\(\int (a+b \sin (d+e x)) (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)) \, dx\) [500]

Optimal result

Integrand size = 37, antiderivative size = 109 \[ \int (a+b \sin (d+e x)) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right ) \, dx=\frac {1}{2} a \left (a^2+4 b^2\right ) x+\frac {\left (a^4-8 a^2 b^2-3 b^4\right ) \cos (d+e x)}{3 b e}+\frac {a \left (a^2-6 b^2\right ) \cos (d+e x) \sin (d+e x)}{6 e}-\frac {a^2 \cos (d+e x) (a+b \sin (d+e x))^2}{3 b e} \]

[Out]

1/2*a*(a^2+4*b^2)*x+1/3*(a^4-8*a^2*b^2-3*b^4)*cos(e*x+d)/b/e+1/6*a*(a^2-6*b^2)*cos(e*x+d)*sin(e*x+d)/e-1/3*a^2
*cos(e*x+d)*(a+b*sin(e*x+d))^2/b/e

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {3102, 2813} \[ \int (a+b \sin (d+e x)) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right ) \, dx=\frac {a \left (a^2-6 b^2\right ) \sin (d+e x) \cos (d+e x)}{6 e}+\frac {1}{2} a x \left (a^2+4 b^2\right )-\frac {a^2 \cos (d+e x) (a+b \sin (d+e x))^2}{3 b e}+\frac {\left (a^4-8 a^2 b^2-3 b^4\right ) \cos (d+e x)}{3 b e} \]

[In]

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

[Out]

(a*(a^2 + 4*b^2)*x)/2 + ((a^4 - 8*a^2*b^2 - 3*b^4)*Cos[d + e*x])/(3*b*e) + (a*(a^2 - 6*b^2)*Cos[d + e*x]*Sin[d
+ e*x])/(6*e) - (a^2*Cos[d + e*x]*(a + b*Sin[d + e*x])^2)/(3*b*e)

Rule 2813

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]

Rule 3102

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {a^2 \cos (d+e x) (a+b \sin (d+e x))^2}{3 b e}+\frac {\int (a+b \sin (d+e x)) \left (b \left (2 a^2+3 b^2\right )-a \left (a^2-6 b^2\right ) \sin (d+e x)\right ) \, dx}{3 b} \\ & = \frac {1}{2} a \left (a^2+4 b^2\right ) x+\frac {\left (a^4-8 a^2 b^2-3 b^4\right ) \cos (d+e x)}{3 b e}+\frac {a \left (a^2-6 b^2\right ) \cos (d+e x) \sin (d+e x)}{6 e}-\frac {a^2 \cos (d+e x) (a+b \sin (d+e x))^2}{3 b e} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.12 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.71 \[ \int (a+b \sin (d+e x)) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right ) \, dx=\frac {-3 b \left (11 a^2+4 b^2\right ) \cos (d+e x)+a \left (6 \left (a^2+4 b^2\right ) (d+e x)+a b \cos (3 (d+e x))-3 \left (a^2+2 b^2\right ) \sin (2 (d+e x))\right )}{12 e} \]

[In]

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

[Out]

(-3*b*(11*a^2 + 4*b^2)*Cos[d + e*x] + a*(6*(a^2 + 4*b^2)*(d + e*x) + a*b*Cos[3*(d + e*x)] - 3*(a^2 + 2*b^2)*Si
n[2*(d + e*x)]))/(12*e)

Maple [A] (verified)

Time = 1.09 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.83

method	result	size
parts	\(a \,b^{2} x -\frac {\left (2 a^{2} b +b^{3}\right ) \cos \left (e x +d \right )}{e}+\frac {\left (a^{3}+2 a \,b^{2}\right ) \left (-\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )}{e}-\frac {a^{2} b \left (2+\sin \left (e x +d \right )^{2}\right ) \cos \left (e x +d \right )}{3 e}\)	\(90\)
risch	\(\frac {a^{3} x}{2}+2 a \,b^{2} x -\frac {11 b \cos \left (e x +d \right ) a^{2}}{4 e}-\frac {b^{3} \cos \left (e x +d \right )}{e}+\frac {b \,a^{2} \cos \left (3 e x +3 d \right )}{12 e}-\frac {a^{3} \sin \left (2 e x +2 d \right )}{4 e}-\frac {a \sin \left (2 e x +2 d \right ) b^{2}}{2 e}\)	\(97\)
parallelrisch	\(\frac {6 a^{3} e x +24 a \,b^{2} e x +a^{2} b \cos \left (3 e x +3 d \right )-3 \sin \left (2 e x +2 d \right ) a^{3}-6 \sin \left (2 e x +2 d \right ) a \,b^{2}-33 \cos \left (e x +d \right ) a^{2} b -12 b^{3} \cos \left (e x +d \right )-32 a^{2} b -12 b^{3}}{12 e}\)	\(99\)
derivativedivides	\(\frac {-\frac {a^{2} b \left (2+\sin \left (e x +d \right )^{2}\right ) \cos \left (e x +d \right )}{3}+a^{3} \left (-\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )+2 a \,b^{2} \left (-\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )-2 \cos \left (e x +d \right ) a^{2} b -b^{3} \cos \left (e x +d \right )+a \,b^{2} \left (e x +d \right )}{e}\)	\(115\)
default	\(\frac {-\frac {a^{2} b \left (2+\sin \left (e x +d \right )^{2}\right ) \cos \left (e x +d \right )}{3}+a^{3} \left (-\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )+2 a \,b^{2} \left (-\frac {\cos \left (e x +d \right ) \sin \left (e x +d \right )}{2}+\frac {e x}{2}+\frac {d}{2}\right )-2 \cos \left (e x +d \right ) a^{2} b -b^{3} \cos \left (e x +d \right )+a \,b^{2} \left (e x +d \right )}{e}\)	\(115\)
norman	\(\frac {\left (2 a \,b^{2}+\frac {1}{2} a^{3}\right ) x +\left (\frac {3}{2} a^{3}+6 a \,b^{2}\right ) x \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}+\left (\frac {3}{2} a^{3}+6 a \,b^{2}\right ) x \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4}+\left (2 a \,b^{2}+\frac {1}{2} a^{3}\right ) x \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{6}+\frac {a \left (a^{2}+2 b^{2}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{5}}{e}-\frac {16 a^{2} b +6 b^{3}}{3 e}-\frac {\left (4 a^{2} b +2 b^{3}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4}}{e}-\frac {\left (12 a^{2} b +4 b^{3}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{e}-\frac {a \left (a^{2}+2 b^{2}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{e}}{\left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}\right )^{3}}\)	\(229\)

[In]

int((a+b*sin(e*x+d))*(b^2+2*a*b*sin(e*x+d)+a^2*sin(e*x+d)^2),x,method=_RETURNVERBOSE)

[Out]

a*b^2*x-(2*a^2*b+b^3)/e*cos(e*x+d)+(a^3+2*a*b^2)/e*(-1/2*cos(e*x+d)*sin(e*x+d)+1/2*e*x+1/2*d)-1/3*a^2*b/e*(2+s
in(e*x+d)^2)*cos(e*x+d)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.70 \[ \int (a+b \sin (d+e x)) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right ) \, dx=\frac {2 \, a^{2} b \cos \left (e x + d\right )^{3} + 3 \, {\left (a^{3} + 4 \, a b^{2}\right )} e x - 3 \, {\left (a^{3} + 2 \, a b^{2}\right )} \cos \left (e x + d\right ) \sin \left (e x + d\right ) - 6 \, {\left (3 \, a^{2} b + b^{3}\right )} \cos \left (e x + d\right )}{6 \, e} \]

[In]

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

[Out]

1/6*(2*a^2*b*cos(e*x + d)^3 + 3*(a^3 + 4*a*b^2)*e*x - 3*(a^3 + 2*a*b^2)*cos(e*x + d)*sin(e*x + d) - 6*(3*a^2*b
+ b^3)*cos(e*x + d))/e

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 204 vs. \(2 (95) = 190\).

Time = 0.14 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.87 \[ \int (a+b \sin (d+e x)) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right ) \, dx=\begin {cases} \frac {a^{3} x \sin ^{2}{\left (d + e x \right )}}{2} + \frac {a^{3} x \cos ^{2}{\left (d + e x \right )}}{2} - \frac {a^{3} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{2 e} - \frac {a^{2} b \sin ^{2}{\left (d + e x \right )} \cos {\left (d + e x \right )}}{e} - \frac {2 a^{2} b \cos ^{3}{\left (d + e x \right )}}{3 e} - \frac {2 a^{2} b \cos {\left (d + e x \right )}}{e} + a b^{2} x \sin ^{2}{\left (d + e x \right )} + a b^{2} x \cos ^{2}{\left (d + e x \right )} + a b^{2} x - \frac {a b^{2} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{e} - \frac {b^{3} \cos {\left (d + e x \right )}}{e} & \text {for}\: e \neq 0 \\x \left (a + b \sin {\left (d \right )}\right ) \left (a^{2} \sin ^{2}{\left (d \right )} + 2 a b \sin {\left (d \right )} + b^{2}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a**3*x*sin(d + e*x)**2/2 + a**3*x*cos(d + e*x)**2/2 - a**3*sin(d + e*x)*cos(d + e*x)/(2*e) - a**2*b
*sin(d + e*x)**2*cos(d + e*x)/e - 2*a**2*b*cos(d + e*x)**3/(3*e) - 2*a**2*b*cos(d + e*x)/e + a*b**2*x*sin(d +
e*x)**2 + a*b**2*x*cos(d + e*x)**2 + a*b**2*x - a*b**2*sin(d + e*x)*cos(d + e*x)/e - b**3*cos(d + e*x)/e, Ne(e
, 0)), (x*(a + b*sin(d))*(a**2*sin(d)**2 + 2*a*b*sin(d) + b**2), True))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.03 \[ \int (a+b \sin (d+e x)) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right ) \, dx=\frac {3 \, {\left (2 \, e x + 2 \, d - \sin \left (2 \, e x + 2 \, d\right )\right )} a^{3} + 4 \, {\left (\cos \left (e x + d\right )^{3} - 3 \, \cos \left (e x + d\right )\right )} a^{2} b + 6 \, {\left (2 \, e x + 2 \, d - \sin \left (2 \, e x + 2 \, d\right )\right )} a b^{2} + 12 \, {\left (e x + d\right )} a b^{2} - 24 \, a^{2} b \cos \left (e x + d\right ) - 12 \, b^{3} \cos \left (e x + d\right )}{12 \, e} \]

[In]

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

[Out]

1/12*(3*(2*e*x + 2*d - sin(2*e*x + 2*d))*a^3 + 4*(cos(e*x + d)^3 - 3*cos(e*x + d))*a^2*b + 6*(2*e*x + 2*d - si
n(2*e*x + 2*d))*a*b^2 + 12*(e*x + d)*a*b^2 - 24*a^2*b*cos(e*x + d) - 12*b^3*cos(e*x + d))/e

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.72 \[ \int (a+b \sin (d+e x)) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right ) \, dx=\frac {a^{2} b \cos \left (3 \, e x + 3 \, d\right )}{12 \, e} + \frac {1}{2} \, {\left (a^{3} + 4 \, a b^{2}\right )} x - \frac {{\left (11 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (e x + d\right )}{4 \, e} - \frac {{\left (a^{3} + 2 \, a b^{2}\right )} \sin \left (2 \, e x + 2 \, d\right )}{4 \, e} \]

[In]

integrate((a+b*sin(e*x+d))*(b^2+2*a*b*sin(e*x+d)+a^2*sin(e*x+d)^2),x, algorithm="giac")

[Out]

1/12*a^2*b*cos(3*e*x + 3*d)/e + 1/2*(a^3 + 4*a*b^2)*x - 1/4*(11*a^2*b + 4*b^3)*cos(e*x + d)/e - 1/4*(a^3 + 2*a
*b^2)*sin(2*e*x + 2*d)/e

Mupad [B] (verification not implemented)

Time = 27.24 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.81 \[ \int (a+b \sin (d+e x)) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right ) \, dx=-\frac {6\,b^3\,\cos \left (d+e\,x\right )+\frac {3\,a^3\,\sin \left (2\,d+2\,e\,x\right )}{2}-\frac {a^2\,b\,\cos \left (3\,d+3\,e\,x\right )}{2}+3\,a\,b^2\,\sin \left (2\,d+2\,e\,x\right )+\frac {33\,a^2\,b\,\cos \left (d+e\,x\right )}{2}-3\,a^3\,e\,x-12\,a\,b^2\,e\,x}{6\,e} \]

[In]

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

[Out]

-(6*b^3*cos(d + e*x) + (3*a^3*sin(2*d + 2*e*x))/2 - (a^2*b*cos(3*d + 3*e*x))/2 + 3*a*b^2*sin(2*d + 2*e*x) + (3
3*a^2*b*cos(d + e*x))/2 - 3*a^3*e*x - 12*a*b^2*e*x)/(6*e)