3.446 \(\int (a+a \sin (e+f x))^3 (c+d \sin (e+f x)) \, dx\)

Optimal. Leaf size=110 \[ \frac {a^3 (c+3 d) \cos ^3(e+f x)}{3 f}-\frac {4 a^3 (c+d) \cos (e+f x)}{f}-\frac {3 a^3 (4 c+5 d) \sin (e+f x) \cos (e+f x)}{8 f}+\frac {5}{8} a^3 x (4 c+3 d)-\frac {a^3 d \sin ^3(e+f x) \cos (e+f x)}{4 f} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 117, normalized size of antiderivative = 1.06, 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} \[ \frac {a^3 (4 c+3 d) \cos ^3(e+f x)}{12 f}-\frac {a^3 (4 c+3 d) \cos (e+f x)}{f}-\frac {3 a^3 (4 c+3 d) \sin (e+f x) \cos (e+f x)}{8 f}+\frac {5}{8} a^3 x (4 c+3 d)-\frac {d \cos (e+f x) (a \sin (e+f x)+a)^3}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a^3*(4*c + 3*d)*x)/8 - (a^3*(4*c + 3*d)*Cos[e + f*x])/f + (a^3*(4*c + 3*d)*Cos[e + f*x]^3)/(12*f) - (3*a^3*
(4*c + 3*d)*Cos[e + f*x]*Sin[e + f*x])/(8*f) - (d*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 (c+d \sin (e+f x)) \, dx &=-\frac {d \cos (e+f x) (a+a \sin (e+f x))^3}{4 f}+\frac {1}{4} (4 c+3 d) \int (a+a \sin (e+f x))^3 \, dx\\ &=-\frac {d \cos (e+f x) (a+a \sin (e+f x))^3}{4 f}+\frac {1}{4} (4 c+3 d) \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 c+3 d) x-\frac {d \cos (e+f x) (a+a \sin (e+f x))^3}{4 f}+\frac {1}{4} \left (a^3 (4 c+3 d)\right ) \int \sin ^3(e+f x) \, dx+\frac {1}{4} \left (3 a^3 (4 c+3 d)\right ) \int \sin (e+f x) \, dx+\frac {1}{4} \left (3 a^3 (4 c+3 d)\right ) \int \sin ^2(e+f x) \, dx\\ &=\frac {1}{4} a^3 (4 c+3 d) x-\frac {3 a^3 (4 c+3 d) \cos (e+f x)}{4 f}-\frac {3 a^3 (4 c+3 d) \cos (e+f x) \sin (e+f x)}{8 f}-\frac {d \cos (e+f x) (a+a \sin (e+f x))^3}{4 f}+\frac {1}{8} \left (3 a^3 (4 c+3 d)\right ) \int 1 \, dx-\frac {\left (a^3 (4 c+3 d)\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 c+3 d) x-\frac {a^3 (4 c+3 d) \cos (e+f x)}{f}+\frac {a^3 (4 c+3 d) \cos ^3(e+f x)}{12 f}-\frac {3 a^3 (4 c+3 d) \cos (e+f x) \sin (e+f x)}{8 f}-\frac {d \cos (e+f x) (a+a \sin (e+f x))^3}{4 f}\\ \end {align*}

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Mathematica [A]  time = 0.52, size = 120, normalized size = 1.09 \[ -\frac {a^3 \cos (e+f x) \left (30 (4 c+3 d) \sin ^{-1}\left (\frac {\sqrt {1-\sin (e+f x)}}{\sqrt {2}}\right )+\sqrt {\cos ^2(e+f x)} \left (8 (c+3 d) \sin ^2(e+f x)+9 (4 c+5 d) \sin (e+f x)+88 c+6 d \sin ^3(e+f x)+72 d\right )\right )}{24 f \sqrt {\cos ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 108, normalized size = 0.98 \[ \frac {8 \, {\left (a^{3} c + 3 \, a^{3} d\right )} \cos \left (f x + e\right )^{3} + 15 \, {\left (4 \, a^{3} c + 3 \, a^{3} d\right )} f x - 96 \, {\left (a^{3} c + a^{3} d\right )} \cos \left (f x + e\right ) + 3 \, {\left (2 \, a^{3} d \cos \left (f x + e\right )^{3} - {\left (12 \, a^{3} c + 17 \, a^{3} d\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{24 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.20, size = 138, normalized size = 1.25 \[ a^{3} c x - \frac {a^{3} d \cos \left (f x + e\right )}{f} + \frac {a^{3} d \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} + \frac {3}{8} \, {\left (4 \, a^{3} c + 5 \, a^{3} d\right )} x + \frac {{\left (a^{3} c + 3 \, a^{3} d\right )} \cos \left (3 \, f x + 3 \, e\right )}{12 \, f} - \frac {3 \, {\left (5 \, a^{3} c + 3 \, a^{3} d\right )} \cos \left (f x + e\right )}{4 \, f} - \frac {{\left (3 \, a^{3} c + 4 \, a^{3} d\right )} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.25, size = 178, normalized size = 1.62 \[ \frac {-\frac {a^{3} c \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )}{3}+a^{3} d \left (-\frac {\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )+3 a^{3} c \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-a^{3} d \left (2+\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )-3 a^{3} c \cos \left (f x +e \right )+3 a^{3} d \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )+a^{3} c \left (f x +e \right )-a^{3} d \cos \left (f x +e \right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.35, size = 171, normalized size = 1.55 \[ \frac {32 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{3} c + 72 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{3} c + 96 \, {\left (f x + e\right )} a^{3} c + 96 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} a^{3} d + 3 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} a^{3} d + 72 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} a^{3} d - 288 \, a^{3} c \cos \left (f x + e\right ) - 96 \, a^{3} d \cos \left (f x + e\right )}{96 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 8.07, size = 330, normalized size = 3.00 \[ \frac {5\,a^3\,\mathrm {atan}\left (\frac {5\,a^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (4\,c+3\,d\right )}{4\,\left (5\,a^3\,c+\frac {15\,a^3\,d}{4}\right )}\right )\,\left (4\,c+3\,d\right )}{4\,f}-\frac {5\,a^3\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )-\frac {f\,x}{2}\right )\,\left (4\,c+3\,d\right )}{4\,f}-\frac {\frac {22\,a^3\,c}{3}+6\,a^3\,d+\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (3\,a^3\,c+\frac {15\,a^3\,d}{4}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (6\,a^3\,c+2\,a^3\,d\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7\,\left (3\,a^3\,c+\frac {15\,a^3\,d}{4}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (3\,a^3\,c+\frac {23\,a^3\,d}{4}\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\left (3\,a^3\,c+\frac {23\,a^3\,d}{4}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (22\,a^3\,c+18\,a^3\,d\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {70\,a^3\,c}{3}+22\,a^3\,d\right )}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.00, size = 371, normalized size = 3.37 \[ \begin {cases} \frac {3 a^{3} c x \sin ^{2}{\left (e + f x \right )}}{2} + \frac {3 a^{3} c x \cos ^{2}{\left (e + f x \right )}}{2} + a^{3} c x - \frac {a^{3} c \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {3 a^{3} c \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} - \frac {2 a^{3} c \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac {3 a^{3} c \cos {\left (e + f x \right )}}{f} + \frac {3 a^{3} d x \sin ^{4}{\left (e + f x \right )}}{8} + \frac {3 a^{3} d x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac {3 a^{3} d x \sin ^{2}{\left (e + f x \right )}}{2} + \frac {3 a^{3} d x \cos ^{4}{\left (e + f x \right )}}{8} + \frac {3 a^{3} d x \cos ^{2}{\left (e + f x \right )}}{2} - \frac {5 a^{3} d \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{8 f} - \frac {3 a^{3} d \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {3 a^{3} d \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} - \frac {3 a^{3} d \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} - \frac {2 a^{3} d \cos ^{3}{\left (e + f x \right )}}{f} - \frac {a^{3} d \cos {\left (e + f x \right )}}{f} & \text {for}\: f \neq 0 \\x \left (c + d \sin {\relax (e )}\right ) \left (a \sin {\relax (e )} + a\right )^{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**3*(c+d*sin(f*x+e)),x)

[Out]

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

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