3.70 \(\int (e x)^m \sin ^4(d (a+b \log (c x^n))) \, dx\)

Optimal. Leaf size=337 \[ \frac {(m+1) (e x)^{m+1} \sin ^4\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e \left (16 b^2 d^2 n^2+(m+1)^2\right )}+\frac {12 b^2 d^2 (m+1) n^2 (e x)^{m+1} \sin ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e \left (4 b^2 d^2 n^2+(m+1)^2\right ) \left (16 b^2 d^2 n^2+(m+1)^2\right )}-\frac {4 b d n (e x)^{m+1} \sin ^3\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \cos \left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e \left (16 b^2 d^2 n^2+(m+1)^2\right )}-\frac {24 b^3 d^3 n^3 (e x)^{m+1} \sin \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \cos \left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e \left (4 b^2 d^2 n^2+(m+1)^2\right ) \left (16 b^2 d^2 n^2+(m+1)^2\right )}+\frac {24 b^4 d^4 n^4 (e x)^{m+1}}{e (m+1) \left (4 b^2 d^2 n^2+(m+1)^2\right ) \left (16 b^2 d^2 n^2+(m+1)^2\right )} \]

[Out]

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)

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Rubi [A]  time = 0.17, antiderivative size = 337, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {4487, 32} \[ \frac {(m+1) (e x)^{m+1} \sin ^4\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e \left (16 b^2 d^2 n^2+(m+1)^2\right )}+\frac {12 b^2 d^2 (m+1) n^2 (e x)^{m+1} \sin ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e \left (4 b^2 d^2 n^2+(m+1)^2\right ) \left (16 b^2 d^2 n^2+(m+1)^2\right )}-\frac {4 b d n (e x)^{m+1} \sin ^3\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \cos \left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e \left (16 b^2 d^2 n^2+(m+1)^2\right )}-\frac {24 b^3 d^3 n^3 (e x)^{m+1} \sin \left (d \left (a+b \log \left (c x^n\right )\right )\right ) \cos \left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e \left (4 b^2 d^2 n^2+(m+1)^2\right ) \left (16 b^2 d^2 n^2+(m+1)^2\right )}+\frac {24 b^4 d^4 n^4 (e x)^{m+1}}{e (m+1) \left (4 b^2 d^2 n^2+(m+1)^2\right ) \left (16 b^2 d^2 n^2+(m+1)^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(e*x)^m*Sin[d*(a + b*Log[c*x^n])]^4,x]

[Out]

(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*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)^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*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)) - (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))

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 4487

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 (e x)^m \sin ^4\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx &=-\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 )}+\frac {\left (12 b^2 d^2 n^2\right ) \int (e x)^m \sin ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx}{(1+m)^2+16 b^2 d^2 n^2}\\ &=-\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 )}+\frac {\left (24 b^4 d^4 n^4\right ) \int (e x)^m \, dx}{(1+m)^4+20 b^2 d^2 (1+m)^2 n^2+64 b^4 d^4 n^4}\\ &=\frac {24 b^4 d^4 n^4 (e x)^{1+m}}{e (1+m) \left ((1+m)^4+20 b^2 d^2 (1+m)^2 n^2+64 b^4 d^4 n^4\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 )}\\ \end {align*}

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Mathematica [A]  time = 2.00, size = 341, normalized size = 1.01 \[ \frac {1}{8} x (e x)^m \left (\frac {4 \sin (2 b d n \log (x)) \left ((m+1) \sin \left (2 d \left (a+b \log \left (c x^n\right )-b n \log (x)\right )\right )-2 b d n \cos \left (2 d \left (a+b \log \left (c x^n\right )-b n \log (x)\right )\right )\right )}{4 b^2 d^2 n^2+m^2+2 m+1}-\frac {4 \cos (2 b d n \log (x)) \left ((m+1) \cos \left (2 d \left (a+b \log \left (c x^n\right )-b n \log (x)\right )\right )+2 b d n \sin \left (2 d \left (a+b \log \left (c x^n\right )-b n \log (x)\right )\right )\right )}{4 b^2 d^2 n^2+m^2+2 m+1}-\frac {\sin (4 b d n \log (x)) \left ((m+1) \sin \left (4 d \left (a+b \log \left (c x^n\right )-b n \log (x)\right )\right )-4 b d n \cos \left (4 d \left (a+b \log \left (c x^n\right )-b n \log (x)\right )\right )\right )}{16 b^2 d^2 n^2+m^2+2 m+1}+\frac {\cos (4 b d n \log (x)) \left ((m+1) \cos \left (4 d \left (a+b \log \left (c x^n\right )-b n \log (x)\right )\right )+4 b d n \sin \left (4 d \left (a+b \log \left (c x^n\right )-b n \log (x)\right )\right )\right )}{16 b^2 d^2 n^2+m^2+2 m+1}+\frac {3}{m+1}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(e*x)^m*Sin[d*(a + b*Log[c*x^n])]^4,x]

[Out]

(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)*Si
n[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

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fricas [A]  time = 0.57, size = 467, normalized size = 1.39 \[ \frac {4 \, {\left ({\left (4 \, {\left (b^{3} d^{3} m + b^{3} d^{3}\right )} n^{3} + {\left (b d m^{3} + 3 \, b d m^{2} + 3 \, b d m + b d\right )} n\right )} x \cos \left (b d n \log \relax (x) + b d \log \relax (c) + a d\right )^{3} - {\left (10 \, {\left (b^{3} d^{3} m + b^{3} d^{3}\right )} n^{3} + {\left (b d m^{3} + 3 \, b d m^{2} + 3 \, b d m + b d\right )} n\right )} x \cos \left (b d n \log \relax (x) + b d \log \relax (c) + a d\right )\right )} e^{\left (m \log \relax (e) + m \log \relax (x)\right )} \sin \left (b d n \log \relax (x) + b d \log \relax (c) + a d\right ) + {\left ({\left (m^{4} + 4 \, m^{3} + 4 \, {\left (b^{2} d^{2} m^{2} + 2 \, b^{2} d^{2} m + b^{2} d^{2}\right )} n^{2} + 6 \, m^{2} + 4 \, m + 1\right )} x \cos \left (b d n \log \relax (x) + b d \log \relax (c) + a d\right )^{4} - 2 \, {\left (m^{4} + 4 \, m^{3} + 10 \, {\left (b^{2} d^{2} m^{2} + 2 \, b^{2} d^{2} m + b^{2} d^{2}\right )} n^{2} + 6 \, m^{2} + 4 \, m + 1\right )} x \cos \left (b d n \log \relax (x) + b d \log \relax (c) + a d\right )^{2} + {\left (24 \, b^{4} d^{4} n^{4} + m^{4} + 4 \, m^{3} + 16 \, {\left (b^{2} d^{2} m^{2} + 2 \, b^{2} d^{2} m + b^{2} d^{2}\right )} n^{2} + 6 \, m^{2} + 4 \, m + 1\right )} x\right )} e^{\left (m \log \relax (e) + m \log \relax (x)\right )}}{m^{5} + 64 \, {\left (b^{4} d^{4} m + b^{4} d^{4}\right )} n^{4} + 5 \, m^{4} + 10 \, m^{3} + 20 \, {\left (b^{2} d^{2} m^{3} + 3 \, b^{2} d^{2} m^{2} + 3 \, b^{2} d^{2} m + b^{2} d^{2}\right )} n^{2} + 10 \, m^{2} + 5 \, m + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(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)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*sin(d*(a+b*log(c*x^n)))^4,x, algorithm="giac")

[Out]

Timed out

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maple [F]  time = 0.16, size = 0, normalized size = 0.00 \[ \int \left (e x \right )^{m} \left (\sin ^{4}\left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*sin(d*(a+b*log(c*x^n)))^4,x, algorithm="maxima")

[Out]

Timed out

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mupad [B]  time = 4.04, size = 175, normalized size = 0.52 \[ \frac {3\,x\,{\left (e\,x\right )}^m}{8\,m+8}-\frac {x\,{\mathrm {e}}^{a\,d\,2{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,d\,2{}\mathrm {i}}\,{\left (e\,x\right )}^m}{4\,m+4+b\,d\,n\,8{}\mathrm {i}}-\frac {x\,{\mathrm {e}}^{-a\,d\,2{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{b\,d\,2{}\mathrm {i}}}\,{\left (e\,x\right )}^m\,1{}\mathrm {i}}{m\,4{}\mathrm {i}+8\,b\,d\,n+4{}\mathrm {i}}+\frac {x\,{\mathrm {e}}^{a\,d\,4{}\mathrm {i}}\,{\left (c\,x^n\right )}^{b\,d\,4{}\mathrm {i}}\,{\left (e\,x\right )}^m}{16\,m+16+b\,d\,n\,64{}\mathrm {i}}+\frac {x\,{\mathrm {e}}^{-a\,d\,4{}\mathrm {i}}\,\frac {1}{{\left (c\,x^n\right )}^{b\,d\,4{}\mathrm {i}}}\,{\left (e\,x\right )}^m\,1{}\mathrm {i}}{m\,16{}\mathrm {i}+64\,b\,d\,n+16{}\mathrm {i}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*(a + b*log(c*x^n)))^4*(e*x)^m,x)

[Out]

(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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\begin {cases} \frac {\log {\relax (x )} \cos {\left (2 a d \right )}}{e} & \text {for}\: b = 0 \wedge m = -1 \\\int \left (e x\right )^{m} \cos {\left (- 2 a d + \frac {i m \log {\left (c x^{n} \right )}}{n} + \frac {i \log {\left (c x^{n} \right )}}{n} \right )}\, dx & \text {for}\: b = - \frac {i \left (m + 1\right )}{2 d n} \\\int \left (e x\right )^{m} \cos {\left (2 a d + \frac {i m \log {\left (c x^{n} \right )}}{n} + \frac {i \log {\left (c x^{n} \right )}}{n} \right )}\, dx & \text {for}\: b = \frac {i \left (m + 1\right )}{2 d n} \\\frac {2 b d e^{m} n x x^{m} \sin {\left (2 a d + 2 b d n \log {\relax (x )} + 2 b d \log {\relax (c )} \right )}}{4 b^{2} d^{2} n^{2} + m^{2} + 2 m + 1} + \frac {e^{m} m x x^{m} \cos {\left (2 a d + 2 b d n \log {\relax (x )} + 2 b d \log {\relax (c )} \right )}}{4 b^{2} d^{2} n^{2} + m^{2} + 2 m + 1} + \frac {e^{m} x x^{m} \cos {\left (2 a d + 2 b d n \log {\relax (x )} + 2 b d \log {\relax (c )} \right )}}{4 b^{2} d^{2} n^{2} + m^{2} + 2 m + 1} & \text {otherwise} \end {cases}}{2} + \frac {\begin {cases} \frac {\log {\relax (x )} \cos {\left (4 a d \right )}}{e} & \text {for}\: b = 0 \wedge m = -1 \\\int \left (e x\right )^{m} \cos {\left (- 4 a d + \frac {i m \log {\left (c x^{n} \right )}}{n} + \frac {i \log {\left (c x^{n} \right )}}{n} \right )}\, dx & \text {for}\: b = - \frac {i \left (m + 1\right )}{4 d n} \\\int \left (e x\right )^{m} \cos {\left (4 a d + \frac {i m \log {\left (c x^{n} \right )}}{n} + \frac {i \log {\left (c x^{n} \right )}}{n} \right )}\, dx & \text {for}\: b = \frac {i \left (m + 1\right )}{4 d n} \\\frac {4 b d e^{m} n x x^{m} \sin {\left (4 a d + 4 b d n \log {\relax (x )} + 4 b d \log {\relax (c )} \right )}}{16 b^{2} d^{2} n^{2} + m^{2} + 2 m + 1} + \frac {e^{m} m x x^{m} \cos {\left (4 a d + 4 b d n \log {\relax (x )} + 4 b d \log {\relax (c )} \right )}}{16 b^{2} d^{2} n^{2} + m^{2} + 2 m + 1} + \frac {e^{m} x x^{m} \cos {\left (4 a d + 4 b d n \log {\relax (x )} + 4 b d \log {\relax (c )} \right )}}{16 b^{2} d^{2} n^{2} + m^{2} + 2 m + 1} & \text {otherwise} \end {cases}}{8} + \frac {3 \left (\begin {cases} \frac {\left (e x\right )^{m + 1}}{m + 1} & \text {for}\: m \neq -1 \\\log {\left (e x \right )} & \text {otherwise} \end {cases}\right )}{8 e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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*e**m*n*x*x**m*sin(2*a*d + 2*b*d*n*log(x) + 2*b*d*log(c))/(4*b**2*
d**2*n**2 + m**2 + 2*m + 1) + e**m*m*x*x**m*cos(2*a*d + 2*b*d*n*log(x) + 2*b*d*log(c))/(4*b**2*d**2*n**2 + m**
2 + 2*m + 1) + e**m*x*x**m*cos(2*a*d + 2*b*d*n*log(x) + 2*b*d*log(c))/(4*b**2*d**2*n**2 + m**2 + 2*m + 1), Tru
e))/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*lo
g(c*x**n)/n), x), Eq(b, I*(m + 1)/(4*d*n))), (4*b*d*e**m*n*x*x**m*sin(4*a*d + 4*b*d*n*log(x) + 4*b*d*log(c))/(
16*b**2*d**2*n**2 + m**2 + 2*m + 1) + e**m*m*x*x**m*cos(4*a*d + 4*b*d*n*log(x) + 4*b*d*log(c))/(16*b**2*d**2*n
**2 + m**2 + 2*m + 1) + e**m*x*x**m*cos(4*a*d + 4*b*d*n*log(x) + 4*b*d*log(c))/(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)

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