3.1.20 \(\int x \sin ^4(a+b \log (c x^n)) \, dx\) [20]

Optimal. Leaf size=210 \[ \frac {3 b^4 n^4 x^2}{4 \left (1+5 b^2 n^2+4 b^4 n^4\right )}-\frac {3 b^3 n^3 x^2 \cos \left (a+b \log \left (c x^n\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )}{2 \left (1+5 b^2 n^2+4 b^4 n^4\right )}+\frac {3 b^2 n^2 x^2 \sin ^2\left (a+b \log \left (c x^n\right )\right )}{2 \left (1+5 b^2 n^2+4 b^4 n^4\right )}-\frac {b n x^2 \cos \left (a+b \log \left (c x^n\right )\right ) \sin ^3\left (a+b \log \left (c x^n\right )\right )}{1+4 b^2 n^2}+\frac {x^2 \sin ^4\left (a+b \log \left (c x^n\right )\right )}{2 \left (1+4 b^2 n^2\right )} \]

[Out]

3/4*b^4*n^4*x^2/(4*b^4*n^4+5*b^2*n^2+1)-3/2*b^3*n^3*x^2*cos(a+b*ln(c*x^n))*sin(a+b*ln(c*x^n))/(4*b^4*n^4+5*b^2
*n^2+1)+3/2*b^2*n^2*x^2*sin(a+b*ln(c*x^n))^2/(4*b^4*n^4+5*b^2*n^2+1)-b*n*x^2*cos(a+b*ln(c*x^n))*sin(a+b*ln(c*x
^n))^3/(4*b^2*n^2+1)+1/2*x^2*sin(a+b*ln(c*x^n))^4/(4*b^2*n^2+1)

________________________________________________________________________________________

Rubi [A]
time = 0.04, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {4575, 30} \begin {gather*} \frac {x^2 \sin ^4\left (a+b \log \left (c x^n\right )\right )}{2 \left (4 b^2 n^2+1\right )}-\frac {b n x^2 \sin ^3\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+1}+\frac {3 b^2 n^2 x^2 \sin ^2\left (a+b \log \left (c x^n\right )\right )}{2 \left (4 b^4 n^4+5 b^2 n^2+1\right )}-\frac {3 b^3 n^3 x^2 \sin \left (a+b \log \left (c x^n\right )\right ) \cos \left (a+b \log \left (c x^n\right )\right )}{2 \left (4 b^4 n^4+5 b^2 n^2+1\right )}+\frac {3 b^4 n^4 x^2}{4 \left (4 b^4 n^4+5 b^2 n^2+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Sin[a + b*Log[c*x^n]]^4,x]

[Out]

(3*b^4*n^4*x^2)/(4*(1 + 5*b^2*n^2 + 4*b^4*n^4)) - (3*b^3*n^3*x^2*Cos[a + b*Log[c*x^n]]*Sin[a + b*Log[c*x^n]])/
(2*(1 + 5*b^2*n^2 + 4*b^4*n^4)) + (3*b^2*n^2*x^2*Sin[a + b*Log[c*x^n]]^2)/(2*(1 + 5*b^2*n^2 + 4*b^4*n^4)) - (b
*n*x^2*Cos[a + b*Log[c*x^n]]*Sin[a + b*Log[c*x^n]]^3)/(1 + 4*b^2*n^2) + (x^2*Sin[a + b*Log[c*x^n]]^4)/(2*(1 +
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 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 ^4\left (a+b \log \left (c x^n\right )\right ) \, dx &=-\frac {b n x^2 \cos \left (a+b \log \left (c x^n\right )\right ) \sin ^3\left (a+b \log \left (c x^n\right )\right )}{1+4 b^2 n^2}+\frac {x^2 \sin ^4\left (a+b \log \left (c x^n\right )\right )}{2 \left (1+4 b^2 n^2\right )}+\frac {\left (3 b^2 n^2\right ) \int x \sin ^2\left (a+b \log \left (c x^n\right )\right ) \, dx}{1+4 b^2 n^2}\\ &=-\frac {3 b^3 n^3 x^2 \cos \left (a+b \log \left (c x^n\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )}{2 \left (1+5 b^2 n^2+4 b^4 n^4\right )}+\frac {3 b^2 n^2 x^2 \sin ^2\left (a+b \log \left (c x^n\right )\right )}{2 \left (1+5 b^2 n^2+4 b^4 n^4\right )}-\frac {b n x^2 \cos \left (a+b \log \left (c x^n\right )\right ) \sin ^3\left (a+b \log \left (c x^n\right )\right )}{1+4 b^2 n^2}+\frac {x^2 \sin ^4\left (a+b \log \left (c x^n\right )\right )}{2 \left (1+4 b^2 n^2\right )}+\frac {\left (3 b^4 n^4\right ) \int x \, dx}{2 \left (1+5 b^2 n^2+4 b^4 n^4\right )}\\ &=\frac {3 b^4 n^4 x^2}{4 \left (1+5 b^2 n^2+4 b^4 n^4\right )}-\frac {3 b^3 n^3 x^2 \cos \left (a+b \log \left (c x^n\right )\right ) \sin \left (a+b \log \left (c x^n\right )\right )}{2 \left (1+5 b^2 n^2+4 b^4 n^4\right )}+\frac {3 b^2 n^2 x^2 \sin ^2\left (a+b \log \left (c x^n\right )\right )}{2 \left (1+5 b^2 n^2+4 b^4 n^4\right )}-\frac {b n x^2 \cos \left (a+b \log \left (c x^n\right )\right ) \sin ^3\left (a+b \log \left (c x^n\right )\right )}{1+4 b^2 n^2}+\frac {x^2 \sin ^4\left (a+b \log \left (c x^n\right )\right )}{2 \left (1+4 b^2 n^2\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.49, size = 169, normalized size = 0.80 \begin {gather*} \frac {x^2 \left (3+15 b^2 n^2+12 b^4 n^4-4 \left (1+4 b^2 n^2\right ) \cos \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+\left (1+b^2 n^2\right ) \cos \left (4 \left (a+b \log \left (c x^n\right )\right )\right )-4 b n \sin \left (2 \left (a+b \log \left (c x^n\right )\right )\right )-16 b^3 n^3 \sin \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+2 b n \sin \left (4 \left (a+b \log \left (c x^n\right )\right )\right )+2 b^3 n^3 \sin \left (4 \left (a+b \log \left (c x^n\right )\right )\right )\right )}{16 \left (1+5 b^2 n^2+4 b^4 n^4\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*Sin[a + b*Log[c*x^n]]^4,x]

[Out]

(x^2*(3 + 15*b^2*n^2 + 12*b^4*n^4 - 4*(1 + 4*b^2*n^2)*Cos[2*(a + b*Log[c*x^n])] + (1 + b^2*n^2)*Cos[4*(a + b*L
og[c*x^n])] - 4*b*n*Sin[2*(a + b*Log[c*x^n])] - 16*b^3*n^3*Sin[2*(a + b*Log[c*x^n])] + 2*b*n*Sin[4*(a + b*Log[
c*x^n])] + 2*b^3*n^3*Sin[4*(a + b*Log[c*x^n])]))/(16*(1 + 5*b^2*n^2 + 4*b^4*n^4))

________________________________________________________________________________________

Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int x \left (\sin ^{4}\left (a +b \ln \left (c \,x^{n}\right )\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sin(a+b*ln(c*x^n))^4,x)

[Out]

int(x*sin(a+b*ln(c*x^n))^4,x)

________________________________________________________________________________________

Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 1085 vs. \(2 (202) = 404\).
time = 0.32, size = 1085, normalized size = 5.17 \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))^4,x, algorithm="maxima")

[Out]

1/32*((2*(b^3*cos(4*b*log(c))*sin(8*b*log(c)) - b^3*cos(8*b*log(c))*sin(4*b*log(c)) + b^3*sin(4*b*log(c)))*n^3
+ (b^2*cos(8*b*log(c))*cos(4*b*log(c)) + b^2*sin(8*b*log(c))*sin(4*b*log(c)) + b^2*cos(4*b*log(c)))*n^2 + 2*(
b*cos(4*b*log(c))*sin(8*b*log(c)) - b*cos(8*b*log(c))*sin(4*b*log(c)) + b*sin(4*b*log(c)))*n + cos(8*b*log(c))
*cos(4*b*log(c)) + sin(8*b*log(c))*sin(4*b*log(c)) + cos(4*b*log(c)))*x^2*cos(4*b*log(x^n) + 4*a) - 4*(4*(b^3*
cos(4*b*log(c))*sin(6*b*log(c)) - b^3*cos(6*b*log(c))*sin(4*b*log(c)) + b^3*cos(2*b*log(c))*sin(4*b*log(c)) -
b^3*cos(4*b*log(c))*sin(2*b*log(c)))*n^3 + 4*(b^2*cos(6*b*log(c))*cos(4*b*log(c)) + b^2*cos(4*b*log(c))*cos(2*
b*log(c)) + b^2*sin(6*b*log(c))*sin(4*b*log(c)) + b^2*sin(4*b*log(c))*sin(2*b*log(c)))*n^2 + (b*cos(4*b*log(c)
)*sin(6*b*log(c)) - b*cos(6*b*log(c))*sin(4*b*log(c)) + b*cos(2*b*log(c))*sin(4*b*log(c)) - b*cos(4*b*log(c))*
sin(2*b*log(c)))*n + cos(6*b*log(c))*cos(4*b*log(c)) + cos(4*b*log(c))*cos(2*b*log(c)) + sin(6*b*log(c))*sin(4
*b*log(c)) + sin(4*b*log(c))*sin(2*b*log(c)))*x^2*cos(2*b*log(x^n) + 2*a) + (2*(b^3*cos(8*b*log(c))*cos(4*b*lo
g(c)) + b^3*sin(8*b*log(c))*sin(4*b*log(c)) + b^3*cos(4*b*log(c)))*n^3 - (b^2*cos(4*b*log(c))*sin(8*b*log(c))
- b^2*cos(8*b*log(c))*sin(4*b*log(c)) + b^2*sin(4*b*log(c)))*n^2 + 2*(b*cos(8*b*log(c))*cos(4*b*log(c)) + b*si
n(8*b*log(c))*sin(4*b*log(c)) + b*cos(4*b*log(c)))*n - cos(4*b*log(c))*sin(8*b*log(c)) + cos(8*b*log(c))*sin(4
*b*log(c)) - sin(4*b*log(c)))*x^2*sin(4*b*log(x^n) + 4*a) - 4*(4*(b^3*cos(6*b*log(c))*cos(4*b*log(c)) + b^3*co
s(4*b*log(c))*cos(2*b*log(c)) + b^3*sin(6*b*log(c))*sin(4*b*log(c)) + b^3*sin(4*b*log(c))*sin(2*b*log(c)))*n^3
- 4*(b^2*cos(4*b*log(c))*sin(6*b*log(c)) - b^2*cos(6*b*log(c))*sin(4*b*log(c)) + b^2*cos(2*b*log(c))*sin(4*b*
log(c)) - b^2*cos(4*b*log(c))*sin(2*b*log(c)))*n^2 + (b*cos(6*b*log(c))*cos(4*b*log(c)) + b*cos(4*b*log(c))*co
s(2*b*log(c)) + b*sin(6*b*log(c))*sin(4*b*log(c)) + b*sin(4*b*log(c))*sin(2*b*log(c)))*n - cos(4*b*log(c))*sin
(6*b*log(c)) + cos(6*b*log(c))*sin(4*b*log(c)) - cos(2*b*log(c))*sin(4*b*log(c)) + cos(4*b*log(c))*sin(2*b*log
(c)))*x^2*sin(2*b*log(x^n) + 2*a) + 6*(4*(b^4*cos(4*b*log(c))^2 + b^4*sin(4*b*log(c))^2)*n^4 + 5*(b^2*cos(4*b*
log(c))^2 + b^2*sin(4*b*log(c))^2)*n^2 + cos(4*b*log(c))^2 + sin(4*b*log(c))^2)*x^2)/(4*(b^4*cos(4*b*log(c))^2
+ b^4*sin(4*b*log(c))^2)*n^4 + 5*(b^2*cos(4*b*log(c))^2 + b^2*sin(4*b*log(c))^2)*n^2 + cos(4*b*log(c))^2 + si
n(4*b*log(c))^2)

________________________________________________________________________________________

Fricas [A]
time = 1.36, size = 177, normalized size = 0.84 \begin {gather*} \frac {2 \, {\left (b^{2} n^{2} + 1\right )} x^{2} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{4} - 2 \, {\left (5 \, b^{2} n^{2} + 2\right )} x^{2} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} + {\left (3 \, b^{4} n^{4} + 8 \, b^{2} n^{2} + 2\right )} x^{2} + 2 \, {\left (2 \, {\left (b^{3} n^{3} + b n\right )} x^{2} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{3} - {\left (5 \, b^{3} n^{3} + 2 \, b n\right )} x^{2} \cos \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )\right )} \sin \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}{4 \, {\left (4 \, b^{4} n^{4} + 5 \, b^{2} n^{2} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sin(a+b*log(c*x^n))^4,x, algorithm="fricas")

[Out]

1/4*(2*(b^2*n^2 + 1)*x^2*cos(b*n*log(x) + b*log(c) + a)^4 - 2*(5*b^2*n^2 + 2)*x^2*cos(b*n*log(x) + b*log(c) +
a)^2 + (3*b^4*n^4 + 8*b^2*n^2 + 2)*x^2 + 2*(2*(b^3*n^3 + b*n)*x^2*cos(b*n*log(x) + b*log(c) + a)^3 - (5*b^3*n^
3 + 2*b*n)*x^2*cos(b*n*log(x) + b*log(c) + a))*sin(b*n*log(x) + b*log(c) + a))/(4*b^4*n^4 + 5*b^2*n^2 + 1)

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sin(a+b*ln(c*x**n))**4,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 16984 vs. \(2 (202) = 404\).
time = 1.02, size = 16984, normalized size = 80.88 \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))^4,x, algorithm="giac")

[Out]

3/16*x^2 + 1/32*(32*b^3*n^3*x^2*e^(pi*b*n*sgn(x) - pi*b*n + pi*b*sgn(c) - pi*b)*tan(2*b*n*log(abs(x)) + 2*b*lo
g(abs(c)))^2*tan(b*n*log(abs(x)) + b*log(abs(c)))^2*tan(2*a)^2*tan(a) + 32*b^3*n^3*x^2*e^(-pi*b*n*sgn(x) + pi*
b*n - pi*b*sgn(c) + pi*b)*tan(2*b*n*log(abs(x)) + 2*b*log(abs(c)))^2*tan(b*n*log(abs(x)) + b*log(abs(c)))^2*ta
n(2*a)^2*tan(a) - 4*b^3*n^3*x^2*e^(2*pi*b*n*sgn(x) - 2*pi*b*n + 2*pi*b*sgn(c) - 2*pi*b)*tan(2*b*n*log(abs(x))
+ 2*b*log(abs(c)))^2*tan(b*n*log(abs(x)) + b*log(abs(c)))^2*tan(2*a)*tan(a)^2 - 4*b^3*n^3*x^2*e^(-2*pi*b*n*sgn
(x) + 2*pi*b*n - 2*pi*b*sgn(c) + 2*pi*b)*tan(2*b*n*log(abs(x)) + 2*b*log(abs(c)))^2*tan(b*n*log(abs(x)) + b*lo
g(abs(c)))^2*tan(2*a)*tan(a)^2 + 32*b^3*n^3*x^2*e^(pi*b*n*sgn(x) - pi*b*n + pi*b*sgn(c) - pi*b)*tan(2*b*n*log(
abs(x)) + 2*b*log(abs(c)))^2*tan(b*n*log(abs(x)) + b*log(abs(c)))*tan(2*a)^2*tan(a)^2 + 32*b^3*n^3*x^2*e^(-pi*
b*n*sgn(x) + pi*b*n - pi*b*sgn(c) + pi*b)*tan(2*b*n*log(abs(x)) + 2*b*log(abs(c)))^2*tan(b*n*log(abs(x)) + b*l
og(abs(c)))*tan(2*a)^2*tan(a)^2 - 4*b^3*n^3*x^2*e^(2*pi*b*n*sgn(x) - 2*pi*b*n + 2*pi*b*sgn(c) - 2*pi*b)*tan(2*
b*n*log(abs(x)) + 2*b*log(abs(c)))*tan(b*n*log(abs(x)) + b*log(abs(c)))^2*tan(2*a)^2*tan(a)^2 - 4*b^3*n^3*x^2*
e^(-2*pi*b*n*sgn(x) + 2*pi*b*n - 2*pi*b*sgn(c) + 2*pi*b)*tan(2*b*n*log(abs(x)) + 2*b*log(abs(c)))*tan(b*n*log(
abs(x)) + b*log(abs(c)))^2*tan(2*a)^2*tan(a)^2 + b^2*n^2*x^2*e^(2*pi*b*n*sgn(x) - 2*pi*b*n + 2*pi*b*sgn(c) - 2
*pi*b)*tan(2*b*n*log(abs(x)) + 2*b*log(abs(c)))^2*tan(b*n*log(abs(x)) + b*log(abs(c)))^2*tan(2*a)^2*tan(a)^2 -
16*b^2*n^2*x^2*e^(pi*b*n*sgn(x) - pi*b*n + pi*b*sgn(c) - pi*b)*tan(2*b*n*log(abs(x)) + 2*b*log(abs(c)))^2*tan
(b*n*log(abs(x)) + b*log(abs(c)))^2*tan(2*a)^2*tan(a)^2 - 16*b^2*n^2*x^2*e^(-pi*b*n*sgn(x) + pi*b*n - pi*b*sgn
(c) + pi*b)*tan(2*b*n*log(abs(x)) + 2*b*log(abs(c)))^2*tan(b*n*log(abs(x)) + b*log(abs(c)))^2*tan(2*a)^2*tan(a
)^2 + b^2*n^2*x^2*e^(-2*pi*b*n*sgn(x) + 2*pi*b*n - 2*pi*b*sgn(c) + 2*pi*b)*tan(2*b*n*log(abs(x)) + 2*b*log(abs
(c)))^2*tan(b*n*log(abs(x)) + b*log(abs(c)))^2*tan(2*a)^2*tan(a)^2 - 4*b^3*n^3*x^2*e^(2*pi*b*n*sgn(x) - 2*pi*b
*n + 2*pi*b*sgn(c) - 2*pi*b)*tan(2*b*n*log(abs(x)) + 2*b*log(abs(c)))^2*tan(b*n*log(abs(x)) + b*log(abs(c)))^2
*tan(2*a) - 4*b^3*n^3*x^2*e^(-2*pi*b*n*sgn(x) + 2*pi*b*n - 2*pi*b*sgn(c) + 2*pi*b)*tan(2*b*n*log(abs(x)) + 2*b
*log(abs(c)))^2*tan(b*n*log(abs(x)) + b*log(abs(c)))^2*tan(2*a) - 32*b^3*n^3*x^2*e^(pi*b*n*sgn(x) - pi*b*n + p
i*b*sgn(c) - pi*b)*tan(2*b*n*log(abs(x)) + 2*b*log(abs(c)))^2*tan(b*n*log(abs(x)) + b*log(abs(c)))*tan(2*a)^2
- 32*b^3*n^3*x^2*e^(-pi*b*n*sgn(x) + pi*b*n - pi*b*sgn(c) + pi*b)*tan(2*b*n*log(abs(x)) + 2*b*log(abs(c)))^2*t
an(b*n*log(abs(x)) + b*log(abs(c)))*tan(2*a)^2 - 4*b^3*n^3*x^2*e^(2*pi*b*n*sgn(x) - 2*pi*b*n + 2*pi*b*sgn(c) -
2*pi*b)*tan(2*b*n*log(abs(x)) + 2*b*log(abs(c)))*tan(b*n*log(abs(x)) + b*log(abs(c)))^2*tan(2*a)^2 - 4*b^3*n^
3*x^2*e^(-2*pi*b*n*sgn(x) + 2*pi*b*n - 2*pi*b*sgn(c) + 2*pi*b)*tan(2*b*n*log(abs(x)) + 2*b*log(abs(c)))*tan(b*
n*log(abs(x)) + b*log(abs(c)))^2*tan(2*a)^2 + 32*b^3*n^3*x^2*e^(pi*b*n*sgn(x) - pi*b*n + pi*b*sgn(c) - pi*b)*t
an(2*b*n*log(abs(x)) + 2*b*log(abs(c)))^2*tan(b*n*log(abs(x)) + b*log(abs(c)))^2*tan(a) + 32*b^3*n^3*x^2*e^(-p
i*b*n*sgn(x) + pi*b*n - pi*b*sgn(c) + pi*b)*tan(2*b*n*log(abs(x)) + 2*b*log(abs(c)))^2*tan(b*n*log(abs(x)) + b
*log(abs(c)))^2*tan(a) - 32*b^3*n^3*x^2*e^(pi*b*n*sgn(x) - pi*b*n + pi*b*sgn(c) - pi*b)*tan(2*b*n*log(abs(x))
+ 2*b*log(abs(c)))^2*tan(2*a)^2*tan(a) - 32*b^3*n^3*x^2*e^(-pi*b*n*sgn(x) + pi*b*n - pi*b*sgn(c) + pi*b)*tan(2
*b*n*log(abs(x)) + 2*b*log(abs(c)))^2*tan(2*a)^2*tan(a) + 32*b^3*n^3*x^2*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(2*a)^2*tan(a) + 32*b^3*n^3*x^2*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(2*a)^2*tan(a) + 32*b^3*n^3*x^2*e^(pi*b*n*sg
n(x) - pi*b*n + pi*b*sgn(c) - pi*b)*tan(2*b*n*log(abs(x)) + 2*b*log(abs(c)))^2*tan(b*n*log(abs(x)) + b*log(abs
(c)))*tan(a)^2 + 32*b^3*n^3*x^2*e^(-pi*b*n*sgn(x) + pi*b*n - pi*b*sgn(c) + pi*b)*tan(2*b*n*log(abs(x)) + 2*b*l
og(abs(c)))^2*tan(b*n*log(abs(x)) + b*log(abs(c)))*tan(a)^2 + 4*b^3*n^3*x^2*e^(2*pi*b*n*sgn(x) - 2*pi*b*n + 2*
pi*b*sgn(c) - 2*pi*b)*tan(2*b*n*log(abs(x)) + 2*b*log(abs(c)))*tan(b*n*log(abs(x)) + b*log(abs(c)))^2*tan(a)^2
+ 4*b^3*n^3*x^2*e^(-2*pi*b*n*sgn(x) + 2*pi*b*n - 2*pi*b*sgn(c) + 2*pi*b)*tan(2*b*n*log(abs(x)) + 2*b*log(abs(
c)))*tan(b*n*log(abs(x)) + b*log(abs(c)))^2*tan(a)^2 - 4*b^3*n^3*x^2*e^(2*pi*b*n*sgn(x) - 2*pi*b*n + 2*pi*b*sg
n(c) - 2*pi*b)*tan(2*b*n*log(abs(x)) + 2*b*log(abs(c)))^2*tan(2*a)*tan(a)^2 - 4*b^3*n^3*x^2*e^(-2*pi*b*n*sgn(x
) + 2*pi*b*n - 2*pi*b*sgn(c) + 2*pi*b)*tan(2*b*n*log(abs(x)) + 2*b*log(abs(c)))^2*tan(2*a)*tan(a)^2 + 4*b^3*n^
3*x^2*e^(2*pi*b*n*sgn(x) - 2*pi*b*n + 2*pi*b*sgn(c) - 2*pi*b)*tan(b*n*log(abs(x)) + b*log(abs(c)))^2*tan(2*a)*
tan(a)^2 + 4*b^3*n^3*x^2*e^(-2*pi*b*n*sgn(x) + 2*pi*b*n - 2*pi*b*sgn(c) + 2*pi*b)*tan(b*n*log(abs(x)) + b*log(
abs(c)))^2*tan(2*a)*tan(a)^2 - 4*b^3*n^3*x^2*e^...

________________________________________________________________________________________

Mupad [B]
time = 3.04, size = 127, normalized size = 0.60 \begin {gather*} \frac {3\,x^2}{16}-\frac {x^2\,{\mathrm {e}}^{-a\,2{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{b\,2{}\mathrm {i}}}\,1{}\mathrm {i}}{8\,b\,n+8{}\mathrm {i}}-\frac {x^2\,{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,2{}\mathrm {i}}}{8+b\,n\,8{}\mathrm {i}}+\frac {x^2\,{\mathrm {e}}^{-a\,4{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{b\,4{}\mathrm {i}}}\,1{}\mathrm {i}}{64\,b\,n+32{}\mathrm {i}}+\frac {x^2\,{\mathrm {e}}^{a\,4{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,4{}\mathrm {i}}}{32+b\,n\,64{}\mathrm {i}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sin(a + b*log(c*x^n))^4,x)

[Out]

(3*x^2)/16 - (x^2*exp(-a*2i)/(c*x^n)^(b*2i)*1i)/(8*b*n + 8i) - (x^2*exp(a*2i)*(c*x^n)^(b*2i))/(b*n*8i + 8) + (
x^2*exp(-a*4i)/(c*x^n)^(b*4i)*1i)/(64*b*n + 32i) + (x^2*exp(a*4i)*(c*x^n)^(b*4i))/(b*n*64i + 32)

________________________________________________________________________________________