\(\int \frac {a+b \log (c (d+e \sqrt {x})^n)}{x} \, dx\) [404]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.05 (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} \[ \int \frac {a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{x} \, dx=2 \log \left (-\frac {e \sqrt {x}}{d}\right ) \left (a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )\right )+2 b n \operatorname {PolyLog}\left (2,\frac {\sqrt {x} e}{d}+1\right ) \]

[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*Sqrt[x])/d)] + 2*b*n*PolyLog[2, 1 + (e*Sqrt[x])/d]

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

Mathematica [A] (verified)

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

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

Maple [F]

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

[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]

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

[In]

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

[Out]

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

Sympy [F]

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

[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. 108 vs. \(2 (44) = 88\).

Time = 0.35 (sec) , antiderivative size = 108, normalized size of antiderivative = 2.12 \[ \int \frac {a+b \log \left (c \left (d+e \sqrt {x}\right )^n\right )}{x} \, dx=-2 \, {\left (\log \left (\frac {e \sqrt {x}}{d} + 1\right ) \log \left (\sqrt {x}\right ) + {\rm Li}_2\left (-\frac {e \sqrt {x}}{d}\right )\right )} b n + \frac {2 \, {\left (b e n \sqrt {x} \log \left (\sqrt {x}\right ) - b e n \sqrt {x}\right )}}{d} + \frac {b d \log \left ({\left (e \sqrt {x} + d\right )}^{n}\right ) \log \left (x\right ) + {\left (b d \log \left (c\right ) + a d\right )} \log \left (x\right ) - \frac {b e n x \log \left (x\right ) - 2 \, b e n x}{\sqrt {x}}}{d} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[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)