3.1.14 \(\int x \sin ^3(a+b \log (c x^n)) \, dx\) [14]

Optimal. Leaf size=158 \[ -\frac {6 b^3 n^3 x^2 \cos \left (a+b \log \left (c x^n\right )\right )}{16+40 b^2 n^2+9 b^4 n^4}+\frac {12 b^2 n^2 x^2 \sin \left (a+b \log \left (c x^n\right )\right )}{16+40 b^2 n^2+9 b^4 n^4}-\frac {3 b n x^2 \cos \left (a+b \log \left (c x^n\right )\right ) \sin ^2\left (a+b \log \left (c x^n\right )\right )}{4+9 b^2 n^2}+\frac {2 x^2 \sin ^3\left (a+b \log \left (c x^n\right )\right )}{4+9 b^2 n^2} \]

[Out]

-6*b^3*n^3*x^2*cos(a+b*ln(c*x^n))/(9*b^4*n^4+40*b^2*n^2+16)+12*b^2*n^2*x^2*sin(a+b*ln(c*x^n))/(9*b^4*n^4+40*b^
2*n^2+16)-3*b*n*x^2*cos(a+b*ln(c*x^n))*sin(a+b*ln(c*x^n))^2/(9*b^2*n^2+4)+2*x^2*sin(a+b*ln(c*x^n))^3/(9*b^2*n^
2+4)

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Rubi [A]
time = 0.03, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {4575, 4573} \begin {gather*} \frac {2 x^2 \sin ^3\left (a+b \log \left (c x^n\right )\right )}{9 b^2 n^2+4}-\frac {3 b n x^2 \sin ^2\left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{9 b^2 n^2+4}+\frac {12 b^2 n^2 x^2 \sin \left (a+b \log \left (c x^n\right )\right )}{9 b^4 n^4+40 b^2 n^2+16}-\frac {6 b^3 n^3 x^2 \cos \left (a+b \log \left (c x^n\right )\right )}{9 b^4 n^4+40 b^2 n^2+16} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Sin[a + b*Log[c*x^n]]^3,x]

[Out]

(-6*b^3*n^3*x^2*Cos[a + b*Log[c*x^n]])/(16 + 40*b^2*n^2 + 9*b^4*n^4) + (12*b^2*n^2*x^2*Sin[a + b*Log[c*x^n]])/
(16 + 40*b^2*n^2 + 9*b^4*n^4) - (3*b*n*x^2*Cos[a + b*Log[c*x^n]]*Sin[a + b*Log[c*x^n]]^2)/(4 + 9*b^2*n^2) + (2
*x^2*Sin[a + b*Log[c*x^n]]^3)/(4 + 9*b^2*n^2)

Rule 4573

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

Rule 4575

Int[((e_.)*(x_))^(m_.)*Sin[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_), x_Symbol] :> Simp[(m + 1)*(e*x)^
(m + 1)*(Sin[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*Sin[d*(a + b*Log[c*x^n])]^(p - 2), x], x] - Simp[b*d*n*p*(e*x)^(m + 1
)*Cos[d*(a + b*Log[c*x^n])]*(Sin[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 \sin ^3\left (a+b \log \left (c x^n\right )\right ) \, dx &=-\frac {3 b n x^2 \cos \left (a+b \log \left (c x^n\right )\right ) \sin ^2\left (a+b \log \left (c x^n\right )\right )}{4+9 b^2 n^2}+\frac {2 x^2 \sin ^3\left (a+b \log \left (c x^n\right )\right )}{4+9 b^2 n^2}+\frac {\left (6 b^2 n^2\right ) \int x \sin \left (a+b \log \left (c x^n\right )\right ) \, dx}{4+9 b^2 n^2}\\ &=-\frac {6 b^3 n^3 x^2 \cos \left (a+b \log \left (c x^n\right )\right )}{16+40 b^2 n^2+9 b^4 n^4}+\frac {12 b^2 n^2 x^2 \sin \left (a+b \log \left (c x^n\right )\right )}{16+40 b^2 n^2+9 b^4 n^4}-\frac {3 b n x^2 \cos \left (a+b \log \left (c x^n\right )\right ) \sin ^2\left (a+b \log \left (c x^n\right )\right )}{4+9 b^2 n^2}+\frac {2 x^2 \sin ^3\left (a+b \log \left (c x^n\right )\right )}{4+9 b^2 n^2}\\ \end {align*}

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Mathematica [A]
time = 0.55, size = 125, normalized size = 0.79 \begin {gather*} \frac {x^2 \left (-3 b n \left (4+9 b^2 n^2\right ) \cos \left (a+b \log \left (c x^n\right )\right )+3 b n \left (4+b^2 n^2\right ) \cos \left (3 \left (a+b \log \left (c x^n\right )\right )\right )-4 \left (-4-13 b^2 n^2+\left (4+b^2 n^2\right ) \cos \left (2 \left (a+b \log \left (c x^n\right )\right )\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )\right )}{4 \left (16+40 b^2 n^2+9 b^4 n^4\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*Sin[a + b*Log[c*x^n]]^3,x]

[Out]

(x^2*(-3*b*n*(4 + 9*b^2*n^2)*Cos[a + b*Log[c*x^n]] + 3*b*n*(4 + b^2*n^2)*Cos[3*(a + b*Log[c*x^n])] - 4*(-4 - 1
3*b^2*n^2 + (4 + b^2*n^2)*Cos[2*(a + b*Log[c*x^n])])*Sin[a + b*Log[c*x^n]]))/(4*(16 + 40*b^2*n^2 + 9*b^4*n^4))

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Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int x \left (\sin ^{3}\left (a +b \ln \left (c \,x^{n}\right )\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sin(a+b*ln(c*x^n))^3,x)

[Out]

int(x*sin(a+b*ln(c*x^n))^3,x)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 1016 vs. \(2 (158) = 316\).
time = 0.31, size = 1016, normalized size = 6.43 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sin(a+b*log(c*x^n))^3,x, algorithm="maxima")

[Out]

1/8*((3*(b^3*cos(6*b*log(c))*cos(3*b*log(c)) + b^3*sin(6*b*log(c))*sin(3*b*log(c)) + b^3*cos(3*b*log(c)))*n^3
- 2*(b^2*cos(3*b*log(c))*sin(6*b*log(c)) - b^2*cos(6*b*log(c))*sin(3*b*log(c)) + b^2*sin(3*b*log(c)))*n^2 + 12
*(b*cos(6*b*log(c))*cos(3*b*log(c)) + b*sin(6*b*log(c))*sin(3*b*log(c)) + b*cos(3*b*log(c)))*n - 8*cos(3*b*log
(c))*sin(6*b*log(c)) + 8*cos(6*b*log(c))*sin(3*b*log(c)) - 8*sin(3*b*log(c)))*x^2*cos(3*b*log(x^n) + 3*a) - 3*
(9*(b^3*cos(4*b*log(c))*cos(3*b*log(c)) + b^3*cos(3*b*log(c))*cos(2*b*log(c)) + b^3*sin(4*b*log(c))*sin(3*b*lo
g(c)) + b^3*sin(3*b*log(c))*sin(2*b*log(c)))*n^3 - 18*(b^2*cos(3*b*log(c))*sin(4*b*log(c)) - b^2*cos(4*b*log(c
))*sin(3*b*log(c)) + b^2*cos(2*b*log(c))*sin(3*b*log(c)) - b^2*cos(3*b*log(c))*sin(2*b*log(c)))*n^2 + 4*(b*cos
(4*b*log(c))*cos(3*b*log(c)) + b*cos(3*b*log(c))*cos(2*b*log(c)) + b*sin(4*b*log(c))*sin(3*b*log(c)) + b*sin(3
*b*log(c))*sin(2*b*log(c)))*n - 8*cos(3*b*log(c))*sin(4*b*log(c)) + 8*cos(4*b*log(c))*sin(3*b*log(c)) - 8*cos(
2*b*log(c))*sin(3*b*log(c)) + 8*cos(3*b*log(c))*sin(2*b*log(c)))*x^2*cos(b*log(x^n) + a) - (3*(b^3*cos(3*b*log
(c))*sin(6*b*log(c)) - b^3*cos(6*b*log(c))*sin(3*b*log(c)) + b^3*sin(3*b*log(c)))*n^3 + 2*(b^2*cos(6*b*log(c))
*cos(3*b*log(c)) + b^2*sin(6*b*log(c))*sin(3*b*log(c)) + b^2*cos(3*b*log(c)))*n^2 + 12*(b*cos(3*b*log(c))*sin(
6*b*log(c)) - b*cos(6*b*log(c))*sin(3*b*log(c)) + b*sin(3*b*log(c)))*n + 8*cos(6*b*log(c))*cos(3*b*log(c)) + 8
*sin(6*b*log(c))*sin(3*b*log(c)) + 8*cos(3*b*log(c)))*x^2*sin(3*b*log(x^n) + 3*a) + 3*(9*(b^3*cos(3*b*log(c))*
sin(4*b*log(c)) - b^3*cos(4*b*log(c))*sin(3*b*log(c)) + b^3*cos(2*b*log(c))*sin(3*b*log(c)) - b^3*cos(3*b*log(
c))*sin(2*b*log(c)))*n^3 + 18*(b^2*cos(4*b*log(c))*cos(3*b*log(c)) + b^2*cos(3*b*log(c))*cos(2*b*log(c)) + b^2
*sin(4*b*log(c))*sin(3*b*log(c)) + b^2*sin(3*b*log(c))*sin(2*b*log(c)))*n^2 + 4*(b*cos(3*b*log(c))*sin(4*b*log
(c)) - b*cos(4*b*log(c))*sin(3*b*log(c)) + b*cos(2*b*log(c))*sin(3*b*log(c)) - b*cos(3*b*log(c))*sin(2*b*log(c
)))*n + 8*cos(4*b*log(c))*cos(3*b*log(c)) + 8*cos(3*b*log(c))*cos(2*b*log(c)) + 8*sin(4*b*log(c))*sin(3*b*log(
c)) + 8*sin(3*b*log(c))*sin(2*b*log(c)))*x^2*sin(b*log(x^n) + a))/(9*(b^4*cos(3*b*log(c))^2 + b^4*sin(3*b*log(
c))^2)*n^4 + 40*(b^2*cos(3*b*log(c))^2 + b^2*sin(3*b*log(c))^2)*n^2 + 16*cos(3*b*log(c))^2 + 16*sin(3*b*log(c)
)^2)

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Fricas [A]
time = 3.21, size = 140, normalized size = 0.89 \begin {gather*} \frac {3 \, {\left (b^{3} n^{3} + 4 \, b n\right )} x^{2} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{3} - 3 \, {\left (3 \, b^{3} n^{3} + 4 \, b n\right )} x^{2} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) - 2 \, {\left ({\left (b^{2} n^{2} + 4\right )} x^{2} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} - {\left (7 \, b^{2} n^{2} + 4\right )} x^{2}\right )} \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}{9 \, b^{4} n^{4} + 40 \, b^{2} n^{2} + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sin(a+b*log(c*x^n))^3,x, algorithm="fricas")

[Out]

(3*(b^3*n^3 + 4*b*n)*x^2*cos(b*n*log(x) + b*log(c) + a)^3 - 3*(3*b^3*n^3 + 4*b*n)*x^2*cos(b*n*log(x) + b*log(c
) + a) - 2*((b^2*n^2 + 4)*x^2*cos(b*n*log(x) + b*log(c) + a)^2 - (7*b^2*n^2 + 4)*x^2)*sin(b*n*log(x) + b*log(c
) + a))/(9*b^4*n^4 + 40*b^2*n^2 + 16)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \begin {cases} \int x \sin ^{3}{\left (a - \frac {2 i \log {\left (c x^{n} \right )}}{n} \right )}\, dx & \text {for}\: b = - \frac {2 i}{n} \\\int x \sin ^{3}{\left (a - \frac {2 i \log {\left (c x^{n} \right )}}{3 n} \right )}\, dx & \text {for}\: b = - \frac {2 i}{3 n} \\\int x \sin ^{3}{\left (a + \frac {2 i \log {\left (c x^{n} \right )}}{3 n} \right )}\, dx & \text {for}\: b = \frac {2 i}{3 n} \\\int x \sin ^{3}{\left (a + \frac {2 i \log {\left (c x^{n} \right )}}{n} \right )}\, dx & \text {for}\: b = \frac {2 i}{n} \\- \frac {9 b^{3} n^{3} x^{2} \sin ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )} \cos {\left (a + b \log {\left (c x^{n} \right )} \right )}}{9 b^{4} n^{4} + 40 b^{2} n^{2} + 16} - \frac {6 b^{3} n^{3} x^{2} \cos ^{3}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{9 b^{4} n^{4} + 40 b^{2} n^{2} + 16} + \frac {14 b^{2} n^{2} x^{2} \sin ^{3}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{9 b^{4} n^{4} + 40 b^{2} n^{2} + 16} + \frac {12 b^{2} n^{2} x^{2} \sin {\left (a + b \log {\left (c x^{n} \right )} \right )} \cos ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{9 b^{4} n^{4} + 40 b^{2} n^{2} + 16} - \frac {12 b n x^{2} \sin ^{2}{\left (a + b \log {\left (c x^{n} \right )} \right )} \cos {\left (a + b \log {\left (c x^{n} \right )} \right )}}{9 b^{4} n^{4} + 40 b^{2} n^{2} + 16} + \frac {8 x^{2} \sin ^{3}{\left (a + b \log {\left (c x^{n} \right )} \right )}}{9 b^{4} n^{4} + 40 b^{2} n^{2} + 16} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sin(a+b*ln(c*x**n))**3,x)

[Out]

Piecewise((Integral(x*sin(a - 2*I*log(c*x**n)/n)**3, x), Eq(b, -2*I/n)), (Integral(x*sin(a - 2*I*log(c*x**n)/(
3*n))**3, x), Eq(b, -2*I/(3*n))), (Integral(x*sin(a + 2*I*log(c*x**n)/(3*n))**3, x), Eq(b, 2*I/(3*n))), (Integ
ral(x*sin(a + 2*I*log(c*x**n)/n)**3, x), Eq(b, 2*I/n)), (-9*b**3*n**3*x**2*sin(a + b*log(c*x**n))**2*cos(a + b
*log(c*x**n))/(9*b**4*n**4 + 40*b**2*n**2 + 16) - 6*b**3*n**3*x**2*cos(a + b*log(c*x**n))**3/(9*b**4*n**4 + 40
*b**2*n**2 + 16) + 14*b**2*n**2*x**2*sin(a + b*log(c*x**n))**3/(9*b**4*n**4 + 40*b**2*n**2 + 16) + 12*b**2*n**
2*x**2*sin(a + b*log(c*x**n))*cos(a + b*log(c*x**n))**2/(9*b**4*n**4 + 40*b**2*n**2 + 16) - 12*b*n*x**2*sin(a
+ b*log(c*x**n))**2*cos(a + b*log(c*x**n))/(9*b**4*n**4 + 40*b**2*n**2 + 16) + 8*x**2*sin(a + b*log(c*x**n))**
3/(9*b**4*n**4 + 40*b**2*n**2 + 16), True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 18117 vs. \(2 (158) = 316\).
time = 1.14, size = 18117, normalized size = 114.66 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sin(a+b*log(c*x^n))^3,x, algorithm="giac")

[Out]

1/8*(3*b^3*n^3*x^2*e^(3/2*pi*b*n*sgn(x) - 3/2*pi*b*n + 3/2*pi*b*sgn(c) - 3/2*pi*b)*tan(3/2*b*n*log(abs(x)) + 3
/2*b*log(abs(c)))^2*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))^2*tan(3/2*a)^2*tan(1/2*a)^2 - 27*b^3*n^3*x^2*
e^(1/2*pi*b*n*sgn(x) - 1/2*pi*b*n + 1/2*pi*b*sgn(c) - 1/2*pi*b)*tan(3/2*b*n*log(abs(x)) + 3/2*b*log(abs(c)))^2
*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))^2*tan(3/2*a)^2*tan(1/2*a)^2 - 27*b^3*n^3*x^2*e^(-1/2*pi*b*n*sgn(
x) + 1/2*pi*b*n - 1/2*pi*b*sgn(c) + 1/2*pi*b)*tan(3/2*b*n*log(abs(x)) + 3/2*b*log(abs(c)))^2*tan(1/2*b*n*log(a
bs(x)) + 1/2*b*log(abs(c)))^2*tan(3/2*a)^2*tan(1/2*a)^2 + 3*b^3*n^3*x^2*e^(-3/2*pi*b*n*sgn(x) + 3/2*pi*b*n - 3
/2*pi*b*sgn(c) + 3/2*pi*b)*tan(3/2*b*n*log(abs(x)) + 3/2*b*log(abs(c)))^2*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(
abs(c)))^2*tan(3/2*a)^2*tan(1/2*a)^2 + 3*b^3*n^3*x^2*e^(3/2*pi*b*n*sgn(x) - 3/2*pi*b*n + 3/2*pi*b*sgn(c) - 3/2
*pi*b)*tan(3/2*b*n*log(abs(x)) + 3/2*b*log(abs(c)))^2*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))^2*tan(3/2*a
)^2 + 27*b^3*n^3*x^2*e^(1/2*pi*b*n*sgn(x) - 1/2*pi*b*n + 1/2*pi*b*sgn(c) - 1/2*pi*b)*tan(3/2*b*n*log(abs(x)) +
3/2*b*log(abs(c)))^2*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))^2*tan(3/2*a)^2 + 27*b^3*n^3*x^2*e^(-1/2*pi*
b*n*sgn(x) + 1/2*pi*b*n - 1/2*pi*b*sgn(c) + 1/2*pi*b)*tan(3/2*b*n*log(abs(x)) + 3/2*b*log(abs(c)))^2*tan(1/2*b
*n*log(abs(x)) + 1/2*b*log(abs(c)))^2*tan(3/2*a)^2 + 3*b^3*n^3*x^2*e^(-3/2*pi*b*n*sgn(x) + 3/2*pi*b*n - 3/2*pi
*b*sgn(c) + 3/2*pi*b)*tan(3/2*b*n*log(abs(x)) + 3/2*b*log(abs(c)))^2*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c
)))^2*tan(3/2*a)^2 + 108*b^3*n^3*x^2*e^(1/2*pi*b*n*sgn(x) - 1/2*pi*b*n + 1/2*pi*b*sgn(c) - 1/2*pi*b)*tan(3/2*b
*n*log(abs(x)) + 3/2*b*log(abs(c)))^2*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))*tan(3/2*a)^2*tan(1/2*a) + 1
08*b^3*n^3*x^2*e^(-1/2*pi*b*n*sgn(x) + 1/2*pi*b*n - 1/2*pi*b*sgn(c) + 1/2*pi*b)*tan(3/2*b*n*log(abs(x)) + 3/2*
b*log(abs(c)))^2*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))*tan(3/2*a)^2*tan(1/2*a) - 3*b^3*n^3*x^2*e^(3/2*p
i*b*n*sgn(x) - 3/2*pi*b*n + 3/2*pi*b*sgn(c) - 3/2*pi*b)*tan(3/2*b*n*log(abs(x)) + 3/2*b*log(abs(c)))^2*tan(1/2
*b*n*log(abs(x)) + 1/2*b*log(abs(c)))^2*tan(1/2*a)^2 - 27*b^3*n^3*x^2*e^(1/2*pi*b*n*sgn(x) - 1/2*pi*b*n + 1/2*
pi*b*sgn(c) - 1/2*pi*b)*tan(3/2*b*n*log(abs(x)) + 3/2*b*log(abs(c)))^2*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs
(c)))^2*tan(1/2*a)^2 - 27*b^3*n^3*x^2*e^(-1/2*pi*b*n*sgn(x) + 1/2*pi*b*n - 1/2*pi*b*sgn(c) + 1/2*pi*b)*tan(3/2
*b*n*log(abs(x)) + 3/2*b*log(abs(c)))^2*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))^2*tan(1/2*a)^2 - 3*b^3*n^
3*x^2*e^(-3/2*pi*b*n*sgn(x) + 3/2*pi*b*n - 3/2*pi*b*sgn(c) + 3/2*pi*b)*tan(3/2*b*n*log(abs(x)) + 3/2*b*log(abs
(c)))^2*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))^2*tan(1/2*a)^2 - 12*b^3*n^3*x^2*e^(3/2*pi*b*n*sgn(x) - 3/
2*pi*b*n + 3/2*pi*b*sgn(c) - 3/2*pi*b)*tan(3/2*b*n*log(abs(x)) + 3/2*b*log(abs(c)))*tan(1/2*b*n*log(abs(x)) +
1/2*b*log(abs(c)))^2*tan(3/2*a)*tan(1/2*a)^2 - 12*b^3*n^3*x^2*e^(-3/2*pi*b*n*sgn(x) + 3/2*pi*b*n - 3/2*pi*b*sg
n(c) + 3/2*pi*b)*tan(3/2*b*n*log(abs(x)) + 3/2*b*log(abs(c)))*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))^2*t
an(3/2*a)*tan(1/2*a)^2 + 3*b^3*n^3*x^2*e^(3/2*pi*b*n*sgn(x) - 3/2*pi*b*n + 3/2*pi*b*sgn(c) - 3/2*pi*b)*tan(3/2
*b*n*log(abs(x)) + 3/2*b*log(abs(c)))^2*tan(3/2*a)^2*tan(1/2*a)^2 + 27*b^3*n^3*x^2*e^(1/2*pi*b*n*sgn(x) - 1/2*
pi*b*n + 1/2*pi*b*sgn(c) - 1/2*pi*b)*tan(3/2*b*n*log(abs(x)) + 3/2*b*log(abs(c)))^2*tan(3/2*a)^2*tan(1/2*a)^2
+ 27*b^3*n^3*x^2*e^(-1/2*pi*b*n*sgn(x) + 1/2*pi*b*n - 1/2*pi*b*sgn(c) + 1/2*pi*b)*tan(3/2*b*n*log(abs(x)) + 3/
2*b*log(abs(c)))^2*tan(3/2*a)^2*tan(1/2*a)^2 + 3*b^3*n^3*x^2*e^(-3/2*pi*b*n*sgn(x) + 3/2*pi*b*n - 3/2*pi*b*sgn
(c) + 3/2*pi*b)*tan(3/2*b*n*log(abs(x)) + 3/2*b*log(abs(c)))^2*tan(3/2*a)^2*tan(1/2*a)^2 - 3*b^3*n^3*x^2*e^(3/
2*pi*b*n*sgn(x) - 3/2*pi*b*n + 3/2*pi*b*sgn(c) - 3/2*pi*b)*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))^2*tan(
3/2*a)^2*tan(1/2*a)^2 - 27*b^3*n^3*x^2*e^(1/2*pi*b*n*sgn(x) - 1/2*pi*b*n + 1/2*pi*b*sgn(c) - 1/2*pi*b)*tan(1/2
*b*n*log(abs(x)) + 1/2*b*log(abs(c)))^2*tan(3/2*a)^2*tan(1/2*a)^2 - 27*b^3*n^3*x^2*e^(-1/2*pi*b*n*sgn(x) + 1/2
*pi*b*n - 1/2*pi*b*sgn(c) + 1/2*pi*b)*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))^2*tan(3/2*a)^2*tan(1/2*a)^2
- 3*b^3*n^3*x^2*e^(-3/2*pi*b*n*sgn(x) + 3/2*pi*b*n - 3/2*pi*b*sgn(c) + 3/2*pi*b)*tan(1/2*b*n*log(abs(x)) + 1/
2*b*log(abs(c)))^2*tan(3/2*a)^2*tan(1/2*a)^2 - 108*b^2*n^2*x^2*e^(1/2*pi*b*n*sgn(x) - 1/2*pi*b*n + 1/2*pi*b*sg
n(c) - 1/2*pi*b)*tan(3/2*b*n*log(abs(x)) + 3/2*b*log(abs(c)))^2*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))^2
*tan(3/2*a)^2*tan(1/2*a) - 108*b^2*n^2*x^2*e^(-1/2*pi*b*n*sgn(x) + 1/2*pi*b*n - 1/2*pi*b*sgn(c) + 1/2*pi*b)*ta
n(3/2*b*n*log(abs(x)) + 3/2*b*log(abs(c)))^2*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))^2*tan(3/2*a)^2*tan(1
/2*a) + 4*b^2*n^2*x^2*e^(3/2*pi*b*n*sgn(x) - 3/2*pi*b*n + 3/2*pi*b*sgn(c) - 3/2*pi*b)*tan(3/2*b*n*log(abs(x))
+ 3/2*b*log(abs(c)))^2*tan(1/2*b*n*log(abs(x)) + 1/2*b*log(abs(c)))^2*tan(3/2*a)*tan(1/2*a)^2 + 4*b^2*n^2*x^2*
e^(-3/2*pi*b*n*sgn(x) + 3/2*pi*b*n - 3/2*pi*b*s...

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Mupad [B]
time = 3.05, size = 122, normalized size = 0.77 \begin {gather*} -\frac {x^2\,{\mathrm {e}}^{-a\,1{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{b\,1{}\mathrm {i}}}\,3{}\mathrm {i}}{-16+b\,n\,8{}\mathrm {i}}-\frac {3\,x^2\,{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,1{}\mathrm {i}}}{8\,b\,n-16{}\mathrm {i}}+\frac {x^2\,{\mathrm {e}}^{-a\,3{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{b\,3{}\mathrm {i}}}\,1{}\mathrm {i}}{-16+b\,n\,24{}\mathrm {i}}+\frac {x^2\,{\mathrm {e}}^{a\,3{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,3{}\mathrm {i}}}{24\,b\,n-16{}\mathrm {i}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sin(a + b*log(c*x^n))^3,x)

[Out]

(x^2*exp(-a*3i)/(c*x^n)^(b*3i)*1i)/(b*n*24i - 16) - (3*x^2*exp(a*1i)*(c*x^n)^(b*1i))/(8*b*n - 16i) - (x^2*exp(
-a*1i)/(c*x^n)^(b*1i)*3i)/(b*n*8i - 16) + (x^2*exp(a*3i)*(c*x^n)^(b*3i))/(24*b*n - 16i)

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