3.909 \(\int (1+2 x)^3 \sin ^2(1+2 x) \, dx\)

Optimal. Leaf size=99 \[ -\frac{3 x^2}{4}+\frac{1}{16} (2 x+1)^4-\frac{3 x}{4}+\frac{3}{8} (2 x+1)^2 \sin ^2(2 x+1)-\frac{3}{16} \sin ^2(2 x+1)-\frac{1}{4} (2 x+1)^3 \sin (2 x+1) \cos (2 x+1)+\frac{3}{8} (2 x+1) \sin (2 x+1) \cos (2 x+1) \]

[Out]

(-3*x)/4 - (3*x^2)/4 + (1 + 2*x)^4/16 + (3*(1 + 2*x)*Cos[1 + 2*x]*Sin[1 + 2*x])/8 - ((1 + 2*x)^3*Cos[1 + 2*x]*
Sin[1 + 2*x])/4 - (3*Sin[1 + 2*x]^2)/16 + (3*(1 + 2*x)^2*Sin[1 + 2*x]^2)/8

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Rubi [A]  time = 0.0595834, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {3311, 32, 3310} \[ -\frac{3 x^2}{4}+\frac{1}{16} (2 x+1)^4-\frac{3 x}{4}+\frac{3}{8} (2 x+1)^2 \sin ^2(2 x+1)-\frac{3}{16} \sin ^2(2 x+1)-\frac{1}{4} (2 x+1)^3 \sin (2 x+1) \cos (2 x+1)+\frac{3}{8} (2 x+1) \sin (2 x+1) \cos (2 x+1) \]

Antiderivative was successfully verified.

[In]

Int[(1 + 2*x)^3*Sin[1 + 2*x]^2,x]

[Out]

(-3*x)/4 - (3*x^2)/4 + (1 + 2*x)^4/16 + (3*(1 + 2*x)*Cos[1 + 2*x]*Sin[1 + 2*x])/8 - ((1 + 2*x)^3*Cos[1 + 2*x]*
Sin[1 + 2*x])/4 - (3*Sin[1 + 2*x]^2)/16 + (3*(1 + 2*x)^2*Sin[1 + 2*x]^2)/8

Rule 3311

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*m*(c + d*x)^(m - 1)*(
b*Sin[e + f*x])^n)/(f^2*n^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[(d^2*m*(m - 1))/(f^2*n^2), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[(b*(c + d*x)^m*Cos[e +
f*x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 3310

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*(b*Sin[e + f*x])^n)/(f^2*n
^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[(b*(c + d*x)*Cos[e + f*
x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

Rubi steps

\begin{align*} \int (1+2 x)^3 \sin ^2(1+2 x) \, dx &=-\frac{1}{4} (1+2 x)^3 \cos (1+2 x) \sin (1+2 x)+\frac{3}{8} (1+2 x)^2 \sin ^2(1+2 x)+\frac{1}{2} \int (1+2 x)^3 \, dx-\frac{3}{2} \int (1+2 x) \sin ^2(1+2 x) \, dx\\ &=\frac{1}{16} (1+2 x)^4+\frac{3}{8} (1+2 x) \cos (1+2 x) \sin (1+2 x)-\frac{1}{4} (1+2 x)^3 \cos (1+2 x) \sin (1+2 x)-\frac{3}{16} \sin ^2(1+2 x)+\frac{3}{8} (1+2 x)^2 \sin ^2(1+2 x)-\frac{3}{4} \int (1+2 x) \, dx\\ &=-\frac{3 x}{4}-\frac{3 x^2}{4}+\frac{1}{16} (1+2 x)^4+\frac{3}{8} (1+2 x) \cos (1+2 x) \sin (1+2 x)-\frac{1}{4} (1+2 x)^3 \cos (1+2 x) \sin (1+2 x)-\frac{3}{16} \sin ^2(1+2 x)+\frac{3}{8} (1+2 x)^2 \sin ^2(1+2 x)\\ \end{align*}

Mathematica [A]  time = 0.229936, size = 55, normalized size = 0.56 \[ \frac{1}{32} \left (2 (2 x+1) \left (\left (-8 x^2-8 x+1\right ) \sin (4 x+2)+(2 x+1)^3\right )-3 \left (8 x^2+8 x+1\right ) \cos (4 x+2)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + 2*x)^3*Sin[1 + 2*x]^2,x]

[Out]

(-3*(1 + 8*x + 8*x^2)*Cos[2 + 4*x] + 2*(1 + 2*x)*((1 + 2*x)^3 + (1 - 8*x - 8*x^2)*Sin[2 + 4*x]))/32

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Maple [A]  time = 0.019, size = 97, normalized size = 1. \begin{align*}{\frac{ \left ( 1+2\,x \right ) ^{3}}{2} \left ( -{\frac{\sin \left ( 1+2\,x \right ) \cos \left ( 1+2\,x \right ) }{2}}+x+{\frac{1}{2}} \right ) }-{\frac{3\, \left ( \cos \left ( 1+2\,x \right ) \right ) ^{2} \left ( 1+2\,x \right ) ^{2}}{8}}+{\frac{3+6\,x}{4} \left ({\frac{\sin \left ( 1+2\,x \right ) \cos \left ( 1+2\,x \right ) }{2}}+x+{\frac{1}{2}} \right ) }-{\frac{3\, \left ( 1+2\,x \right ) ^{2}}{16}}-{\frac{3\, \left ( \sin \left ( 1+2\,x \right ) \right ) ^{2}}{16}}-{\frac{3\, \left ( 1+2\,x \right ) ^{4}}{16}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+2*x)^3*sin(1+2*x)^2,x)

[Out]

1/2*(1+2*x)^3*(-1/2*sin(1+2*x)*cos(1+2*x)+x+1/2)-3/8*cos(1+2*x)^2*(1+2*x)^2+3/4*(1+2*x)*(1/2*sin(1+2*x)*cos(1+
2*x)+x+1/2)-3/16*(1+2*x)^2-3/16*sin(1+2*x)^2-3/16*(1+2*x)^4

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Maxima [A]  time = 0.983696, size = 69, normalized size = 0.7 \begin{align*} \frac{1}{16} \,{\left (2 \, x + 1\right )}^{4} - \frac{3}{32} \,{\left (2 \,{\left (2 \, x + 1\right )}^{2} - 1\right )} \cos \left (4 \, x + 2\right ) - \frac{1}{16} \,{\left (2 \,{\left (2 \, x + 1\right )}^{3} - 6 \, x - 3\right )} \sin \left (4 \, x + 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)^3*sin(1+2*x)^2,x, algorithm="maxima")

[Out]

1/16*(2*x + 1)^4 - 3/32*(2*(2*x + 1)^2 - 1)*cos(4*x + 2) - 1/16*(2*(2*x + 1)^3 - 6*x - 3)*sin(4*x + 2)

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Fricas [A]  time = 2.15695, size = 177, normalized size = 1.79 \begin{align*} x^{4} + 2 \, x^{3} - \frac{3}{16} \,{\left (8 \, x^{2} + 8 \, x + 1\right )} \cos \left (2 \, x + 1\right )^{2} - \frac{1}{8} \,{\left (16 \, x^{3} + 24 \, x^{2} + 6 \, x - 1\right )} \cos \left (2 \, x + 1\right ) \sin \left (2 \, x + 1\right ) + \frac{9}{4} \, x^{2} + \frac{5}{4} \, x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)^3*sin(1+2*x)^2,x, algorithm="fricas")

[Out]

x^4 + 2*x^3 - 3/16*(8*x^2 + 8*x + 1)*cos(2*x + 1)^2 - 1/8*(16*x^3 + 24*x^2 + 6*x - 1)*cos(2*x + 1)*sin(2*x + 1
) + 9/4*x^2 + 5/4*x

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Sympy [B]  time = 1.31698, size = 189, normalized size = 1.91 \begin{align*} x^{4} \sin ^{2}{\left (2 x + 1 \right )} + x^{4} \cos ^{2}{\left (2 x + 1 \right )} + 2 x^{3} \sin ^{2}{\left (2 x + 1 \right )} - 2 x^{3} \sin{\left (2 x + 1 \right )} \cos{\left (2 x + 1 \right )} + 2 x^{3} \cos ^{2}{\left (2 x + 1 \right )} + \frac{9 x^{2} \sin ^{2}{\left (2 x + 1 \right )}}{4} - 3 x^{2} \sin{\left (2 x + 1 \right )} \cos{\left (2 x + 1 \right )} + \frac{3 x^{2} \cos ^{2}{\left (2 x + 1 \right )}}{4} + \frac{5 x \sin ^{2}{\left (2 x + 1 \right )}}{4} - \frac{3 x \sin{\left (2 x + 1 \right )} \cos{\left (2 x + 1 \right )}}{4} - \frac{x \cos ^{2}{\left (2 x + 1 \right )}}{4} + \frac{\sin{\left (2 x + 1 \right )} \cos{\left (2 x + 1 \right )}}{8} - \frac{3 \cos ^{2}{\left (2 x + 1 \right )}}{16} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)**3*sin(1+2*x)**2,x)

[Out]

x**4*sin(2*x + 1)**2 + x**4*cos(2*x + 1)**2 + 2*x**3*sin(2*x + 1)**2 - 2*x**3*sin(2*x + 1)*cos(2*x + 1) + 2*x*
*3*cos(2*x + 1)**2 + 9*x**2*sin(2*x + 1)**2/4 - 3*x**2*sin(2*x + 1)*cos(2*x + 1) + 3*x**2*cos(2*x + 1)**2/4 +
5*x*sin(2*x + 1)**2/4 - 3*x*sin(2*x + 1)*cos(2*x + 1)/4 - x*cos(2*x + 1)**2/4 + sin(2*x + 1)*cos(2*x + 1)/8 -
3*cos(2*x + 1)**2/16

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Giac [A]  time = 1.0762, size = 78, normalized size = 0.79 \begin{align*} x^{4} + 2 \, x^{3} + \frac{3}{2} \, x^{2} - \frac{3}{32} \,{\left (8 \, x^{2} + 8 \, x + 1\right )} \cos \left (4 \, x + 2\right ) - \frac{1}{16} \,{\left (16 \, x^{3} + 24 \, x^{2} + 6 \, x - 1\right )} \sin \left (4 \, x + 2\right ) + \frac{1}{2} \, x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*x)^3*sin(1+2*x)^2,x, algorithm="giac")

[Out]

x^4 + 2*x^3 + 3/2*x^2 - 3/32*(8*x^2 + 8*x + 1)*cos(4*x + 2) - 1/16*(16*x^3 + 24*x^2 + 6*x - 1)*sin(4*x + 2) +
1/2*x