\(\int (1+2 x)^3 \sin ^2(1+2 x) \, dx\) [909]

Optimal result

Integrand size = 16, antiderivative size = 99 \[ \int (1+2 x)^3 \sin ^2(1+2 x) \, dx=-\frac {3 x}{4}-\frac {3 x^2}{4}+\frac {1}{16} (1+2 x)^4+\frac {3}{8} (1+2 x) \cos (1+2 x) \sin (1+2 x)-\frac {1}{4} (1+2 x)^3 \cos (1+2 x) \sin (1+2 x)-\frac {3}{16} \sin ^2(1+2 x)+\frac {3}{8} (1+2 x)^2 \sin ^2(1+2 x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {3392, 32, 3391} \[ \int (1+2 x)^3 \sin ^2(1+2 x) \, dx=-\frac {3 x^2}{4}+\frac {1}{16} (2 x+1)^4-\frac {3 x}{4}+\frac {3}{8} (2 x+1)^2 \sin ^2(2 x+1)-\frac {3}{16} \sin ^2(2 x+1)-\frac {1}{4} (2 x+1)^3 \sin (2 x+1) \cos (2 x+1)+\frac {3}{8} (2 x+1) \sin (2 x+1) \cos (2 x+1) \]

[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 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 3391

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]

Rule 3392

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]

Rubi steps \begin{align*} \text {integral}& = -\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] (verified)

Time = 0.15 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.56 \[ \int (1+2 x)^3 \sin ^2(1+2 x) \, dx=\frac {1}{32} \left (-3 \left (1+8 x+8 x^2\right ) \cos (2+4 x)+2 (1+2 x) \left ((1+2 x)^3+\left (1-8 x-8 x^2\right ) \sin (2+4 x)\right )\right ) \]

[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

Maple [A] (verified)

Time = 1.62 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.61

method	result	size
risch	\(x^{4}+2 x^{3}+\frac {3 x^{2}}{2}+\frac {x}{2}+\frac {1}{16}-\frac {3 \left (8 x^{2}+8 x +1\right ) \cos \left (2+4 x \right )}{32}-\frac {\left (16 x^{3}+24 x^{2}+6 x -1\right ) \sin \left (2+4 x \right )}{16}\)	\(60\)
derivativedivides	\(\frac {\left (1+2 x \right )^{3} \left (-\frac {\cos \left (1+2 x \right ) \sin \left (1+2 x \right )}{2}+\frac {1}{2}+x \right )}{2}-\frac {3 \left (1+2 x \right )^{2} \cos \left (1+2 x \right )^{2}}{8}+\frac {3 \left (1+2 x \right ) \left (\frac {\cos \left (1+2 x \right ) \sin \left (1+2 x \right )}{2}+\frac {1}{2}+x \right )}{4}-\frac {3 \left (1+2 x \right )^{2}}{16}-\frac {3 \sin \left (1+2 x \right )^{2}}{16}-\frac {3 \left (1+2 x \right )^{4}}{16}\)	\(97\)
default	\(\frac {\left (1+2 x \right )^{3} \left (-\frac {\cos \left (1+2 x \right ) \sin \left (1+2 x \right )}{2}+\frac {1}{2}+x \right )}{2}-\frac {3 \left (1+2 x \right )^{2} \cos \left (1+2 x \right )^{2}}{8}+\frac {3 \left (1+2 x \right ) \left (\frac {\cos \left (1+2 x \right ) \sin \left (1+2 x \right )}{2}+\frac {1}{2}+x \right )}{4}-\frac {3 \left (1+2 x \right )^{2}}{16}-\frac {3 \sin \left (1+2 x \right )^{2}}{16}-\frac {3 \left (1+2 x \right )^{4}}{16}\)	\(97\)
norman	\(\frac {x^{4}+x^{4} \tan \left (x +\frac {1}{2}\right )^{4}+\frac {3 \tan \left (x +\frac {1}{2}\right )^{2}}{4}-\frac {x}{4}+\frac {3 x^{2}}{4}+2 x^{3}-\frac {\tan \left (x +\frac {1}{2}\right )^{3}}{4}-\frac {3 x \tan \left (x +\frac {1}{2}\right )}{2}+\frac {11 x \tan \left (x +\frac {1}{2}\right )^{2}}{2}+\frac {3 x \tan \left (x +\frac {1}{2}\right )^{3}}{2}-\frac {x \tan \left (x +\frac {1}{2}\right )^{4}}{4}-6 x^{2} \tan \left (x +\frac {1}{2}\right )+\frac {15 x^{2} \tan \left (x +\frac {1}{2}\right )^{2}}{2}+6 x^{2} \tan \left (x +\frac {1}{2}\right )^{3}+\frac {3 x^{2} \tan \left (x +\frac {1}{2}\right )^{4}}{4}-4 x^{3} \tan \left (x +\frac {1}{2}\right )+4 x^{3} \tan \left (x +\frac {1}{2}\right )^{2}+4 x^{3} \tan \left (x +\frac {1}{2}\right )^{3}+2 x^{3} \tan \left (x +\frac {1}{2}\right )^{4}+2 x^{4} \tan \left (x +\frac {1}{2}\right )^{2}+\frac {\tan \left (x +\frac {1}{2}\right )}{4}}{\left (1+\tan \left (x +\frac {1}{2}\right )^{2}\right )^{2}}\)	\(190\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.67 \[ \int (1+2 x)^3 \sin ^2(1+2 x) \, dx=x^{4} + 2 \, x^{3} - \frac {3}{16} \, {\left (8 \, x^{2} + 8 \, x + 1\right )} \cos \left (2 \, x + 1\right )^{2} - \frac {1}{8} \, {\left (16 \, x^{3} + 24 \, x^{2} + 6 \, x - 1\right )} \cos \left (2 \, x + 1\right ) \sin \left (2 \, x + 1\right ) + \frac {9}{4} \, x^{2} + \frac {5}{4} \, x \]

[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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (94) = 188\).

Time = 0.19 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.91 \[ \int (1+2 x)^3 \sin ^2(1+2 x) \, dx=x^{4} \sin ^{2}{\left (2 x + 1 \right )} + x^{4} \cos ^{2}{\left (2 x + 1 \right )} + 2 x^{3} \sin ^{2}{\left (2 x + 1 \right )} - 2 x^{3} \sin {\left (2 x + 1 \right )} \cos {\left (2 x + 1 \right )} + 2 x^{3} \cos ^{2}{\left (2 x + 1 \right )} + \frac {9 x^{2} \sin ^{2}{\left (2 x + 1 \right )}}{4} - 3 x^{2} \sin {\left (2 x + 1 \right )} \cos {\left (2 x + 1 \right )} + \frac {3 x^{2} \cos ^{2}{\left (2 x + 1 \right )}}{4} + \frac {5 x \sin ^{2}{\left (2 x + 1 \right )}}{4} - \frac {3 x \sin {\left (2 x + 1 \right )} \cos {\left (2 x + 1 \right )}}{4} - \frac {x \cos ^{2}{\left (2 x + 1 \right )}}{4} + \frac {3 \sin ^{2}{\left (2 x + 1 \right )}}{16} + \frac {\sin {\left (2 x + 1 \right )} \cos {\left (2 x + 1 \right )}}{8} \]

[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 + 3*sin(2*x + 1)**2/16 + sin(2*x
+ 1)*cos(2*x + 1)/8

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.52 \[ \int (1+2 x)^3 \sin ^2(1+2 x) \, dx=\frac {1}{16} \, {\left (2 \, x + 1\right )}^{4} - \frac {3}{32} \, {\left (2 \, {\left (2 \, x + 1\right )}^{2} - 1\right )} \cos \left (4 \, x + 2\right ) - \frac {1}{16} \, {\left (2 \, {\left (2 \, x + 1\right )}^{3} - 6 \, x - 3\right )} \sin \left (4 \, x + 2\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.59 \[ \int (1+2 x)^3 \sin ^2(1+2 x) \, dx=x^{4} + 2 \, x^{3} + \frac {3}{2} \, x^{2} - \frac {3}{32} \, {\left (8 \, x^{2} + 8 \, x + 1\right )} \cos \left (4 \, x + 2\right ) - \frac {1}{16} \, {\left (16 \, x^{3} + 24 \, x^{2} + 6 \, x - 1\right )} \sin \left (4 \, x + 2\right ) + \frac {1}{2} \, x \]

[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

Mupad [B] (verification not implemented)

Time = 26.54 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.70 \[ \int (1+2 x)^3 \sin ^2(1+2 x) \, dx=\frac {3\,\sin \left (4\,x+2\right )\,\left (2\,x+1\right )}{16}-\frac {3\,{\sin \left (2\,x+1\right )}^2}{16}+\frac {{\left (2\,x+1\right )}^4}{16}-\frac {\sin \left (4\,x+2\right )\,{\left (2\,x+1\right )}^3}{8}+\frac {3\,{\left (2\,x+1\right )}^2\,\left (2\,{\sin \left (2\,x+1\right )}^2-1\right )}{16} \]

[In]

int(sin(2*x + 1)^2*(2*x + 1)^3,x)

[Out]

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