3.175 \(\int (a+b \sin (e+f x))^4 \, dx\)

Optimal. Leaf size=137 \[ -\frac{a b \left (19 a^2+16 b^2\right ) \cos (e+f x)}{6 f}-\frac{b^2 \left (26 a^2+9 b^2\right ) \sin (e+f x) \cos (e+f x)}{24 f}+\frac{1}{8} x \left (24 a^2 b^2+8 a^4+3 b^4\right )-\frac{b \cos (e+f x) (a+b \sin (e+f x))^3}{4 f}-\frac{7 a b \cos (e+f x) (a+b \sin (e+f x))^2}{12 f} \]

[Out]

((8*a^4 + 24*a^2*b^2 + 3*b^4)*x)/8 - (a*b*(19*a^2 + 16*b^2)*Cos[e + f*x])/(6*f) - (b^2*(26*a^2 + 9*b^2)*Cos[e
+ f*x]*Sin[e + f*x])/(24*f) - (7*a*b*Cos[e + f*x]*(a + b*Sin[e + f*x])^2)/(12*f) - (b*Cos[e + f*x]*(a + b*Sin[
e + f*x])^3)/(4*f)

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Rubi [A]  time = 0.145565, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2656, 2753, 2734} \[ -\frac{a b \left (19 a^2+16 b^2\right ) \cos (e+f x)}{6 f}-\frac{b^2 \left (26 a^2+9 b^2\right ) \sin (e+f x) \cos (e+f x)}{24 f}+\frac{1}{8} x \left (24 a^2 b^2+8 a^4+3 b^4\right )-\frac{b \cos (e+f x) (a+b \sin (e+f x))^3}{4 f}-\frac{7 a b \cos (e+f x) (a+b \sin (e+f x))^2}{12 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[e + f*x])^4,x]

[Out]

((8*a^4 + 24*a^2*b^2 + 3*b^4)*x)/8 - (a*b*(19*a^2 + 16*b^2)*Cos[e + f*x])/(6*f) - (b^2*(26*a^2 + 9*b^2)*Cos[e
+ f*x]*Sin[e + f*x])/(24*f) - (7*a*b*Cos[e + f*x]*(a + b*Sin[e + f*x])^2)/(12*f) - (b*Cos[e + f*x]*(a + b*Sin[
e + f*x])^3)/(4*f)

Rule 2656

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

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]

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]

Rubi steps

\begin{align*} \int (a+b \sin (e+f x))^4 \, dx &=-\frac{b \cos (e+f x) (a+b \sin (e+f x))^3}{4 f}+\frac{1}{4} \int (a+b \sin (e+f x))^2 \left (4 a^2+3 b^2+7 a b \sin (e+f x)\right ) \, dx\\ &=-\frac{7 a b \cos (e+f x) (a+b \sin (e+f x))^2}{12 f}-\frac{b \cos (e+f x) (a+b \sin (e+f x))^3}{4 f}+\frac{1}{12} \int (a+b \sin (e+f x)) \left (a \left (12 a^2+23 b^2\right )+b \left (26 a^2+9 b^2\right ) \sin (e+f x)\right ) \, dx\\ &=\frac{1}{8} \left (8 a^4+24 a^2 b^2+3 b^4\right ) x-\frac{a b \left (19 a^2+16 b^2\right ) \cos (e+f x)}{6 f}-\frac{b^2 \left (26 a^2+9 b^2\right ) \cos (e+f x) \sin (e+f x)}{24 f}-\frac{7 a b \cos (e+f x) (a+b \sin (e+f x))^2}{12 f}-\frac{b \cos (e+f x) (a+b \sin (e+f x))^3}{4 f}\\ \end{align*}

Mathematica [A]  time = 0.358763, size = 106, normalized size = 0.77 \[ \frac{3 \left (4 \left (24 a^2 b^2+8 a^4+3 b^4\right ) (e+f x)-8 \left (6 a^2 b^2+b^4\right ) \sin (2 (e+f x))+b^4 \sin (4 (e+f x))\right )-96 a b \left (4 a^2+3 b^2\right ) \cos (e+f x)+32 a b^3 \cos (3 (e+f x))}{96 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[e + f*x])^4,x]

[Out]

(-96*a*b*(4*a^2 + 3*b^2)*Cos[e + f*x] + 32*a*b^3*Cos[3*(e + f*x)] + 3*(4*(8*a^4 + 24*a^2*b^2 + 3*b^4)*(e + f*x
) - 8*(6*a^2*b^2 + b^4)*Sin[2*(e + f*x)] + b^4*Sin[4*(e + f*x)]))/(96*f)

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Maple [A]  time = 0.023, size = 116, normalized size = 0.9 \begin{align*}{\frac{1}{f} \left ({b}^{4} \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\,a{b}^{3} \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}+6\,{a}^{2}{b}^{2} \left ( -1/2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +1/2\,fx+e/2 \right ) -4\,{a}^{3}b\cos \left ( fx+e \right ) +{a}^{4} \left ( fx+e \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(f*x+e))^4,x)

[Out]

1/f*(b^4*(-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*a*b^3*(2+sin(f*x+e)^2)*cos(f*x+e)+6
*a^2*b^2*(-1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*e)-4*a^3*b*cos(f*x+e)+a^4*(f*x+e))

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Maxima [A]  time = 2.40591, size = 153, normalized size = 1.12 \begin{align*} a^{4} x + \frac{3 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{2} b^{2}}{2 \, f} + \frac{4 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a b^{3}}{3 \, f} + \frac{{\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^{4}}{32 \, f} - \frac{4 \, a^{3} b \cos \left (f x + e\right )}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.78019, size = 244, normalized size = 1.78 \begin{align*} \frac{32 \, a b^{3} \cos \left (f x + e\right )^{3} + 3 \,{\left (8 \, a^{4} + 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} f x - 96 \,{\left (a^{3} b + a b^{3}\right )} \cos \left (f x + e\right ) + 3 \,{\left (2 \, b^{4} \cos \left (f x + e\right )^{3} -{\left (24 \, a^{2} b^{2} + 5 \, b^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{24 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))^4,x, algorithm="fricas")

[Out]

1/24*(32*a*b^3*cos(f*x + e)^3 + 3*(8*a^4 + 24*a^2*b^2 + 3*b^4)*f*x - 96*(a^3*b + a*b^3)*cos(f*x + e) + 3*(2*b^
4*cos(f*x + e)^3 - (24*a^2*b^2 + 5*b^4)*cos(f*x + e))*sin(f*x + e))/f

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Sympy [A]  time = 2.11111, size = 240, normalized size = 1.75 \begin{align*} \begin{cases} a^{4} x - \frac{4 a^{3} b \cos{\left (e + f x \right )}}{f} + 3 a^{2} b^{2} x \sin ^{2}{\left (e + f x \right )} + 3 a^{2} b^{2} x \cos ^{2}{\left (e + f x \right )} - \frac{3 a^{2} b^{2} \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{4 a b^{3} \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{8 a b^{3} \cos ^{3}{\left (e + f x \right )}}{3 f} + \frac{3 b^{4} x \sin ^{4}{\left (e + f x \right )}}{8} + \frac{3 b^{4} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac{3 b^{4} x \cos ^{4}{\left (e + f x \right )}}{8} - \frac{5 b^{4} \sin ^{3}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{8 f} - \frac{3 b^{4} \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{\left (e \right )}\right )^{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))**4,x)

[Out]

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

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Giac [A]  time = 1.5568, size = 151, normalized size = 1.1 \begin{align*} \frac{a b^{3} \cos \left (3 \, f x + 3 \, e\right )}{3 \, f} + \frac{b^{4} \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} + \frac{1}{8} \,{\left (8 \, a^{4} + 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} x - \frac{{\left (4 \, a^{3} b + 3 \, a b^{3}\right )} \cos \left (f x + e\right )}{f} - \frac{{\left (6 \, a^{2} b^{2} + b^{4}\right )} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))^4,x, algorithm="giac")

[Out]

1/3*a*b^3*cos(3*f*x + 3*e)/f + 1/32*b^4*sin(4*f*x + 4*e)/f + 1/8*(8*a^4 + 24*a^2*b^2 + 3*b^4)*x - (4*a^3*b + 3
*a*b^3)*cos(f*x + e)/f - 1/4*(6*a^2*b^2 + b^4)*sin(2*f*x + 2*e)/f