3.91 \(\int x^2 \cos ^2(a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=97 \[ \frac {3 x^3 \cos ^2\left (a+b \log \left (c x^n\right )\right )}{4 b^2 n^2+9}+\frac {2 b n x^3 \sin \left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{4 b^2 n^2+9}+\frac {2 b^2 n^2 x^3}{3 \left (4 b^2 n^2+9\right )} \]

[Out]

2/3*b^2*n^2*x^3/(4*b^2*n^2+9)+3*x^3*cos(a+b*ln(c*x^n))^2/(4*b^2*n^2+9)+2*b*n*x^3*cos(a+b*ln(c*x^n))*sin(a+b*ln
(c*x^n))/(4*b^2*n^2+9)

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Rubi [A]  time = 0.03, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {4488, 30} \[ \frac {3 x^3 \cos ^2\left (a+b \log \left (c x^n\right )\right )}{4 b^2 n^2+9}+\frac {2 b n x^3 \sin \left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{4 b^2 n^2+9}+\frac {2 b^2 n^2 x^3}{3 \left (4 b^2 n^2+9\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Cos[a + b*Log[c*x^n]]^2,x]

[Out]

(2*b^2*n^2*x^3)/(3*(9 + 4*b^2*n^2)) + (3*x^3*Cos[a + b*Log[c*x^n]]^2)/(9 + 4*b^2*n^2) + (2*b*n*x^3*Cos[a + b*L
og[c*x^n]]*Sin[a + b*Log[c*x^n]])/(9 + 4*b^2*n^2)

Rule 30

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

Rule 4488

Int[Cos[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_)*((e_.)*(x_))^(m_.), x_Symbol] :> Simp[((m + 1)*(e*x)
^(m + 1)*Cos[d*(a + b*Log[c*x^n])]^p)/(b^2*d^2*e*n^2*p^2 + e*(m + 1)^2), x] + (Dist[(b^2*d^2*n^2*p*(p - 1))/(b
^2*d^2*n^2*p^2 + (m + 1)^2), Int[(e*x)^m*Cos[d*(a + b*Log[c*x^n])]^(p - 2), x], x] + Simp[(b*d*n*p*(e*x)^(m +
1)*Sin[d*(a + b*Log[c*x^n])]*Cos[d*(a + b*Log[c*x^n])]^(p - 1))/(b^2*d^2*e*n^2*p^2 + e*(m + 1)^2), x]) /; Free
Q[{a, b, c, d, e, m, n}, x] && IGtQ[p, 1] && NeQ[b^2*d^2*n^2*p^2 + (m + 1)^2, 0]

Rubi steps

\begin {align*} \int x^2 \cos ^2\left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac {3 x^3 \cos ^2\left (a+b \log \left (c x^n\right )\right )}{9+4 b^2 n^2}+\frac {2 b n x^3 \cos \left (a+b \log \left (c x^n\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )}{9+4 b^2 n^2}+\frac {\left (2 b^2 n^2\right ) \int x^2 \, dx}{9+4 b^2 n^2}\\ &=\frac {2 b^2 n^2 x^3}{3 \left (9+4 b^2 n^2\right )}+\frac {3 x^3 \cos ^2\left (a+b \log \left (c x^n\right )\right )}{9+4 b^2 n^2}+\frac {2 b n x^3 \cos \left (a+b \log \left (c x^n\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )}{9+4 b^2 n^2}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 61, normalized size = 0.63 \[ \frac {x^3 \left (6 b n \sin \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+9 \cos \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+4 b^2 n^2+9\right )}{6 \left (4 b^2 n^2+9\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Cos[a + b*Log[c*x^n]]^2,x]

[Out]

(x^3*(9 + 4*b^2*n^2 + 9*Cos[2*(a + b*Log[c*x^n])] + 6*b*n*Sin[2*(a + b*Log[c*x^n])]))/(6*(9 + 4*b^2*n^2))

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fricas [A]  time = 0.68, size = 76, normalized size = 0.78 \[ \frac {2 \, b^{2} n^{2} x^{3} + 6 \, b n x^{3} \cos \left (b n \log \relax (x) + b \log \relax (c) + a\right ) \sin \left (b n \log \relax (x) + b \log \relax (c) + a\right ) + 9 \, x^{3} \cos \left (b n \log \relax (x) + b \log \relax (c) + a\right )^{2}}{3 \, {\left (4 \, b^{2} n^{2} + 9\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cos(a+b*log(c*x^n))^2,x, algorithm="fricas")

[Out]

1/3*(2*b^2*n^2*x^3 + 6*b*n*x^3*cos(b*n*log(x) + b*log(c) + a)*sin(b*n*log(x) + b*log(c) + a) + 9*x^3*cos(b*n*l
og(x) + b*log(c) + a)^2)/(4*b^2*n^2 + 9)

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giac [B]  time = 0.53, size = 833, normalized size = 8.59 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cos(a+b*log(c*x^n))^2,x, algorithm="giac")

[Out]

1/6*x^3 - 1/4*(4*b*n*x^3*e^(pi*b*n*sgn(x) - pi*b*n + pi*b*sgn(c) - pi*b)*tan(b*n*log(abs(x)) + b*log(abs(c)))^
2*tan(a) + 4*b*n*x^3*e^(-pi*b*n*sgn(x) + pi*b*n - pi*b*sgn(c) + pi*b)*tan(b*n*log(abs(x)) + b*log(abs(c)))^2*t
an(a) + 4*b*n*x^3*e^(pi*b*n*sgn(x) - pi*b*n + pi*b*sgn(c) - pi*b)*tan(b*n*log(abs(x)) + b*log(abs(c)))*tan(a)^
2 + 4*b*n*x^3*e^(-pi*b*n*sgn(x) + pi*b*n - pi*b*sgn(c) + pi*b)*tan(b*n*log(abs(x)) + b*log(abs(c)))*tan(a)^2 -
3*x^3*e^(pi*b*n*sgn(x) - pi*b*n + pi*b*sgn(c) - pi*b)*tan(b*n*log(abs(x)) + b*log(abs(c)))^2*tan(a)^2 - 3*x^3
*e^(-pi*b*n*sgn(x) + pi*b*n - pi*b*sgn(c) + pi*b)*tan(b*n*log(abs(x)) + b*log(abs(c)))^2*tan(a)^2 - 4*b*n*x^3*
e^(pi*b*n*sgn(x) - pi*b*n + pi*b*sgn(c) - pi*b)*tan(b*n*log(abs(x)) + b*log(abs(c))) - 4*b*n*x^3*e^(-pi*b*n*sg
n(x) + pi*b*n - pi*b*sgn(c) + pi*b)*tan(b*n*log(abs(x)) + b*log(abs(c))) - 4*b*n*x^3*e^(pi*b*n*sgn(x) - pi*b*n
+ pi*b*sgn(c) - pi*b)*tan(a) - 4*b*n*x^3*e^(-pi*b*n*sgn(x) + pi*b*n - pi*b*sgn(c) + pi*b)*tan(a) + 3*x^3*e^(p
i*b*n*sgn(x) - pi*b*n + pi*b*sgn(c) - pi*b)*tan(b*n*log(abs(x)) + b*log(abs(c)))^2 + 3*x^3*e^(-pi*b*n*sgn(x) +
pi*b*n - pi*b*sgn(c) + pi*b)*tan(b*n*log(abs(x)) + b*log(abs(c)))^2 + 12*x^3*e^(pi*b*n*sgn(x) - pi*b*n + pi*b
*sgn(c) - pi*b)*tan(b*n*log(abs(x)) + b*log(abs(c)))*tan(a) + 12*x^3*e^(-pi*b*n*sgn(x) + pi*b*n - pi*b*sgn(c)
+ pi*b)*tan(b*n*log(abs(x)) + b*log(abs(c)))*tan(a) + 3*x^3*e^(pi*b*n*sgn(x) - pi*b*n + pi*b*sgn(c) - pi*b)*ta
n(a)^2 + 3*x^3*e^(-pi*b*n*sgn(x) + pi*b*n - pi*b*sgn(c) + pi*b)*tan(a)^2 - 3*x^3*e^(pi*b*n*sgn(x) - pi*b*n + p
i*b*sgn(c) - pi*b) - 3*x^3*e^(-pi*b*n*sgn(x) + pi*b*n - pi*b*sgn(c) + pi*b))/(4*b^2*n^2*tan(b*n*log(abs(x)) +
b*log(abs(c)))^2*tan(a)^2 + 4*b^2*n^2*tan(b*n*log(abs(x)) + b*log(abs(c)))^2 + 4*b^2*n^2*tan(a)^2 + 4*b^2*n^2
+ 9*tan(b*n*log(abs(x)) + b*log(abs(c)))^2*tan(a)^2 + 9*tan(b*n*log(abs(x)) + b*log(abs(c)))^2 + 9*tan(a)^2 +
9)

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int x^{2} \left (\cos ^{2}\left (a +b \ln \left (c \,x^{n}\right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*cos(a+b*ln(c*x^n))^2,x)

[Out]

int(x^2*cos(a+b*ln(c*x^n))^2,x)

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maxima [B]  time = 0.38, size = 301, normalized size = 3.10 \[ \frac {3 \, {\left (2 \, {\left (b \cos \left (2 \, b \log \relax (c)\right ) \sin \left (4 \, b \log \relax (c)\right ) - b \cos \left (4 \, b \log \relax (c)\right ) \sin \left (2 \, b \log \relax (c)\right ) + b \sin \left (2 \, b \log \relax (c)\right )\right )} n + 3 \, \cos \left (4 \, b \log \relax (c)\right ) \cos \left (2 \, b \log \relax (c)\right ) + 3 \, \sin \left (4 \, b \log \relax (c)\right ) \sin \left (2 \, b \log \relax (c)\right ) + 3 \, \cos \left (2 \, b \log \relax (c)\right )\right )} x^{3} \cos \left (2 \, b \log \left (x^{n}\right ) + 2 \, a\right ) + 3 \, {\left (2 \, {\left (b \cos \left (4 \, b \log \relax (c)\right ) \cos \left (2 \, b \log \relax (c)\right ) + b \sin \left (4 \, b \log \relax (c)\right ) \sin \left (2 \, b \log \relax (c)\right ) + b \cos \left (2 \, b \log \relax (c)\right )\right )} n - 3 \, \cos \left (2 \, b \log \relax (c)\right ) \sin \left (4 \, b \log \relax (c)\right ) + 3 \, \cos \left (4 \, b \log \relax (c)\right ) \sin \left (2 \, b \log \relax (c)\right ) - 3 \, \sin \left (2 \, b \log \relax (c)\right )\right )} x^{3} \sin \left (2 \, b \log \left (x^{n}\right ) + 2 \, a\right ) + 2 \, {\left (4 \, {\left (b^{2} \cos \left (2 \, b \log \relax (c)\right )^{2} + b^{2} \sin \left (2 \, b \log \relax (c)\right )^{2}\right )} n^{2} + 9 \, \cos \left (2 \, b \log \relax (c)\right )^{2} + 9 \, \sin \left (2 \, b \log \relax (c)\right )^{2}\right )} x^{3}}{12 \, {\left (4 \, {\left (b^{2} \cos \left (2 \, b \log \relax (c)\right )^{2} + b^{2} \sin \left (2 \, b \log \relax (c)\right )^{2}\right )} n^{2} + 9 \, \cos \left (2 \, b \log \relax (c)\right )^{2} + 9 \, \sin \left (2 \, b \log \relax (c)\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cos(a+b*log(c*x^n))^2,x, algorithm="maxima")

[Out]

1/12*(3*(2*(b*cos(2*b*log(c))*sin(4*b*log(c)) - b*cos(4*b*log(c))*sin(2*b*log(c)) + b*sin(2*b*log(c)))*n + 3*c
os(4*b*log(c))*cos(2*b*log(c)) + 3*sin(4*b*log(c))*sin(2*b*log(c)) + 3*cos(2*b*log(c)))*x^3*cos(2*b*log(x^n) +
2*a) + 3*(2*(b*cos(4*b*log(c))*cos(2*b*log(c)) + b*sin(4*b*log(c))*sin(2*b*log(c)) + b*cos(2*b*log(c)))*n - 3
*cos(2*b*log(c))*sin(4*b*log(c)) + 3*cos(4*b*log(c))*sin(2*b*log(c)) - 3*sin(2*b*log(c)))*x^3*sin(2*b*log(x^n)
+ 2*a) + 2*(4*(b^2*cos(2*b*log(c))^2 + b^2*sin(2*b*log(c))^2)*n^2 + 9*cos(2*b*log(c))^2 + 9*sin(2*b*log(c))^2
)*x^3)/(4*(b^2*cos(2*b*log(c))^2 + b^2*sin(2*b*log(c))^2)*n^2 + 9*cos(2*b*log(c))^2 + 9*sin(2*b*log(c))^2)

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mupad [B]  time = 2.70, size = 66, normalized size = 0.68 \[ \frac {x^3}{6}+\frac {x^3\,{\mathrm {e}}^{-a\,2{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{b\,2{}\mathrm {i}}}\,1{}\mathrm {i}}{8\,b\,n+12{}\mathrm {i}}+\frac {x^3\,{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,2{}\mathrm {i}}}{12+b\,n\,8{}\mathrm {i}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*cos(a + b*log(c*x^n))^2,x)

[Out]

x^3/6 + (x^3*exp(-a*2i)/(c*x^n)^(b*2i)*1i)/(8*b*n + 12i) + (x^3*exp(a*2i)*(c*x^n)^(b*2i))/(b*n*8i + 12)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \begin {cases} \int x^{2} \cos ^{2}{\left (a - \frac {3 i \log {\left (c x^{n} \right )}}{2 n} \right )}\, dx & \text {for}\: b = - \frac {3 i}{2 n} \\\int x^{2} \cos ^{2}{\left (a + \frac {3 i \log {\left (c x^{n} \right )}}{2 n} \right )}\, dx & \text {for}\: b = \frac {3 i}{2 n} \\\frac {2 b^{2} n^{2} x^{3} \sin ^{2}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{12 b^{2} n^{2} + 27} + \frac {2 b^{2} n^{2} x^{3} \cos ^{2}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{12 b^{2} n^{2} + 27} + \frac {6 b n x^{3} \sin {\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )} \cos {\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{12 b^{2} n^{2} + 27} + \frac {9 x^{3} \cos ^{2}{\left (a + b n \log {\relax (x )} + b \log {\relax (c )} \right )}}{12 b^{2} n^{2} + 27} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*cos(a+b*ln(c*x**n))**2,x)

[Out]

Piecewise((Integral(x**2*cos(a - 3*I*log(c*x**n)/(2*n))**2, x), Eq(b, -3*I/(2*n))), (Integral(x**2*cos(a + 3*I
*log(c*x**n)/(2*n))**2, x), Eq(b, 3*I/(2*n))), (2*b**2*n**2*x**3*sin(a + b*n*log(x) + b*log(c))**2/(12*b**2*n*
*2 + 27) + 2*b**2*n**2*x**3*cos(a + b*n*log(x) + b*log(c))**2/(12*b**2*n**2 + 27) + 6*b*n*x**3*sin(a + b*n*log
(x) + b*log(c))*cos(a + b*n*log(x) + b*log(c))/(12*b**2*n**2 + 27) + 9*x**3*cos(a + b*n*log(x) + b*log(c))**2/
(12*b**2*n**2 + 27), True))

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