\(\int (e x)^m \text {erf}(d (a+b \log (c x^n))) \, dx\) [46]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [F]
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 19, antiderivative size = 125 \[ \int (e x)^m \text {erf}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {(e x)^{1+m} \text {erf}\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e (1+m)}+\frac {e^{\frac {(1+m) \left (1+m-4 a b d^2 n\right )}{4 b^2 d^2 n^2}} x (e x)^m \left (c x^n\right )^{-\frac {1+m}{n}} \text {erf}\left (\frac {1+m-2 a b d^2 n-2 b^2 d^2 n \log \left (c x^n\right )}{2 b d n}\right )}{1+m} \] Output:

(e*x)^(1+m)*erf(d*(a+b*ln(c*x^n)))/e/(1+m)+exp(1/4*(1+m)*(-4*a*b*d^2*n+m+1 
)/b^2/d^2/n^2)*x*(e*x)^m*erf(1/2*(1+m-2*a*b*d^2*n-2*b^2*d^2*n*ln(c*x^n))/b 
/d/n)/(1+m)/((c*x^n)^((1+m)/n))
 

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.02 \[ \int (e x)^m \text {erf}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {(e x)^m \left (x \text {erf}\left (d \left (a+b \log \left (c x^n\right )\right )\right )-e^{\frac {(1+m) \left (1+m-4 a b d^2 n+4 b^2 d^2 n^2 \log (x)-4 b^2 d^2 n \log \left (c x^n\right )\right )}{4 b^2 d^2 n^2}} x^{-m} \text {erf}\left (a d-\frac {1+m-2 b^2 d^2 n \log \left (c x^n\right )}{2 b d n}\right )\right )}{1+m} \] Input:

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

Output:

((e*x)^m*(x*Erf[d*(a + b*Log[c*x^n])] - (E^(((1 + m)*(1 + m - 4*a*b*d^2*n 
+ 4*b^2*d^2*n^2*Log[x] - 4*b^2*d^2*n*Log[c*x^n]))/(4*b^2*d^2*n^2))*Erf[a*d 
 - (1 + m - 2*b^2*d^2*n*Log[c*x^n])/(2*b*d*n)])/x^m))/(1 + m)
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.13, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {6955, 2712, 2706, 2664, 2634}

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.

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

\(\Big \downarrow \) 6955

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

\(\Big \downarrow \) 2712

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

\(\Big \downarrow \) 2706

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

\(\Big \downarrow \) 2664

\(\displaystyle \frac {(e x)^{m+1} \text {erf}\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e (m+1)}-\frac {2 b d x (e x)^m \exp \left (\frac {(m+1) \left (-4 a b d^2 n+m+1\right )}{4 b^2 d^2 n^2}\right ) \left (c x^n\right )^{-\frac {-2 a b d^2 n+m+1}{n}-2 a b d^2} \int \exp \left (-\frac {\left (-2 a b n d^2-2 b^2 n \log \left (c x^n\right ) d^2+m+1\right )^2}{4 b^2 d^2 n^2}\right )d\log \left (c x^n\right )}{\sqrt {\pi } (m+1)}\)

\(\Big \downarrow \) 2634

\(\displaystyle \frac {x (e x)^m \exp \left (\frac {(m+1) \left (-4 a b d^2 n+m+1\right )}{4 b^2 d^2 n^2}\right ) \left (c x^n\right )^{-\frac {-2 a b d^2 n+m+1}{n}-2 a b d^2} \text {erf}\left (\frac {-2 a b d^2 n-2 b^2 d^2 n \log \left (c x^n\right )+m+1}{2 b d n}\right )}{m+1}+\frac {(e x)^{m+1} \text {erf}\left (d \left (a+b \log \left (c x^n\right )\right )\right )}{e (m+1)}\)

Input:

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

Output:

((e*x)^(1 + m)*Erf[d*(a + b*Log[c*x^n])])/(e*(1 + m)) + (E^(((1 + m)*(1 + 
m - 4*a*b*d^2*n))/(4*b^2*d^2*n^2))*x*(e*x)^m*(c*x^n)^(-2*a*b*d^2 - (1 + m 
- 2*a*b*d^2*n)/n)*Erf[(1 + m - 2*a*b*d^2*n - 2*b^2*d^2*n*Log[c*x^n])/(2*b* 
d*n)])/(1 + m)
 

Defintions of rubi rules used

rule 2634
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt 
[Pi]*(Erf[(c + d*x)*Rt[(-b)*Log[F], 2]]/(2*d*Rt[(-b)*Log[F], 2])), x] /; Fr 
eeQ[{F, a, b, c, d}, x] && NegQ[b]
 

rule 2664
Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[F^(a - b^2/ 
(4*c))   Int[F^((b + 2*c*x)^2/(4*c)), x], x] /; FreeQ[{F, a, b, c}, x]
 

rule 2706
Int[(F_)^(((a_.) + Log[(c_.)*((d_.) + (e_.)*(x_))^(n_.)]^2*(b_.))*(f_.))*(( 
g_.) + (h_.)*(x_))^(m_.), x_Symbol] :> Simp[(g + h*x)^(m + 1)/(h*n*(c*(d + 
e*x)^n)^((m + 1)/n))   Subst[Int[E^(a*f*Log[F] + ((m + 1)*x)/n + b*f*Log[F] 
*x^2), x], x, Log[c*(d + e*x)^n]], x] /; FreeQ[{F, a, b, c, d, e, f, g, h, 
m, n}, x] && EqQ[e*g - d*h, 0]
 

rule 2712
Int[(F_)^(((a_.) + Log[(c_.)*((d_.) + (e_.)*(x_))^(n_.)]*(b_.))^2*(f_.))*(( 
g_.) + (h_.)*(x_))^(m_.), x_Symbol] :> Simp[(g + h*x)^m*((c*(d + e*x)^n)^(2 
*a*b*f*Log[F])/(d + e*x)^(m + 2*a*b*f*n*Log[F]))*Int[(d + e*x)^(m + 2*a*b*f 
*n*Log[F])*F^(a^2*f + b^2*f*Log[c*(d + e*x)^n]^2), x], x] /; FreeQ[{F, a, b 
, c, d, e, f, g, h, m, n}, x] && EqQ[e*g - d*h, 0]
 

rule 6955
Int[Erf[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]*((e_.)*(x_))^(m_.), x_ 
Symbol] :> Simp[(e*x)^(m + 1)*(Erf[d*(a + b*Log[c*x^n])]/(e*(m + 1))), x] - 
 Simp[2*b*d*(n/(Sqrt[Pi]*(m + 1)))   Int[(e*x)^m/E^(d*(a + b*Log[c*x^n]))^2 
, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[m, -1]
 
Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.44 \[ \int (e x)^m \text {erf}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {x \operatorname {erf}\left (b d \log \left (c x^{n}\right ) + a d\right ) e^{\left (m \log \left (e\right ) + m \log \left (x\right )\right )} - \sqrt {b^{2} d^{2} n^{2}} \operatorname {erf}\left (\frac {{\left (2 \, b^{2} d^{2} n^{2} \log \left (x\right ) + 2 \, b^{2} d^{2} n \log \left (c\right ) + 2 \, a b d^{2} n - m - 1\right )} \sqrt {b^{2} d^{2} n^{2}}}{2 \, b^{2} d^{2} n^{2}}\right ) e^{\left (\frac {4 \, b^{2} d^{2} m n^{2} \log \left (e\right ) - 4 \, {\left (b^{2} d^{2} m + b^{2} d^{2}\right )} n \log \left (c\right ) + m^{2} - 4 \, {\left (a b d^{2} m + a b d^{2}\right )} n + 2 \, m + 1}{4 \, b^{2} d^{2} n^{2}}\right )}}{m + 1} \] Input:

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

Output:

(x*erf(b*d*log(c*x^n) + a*d)*e^(m*log(e) + m*log(x)) - sqrt(b^2*d^2*n^2)*e 
rf(1/2*(2*b^2*d^2*n^2*log(x) + 2*b^2*d^2*n*log(c) + 2*a*b*d^2*n - m - 1)*s 
qrt(b^2*d^2*n^2)/(b^2*d^2*n^2))*e^(1/4*(4*b^2*d^2*m*n^2*log(e) - 4*(b^2*d^ 
2*m + b^2*d^2)*n*log(c) + m^2 - 4*(a*b*d^2*m + a*b*d^2)*n + 2*m + 1)/(b^2* 
d^2*n^2)))/(m + 1)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

e^m*x*x^m*erf(b*d*log(x^n) + (b*log(c) + a)*d)/(m + 1) - 2*b*d*e^m*n*integ 
rate(e^(-b^2*d^2*log(c)^2 - 2*b^2*d^2*log(c)*log(x^n) - b^2*d^2*log(x^n)^2 
 - 2*a*b*d^2*log(x^n) - a^2*d^2 + m*log(x)), x)/(sqrt(pi)*c^(2*a*b*d^2)*(m 
 + 1))
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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