\(\int e^{\log ^2((d+e x)^n)} (d+e x)^m \, dx\) [586]

Optimal result

Integrand size = 20, antiderivative size = 76 \[ \int e^{\log ^2\left ((d+e x)^n\right )} (d+e x)^m \, dx=\frac {e^{-\frac {(1+m)^2}{4 n^2}} \sqrt {\pi } (d+e x)^{1+m} \left ((d+e x)^n\right )^{-\frac {1+m}{n}} \text {erfi}\left (\frac {1+m+2 n \log \left ((d+e x)^n\right )}{2 n}\right )}{2 e n} \]

[Out]

1/2*(e*x+d)^(1+m)*erfi(1/2*(1+m+2*n*ln((e*x+d)^n))/n)*Pi^(1/2)/e/exp(1/4*(1+m)^2/n^2)/n/(((e*x+d)^n)^((1+m)/n)
)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2308, 2266, 2235} \[ \int e^{\log ^2\left ((d+e x)^n\right )} (d+e x)^m \, dx=\frac {\sqrt {\pi } e^{-\frac {(m+1)^2}{4 n^2}} (d+e x)^{m+1} \left ((d+e x)^n\right )^{-\frac {m+1}{n}} \text {erfi}\left (\frac {2 n \log \left ((d+e x)^n\right )+m+1}{2 n}\right )}{2 e n} \]

[In]

Int[E^Log[(d + e*x)^n]^2*(d + e*x)^m,x]

[Out]

(Sqrt[Pi]*(d + e*x)^(1 + m)*Erfi[(1 + m + 2*n*Log[(d + e*x)^n])/(2*n)])/(2*e*E^((1 + m)^2/(4*n^2))*n*((d + e*x
)^n)^((1 + m)/n))

Rule 2235

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2
]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2266

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

Rule 2308

Int[(F_)^(((a_.) + Log[(c_.)*((d_.) + (e_.)*(x_))^(n_.)]^2*(b_.))*(f_.))*((g_.) + (h_.)*(x_))^(m_.), x_Symbol]
:> Dist[(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*Lo
g[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\left ((d+e x)^{1+m} \left ((d+e x)^n\right )^{-\frac {1+m}{n}}\right ) \text {Subst}\left (\int e^{\frac {(1+m) x}{n}+x^2} \, dx,x,\log \left ((d+e x)^n\right )\right )}{e n} \\ & = \frac {\left (e^{-\frac {(1+m)^2}{4 n^2}} (d+e x)^{1+m} \left ((d+e x)^n\right )^{-\frac {1+m}{n}}\right ) \text {Subst}\left (\int e^{\frac {1}{4} \left (\frac {1+m}{n}+2 x\right )^2} \, dx,x,\log \left ((d+e x)^n\right )\right )}{e n} \\ & = \frac {e^{-\frac {(1+m)^2}{4 n^2}} \sqrt {\pi } (d+e x)^{1+m} \left ((d+e x)^n\right )^{-\frac {1+m}{n}} \text {erfi}\left (\frac {1+m+2 n \log \left ((d+e x)^n\right )}{2 n}\right )}{2 e n} \\ \end{align*}

Mathematica [F]

\[ \int e^{\log ^2\left ((d+e x)^n\right )} (d+e x)^m \, dx=\int e^{\log ^2\left ((d+e x)^n\right )} (d+e x)^m \, dx \]

[In]

Integrate[E^Log[(d + e*x)^n]^2*(d + e*x)^m,x]

[Out]

Integrate[E^Log[(d + e*x)^n]^2*(d + e*x)^m, x]

Maple [F]

\[\int {\mathrm e}^{\ln \left (\left (e x +d \right )^{n}\right )^{2}} \left (e x +d \right )^{m}d x\]

[In]

int(exp(ln((e*x+d)^n)^2)*(e*x+d)^m,x)

[Out]

int(exp(ln((e*x+d)^n)^2)*(e*x+d)^m,x)

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.78 \[ \int e^{\log ^2\left ((d+e x)^n\right )} (d+e x)^m \, dx=-\frac {\sqrt {\pi } \sqrt {-n^{2}} \operatorname {erf}\left (\frac {{\left (2 \, n^{2} \log \left (e x + d\right ) + m + 1\right )} \sqrt {-n^{2}}}{2 \, n^{2}}\right ) e^{\left (-\frac {m^{2} + 2 \, m + 1}{4 \, n^{2}}\right )}}{2 \, e n} \]

[In]

integrate(exp(log((e*x+d)^n)^2)*(e*x+d)^m,x, algorithm="fricas")

[Out]

-1/2*sqrt(pi)*sqrt(-n^2)*erf(1/2*(2*n^2*log(e*x + d) + m + 1)*sqrt(-n^2)/n^2)*e^(-1/4*(m^2 + 2*m + 1)/n^2)/(e*
n)

Sympy [F]

\[ \int e^{\log ^2\left ((d+e x)^n\right )} (d+e x)^m \, dx=\int \left (d + e x\right )^{m} e^{\log {\left (\left (d + e x\right )^{n} \right )}^{2}}\, dx \]

[In]

integrate(exp(ln((e*x+d)**n)**2)*(e*x+d)**m,x)

[Out]

Integral((d + e*x)**m*exp(log((d + e*x)**n)**2), x)

Maxima [F]

\[ \int e^{\log ^2\left ((d+e x)^n\right )} (d+e x)^m \, dx=\int { {\left (e x + d\right )}^{m} e^{\left (\log \left ({\left (e x + d\right )}^{n}\right )^{2}\right )} \,d x } \]

[In]

integrate(exp(log((e*x+d)^n)^2)*(e*x+d)^m,x, algorithm="maxima")

[Out]

integrate((e*x + d)^m*e^(log((e*x + d)^n)^2), x)

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.31 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.71 \[ \int e^{\log ^2\left ((d+e x)^n\right )} (d+e x)^m \, dx=-\frac {i \, \sqrt {\pi } \operatorname {erf}\left (i \, n \log \left (e x + d\right ) + \frac {i \, m}{2 \, n} + \frac {i}{2 \, n}\right ) e^{\left (-\frac {m^{2}}{4 \, n^{2}} - \frac {m}{2 \, n^{2}} - \frac {1}{4 \, n^{2}}\right )}}{2 \, e n} \]

[In]

integrate(exp(log((e*x+d)^n)^2)*(e*x+d)^m,x, algorithm="giac")

[Out]

-1/2*I*sqrt(pi)*erf(I*n*log(e*x + d) + 1/2*I*m/n + 1/2*I/n)*e^(-1/4*m^2/n^2 - 1/2*m/n^2 - 1/4/n^2)/(e*n)

Mupad [F(-1)]

Timed out. \[ \int e^{\log ^2\left ((d+e x)^n\right )} (d+e x)^m \, dx=\int {\mathrm {e}}^{{\ln \left ({\left (d+e\,x\right )}^n\right )}^2}\,{\left (d+e\,x\right )}^m \,d x \]

[In]

int(exp(log((d + e*x)^n)^2)*(d + e*x)^m,x)

[Out]

int(exp(log((d + e*x)^n)^2)*(d + e*x)^m, x)