3.167 \(\int \sin ^2(e+f x) (a+b \sin (e+f x))^3 \, dx\)

Optimal. Leaf size=160 \[ \frac{b \left (15 a^2+4 b^2\right ) \cos ^3(e+f x)}{15 f}-\frac{b \left (15 a^2+4 b^2\right ) \cos (e+f x)}{5 f}-\frac{a \left (4 a^2+9 b^2\right ) \sin (e+f x) \cos (e+f x)}{8 f}+\frac{1}{8} a x \left (4 a^2+9 b^2\right )-\frac{11 a b^2 \sin ^3(e+f x) \cos (e+f x)}{20 f}-\frac{b^2 \sin ^3(e+f x) \cos (e+f x) (a+b \sin (e+f x))}{5 f} \]

[Out]

(a*(4*a^2 + 9*b^2)*x)/8 - (b*(15*a^2 + 4*b^2)*Cos[e + f*x])/(5*f) + (b*(15*a^2 + 4*b^2)*Cos[e + f*x]^3)/(15*f)
- (a*(4*a^2 + 9*b^2)*Cos[e + f*x]*Sin[e + f*x])/(8*f) - (11*a*b^2*Cos[e + f*x]*Sin[e + f*x]^3)/(20*f) - (b^2*
Cos[e + f*x]*Sin[e + f*x]^3*(a + b*Sin[e + f*x]))/(5*f)

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Rubi [A]  time = 0.215738, antiderivative size = 180, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2791, 2753, 2734} \[ \frac{\left (-52 a^2 b^2+3 a^4-16 b^4\right ) \cos (e+f x)}{30 b f}+\frac{\left (3 a^2-16 b^2\right ) \cos (e+f x) (a+b \sin (e+f x))^2}{60 b f}+\frac{a \left (6 a^2-71 b^2\right ) \sin (e+f x) \cos (e+f x)}{120 f}+\frac{1}{8} a x \left (4 a^2+9 b^2\right )-\frac{\cos (e+f x) (a+b \sin (e+f x))^4}{5 b f}+\frac{a \cos (e+f x) (a+b \sin (e+f x))^3}{20 b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(4*a^2 + 9*b^2)*x)/8 + ((3*a^4 - 52*a^2*b^2 - 16*b^4)*Cos[e + f*x])/(30*b*f) + (a*(6*a^2 - 71*b^2)*Cos[e +
f*x]*Sin[e + f*x])/(120*f) + ((3*a^2 - 16*b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^2)/(60*b*f) + (a*Cos[e + f*x]
*(a + b*Sin[e + f*x])^3)/(20*b*f) - (Cos[e + f*x]*(a + b*Sin[e + f*x])^4)/(5*b*f)

Rule 2791

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

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 \sin ^2(e+f x) (a+b \sin (e+f x))^3 \, dx &=-\frac{\cos (e+f x) (a+b \sin (e+f x))^4}{5 b f}+\frac{\int (4 b-a \sin (e+f x)) (a+b \sin (e+f x))^3 \, dx}{5 b}\\ &=\frac{a \cos (e+f x) (a+b \sin (e+f x))^3}{20 b f}-\frac{\cos (e+f x) (a+b \sin (e+f x))^4}{5 b f}+\frac{\int (a+b \sin (e+f x))^2 \left (13 a b-\left (3 a^2-16 b^2\right ) \sin (e+f x)\right ) \, dx}{20 b}\\ &=\frac{\left (3 a^2-16 b^2\right ) \cos (e+f x) (a+b \sin (e+f x))^2}{60 b f}+\frac{a \cos (e+f x) (a+b \sin (e+f x))^3}{20 b f}-\frac{\cos (e+f x) (a+b \sin (e+f x))^4}{5 b f}+\frac{\int (a+b \sin (e+f x)) \left (b \left (33 a^2+32 b^2\right )-a \left (6 a^2-71 b^2\right ) \sin (e+f x)\right ) \, dx}{60 b}\\ &=\frac{1}{8} a \left (4 a^2+9 b^2\right ) x+\frac{\left (3 a^4-52 a^2 b^2-16 b^4\right ) \cos (e+f x)}{30 b f}+\frac{a \left (6 a^2-71 b^2\right ) \cos (e+f x) \sin (e+f x)}{120 f}+\frac{\left (3 a^2-16 b^2\right ) \cos (e+f x) (a+b \sin (e+f x))^2}{60 b f}+\frac{a \cos (e+f x) (a+b \sin (e+f x))^3}{20 b f}-\frac{\cos (e+f x) (a+b \sin (e+f x))^4}{5 b f}\\ \end{align*}

Mathematica [A]  time = 0.635079, size = 117, normalized size = 0.73 \[ \frac{15 a \left (4 \left (4 a^2+9 b^2\right ) (e+f x)-8 \left (a^2+3 b^2\right ) \sin (2 (e+f x))+3 b^2 \sin (4 (e+f x))\right )-60 b \left (18 a^2+5 b^2\right ) \cos (e+f x)+10 \left (12 a^2 b+5 b^3\right ) \cos (3 (e+f x))-6 b^3 \cos (5 (e+f x))}{480 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-60*b*(18*a^2 + 5*b^2)*Cos[e + f*x] + 10*(12*a^2*b + 5*b^3)*Cos[3*(e + f*x)] - 6*b^3*Cos[5*(e + f*x)] + 15*a*
(4*(4*a^2 + 9*b^2)*(e + f*x) - 8*(a^2 + 3*b^2)*Sin[2*(e + f*x)] + 3*b^2*Sin[4*(e + f*x)]))/(480*f)

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Maple [A]  time = 0.027, size = 124, normalized size = 0.8 \begin{align*}{\frac{1}{f} \left ( -{\frac{{b}^{3}\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 ) }+3\,a{b}^{2} \left ( -1/4\, \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{3}+3/2\,\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) +3/8\,fx+3/8\,e \right ) -{a}^{2}b \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) +{a}^{3} \left ( -{\frac{\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{2}}+{\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.73668, size = 163, normalized size = 1.02 \begin{align*} \frac{120 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{3} + 480 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{2} b + 45 \,{\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 b^{2} - 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 )} b^{3}}{480 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^2*(a+b*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

1/480*(120*(2*f*x + 2*e - sin(2*f*x + 2*e))*a^3 + 480*(cos(f*x + e)^3 - 3*cos(f*x + e))*a^2*b + 45*(12*f*x + 1
2*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*a*b^2 - 32*(3*cos(f*x + e)^5 - 10*cos(f*x + e)^3 + 15*cos(f*x + e
))*b^3)/f

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Fricas [A]  time = 1.67325, size = 285, normalized size = 1.78 \begin{align*} -\frac{24 \, b^{3} \cos \left (f x + e\right )^{5} - 40 \,{\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (f x + e\right )^{3} - 15 \,{\left (4 \, a^{3} + 9 \, a b^{2}\right )} f x + 120 \,{\left (3 \, a^{2} b + b^{3}\right )} \cos \left (f x + e\right ) - 15 \,{\left (6 \, a b^{2} \cos \left (f x + e\right )^{3} -{\left (4 \, a^{3} + 15 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{120 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^2*(a+b*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

-1/120*(24*b^3*cos(f*x + e)^5 - 40*(3*a^2*b + 2*b^3)*cos(f*x + e)^3 - 15*(4*a^3 + 9*a*b^2)*f*x + 120*(3*a^2*b
+ b^3)*cos(f*x + e) - 15*(6*a*b^2*cos(f*x + e)^3 - (4*a^3 + 15*a*b^2)*cos(f*x + e))*sin(f*x + e))/f

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Sympy [A]  time = 3.52925, size = 284, normalized size = 1.78 \begin{align*} \begin{cases} \frac{a^{3} x \sin ^{2}{\left (e + f x \right )}}{2} + \frac{a^{3} x \cos ^{2}{\left (e + f x \right )}}{2} - \frac{a^{3} \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} - \frac{3 a^{2} b \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{2 a^{2} b \cos ^{3}{\left (e + f x \right )}}{f} + \frac{9 a b^{2} x \sin ^{4}{\left (e + f x \right )}}{8} + \frac{9 a b^{2} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac{9 a b^{2} x \cos ^{4}{\left (e + f x \right )}}{8} - \frac{15 a b^{2} \sin ^{3}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{8 f} - \frac{9 a b^{2} \sin{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} - \frac{b^{3} \sin ^{4}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{4 b^{3} \sin ^{2}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac{8 b^{3} \cos ^{5}{\left (e + f x \right )}}{15 f} & \text{for}\: f \neq 0 \\x \left (a + b \sin{\left (e \right )}\right )^{3} \sin ^{2}{\left (e \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)**2*(a+b*sin(f*x+e))**3,x)

[Out]

Piecewise((a**3*x*sin(e + f*x)**2/2 + a**3*x*cos(e + f*x)**2/2 - a**3*sin(e + f*x)*cos(e + f*x)/(2*f) - 3*a**2
*b*sin(e + f*x)**2*cos(e + f*x)/f - 2*a**2*b*cos(e + f*x)**3/f + 9*a*b**2*x*sin(e + f*x)**4/8 + 9*a*b**2*x*sin
(e + f*x)**2*cos(e + f*x)**2/4 + 9*a*b**2*x*cos(e + f*x)**4/8 - 15*a*b**2*sin(e + f*x)**3*cos(e + f*x)/(8*f) -
9*a*b**2*sin(e + f*x)*cos(e + f*x)**3/(8*f) - b**3*sin(e + f*x)**4*cos(e + f*x)/f - 4*b**3*sin(e + f*x)**2*co
s(e + f*x)**3/(3*f) - 8*b**3*cos(e + f*x)**5/(15*f), Ne(f, 0)), (x*(a + b*sin(e))**3*sin(e)**2, True))

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Giac [A]  time = 1.71505, size = 174, normalized size = 1.09 \begin{align*} -\frac{b^{3} \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} + \frac{3 \, a b^{2} \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} + \frac{1}{8} \,{\left (4 \, a^{3} + 9 \, a b^{2}\right )} x + \frac{{\left (12 \, a^{2} b + 5 \, b^{3}\right )} \cos \left (3 \, f x + 3 \, e\right )}{48 \, f} - \frac{{\left (18 \, a^{2} b + 5 \, b^{3}\right )} \cos \left (f x + e\right )}{8 \, f} - \frac{{\left (a^{3} + 3 \, a b^{2}\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(sin(f*x+e)^2*(a+b*sin(f*x+e))^3,x, algorithm="giac")

[Out]

-1/80*b^3*cos(5*f*x + 5*e)/f + 3/32*a*b^2*sin(4*f*x + 4*e)/f + 1/8*(4*a^3 + 9*a*b^2)*x + 1/48*(12*a^2*b + 5*b^
3)*cos(3*f*x + 3*e)/f - 1/8*(18*a^2*b + 5*b^3)*cos(f*x + e)/f - 1/4*(a^3 + 3*a*b^2)*sin(2*f*x + 2*e)/f