3.238 \(\int (a+a \sin (e+f x))^2 (c-c \sin (e+f x))^3 \, dx\)

Optimal. Leaf size=85 \[ \frac {a^2 c^3 \cos ^5(e+f x)}{5 f}+\frac {a^2 c^3 \sin (e+f x) \cos ^3(e+f x)}{4 f}+\frac {3 a^2 c^3 \sin (e+f x) \cos (e+f x)}{8 f}+\frac {3}{8} a^2 c^3 x \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2736, 2669, 2635, 8} \[ \frac {a^2 c^3 \cos ^5(e+f x)}{5 f}+\frac {a^2 c^3 \sin (e+f x) \cos ^3(e+f x)}{4 f}+\frac {3 a^2 c^3 \sin (e+f x) \cos (e+f x)}{8 f}+\frac {3}{8} a^2 c^3 x \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^3,x]

[Out]

(3*a^2*c^3*x)/8 + (a^2*c^3*Cos[e + f*x]^5)/(5*f) + (3*a^2*c^3*Cos[e + f*x]*Sin[e + f*x])/(8*f) + (a^2*c^3*Cos[
e + f*x]^3*Sin[e + f*x])/(4*f)

Rule 8

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

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 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 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]))

Rubi steps

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

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Mathematica [A]  time = 1.64, size = 69, normalized size = 0.81 \[ \frac {a^2 c^3 (40 \sin (2 (e+f x))+5 \sin (4 (e+f x))+20 \cos (e+f x)+10 \cos (3 (e+f x))+2 \cos (5 (e+f x))+60 e+60 f x)}{160 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^3,x]

[Out]

(a^2*c^3*(60*e + 60*f*x + 20*Cos[e + f*x] + 10*Cos[3*(e + f*x)] + 2*Cos[5*(e + f*x)] + 40*Sin[2*(e + f*x)] + 5
*Sin[4*(e + f*x)]))/(160*f)

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fricas [A]  time = 0.47, size = 71, normalized size = 0.84 \[ \frac {8 \, a^{2} c^{3} \cos \left (f x + e\right )^{5} + 15 \, a^{2} c^{3} f x + 5 \, {\left (2 \, a^{2} c^{3} \cos \left (f x + e\right )^{3} + 3 \, a^{2} c^{3} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{40 \, f} \]

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))^3,x, algorithm="fricas")

[Out]

1/40*(8*a^2*c^3*cos(f*x + e)^5 + 15*a^2*c^3*f*x + 5*(2*a^2*c^3*cos(f*x + e)^3 + 3*a^2*c^3*cos(f*x + e))*sin(f*
x + e))/f

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giac [A]  time = 0.20, size = 112, normalized size = 1.32 \[ \frac {3}{8} \, a^{2} c^{3} x + \frac {a^{2} c^{3} \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} + \frac {a^{2} c^{3} \cos \left (3 \, f x + 3 \, e\right )}{16 \, f} + \frac {a^{2} c^{3} \cos \left (f x + e\right )}{8 \, f} + \frac {a^{2} c^{3} \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} + \frac {a^{2} c^{3} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \]

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))^3,x, algorithm="giac")

[Out]

3/8*a^2*c^3*x + 1/80*a^2*c^3*cos(5*f*x + 5*e)/f + 1/16*a^2*c^3*cos(3*f*x + 3*e)/f + 1/8*a^2*c^3*cos(f*x + e)/f
+ 1/32*a^2*c^3*sin(4*f*x + 4*e)/f + 1/4*a^2*c^3*sin(2*f*x + 2*e)/f

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maple [B]  time = 0.29, size = 159, normalized size = 1.87 \[ \frac {\frac {c^{3} a^{2} \left (\frac {8}{3}+\sin ^{4}\left (f x +e \right )+\frac {4 \left (\sin ^{2}\left (f x +e \right )\right )}{3}\right ) \cos \left (f x +e \right )}{5}+c^{3} a^{2} \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-\frac {2 c^{3} a^{2} \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}-2 c^{3} a^{2} \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )+c^{3} a^{2} \cos \left (f x +e \right )+a^{2} c^{3} \left (f x +e \right )}{f} \]

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))^3,x)

[Out]

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

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maxima [B]  time = 1.00, size = 158, normalized size = 1.86 \[ \frac {32 \, {\left (3 \, \cos \left (f x + e\right )^{5} - 10 \, \cos \left (f x + e\right )^{3} + 15 \, \cos \left (f x + e\right )\right )} a^{2} c^{3} + 320 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{2} c^{3} + 15 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} c^{3} - 240 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} c^{3} + 480 \, {\left (f x + e\right )} a^{2} c^{3} + 480 \, a^{2} c^{3} \cos \left (f x + e\right )}{480 \, f} \]

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))^3,x, algorithm="maxima")

[Out]

1/480*(32*(3*cos(f*x + e)^5 - 10*cos(f*x + e)^3 + 15*cos(f*x + e))*a^2*c^3 + 320*(cos(f*x + e)^3 - 3*cos(f*x +
e))*a^2*c^3 + 15*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*a^2*c^3 - 240*(2*f*x + 2*e - sin(2*f
*x + 2*e))*a^2*c^3 + 480*(f*x + e)*a^2*c^3 + 480*a^2*c^3*cos(f*x + e))/f

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mupad [B]  time = 10.00, size = 220, normalized size = 2.59 \[ \frac {3\,a^2\,c^3\,x}{8}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8\,\left (\frac {a^2\,c^3\,\left (75\,e+75\,f\,x+80\right )}{40}-\frac {15\,a^2\,c^3\,\left (e+f\,x\right )}{8}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {a^2\,c^3\,\left (150\,e+150\,f\,x+160\right )}{40}-\frac {15\,a^2\,c^3\,\left (e+f\,x\right )}{4}\right )+\frac {a^2\,c^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{2}-\frac {a^2\,c^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7}{2}-\frac {5\,a^2\,c^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9}{4}+\frac {a^2\,c^3\,\left (15\,e+15\,f\,x+16\right )}{40}+\frac {5\,a^2\,c^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{4}-\frac {3\,a^2\,c^3\,\left (e+f\,x\right )}{8}}{f\,{\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(e + f*x))^2*(c - c*sin(e + f*x))^3,x)

[Out]

(3*a^2*c^3*x)/8 + (tan(e/2 + (f*x)/2)^8*((a^2*c^3*(75*e + 75*f*x + 80))/40 - (15*a^2*c^3*(e + f*x))/8) + tan(e
/2 + (f*x)/2)^4*((a^2*c^3*(150*e + 150*f*x + 160))/40 - (15*a^2*c^3*(e + f*x))/4) + (a^2*c^3*tan(e/2 + (f*x)/2
)^3)/2 - (a^2*c^3*tan(e/2 + (f*x)/2)^7)/2 - (5*a^2*c^3*tan(e/2 + (f*x)/2)^9)/4 + (a^2*c^3*(15*e + 15*f*x + 16)
)/40 + (5*a^2*c^3*tan(e/2 + (f*x)/2))/4 - (3*a^2*c^3*(e + f*x))/8)/(f*(tan(e/2 + (f*x)/2)^2 + 1)^5)

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sympy [A]  time = 3.04, size = 340, normalized size = 4.00 \[ \begin {cases} \frac {3 a^{2} c^{3} x \sin ^{4}{\left (e + f x \right )}}{8} + \frac {3 a^{2} c^{3} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} - a^{2} c^{3} x \sin ^{2}{\left (e + f x \right )} + \frac {3 a^{2} c^{3} x \cos ^{4}{\left (e + f x \right )}}{8} - a^{2} c^{3} x \cos ^{2}{\left (e + f x \right )} + a^{2} c^{3} x + \frac {a^{2} c^{3} \sin ^{4}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {5 a^{2} c^{3} \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{8 f} + \frac {4 a^{2} c^{3} \sin ^{2}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac {2 a^{2} c^{3} \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {3 a^{2} c^{3} \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} + \frac {a^{2} c^{3} \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} + \frac {8 a^{2} c^{3} \cos ^{5}{\left (e + f x \right )}}{15 f} - \frac {4 a^{2} c^{3} \cos ^{3}{\left (e + f x \right )}}{3 f} + \frac {a^{2} c^{3} \cos {\left (e + f x \right )}}{f} & \text {for}\: f \neq 0 \\x \left (a \sin {\relax (e )} + a\right )^{2} \left (- c \sin {\relax (e )} + c\right )^{3} & \text {otherwise} \end {cases} \]

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))**3,x)

[Out]

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

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