\(\int (e x)^m \text {Si}(d (a+b \log (c x^n))) \, dx\) [38]

Optimal result

Integrand size = 19, antiderivative size = 176 \[ \int (e x)^m \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=-\frac {i e^{-\frac {a (1+m)}{b n}} x (e x)^m \left (c x^n\right )^{-\frac {1+m}{n}} \operatorname {ExpIntegralEi}\left (\frac {(1+m-i b d n) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )}{2 (1+m)}+\frac {i e^{-\frac {a (1+m)}{b n}} x (e x)^m \left (c x^n\right )^{-\frac {1+m}{n}} \operatorname {ExpIntegralEi}\left (\frac {(1+m+i b d n) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )}{2 (1+m)}+\frac {(e x)^{1+m} \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e (1+m)} \] Output:

-1/2*I*x*(e*x)^m*Ei((1+m-I*b*d*n)*(a+b*ln(c*x^n))/b/n)/exp(a*(1+m)/b/n)/(1 
+m)/((c*x^n)^((1+m)/n))+1/2*I*x*(e*x)^m*Ei((1+m+I*b*d*n)*(a+b*ln(c*x^n))/b 
/n)/exp(a*(1+m)/b/n)/(1+m)/((c*x^n)^((1+m)/n))+(e*x)^(1+m)*Si(d*(a+b*ln(c* 
x^n)))/e/(1+m)
 

Mathematica [A] (verified)

Time = 1.99 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.73 \[ \int (e x)^m \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {(e x)^m \left (-i e^{-\frac {(1+m) \left (a-b n \log (x)+b \log \left (c x^n\right )\right )}{b n}} x^{-m} \left (\operatorname {ExpIntegralEi}\left (\frac {(1+m-i b d n) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )-\operatorname {ExpIntegralEi}\left (\frac {(1+m+i b d n) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )\right )+2 x \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right )\right )}{2 (1+m)} \] Input:

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

Output:

((e*x)^m*(((-I)*(ExpIntegralEi[((1 + m - I*b*d*n)*(a + b*Log[c*x^n]))/(b*n 
)] - ExpIntegralEi[((1 + m + I*b*d*n)*(a + b*Log[c*x^n]))/(b*n)]))/(E^(((1 
 + m)*(a - b*n*Log[x] + b*Log[c*x^n]))/(b*n))*x^m) + 2*x*SinIntegral[d*(a 
+ b*Log[c*x^n])]))/(2*(1 + m))
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.39, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {7080, 27, 5000, 2747, 2609}

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.

\(\displaystyle \int (e x)^m \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx\)

\(\Big \downarrow \) 7080

\(\displaystyle \frac {(e x)^{m+1} \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e (m+1)}-\frac {b d n \int \frac {(e x)^m \sin \left (d \left (a+b \log \left (c x^n\right )\right )\right )}{d \left (a+b \log \left (c x^n\right )\right )}dx}{m+1}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(e x)^{m+1} \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e (m+1)}-\frac {b n \int \frac {(e x)^m \sin \left (d \left (a+b \log \left (c x^n\right )\right )\right )}{a+b \log \left (c x^n\right )}dx}{m+1}\)

\(\Big \downarrow \) 5000

\(\displaystyle \frac {(e x)^{m+1} \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e (m+1)}-\frac {b n \left (\frac {1}{2} i e^{-i a d} (e x)^m \left (c x^n\right )^{-i b d} x^{-m+i b d n} \int \frac {x^{m-i b d n}}{a+b \log \left (c x^n\right )}dx-\frac {1}{2} i e^{i a d} (e x)^m \left (c x^n\right )^{i b d} x^{-m-i b d n} \int \frac {x^{m+i b d n}}{a+b \log \left (c x^n\right )}dx\right )}{m+1}\)

\(\Big \downarrow \) 2747

\(\displaystyle \frac {(e x)^{m+1} \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e (m+1)}-\frac {b n \left (\frac {i x e^{-i a d} (e x)^m \left (c x^n\right )^{-\frac {-i b d n+m+1}{n}-i b d} \int \frac {\left (c x^n\right )^{\frac {m-i b d n+1}{n}}}{a+b \log \left (c x^n\right )}d\log \left (c x^n\right )}{2 n}-\frac {i x e^{i a d} (e x)^m \left (c x^n\right )^{i b d-\frac {i b d n+m+1}{n}} \int \frac {\left (c x^n\right )^{\frac {m+i b d n+1}{n}}}{a+b \log \left (c x^n\right )}d\log \left (c x^n\right )}{2 n}\right )}{m+1}\)

\(\Big \downarrow \) 2609

\(\displaystyle \frac {(e x)^{m+1} \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e (m+1)}-\frac {b n \left (\frac {i x (e x)^m e^{-\frac {a (-i b d n+m+1)}{b n}-i a d} \left (c x^n\right )^{-\frac {-i b d n+m+1}{n}-i b d} \operatorname {ExpIntegralEi}\left (\frac {(m-i b d n+1) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )}{2 b n}-\frac {i x (e x)^m e^{i a d-\frac {a (i b d n+m+1)}{b n}} \left (c x^n\right )^{i b d-\frac {i b d n+m+1}{n}} \operatorname {ExpIntegralEi}\left (\frac {(m+i b d n+1) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )}{2 b n}\right )}{m+1}\)

Input:

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

Output:

-((b*n*(((I/2)*E^((-I)*a*d - (a*(1 + m - I*b*d*n))/(b*n))*x*(e*x)^m*(c*x^n 
)^((-I)*b*d - (1 + m - I*b*d*n)/n)*ExpIntegralEi[((1 + m - I*b*d*n)*(a + b 
*Log[c*x^n]))/(b*n)])/(b*n) - ((I/2)*E^(I*a*d - (a*(1 + m + I*b*d*n))/(b*n 
))*x*(e*x)^m*(c*x^n)^(I*b*d - (1 + m + I*b*d*n)/n)*ExpIntegralEi[((1 + m + 
 I*b*d*n)*(a + b*Log[c*x^n]))/(b*n)])/(b*n)))/(1 + m)) + ((e*x)^(1 + m)*Si 
nIntegral[d*(a + b*Log[c*x^n])])/(e*(1 + m))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2609

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Si 
mp[(F^(g*(e - c*(f/d)))/d)*ExpIntegralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; F 
reeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]
 

rule 2747

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol 
] :> Simp[(d*x)^(m + 1)/(d*n*(c*x^n)^((m + 1)/n))   Subst[Int[E^(((m + 1)/n 
)*x)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, m, n, p}, x]
 

rule 5000

Int[(((e_.) + Log[(g_.)*(x_)^(m_.)]*(f_.))*(h_.))^(q_.)*((i_.)*(x_))^(r_.)* 
Sin[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)], x_Symbol] :> Simp[(I*(i*x 
)^r*(1/((c*x^n)^(I*b*d)*(2*x^(r - I*b*d*n)))))/E^(I*a*d)   Int[x^(r - I*b*d 
*n)*(h*(e + f*Log[g*x^m]))^q, x], x] - Simp[I*E^(I*a*d)*(i*x)^r*((c*x^n)^(I 
*b*d)/(2*x^(r + I*b*d*n)))   Int[x^(r + I*b*d*n)*(h*(e + f*Log[g*x^m]))^q, 
x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, m, n, q, r}, x]
 

rule 7080

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

Maple [F]

\[\int \left (e x \right )^{m} \operatorname {Si}\left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )d x\]

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.02 \[ \int (e x)^m \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {2 \, x e^{\left (m \log \left (e\right ) + m \log \left (x\right )\right )} \operatorname {Si}\left (b d \log \left (c x^{n}\right ) + a d\right ) + {\left (i \, {\rm Ei}\left (\frac {i \, a b d n + a m + {\left (i \, b^{2} d n + b m + b\right )} \log \left (c\right ) + {\left (i \, b^{2} d n^{2} + {\left (b m + b\right )} n\right )} \log \left (x\right ) + a}{b n}\right ) - i \, {\rm Ei}\left (\frac {-i \, a b d n + a m + {\left (-i \, b^{2} d n + b m + b\right )} \log \left (c\right ) + {\left (-i \, b^{2} d n^{2} + {\left (b m + b\right )} n\right )} \log \left (x\right ) + a}{b n}\right )\right )} e^{\left (\frac {b m n \log \left (e\right ) - a m - {\left (b m + b\right )} \log \left (c\right ) - a}{b n}\right )}}{2 \, {\left (m + 1\right )}} \] Input:

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

Output:

1/2*(2*x*e^(m*log(e) + m*log(x))*sin_integral(b*d*log(c*x^n) + a*d) + (I*E 
i((I*a*b*d*n + a*m + (I*b^2*d*n + b*m + b)*log(c) + (I*b^2*d*n^2 + (b*m + 
b)*n)*log(x) + a)/(b*n)) - I*Ei((-I*a*b*d*n + a*m + (-I*b^2*d*n + b*m + b) 
*log(c) + (-I*b^2*d*n^2 + (b*m + b)*n)*log(x) + a)/(b*n)))*e^((b*m*n*log(e 
) - a*m - (b*m + b)*log(c) - a)/(b*n)))/(m + 1)
 

Sympy [F]

\[ \int (e x)^m \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int \left (e x\right )^{m} \operatorname {Si}{\left (a d + b d \log {\left (c x^{n} \right )} \right )}\, dx \] Input:

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

Output:

Integral((e*x)**m*Si(a*d + b*d*log(c*x**n)), x)
 

Maxima [F]

\[ \int (e x)^m \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { \left (e x\right )^{m} \operatorname {Si}\left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right ) \,d x } \] Input:

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

Output:

integrate((e*x)^m*sin_integral((b*log(c*x^n) + a)*d), x)
 

Giac [F(-1)]

Timed out. \[ \int (e x)^m \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int (e x)^m \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int \mathrm {sinint}\left (d\,\left (a+b\,\ln \left (c\,x^n\right )\right )\right )\,{\left (e\,x\right )}^m \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int (e x)^m \text {Si}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=e^{m} \left (\int x^{m} \mathit {si} \left (\mathrm {log}\left (x^{n} c \right ) b d +a d \right )d x \right ) \] Input:

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

Output:

e**m*int(x**m*si(log(x**n*c)*b*d + a*d),x)