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

Optimal. Leaf size=142 \[ -\frac{a (2 A-5 B) \cos (e+f x)}{105 f \left (c^4-c^4 \sin (e+f x)\right )}-\frac{a (2 A-5 B) \cos (e+f x)}{105 f \left (c^2-c^2 \sin (e+f x)\right )^2}-\frac{a (A+15 B) \cos (e+f x)}{35 c f (c-c \sin (e+f x))^3}+\frac{2 a (A+B) \cos (e+f x)}{7 f (c-c \sin (e+f x))^4} \]

[Out]

(2*a*(A + B)*Cos[e + f*x])/(7*f*(c - c*Sin[e + f*x])^4) - (a*(A + 15*B)*Cos[e + f*x])/(35*c*f*(c - c*Sin[e + f
*x])^3) - (a*(2*A - 5*B)*Cos[e + f*x])/(105*f*(c^2 - c^2*Sin[e + f*x])^2) - (a*(2*A - 5*B)*Cos[e + f*x])/(105*
f*(c^4 - c^4*Sin[e + f*x]))

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Rubi [A]  time = 0.285445, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {2967, 2857, 2750, 2650, 2648} \[ -\frac{a (2 A-5 B) \cos (e+f x)}{105 f \left (c^4-c^4 \sin (e+f x)\right )}-\frac{a (2 A-5 B) \cos (e+f x)}{105 f \left (c^2-c^2 \sin (e+f x)\right )^2}-\frac{a (A+15 B) \cos (e+f x)}{35 c f (c-c \sin (e+f x))^3}+\frac{2 a (A+B) \cos (e+f x)}{7 f (c-c \sin (e+f x))^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(A + B)*Cos[e + f*x])/(7*f*(c - c*Sin[e + f*x])^4) - (a*(A + 15*B)*Cos[e + f*x])/(35*c*f*(c - c*Sin[e + f
*x])^3) - (a*(2*A - 5*B)*Cos[e + f*x])/(105*f*(c^2 - c^2*Sin[e + f*x])^2) - (a*(2*A - 5*B)*Cos[e + f*x])/(105*
f*(c^4 - c^4*Sin[e + f*x]))

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 2857

Int[cos[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_
)]), x_Symbol] :> Simp[(2*(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(2*m + 3)), x] + Dist[
1/(b^3*(2*m + 3)), Int[(a + b*Sin[e + f*x])^(m + 2)*(b*c + 2*a*d*(m + 1) - b*d*(2*m + 3)*Sin[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -3/2]

Rule 2750

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((b
*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)
), Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 -
b^2, 0] && LtQ[m, -2^(-1)]

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4} \, dx &=(a c) \int \frac{\cos ^2(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^5} \, dx\\ &=\frac{2 a (A+B) \cos (e+f x)}{7 f (c-c \sin (e+f x))^4}+\frac{a \int \frac{-A c-8 B c-7 B c \sin (e+f x)}{(c-c \sin (e+f x))^3} \, dx}{7 c^2}\\ &=\frac{2 a (A+B) \cos (e+f x)}{7 f (c-c \sin (e+f x))^4}-\frac{a (A+15 B) \cos (e+f x)}{35 c f (c-c \sin (e+f x))^3}-\frac{(a (2 A-5 B)) \int \frac{1}{(c-c \sin (e+f x))^2} \, dx}{35 c^2}\\ &=\frac{2 a (A+B) \cos (e+f x)}{7 f (c-c \sin (e+f x))^4}-\frac{a (A+15 B) \cos (e+f x)}{35 c f (c-c \sin (e+f x))^3}-\frac{a (2 A-5 B) \cos (e+f x)}{105 f \left (c^2-c^2 \sin (e+f x)\right )^2}-\frac{(a (2 A-5 B)) \int \frac{1}{c-c \sin (e+f x)} \, dx}{105 c^3}\\ &=\frac{2 a (A+B) \cos (e+f x)}{7 f (c-c \sin (e+f x))^4}-\frac{a (A+15 B) \cos (e+f x)}{35 c f (c-c \sin (e+f x))^3}-\frac{a (2 A-5 B) \cos (e+f x)}{105 f \left (c^2-c^2 \sin (e+f x)\right )^2}-\frac{a (2 A-5 B) \cos (e+f x)}{105 f \left (c^4-c^4 \sin (e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.825032, size = 174, normalized size = 1.23 \[ \frac{a \left (35 (4 A-B) \cos \left (e+\frac{f x}{2}\right )+14 A \sin \left (2 e+\frac{5 f x}{2}\right )-42 A \cos \left (e+\frac{3 f x}{2}\right )+2 A \cos \left (3 e+\frac{7 f x}{2}\right )+70 A \sin \left (\frac{f x}{2}\right )+105 B \sin \left (2 e+\frac{3 f x}{2}\right )-35 B \sin \left (2 e+\frac{5 f x}{2}\right )-5 B \cos \left (3 e+\frac{7 f x}{2}\right )+140 B \sin \left (\frac{f x}{2}\right )\right )}{420 c^4 f \left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^7} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(35*(4*A - B)*Cos[e + (f*x)/2] - 42*A*Cos[e + (3*f*x)/2] + 2*A*Cos[3*e + (7*f*x)/2] - 5*B*Cos[3*e + (7*f*x)
/2] + 70*A*Sin[(f*x)/2] + 140*B*Sin[(f*x)/2] + 105*B*Sin[2*e + (3*f*x)/2] + 14*A*Sin[2*e + (5*f*x)/2] - 35*B*S
in[2*e + (5*f*x)/2]))/(420*c^4*f*(Cos[e/2] - Sin[e/2])*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7)

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Maple [A]  time = 0.119, size = 159, normalized size = 1.1 \begin{align*} 2\,{\frac{a}{f{c}^{4}} \left ( -1/6\,{\frac{48\,A+48\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{6}}}-1/4\,{\frac{56\,A+40\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{4}}}-1/5\,{\frac{68\,A+60\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{5}}}-1/2\,{\frac{8\,A+2\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{2}}}-1/3\,{\frac{28\,A+14\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{3}}}-1/7\,{\frac{16\,A+16\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{7}}}-{\frac{A}{\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+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^4,x)

[Out]

2/f*a/c^4*(-1/6*(48*A+48*B)/(tan(1/2*f*x+1/2*e)-1)^6-1/4*(56*A+40*B)/(tan(1/2*f*x+1/2*e)-1)^4-1/5*(68*A+60*B)/
(tan(1/2*f*x+1/2*e)-1)^5-1/2*(8*A+2*B)/(tan(1/2*f*x+1/2*e)-1)^2-1/3*(28*A+14*B)/(tan(1/2*f*x+1/2*e)-1)^3-1/7*(
16*A+16*B)/(tan(1/2*f*x+1/2*e)-1)^7-A/(tan(1/2*f*x+1/2*e)-1))

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Maxima [B]  time = 1.09437, size = 1458, normalized size = 10.27 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^4,x, algorithm="maxima")

[Out]

2/105*(A*a*(91*sin(f*x + e)/(cos(f*x + e) + 1) - 168*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 280*sin(f*x + e)^3/
(cos(f*x + e) + 1)^3 - 175*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 105*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 13)
/(c^4 - 7*c^4*sin(f*x + e)/(cos(f*x + e) + 1) + 21*c^4*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 35*c^4*sin(f*x +
e)^3/(cos(f*x + e) + 1)^3 + 35*c^4*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 21*c^4*sin(f*x + e)^5/(cos(f*x + e) +
1)^5 + 7*c^4*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - c^4*sin(f*x + e)^7/(cos(f*x + e) + 1)^7) + B*a*(91*sin(f*x
+ e)/(cos(f*x + e) + 1) - 168*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 280*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 -
175*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 105*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 13)/(c^4 - 7*c^4*sin(f*x
+ e)/(cos(f*x + e) + 1) + 21*c^4*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 35*c^4*sin(f*x + e)^3/(cos(f*x + e) + 1
)^3 + 35*c^4*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 21*c^4*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 7*c^4*sin(f*x
+ e)^6/(cos(f*x + e) + 1)^6 - c^4*sin(f*x + e)^7/(cos(f*x + e) + 1)^7) - 3*A*a*(49*sin(f*x + e)/(cos(f*x + e)
+ 1) - 147*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 210*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 210*sin(f*x + e)^4/
(cos(f*x + e) + 1)^4 + 105*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 35*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 12)/
(c^4 - 7*c^4*sin(f*x + e)/(cos(f*x + e) + 1) + 21*c^4*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 35*c^4*sin(f*x + e
)^3/(cos(f*x + e) + 1)^3 + 35*c^4*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 21*c^4*sin(f*x + e)^5/(cos(f*x + e) +
1)^5 + 7*c^4*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - c^4*sin(f*x + e)^7/(cos(f*x + e) + 1)^7) - 4*B*a*(14*sin(f*
x + e)/(cos(f*x + e) + 1) - 42*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 35*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 -
35*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 2)/(c^4 - 7*c^4*sin(f*x + e)/(cos(f*x + e) + 1) + 21*c^4*sin(f*x + e)
^2/(cos(f*x + e) + 1)^2 - 35*c^4*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 35*c^4*sin(f*x + e)^4/(cos(f*x + e) + 1
)^4 - 21*c^4*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 7*c^4*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - c^4*sin(f*x + e
)^7/(cos(f*x + e) + 1)^7))/f

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Fricas [A]  time = 1.3599, size = 637, normalized size = 4.49 \begin{align*} \frac{{\left (2 \, A - 5 \, B\right )} a \cos \left (f x + e\right )^{4} + 4 \,{\left (2 \, A - 5 \, B\right )} a \cos \left (f x + e\right )^{3} - 3 \,{\left (3 \, A + 10 \, B\right )} a \cos \left (f x + e\right )^{2} + 15 \,{\left (A + B\right )} a \cos \left (f x + e\right ) + 30 \,{\left (A + B\right )} a -{\left ({\left (2 \, A - 5 \, B\right )} a \cos \left (f x + e\right )^{3} - 3 \,{\left (2 \, A - 5 \, B\right )} a \cos \left (f x + e\right )^{2} - 15 \,{\left (A + B\right )} a \cos \left (f x + e\right ) - 30 \,{\left (A + B\right )} a\right )} \sin \left (f x + e\right )}{105 \,{\left (c^{4} f \cos \left (f x + e\right )^{4} - 3 \, c^{4} f \cos \left (f x + e\right )^{3} - 8 \, c^{4} f \cos \left (f x + e\right )^{2} + 4 \, c^{4} f \cos \left (f x + e\right ) + 8 \, c^{4} f +{\left (c^{4} f \cos \left (f x + e\right )^{3} + 4 \, c^{4} f \cos \left (f x + e\right )^{2} - 4 \, c^{4} f \cos \left (f x + e\right ) - 8 \, c^{4} f\right )} \sin \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^4,x, algorithm="fricas")

[Out]

1/105*((2*A - 5*B)*a*cos(f*x + e)^4 + 4*(2*A - 5*B)*a*cos(f*x + e)^3 - 3*(3*A + 10*B)*a*cos(f*x + e)^2 + 15*(A
+ B)*a*cos(f*x + e) + 30*(A + B)*a - ((2*A - 5*B)*a*cos(f*x + e)^3 - 3*(2*A - 5*B)*a*cos(f*x + e)^2 - 15*(A +
B)*a*cos(f*x + e) - 30*(A + B)*a)*sin(f*x + e))/(c^4*f*cos(f*x + e)^4 - 3*c^4*f*cos(f*x + e)^3 - 8*c^4*f*cos(
f*x + e)^2 + 4*c^4*f*cos(f*x + e) + 8*c^4*f + (c^4*f*cos(f*x + e)^3 + 4*c^4*f*cos(f*x + e)^2 - 4*c^4*f*cos(f*x
+ e) - 8*c^4*f)*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+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**4,x)

[Out]

Timed out

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Giac [A]  time = 1.19253, size = 252, normalized size = 1.77 \begin{align*} -\frac{2 \,{\left (105 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 210 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 105 \, B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 455 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 35 \, B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 350 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 140 \, B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 273 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 56 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 35 \, B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 23 \, A a - 5 \, B a\right )}}{105 \, c^{4} f{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^4,x, algorithm="giac")

[Out]

-2/105*(105*A*a*tan(1/2*f*x + 1/2*e)^6 - 210*A*a*tan(1/2*f*x + 1/2*e)^5 + 105*B*a*tan(1/2*f*x + 1/2*e)^5 + 455
*A*a*tan(1/2*f*x + 1/2*e)^4 - 35*B*a*tan(1/2*f*x + 1/2*e)^4 - 350*A*a*tan(1/2*f*x + 1/2*e)^3 + 140*B*a*tan(1/2
*f*x + 1/2*e)^3 + 273*A*a*tan(1/2*f*x + 1/2*e)^2 - 56*A*a*tan(1/2*f*x + 1/2*e) + 35*B*a*tan(1/2*f*x + 1/2*e) +
23*A*a - 5*B*a)/(c^4*f*(tan(1/2*f*x + 1/2*e) - 1)^7)