3.500 \(\int (a+b \sin (d+e x)) (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)) \, dx\)

Optimal. Leaf size=109 \[ \frac{\left (-8 a^2 b^2+a^4-3 b^4\right ) \cos (d+e x)}{3 b e}+\frac{a \left (a^2-6 b^2\right ) \sin (d+e x) \cos (d+e x)}{6 e}+\frac{1}{2} a x \left (a^2+4 b^2\right )-\frac{a^2 \cos (d+e x) (a+b \sin (d+e x))^2}{3 b e} \]

[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)

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Rubi [A]  time = 0.0987093, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.054, Rules used = {3023, 2734} \[ \frac{\left (-8 a^2 b^2+a^4-3 b^4\right ) \cos (d+e x)}{3 b e}+\frac{a \left (a^2-6 b^2\right ) \sin (d+e x) \cos (d+e x)}{6 e}+\frac{1}{2} a x \left (a^2+4 b^2\right )-\frac{a^2 \cos (d+e x) (a+b \sin (d+e x))^2}{3 b e} \]

Antiderivative was successfully verified.

[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 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 2734

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]

Rubi steps

\begin{align*} \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 \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]  time = 0.296101, size = 77, normalized size = 0.71 \[ \frac{a \left (6 \left (a^2+4 b^2\right ) (d+e x)-3 \left (a^2+2 b^2\right ) \sin (2 (d+e x))+a b \cos (3 (d+e x))\right )-3 b \left (11 a^2+4 b^2\right ) \cos (d+e x)}{12 e} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.024, size = 115, normalized size = 1.1 \begin{align*}{\frac{1}{e} \left ( -{\frac{{a}^{2}b \left ( 2+ \left ( \sin \left ( ex+d \right ) \right ) ^{2} \right ) \cos \left ( ex+d \right ) }{3}}+{a}^{3} \left ( -{\frac{\sin \left ( ex+d \right ) \cos \left ( ex+d \right ) }{2}}+{\frac{ex}{2}}+{\frac{d}{2}} \right ) +2\,a{b}^{2} \left ( -1/2\,\sin \left ( ex+d \right ) \cos \left ( ex+d \right ) +1/2\,ex+d/2 \right ) -2\,\cos \left ( ex+d \right ){a}^{2}b-\cos \left ( ex+d \right ){b}^{3}+a{b}^{2} \left ( ex+d \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

[Out]

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

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Maxima [A]  time = 0.991491, size = 151, normalized size = 1.39 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.74264, size = 182, normalized size = 1.67 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [A]  time = 0.781657, size = 204, normalized size = 1.87 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Giac [A]  time = 1.13899, size = 107, normalized size = 0.98 \begin{align*} \frac{1}{12} \, a^{2} b \cos \left (3 \, x e + 3 \, d\right ) e^{\left (-1\right )} - \frac{1}{4} \,{\left (11 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (x e + d\right ) e^{\left (-1\right )} - \frac{1}{4} \,{\left (a^{3} + 2 \, a b^{2}\right )} e^{\left (-1\right )} \sin \left (2 \, x e + 2 \, d\right ) + \frac{1}{2} \,{\left (a^{3} + 4 \, a b^{2}\right )} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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*x*e + 3*d)*e^(-1) - 1/4*(11*a^2*b + 4*b^3)*cos(x*e + d)*e^(-1) - 1/4*(a^3 + 2*a*b^2)*e^(-1)*s
in(2*x*e + 2*d) + 1/2*(a^3 + 4*a*b^2)*x