e)/(cos(f*x + e) + 1) + 14*a*c^4*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 14*a*c^4*sin(f*
x + e)^3/(cos(f*x + e) + 1)^3 + 14*a*c^4*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 14*a*c^4*sin(f*x + e)^6/(cos(f*
x + e) + 1)^6 + 6*a*c^4*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - a*c^4*sin(f*x + e)^8/(cos(f*x + e) + 1)^8))/f

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Fricas [A]  time = 1.30648, size = 350, normalized size = 2.46 \begin{align*} \frac{2 \,{\left (4 \, A - 3 \, B\right )} \cos \left (f x + e\right )^{4} - 9 \,{\left (4 \, A - 3 \, B\right )} \cos \left (f x + e\right )^{2} +{\left (6 \,{\left (4 \, A - 3 \, B\right )} \cos \left (f x + e\right )^{2} - 20 \, A + 15 \, B\right )} \sin \left (f x + e\right ) + 15 \, A - 20 \, B}{35 \,{\left (3 \, a c^{4} f \cos \left (f x + e\right )^{3} - 4 \, a c^{4} f \cos \left (f x + e\right ) -{\left (a c^{4} f \cos \left (f x + e\right )^{3} - 4 \, a c^{4} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\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))/(c-c*sin(f*x+e))^4,x, algorithm="fricas")

[Out]

1/35*(2*(4*A - 3*B)*cos(f*x + e)^4 - 9*(4*A - 3*B)*cos(f*x + e)^2 + (6*(4*A - 3*B)*cos(f*x + e)^2 - 20*A + 15*
B)*sin(f*x + e) + 15*A - 20*B)/(3*a*c^4*f*cos(f*x + e)^3 - 4*a*c^4*f*cos(f*x + e) - (a*c^4*f*cos(f*x + e)^3 -
4*a*c^4*f*cos(f*x + e))*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+B*sin(f*x+e))/(a+a*sin(f*x+e))/(c-c*sin(f*x+e))**4,x)

[Out]

Timed out

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Giac [A]  time = 1.19567, size = 320, normalized size = 2.25 \begin{align*} -\frac{\frac{35 \,{\left (A - B\right )}}{a c^{4}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}} + \frac{525 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} + 35 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 1960 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 280 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 4025 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 665 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 4480 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 1120 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 3143 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 791 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1176 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 392 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 243 \, A - 51 \, B}{a c^{4}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}^{7}}}{280 \, 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))/(c-c*sin(f*x+e))^4,x, algorithm="giac")

[Out]

-1/280*(35*(A - B)/(a*c^4*(tan(1/2*f*x + 1/2*e) + 1)) + (525*A*tan(1/2*f*x + 1/2*e)^6 + 35*B*tan(1/2*f*x + 1/2
*e)^6 - 1960*A*tan(1/2*f*x + 1/2*e)^5 + 280*B*tan(1/2*f*x + 1/2*e)^5 + 4025*A*tan(1/2*f*x + 1/2*e)^4 - 665*B*t
an(1/2*f*x + 1/2*e)^4 - 4480*A*tan(1/2*f*x + 1/2*e)^3 + 1120*B*tan(1/2*f*x + 1/2*e)^3 + 3143*A*tan(1/2*f*x + 1
/2*e)^2 - 791*B*tan(1/2*f*x + 1/2*e)^2 - 1176*A*tan(1/2*f*x + 1/2*e) + 392*B*tan(1/2*f*x + 1/2*e) + 243*A - 51
*B)/(a*c^4*(tan(1/2*f*x + 1/2*e) - 1)^7))/f