3.237 \(\int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^4 \, dx\)

Optimal. Leaf size=118 \[ \frac{7 a^2 c^4 \cos ^5(e+f x)}{30 f}+\frac{a^2 \cos ^5(e+f x) \left (c^4-c^4 \sin (e+f x)\right )}{6 f}+\frac{7 a^2 c^4 \sin (e+f x) \cos ^3(e+f x)}{24 f}+\frac{7 a^2 c^4 \sin (e+f x) \cos (e+f x)}{16 f}+\frac{7}{16} a^2 c^4 x \]

[Out]

(7*a^2*c^4*x)/16 + (7*a^2*c^4*Cos[e + f*x]^5)/(30*f) + (7*a^2*c^4*Cos[e + f*x]*Sin[e + f*x])/(16*f) + (7*a^2*c
^4*Cos[e + f*x]^3*Sin[e + f*x])/(24*f) + (a^2*Cos[e + f*x]^5*(c^4 - c^4*Sin[e + f*x]))/(6*f)

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Rubi [A]  time = 0.1447, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{7 a^2 c^4 \cos ^5(e+f x)}{30 f}+\frac{a^2 \cos ^5(e+f x) \left (c^4-c^4 \sin (e+f x)\right )}{6 f}+\frac{7 a^2 c^4 \sin (e+f x) \cos ^3(e+f x)}{24 f}+\frac{7 a^2 c^4 \sin (e+f x) \cos (e+f x)}{16 f}+\frac{7}{16} a^2 c^4 x \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^4,x]

[Out]

(7*a^2*c^4*x)/16 + (7*a^2*c^4*Cos[e + f*x]^5)/(30*f) + (7*a^2*c^4*Cos[e + f*x]*Sin[e + f*x])/(16*f) + (7*a^2*c
^4*Cos[e + f*x]^3*Sin[e + f*x])/(24*f) + (a^2*Cos[e + f*x]^5*(c^4 - c^4*Sin[e + f*x]))/(6*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))^4 \, dx &=\left (a^2 c^2\right ) \int \cos ^4(e+f x) (c-c \sin (e+f x))^2 \, dx\\ &=\frac{a^2 \cos ^5(e+f x) \left (c^4-c^4 \sin (e+f x)\right )}{6 f}+\frac{1}{6} \left (7 a^2 c^3\right ) \int \cos ^4(e+f x) (c-c \sin (e+f x)) \, dx\\ &=\frac{7 a^2 c^4 \cos ^5(e+f x)}{30 f}+\frac{a^2 \cos ^5(e+f x) \left (c^4-c^4 \sin (e+f x)\right )}{6 f}+\frac{1}{6} \left (7 a^2 c^4\right ) \int \cos ^4(e+f x) \, dx\\ &=\frac{7 a^2 c^4 \cos ^5(e+f x)}{30 f}+\frac{7 a^2 c^4 \cos ^3(e+f x) \sin (e+f x)}{24 f}+\frac{a^2 \cos ^5(e+f x) \left (c^4-c^4 \sin (e+f x)\right )}{6 f}+\frac{1}{8} \left (7 a^2 c^4\right ) \int \cos ^2(e+f x) \, dx\\ &=\frac{7 a^2 c^4 \cos ^5(e+f x)}{30 f}+\frac{7 a^2 c^4 \cos (e+f x) \sin (e+f x)}{16 f}+\frac{7 a^2 c^4 \cos ^3(e+f x) \sin (e+f x)}{24 f}+\frac{a^2 \cos ^5(e+f x) \left (c^4-c^4 \sin (e+f x)\right )}{6 f}+\frac{1}{16} \left (7 a^2 c^4\right ) \int 1 \, dx\\ &=\frac{7}{16} a^2 c^4 x+\frac{7 a^2 c^4 \cos ^5(e+f x)}{30 f}+\frac{7 a^2 c^4 \cos (e+f x) \sin (e+f x)}{16 f}+\frac{7 a^2 c^4 \cos ^3(e+f x) \sin (e+f x)}{24 f}+\frac{a^2 \cos ^5(e+f x) \left (c^4-c^4 \sin (e+f x)\right )}{6 f}\\ \end{align*}

Mathematica [A]  time = 0.745638, size = 79, normalized size = 0.67 \[ \frac{a^2 c^4 (255 \sin (2 (e+f x))+15 \sin (4 (e+f x))-5 \sin (6 (e+f x))+240 \cos (e+f x)+120 \cos (3 (e+f x))+24 \cos (5 (e+f x))+420 e+420 f x)}{960 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^4,x]

[Out]

(a^2*c^4*(420*e + 420*f*x + 240*Cos[e + f*x] + 120*Cos[3*(e + f*x)] + 24*Cos[5*(e + f*x)] + 255*Sin[2*(e + f*x
)] + 15*Sin[4*(e + f*x)] - 5*Sin[6*(e + f*x)]))/(960*f)

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Maple [A]  time = 0.022, size = 211, normalized size = 1.8 \begin{align*}{\frac{1}{f} \left ({c}^{4}{a}^{2} \left ( -{\frac{\cos \left ( fx+e \right ) }{6} \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{5}+{\frac{5\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{4}}+{\frac{15\,\sin \left ( fx+e \right ) }{8}} \right ) }+{\frac{5\,fx}{16}}+{\frac{5\,e}{16}} \right ) +{\frac{2\,{c}^{4}{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 ) }-{c}^{4}{a}^{2} \left ( -{\frac{\cos \left ( fx+e \right ) }{4} \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{3}+{\frac{3\,\sin \left ( fx+e \right ) }{2}} \right ) }+{\frac{3\,fx}{8}}+{\frac{3\,e}{8}} \right ) -{\frac{4\,{c}^{4}{a}^{2} \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}-{c}^{4}{a}^{2} \left ( -{\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) +2\,{c}^{4}{a}^{2}\cos \left ( fx+e \right ) +{c}^{4}{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))^4,x)

[Out]

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

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Maxima [A]  time = 1.14419, size = 282, normalized size = 2.39 \begin{align*} \frac{128 \,{\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^{4} + 1280 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{2} c^{4} + 5 \,{\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^{4} - 30 \,{\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^{4} - 240 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} c^{4} + 960 \,{\left (f x + e\right )} a^{2} c^{4} + 1920 \, a^{2} c^{4} \cos \left (f x + e\right )}{960 \, 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))^4,x, algorithm="maxima")

[Out]

1/960*(128*(3*cos(f*x + e)^5 - 10*cos(f*x + e)^3 + 15*cos(f*x + e))*a^2*c^4 + 1280*(cos(f*x + e)^3 - 3*cos(f*x
+ e))*a^2*c^4 + 5*(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^4 -
30*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*a^2*c^4 - 240*(2*f*x + 2*e - sin(2*f*x + 2*e))*a^2
*c^4 + 960*(f*x + e)*a^2*c^4 + 1920*a^2*c^4*cos(f*x + e))/f

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Fricas [A]  time = 1.44834, size = 207, normalized size = 1.75 \begin{align*} \frac{96 \, a^{2} c^{4} \cos \left (f x + e\right )^{5} + 105 \, a^{2} c^{4} f x - 5 \,{\left (8 \, a^{2} c^{4} \cos \left (f x + e\right )^{5} - 14 \, a^{2} c^{4} \cos \left (f x + e\right )^{3} - 21 \, a^{2} c^{4} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{240 \, 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))^4,x, algorithm="fricas")

[Out]

1/240*(96*a^2*c^4*cos(f*x + e)^5 + 105*a^2*c^4*f*x - 5*(8*a^2*c^4*cos(f*x + e)^5 - 14*a^2*c^4*cos(f*x + e)^3 -
21*a^2*c^4*cos(f*x + e))*sin(f*x + e))/f

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Sympy [A]  time = 10.0939, size = 530, normalized size = 4.49 \begin{align*} \begin{cases} \frac{5 a^{2} c^{4} x \sin ^{6}{\left (e + f x \right )}}{16} + \frac{15 a^{2} c^{4} x \sin ^{4}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{16} - \frac{3 a^{2} c^{4} x \sin ^{4}{\left (e + f x \right )}}{8} + \frac{15 a^{2} c^{4} x \sin ^{2}{\left (e + f x \right )} \cos ^{4}{\left (e + f x \right )}}{16} - \frac{3 a^{2} c^{4} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} - \frac{a^{2} c^{4} x \sin ^{2}{\left (e + f x \right )}}{2} + \frac{5 a^{2} c^{4} x \cos ^{6}{\left (e + f x \right )}}{16} - \frac{3 a^{2} c^{4} x \cos ^{4}{\left (e + f x \right )}}{8} - \frac{a^{2} c^{4} x \cos ^{2}{\left (e + f x \right )}}{2} + a^{2} c^{4} x - \frac{11 a^{2} c^{4} \sin ^{5}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{16 f} + \frac{2 a^{2} c^{4} \sin ^{4}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{5 a^{2} c^{4} \sin ^{3}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{6 f} + \frac{5 a^{2} c^{4} \sin ^{3}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{8 f} + \frac{8 a^{2} c^{4} \sin ^{2}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac{4 a^{2} c^{4} \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{5 a^{2} c^{4} \sin{\left (e + f x \right )} \cos ^{5}{\left (e + f x \right )}}{16 f} + \frac{3 a^{2} c^{4} \sin{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} + \frac{a^{2} c^{4} \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} + \frac{16 a^{2} c^{4} \cos ^{5}{\left (e + f x \right )}}{15 f} - \frac{8 a^{2} c^{4} \cos ^{3}{\left (e + f x \right )}}{3 f} + \frac{2 a^{2} c^{4} \cos{\left (e + f x \right )}}{f} & \text{for}\: f \neq 0 \\x \left (a \sin{\left (e \right )} + a\right )^{2} \left (- c \sin{\left (e \right )} + c\right )^{4} & \text{otherwise} \end{cases} \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))**4,x)

[Out]

Piecewise((5*a**2*c**4*x*sin(e + f*x)**6/16 + 15*a**2*c**4*x*sin(e + f*x)**4*cos(e + f*x)**2/16 - 3*a**2*c**4*
x*sin(e + f*x)**4/8 + 15*a**2*c**4*x*sin(e + f*x)**2*cos(e + f*x)**4/16 - 3*a**2*c**4*x*sin(e + f*x)**2*cos(e
+ f*x)**2/4 - a**2*c**4*x*sin(e + f*x)**2/2 + 5*a**2*c**4*x*cos(e + f*x)**6/16 - 3*a**2*c**4*x*cos(e + f*x)**4
/8 - a**2*c**4*x*cos(e + f*x)**2/2 + a**2*c**4*x - 11*a**2*c**4*sin(e + f*x)**5*cos(e + f*x)/(16*f) + 2*a**2*c
**4*sin(e + f*x)**4*cos(e + f*x)/f - 5*a**2*c**4*sin(e + f*x)**3*cos(e + f*x)**3/(6*f) + 5*a**2*c**4*sin(e + f
*x)**3*cos(e + f*x)/(8*f) + 8*a**2*c**4*sin(e + f*x)**2*cos(e + f*x)**3/(3*f) - 4*a**2*c**4*sin(e + f*x)**2*co
s(e + f*x)/f - 5*a**2*c**4*sin(e + f*x)*cos(e + f*x)**5/(16*f) + 3*a**2*c**4*sin(e + f*x)*cos(e + f*x)**3/(8*f
) + a**2*c**4*sin(e + f*x)*cos(e + f*x)/(2*f) + 16*a**2*c**4*cos(e + f*x)**5/(15*f) - 8*a**2*c**4*cos(e + f*x)
**3/(3*f) + 2*a**2*c**4*cos(e + f*x)/f, Ne(f, 0)), (x*(a*sin(e) + a)**2*(-c*sin(e) + c)**4, True))

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Giac [A]  time = 1.66635, size = 180, normalized size = 1.53 \begin{align*} \frac{7}{16} \, a^{2} c^{4} x + \frac{a^{2} c^{4} \cos \left (5 \, f x + 5 \, e\right )}{40 \, f} + \frac{a^{2} c^{4} \cos \left (3 \, f x + 3 \, e\right )}{8 \, f} + \frac{a^{2} c^{4} \cos \left (f x + e\right )}{4 \, f} - \frac{a^{2} c^{4} \sin \left (6 \, f x + 6 \, e\right )}{192 \, f} + \frac{a^{2} c^{4} \sin \left (4 \, f x + 4 \, e\right )}{64 \, f} + \frac{17 \, a^{2} c^{4} \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))^4,x, algorithm="giac")

[Out]

7/16*a^2*c^4*x + 1/40*a^2*c^4*cos(5*f*x + 5*e)/f + 1/8*a^2*c^4*cos(3*f*x + 3*e)/f + 1/4*a^2*c^4*cos(f*x + e)/f
- 1/192*a^2*c^4*sin(6*f*x + 6*e)/f + 1/64*a^2*c^4*sin(4*f*x + 4*e)/f + 17/64*a^2*c^4*sin(2*f*x + 2*e)/f