Integrand size = 21, antiderivative size = 337 \[ \int (e x)^m \sin ^4\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {24 b^4 d^4 n^4 (e x)^{1+m}}{e (1+m) \left ((1+m)^2+4 b^2 d^2 n^2\right ) \left ((1+m)^2+16 b^2 d^2 n^2\right )}-\frac {24 b^3 d^3 n^3 (e x)^{1+m} \cos \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \sin \left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e \left ((1+m)^2+4 b^2 d^2 n^2\right ) \left ((1+m)^2+16 b^2 d^2 n^2\right )}+\frac {12 b^2 d^2 (1+m) n^2 (e x)^{1+m} \sin ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e \left ((1+m)^2+4 b^2 d^2 n^2\right ) \left ((1+m)^2+16 b^2 d^2 n^2\right )}-\frac {4 b d n (e x)^{1+m} \cos \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \sin ^3\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e \left ((1+m)^2+16 b^2 d^2 n^2\right )}+\frac {(1+m) (e x)^{1+m} \sin ^4\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e \left ((1+m)^2+16 b^2 d^2 n^2\right )} \] Output:
24*b^4*d^4*n^4*(e*x)^(1+m)/e/(1+m)/((1+m)^2+4*b^2*d^2*n^2)/((1+m)^2+16*b^2 *d^2*n^2)-24*b^3*d^3*n^3*(e*x)^(1+m)*cos(d*(a+b*ln(c*x^n)))*sin(d*(a+b*ln( c*x^n)))/e/((1+m)^2+4*b^2*d^2*n^2)/((1+m)^2+16*b^2*d^2*n^2)+12*b^2*d^2*(1+ m)*n^2*(e*x)^(1+m)*sin(d*(a+b*ln(c*x^n)))^2/e/((1+m)^2+4*b^2*d^2*n^2)/((1+ m)^2+16*b^2*d^2*n^2)-4*b*d*n*(e*x)^(1+m)*cos(d*(a+b*ln(c*x^n)))*sin(d*(a+b *ln(c*x^n)))^3/e/((1+m)^2+16*b^2*d^2*n^2)+(1+m)*(e*x)^(1+m)*sin(d*(a+b*ln( c*x^n)))^4/e/((1+m)^2+16*b^2*d^2*n^2)
Time = 1.39 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.01 \[ \int (e x)^m \sin ^4\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {1}{8} x (e x)^m \left (\frac {3}{1+m}+\frac {4 \sin (2 b d n \log (x)) \left (-2 b d n \cos \left (2 d \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )+(1+m) \sin \left (2 d \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )\right )}{1+2 m+m^2+4 b^2 d^2 n^2}-\frac {4 \cos (2 b d n \log (x)) \left ((1+m) \cos \left (2 d \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )+2 b d n \sin \left (2 d \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )\right )}{1+2 m+m^2+4 b^2 d^2 n^2}-\frac {\sin (4 b d n \log (x)) \left (-4 b d n \cos \left (4 d \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )+(1+m) \sin \left (4 d \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )\right )}{1+2 m+m^2+16 b^2 d^2 n^2}+\frac {\cos (4 b d n \log (x)) \left ((1+m) \cos \left (4 d \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )+4 b d n \sin \left (4 d \left (a-b n \log (x)+b \log \left (c x^n\right )\right )\right )\right )}{1+2 m+m^2+16 b^2 d^2 n^2}\right ) \] Input:
Integrate[(e*x)^m*Sin[d*(a + b*Log[c*x^n])]^4,x]
Output:
(x*(e*x)^m*(3/(1 + m) + (4*Sin[2*b*d*n*Log[x]]*(-2*b*d*n*Cos[2*d*(a - b*n* Log[x] + b*Log[c*x^n])] + (1 + m)*Sin[2*d*(a - b*n*Log[x] + b*Log[c*x^n])] ))/(1 + 2*m + m^2 + 4*b^2*d^2*n^2) - (4*Cos[2*b*d*n*Log[x]]*((1 + m)*Cos[2 *d*(a - b*n*Log[x] + b*Log[c*x^n])] + 2*b*d*n*Sin[2*d*(a - b*n*Log[x] + b* Log[c*x^n])]))/(1 + 2*m + m^2 + 4*b^2*d^2*n^2) - (Sin[4*b*d*n*Log[x]]*(-4* b*d*n*Cos[4*d*(a - b*n*Log[x] + b*Log[c*x^n])] + (1 + m)*Sin[4*d*(a - b*n* Log[x] + b*Log[c*x^n])]))/(1 + 2*m + m^2 + 16*b^2*d^2*n^2) + (Cos[4*b*d*n* Log[x]]*((1 + m)*Cos[4*d*(a - b*n*Log[x] + b*Log[c*x^n])] + 4*b*d*n*Sin[4* d*(a - b*n*Log[x] + b*Log[c*x^n])]))/(1 + 2*m + m^2 + 16*b^2*d^2*n^2)))/8
Time = 0.49 (sec) , antiderivative size = 295, normalized size of antiderivative = 0.88, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4990, 4990, 17}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int (e x)^m \sin ^4\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx\) |
\(\Big \downarrow \) 4990 |
\(\displaystyle \frac {12 b^2 d^2 n^2 \int (e x)^m \sin ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right )dx}{16 b^2 d^2 n^2+(m+1)^2}+\frac {(m+1) (e x)^{m+1} \sin ^4\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e \left (16 b^2 d^2 n^2+(m+1)^2\right )}-\frac {4 b d n (e x)^{m+1} \sin ^3\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \cos \left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e \left (16 b^2 d^2 n^2+(m+1)^2\right )}\) |
\(\Big \downarrow \) 4990 |
\(\displaystyle \frac {12 b^2 d^2 n^2 \left (\frac {2 b^2 d^2 n^2 \int (e x)^mdx}{4 b^2 d^2 n^2+(m+1)^2}+\frac {(m+1) (e x)^{m+1} \sin ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e \left (4 b^2 d^2 n^2+(m+1)^2\right )}-\frac {2 b d n (e x)^{m+1} \sin \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \cos \left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e \left (4 b^2 d^2 n^2+(m+1)^2\right )}\right )}{16 b^2 d^2 n^2+(m+1)^2}+\frac {(m+1) (e x)^{m+1} \sin ^4\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e \left (16 b^2 d^2 n^2+(m+1)^2\right )}-\frac {4 b d n (e x)^{m+1} \sin ^3\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \cos \left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e \left (16 b^2 d^2 n^2+(m+1)^2\right )}\) |
\(\Big \downarrow \) 17 |
\(\displaystyle \frac {(m+1) (e x)^{m+1} \sin ^4\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e \left (16 b^2 d^2 n^2+(m+1)^2\right )}-\frac {4 b d n (e x)^{m+1} \sin ^3\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \cos \left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e \left (16 b^2 d^2 n^2+(m+1)^2\right )}+\frac {12 b^2 d^2 n^2 \left (\frac {(m+1) (e x)^{m+1} \sin ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e \left (4 b^2 d^2 n^2+(m+1)^2\right )}-\frac {2 b d n (e x)^{m+1} \sin \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \cos \left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e \left (4 b^2 d^2 n^2+(m+1)^2\right )}+\frac {2 b^2 d^2 n^2 (e x)^{m+1}}{e (m+1) \left (4 b^2 d^2 n^2+(m+1)^2\right )}\right )}{16 b^2 d^2 n^2+(m+1)^2}\) |
Input:
Int[(e*x)^m*Sin[d*(a + b*Log[c*x^n])]^4,x]
Output:
(-4*b*d*n*(e*x)^(1 + m)*Cos[d*(a + b*Log[c*x^n])]*Sin[d*(a + b*Log[c*x^n]) ]^3)/(e*((1 + m)^2 + 16*b^2*d^2*n^2)) + ((1 + m)*(e*x)^(1 + m)*Sin[d*(a + b*Log[c*x^n])]^4)/(e*((1 + m)^2 + 16*b^2*d^2*n^2)) + (12*b^2*d^2*n^2*((2*b ^2*d^2*n^2*(e*x)^(1 + m))/(e*(1 + m)*((1 + m)^2 + 4*b^2*d^2*n^2)) - (2*b*d *n*(e*x)^(1 + m)*Cos[d*(a + b*Log[c*x^n])]*Sin[d*(a + b*Log[c*x^n])])/(e*( (1 + m)^2 + 4*b^2*d^2*n^2)) + ((1 + m)*(e*x)^(1 + m)*Sin[d*(a + b*Log[c*x^ n])]^2)/(e*((1 + m)^2 + 4*b^2*d^2*n^2))))/((1 + m)^2 + 16*b^2*d^2*n^2)
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
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] + (-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] + Simp[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]) /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 1] && NeQ[b^2*d^2*n^2*p^2 + (m + 1)^2, 0]
\[\int \left (e x \right )^{m} {\sin \left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )}^{4}d x\]
Input:
int((e*x)^m*sin(d*(a+b*ln(c*x^n)))^4,x)
Output:
int((e*x)^m*sin(d*(a+b*ln(c*x^n)))^4,x)
Time = 0.13 (sec) , antiderivative size = 467, normalized size of antiderivative = 1.39 \[ \int (e x)^m \sin ^4\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {4 \, {\left ({\left (4 \, {\left (b^{3} d^{3} m + b^{3} d^{3}\right )} n^{3} + {\left (b d m^{3} + 3 \, b d m^{2} + 3 \, b d m + b d\right )} n\right )} x \cos \left (b d n \log \left (x\right ) + b d \log \left (c\right ) + a d\right )^{3} - {\left (10 \, {\left (b^{3} d^{3} m + b^{3} d^{3}\right )} n^{3} + {\left (b d m^{3} + 3 \, b d m^{2} + 3 \, b d m + b d\right )} n\right )} x \cos \left (b d n \log \left (x\right ) + b d \log \left (c\right ) + a d\right )\right )} e^{\left (m \log \left (e\right ) + m \log \left (x\right )\right )} \sin \left (b d n \log \left (x\right ) + b d \log \left (c\right ) + a d\right ) + {\left ({\left (m^{4} + 4 \, m^{3} + 4 \, {\left (b^{2} d^{2} m^{2} + 2 \, b^{2} d^{2} m + b^{2} d^{2}\right )} n^{2} + 6 \, m^{2} + 4 \, m + 1\right )} x \cos \left (b d n \log \left (x\right ) + b d \log \left (c\right ) + a d\right )^{4} - 2 \, {\left (m^{4} + 4 \, m^{3} + 10 \, {\left (b^{2} d^{2} m^{2} + 2 \, b^{2} d^{2} m + b^{2} d^{2}\right )} n^{2} + 6 \, m^{2} + 4 \, m + 1\right )} x \cos \left (b d n \log \left (x\right ) + b d \log \left (c\right ) + a d\right )^{2} + {\left (24 \, b^{4} d^{4} n^{4} + m^{4} + 4 \, m^{3} + 16 \, {\left (b^{2} d^{2} m^{2} + 2 \, b^{2} d^{2} m + b^{2} d^{2}\right )} n^{2} + 6 \, m^{2} + 4 \, m + 1\right )} x\right )} e^{\left (m \log \left (e\right ) + m \log \left (x\right )\right )}}{m^{5} + 64 \, {\left (b^{4} d^{4} m + b^{4} d^{4}\right )} n^{4} + 5 \, m^{4} + 10 \, m^{3} + 20 \, {\left (b^{2} d^{2} m^{3} + 3 \, b^{2} d^{2} m^{2} + 3 \, b^{2} d^{2} m + b^{2} d^{2}\right )} n^{2} + 10 \, m^{2} + 5 \, m + 1} \] Input:
integrate((e*x)^m*sin(d*(a+b*log(c*x^n)))^4,x, algorithm="fricas")
Output:
(4*((4*(b^3*d^3*m + b^3*d^3)*n^3 + (b*d*m^3 + 3*b*d*m^2 + 3*b*d*m + b*d)*n )*x*cos(b*d*n*log(x) + b*d*log(c) + a*d)^3 - (10*(b^3*d^3*m + b^3*d^3)*n^3 + (b*d*m^3 + 3*b*d*m^2 + 3*b*d*m + b*d)*n)*x*cos(b*d*n*log(x) + b*d*log(c ) + a*d))*e^(m*log(e) + m*log(x))*sin(b*d*n*log(x) + b*d*log(c) + a*d) + ( (m^4 + 4*m^3 + 4*(b^2*d^2*m^2 + 2*b^2*d^2*m + b^2*d^2)*n^2 + 6*m^2 + 4*m + 1)*x*cos(b*d*n*log(x) + b*d*log(c) + a*d)^4 - 2*(m^4 + 4*m^3 + 10*(b^2*d^ 2*m^2 + 2*b^2*d^2*m + b^2*d^2)*n^2 + 6*m^2 + 4*m + 1)*x*cos(b*d*n*log(x) + b*d*log(c) + a*d)^2 + (24*b^4*d^4*n^4 + m^4 + 4*m^3 + 16*(b^2*d^2*m^2 + 2 *b^2*d^2*m + b^2*d^2)*n^2 + 6*m^2 + 4*m + 1)*x)*e^(m*log(e) + m*log(x)))/( m^5 + 64*(b^4*d^4*m + b^4*d^4)*n^4 + 5*m^4 + 10*m^3 + 20*(b^2*d^2*m^3 + 3* b^2*d^2*m^2 + 3*b^2*d^2*m + b^2*d^2)*n^2 + 10*m^2 + 5*m + 1)
\[ \int (e x)^m \sin ^4\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx =\text {Too large to display} \] Input:
integrate((e*x)**m*sin(d*(a+b*ln(c*x**n)))**4,x)
Output:
-Piecewise((log(x)*cos(2*a*d)/e, Eq(b, 0) & Eq(m, -1)), (Integral((e*x)**m *cos(-2*a*d + I*m*log(c*x**n)/n + I*log(c*x**n)/n), x), Eq(b, -I*(m + 1)/( 2*d*n))), (Integral((e*x)**m*cos(2*a*d + I*m*log(c*x**n)/n + I*log(c*x**n) /n), x), Eq(b, I*(m + 1)/(2*d*n))), (2*b*d*n*x*(e*x)**m*sin(2*a*d + 2*b*d* log(c*x**n))/(4*b**2*d**2*n**2 + m**2 + 2*m + 1) + m*x*(e*x)**m*cos(2*a*d + 2*b*d*log(c*x**n))/(4*b**2*d**2*n**2 + m**2 + 2*m + 1) + x*(e*x)**m*cos( 2*a*d + 2*b*d*log(c*x**n))/(4*b**2*d**2*n**2 + m**2 + 2*m + 1), True))/2 + Piecewise((log(x)*cos(4*a*d)/e, Eq(b, 0) & Eq(m, -1)), (Integral((e*x)**m *cos(-4*a*d + I*m*log(c*x**n)/n + I*log(c*x**n)/n), x), Eq(b, -I*(m + 1)/( 4*d*n))), (Integral((e*x)**m*cos(4*a*d + I*m*log(c*x**n)/n + I*log(c*x**n) /n), x), Eq(b, I*(m + 1)/(4*d*n))), (4*b*d*n*x*(e*x)**m*sin(4*a*d + 4*b*d* log(c*x**n))/(16*b**2*d**2*n**2 + m**2 + 2*m + 1) + m*x*(e*x)**m*cos(4*a*d + 4*b*d*log(c*x**n))/(16*b**2*d**2*n**2 + m**2 + 2*m + 1) + x*(e*x)**m*co s(4*a*d + 4*b*d*log(c*x**n))/(16*b**2*d**2*n**2 + m**2 + 2*m + 1), True))/ 8 + 3*Piecewise(((e*x)**(m + 1)/(m + 1), Ne(m, -1)), (log(e*x), True))/(8* e)
Leaf count of result is larger than twice the leaf count of optimal. 16932 vs. \(2 (337) = 674\).
Time = 0.86 (sec) , antiderivative size = 16932, normalized size of antiderivative = 50.24 \[ \int (e x)^m \sin ^4\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\text {Too large to display} \] Input:
integrate((e*x)^m*sin(d*(a+b*log(c*x^n)))^4,x, algorithm="maxima")
Output:
1/16*(((((cos(8*a*d)*cos(4*a*d) + sin(8*a*d)*sin(4*a*d))*cos(4*b*d*log(c)) + (cos(4*a*d)*sin(8*a*d) - cos(8*a*d)*sin(4*a*d))*sin(4*b*d*log(c)))*cos( 8*b*d*log(c)) + cos(4*b*d*log(c))*cos(4*a*d) - ((cos(4*a*d)*sin(8*a*d) - c os(8*a*d)*sin(4*a*d))*cos(4*b*d*log(c)) - (cos(8*a*d)*cos(4*a*d) + sin(8*a *d)*sin(4*a*d))*sin(4*b*d*log(c)))*sin(8*b*d*log(c)) - sin(4*b*d*log(c))*s in(4*a*d))*e^m*m^4 + 4*(((cos(8*a*d)*cos(4*a*d) + sin(8*a*d)*sin(4*a*d))*c os(4*b*d*log(c)) + (cos(4*a*d)*sin(8*a*d) - cos(8*a*d)*sin(4*a*d))*sin(4*b *d*log(c)))*cos(8*b*d*log(c)) + cos(4*b*d*log(c))*cos(4*a*d) - ((cos(4*a*d )*sin(8*a*d) - cos(8*a*d)*sin(4*a*d))*cos(4*b*d*log(c)) - (cos(8*a*d)*cos( 4*a*d) + sin(8*a*d)*sin(4*a*d))*sin(4*b*d*log(c)))*sin(8*b*d*log(c)) - sin (4*b*d*log(c))*sin(4*a*d))*e^m*m^3 + 6*(((cos(8*a*d)*cos(4*a*d) + sin(8*a* d)*sin(4*a*d))*cos(4*b*d*log(c)) + (cos(4*a*d)*sin(8*a*d) - cos(8*a*d)*sin (4*a*d))*sin(4*b*d*log(c)))*cos(8*b*d*log(c)) + cos(4*b*d*log(c))*cos(4*a* d) - ((cos(4*a*d)*sin(8*a*d) - cos(8*a*d)*sin(4*a*d))*cos(4*b*d*log(c)) - (cos(8*a*d)*cos(4*a*d) + sin(8*a*d)*sin(4*a*d))*sin(4*b*d*log(c)))*sin(8*b *d*log(c)) - sin(4*b*d*log(c))*sin(4*a*d))*e^m*m^2 + 16*((b^3*d^3*cos(4*a* d)*sin(4*b*d*log(c)) + b^3*d^3*cos(4*b*d*log(c))*sin(4*a*d) + ((b^3*d^3*co s(4*a*d)*sin(8*a*d) - b^3*d^3*cos(8*a*d)*sin(4*a*d))*cos(4*b*d*log(c)) - ( b^3*d^3*cos(8*a*d)*cos(4*a*d) + b^3*d^3*sin(8*a*d)*sin(4*a*d))*sin(4*b*d*l og(c)))*cos(8*b*d*log(c)) + ((b^3*d^3*cos(8*a*d)*cos(4*a*d) + b^3*d^3*s...
Leaf count of result is larger than twice the leaf count of optimal. 706991 vs. \(2 (337) = 674\).
Time = 18.84 (sec) , antiderivative size = 706991, normalized size of antiderivative = 2097.90 \[ \int (e x)^m \sin ^4\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\text {Too large to display} \] Input:
integrate((e*x)^m*sin(d*(a+b*log(c*x^n)))^4,x, algorithm="giac")
Output:
-1/16*(384*(abs(e)*abs(x))^m*b^4*d^4*n^4*x*tan(2*b*d*n*log(abs(x)) + 2*b*d *log(abs(c)))^2*tan(b*d*n*log(abs(x)) + b*d*log(abs(c)))^2*tan(pi*m*floor( -1/4*sgn(e) - 1/4*sgn(x) + 1) + 1/4*pi*m*sgn(e) + 1/4*pi*m*sgn(x) - 1/2*pi *m)^2*tan(1/4*pi*m*sgn(e) + 1/4*pi*m*sgn(x) - 1/2*pi*m)^2*tan(2*a*d)^2*tan (a*d)^2 + 384*(abs(e)*abs(x))^m*b^4*d^4*n^4*x*tan(2*b*d*n*log(abs(x)) + 2* b*d*log(abs(c)))^2*tan(b*d*n*log(abs(x)) + b*d*log(abs(c)))^2*tan(pi*m*flo or(-1/4*sgn(e) - 1/4*sgn(x) + 1) + 1/4*pi*m*sgn(e) + 1/4*pi*m*sgn(x) - 1/2 *pi*m)^2*tan(1/4*pi*m*sgn(e) + 1/4*pi*m*sgn(x) - 1/2*pi*m)^2*tan(2*a*d)^2 + 384*(abs(e)*abs(x))^m*b^4*d^4*n^4*x*tan(2*b*d*n*log(abs(x)) + 2*b*d*log( abs(c)))^2*tan(b*d*n*log(abs(x)) + b*d*log(abs(c)))^2*tan(pi*m*floor(-1/4* sgn(e) - 1/4*sgn(x) + 1) + 1/4*pi*m*sgn(e) + 1/4*pi*m*sgn(x) - 1/2*pi*m)^2 *tan(1/4*pi*m*sgn(e) + 1/4*pi*m*sgn(x) - 1/2*pi*m)^2*tan(a*d)^2 + 384*(abs (e)*abs(x))^m*b^4*d^4*n^4*x*tan(2*b*d*n*log(abs(x)) + 2*b*d*log(abs(c)))^2 *tan(b*d*n*log(abs(x)) + b*d*log(abs(c)))^2*tan(pi*m*floor(-1/4*sgn(e) - 1 /4*sgn(x) + 1) + 1/4*pi*m*sgn(e) + 1/4*pi*m*sgn(x) - 1/2*pi*m)^2*tan(2*a*d )^2*tan(a*d)^2 - 384*(abs(e)*abs(x))^m*b^4*d^4*n^4*x*tan(2*b*d*n*log(abs(x )) + 2*b*d*log(abs(c)))^2*tan(b*d*n*log(abs(x)) + b*d*log(abs(c)))^2*tan(1 /4*pi*m*sgn(e) + 1/4*pi*m*sgn(x) - 1/2*pi*m)^2*tan(2*a*d)^2*tan(a*d)^2 + 3 84*(abs(e)*abs(x))^m*b^4*d^4*n^4*x*tan(2*b*d*n*log(abs(x)) + 2*b*d*log(abs (c)))^2*tan(pi*m*floor(-1/4*sgn(e) - 1/4*sgn(x) + 1) + 1/4*pi*m*sgn(e) ...
Time = 22.58 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.52 \[ \int (e x)^m \sin ^4\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {3\,x\,{\left (e\,x\right )}^m}{8\,m+8}-\frac {x\,{\mathrm {e}}^{a\,d\,2{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,d\,2{}\mathrm {i}}\,{\left (e\,x\right )}^m}{4\,m+4+b\,d\,n\,8{}\mathrm {i}}-\frac {x\,{\mathrm {e}}^{-a\,d\,2{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{b\,d\,2{}\mathrm {i}}}\,{\left (e\,x\right )}^m\,1{}\mathrm {i}}{m\,4{}\mathrm {i}+8\,b\,d\,n+4{}\mathrm {i}}+\frac {x\,{\mathrm {e}}^{a\,d\,4{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,d\,4{}\mathrm {i}}\,{\left (e\,x\right )}^m}{16\,m+16+b\,d\,n\,64{}\mathrm {i}}+\frac {x\,{\mathrm {e}}^{-a\,d\,4{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{b\,d\,4{}\mathrm {i}}}\,{\left (e\,x\right )}^m\,1{}\mathrm {i}}{m\,16{}\mathrm {i}+64\,b\,d\,n+16{}\mathrm {i}} \] Input:
int(sin(d*(a + b*log(c*x^n)))^4*(e*x)^m,x)
Output:
(3*x*(e*x)^m)/(8*m + 8) - (x*exp(a*d*2i)*(c*x^n)^(b*d*2i)*(e*x)^m)/(4*m + b*d*n*8i + 4) - (x*exp(-a*d*2i)/(c*x^n)^(b*d*2i)*(e*x)^m*1i)/(m*4i + 8*b*d *n + 4i) + (x*exp(a*d*4i)*(c*x^n)^(b*d*4i)*(e*x)^m)/(16*m + b*d*n*64i + 16 ) + (x*exp(-a*d*4i)/(c*x^n)^(b*d*4i)*(e*x)^m*1i)/(m*16i + 64*b*d*n + 16i)
Time = 0.19 (sec) , antiderivative size = 695, normalized size of antiderivative = 2.06 \[ \int (e x)^m \sin ^4\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx =\text {Too large to display} \] Input:
int((e*x)^m*sin(d*(a+b*log(c*x^n)))^4,x)
Output:
(x**m*e**m*x*( - 16*cos(log(x**n*c)*b*d + a*d)*sin(log(x**n*c)*b*d + a*d)* *3*b**3*d**3*m*n**3 - 16*cos(log(x**n*c)*b*d + a*d)*sin(log(x**n*c)*b*d + a*d)**3*b**3*d**3*n**3 - 4*cos(log(x**n*c)*b*d + a*d)*sin(log(x**n*c)*b*d + a*d)**3*b*d*m**3*n - 12*cos(log(x**n*c)*b*d + a*d)*sin(log(x**n*c)*b*d + a*d)**3*b*d*m**2*n - 12*cos(log(x**n*c)*b*d + a*d)*sin(log(x**n*c)*b*d + a*d)**3*b*d*m*n - 4*cos(log(x**n*c)*b*d + a*d)*sin(log(x**n*c)*b*d + a*d)* *3*b*d*n - 24*cos(log(x**n*c)*b*d + a*d)*sin(log(x**n*c)*b*d + a*d)*b**3*d **3*m*n**3 - 24*cos(log(x**n*c)*b*d + a*d)*sin(log(x**n*c)*b*d + a*d)*b**3 *d**3*n**3 + 4*sin(log(x**n*c)*b*d + a*d)**4*b**2*d**2*m**2*n**2 + 8*sin(l og(x**n*c)*b*d + a*d)**4*b**2*d**2*m*n**2 + 4*sin(log(x**n*c)*b*d + a*d)** 4*b**2*d**2*n**2 + sin(log(x**n*c)*b*d + a*d)**4*m**4 + 4*sin(log(x**n*c)* b*d + a*d)**4*m**3 + 6*sin(log(x**n*c)*b*d + a*d)**4*m**2 + 4*sin(log(x**n *c)*b*d + a*d)**4*m + sin(log(x**n*c)*b*d + a*d)**4 + 12*sin(log(x**n*c)*b *d + a*d)**2*b**2*d**2*m**2*n**2 + 24*sin(log(x**n*c)*b*d + a*d)**2*b**2*d **2*m*n**2 + 12*sin(log(x**n*c)*b*d + a*d)**2*b**2*d**2*n**2 + 24*b**4*d** 4*n**4))/(64*b**4*d**4*m*n**4 + 64*b**4*d**4*n**4 + 20*b**2*d**2*m**3*n**2 + 60*b**2*d**2*m**2*n**2 + 60*b**2*d**2*m*n**2 + 20*b**2*d**2*n**2 + m**5 + 5*m**4 + 10*m**3 + 10*m**2 + 5*m + 1)