3.485 \(\int x^2 \sin ^6(x) \, dx\)

Optimal. Leaf size=105 \[ \frac {5 x^3}{48}-\frac {1}{6} x^2 \sin ^5(x) \cos (x)-\frac {5}{24} x^2 \sin ^3(x) \cos (x)-\frac {5}{16} x^2 \sin (x) \cos (x)-\frac {245 x}{1152}+\frac {1}{18} x \sin ^6(x)+\frac {5}{48} x \sin ^4(x)+\frac {5}{16} x \sin ^2(x)+\frac {1}{108} \sin ^5(x) \cos (x)+\frac {65 \sin ^3(x) \cos (x)}{1728}+\frac {245 \sin (x) \cos (x)}{1152} \]

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Rubi [A]  time = 0.11, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3311, 30, 2635, 8} \[ \frac {5 x^3}{48}-\frac {1}{6} x^2 \sin ^5(x) \cos (x)-\frac {5}{24} x^2 \sin ^3(x) \cos (x)-\frac {5}{16} x^2 \sin (x) \cos (x)-\frac {245 x}{1152}+\frac {1}{18} x \sin ^6(x)+\frac {5}{48} x \sin ^4(x)+\frac {5}{16} x \sin ^2(x)+\frac {1}{108} \sin ^5(x) \cos (x)+\frac {65 \sin ^3(x) \cos (x)}{1728}+\frac {245 \sin (x) \cos (x)}{1152} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sin[x]^6,x]

[Out]

(-245*x)/1152 + (5*x^3)/48 + (245*Cos[x]*Sin[x])/1152 - (5*x^2*Cos[x]*Sin[x])/16 + (5*x*Sin[x]^2)/16 + (65*Cos
[x]*Sin[x]^3)/1728 - (5*x^2*Cos[x]*Sin[x]^3)/24 + (5*x*Sin[x]^4)/48 + (Cos[x]*Sin[x]^5)/108 - (x^2*Cos[x]*Sin[
x]^5)/6 + (x*Sin[x]^6)/18

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2635

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

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 x^2 \sin ^6(x) \, dx &=-\frac {1}{6} x^2 \cos (x) \sin ^5(x)+\frac {1}{18} x \sin ^6(x)-\frac {1}{18} \int \sin ^6(x) \, dx+\frac {5}{6} \int x^2 \sin ^4(x) \, dx\\ &=-\frac {5}{24} x^2 \cos (x) \sin ^3(x)+\frac {5}{48} x \sin ^4(x)+\frac {1}{108} \cos (x) \sin ^5(x)-\frac {1}{6} x^2 \cos (x) \sin ^5(x)+\frac {1}{18} x \sin ^6(x)-\frac {5}{108} \int \sin ^4(x) \, dx-\frac {5}{48} \int \sin ^4(x) \, dx+\frac {5}{8} \int x^2 \sin ^2(x) \, dx\\ &=-\frac {5}{16} x^2 \cos (x) \sin (x)+\frac {5}{16} x \sin ^2(x)+\frac {65 \cos (x) \sin ^3(x)}{1728}-\frac {5}{24} x^2 \cos (x) \sin ^3(x)+\frac {5}{48} x \sin ^4(x)+\frac {1}{108} \cos (x) \sin ^5(x)-\frac {1}{6} x^2 \cos (x) \sin ^5(x)+\frac {1}{18} x \sin ^6(x)-\frac {5}{144} \int \sin ^2(x) \, dx-\frac {5}{64} \int \sin ^2(x) \, dx+\frac {5 \int x^2 \, dx}{16}-\frac {5}{16} \int \sin ^2(x) \, dx\\ &=\frac {5 x^3}{48}+\frac {245 \cos (x) \sin (x)}{1152}-\frac {5}{16} x^2 \cos (x) \sin (x)+\frac {5}{16} x \sin ^2(x)+\frac {65 \cos (x) \sin ^3(x)}{1728}-\frac {5}{24} x^2 \cos (x) \sin ^3(x)+\frac {5}{48} x \sin ^4(x)+\frac {1}{108} \cos (x) \sin ^5(x)-\frac {1}{6} x^2 \cos (x) \sin ^5(x)+\frac {1}{18} x \sin ^6(x)-\frac {5 \int 1 \, dx}{288}-\frac {5 \int 1 \, dx}{128}-\frac {5 \int 1 \, dx}{32}\\ &=-\frac {245 x}{1152}+\frac {5 x^3}{48}+\frac {245 \cos (x) \sin (x)}{1152}-\frac {5}{16} x^2 \cos (x) \sin (x)+\frac {5}{16} x \sin ^2(x)+\frac {65 \cos (x) \sin ^3(x)}{1728}-\frac {5}{24} x^2 \cos (x) \sin ^3(x)+\frac {5}{48} x \sin ^4(x)+\frac {1}{108} \cos (x) \sin ^5(x)-\frac {1}{6} x^2 \cos (x) \sin ^5(x)+\frac {1}{18} x \sin ^6(x)\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 70, normalized size = 0.67 \[ \frac {1440 x^3-1620 \left (2 x^2-1\right ) \sin (2 x)+81 \left (8 x^2-1\right ) \sin (4 x)-4 \left (18 x^2-1\right ) \sin (6 x)-3240 x \cos (2 x)+324 x \cos (4 x)-24 x \cos (6 x)}{13824} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sin[x]^6,x]

[Out]

(1440*x^3 - 3240*x*Cos[2*x] + 324*x*Cos[4*x] - 24*x*Cos[6*x] - 1620*(-1 + 2*x^2)*Sin[2*x] + 81*(-1 + 8*x^2)*Si
n[4*x] - 4*(-1 + 18*x^2)*Sin[6*x])/13824

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^2 \sin ^6(x) \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[x^2*Sin[x]^6,x]

[Out]

Could not integrate

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fricas [A]  time = 1.26, size = 72, normalized size = 0.69 \[ -\frac {1}{18} \, x \cos \relax (x)^{6} + \frac {13}{48} \, x \cos \relax (x)^{4} + \frac {5}{48} \, x^{3} - \frac {11}{16} \, x \cos \relax (x)^{2} - \frac {1}{3456} \, {\left (32 \, {\left (18 \, x^{2} - 1\right )} \cos \relax (x)^{5} - 2 \, {\left (936 \, x^{2} - 97\right )} \cos \relax (x)^{3} + 3 \, {\left (792 \, x^{2} - 299\right )} \cos \relax (x)\right )} \sin \relax (x) + \frac {299}{1152} \, x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*sin(x)^6,x, algorithm="fricas")

[Out]

-1/18*x*cos(x)^6 + 13/48*x*cos(x)^4 + 5/48*x^3 - 11/16*x*cos(x)^2 - 1/3456*(32*(18*x^2 - 1)*cos(x)^5 - 2*(936*
x^2 - 97)*cos(x)^3 + 3*(792*x^2 - 299)*cos(x))*sin(x) + 299/1152*x

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giac [A]  time = 0.59, size = 66, normalized size = 0.63 \[ \frac {5}{48} \, x^{3} - \frac {1}{576} \, x \cos \left (6 \, x\right ) + \frac {3}{128} \, x \cos \left (4 \, x\right ) - \frac {15}{64} \, x \cos \left (2 \, x\right ) - \frac {1}{3456} \, {\left (18 \, x^{2} - 1\right )} \sin \left (6 \, x\right ) + \frac {3}{512} \, {\left (8 \, x^{2} - 1\right )} \sin \left (4 \, x\right ) - \frac {15}{128} \, {\left (2 \, x^{2} - 1\right )} \sin \left (2 \, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*sin(x)^6,x, algorithm="giac")

[Out]

5/48*x^3 - 1/576*x*cos(6*x) + 3/128*x*cos(4*x) - 15/64*x*cos(2*x) - 1/3456*(18*x^2 - 1)*sin(6*x) + 3/512*(8*x^
2 - 1)*sin(4*x) - 15/128*(2*x^2 - 1)*sin(2*x)

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maple [A]  time = 0.42, size = 67, normalized size = 0.64

method	result	size
risch	\(\frac {5 x^{3}}{48}-\frac {x \cos \left (6 x \right )}{576}-\frac {\left (18 x^{2}-1\right ) \sin \left (6 x \right )}{3456}+\frac {3 x \cos \left (4 x \right )}{128}+\frac {3 \left (8 x^{2}-1\right ) \sin \left (4 x \right )}{512}-\frac {15 x \cos \left (2 x \right )}{64}-\frac {15 \left (2 x^{2}-1\right ) \sin \left (2 x \right )}{128}\)	\(67\)
default	\(x^{2} \left (-\frac {\left (\sin ^{5}\relax (x )+\frac {5 \left (\sin ^{3}\relax (x )\right )}{4}+\frac {15 \sin \relax (x )}{8}\right ) \cos \relax (x )}{6}+\frac {5 x}{16}\right )+\frac {x \left (\sin ^{6}\relax (x )\right )}{18}+\frac {\left (\sin ^{5}\relax (x )+\frac {5 \left (\sin ^{3}\relax (x )\right )}{4}+\frac {15 \sin \relax (x )}{8}\right ) \cos \relax (x )}{108}+\frac {115 x}{1152}+\frac {5 x \left (\sin ^{4}\relax (x )\right )}{48}+\frac {5 \left (\sin ^{3}\relax (x )+\frac {3 \sin \relax (x )}{2}\right ) \cos \relax (x )}{192}-\frac {5 x \left (\cos ^{2}\relax (x )\right )}{16}+\frac {5 \cos \relax (x ) \sin \relax (x )}{32}-\frac {5 x^{3}}{24}\)	\(96\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*sin(x)^6,x,method=_RETURNVERBOSE)

[Out]

5/48*x^3-1/576*x*cos(6*x)-1/3456*(18*x^2-1)*sin(6*x)+3/128*x*cos(4*x)+3/512*(8*x^2-1)*sin(4*x)-15/64*x*cos(2*x
)-15/128*(2*x^2-1)*sin(2*x)

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maxima [A]  time = 0.56, size = 66, normalized size = 0.63 \[ \frac {5}{48} \, x^{3} - \frac {1}{576} \, x \cos \left (6 \, x\right ) + \frac {3}{128} \, x \cos \left (4 \, x\right ) - \frac {15}{64} \, x \cos \left (2 \, x\right ) - \frac {1}{3456} \, {\left (18 \, x^{2} - 1\right )} \sin \left (6 \, x\right ) + \frac {3}{512} \, {\left (8 \, x^{2} - 1\right )} \sin \left (4 \, x\right ) - \frac {15}{128} \, {\left (2 \, x^{2} - 1\right )} \sin \left (2 \, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*sin(x)^6,x, algorithm="maxima")

[Out]

5/48*x^3 - 1/576*x*cos(6*x) + 3/128*x*cos(4*x) - 15/64*x*cos(2*x) - 1/3456*(18*x^2 - 1)*sin(6*x) + 3/512*(8*x^
2 - 1)*sin(4*x) - 15/128*(2*x^2 - 1)*sin(2*x)

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mupad [B]  time = 0.40, size = 88, normalized size = 0.84 \[ \frac {15\,\sin \left (2\,x\right )}{128}-\frac {3\,\sin \left (4\,x\right )}{512}+\frac {\sin \left (6\,x\right )}{3456}-\frac {3\,x\,\left (2\,{\sin \left (2\,x\right )}^2-1\right )}{128}+\frac {x\,\left (2\,{\sin \left (3\,x\right )}^2-1\right )}{576}-\frac {15\,x^2\,\sin \left (2\,x\right )}{64}+\frac {3\,x^2\,\sin \left (4\,x\right )}{64}-\frac {x^2\,\sin \left (6\,x\right )}{192}+\frac {5\,x^3}{48}+\frac {15\,x\,\left (2\,{\sin \relax (x)}^2-1\right )}{64} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*sin(x)^6,x)

[Out]

(15*sin(2*x))/128 - (3*sin(4*x))/512 + sin(6*x)/3456 - (3*x*(2*sin(2*x)^2 - 1))/128 + (x*(2*sin(3*x)^2 - 1))/5
76 - (15*x^2*sin(2*x))/64 + (3*x^2*sin(4*x))/64 - (x^2*sin(6*x))/192 + (5*x^3)/48 + (15*x*(2*sin(x)^2 - 1))/64

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sympy [A]  time = 5.90, size = 192, normalized size = 1.83 \[ \frac {5 x^{3} \sin ^{6}{\relax (x )}}{48} + \frac {5 x^{3} \sin ^{4}{\relax (x )} \cos ^{2}{\relax (x )}}{16} + \frac {5 x^{3} \sin ^{2}{\relax (x )} \cos ^{4}{\relax (x )}}{16} + \frac {5 x^{3} \cos ^{6}{\relax (x )}}{48} - \frac {11 x^{2} \sin ^{5}{\relax (x )} \cos {\relax (x )}}{16} - \frac {5 x^{2} \sin ^{3}{\relax (x )} \cos ^{3}{\relax (x )}}{6} - \frac {5 x^{2} \sin {\relax (x )} \cos ^{5}{\relax (x )}}{16} + \frac {299 x \sin ^{6}{\relax (x )}}{1152} + \frac {35 x \sin ^{4}{\relax (x )} \cos ^{2}{\relax (x )}}{384} - \frac {125 x \sin ^{2}{\relax (x )} \cos ^{4}{\relax (x )}}{384} - \frac {245 x \cos ^{6}{\relax (x )}}{1152} + \frac {299 \sin ^{5}{\relax (x )} \cos {\relax (x )}}{1152} + \frac {25 \sin ^{3}{\relax (x )} \cos ^{3}{\relax (x )}}{54} + \frac {245 \sin {\relax (x )} \cos ^{5}{\relax (x )}}{1152} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*sin(x)**6,x)

[Out]

5*x**3*sin(x)**6/48 + 5*x**3*sin(x)**4*cos(x)**2/16 + 5*x**3*sin(x)**2*cos(x)**4/16 + 5*x**3*cos(x)**6/48 - 11
*x**2*sin(x)**5*cos(x)/16 - 5*x**2*sin(x)**3*cos(x)**3/6 - 5*x**2*sin(x)*cos(x)**5/16 + 299*x*sin(x)**6/1152 +
35*x*sin(x)**4*cos(x)**2/384 - 125*x*sin(x)**2*cos(x)**4/384 - 245*x*cos(x)**6/1152 + 299*sin(x)**5*cos(x)/11
52 + 25*sin(x)**3*cos(x)**3/54 + 245*sin(x)*cos(x)**5/1152

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