3.236 \(\int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^5 \, dx\)

Optimal. Leaf size=152 \[ \frac{3 a^2 c^5 \cos ^5(e+f x)}{10 f}+\frac{3 a^2 c^5 \sin (e+f x) \cos ^3(e+f x)}{8 f}+\frac{a^2 c^3 \cos ^5(e+f x) (c-c \sin (e+f x))^2}{7 f}+\frac{3 a^2 \cos ^5(e+f x) \left (c^5-c^5 \sin (e+f x)\right )}{14 f}+\frac{9 a^2 c^5 \sin (e+f x) \cos (e+f x)}{16 f}+\frac{9}{16} a^2 c^5 x \]

[Out]

(9*a^2*c^5*x)/16 + (3*a^2*c^5*Cos[e + f*x]^5)/(10*f) + (9*a^2*c^5*Cos[e + f*x]*Sin[e + f*x])/(16*f) + (3*a^2*c
^5*Cos[e + f*x]^3*Sin[e + f*x])/(8*f) + (a^2*c^3*Cos[e + f*x]^5*(c - c*Sin[e + f*x])^2)/(7*f) + (3*a^2*Cos[e +
f*x]^5*(c^5 - c^5*Sin[e + f*x]))/(14*f)

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Rubi [A]  time = 0.196606, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {2736, 2678, 2669, 2635, 8} \[ \frac{3 a^2 c^5 \cos ^5(e+f x)}{10 f}+\frac{3 a^2 c^5 \sin (e+f x) \cos ^3(e+f x)}{8 f}+\frac{a^2 c^3 \cos ^5(e+f x) (c-c \sin (e+f x))^2}{7 f}+\frac{3 a^2 \cos ^5(e+f x) \left (c^5-c^5 \sin (e+f x)\right )}{14 f}+\frac{9 a^2 c^5 \sin (e+f x) \cos (e+f x)}{16 f}+\frac{9}{16} a^2 c^5 x \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^5,x]

[Out]

(9*a^2*c^5*x)/16 + (3*a^2*c^5*Cos[e + f*x]^5)/(10*f) + (9*a^2*c^5*Cos[e + f*x]*Sin[e + f*x])/(16*f) + (3*a^2*c
^5*Cos[e + f*x]^3*Sin[e + f*x])/(8*f) + (a^2*c^3*Cos[e + f*x]^5*(c - c*Sin[e + f*x])^2)/(7*f) + (3*a^2*Cos[e +
f*x]^5*(c^5 - c^5*Sin[e + f*x]))/(14*f)

Rule 2736

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0,
n, m] || LtQ[m, n, 0]))

Rule 2678

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 - 1))/(f*g*(m + p)), x] + Dist[(a*(2*m + p - 1))/(m + p), 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]
&& GtQ[m, 0] && NeQ[m + p, 0] && IntegersQ[2*m, 2*p]

Rule 2669

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b*(g*Cos[
e + f*x])^(p + 1))/(f*g*(p + 1)), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x]
&& (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

Rule 2635

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

Rule 8

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

Rubi steps

\begin{align*} \int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^5 \, dx &=\left (a^2 c^2\right ) \int \cos ^4(e+f x) (c-c \sin (e+f x))^3 \, dx\\ &=\frac{a^2 c^3 \cos ^5(e+f x) (c-c \sin (e+f x))^2}{7 f}+\frac{1}{7} \left (9 a^2 c^3\right ) \int \cos ^4(e+f x) (c-c \sin (e+f x))^2 \, dx\\ &=\frac{a^2 c^3 \cos ^5(e+f x) (c-c \sin (e+f x))^2}{7 f}+\frac{3 a^2 \cos ^5(e+f x) \left (c^5-c^5 \sin (e+f x)\right )}{14 f}+\frac{1}{2} \left (3 a^2 c^4\right ) \int \cos ^4(e+f x) (c-c \sin (e+f x)) \, dx\\ &=\frac{3 a^2 c^5 \cos ^5(e+f x)}{10 f}+\frac{a^2 c^3 \cos ^5(e+f x) (c-c \sin (e+f x))^2}{7 f}+\frac{3 a^2 \cos ^5(e+f x) \left (c^5-c^5 \sin (e+f x)\right )}{14 f}+\frac{1}{2} \left (3 a^2 c^5\right ) \int \cos ^4(e+f x) \, dx\\ &=\frac{3 a^2 c^5 \cos ^5(e+f x)}{10 f}+\frac{3 a^2 c^5 \cos ^3(e+f x) \sin (e+f x)}{8 f}+\frac{a^2 c^3 \cos ^5(e+f x) (c-c \sin (e+f x))^2}{7 f}+\frac{3 a^2 \cos ^5(e+f x) \left (c^5-c^5 \sin (e+f x)\right )}{14 f}+\frac{1}{8} \left (9 a^2 c^5\right ) \int \cos ^2(e+f x) \, dx\\ &=\frac{3 a^2 c^5 \cos ^5(e+f x)}{10 f}+\frac{9 a^2 c^5 \cos (e+f x) \sin (e+f x)}{16 f}+\frac{3 a^2 c^5 \cos ^3(e+f x) \sin (e+f x)}{8 f}+\frac{a^2 c^3 \cos ^5(e+f x) (c-c \sin (e+f x))^2}{7 f}+\frac{3 a^2 \cos ^5(e+f x) \left (c^5-c^5 \sin (e+f x)\right )}{14 f}+\frac{1}{16} \left (9 a^2 c^5\right ) \int 1 \, dx\\ &=\frac{9}{16} a^2 c^5 x+\frac{3 a^2 c^5 \cos ^5(e+f x)}{10 f}+\frac{9 a^2 c^5 \cos (e+f x) \sin (e+f x)}{16 f}+\frac{3 a^2 c^5 \cos ^3(e+f x) \sin (e+f x)}{8 f}+\frac{a^2 c^3 \cos ^5(e+f x) (c-c \sin (e+f x))^2}{7 f}+\frac{3 a^2 \cos ^5(e+f x) \left (c^5-c^5 \sin (e+f x)\right )}{14 f}\\ \end{align*}

Mathematica [A]  time = 1.12437, size = 89, normalized size = 0.59 \[ \frac{a^2 c^5 (665 \sin (2 (e+f x))-35 \sin (4 (e+f x))-35 \sin (6 (e+f x))+945 \cos (e+f x)+455 \cos (3 (e+f x))+77 \cos (5 (e+f x))-5 \cos (7 (e+f x))+1260 e+1260 f x)}{2240 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^5,x]

[Out]

(a^2*c^5*(1260*e + 1260*f*x + 945*Cos[e + f*x] + 455*Cos[3*(e + f*x)] + 77*Cos[5*(e + f*x)] - 5*Cos[7*(e + f*x
)] + 665*Sin[2*(e + f*x)] - 35*Sin[4*(e + f*x)] - 35*Sin[6*(e + f*x)]))/(2240*f)

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Maple [A]  time = 0.022, size = 255, normalized size = 1.7 \begin{align*}{\frac{1}{f} \left ({\frac{{c}^{5}{a}^{2}\cos \left ( fx+e \right ) }{7} \left ({\frac{16}{5}}+ \left ( \sin \left ( fx+e \right ) \right ) ^{6}+{\frac{6\, \left ( \sin \left ( fx+e \right ) \right ) ^{4}}{5}}+{\frac{8\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{5}} \right ) }+3\,{c}^{5}{a}^{2} \left ( -1/6\, \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{5}+5/4\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}+{\frac{15\,\sin \left ( fx+e \right ) }{8}} \right ) \cos \left ( fx+e \right ) +{\frac{5\,fx}{16}}+{\frac{5\,e}{16}} \right ) +{\frac{{c}^{5}{a}^{2}\cos \left ( fx+e \right ) }{5} \left ({\frac{8}{3}}+ \left ( \sin \left ( fx+e \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{3}} \right ) }-5\,{c}^{5}{a}^{2} \left ( -1/4\, \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{3}+3/2\,\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) +3/8\,fx+3/8\,e \right ) -{\frac{5\,{c}^{5}{a}^{2} \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}+{c}^{5}{a}^{2} \left ( -{\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) +3\,{c}^{5}{a}^{2}\cos \left ( fx+e \right ) +{c}^{5}{a}^{2} \left ( fx+e \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^2*(c-c*sin(f*x+e))^5,x)

[Out]

1/f*(1/7*c^5*a^2*(16/5+sin(f*x+e)^6+6/5*sin(f*x+e)^4+8/5*sin(f*x+e)^2)*cos(f*x+e)+3*c^5*a^2*(-1/6*(sin(f*x+e)^
5+5/4*sin(f*x+e)^3+15/8*sin(f*x+e))*cos(f*x+e)+5/16*f*x+5/16*e)+1/5*c^5*a^2*(8/3+sin(f*x+e)^4+4/3*sin(f*x+e)^2
)*cos(f*x+e)-5*c^5*a^2*(-1/4*(sin(f*x+e)^3+3/2*sin(f*x+e))*cos(f*x+e)+3/8*f*x+3/8*e)-5/3*c^5*a^2*(2+sin(f*x+e)
^2)*cos(f*x+e)+c^5*a^2*(-1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*e)+3*c^5*a^2*cos(f*x+e)+c^5*a^2*(f*x+e))

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Maxima [A]  time = 1.18755, size = 346, normalized size = 2.28 \begin{align*} -\frac{192 \,{\left (5 \, \cos \left (f x + e\right )^{7} - 21 \, \cos \left (f x + e\right )^{5} + 35 \, \cos \left (f x + e\right )^{3} - 35 \, \cos \left (f x + e\right )\right )} a^{2} c^{5} - 448 \,{\left (3 \, \cos \left (f x + e\right )^{5} - 10 \, \cos \left (f x + e\right )^{3} + 15 \, \cos \left (f x + e\right )\right )} a^{2} c^{5} - 11200 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{2} c^{5} - 105 \,{\left (4 \, \sin \left (2 \, f x + 2 \, e\right )^{3} + 60 \, f x + 60 \, e + 9 \, \sin \left (4 \, f x + 4 \, e\right ) - 48 \, \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} c^{5} + 1050 \,{\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} c^{5} - 1680 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} c^{5} - 6720 \,{\left (f x + e\right )} a^{2} c^{5} - 20160 \, a^{2} c^{5} \cos \left (f x + e\right )}{6720 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2*(c-c*sin(f*x+e))^5,x, algorithm="maxima")

[Out]

-1/6720*(192*(5*cos(f*x + e)^7 - 21*cos(f*x + e)^5 + 35*cos(f*x + e)^3 - 35*cos(f*x + e))*a^2*c^5 - 448*(3*cos
(f*x + e)^5 - 10*cos(f*x + e)^3 + 15*cos(f*x + e))*a^2*c^5 - 11200*(cos(f*x + e)^3 - 3*cos(f*x + e))*a^2*c^5 -
105*(4*sin(2*f*x + 2*e)^3 + 60*f*x + 60*e + 9*sin(4*f*x + 4*e) - 48*sin(2*f*x + 2*e))*a^2*c^5 + 1050*(12*f*x
+ 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*a^2*c^5 - 1680*(2*f*x + 2*e - sin(2*f*x + 2*e))*a^2*c^5 - 6720
*(f*x + e)*a^2*c^5 - 20160*a^2*c^5*cos(f*x + e))/f

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Fricas [A]  time = 1.45072, size = 246, normalized size = 1.62 \begin{align*} -\frac{80 \, a^{2} c^{5} \cos \left (f x + e\right )^{7} - 448 \, a^{2} c^{5} \cos \left (f x + e\right )^{5} - 315 \, a^{2} c^{5} f x + 35 \,{\left (8 \, a^{2} c^{5} \cos \left (f x + e\right )^{5} - 6 \, a^{2} c^{5} \cos \left (f x + e\right )^{3} - 9 \, a^{2} c^{5} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{560 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2*(c-c*sin(f*x+e))^5,x, algorithm="fricas")

[Out]

-1/560*(80*a^2*c^5*cos(f*x + e)^7 - 448*a^2*c^5*cos(f*x + e)^5 - 315*a^2*c^5*f*x + 35*(8*a^2*c^5*cos(f*x + e)^
5 - 6*a^2*c^5*cos(f*x + e)^3 - 9*a^2*c^5*cos(f*x + e))*sin(f*x + e))/f

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Sympy [A]  time = 19.7392, size = 629, normalized size = 4.14 \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))**2*(c-c*sin(f*x+e))**5,x)

[Out]

Piecewise((15*a**2*c**5*x*sin(e + f*x)**6/16 + 45*a**2*c**5*x*sin(e + f*x)**4*cos(e + f*x)**2/16 - 15*a**2*c**
5*x*sin(e + f*x)**4/8 + 45*a**2*c**5*x*sin(e + f*x)**2*cos(e + f*x)**4/16 - 15*a**2*c**5*x*sin(e + f*x)**2*cos
(e + f*x)**2/4 + a**2*c**5*x*sin(e + f*x)**2/2 + 15*a**2*c**5*x*cos(e + f*x)**6/16 - 15*a**2*c**5*x*cos(e + f*
x)**4/8 + a**2*c**5*x*cos(e + f*x)**2/2 + a**2*c**5*x + a**2*c**5*sin(e + f*x)**6*cos(e + f*x)/f - 33*a**2*c**
5*sin(e + f*x)**5*cos(e + f*x)/(16*f) + 2*a**2*c**5*sin(e + f*x)**4*cos(e + f*x)**3/f + a**2*c**5*sin(e + f*x)
**4*cos(e + f*x)/f - 5*a**2*c**5*sin(e + f*x)**3*cos(e + f*x)**3/(2*f) + 25*a**2*c**5*sin(e + f*x)**3*cos(e +
f*x)/(8*f) + 8*a**2*c**5*sin(e + f*x)**2*cos(e + f*x)**5/(5*f) + 4*a**2*c**5*sin(e + f*x)**2*cos(e + f*x)**3/(
3*f) - 5*a**2*c**5*sin(e + f*x)**2*cos(e + f*x)/f - 15*a**2*c**5*sin(e + f*x)*cos(e + f*x)**5/(16*f) + 15*a**2
*c**5*sin(e + f*x)*cos(e + f*x)**3/(8*f) - a**2*c**5*sin(e + f*x)*cos(e + f*x)/(2*f) + 16*a**2*c**5*cos(e + f*
x)**7/(35*f) + 8*a**2*c**5*cos(e + f*x)**5/(15*f) - 10*a**2*c**5*cos(e + f*x)**3/(3*f) + 3*a**2*c**5*cos(e + f
*x)/f, Ne(f, 0)), (x*(a*sin(e) + a)**2*(-c*sin(e) + c)**5, True))

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Giac [A]  time = 1.65352, size = 208, normalized size = 1.37 \begin{align*} \frac{9}{16} \, a^{2} c^{5} x - \frac{a^{2} c^{5} \cos \left (7 \, f x + 7 \, e\right )}{448 \, f} + \frac{11 \, a^{2} c^{5} \cos \left (5 \, f x + 5 \, e\right )}{320 \, f} + \frac{13 \, a^{2} c^{5} \cos \left (3 \, f x + 3 \, e\right )}{64 \, f} + \frac{27 \, a^{2} c^{5} \cos \left (f x + e\right )}{64 \, f} - \frac{a^{2} c^{5} \sin \left (6 \, f x + 6 \, e\right )}{64 \, f} - \frac{a^{2} c^{5} \sin \left (4 \, f x + 4 \, e\right )}{64 \, f} + \frac{19 \, a^{2} c^{5} \sin \left (2 \, f x + 2 \, e\right )}{64 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2*(c-c*sin(f*x+e))^5,x, algorithm="giac")

[Out]

9/16*a^2*c^5*x - 1/448*a^2*c^5*cos(7*f*x + 7*e)/f + 11/320*a^2*c^5*cos(5*f*x + 5*e)/f + 13/64*a^2*c^5*cos(3*f*
x + 3*e)/f + 27/64*a^2*c^5*cos(f*x + e)/f - 1/64*a^2*c^5*sin(6*f*x + 6*e)/f - 1/64*a^2*c^5*sin(4*f*x + 4*e)/f
+ 19/64*a^2*c^5*sin(2*f*x + 2*e)/f