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*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))

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Rubi [A]  time = 0.16951, antiderivative size = 337, normalized size of antiderivative = 1., 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 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]

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]

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*}

Mathematica [A]  time = 1.90931, 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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Maple [F]  time = 0.115, size = 0, normalized size = 0. \begin{align*} \int \left ( ex \right ) ^{m} \left ( \sin \left ( d \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \right ) \right ) ^{4}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Fricas [A]  time = 0.590223, size = 1083, normalized size = 3.21 \begin{align*} \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} \end{align*}

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

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]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

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]

Exception raised: NotImplementedError