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

Optimal. Leaf size=171 \[ \frac{a \left (a^2+6 b^2\right ) \cos ^3(e+f x)}{3 f}-\frac{a \left (a^2+3 b^2\right ) \cos (e+f x)}{f}-\frac{b \left (18 a^2+5 b^2\right ) \sin ^3(e+f x) \cos (e+f x)}{24 f}-\frac{b \left (18 a^2+5 b^2\right ) \sin (e+f x) \cos (e+f x)}{16 f}+\frac{1}{16} b x \left (18 a^2+5 b^2\right )-\frac{3 a b^2 \cos ^5(e+f x)}{5 f}-\frac{b^3 \sin ^5(e+f x) \cos (e+f x)}{6 f} \]

[Out]

(b*(18*a^2 + 5*b^2)*x)/16 - (a*(a^2 + 3*b^2)*Cos[e + f*x])/f + (a*(a^2 + 6*b^2)*Cos[e + f*x]^3)/(3*f) - (3*a*b
^2*Cos[e + f*x]^5)/(5*f) - (b*(18*a^2 + 5*b^2)*Cos[e + f*x]*Sin[e + f*x])/(16*f) - (b*(18*a^2 + 5*b^2)*Cos[e +
f*x]*Sin[e + f*x]^3)/(24*f) - (b^3*Cos[e + f*x]*Sin[e + f*x]^5)/(6*f)

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Rubi [A]  time = 0.209467, antiderivative size = 193, normalized size of antiderivative = 1.13, number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2793, 3023, 2748, 2633, 2635, 8} \[ \frac{a \left (5 a^2+12 b^2\right ) \cos ^3(e+f x)}{15 f}-\frac{a \left (5 a^2+12 b^2\right ) \cos (e+f x)}{5 f}-\frac{b \left (18 a^2+5 b^2\right ) \sin ^3(e+f x) \cos (e+f x)}{24 f}-\frac{b \left (18 a^2+5 b^2\right ) \sin (e+f x) \cos (e+f x)}{16 f}+\frac{1}{16} b x \left (18 a^2+5 b^2\right )-\frac{13 a b^2 \sin ^4(e+f x) \cos (e+f x)}{30 f}-\frac{b^2 \sin ^4(e+f x) \cos (e+f x) (a+b \sin (e+f x))}{6 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(18*a^2 + 5*b^2)*x)/16 - (a*(5*a^2 + 12*b^2)*Cos[e + f*x])/(5*f) + (a*(5*a^2 + 12*b^2)*Cos[e + f*x]^3)/(15*
f) - (b*(18*a^2 + 5*b^2)*Cos[e + f*x]*Sin[e + f*x])/(16*f) - (b*(18*a^2 + 5*b^2)*Cos[e + f*x]*Sin[e + f*x]^3)/
(24*f) - (13*a*b^2*Cos[e + f*x]*Sin[e + f*x]^4)/(30*f) - (b^2*Cos[e + f*x]*Sin[e + f*x]^4*(a + b*Sin[e + f*x])
)/(6*f)

Rule 2793

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d
*(m + n)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d*(m + n) + b^2*(b*c*(m - 2) + a*d
*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n -
2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] ||
(EqQ[a, 0] && NeQ[c, 0])))

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*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[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

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 8

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

Rubi steps

\begin{align*} \int \sin ^3(e+f x) (a+b \sin (e+f x))^3 \, dx &=-\frac{b^2 \cos (e+f x) \sin ^4(e+f x) (a+b \sin (e+f x))}{6 f}+\frac{1}{6} \int \sin ^3(e+f x) \left (2 a \left (3 a^2+2 b^2\right )+b \left (18 a^2+5 b^2\right ) \sin (e+f x)+13 a b^2 \sin ^2(e+f x)\right ) \, dx\\ &=-\frac{13 a b^2 \cos (e+f x) \sin ^4(e+f x)}{30 f}-\frac{b^2 \cos (e+f x) \sin ^4(e+f x) (a+b \sin (e+f x))}{6 f}+\frac{1}{30} \int \sin ^3(e+f x) \left (6 a \left (5 a^2+12 b^2\right )+5 b \left (18 a^2+5 b^2\right ) \sin (e+f x)\right ) \, dx\\ &=-\frac{13 a b^2 \cos (e+f x) \sin ^4(e+f x)}{30 f}-\frac{b^2 \cos (e+f x) \sin ^4(e+f x) (a+b \sin (e+f x))}{6 f}+\frac{1}{6} \left (b \left (18 a^2+5 b^2\right )\right ) \int \sin ^4(e+f x) \, dx+\frac{1}{5} \left (a \left (5 a^2+12 b^2\right )\right ) \int \sin ^3(e+f x) \, dx\\ &=-\frac{b \left (18 a^2+5 b^2\right ) \cos (e+f x) \sin ^3(e+f x)}{24 f}-\frac{13 a b^2 \cos (e+f x) \sin ^4(e+f x)}{30 f}-\frac{b^2 \cos (e+f x) \sin ^4(e+f x) (a+b \sin (e+f x))}{6 f}+\frac{1}{8} \left (b \left (18 a^2+5 b^2\right )\right ) \int \sin ^2(e+f x) \, dx-\frac{\left (a \left (5 a^2+12 b^2\right )\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (e+f x)\right )}{5 f}\\ &=-\frac{a \left (5 a^2+12 b^2\right ) \cos (e+f x)}{5 f}+\frac{a \left (5 a^2+12 b^2\right ) \cos ^3(e+f x)}{15 f}-\frac{b \left (18 a^2+5 b^2\right ) \cos (e+f x) \sin (e+f x)}{16 f}-\frac{b \left (18 a^2+5 b^2\right ) \cos (e+f x) \sin ^3(e+f x)}{24 f}-\frac{13 a b^2 \cos (e+f x) \sin ^4(e+f x)}{30 f}-\frac{b^2 \cos (e+f x) \sin ^4(e+f x) (a+b \sin (e+f x))}{6 f}+\frac{1}{16} \left (b \left (18 a^2+5 b^2\right )\right ) \int 1 \, dx\\ &=\frac{1}{16} b \left (18 a^2+5 b^2\right ) x-\frac{a \left (5 a^2+12 b^2\right ) \cos (e+f x)}{5 f}+\frac{a \left (5 a^2+12 b^2\right ) \cos ^3(e+f x)}{15 f}-\frac{b \left (18 a^2+5 b^2\right ) \cos (e+f x) \sin (e+f x)}{16 f}-\frac{b \left (18 a^2+5 b^2\right ) \cos (e+f x) \sin ^3(e+f x)}{24 f}-\frac{13 a b^2 \cos (e+f x) \sin ^4(e+f x)}{30 f}-\frac{b^2 \cos (e+f x) \sin ^4(e+f x) (a+b \sin (e+f x))}{6 f}\\ \end{align*}

Mathematica [A]  time = 0.700237, size = 147, normalized size = 0.86 \[ \frac{-360 a \left (2 a^2+5 b^2\right ) \cos (e+f x)+20 \left (4 a^3+15 a b^2\right ) \cos (3 (e+f x))+b \left (5 \left (-9 \left (16 a^2+5 b^2\right ) \sin (2 (e+f x))+9 \left (2 a^2+b^2\right ) \sin (4 (e+f x))+216 a^2 e+216 a^2 f x-b^2 \sin (6 (e+f x))+60 b^2 e+60 b^2 f x\right )-36 a b \cos (5 (e+f x))\right )}{960 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-360*a*(2*a^2 + 5*b^2)*Cos[e + f*x] + 20*(4*a^3 + 15*a*b^2)*Cos[3*(e + f*x)] + b*(-36*a*b*Cos[5*(e + f*x)] +
5*(216*a^2*e + 60*b^2*e + 216*a^2*f*x + 60*b^2*f*x - 9*(16*a^2 + 5*b^2)*Sin[2*(e + f*x)] + 9*(2*a^2 + b^2)*Sin
[4*(e + f*x)] - b^2*Sin[6*(e + f*x)])))/(960*f)

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Maple [A]  time = 0.025, size = 145, normalized size = 0.9 \begin{align*}{\frac{1}{f} \left ({b}^{3} \left ( -{\frac{\cos \left ( fx+e \right ) }{6} \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{5}+{\frac{5\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}}{4}}+{\frac{15\,\sin \left ( fx+e \right ) }{8}} \right ) }+{\frac{5\,fx}{16}}+{\frac{5\,e}{16}} \right ) -{\frac{3\,a{b}^{2}\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}^{2}b \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 ) -{\frac{{a}^{3} \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(b^3*(-1/6*(sin(f*x+e)^5+5/4*sin(f*x+e)^3+15/8*sin(f*x+e))*cos(f*x+e)+5/16*f*x+5/16*e)-3/5*a*b^2*(8/3+sin(
f*x+e)^4+4/3*sin(f*x+e)^2)*cos(f*x+e)+3*a^2*b*(-1/4*(sin(f*x+e)^3+3/2*sin(f*x+e))*cos(f*x+e)+3/8*f*x+3/8*e)-1/
3*a^3*(2+sin(f*x+e)^2)*cos(f*x+e))

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Maxima [A]  time = 2.55476, size = 196, normalized size = 1.15 \begin{align*} \frac{320 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{3} + 90 \,{\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} b - 192 \,{\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 b^{2} + 5 \,{\left (4 \, \sin \left (2 \, f x + 2 \, e\right )^{3} + 60 \, f x + 60 \, e + 9 \, \sin \left (4 \, f x + 4 \, e\right ) - 48 \, \sin \left (2 \, f x + 2 \, e\right )\right )} b^{3}}{960 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/960*(320*(cos(f*x + e)^3 - 3*cos(f*x + e))*a^3 + 90*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*
a^2*b - 192*(3*cos(f*x + e)^5 - 10*cos(f*x + e)^3 + 15*cos(f*x + e))*a*b^2 + 5*(4*sin(2*f*x + 2*e)^3 + 60*f*x
+ 60*e + 9*sin(4*f*x + 4*e) - 48*sin(2*f*x + 2*e))*b^3)/f

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Fricas [A]  time = 1.73995, size = 340, normalized size = 1.99 \begin{align*} -\frac{144 \, a b^{2} \cos \left (f x + e\right )^{5} - 80 \,{\left (a^{3} + 6 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - 15 \,{\left (18 \, a^{2} b + 5 \, b^{3}\right )} f x + 240 \,{\left (a^{3} + 3 \, a b^{2}\right )} \cos \left (f x + e\right ) + 5 \,{\left (8 \, b^{3} \cos \left (f x + e\right )^{5} - 2 \,{\left (18 \, a^{2} b + 13 \, b^{3}\right )} \cos \left (f x + e\right )^{3} + 3 \,{\left (30 \, a^{2} b + 11 \, b^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{240 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*(144*a*b^2*cos(f*x + e)^5 - 80*(a^3 + 6*a*b^2)*cos(f*x + e)^3 - 15*(18*a^2*b + 5*b^3)*f*x + 240*(a^3 +
3*a*b^2)*cos(f*x + e) + 5*(8*b^3*cos(f*x + e)^5 - 2*(18*a^2*b + 13*b^3)*cos(f*x + e)^3 + 3*(30*a^2*b + 11*b^3)
*cos(f*x + e))*sin(f*x + e))/f

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Sympy [A]  time = 6.07667, size = 393, normalized size = 2.3 \begin{align*} \begin{cases} - \frac{a^{3} \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{2 a^{3} \cos ^{3}{\left (e + f x \right )}}{3 f} + \frac{9 a^{2} b x \sin ^{4}{\left (e + f x \right )}}{8} + \frac{9 a^{2} b x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac{9 a^{2} b x \cos ^{4}{\left (e + f x \right )}}{8} - \frac{15 a^{2} b \sin ^{3}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{8 f} - \frac{9 a^{2} b \sin{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} - \frac{3 a b^{2} \sin ^{4}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{4 a b^{2} \sin ^{2}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{f} - \frac{8 a b^{2} \cos ^{5}{\left (e + f x \right )}}{5 f} + \frac{5 b^{3} x \sin ^{6}{\left (e + f x \right )}}{16} + \frac{15 b^{3} x \sin ^{4}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{16} + \frac{15 b^{3} x \sin ^{2}{\left (e + f x \right )} \cos ^{4}{\left (e + f x \right )}}{16} + \frac{5 b^{3} x \cos ^{6}{\left (e + f x \right )}}{16} - \frac{11 b^{3} \sin ^{5}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{16 f} - \frac{5 b^{3} \sin ^{3}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{6 f} - \frac{5 b^{3} \sin{\left (e + f x \right )} \cos ^{5}{\left (e + f x \right )}}{16 f} & \text{for}\: f \neq 0 \\x \left (a + b \sin{\left (e \right )}\right )^{3} \sin ^{3}{\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)**3*(a+b*sin(f*x+e))**3,x)

[Out]

Piecewise((-a**3*sin(e + f*x)**2*cos(e + f*x)/f - 2*a**3*cos(e + f*x)**3/(3*f) + 9*a**2*b*x*sin(e + f*x)**4/8
+ 9*a**2*b*x*sin(e + f*x)**2*cos(e + f*x)**2/4 + 9*a**2*b*x*cos(e + f*x)**4/8 - 15*a**2*b*sin(e + f*x)**3*cos(
e + f*x)/(8*f) - 9*a**2*b*sin(e + f*x)*cos(e + f*x)**3/(8*f) - 3*a*b**2*sin(e + f*x)**4*cos(e + f*x)/f - 4*a*b
**2*sin(e + f*x)**2*cos(e + f*x)**3/f - 8*a*b**2*cos(e + f*x)**5/(5*f) + 5*b**3*x*sin(e + f*x)**6/16 + 15*b**3
*x*sin(e + f*x)**4*cos(e + f*x)**2/16 + 15*b**3*x*sin(e + f*x)**2*cos(e + f*x)**4/16 + 5*b**3*x*cos(e + f*x)**
6/16 - 11*b**3*sin(e + f*x)**5*cos(e + f*x)/(16*f) - 5*b**3*sin(e + f*x)**3*cos(e + f*x)**3/(6*f) - 5*b**3*sin
(e + f*x)*cos(e + f*x)**5/(16*f), Ne(f, 0)), (x*(a + b*sin(e))**3*sin(e)**3, True))

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Giac [A]  time = 2.21464, size = 243, normalized size = 1.42 \begin{align*} -\frac{3 \, a b^{2} \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} - \frac{b^{3} \sin \left (6 \, f x + 6 \, e\right )}{192 \, f} + \frac{1}{16} \,{\left (18 \, a^{2} b + 5 \, b^{3}\right )} x + \frac{{\left (4 \, a^{3} + 15 \, a b^{2}\right )} \cos \left (3 \, f x + 3 \, e\right )}{48 \, f} - \frac{{\left (2 \, a^{3} + 9 \, a b^{2}\right )} \cos \left (f x + e\right )}{8 \, f} - \frac{{\left (2 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (f x + e\right )}{4 \, f} + \frac{3 \,{\left (2 \, a^{2} b + b^{3}\right )} \sin \left (4 \, f x + 4 \, e\right )}{64 \, f} - \frac{3 \,{\left (16 \, a^{2} b + 5 \, b^{3}\right )} \sin \left (2 \, f x + 2 \, e\right )}{64 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/80*a*b^2*cos(5*f*x + 5*e)/f - 1/192*b^3*sin(6*f*x + 6*e)/f + 1/16*(18*a^2*b + 5*b^3)*x + 1/48*(4*a^3 + 15*a
*b^2)*cos(3*f*x + 3*e)/f - 1/8*(2*a^3 + 9*a*b^2)*cos(f*x + e)/f - 1/4*(2*a^3 + 3*a*b^2)*cos(f*x + e)/f + 3/64*
(2*a^2*b + b^3)*sin(4*f*x + 4*e)/f - 3/64*(16*a^2*b + 5*b^3)*sin(2*f*x + 2*e)/f