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]

5/8*a^3*(4*A+3*B)*x-5/6*a^3*(4*A+3*B)*cos(f*x+e)/f-5/24*a^3*(4*A+3*B)*cos(f*x+e)*sin(f*x+e)/f-1/12*a*(4*A+3*B)
*cos(f*x+e)*(a+a*sin(f*x+e))^2/f-1/4*B*cos(f*x+e)*(a+a*sin(f*x+e))^3/f

________________________________________________________________________________________

Rubi [A]  time = 0.10, 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 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]

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 2638

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

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 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)]

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 = 0.50, 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]

-1/24*(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)))/(f*Sqrt[Cos[e + f*x]^2]
)

________________________________________________________________________________________

fricas [A]  time = 0.45, size = 93, normalized size = 0.73 \[ \frac {8 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (f x + e\right )^{3} + 15 \, {\left (4 \, A + 3 \, B\right )} a^{3} f x - 96 \, {\left (A + B\right )} a^{3} \cos \left (f x + e\right ) + 3 \, {\left (2 \, B a^{3} \cos \left (f x + e\right )^{3} - {\left (12 \, A + 17 \, B\right )} a^{3} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{24 \, f} \]

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

________________________________________________________________________________________

giac [A]  time = 0.17, size = 116, normalized size = 0.91 \[ \frac {B a^{3} \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} + \frac {5}{8} \, {\left (4 \, A a^{3} + 3 \, B a^{3}\right )} x + \frac {{\left (A a^{3} + 3 \, B a^{3}\right )} \cos \left (3 \, f x + 3 \, e\right )}{12 \, f} - \frac {{\left (15 \, A a^{3} + 13 \, B a^{3}\right )} \cos \left (f x + e\right )}{4 \, f} - \frac {{\left (3 \, A a^{3} + 4 \, B a^{3}\right )} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \]

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

________________________________________________________________________________________

maple [A]  time = 0.41, size = 178, normalized size = 1.40 \[ \frac {-\frac {a^{3} A \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}+B \,a^{3} \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )+3 a^{3} A \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-B \,a^{3} \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )-3 a^{3} A \cos \left (f x +e \right )+3 B \,a^{3} \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )+a^{3} A \left (f x +e \right )-B \,a^{3} \cos \left (f x +e \right )}{f} \]

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))

________________________________________________________________________________________

maxima [A]  time = 0.38, size = 171, normalized size = 1.35 \[ \frac {32 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} A a^{3} + 72 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} A a^{3} + 96 \, {\left (f x + e\right )} A a^{3} + 96 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} B a^{3} + 3 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} B a^{3} + 72 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} B a^{3} - 288 \, A a^{3} \cos \left (f x + e\right ) - 96 \, B a^{3} \cos \left (f x + e\right )}{96 \, f} \]

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

________________________________________________________________________________________

mupad [B]  time = 14.26, size = 330, normalized size = 2.60 \[ \frac {5\,a^3\,\mathrm {atan}\left (\frac {5\,a^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (4\,A+3\,B\right )}{4\,\left (5\,A\,a^3+\frac {15\,B\,a^3}{4}\right )}\right )\,\left (4\,A+3\,B\right )}{4\,f}-\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (3\,A\,a^3+\frac {15\,B\,a^3}{4}\right )+\frac {22\,A\,a^3}{3}+6\,B\,a^3+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (6\,A\,a^3+2\,B\,a^3\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7\,\left (3\,A\,a^3+\frac {15\,B\,a^3}{4}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (3\,A\,a^3+\frac {23\,B\,a^3}{4}\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\left (3\,A\,a^3+\frac {23\,B\,a^3}{4}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (22\,A\,a^3+18\,B\,a^3\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {70\,A\,a^3}{3}+22\,B\,a^3\right )}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}-\frac {5\,a^3\,\left (4\,A+3\,B\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )-\frac {f\,x}{2}\right )}{4\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*a^3*atan((5*a^3*tan(e/2 + (f*x)/2)*(4*A + 3*B))/(4*(5*A*a^3 + (15*B*a^3)/4)))*(4*A + 3*B))/(4*f) - (tan(e/2
+ (f*x)/2)*(3*A*a^3 + (15*B*a^3)/4) + (22*A*a^3)/3 + 6*B*a^3 + tan(e/2 + (f*x)/2)^6*(6*A*a^3 + 2*B*a^3) - tan
(e/2 + (f*x)/2)^7*(3*A*a^3 + (15*B*a^3)/4) + tan(e/2 + (f*x)/2)^3*(3*A*a^3 + (23*B*a^3)/4) - tan(e/2 + (f*x)/2
)^5*(3*A*a^3 + (23*B*a^3)/4) + tan(e/2 + (f*x)/2)^4*(22*A*a^3 + 18*B*a^3) + tan(e/2 + (f*x)/2)^2*((70*A*a^3)/3
+ 22*B*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)) - (5*a^3*(4*A + 3*B)*(atan(tan(e/2 + (f*x)/2)) - (f*x)/2))/(4*f)

________________________________________________________________________________________

sympy [A]  time = 2.08, size = 371, normalized size = 2.92 \[ \begin {cases} \frac {3 A a^{3} x \sin ^{2}{\left (e + f x \right )}}{2} + \frac {3 A a^{3} x \cos ^{2}{\left (e + f x \right )}}{2} + A a^{3} x - \frac {A a^{3} \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {3 A a^{3} \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} - \frac {2 A a^{3} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac {3 A a^{3} \cos {\left (e + f x \right )}}{f} + \frac {3 B a^{3} x \sin ^{4}{\left (e + f x \right )}}{8} + \frac {3 B a^{3} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac {3 B a^{3} x \sin ^{2}{\left (e + f x \right )}}{2} + \frac {3 B a^{3} x \cos ^{4}{\left (e + f x \right )}}{8} + \frac {3 B a^{3} x \cos ^{2}{\left (e + f x \right )}}{2} - \frac {5 B a^{3} \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{8 f} - \frac {3 B a^{3} \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {3 B a^{3} \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} - \frac {3 B a^{3} \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} - \frac {2 B a^{3} \cos ^{3}{\left (e + f x \right )}}{f} - \frac {B a^{3} \cos {\left (e + f x \right )}}{f} & \text {for}\: f \neq 0 \\x \left (A + B \sin {\relax (e )}\right ) \left (a \sin {\relax (e )} + a\right )^{3} & \text {otherwise} \end {cases} \]

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))

________________________________________________________________________________________