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

Optimal. Leaf size=82 \[ -\frac {5 a^3 c \cos ^3(e+f x)}{12 f}-\frac {c \cos ^3(e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{4 f}+\frac {5 a^3 c \sin (e+f x) \cos (e+f x)}{8 f}+\frac {5}{8} a^3 c x \]

[Out]

5/8*a^3*c*x-5/12*a^3*c*cos(f*x+e)^3/f+5/8*a^3*c*cos(f*x+e)*sin(f*x+e)/f-1/4*c*cos(f*x+e)^3*(a^3+a^3*sin(f*x+e)
)/f

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Rubi [A]  time = 0.10, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {2736, 2678, 2669, 2635, 8} \[ -\frac {5 a^3 c \cos ^3(e+f x)}{12 f}-\frac {c \cos ^3(e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{4 f}+\frac {5 a^3 c \sin (e+f x) \cos (e+f x)}{8 f}+\frac {5}{8} a^3 c x \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a^3*c*x)/8 - (5*a^3*c*Cos[e + f*x]^3)/(12*f) + (5*a^3*c*Cos[e + f*x]*Sin[e + f*x])/(8*f) - (c*Cos[e + f*x]^
3*(a^3 + a^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 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 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))^3 (c-c \sin (e+f x)) \, dx &=(a c) \int \cos ^2(e+f x) (a+a \sin (e+f x))^2 \, dx\\ &=-\frac {c \cos ^3(e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{4 f}+\frac {1}{4} \left (5 a^2 c\right ) \int \cos ^2(e+f x) (a+a \sin (e+f x)) \, dx\\ &=-\frac {5 a^3 c \cos ^3(e+f x)}{12 f}-\frac {c \cos ^3(e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{4 f}+\frac {1}{4} \left (5 a^3 c\right ) \int \cos ^2(e+f x) \, dx\\ &=-\frac {5 a^3 c \cos ^3(e+f x)}{12 f}+\frac {5 a^3 c \cos (e+f x) \sin (e+f x)}{8 f}-\frac {c \cos ^3(e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{4 f}+\frac {1}{8} \left (5 a^3 c\right ) \int 1 \, dx\\ &=\frac {5}{8} a^3 c x-\frac {5 a^3 c \cos ^3(e+f x)}{12 f}+\frac {5 a^3 c \cos (e+f x) \sin (e+f x)}{8 f}-\frac {c \cos ^3(e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{4 f}\\ \end {align*}

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Mathematica [A]  time = 0.38, size = 54, normalized size = 0.66 \[ \frac {a^3 c (24 \sin (2 (e+f x))-3 \sin (4 (e+f x))-48 \cos (e+f x)-16 \cos (3 (e+f x))+60 f x)}{96 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*c*(60*f*x - 48*Cos[e + f*x] - 16*Cos[3*(e + f*x)] + 24*Sin[2*(e + f*x)] - 3*Sin[4*(e + f*x)]))/(96*f)

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fricas [A]  time = 0.48, size = 63, normalized size = 0.77 \[ -\frac {16 \, a^{3} c \cos \left (f x + e\right )^{3} - 15 \, a^{3} c f x + 3 \, {\left (2 \, a^{3} c \cos \left (f x + e\right )^{3} - 5 \, a^{3} c \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{24 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3*(c-c*sin(f*x+e)),x, algorithm="fricas")

[Out]

-1/24*(16*a^3*c*cos(f*x + e)^3 - 15*a^3*c*f*x + 3*(2*a^3*c*cos(f*x + e)^3 - 5*a^3*c*cos(f*x + e))*sin(f*x + e)
)/f

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giac [A]  time = 0.42, size = 81, normalized size = 0.99 \[ \frac {5}{8} \, a^{3} c x - \frac {a^{3} c \cos \left (3 \, f x + 3 \, e\right )}{6 \, f} - \frac {a^{3} c \cos \left (f x + e\right )}{2 \, f} - \frac {a^{3} c \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} + \frac {a^{3} c \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3*(c-c*sin(f*x+e)),x, algorithm="giac")

[Out]

5/8*a^3*c*x - 1/6*a^3*c*cos(3*f*x + 3*e)/f - 1/2*a^3*c*cos(f*x + e)/f - 1/32*a^3*c*sin(4*f*x + 4*e)/f + 1/4*a^
3*c*sin(2*f*x + 2*e)/f

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maple [A]  time = 0.24, size = 89, normalized size = 1.09 \[ \frac {-a^{3} c \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )+\frac {2 a^{3} c \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}-2 a^{3} c \cos \left (f x +e \right )+a^{3} c \left (f x +e \right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.83, size = 86, normalized size = 1.05 \[ -\frac {64 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{3} c + 3 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} a^{3} c - 96 \, {\left (f x + e\right )} a^{3} c + 192 \, a^{3} c \cos \left (f x + e\right )}{96 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3*(c-c*sin(f*x+e)),x, algorithm="maxima")

[Out]

-1/96*(64*(cos(f*x + e)^3 - 3*cos(f*x + e))*a^3*c + 3*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*
a^3*c - 96*(f*x + e)*a^3*c + 192*a^3*c*cos(f*x + e))/f

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mupad [B]  time = 8.81, size = 250, normalized size = 3.05 \[ \frac {5\,a^3\,c\,x}{8}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {a^3\,c\,\left (15\,e+15\,f\,x\right )}{6}-\frac {a^3\,c\,\left (60\,e+60\,f\,x-32\right )}{24}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (\frac {a^3\,c\,\left (15\,e+15\,f\,x\right )}{6}-\frac {a^3\,c\,\left (60\,e+60\,f\,x-96\right )}{24}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {a^3\,c\,\left (15\,e+15\,f\,x\right )}{4}-\frac {a^3\,c\,\left (90\,e+90\,f\,x-96\right )}{24}\right )-\frac {3\,a^3\,c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{4}-\frac {11\,a^3\,c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{4}+\frac {11\,a^3\,c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{4}+\frac {3\,a^3\,c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7}{4}+\frac {a^3\,c\,\left (15\,e+15\,f\,x\right )}{24}-\frac {a^3\,c\,\left (15\,e+15\,f\,x-32\right )}{24}}{f\,{\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*a^3*c*x)/8 - (tan(e/2 + (f*x)/2)^2*((a^3*c*(15*e + 15*f*x))/6 - (a^3*c*(60*e + 60*f*x - 32))/24) + tan(e/2
+ (f*x)/2)^6*((a^3*c*(15*e + 15*f*x))/6 - (a^3*c*(60*e + 60*f*x - 96))/24) + tan(e/2 + (f*x)/2)^4*((a^3*c*(15*
e + 15*f*x))/4 - (a^3*c*(90*e + 90*f*x - 96))/24) - (3*a^3*c*tan(e/2 + (f*x)/2))/4 - (11*a^3*c*tan(e/2 + (f*x)
/2)^3)/4 + (11*a^3*c*tan(e/2 + (f*x)/2)^5)/4 + (3*a^3*c*tan(e/2 + (f*x)/2)^7)/4 + (a^3*c*(15*e + 15*f*x))/24 -
(a^3*c*(15*e + 15*f*x - 32))/24)/(f*(tan(e/2 + (f*x)/2)^2 + 1)^4)

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sympy [A]  time = 1.49, size = 196, normalized size = 2.39 \[ \begin {cases} - \frac {3 a^{3} c x \sin ^{4}{\left (e + f x \right )}}{8} - \frac {3 a^{3} c x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} - \frac {3 a^{3} c x \cos ^{4}{\left (e + f x \right )}}{8} + a^{3} c x + \frac {5 a^{3} c \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{8 f} + \frac {2 a^{3} c \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} + \frac {3 a^{3} c \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} + \frac {4 a^{3} c \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac {2 a^{3} c \cos {\left (e + f x \right )}}{f} & \text {for}\: f \neq 0 \\x \left (a \sin {\relax (e )} + a\right )^{3} \left (- c \sin {\relax (e )} + c\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**3*(c-c*sin(f*x+e)),x)

[Out]

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

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