3.483 \(\int x^2 \cos ^5(x) \, dx\)

Optimal. Leaf size=83 \[ \frac {8}{15} x^2 \sin (x)+\frac {1}{5} x^2 \sin (x) \cos ^4(x)+\frac {4}{15} x^2 \sin (x) \cos ^2(x)-\frac {2 \sin ^5(x)}{125}+\frac {76 \sin ^3(x)}{675}-\frac {298 \sin (x)}{225}+\frac {2}{25} x \cos ^5(x)+\frac {8}{45} x \cos ^3(x)+\frac {16}{15} x \cos (x) \]

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Rubi [A]  time = 0.09, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3311, 3296, 2637, 2633} \[ \frac {8}{15} x^2 \sin (x)+\frac {1}{5} x^2 \sin (x) \cos ^4(x)+\frac {4}{15} x^2 \sin (x) \cos ^2(x)-\frac {2 \sin ^5(x)}{125}+\frac {76 \sin ^3(x)}{675}-\frac {298 \sin (x)}{225}+\frac {2}{25} x \cos ^5(x)+\frac {8}{45} x \cos ^3(x)+\frac {16}{15} x \cos (x) \]

Antiderivative was successfully verified.

[In]

Int[x^2*Cos[x]^5,x]

[Out]

(16*x*Cos[x])/15 + (8*x*Cos[x]^3)/45 + (2*x*Cos[x]^5)/25 - (298*Sin[x])/225 + (8*x^2*Sin[x])/15 + (4*x^2*Cos[x
]^2*Sin[x])/15 + (x^2*Cos[x]^4*Sin[x])/5 + (76*Sin[x]^3)/675 - (2*Sin[x]^5)/125

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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[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 x^2 \cos ^5(x) \, dx &=\frac {2}{25} x \cos ^5(x)+\frac {1}{5} x^2 \cos ^4(x) \sin (x)-\frac {2}{25} \int \cos ^5(x) \, dx+\frac {4}{5} \int x^2 \cos ^3(x) \, dx\\ &=\frac {8}{45} x \cos ^3(x)+\frac {2}{25} x \cos ^5(x)+\frac {4}{15} x^2 \cos ^2(x) \sin (x)+\frac {1}{5} x^2 \cos ^4(x) \sin (x)+\frac {2}{25} \operatorname {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (x)\right )-\frac {8}{45} \int \cos ^3(x) \, dx+\frac {8}{15} \int x^2 \cos (x) \, dx\\ &=\frac {8}{45} x \cos ^3(x)+\frac {2}{25} x \cos ^5(x)-\frac {2 \sin (x)}{25}+\frac {8}{15} x^2 \sin (x)+\frac {4}{15} x^2 \cos ^2(x) \sin (x)+\frac {1}{5} x^2 \cos ^4(x) \sin (x)+\frac {4 \sin ^3(x)}{75}-\frac {2 \sin ^5(x)}{125}+\frac {8}{45} \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (x)\right )-\frac {16}{15} \int x \sin (x) \, dx\\ &=\frac {16}{15} x \cos (x)+\frac {8}{45} x \cos ^3(x)+\frac {2}{25} x \cos ^5(x)-\frac {58 \sin (x)}{225}+\frac {8}{15} x^2 \sin (x)+\frac {4}{15} x^2 \cos ^2(x) \sin (x)+\frac {1}{5} x^2 \cos ^4(x) \sin (x)+\frac {76 \sin ^3(x)}{675}-\frac {2 \sin ^5(x)}{125}-\frac {16}{15} \int \cos (x) \, dx\\ &=\frac {16}{15} x \cos (x)+\frac {8}{45} x \cos ^3(x)+\frac {2}{25} x \cos ^5(x)-\frac {298 \sin (x)}{225}+\frac {8}{15} x^2 \sin (x)+\frac {4}{15} x^2 \cos ^2(x) \sin (x)+\frac {1}{5} x^2 \cos ^4(x) \sin (x)+\frac {76 \sin ^3(x)}{675}-\frac {2 \sin ^5(x)}{125}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 67, normalized size = 0.81 \[ \frac {5}{8} \left (x^2-2\right ) \sin (x)+\frac {5}{432} \left (9 x^2-2\right ) \sin (3 x)+\frac {\left (25 x^2-2\right ) \sin (5 x)}{2000}+\frac {5}{4} x \cos (x)+\frac {5}{72} x \cos (3 x)+\frac {1}{200} x \cos (5 x) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Cos[x]^5,x]

[Out]

(5*x*Cos[x])/4 + (5*x*Cos[3*x])/72 + (x*Cos[5*x])/200 + (5*(-2 + x^2)*Sin[x])/8 + (5*(-2 + 9*x^2)*Sin[3*x])/43
2 + ((-2 + 25*x^2)*Sin[5*x])/2000

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^2 \cos ^5(x) \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[x^2*Cos[x]^5,x]

[Out]

Could not integrate

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fricas [A]  time = 1.12, size = 57, normalized size = 0.69 \[ \frac {2}{25} \, x \cos \relax (x)^{5} + \frac {8}{45} \, x \cos \relax (x)^{3} + \frac {16}{15} \, x \cos \relax (x) + \frac {1}{3375} \, {\left (27 \, {\left (25 \, x^{2} - 2\right )} \cos \relax (x)^{4} + 4 \, {\left (225 \, x^{2} - 68\right )} \cos \relax (x)^{2} + 1800 \, x^{2} - 4144\right )} \sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cos(x)^5,x, algorithm="fricas")

[Out]

2/25*x*cos(x)^5 + 8/45*x*cos(x)^3 + 16/15*x*cos(x) + 1/3375*(27*(25*x^2 - 2)*cos(x)^4 + 4*(225*x^2 - 68)*cos(x
)^2 + 1800*x^2 - 4144)*sin(x)

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giac [A]  time = 0.58, size = 55, normalized size = 0.66 \[ \frac {1}{200} \, x \cos \left (5 \, x\right ) + \frac {5}{72} \, x \cos \left (3 \, x\right ) + \frac {5}{4} \, x \cos \relax (x) + \frac {1}{2000} \, {\left (25 \, x^{2} - 2\right )} \sin \left (5 \, x\right ) + \frac {5}{432} \, {\left (9 \, x^{2} - 2\right )} \sin \left (3 \, x\right ) + \frac {5}{8} \, {\left (x^{2} - 2\right )} \sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cos(x)^5,x, algorithm="giac")

[Out]

1/200*x*cos(5*x) + 5/72*x*cos(3*x) + 5/4*x*cos(x) + 1/2000*(25*x^2 - 2)*sin(5*x) + 5/432*(9*x^2 - 2)*sin(3*x)
+ 5/8*(x^2 - 2)*sin(x)

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maple [A]  time = 0.38, size = 56, normalized size = 0.67

method	result	size
risch	\(\frac {5 x \cos \relax (x )}{4}+\frac {5 \left (x^{2}-2\right ) \sin \relax (x )}{8}+\frac {x \cos \left (5 x \right )}{200}+\frac {\left (25 x^{2}-2\right ) \sin \left (5 x \right )}{2000}+\frac {5 x \cos \left (3 x \right )}{72}+\frac {5 \left (9 x^{2}-2\right ) \sin \left (3 x \right )}{432}\)	\(56\)
default	\(\frac {x^{2} \left (\frac {8}{3}+\cos ^{4}\relax (x )+\frac {4 \left (\cos ^{2}\relax (x )\right )}{3}\right ) \sin \relax (x )}{5}-\frac {16 \sin \relax (x )}{15}+\frac {16 x \cos \relax (x )}{15}+\frac {2 x \left (\cos ^{5}\relax (x )\right )}{25}-\frac {2 \left (\frac {8}{3}+\cos ^{4}\relax (x )+\frac {4 \left (\cos ^{2}\relax (x )\right )}{3}\right ) \sin \relax (x )}{125}+\frac {8 x \left (\cos ^{3}\relax (x )\right )}{45}-\frac {8 \left (2+\cos ^{2}\relax (x )\right ) \sin \relax (x )}{135}\)	\(70\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*cos(x)^5,x,method=_RETURNVERBOSE)

[Out]

5/4*x*cos(x)+5/8*(x^2-2)*sin(x)+1/200*x*cos(5*x)+1/2000*(25*x^2-2)*sin(5*x)+5/72*x*cos(3*x)+5/432*(9*x^2-2)*si
n(3*x)

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maxima [A]  time = 0.51, size = 55, normalized size = 0.66 \[ \frac {1}{200} \, x \cos \left (5 \, x\right ) + \frac {5}{72} \, x \cos \left (3 \, x\right ) + \frac {5}{4} \, x \cos \relax (x) + \frac {1}{2000} \, {\left (25 \, x^{2} - 2\right )} \sin \left (5 \, x\right ) + \frac {5}{432} \, {\left (9 \, x^{2} - 2\right )} \sin \left (3 \, x\right ) + \frac {5}{8} \, {\left (x^{2} - 2\right )} \sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cos(x)^5,x, algorithm="maxima")

[Out]

1/200*x*cos(5*x) + 5/72*x*cos(3*x) + 5/4*x*cos(x) + 1/2000*(25*x^2 - 2)*sin(5*x) + 5/432*(9*x^2 - 2)*sin(3*x)
+ 5/8*(x^2 - 2)*sin(x)

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mupad [B]  time = 0.40, size = 69, normalized size = 0.83 \[ \frac {8\,x\,{\cos \relax (x)}^3}{45}-\frac {4144\,\sin \relax (x)}{3375}+\frac {2\,x\,{\cos \relax (x)}^5}{25}+\frac {8\,x^2\,\sin \relax (x)}{15}-\frac {272\,{\cos \relax (x)}^2\,\sin \relax (x)}{3375}-\frac {2\,{\cos \relax (x)}^4\,\sin \relax (x)}{125}+\frac {16\,x\,\cos \relax (x)}{15}+\frac {4\,x^2\,{\cos \relax (x)}^2\,\sin \relax (x)}{15}+\frac {x^2\,{\cos \relax (x)}^4\,\sin \relax (x)}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*cos(x)^5,x)

[Out]

(8*x*cos(x)^3)/45 - (4144*sin(x))/3375 + (2*x*cos(x)^5)/25 + (8*x^2*sin(x))/15 - (272*cos(x)^2*sin(x))/3375 -
(2*cos(x)^4*sin(x))/125 + (16*x*cos(x))/15 + (4*x^2*cos(x)^2*sin(x))/15 + (x^2*cos(x)^4*sin(x))/5

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sympy [A]  time = 3.49, size = 112, normalized size = 1.35 \[ \frac {8 x^{2} \sin ^{5}{\relax (x )}}{15} + \frac {4 x^{2} \sin ^{3}{\relax (x )} \cos ^{2}{\relax (x )}}{3} + x^{2} \sin {\relax (x )} \cos ^{4}{\relax (x )} + \frac {16 x \sin ^{4}{\relax (x )} \cos {\relax (x )}}{15} + \frac {104 x \sin ^{2}{\relax (x )} \cos ^{3}{\relax (x )}}{45} + \frac {298 x \cos ^{5}{\relax (x )}}{225} - \frac {4144 \sin ^{5}{\relax (x )}}{3375} - \frac {1712 \sin ^{3}{\relax (x )} \cos ^{2}{\relax (x )}}{675} - \frac {298 \sin {\relax (x )} \cos ^{4}{\relax (x )}}{225} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*cos(x)**5,x)

[Out]

8*x**2*sin(x)**5/15 + 4*x**2*sin(x)**3*cos(x)**2/3 + x**2*sin(x)*cos(x)**4 + 16*x*sin(x)**4*cos(x)/15 + 104*x*
sin(x)**2*cos(x)**3/45 + 298*x*cos(x)**5/225 - 4144*sin(x)**5/3375 - 1712*sin(x)**3*cos(x)**2/675 - 298*sin(x)
*cos(x)**4/225

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