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

Optimal. Leaf size=195 \[ -\frac{b \left (69 a^2 b^2+32 a^4+4 b^4\right ) \cos (d+e x)}{10 e}-\frac{\left (5 a^2+4 b^2\right ) \cos (d+e x) (a \sin (d+e x)+b)^3}{20 e}-\frac{b \left (17 a^2+4 b^2\right ) \cos (d+e x) (a \sin (d+e x)+b)^2}{20 e}-\frac{a \left (82 a^2 b^2+15 a^4+8 b^4\right ) \sin (d+e x) \cos (d+e x)}{40 e}+\frac{3}{8} a x \left (12 a^2 b^2+a^4+8 b^4\right )-\frac{b \cos (d+e x) (a \sin (d+e x)+b)^4}{5 e} \]

[Out]

(3*a*(a^4 + 12*a^2*b^2 + 8*b^4)*x)/8 - (b*(32*a^4 + 69*a^2*b^2 + 4*b^4)*Cos[d + e*x])/(10*e) - (a*(15*a^4 + 82
*a^2*b^2 + 8*b^4)*Cos[d + e*x]*Sin[d + e*x])/(40*e) - (b*(17*a^2 + 4*b^2)*Cos[d + e*x]*(b + a*Sin[d + e*x])^2)
/(20*e) - ((5*a^2 + 4*b^2)*Cos[d + e*x]*(b + a*Sin[d + e*x])^3)/(20*e) - (b*Cos[d + e*x]*(b + a*Sin[d + e*x])^
4)/(5*e)

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Rubi [A]  time = 0.392681, antiderivative size = 195, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {3288, 2753, 2734} \[ -\frac{b \left (69 a^2 b^2+32 a^4+4 b^4\right ) \cos (d+e x)}{10 e}-\frac{\left (5 a^2+4 b^2\right ) \cos (d+e x) (a \sin (d+e x)+b)^3}{20 e}-\frac{b \left (17 a^2+4 b^2\right ) \cos (d+e x) (a \sin (d+e x)+b)^2}{20 e}-\frac{a \left (82 a^2 b^2+15 a^4+8 b^4\right ) \sin (d+e x) \cos (d+e x)}{40 e}+\frac{3}{8} a x \left (12 a^2 b^2+a^4+8 b^4\right )-\frac{b \cos (d+e x) (a \sin (d+e x)+b)^4}{5 e} \]

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)^2,x]

[Out]

(3*a*(a^4 + 12*a^2*b^2 + 8*b^4)*x)/8 - (b*(32*a^4 + 69*a^2*b^2 + 4*b^4)*Cos[d + e*x])/(10*e) - (a*(15*a^4 + 82
*a^2*b^2 + 8*b^4)*Cos[d + e*x]*Sin[d + e*x])/(40*e) - (b*(17*a^2 + 4*b^2)*Cos[d + e*x]*(b + a*Sin[d + e*x])^2)
/(20*e) - ((5*a^2 + 4*b^2)*Cos[d + e*x]*(b + a*Sin[d + e*x])^3)/(20*e) - (b*Cos[d + e*x]*(b + a*Sin[d + e*x])^
4)/(5*e)

Rule 3288

Int[((A_) + (B_.)*sin[(d_.) + (e_.)*(x_)])*((a_) + (b_.)*sin[(d_.) + (e_.)*(x_)] + (c_.)*sin[(d_.) + (e_.)*(x_
)]^2)^(n_), x_Symbol] :> Dist[1/(4^n*c^n), Int[(A + B*Sin[d + e*x])*(b + 2*c*Sin[d + e*x])^(2*n), x], x] /; Fr
eeQ[{a, b, c, d, e, A, B}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[n]

Rule 2753

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

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 )^2 \, dx &=\frac{\int \left (2 a b+2 a^2 \sin (d+e x)\right )^4 (a+b \sin (d+e x)) \, dx}{16 a^4}\\ &=-\frac{b \cos (d+e x) (b+a \sin (d+e x))^4}{5 e}+\frac{\int \left (2 a b+2 a^2 \sin (d+e x)\right )^3 \left (18 a^2 b+2 a \left (5 a^2+4 b^2\right ) \sin (d+e x)\right ) \, dx}{80 a^4}\\ &=-\frac{\left (5 a^2+4 b^2\right ) \cos (d+e x) (b+a \sin (d+e x))^3}{20 e}-\frac{b \cos (d+e x) (b+a \sin (d+e x))^4}{5 e}+\frac{\int \left (2 a b+2 a^2 \sin (d+e x)\right )^2 \left (12 a^3 \left (5 a^2+16 b^2\right )+12 a^2 b \left (17 a^2+4 b^2\right ) \sin (d+e x)\right ) \, dx}{320 a^4}\\ &=-\frac{b \left (17 a^2+4 b^2\right ) \cos (d+e x) (b+a \sin (d+e x))^2}{20 e}-\frac{\left (5 a^2+4 b^2\right ) \cos (d+e x) (b+a \sin (d+e x))^3}{20 e}-\frac{b \cos (d+e x) (b+a \sin (d+e x))^4}{5 e}+\frac{\int \left (2 a b+2 a^2 \sin (d+e x)\right ) \left (168 a^4 b \left (7 a^2+8 b^2\right )+24 a^3 \left (15 a^4+82 a^2 b^2+8 b^4\right ) \sin (d+e x)\right ) \, dx}{960 a^4}\\ &=\frac{3}{8} a \left (a^4+12 a^2 b^2+8 b^4\right ) x-\frac{b \left (32 a^4+69 a^2 b^2+4 b^4\right ) \cos (d+e x)}{10 e}-\frac{a \left (15 a^4+82 a^2 b^2+8 b^4\right ) \cos (d+e x) \sin (d+e x)}{40 e}-\frac{b \left (17 a^2+4 b^2\right ) \cos (d+e x) (b+a \sin (d+e x))^2}{20 e}-\frac{\left (5 a^2+4 b^2\right ) \cos (d+e x) (b+a \sin (d+e x))^3}{20 e}-\frac{b \cos (d+e x) (b+a \sin (d+e x))^4}{5 e}\\ \end{align*}

Mathematica [A]  time = 0.942855, size = 149, normalized size = 0.76 \[ \frac{a \left (60 \left (12 a^2 b^2+a^4+8 b^4\right ) (d+e x)-40 \left (10 a^2 b^2+a^4+4 b^4\right ) \sin (2 (d+e x))+5 \left (4 a^2 b^2+a^4\right ) \sin (4 (d+e x))+10 \left (7 a^3 b+8 a b^3\right ) \cos (3 (d+e x))-2 a^3 b \cos (5 (d+e x))\right )-20 b \left (68 a^2 b^2+29 a^4+8 b^4\right ) \cos (d+e x)}{160 e} \]

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)^2,x]

[Out]

(-20*b*(29*a^4 + 68*a^2*b^2 + 8*b^4)*Cos[d + e*x] + a*(60*(a^4 + 12*a^2*b^2 + 8*b^4)*(d + e*x) + 10*(7*a^3*b +
8*a*b^3)*Cos[3*(d + e*x)] - 2*a^3*b*Cos[5*(d + e*x)] - 40*(a^4 + 10*a^2*b^2 + 4*b^4)*Sin[2*(d + e*x)] + 5*(a^
4 + 4*a^2*b^2)*Sin[4*(d + e*x)]))/(160*e)

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Maple [A]  time = 0.035, size = 255, normalized size = 1.3 \begin{align*}{\frac{1}{e} \left ( a{b}^{4} \left ( ex+d \right ) -4\,\cos \left ( ex+d \right ){a}^{2}{b}^{3}+6\,{a}^{3}{b}^{2} \left ( -1/2\,\sin \left ( ex+d \right ) \cos \left ( ex+d \right ) +1/2\,ex+d/2 \right ) -{\frac{4\,{a}^{4}b \left ( 2+ \left ( \sin \left ( ex+d \right ) \right ) ^{2} \right ) \cos \left ( ex+d \right ) }{3}}+{a}^{5} \left ( -{\frac{\cos \left ( ex+d \right ) }{4} \left ( \left ( \sin \left ( ex+d \right ) \right ) ^{3}+{\frac{3\,\sin \left ( ex+d \right ) }{2}} \right ) }+{\frac{3\,ex}{8}}+{\frac{3\,d}{8}} \right ) -\cos \left ( ex+d \right ){b}^{5}+4\,a{b}^{4} \left ( -1/2\,\sin \left ( ex+d \right ) \cos \left ( ex+d \right ) +1/2\,ex+d/2 \right ) -2\,{a}^{2}{b}^{3} \left ( 2+ \left ( \sin \left ( ex+d \right ) \right ) ^{2} \right ) \cos \left ( ex+d \right ) +4\,{a}^{3}{b}^{2} \left ( -1/4\, \left ( \left ( \sin \left ( ex+d \right ) \right ) ^{3}+3/2\,\sin \left ( ex+d \right ) \right ) \cos \left ( ex+d \right ) +3/8\,ex+3/8\,d \right ) -{\frac{{a}^{4}b\cos \left ( ex+d \right ) }{5} \left ({\frac{8}{3}}+ \left ( \sin \left ( ex+d \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( ex+d \right ) \right ) ^{2}}{3}} \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)^2,x)

[Out]

1/e*(a*b^4*(e*x+d)-4*cos(e*x+d)*a^2*b^3+6*a^3*b^2*(-1/2*sin(e*x+d)*cos(e*x+d)+1/2*e*x+1/2*d)-4/3*a^4*b*(2+sin(
e*x+d)^2)*cos(e*x+d)+a^5*(-1/4*(sin(e*x+d)^3+3/2*sin(e*x+d))*cos(e*x+d)+3/8*e*x+3/8*d)-cos(e*x+d)*b^5+4*a*b^4*
(-1/2*sin(e*x+d)*cos(e*x+d)+1/2*e*x+1/2*d)-2*a^2*b^3*(2+sin(e*x+d)^2)*cos(e*x+d)+4*a^3*b^2*(-1/4*(sin(e*x+d)^3
+3/2*sin(e*x+d))*cos(e*x+d)+3/8*e*x+3/8*d)-1/5*a^4*b*(8/3+sin(e*x+d)^4+4/3*sin(e*x+d)^2)*cos(e*x+d))

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Maxima [A]  time = 1.03736, size = 332, normalized size = 1.7 \begin{align*} \frac{15 \,{\left (12 \, e x + 12 \, d + \sin \left (4 \, e x + 4 \, d\right ) - 8 \, \sin \left (2 \, e x + 2 \, d\right )\right )} a^{5} - 32 \,{\left (3 \, \cos \left (e x + d\right )^{5} - 10 \, \cos \left (e x + d\right )^{3} + 15 \, \cos \left (e x + d\right )\right )} a^{4} b + 640 \,{\left (\cos \left (e x + d\right )^{3} - 3 \, \cos \left (e x + d\right )\right )} a^{4} b + 60 \,{\left (12 \, e x + 12 \, d + \sin \left (4 \, e x + 4 \, d\right ) - 8 \, \sin \left (2 \, e x + 2 \, d\right )\right )} a^{3} b^{2} + 720 \,{\left (2 \, e x + 2 \, d - \sin \left (2 \, e x + 2 \, d\right )\right )} a^{3} b^{2} + 960 \,{\left (\cos \left (e x + d\right )^{3} - 3 \, \cos \left (e x + d\right )\right )} a^{2} b^{3} + 480 \,{\left (2 \, e x + 2 \, d - \sin \left (2 \, e x + 2 \, d\right )\right )} a b^{4} + 480 \,{\left (e x + d\right )} a b^{4} - 1920 \, a^{2} b^{3} \cos \left (e x + d\right ) - 480 \, b^{5} \cos \left (e x + d\right )}{480 \, 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)^2,x, algorithm="maxima")

[Out]

1/480*(15*(12*e*x + 12*d + sin(4*e*x + 4*d) - 8*sin(2*e*x + 2*d))*a^5 - 32*(3*cos(e*x + d)^5 - 10*cos(e*x + d)
^3 + 15*cos(e*x + d))*a^4*b + 640*(cos(e*x + d)^3 - 3*cos(e*x + d))*a^4*b + 60*(12*e*x + 12*d + sin(4*e*x + 4*
d) - 8*sin(2*e*x + 2*d))*a^3*b^2 + 720*(2*e*x + 2*d - sin(2*e*x + 2*d))*a^3*b^2 + 960*(cos(e*x + d)^3 - 3*cos(
e*x + d))*a^2*b^3 + 480*(2*e*x + 2*d - sin(2*e*x + 2*d))*a*b^4 + 480*(e*x + d)*a*b^4 - 1920*a^2*b^3*cos(e*x +
d) - 480*b^5*cos(e*x + d))/e

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Fricas [A]  time = 1.82638, size = 348, normalized size = 1.78 \begin{align*} -\frac{8 \, a^{4} b \cos \left (e x + d\right )^{5} - 80 \,{\left (a^{4} b + a^{2} b^{3}\right )} \cos \left (e x + d\right )^{3} - 15 \,{\left (a^{5} + 12 \, a^{3} b^{2} + 8 \, a b^{4}\right )} e x + 40 \,{\left (5 \, a^{4} b + 10 \, a^{2} b^{3} + b^{5}\right )} \cos \left (e x + d\right ) - 5 \,{\left (2 \,{\left (a^{5} + 4 \, a^{3} b^{2}\right )} \cos \left (e x + d\right )^{3} -{\left (5 \, a^{5} + 44 \, a^{3} b^{2} + 16 \, a b^{4}\right )} \cos \left (e x + d\right )\right )} \sin \left (e x + d\right )}{40 \, 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)^2,x, algorithm="fricas")

[Out]

-1/40*(8*a^4*b*cos(e*x + d)^5 - 80*(a^4*b + a^2*b^3)*cos(e*x + d)^3 - 15*(a^5 + 12*a^3*b^2 + 8*a*b^4)*e*x + 40
*(5*a^4*b + 10*a^2*b^3 + b^5)*cos(e*x + d) - 5*(2*(a^5 + 4*a^3*b^2)*cos(e*x + d)^3 - (5*a^5 + 44*a^3*b^2 + 16*
a*b^4)*cos(e*x + d))*sin(e*x + d))/e

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Sympy [A]  time = 3.51857, size = 566, normalized size = 2.9 \begin{align*} \begin{cases} \frac{3 a^{5} x \sin ^{4}{\left (d + e x \right )}}{8} + \frac{3 a^{5} x \sin ^{2}{\left (d + e x \right )} \cos ^{2}{\left (d + e x \right )}}{4} + \frac{3 a^{5} x \cos ^{4}{\left (d + e x \right )}}{8} - \frac{5 a^{5} \sin ^{3}{\left (d + e x \right )} \cos{\left (d + e x \right )}}{8 e} - \frac{3 a^{5} \sin{\left (d + e x \right )} \cos ^{3}{\left (d + e x \right )}}{8 e} - \frac{a^{4} b \sin ^{4}{\left (d + e x \right )} \cos{\left (d + e x \right )}}{e} - \frac{4 a^{4} b \sin ^{2}{\left (d + e x \right )} \cos ^{3}{\left (d + e x \right )}}{3 e} - \frac{4 a^{4} b \sin ^{2}{\left (d + e x \right )} \cos{\left (d + e x \right )}}{e} - \frac{8 a^{4} b \cos ^{5}{\left (d + e x \right )}}{15 e} - \frac{8 a^{4} b \cos ^{3}{\left (d + e x \right )}}{3 e} + \frac{3 a^{3} b^{2} x \sin ^{4}{\left (d + e x \right )}}{2} + 3 a^{3} b^{2} x \sin ^{2}{\left (d + e x \right )} \cos ^{2}{\left (d + e x \right )} + 3 a^{3} b^{2} x \sin ^{2}{\left (d + e x \right )} + \frac{3 a^{3} b^{2} x \cos ^{4}{\left (d + e x \right )}}{2} + 3 a^{3} b^{2} x \cos ^{2}{\left (d + e x \right )} - \frac{5 a^{3} b^{2} \sin ^{3}{\left (d + e x \right )} \cos{\left (d + e x \right )}}{2 e} - \frac{3 a^{3} b^{2} \sin{\left (d + e x \right )} \cos ^{3}{\left (d + e x \right )}}{2 e} - \frac{3 a^{3} b^{2} \sin{\left (d + e x \right )} \cos{\left (d + e x \right )}}{e} - \frac{6 a^{2} b^{3} \sin ^{2}{\left (d + e x \right )} \cos{\left (d + e x \right )}}{e} - \frac{4 a^{2} b^{3} \cos ^{3}{\left (d + e x \right )}}{e} - \frac{4 a^{2} b^{3} \cos{\left (d + e x \right )}}{e} + 2 a b^{4} x \sin ^{2}{\left (d + e x \right )} + 2 a b^{4} x \cos ^{2}{\left (d + e x \right )} + a b^{4} x - \frac{2 a b^{4} \sin{\left (d + e x \right )} \cos{\left (d + e x \right )}}{e} - \frac{b^{5} \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 )^{2} & \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)**2,x)

[Out]

Piecewise((3*a**5*x*sin(d + e*x)**4/8 + 3*a**5*x*sin(d + e*x)**2*cos(d + e*x)**2/4 + 3*a**5*x*cos(d + e*x)**4/
8 - 5*a**5*sin(d + e*x)**3*cos(d + e*x)/(8*e) - 3*a**5*sin(d + e*x)*cos(d + e*x)**3/(8*e) - a**4*b*sin(d + e*x
)**4*cos(d + e*x)/e - 4*a**4*b*sin(d + e*x)**2*cos(d + e*x)**3/(3*e) - 4*a**4*b*sin(d + e*x)**2*cos(d + e*x)/e
- 8*a**4*b*cos(d + e*x)**5/(15*e) - 8*a**4*b*cos(d + e*x)**3/(3*e) + 3*a**3*b**2*x*sin(d + e*x)**4/2 + 3*a**3
*b**2*x*sin(d + e*x)**2*cos(d + e*x)**2 + 3*a**3*b**2*x*sin(d + e*x)**2 + 3*a**3*b**2*x*cos(d + e*x)**4/2 + 3*
a**3*b**2*x*cos(d + e*x)**2 - 5*a**3*b**2*sin(d + e*x)**3*cos(d + e*x)/(2*e) - 3*a**3*b**2*sin(d + e*x)*cos(d
+ e*x)**3/(2*e) - 3*a**3*b**2*sin(d + e*x)*cos(d + e*x)/e - 6*a**2*b**3*sin(d + e*x)**2*cos(d + e*x)/e - 4*a**
2*b**3*cos(d + e*x)**3/e - 4*a**2*b**3*cos(d + e*x)/e + 2*a*b**4*x*sin(d + e*x)**2 + 2*a*b**4*x*cos(d + e*x)**
2 + a*b**4*x - 2*a*b**4*sin(d + e*x)*cos(d + e*x)/e - b**5*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)**2, True))

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Giac [A]  time = 1.16671, size = 213, normalized size = 1.09 \begin{align*} -\frac{1}{80} \, a^{4} b \cos \left (5 \, x e + 5 \, d\right ) e^{\left (-1\right )} + \frac{1}{16} \,{\left (7 \, a^{4} b + 8 \, a^{2} b^{3}\right )} \cos \left (3 \, x e + 3 \, d\right ) e^{\left (-1\right )} - \frac{1}{8} \,{\left (29 \, a^{4} b + 68 \, a^{2} b^{3} + 8 \, b^{5}\right )} \cos \left (x e + d\right ) e^{\left (-1\right )} + \frac{1}{32} \,{\left (a^{5} + 4 \, a^{3} b^{2}\right )} e^{\left (-1\right )} \sin \left (4 \, x e + 4 \, d\right ) - \frac{1}{4} \,{\left (a^{5} + 10 \, a^{3} b^{2} + 4 \, a b^{4}\right )} e^{\left (-1\right )} \sin \left (2 \, x e + 2 \, d\right ) + \frac{3}{8} \,{\left (a^{5} + 12 \, a^{3} b^{2} + 8 \, a b^{4}\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)^2,x, algorithm="giac")

[Out]

-1/80*a^4*b*cos(5*x*e + 5*d)*e^(-1) + 1/16*(7*a^4*b + 8*a^2*b^3)*cos(3*x*e + 3*d)*e^(-1) - 1/8*(29*a^4*b + 68*
a^2*b^3 + 8*b^5)*cos(x*e + d)*e^(-1) + 1/32*(a^5 + 4*a^3*b^2)*e^(-1)*sin(4*x*e + 4*d) - 1/4*(a^5 + 10*a^3*b^2
+ 4*a*b^4)*e^(-1)*sin(2*x*e + 2*d) + 3/8*(a^5 + 12*a^3*b^2 + 8*a*b^4)*x