\(\int \frac {a+b \log (c (d+\frac {e}{\sqrt {x}})^n)}{x} \, dx\) [425]

Optimal result

Integrand size = 22, antiderivative size = 51 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x} \, dx=-2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right ) \log \left (-\frac {e}{d \sqrt {x}}\right )-2 b n \operatorname {PolyLog}\left (2,1+\frac {e}{d \sqrt {x}}\right ) \]

[Out]

-2*(a+b*ln(c*(d+e/x^(1/2))^n))*ln(-e/d/x^(1/2))-2*b*n*polylog(2,1+e/d/x^(1/2))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2504, 2441, 2352} \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x} \, dx=-2 \log \left (-\frac {e}{d \sqrt {x}}\right ) \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right )-2 b n \operatorname {PolyLog}\left (2,\frac {e}{d \sqrt {x}}+1\right ) \]

[In]

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

[Out]

-2*(a + b*Log[c*(d + e/Sqrt[x])^n])*Log[-(e/(d*Sqrt[x]))] - 2*b*n*PolyLog[2, 1 + e/(d*Sqrt[x])]

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2441

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[Log[e*((f + g
*x)/(e*f - d*g))]*((a + b*Log[c*(d + e*x)^n])/g), x] - Dist[b*e*(n/g), Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2504

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rubi steps \begin{align*} \text {integral}& = -\left (2 \text {Subst}\left (\int \frac {a+b \log \left (c (d+e x)^n\right )}{x} \, dx,x,\frac {1}{\sqrt {x}}\right )\right ) \\ & = -2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right ) \log \left (-\frac {e}{d \sqrt {x}}\right )+(2 b e n) \text {Subst}\left (\int \frac {\log \left (-\frac {e x}{d}\right )}{d+e x} \, dx,x,\frac {1}{\sqrt {x}}\right ) \\ & = -2 \left (a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )\right ) \log \left (-\frac {e}{d \sqrt {x}}\right )-2 b n \text {Li}_2\left (1+\frac {e}{d \sqrt {x}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.04 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x} \, dx=-2 b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right ) \log \left (-\frac {e}{d \sqrt {x}}\right )+a \log (x)-2 b n \operatorname {PolyLog}\left (2,\frac {d+\frac {e}{\sqrt {x}}}{d}\right ) \]

[In]

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

[Out]

-2*b*Log[c*(d + e/Sqrt[x])^n]*Log[-(e/(d*Sqrt[x]))] + a*Log[x] - 2*b*n*PolyLog[2, (d + e/Sqrt[x])/d]

Maple [F]

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

[In]

int((a+b*ln(c*(d+e/x^(1/2))^n))/x,x)

[Out]

int((a+b*ln(c*(d+e/x^(1/2))^n))/x,x)

Fricas [F]

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

[In]

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

[Out]

integral((b*log(c*((d*x + e*sqrt(x))/x)^n) + a)/x, x)

Sympy [F]

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

[In]

integrate((a+b*ln(c*(d+e/x**(1/2))**n))/x,x)

[Out]

Integral((a + b*log(c*(d + e/sqrt(x))**n))/x, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (44) = 88\).

Time = 0.48 (sec) , antiderivative size = 124, normalized size of antiderivative = 2.43 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt {x}}\right )^n\right )}{x} \, dx=-2 \, {\left (\log \left (\frac {d \sqrt {x}}{e} + 1\right ) \log \left (\sqrt {x}\right ) + {\rm Li}_2\left (-\frac {d \sqrt {x}}{e}\right )\right )} b n + \frac {b e n \log \left (x\right )^{2} + 4 \, b d n \sqrt {x} \log \left (x\right ) + 4 \, b e \log \left ({\left (d \sqrt {x} + e\right )}^{n}\right ) \log \left (x\right ) - 4 \, b e \log \left (x\right ) \log \left (x^{\frac {1}{2} \, n}\right ) - 8 \, b d n \sqrt {x} + 4 \, {\left (b e \log \left (c\right ) + a e\right )} \log \left (x\right ) - \frac {4 \, {\left (b d n x \log \left (x\right ) - 2 \, b d n x\right )}}{\sqrt {x}}}{4 \, e} \]

[In]

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

[Out]

-2*(log(d*sqrt(x)/e + 1)*log(sqrt(x)) + dilog(-d*sqrt(x)/e))*b*n + 1/4*(b*e*n*log(x)^2 + 4*b*d*n*sqrt(x)*log(x
) + 4*b*e*log((d*sqrt(x) + e)^n)*log(x) - 4*b*e*log(x)*log(x^(1/2*n)) - 8*b*d*n*sqrt(x) + 4*(b*e*log(c) + a*e)
*log(x) - 4*(b*d*n*x*log(x) - 2*b*d*n*x)/sqrt(x))/e

Giac [F]

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

[In]

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

[Out]

integrate((b*log(c*(d + e/sqrt(x))^n) + a)/x, x)

Mupad [F(-1)]

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

[In]

int((a + b*log(c*(d + e/x^(1/2))^n))/x,x)

[Out]

int((a + b*log(c*(d + e/x^(1/2))^n))/x, x)