\(\int x^m \log ^{\frac {3}{2}}(a x^n) \, dx\) [162]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 14, antiderivative size = 111 \[ \int x^m \log ^{\frac {3}{2}}\left (a x^n\right ) \, dx=\frac {3 n^{3/2} \sqrt {\pi } x^{1+m} \left (a x^n\right )^{-\frac {1+m}{n}} \text {erfi}\left (\frac {\sqrt {1+m} \sqrt {\log \left (a x^n\right )}}{\sqrt {n}}\right )}{4 (1+m)^{5/2}}-\frac {3 n x^{1+m} \sqrt {\log \left (a x^n\right )}}{2 (1+m)^2}+\frac {x^{1+m} \log ^{\frac {3}{2}}\left (a x^n\right )}{1+m} \]

[Out]

x^(1+m)*ln(a*x^n)^(3/2)/(1+m)+3/4*n^(3/2)*x^(1+m)*erfi((1+m)^(1/2)*ln(a*x^n)^(1/2)/n^(1/2))*Pi^(1/2)/(1+m)^(5/
2)/((a*x^n)^((1+m)/n))-3/2*n*x^(1+m)*ln(a*x^n)^(1/2)/(1+m)^2

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2342, 2347, 2211, 2235} \[ \int x^m \log ^{\frac {3}{2}}\left (a x^n\right ) \, dx=\frac {3 \sqrt {\pi } n^{3/2} x^{m+1} \left (a x^n\right )^{-\frac {m+1}{n}} \text {erfi}\left (\frac {\sqrt {m+1} \sqrt {\log \left (a x^n\right )}}{\sqrt {n}}\right )}{4 (m+1)^{5/2}}+\frac {x^{m+1} \log ^{\frac {3}{2}}\left (a x^n\right )}{m+1}-\frac {3 n x^{m+1} \sqrt {\log \left (a x^n\right )}}{2 (m+1)^2} \]

[In]

Int[x^m*Log[a*x^n]^(3/2),x]

[Out]

(3*n^(3/2)*Sqrt[Pi]*x^(1 + m)*Erfi[(Sqrt[1 + m]*Sqrt[Log[a*x^n]])/Sqrt[n]])/(4*(1 + m)^(5/2)*(a*x^n)^((1 + m)/
n)) - (3*n*x^(1 + m)*Sqrt[Log[a*x^n]])/(2*(1 + m)^2) + (x^(1 + m)*Log[a*x^n]^(3/2))/(1 + m)

Rule 2211

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[F^(g*(e - c*(
f/d)) + f*g*(x^2/d)), x], x, Sqrt[c + d*x]], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

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 2342

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Lo
g[c*x^n])^p/(d*(m + 1))), x] - Dist[b*n*(p/(m + 1)), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Rule 2347

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Dist[(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]

Rubi steps \begin{align*} \text {integral}& = \frac {x^{1+m} \log ^{\frac {3}{2}}\left (a x^n\right )}{1+m}-\frac {(3 n) \int x^m \sqrt {\log \left (a x^n\right )} \, dx}{2 (1+m)} \\ & = -\frac {3 n x^{1+m} \sqrt {\log \left (a x^n\right )}}{2 (1+m)^2}+\frac {x^{1+m} \log ^{\frac {3}{2}}\left (a x^n\right )}{1+m}+\frac {\left (3 n^2\right ) \int \frac {x^m}{\sqrt {\log \left (a x^n\right )}} \, dx}{4 (1+m)^2} \\ & = -\frac {3 n x^{1+m} \sqrt {\log \left (a x^n\right )}}{2 (1+m)^2}+\frac {x^{1+m} \log ^{\frac {3}{2}}\left (a x^n\right )}{1+m}+\frac {\left (3 n x^{1+m} \left (a x^n\right )^{-\frac {1+m}{n}}\right ) \text {Subst}\left (\int \frac {e^{\frac {(1+m) x}{n}}}{\sqrt {x}} \, dx,x,\log \left (a x^n\right )\right )}{4 (1+m)^2} \\ & = -\frac {3 n x^{1+m} \sqrt {\log \left (a x^n\right )}}{2 (1+m)^2}+\frac {x^{1+m} \log ^{\frac {3}{2}}\left (a x^n\right )}{1+m}+\frac {\left (3 n x^{1+m} \left (a x^n\right )^{-\frac {1+m}{n}}\right ) \text {Subst}\left (\int e^{\frac {(1+m) x^2}{n}} \, dx,x,\sqrt {\log \left (a x^n\right )}\right )}{2 (1+m)^2} \\ & = \frac {3 n^{3/2} \sqrt {\pi } x^{1+m} \left (a x^n\right )^{-\frac {1+m}{n}} \text {erfi}\left (\frac {\sqrt {1+m} \sqrt {\log \left (a x^n\right )}}{\sqrt {n}}\right )}{4 (1+m)^{5/2}}-\frac {3 n x^{1+m} \sqrt {\log \left (a x^n\right )}}{2 (1+m)^2}+\frac {x^{1+m} \log ^{\frac {3}{2}}\left (a x^n\right )}{1+m} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.91 \[ \int x^m \log ^{\frac {3}{2}}\left (a x^n\right ) \, dx=\frac {x^{1+m} \left (3 n^{3/2} \sqrt {\pi } \left (a x^n\right )^{-\frac {1+m}{n}} \text {erfi}\left (\frac {\sqrt {1+m} \sqrt {\log \left (a x^n\right )}}{\sqrt {n}}\right )+2 \sqrt {1+m} \sqrt {\log \left (a x^n\right )} \left (-3 n+2 (1+m) \log \left (a x^n\right )\right )\right )}{4 (1+m)^{5/2}} \]

[In]

Integrate[x^m*Log[a*x^n]^(3/2),x]

[Out]

(x^(1 + m)*((3*n^(3/2)*Sqrt[Pi]*Erfi[(Sqrt[1 + m]*Sqrt[Log[a*x^n]])/Sqrt[n]])/(a*x^n)^((1 + m)/n) + 2*Sqrt[1 +
 m]*Sqrt[Log[a*x^n]]*(-3*n + 2*(1 + m)*Log[a*x^n])))/(4*(1 + m)^(5/2))

Maple [F]

\[\int x^{m} \ln \left (a \,x^{n}\right )^{\frac {3}{2}}d x\]

[In]

int(x^m*ln(a*x^n)^(3/2),x)

[Out]

int(x^m*ln(a*x^n)^(3/2),x)

Fricas [F]

\[ \int x^m \log ^{\frac {3}{2}}\left (a x^n\right ) \, dx=\int { x^{m} \log \left (a x^{n}\right )^{\frac {3}{2}} \,d x } \]

[In]

integrate(x^m*log(a*x^n)^(3/2),x, algorithm="fricas")

[Out]

integral(x^m*log(a*x^n)^(3/2), x)

Sympy [F]

\[ \int x^m \log ^{\frac {3}{2}}\left (a x^n\right ) \, dx=\int x^{m} \log {\left (a x^{n} \right )}^{\frac {3}{2}}\, dx \]

[In]

integrate(x**m*ln(a*x**n)**(3/2),x)

[Out]

Integral(x**m*log(a*x**n)**(3/2), x)

Maxima [F]

\[ \int x^m \log ^{\frac {3}{2}}\left (a x^n\right ) \, dx=\int { x^{m} \log \left (a x^{n}\right )^{\frac {3}{2}} \,d x } \]

[In]

integrate(x^m*log(a*x^n)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^m*log(a*x^n)^(3/2), x)

Giac [F]

\[ \int x^m \log ^{\frac {3}{2}}\left (a x^n\right ) \, dx=\int { x^{m} \log \left (a x^{n}\right )^{\frac {3}{2}} \,d x } \]

[In]

integrate(x^m*log(a*x^n)^(3/2),x, algorithm="giac")

[Out]

integrate(x^m*log(a*x^n)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int x^m \log ^{\frac {3}{2}}\left (a x^n\right ) \, dx=\int x^m\,{\ln \left (a\,x^n\right )}^{3/2} \,d x \]

[In]

int(x^m*log(a*x^n)^(3/2),x)

[Out]

int(x^m*log(a*x^n)^(3/2), x)