\(\int \frac {\log (x)}{\sqrt {a+b \log (x)}} \, dx\) [301]

Optimal result

Integrand size = 13, antiderivative size = 60 \[ \int \frac {\log (x)}{\sqrt {a+b \log (x)}} \, dx=-\frac {(2 a+b) e^{-\frac {a}{b}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {a+b \log (x)}}{\sqrt {b}}\right )}{2 b^{3/2}}+\frac {x \sqrt {a+b \log (x)}}{b} \]

[Out]

-1/2*(2*a+b)*erfi((a+b*ln(x))^(1/2)/b^(1/2))*Pi^(1/2)/b^(3/2)/exp(a/b)+x*(a+b*ln(x))^(1/2)/b

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2399, 2336, 2211, 2235} \[ \int \frac {\log (x)}{\sqrt {a+b \log (x)}} \, dx=\frac {x \sqrt {a+b \log (x)}}{b}-\frac {\sqrt {\pi } (2 a+b) e^{-\frac {a}{b}} \text {erfi}\left (\frac {\sqrt {a+b \log (x)}}{\sqrt {b}}\right )}{2 b^{3/2}} \]

[In]

Int[Log[x]/Sqrt[a + b*Log[x]],x]

[Out]

-1/2*((2*a + b)*Sqrt[Pi]*Erfi[Sqrt[a + b*Log[x]]/Sqrt[b]])/(b^(3/2)*E^(a/b)) + (x*Sqrt[a + b*Log[x]])/b

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 2336

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

Rule 2399

Int[((A_.) + Log[(c_.)*((d_.) + (e_.)*(x_))^(n_.)]*(B_.))/Sqrt[Log[(c_.)*((d_.) + (e_.)*(x_))^(n_.)]*(b_.) + (
a_)], x_Symbol] :> Simp[B*(d + e*x)*(Sqrt[a + b*Log[c*(d + e*x)^n]]/(b*e)), x] + Dist[(2*A*b - B*(2*a + b*n))/
(2*b), Int[1/Sqrt[a + b*Log[c*(d + e*x)^n]], x], x] /; FreeQ[{a, b, c, d, e, A, B, n}, x]

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.20 \[ \int \frac {\log (x)}{\sqrt {a+b \log (x)}} \, dx=\frac {2 x (a+b \log (x))-(2 a+b) e^{-\frac {a}{b}} \Gamma \left (\frac {1}{2},-\frac {a+b \log (x)}{b}\right ) \sqrt {-\frac {a+b \log (x)}{b}}}{2 b \sqrt {a+b \log (x)}} \]

[In]

Integrate[Log[x]/Sqrt[a + b*Log[x]],x]

[Out]

(2*x*(a + b*Log[x]) - ((2*a + b)*Gamma[1/2, -((a + b*Log[x])/b)]*Sqrt[-((a + b*Log[x])/b)])/E^(a/b))/(2*b*Sqrt
[a + b*Log[x]])

Maple [F]

\[\int \frac {\ln \left (x \right )}{\sqrt {a +b \ln \left (x \right )}}d x\]

[In]

int(ln(x)/(a+b*ln(x))^(1/2),x)

[Out]

int(ln(x)/(a+b*ln(x))^(1/2),x)

Fricas [F(-2)]

Exception generated. \[ \int \frac {\log (x)}{\sqrt {a+b \log (x)}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(log(x)/(a+b*log(x))^(1/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

Sympy [F]

\[ \int \frac {\log (x)}{\sqrt {a+b \log (x)}} \, dx=\int \frac {\log {\left (x \right )}}{\sqrt {a + b \log {\left (x \right )}}}\, dx \]

[In]

integrate(ln(x)/(a+b*ln(x))**(1/2),x)

[Out]

Integral(log(x)/sqrt(a + b*log(x)), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 108 vs. \(2 (47) = 94\).

Time = 0.26 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.80 \[ \int \frac {\log (x)}{\sqrt {a+b \log (x)}} \, dx=-\frac {\frac {2 \, \sqrt {\pi } a \operatorname {erf}\left (\sqrt {b \log \left (x\right ) + a} \sqrt {-\frac {1}{b}}\right ) e^{\left (-\frac {a}{b}\right )}}{\sqrt {-\frac {1}{b}}} + \frac {\sqrt {\pi } b \operatorname {erf}\left (\sqrt {b \log \left (x\right ) + a} \sqrt {-\frac {1}{b}}\right ) e^{\left (-\frac {a}{b}\right )}}{\sqrt {-\frac {1}{b}}} - 2 \, \sqrt {b \log \left (x\right ) + a} b e^{\left (\frac {b \log \left (x\right ) + a}{b} - \frac {a}{b}\right )}}{2 \, b^{2}} \]

[In]

integrate(log(x)/(a+b*log(x))^(1/2),x, algorithm="maxima")

[Out]

-1/2*(2*sqrt(pi)*a*erf(sqrt(b*log(x) + a)*sqrt(-1/b))*e^(-a/b)/sqrt(-1/b) + sqrt(pi)*b*erf(sqrt(b*log(x) + a)*
sqrt(-1/b))*e^(-a/b)/sqrt(-1/b) - 2*sqrt(b*log(x) + a)*b*e^((b*log(x) + a)/b - a/b))/b^2

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.48 \[ \int \frac {\log (x)}{\sqrt {a+b \log (x)}} \, dx=\frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {\sqrt {b \log \left (x\right ) + a} \sqrt {-b}}{b}\right ) e^{\left (-\frac {a}{b}\right )}}{2 \, \sqrt {-b}} + \frac {\sqrt {\pi } a \operatorname {erf}\left (-\frac {\sqrt {b \log \left (x\right ) + a} \sqrt {-b}}{b}\right ) e^{\left (-\frac {a}{b}\right )}}{\sqrt {-b} b} + \frac {\sqrt {b \log \left (x\right ) + a} x}{b} \]

[In]

integrate(log(x)/(a+b*log(x))^(1/2),x, algorithm="giac")

[Out]

1/2*sqrt(pi)*erf(-sqrt(b*log(x) + a)*sqrt(-b)/b)*e^(-a/b)/sqrt(-b) + sqrt(pi)*a*erf(-sqrt(b*log(x) + a)*sqrt(-
b)/b)*e^(-a/b)/(sqrt(-b)*b) + sqrt(b*log(x) + a)*x/b

Mupad [F(-1)]

Timed out. \[ \int \frac {\log (x)}{\sqrt {a+b \log (x)}} \, dx=\int \frac {\ln \left (x\right )}{\sqrt {a+b\,\ln \left (x\right )}} \,d x \]

[In]

int(log(x)/(a + b*log(x))^(1/2),x)

[Out]

int(log(x)/(a + b*log(x))^(1/2), x)