\(\int \text {erfc}(d (a+b \log (c x^n))) \, dx\) [145]

Optimal result

Integrand size = 13, antiderivative size = 92 \[ \int \text {erfc}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=e^{\frac {1-4 a b d^2 n}{4 b^2 d^2 n^2}} x \left (c x^n\right )^{-1/n} \text {erf}\left (\frac {2 a b d^2-\frac {1}{n}+2 b^2 d^2 \log \left (c x^n\right )}{2 b d}\right )+x \text {erfc}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {6533, 2312, 2308, 2266, 2236} \[ \int \text {erfc}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=x \left (c x^n\right )^{-1/n} e^{\frac {1-4 a b d^2 n}{4 b^2 d^2 n^2}} \text {erf}\left (\frac {2 a b d^2+2 b^2 d^2 \log \left (c x^n\right )-\frac {1}{n}}{2 b d}\right )+x \text {erfc}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \]

[In]

Int[Erfc[d*(a + b*Log[c*x^n])],x]

[Out]

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

Rule 2236

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] /; FreeQ[{F, a, b, c, d}, x] && NegQ[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]

Rule 2312

Int[(F_)^(((a_.) + Log[(c_.)*((d_.) + (e_.)*(x_))^(n_.)]*(b_.))^2*(f_.)), x_Symbol] :> Dist[(c*(d + e*x)^n)^(2
*a*b*f*Log[F])/(d + e*x)^(2*a*b*f*n*Log[F]), Int[(d + e*x)^(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, n}, x] &&  !IntegerQ[2*a*b*f*Log[F]]

Rule 6533

Int[Erfc[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)], x_Symbol] :> Simp[x*Erfc[d*(a + b*Log[c*x^n])], x] + Di
st[2*b*d*(n/Sqrt[Pi]), Int[1/E^(d*(a + b*Log[c*x^n]))^2, x], x] /; FreeQ[{a, b, c, d, n}, x]

Rubi steps \begin{align*} \text {integral}& = x \text {erfc}\left (d \left (a+b \log \left (c x^n\right )\right )\right )+\frac {(2 b d n) \int e^{-d^2 \left (a+b \log \left (c x^n\right )\right )^2} \, dx}{\sqrt {\pi }} \\ & = x \text {erfc}\left (d \left (a+b \log \left (c x^n\right )\right )\right )+\frac {\left (2 b d n x^{2 a b d^2 n} \left (c x^n\right )^{-2 a b d^2}\right ) \int e^{-a^2 d^2-b^2 d^2 \log ^2\left (c x^n\right )} x^{-2 a b d^2 n} \, dx}{\sqrt {\pi }} \\ & = x \text {erfc}\left (d \left (a+b \log \left (c x^n\right )\right )\right )+\frac {\left (2 b d x \left (c x^n\right )^{-2 a b d^2-\frac {1-2 a b d^2 n}{n}}\right ) \text {Subst}\left (\int \exp \left (-a^2 d^2+\frac {\left (1-2 a b d^2 n\right ) x}{n}-b^2 d^2 x^2\right ) \, dx,x,\log \left (c x^n\right )\right )}{\sqrt {\pi }} \\ & = x \text {erfc}\left (d \left (a+b \log \left (c x^n\right )\right )\right )+\frac {\left (2 b d e^{\frac {1-4 a b d^2 n}{4 b^2 d^2 n^2}} x \left (c x^n\right )^{-2 a b d^2-\frac {1-2 a b d^2 n}{n}}\right ) \text {Subst}\left (\int \exp \left (-\frac {\left (\frac {1-2 a b d^2 n}{n}-2 b^2 d^2 x\right )^2}{4 b^2 d^2}\right ) \, dx,x,\log \left (c x^n\right )\right )}{\sqrt {\pi }} \\ & = e^{\frac {1-4 a b d^2 n}{4 b^2 d^2 n^2}} x \left (c x^n\right )^{-1/n} \text {erf}\left (\frac {2 a b d^2-\frac {1}{n}+2 b^2 d^2 \log \left (c x^n\right )}{2 b d}\right )+x \text {erfc}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.84 \[ \int \text {erfc}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=e^{\frac {\frac {\frac {1}{d^2}-4 a b n}{b^2}-4 n \log \left (c x^n\right )}{4 n^2}} x \text {erf}\left (a d-\frac {1}{2 b d n}+b d \log \left (c x^n\right )\right )+x \text {erfc}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \]

[In]

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

[Out]

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

Maple [F]

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

[In]

int(erfc(d*(a+b*ln(c*x^n))),x)

[Out]

int(erfc(d*(a+b*ln(c*x^n))),x)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.34 \[ \int \text {erfc}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\sqrt {b^{2} d^{2} n^{2}} \operatorname {erf}\left (\frac {{\left (2 \, b^{2} d^{2} n^{2} \log \left (x\right ) + 2 \, b^{2} d^{2} n \log \left (c\right ) + 2 \, a b d^{2} n - 1\right )} \sqrt {b^{2} d^{2} n^{2}}}{2 \, b^{2} d^{2} n^{2}}\right ) e^{\left (-\frac {4 \, b^{2} d^{2} n \log \left (c\right ) + 4 \, a b d^{2} n - 1}{4 \, b^{2} d^{2} n^{2}}\right )} - x \operatorname {erf}\left (b d \log \left (c x^{n}\right ) + a d\right ) + x \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

integrate(erfc(d*(a+b*ln(c*x**n))),x)

[Out]

Integral(erfc(d*(a + b*log(c*x**n))), x)

Maxima [F]

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

[In]

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

[Out]

integrate(erfc((b*log(c*x^n) + a)*d), x)

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.89 \[ \int \text {erfc}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=-x \operatorname {erf}\left (b d n \log \left (x\right ) + b d \log \left (c\right ) + a d\right ) + x - \frac {\operatorname {erf}\left (-b d n \log \left (x\right ) - b d \log \left (c\right ) - a d + \frac {1}{2 \, b d n}\right ) e^{\left (-\frac {a}{b n} + \frac {1}{4 \, b^{2} d^{2} n^{2}}\right )}}{c^{\left (\frac {1}{n}\right )}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(erfc(d*(a + b*log(c*x^n))),x)

[Out]

int(erfc(d*(a + b*log(c*x^n))), x)