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

Optimal. Leaf size=65 \[ \frac {2}{3} (3 x+2) \sin ^3(x)+4 (3 x+2) \sin (x)-\frac {2}{3} \cos ^3(x)-\frac {2}{3} (3 x+2)^2 \cos (x)+14 \cos (x)-\frac {1}{3} (3 x+2)^2 \sin ^2(x) \cos (x) \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3311, 3296, 2638, 2633} \[ \frac {2}{3} (3 x+2) \sin ^3(x)+4 (3 x+2) \sin (x)-\frac {2}{3} \cos ^3(x)-\frac {2}{3} (3 x+2)^2 \cos (x)+14 \cos (x)-\frac {1}{3} (3 x+2)^2 \sin ^2(x) \cos (x) \]

Antiderivative was successfully verified.

[In]

Int[(2 + 3*x)^2*Sin[x]^3,x]

[Out]

14*Cos[x] - (2*(2 + 3*x)^2*Cos[x])/3 - (2*Cos[x]^3)/3 + 4*(2 + 3*x)*Sin[x] - ((2 + 3*x)^2*Cos[x]*Sin[x]^2)/3 +
(2*(2 + 3*x)*Sin[x]^3)/3

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

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]

Rubi steps

\begin {align*} \int (2+3 x)^2 \sin ^3(x) \, dx &=-\frac {1}{3} (2+3 x)^2 \cos (x) \sin ^2(x)+\frac {2}{3} (2+3 x) \sin ^3(x)+\frac {2}{3} \int (2+3 x)^2 \sin (x) \, dx-2 \int \sin ^3(x) \, dx\\ &=-\frac {2}{3} (2+3 x)^2 \cos (x)-\frac {1}{3} (2+3 x)^2 \cos (x) \sin ^2(x)+\frac {2}{3} (2+3 x) \sin ^3(x)+2 \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (x)\right )+4 \int (2+3 x) \cos (x) \, dx\\ &=2 \cos (x)-\frac {2}{3} (2+3 x)^2 \cos (x)-\frac {2 \cos ^3(x)}{3}+4 (2+3 x) \sin (x)-\frac {1}{3} (2+3 x)^2 \cos (x) \sin ^2(x)+\frac {2}{3} (2+3 x) \sin ^3(x)-12 \int \sin (x) \, dx\\ &=14 \cos (x)-\frac {2}{3} (2+3 x)^2 \cos (x)-\frac {2 \cos ^3(x)}{3}+4 (2+3 x) \sin (x)-\frac {1}{3} (2+3 x)^2 \cos (x) \sin ^2(x)+\frac {2}{3} (2+3 x) \sin ^3(x)\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 50, normalized size = 0.77 \[ \frac {1}{12} \left (-9 \left (9 x^2+12 x-14\right ) \cos (x)+\left (9 x^2+12 x+2\right ) \cos (3 x)-2 (3 x+2) (\sin (3 x)-27 \sin (x))\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + 3*x)^2*Sin[x]^3,x]

[Out]

(-9*(-14 + 12*x + 9*x^2)*Cos[x] + (2 + 12*x + 9*x^2)*Cos[3*x] - 2*(2 + 3*x)*(-27*Sin[x] + Sin[3*x]))/12

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fricas [A]  time = 1.65, size = 50, normalized size = 0.77 \[ \frac {1}{3} \, {\left (9 \, x^{2} + 12 \, x + 2\right )} \cos \relax (x)^{3} - {\left (9 \, x^{2} + 12 \, x - 10\right )} \cos \relax (x) - \frac {2}{3} \, {\left ({\left (3 \, x + 2\right )} \cos \relax (x)^{2} - 21 \, x - 14\right )} \sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(9*x^2 + 12*x + 2)*cos(x)^3 - (9*x^2 + 12*x - 10)*cos(x) - 2/3*((3*x + 2)*cos(x)^2 - 21*x - 14)*sin(x)

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giac [A]  time = 1.98, size = 51, normalized size = 0.78 \[ \frac {1}{12} \, {\left (9 \, x^{2} + 12 \, x + 2\right )} \cos \left (3 \, x\right ) - \frac {3}{4} \, {\left (9 \, x^{2} + 12 \, x - 14\right )} \cos \relax (x) - \frac {1}{6} \, {\left (3 \, x + 2\right )} \sin \left (3 \, x\right ) + \frac {9}{2} \, {\left (3 \, x + 2\right )} \sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(9*x^2 + 12*x + 2)*cos(3*x) - 3/4*(9*x^2 + 12*x - 14)*cos(x) - 1/6*(3*x + 2)*sin(3*x) + 9/2*(3*x + 2)*sin
(x)

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maple [A]  time = 0.03, size = 62, normalized size = 0.95 \[ -3 x^{2} \left (2+\sin ^{2}\relax (x )\right ) \cos \relax (x )+12 \cos \relax (x )+12 x \sin \relax (x )+2 \left (\sin ^{3}\relax (x )\right ) x -\frac {2 \left (2+\sin ^{2}\relax (x )\right ) \cos \relax (x )}{3}-4 x \left (2+\sin ^{2}\relax (x )\right ) \cos \relax (x )+\frac {4 \left (\sin ^{3}\relax (x )\right )}{3}+8 \sin \relax (x ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*x^2*(2+sin(x)^2)*cos(x)+12*cos(x)+12*x*sin(x)+2*sin(x)^3*x-2/3*(2+sin(x)^2)*cos(x)-4*x*(2+sin(x)^2)*cos(x)+
4/3*sin(x)^3+8*sin(x)

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maxima [A]  time = 0.33, size = 66, normalized size = 1.02 \[ \frac {4}{3} \, \cos \relax (x)^{3} + \frac {1}{12} \, {\left (9 \, x^{2} - 2\right )} \cos \left (3 \, x\right ) + x \cos \left (3 \, x\right ) - \frac {27}{4} \, {\left (x^{2} - 2\right )} \cos \relax (x) - 9 \, x \cos \relax (x) - \frac {1}{2} \, x \sin \left (3 \, x\right ) + \frac {27}{2} \, x \sin \relax (x) - 4 \, \cos \relax (x) - \frac {1}{3} \, \sin \left (3 \, x\right ) + 9 \, \sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/3*cos(x)^3 + 1/12*(9*x^2 - 2)*cos(3*x) + x*cos(3*x) - 27/4*(x^2 - 2)*cos(x) - 9*x*cos(x) - 1/2*x*sin(3*x) +
27/2*x*sin(x) - 4*cos(x) - 1/3*sin(3*x) + 9*sin(x)

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mupad [B]  time = 3.03, size = 65, normalized size = 1.00 \[ 10\,\cos \relax (x)+\frac {28\,\sin \relax (x)}{3}-9\,x^2\,\cos \relax (x)+4\,x\,{\cos \relax (x)}^3+\frac {2\,{\cos \relax (x)}^3}{3}+3\,x^2\,{\cos \relax (x)}^3-\frac {4\,{\cos \relax (x)}^2\,\sin \relax (x)}{3}-12\,x\,\cos \relax (x)+14\,x\,\sin \relax (x)-2\,x\,{\cos \relax (x)}^2\,\sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^3*(3*x + 2)^2,x)

[Out]

10*cos(x) + (28*sin(x))/3 - 9*x^2*cos(x) + 4*x*cos(x)^3 + (2*cos(x)^3)/3 + 3*x^2*cos(x)^3 - (4*cos(x)^2*sin(x)
)/3 - 12*x*cos(x) + 14*x*sin(x) - 2*x*cos(x)^2*sin(x)

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sympy [A]  time = 1.22, size = 100, normalized size = 1.54 \[ - 9 x^{2} \sin ^{2}{\relax (x )} \cos {\relax (x )} - 6 x^{2} \cos ^{3}{\relax (x )} + 14 x \sin ^{3}{\relax (x )} - 12 x \sin ^{2}{\relax (x )} \cos {\relax (x )} + 12 x \sin {\relax (x )} \cos ^{2}{\relax (x )} - 8 x \cos ^{3}{\relax (x )} + \frac {28 \sin ^{3}{\relax (x )}}{3} + 10 \sin ^{2}{\relax (x )} \cos {\relax (x )} + 8 \sin {\relax (x )} \cos ^{2}{\relax (x )} + \frac {32 \cos ^{3}{\relax (x )}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9*x**2*sin(x)**2*cos(x) - 6*x**2*cos(x)**3 + 14*x*sin(x)**3 - 12*x*sin(x)**2*cos(x) + 12*x*sin(x)*cos(x)**2 -
8*x*cos(x)**3 + 28*sin(x)**3/3 + 10*sin(x)**2*cos(x) + 8*sin(x)*cos(x)**2 + 32*cos(x)**3/3

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