3.168 \(\int \sin (e+f x) (a+b \sin (e+f x))^3 \, dx\)

Optimal. Leaf size=121 \[ -\frac {a \left (a^2+4 b^2\right ) \cos (e+f x)}{2 f}-\frac {b \left (2 a^2+3 b^2\right ) \sin (e+f x) \cos (e+f x)}{8 f}+\frac {3}{8} b x \left (4 a^2+b^2\right )-\frac {\cos (e+f x) (a+b \sin (e+f x))^3}{4 f}-\frac {a \cos (e+f x) (a+b \sin (e+f x))^2}{4 f} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2753, 2734} \[ -\frac {a \left (a^2+4 b^2\right ) \cos (e+f x)}{2 f}-\frac {b \left (2 a^2+3 b^2\right ) \sin (e+f x) \cos (e+f x)}{8 f}+\frac {3}{8} b x \left (4 a^2+b^2\right )-\frac {\cos (e+f x) (a+b \sin (e+f x))^3}{4 f}-\frac {a \cos (e+f x) (a+b \sin (e+f x))^2}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2734

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

Rule 2753

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

Rubi steps

\begin {align*} \int \sin (e+f x) (a+b \sin (e+f x))^3 \, dx &=-\frac {\cos (e+f x) (a+b \sin (e+f x))^3}{4 f}+\frac {1}{4} \int (3 b+3 a \sin (e+f x)) (a+b \sin (e+f x))^2 \, dx\\ &=-\frac {a \cos (e+f x) (a+b \sin (e+f x))^2}{4 f}-\frac {\cos (e+f x) (a+b \sin (e+f x))^3}{4 f}+\frac {1}{12} \int (a+b \sin (e+f x)) \left (15 a b+3 \left (2 a^2+3 b^2\right ) \sin (e+f x)\right ) \, dx\\ &=\frac {3}{8} b \left (4 a^2+b^2\right ) x-\frac {a \left (a^2+4 b^2\right ) \cos (e+f x)}{2 f}-\frac {b \left (2 a^2+3 b^2\right ) \cos (e+f x) \sin (e+f x)}{8 f}-\frac {a \cos (e+f x) (a+b \sin (e+f x))^2}{4 f}-\frac {\cos (e+f x) (a+b \sin (e+f x))^3}{4 f}\\ \end {align*}

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Mathematica [A]  time = 0.35, size = 100, normalized size = 0.83 \[ \frac {b \left (-8 \left (3 a^2+b^2\right ) \sin (2 (e+f x))+48 a^2 e+48 a^2 f x+8 a b \cos (3 (e+f x))+b^2 \sin (4 (e+f x))+12 b^2 e+12 b^2 f x\right )-8 a \left (4 a^2+9 b^2\right ) \cos (e+f x)}{32 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*a*(4*a^2 + 9*b^2)*Cos[e + f*x] + b*(48*a^2*e + 12*b^2*e + 48*a^2*f*x + 12*b^2*f*x + 8*a*b*Cos[3*(e + f*x)]
- 8*(3*a^2 + b^2)*Sin[2*(e + f*x)] + b^2*Sin[4*(e + f*x)]))/(32*f)

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fricas [A]  time = 0.52, size = 93, normalized size = 0.77 \[ \frac {8 \, a b^{2} \cos \left (f x + e\right )^{3} + 3 \, {\left (4 \, a^{2} b + b^{3}\right )} f x - 8 \, {\left (a^{3} + 3 \, a b^{2}\right )} \cos \left (f x + e\right ) + {\left (2 \, b^{3} \cos \left (f x + e\right )^{3} - {\left (12 \, a^{2} b + 5 \, b^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.39, size = 116, normalized size = 0.96 \[ \frac {a b^{2} \cos \left (3 \, f x + 3 \, e\right )}{4 \, f} - \frac {3 \, a b^{2} \cos \left (f x + e\right )}{4 \, f} + \frac {b^{3} \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} + \frac {3}{8} \, {\left (4 \, a^{2} b + b^{3}\right )} x - \frac {{\left (2 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (f x + e\right )}{2 \, f} - \frac {{\left (3 \, a^{2} b + b^{3}\right )} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.23, size = 104, normalized size = 0.86 \[ \frac {b^{3} \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-a \,b^{2} \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )+3 a^{2} b \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-a^{3} \cos \left (f x +e \right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.51, size = 97, normalized size = 0.80 \[ \frac {24 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} b + 32 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a b^{2} + {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} b^{3} - 32 \, a^{3} \cos \left (f x + e\right )}{32 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(24*(2*f*x + 2*e - sin(2*f*x + 2*e))*a^2*b + 32*(cos(f*x + e)^3 - 3*cos(f*x + e))*a*b^2 + (12*f*x + 12*e
+ sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*b^3 - 32*a^3*cos(f*x + e))/f

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mupad [B]  time = 8.08, size = 313, normalized size = 2.59 \[ \frac {3\,b\,\mathrm {atan}\left (\frac {3\,b\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (4\,a^2+b^2\right )}{4\,\left (3\,a^2\,b+\frac {3\,b^3}{4}\right )}\right )\,\left (4\,a^2+b^2\right )}{4\,f}-\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (3\,a^2\,b+\frac {3\,b^3}{4}\right )+2\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+4\,a\,b^2+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (6\,a^3+12\,a\,b^2\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (6\,a^3+16\,a\,b^2\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7\,\left (3\,a^2\,b+\frac {3\,b^3}{4}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (3\,a^2\,b+\frac {11\,b^3}{4}\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\left (3\,a^2\,b+\frac {11\,b^3}{4}\right )+2\,a^3}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}-\frac {3\,b\,\left (4\,a^2+b^2\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )-\frac {f\,x}{2}\right )}{4\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*b*atan((3*b*tan(e/2 + (f*x)/2)*(4*a^2 + b^2))/(4*(3*a^2*b + (3*b^3)/4)))*(4*a^2 + b^2))/(4*f) - (tan(e/2 +
(f*x)/2)*(3*a^2*b + (3*b^3)/4) + 2*a^3*tan(e/2 + (f*x)/2)^6 + 4*a*b^2 + tan(e/2 + (f*x)/2)^4*(12*a*b^2 + 6*a^3
) + tan(e/2 + (f*x)/2)^2*(16*a*b^2 + 6*a^3) - tan(e/2 + (f*x)/2)^7*(3*a^2*b + (3*b^3)/4) + tan(e/2 + (f*x)/2)^
3*(3*a^2*b + (11*b^3)/4) - tan(e/2 + (f*x)/2)^5*(3*a^2*b + (11*b^3)/4) + 2*a^3)/(f*(4*tan(e/2 + (f*x)/2)^2 + 6
*tan(e/2 + (f*x)/2)^4 + 4*tan(e/2 + (f*x)/2)^6 + tan(e/2 + (f*x)/2)^8 + 1)) - (3*b*(4*a^2 + b^2)*(atan(tan(e/2
+ (f*x)/2)) - (f*x)/2))/(4*f)

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sympy [A]  time = 1.66, size = 233, normalized size = 1.93 \[ \begin {cases} - \frac {a^{3} \cos {\left (e + f x \right )}}{f} + \frac {3 a^{2} b x \sin ^{2}{\left (e + f x \right )}}{2} + \frac {3 a^{2} b x \cos ^{2}{\left (e + f x \right )}}{2} - \frac {3 a^{2} b \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} - \frac {3 a b^{2} \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {2 a b^{2} \cos ^{3}{\left (e + f x \right )}}{f} + \frac {3 b^{3} x \sin ^{4}{\left (e + f x \right )}}{8} + \frac {3 b^{3} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac {3 b^{3} x \cos ^{4}{\left (e + f x \right )}}{8} - \frac {5 b^{3} \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{8 f} - \frac {3 b^{3} \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} & \text {for}\: f \neq 0 \\x \left (a + b \sin {\relax (e )}\right )^{3} \sin {\relax (e )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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