3.261 \(\int (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) \, dx\)

Optimal. Leaf size=127 \[ -\frac{5 a^3 (4 A+3 B) \cos (e+f x)}{6 f}-\frac{5 a^3 (4 A+3 B) \sin (e+f x) \cos (e+f x)}{24 f}+\frac{5}{8} a^3 x (4 A+3 B)-\frac{a (4 A+3 B) \cos (e+f x) (a \sin (e+f x)+a)^2}{12 f}-\frac{B \cos (e+f x) (a \sin (e+f x)+a)^3}{4 f} \]

[Out]

(5*a^3*(4*A + 3*B)*x)/8 - (5*a^3*(4*A + 3*B)*Cos[e + f*x])/(6*f) - (5*a^3*(4*A + 3*B)*Cos[e + f*x]*Sin[e + f*x
])/(24*f) - (a*(4*A + 3*B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^2)/(12*f) - (B*Cos[e + f*x]*(a + a*Sin[e + f*x])^
3)/(4*f)

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Rubi [A]  time = 0.100562, antiderivative size = 117, normalized size of antiderivative = 0.92, number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {2751, 2645, 2638, 2635, 8, 2633} \[ \frac{a^3 (4 A+3 B) \cos ^3(e+f x)}{12 f}-\frac{a^3 (4 A+3 B) \cos (e+f x)}{f}-\frac{3 a^3 (4 A+3 B) \sin (e+f x) \cos (e+f x)}{8 f}+\frac{5}{8} a^3 x (4 A+3 B)-\frac{B \cos (e+f x) (a \sin (e+f x)+a)^3}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a^3*(4*A + 3*B)*x)/8 - (a^3*(4*A + 3*B)*Cos[e + f*x])/f + (a^3*(4*A + 3*B)*Cos[e + f*x]^3)/(12*f) - (3*a^3*
(4*A + 3*B)*Cos[e + f*x]*Sin[e + f*x])/(8*f) - (B*Cos[e + f*x]*(a + a*Sin[e + f*x])^3)/(4*f)

Rule 2751

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[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*Sin
[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m,
-2^(-1)]

Rule 2645

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Int[ExpandTrig[(a + b*sin[c + d*x])^n, x], x] /;
FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] && IGtQ[n, 0]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, 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 8

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

Rubi steps

\begin{align*} \int (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) \, dx &=-\frac{B \cos (e+f x) (a+a \sin (e+f x))^3}{4 f}+\frac{1}{4} (4 A+3 B) \int (a+a \sin (e+f x))^3 \, dx\\ &=-\frac{B \cos (e+f x) (a+a \sin (e+f x))^3}{4 f}+\frac{1}{4} (4 A+3 B) \int \left (a^3+3 a^3 \sin (e+f x)+3 a^3 \sin ^2(e+f x)+a^3 \sin ^3(e+f x)\right ) \, dx\\ &=\frac{1}{4} a^3 (4 A+3 B) x-\frac{B \cos (e+f x) (a+a \sin (e+f x))^3}{4 f}+\frac{1}{4} \left (a^3 (4 A+3 B)\right ) \int \sin ^3(e+f x) \, dx+\frac{1}{4} \left (3 a^3 (4 A+3 B)\right ) \int \sin (e+f x) \, dx+\frac{1}{4} \left (3 a^3 (4 A+3 B)\right ) \int \sin ^2(e+f x) \, dx\\ &=\frac{1}{4} a^3 (4 A+3 B) x-\frac{3 a^3 (4 A+3 B) \cos (e+f x)}{4 f}-\frac{3 a^3 (4 A+3 B) \cos (e+f x) \sin (e+f x)}{8 f}-\frac{B \cos (e+f x) (a+a \sin (e+f x))^3}{4 f}+\frac{1}{8} \left (3 a^3 (4 A+3 B)\right ) \int 1 \, dx-\frac{\left (a^3 (4 A+3 B)\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (e+f x)\right )}{4 f}\\ &=\frac{5}{8} a^3 (4 A+3 B) x-\frac{a^3 (4 A+3 B) \cos (e+f x)}{f}+\frac{a^3 (4 A+3 B) \cos ^3(e+f x)}{12 f}-\frac{3 a^3 (4 A+3 B) \cos (e+f x) \sin (e+f x)}{8 f}-\frac{B \cos (e+f x) (a+a \sin (e+f x))^3}{4 f}\\ \end{align*}

Mathematica [A]  time = 0.48649, size = 120, normalized size = 0.94 \[ -\frac{a^3 \cos (e+f x) \left (30 (4 A+3 B) \sin ^{-1}\left (\frac{\sqrt{1-\sin (e+f x)}}{\sqrt{2}}\right )+\sqrt{\cos ^2(e+f x)} \left (8 (A+3 B) \sin ^2(e+f x)+9 (4 A+5 B) \sin (e+f x)+88 A+6 B \sin ^3(e+f x)+72 B\right )\right )}{24 f \sqrt{\cos ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*Cos[e + f*x]*(30*(4*A + 3*B)*ArcSin[Sqrt[1 - Sin[e + f*x]]/Sqrt[2]] + Sqrt[Cos[e + f*x]^2]*(88*A + 72*B
+ 9*(4*A + 5*B)*Sin[e + f*x] + 8*(A + 3*B)*Sin[e + f*x]^2 + 6*B*Sin[e + f*x]^3)))/(24*f*Sqrt[Cos[e + f*x]^2])

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Maple [A]  time = 0.046, size = 178, normalized size = 1.4 \begin{align*}{\frac{1}{f} \left ( -{\frac{{a}^{3}A \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) }{3}}+B{a}^{3} \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 ) +3\,{a}^{3}A \left ( -1/2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +1/2\,fx+e/2 \right ) -B{a}^{3} \left ( 2+ \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \cos \left ( fx+e \right ) -3\,{a}^{3}A\cos \left ( fx+e \right ) +3\,B{a}^{3} \left ( -1/2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +1/2\,fx+e/2 \right ) +{a}^{3}A \left ( fx+e \right ) -B{a}^{3}\cos \left ( fx+e \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(-1/3*a^3*A*(2+sin(f*x+e)^2)*cos(f*x+e)+B*a^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
)+3*a^3*A*(-1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*e)-B*a^3*(2+sin(f*x+e)^2)*cos(f*x+e)-3*a^3*A*cos(f*x+e)+3*B*
a^3*(-1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*e)+a^3*A*(f*x+e)-B*a^3*cos(f*x+e))

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Maxima [A]  time = 0.966436, size = 231, normalized size = 1.82 \begin{align*} \frac{32 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} A a^{3} + 72 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} A a^{3} + 96 \,{\left (f x + e\right )} A a^{3} + 96 \,{\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} B a^{3} + 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 )} B a^{3} + 72 \,{\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} B a^{3} - 288 \, A a^{3} \cos \left (f x + e\right ) - 96 \, B a^{3} \cos \left (f x + e\right )}{96 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(32*(cos(f*x + e)^3 - 3*cos(f*x + e))*A*a^3 + 72*(2*f*x + 2*e - sin(2*f*x + 2*e))*A*a^3 + 96*(f*x + e)*A*
a^3 + 96*(cos(f*x + e)^3 - 3*cos(f*x + e))*B*a^3 + 3*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*B
*a^3 + 72*(2*f*x + 2*e - sin(2*f*x + 2*e))*B*a^3 - 288*A*a^3*cos(f*x + e) - 96*B*a^3*cos(f*x + e))/f

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Fricas [A]  time = 1.94411, size = 231, normalized size = 1.82 \begin{align*} \frac{8 \,{\left (A + 3 \, B\right )} a^{3} \cos \left (f x + e\right )^{3} + 15 \,{\left (4 \, A + 3 \, B\right )} a^{3} f x - 96 \,{\left (A + B\right )} a^{3} \cos \left (f x + e\right ) + 3 \,{\left (2 \, B a^{3} \cos \left (f x + e\right )^{3} -{\left (12 \, A + 17 \, B\right )} a^{3} \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+a*sin(f*x+e))^3*(A+B*sin(f*x+e)),x, algorithm="fricas")

[Out]

1/24*(8*(A + 3*B)*a^3*cos(f*x + e)^3 + 15*(4*A + 3*B)*a^3*f*x - 96*(A + B)*a^3*cos(f*x + e) + 3*(2*B*a^3*cos(f
*x + e)^3 - (12*A + 17*B)*a^3*cos(f*x + e))*sin(f*x + e))/f

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Sympy [A]  time = 1.96336, size = 371, normalized size = 2.92 \begin{align*} \begin{cases} \frac{3 A a^{3} x \sin ^{2}{\left (e + f x \right )}}{2} + \frac{3 A a^{3} x \cos ^{2}{\left (e + f x \right )}}{2} + A a^{3} x - \frac{A a^{3} \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{3 A a^{3} \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} - \frac{2 A a^{3} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac{3 A a^{3} \cos{\left (e + f x \right )}}{f} + \frac{3 B a^{3} x \sin ^{4}{\left (e + f x \right )}}{8} + \frac{3 B a^{3} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac{3 B a^{3} x \sin ^{2}{\left (e + f x \right )}}{2} + \frac{3 B a^{3} x \cos ^{4}{\left (e + f x \right )}}{8} + \frac{3 B a^{3} x \cos ^{2}{\left (e + f x \right )}}{2} - \frac{5 B a^{3} \sin ^{3}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{8 f} - \frac{3 B a^{3} \sin ^{2}{\left (e + f x \right )} \cos{\left (e + f x \right )}}{f} - \frac{3 B a^{3} \sin{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} - \frac{3 B a^{3} \sin{\left (e + f x \right )} \cos{\left (e + f x \right )}}{2 f} - \frac{2 B a^{3} \cos ^{3}{\left (e + f x \right )}}{f} - \frac{B a^{3} \cos{\left (e + f x \right )}}{f} & \text{for}\: f \neq 0 \\x \left (A + B \sin{\left (e \right )}\right ) \left (a \sin{\left (e \right )} + a\right )^{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.28961, size = 157, normalized size = 1.24 \begin{align*} \frac{B a^{3} \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} + \frac{5}{8} \,{\left (4 \, A a^{3} + 3 \, B a^{3}\right )} x + \frac{{\left (A a^{3} + 3 \, B a^{3}\right )} \cos \left (3 \, f x + 3 \, e\right )}{12 \, f} - \frac{{\left (15 \, A a^{3} + 13 \, B a^{3}\right )} \cos \left (f x + e\right )}{4 \, f} - \frac{{\left (3 \, A a^{3} + 4 \, B a^{3}\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+a*sin(f*x+e))^3*(A+B*sin(f*x+e)),x, algorithm="giac")

[Out]

1/32*B*a^3*sin(4*f*x + 4*e)/f + 5/8*(4*A*a^3 + 3*B*a^3)*x + 1/12*(A*a^3 + 3*B*a^3)*cos(3*f*x + 3*e)/f - 1/4*(1
5*A*a^3 + 13*B*a^3)*cos(f*x + e)/f - 1/4*(3*A*a^3 + 4*B*a^3)*sin(2*f*x + 2*e)/f